Optimal. Leaf size=84 \[ -\frac{i \text{PolyLog}\left (2,e^{2 i \sin ^{-1}\left (c e^{a+b x}\right )}\right )}{2 b}-\frac{i \sin ^{-1}\left (c e^{a+b x}\right )^2}{2 b}+\frac{\sin ^{-1}\left (c e^{a+b x}\right ) \log \left (1-e^{2 i \sin ^{-1}\left (c e^{a+b x}\right )}\right )}{b} \]
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Rubi [A] time = 0.0707198, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {2282, 4625, 3717, 2190, 2279, 2391} \[ -\frac{i \text{PolyLog}\left (2,e^{2 i \sin ^{-1}\left (c e^{a+b x}\right )}\right )}{2 b}-\frac{i \sin ^{-1}\left (c e^{a+b x}\right )^2}{2 b}+\frac{\sin ^{-1}\left (c e^{a+b x}\right ) \log \left (1-e^{2 i \sin ^{-1}\left (c e^{a+b x}\right )}\right )}{b} \]
Antiderivative was successfully verified.
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Rule 2282
Rule 4625
Rule 3717
Rule 2190
Rule 2279
Rule 2391
Rubi steps
\begin{align*} \int \sin ^{-1}\left (c e^{a+b x}\right ) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\sin ^{-1}(c x)}{x} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int x \cot (x) \, dx,x,\sin ^{-1}\left (c e^{a+b x}\right )\right )}{b}\\ &=-\frac{i \sin ^{-1}\left (c e^{a+b x}\right )^2}{2 b}-\frac{(2 i) \operatorname{Subst}\left (\int \frac{e^{2 i x} x}{1-e^{2 i x}} \, dx,x,\sin ^{-1}\left (c e^{a+b x}\right )\right )}{b}\\ &=-\frac{i \sin ^{-1}\left (c e^{a+b x}\right )^2}{2 b}+\frac{\sin ^{-1}\left (c e^{a+b x}\right ) \log \left (1-e^{2 i \sin ^{-1}\left (c e^{a+b x}\right )}\right )}{b}-\frac{\operatorname{Subst}\left (\int \log \left (1-e^{2 i x}\right ) \, dx,x,\sin ^{-1}\left (c e^{a+b x}\right )\right )}{b}\\ &=-\frac{i \sin ^{-1}\left (c e^{a+b x}\right )^2}{2 b}+\frac{\sin ^{-1}\left (c e^{a+b x}\right ) \log \left (1-e^{2 i \sin ^{-1}\left (c e^{a+b x}\right )}\right )}{b}+\frac{i \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{2 i \sin ^{-1}\left (c e^{a+b x}\right )}\right )}{2 b}\\ &=-\frac{i \sin ^{-1}\left (c e^{a+b x}\right )^2}{2 b}+\frac{\sin ^{-1}\left (c e^{a+b x}\right ) \log \left (1-e^{2 i \sin ^{-1}\left (c e^{a+b x}\right )}\right )}{b}-\frac{i \text{Li}_2\left (e^{2 i \sin ^{-1}\left (c e^{a+b x}\right )}\right )}{2 b}\\ \end{align*}
Mathematica [F] time = 0.972079, size = 0, normalized size = 0. \[ \int \sin ^{-1}\left (c e^{a+b x}\right ) \, dx \]
Verification is Not applicable to the result.
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Maple [A] time = 0.025, size = 181, normalized size = 2.2 \begin{align*}{\frac{-{\frac{i}{2}} \left ( \arcsin \left ( c{{\rm e}^{bx+a}} \right ) \right ) ^{2}}{b}}+{\frac{\arcsin \left ( c{{\rm e}^{bx+a}} \right ) }{b}\ln \left ( 1+ic{{\rm e}^{bx+a}}+\sqrt{1-{c}^{2} \left ({{\rm e}^{bx+a}} \right ) ^{2}} \right ) }+{\frac{\arcsin \left ( c{{\rm e}^{bx+a}} \right ) }{b}\ln \left ( 1-ic{{\rm e}^{bx+a}}-\sqrt{1-{c}^{2} \left ({{\rm e}^{bx+a}} \right ) ^{2}} \right ) }-{\frac{i}{b}{\it polylog} \left ( 2,-ic{{\rm e}^{bx+a}}-\sqrt{1-{c}^{2} \left ({{\rm e}^{bx+a}} \right ) ^{2}} \right ) }-{\frac{i}{b}{\it polylog} \left ( 2,ic{{\rm e}^{bx+a}}+\sqrt{1-{c}^{2} \left ({{\rm e}^{bx+a}} \right ) ^{2}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \operatorname{asin}{\left (c e^{a + b x} \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \arcsin \left (c e^{\left (b x + a\right )}\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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