The inverse trigonometric substitution takes a linear argument, a multiple of its quadratic and a power of a + b arcsin(c x); an exponential of i times an inverse function is algebraic; the resonant exponential times a sine is by Euler; a zero discriminant is seen through its spellings

[Rafael-SOWNet]

Rubi's inverse trigonometric families (family 5): 126 → 174 of the 257-problem sample, 0 wrong, 0 error. The independent suites: 1707/1774, 0 wrong, 0 timeouts. Cut from master; independent of #1377 and #1379 (no shared hunks: merge-tree is clean against both).

What it does

The inverse trigonometric substitution takes a linear argument, a multiple of its quadratic, and a power of a + b arcsin(c x). x = sin(u) took only arcsin(x) itself, and Rubi's family is (a + b arcsin(c x))^n beside powers of x and of d − c^2 d x^2. With c x + d: x = (sin(u) − d)/c, dx = cos(u) du/c; a fractional power of a constant multiple k of 1 − (c x + d)^2 is k to that power times the cosine to it (the generic case); a whole power of the quadratic from the second up counts too — parts in x on x (d − c^2 d x^2)^3 (a + b arcsin(c x)) left a seventh-degree polynomial over the root and went past its budget; and a power of a + b arcsin(c x) counts as a power of the inverse — parts in x stalled on x arcsin(a x)^2 at x^2 arcsin(a x)/sqrt(1 − a^2 x^2), where under the sine it is u^2 sin(u) cos(u)/a^2, parts twice against sin(2u). The sign the cotangent and cosecant carry into their root is folded before the question is asked in u.

An exponential of i times an inverse function is algebraic. e^(i arctan(L)) = (1 + i L)/sqrt(1 + L^2), e^(i arcsin(L)) = sqrt(1 − L^2) + i L, e^(i arccos(L)) = L + i sqrt(1 − L^2) (principal branch, real L), and e^(n i f(L)) is that to the nth for rational n. Rubi's e^(i arctan(a x))/sqrt(c + a^2 c x^2) is then (1 + i a x)/(sqrt(c) (1 + a^2 x^2)), a rational function, where under x = tan(u)/a it was e^(i u) sec(u), which nothing in u reads. These integrands are complex-valued, so the harness reports their answers unverifiable on the reals rather than solved — the library checks each at sampled points before giving it.

Two defects that route exposed, both fixed:

• The closed form for P(x) e^(a x) cos(b x) divides by a^2 + b^2, zero for a = i b: the table's four arms and the rule answered NaN for e^(−i arctan(a x))/(c + a^2 c x^2)^(3/2) (which is e^(−i u) cos(u) under the tangent). By Euler, c cos(b x) + d sin(b x) = (c − i d) e^(i b x)/2 + (c + i d) e^(−i b x)/2, and each term beside the polynomial and e^(a x) is a polynomial times one exponential — closed on their own. e^(i x) cos(x) is x/2 + e^(2 i x)/(4 i) now.
• The table's k/(a x^2 + b x + c)^n on a symbolic quadratic is a piecewise on the sign of the discriminant, whose zero was not seen when written in two spellings of one term (4 a^2 b^(−2) against 4 (1/b)^2 a^2, for (x + (1 + i a)/(i b))^2); two of the three arms divide by it, and Simplify made NaN of the whole for x e^(−2 i arctan(a + b x)). A discriminant that is zero once simplified is the perfect square, in both the first-power and the reduction arm.

Measured

Family 5, 15 problems a file, 5 s budget, master 2b4a5256 against this branch, same harness:

Source	Run	master	branch
5.1.2 (d x)^m (a+b arcsin(c x))^n	15	8	15
5.1.4 (f x)^m (d+e x^2)^p (a+b arcsin(c x))^n	15	1	9
5.1.5 Inverse sine functions	15	6	12
5.2.2 (d x)^m (a+b arccos(c x))^n	15	8	15
5.2.4 (f x)^m (d+e x^2)^p (a+b arccos(c x))^n	13	8	9
5.2.5 Inverse cosine functions	15	7	10
5.3.2 (d x)^m (a+b arctan(c x^n))^p	15	11	11
5.3.3 (d+e x)^m (a+b arctan(c x^n))^p	13	11	11
5.3.4 u (a+b arctan(c x))^p	15	6	10
5.3.5 u (a+b arctan(c+d x))^p	15	15	15
5.3.6 Exponentials of inverse tangent	15	0	1
5.3.7 Inverse tangent functions	15	9	9
5.4.1 Inverse cotangent functions	15	14	14
5.4.2 Exponentials of inverse cotangent	6	3	5
5.5.1 u (a+b arcsec(c x))^n	15	1	4
5.5.2 Inverse secant functions	15	8	9
5.6.1 u (a+b arccsc(c x))^n	15	1	4
5.6.2 Inverse cosecant functions	15	9	11
total	257	126	174

48 gained, nothing lost. Left in the family: the secant/cosecant u (a + b arcsec(c x))^n (4/15 — a polynomial beside the arcsecant, where parts leaves x^m/sqrt(1 − 1/(c^2 x^2))), x^m arctan(c x^n)/x^k for n = 2, 3, polynomial × arcsine of high degree (parts in x with a degree-9 polynomial over the root), and the e^(n i arctan) shapes with a complex linear below the bar — 1/((1 + i a x)(1 + a^2 x^2)) is a rational function over Q(i) that the partial fractions decline, a capability of its own.

Four suites green. No gate row is integration.

Part of #718.

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