Optimal. Leaf size=156 \[ \frac{2 i b \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c d}-\frac{2 i b \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c d}-\frac{2 b^2 \text{PolyLog}\left (3,-i e^{i \sin ^{-1}(c x)}\right )}{c d}+\frac{2 b^2 \text{PolyLog}\left (3,i e^{i \sin ^{-1}(c x)}\right )}{c d}-\frac{2 i \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{c d} \]
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Rubi [A] time = 0.127072, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {4657, 4181, 2531, 2282, 6589} \[ \frac{2 i b \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c d}-\frac{2 i b \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c d}-\frac{2 b^2 \text{PolyLog}\left (3,-i e^{i \sin ^{-1}(c x)}\right )}{c d}+\frac{2 b^2 \text{PolyLog}\left (3,i e^{i \sin ^{-1}(c x)}\right )}{c d}-\frac{2 i \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{c d} \]
Antiderivative was successfully verified.
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Rule 4657
Rule 4181
Rule 2531
Rule 2282
Rule 6589
Rubi steps
\begin{align*} \int \frac{\left (a+b \sin ^{-1}(c x)\right )^2}{d-c^2 d x^2} \, dx &=\frac{\operatorname{Subst}\left (\int (a+b x)^2 \sec (x) \, dx,x,\sin ^{-1}(c x)\right )}{c d}\\ &=-\frac{2 i \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c d}-\frac{(2 b) \operatorname{Subst}\left (\int (a+b x) \log \left (1-i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c d}+\frac{(2 b) \operatorname{Subst}\left (\int (a+b x) \log \left (1+i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c d}\\ &=-\frac{2 i \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c d}+\frac{2 i b \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{c d}-\frac{2 i b \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{c d}-\frac{\left (2 i b^2\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (-i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c d}+\frac{\left (2 i b^2\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c d}\\ &=-\frac{2 i \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c d}+\frac{2 i b \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{c d}-\frac{2 i b \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{c d}-\frac{\left (2 b^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{c d}+\frac{\left (2 b^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{c d}\\ &=-\frac{2 i \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c d}+\frac{2 i b \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{c d}-\frac{2 i b \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{c d}-\frac{2 b^2 \text{Li}_3\left (-i e^{i \sin ^{-1}(c x)}\right )}{c d}+\frac{2 b^2 \text{Li}_3\left (i e^{i \sin ^{-1}(c x)}\right )}{c d}\\ \end{align*}
Mathematica [A] time = 0.509011, size = 207, normalized size = 1.33 \[ \frac{4 i b \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )-4 i b \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )-4 b^2 \text{PolyLog}\left (3,-i e^{i \sin ^{-1}(c x)}\right )+4 b^2 \text{PolyLog}\left (3,i e^{i \sin ^{-1}(c x)}\right )+a^2 (-\log (1-c x))+a^2 \log (c x+1)+4 a b \sin ^{-1}(c x) \log \left (1-i e^{i \sin ^{-1}(c x)}\right )-4 a b \sin ^{-1}(c x) \log \left (1+i e^{i \sin ^{-1}(c x)}\right )-4 i b^2 \sin ^{-1}(c x)^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{2 c d} \]
Warning: Unable to verify antiderivative.
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Maple [B] time = 0.106, size = 404, normalized size = 2.6 \begin{align*} -{\frac{{b}^{2} \left ( \arcsin \left ( cx \right ) \right ) ^{2}}{dc}\ln \left ( 1+i \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) \right ) }+{\frac{2\,i{b}^{2}\arcsin \left ( cx \right ) }{dc}{\it polylog} \left ( 2,-i \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) \right ) }-2\,{\frac{{b}^{2}{\it polylog} \left ( 3,-i \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) \right ) }{dc}}+{\frac{{b}^{2} \left ( \arcsin \left ( cx \right ) \right ) ^{2}}{dc}\ln \left ( 1-i \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) \right ) }-{\frac{2\,i{b}^{2}\arcsin \left ( cx \right ) }{dc}{\it polylog} \left ( 2,i \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) \right ) }+2\,{\frac{{b}^{2}{\it polylog} \left ( 3,i \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) \right ) }{dc}}-2\,{\frac{ab\arcsin \left ( cx \right ) \ln \left ( 1+i \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) \right ) }{dc}}+{\frac{2\,iab}{dc}{\it polylog} \left ( 2,-i \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) \right ) }+2\,{\frac{ab\arcsin \left ( cx \right ) \ln \left ( 1-i \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) \right ) }{dc}}-{\frac{2\,iab}{dc}{\it polylog} \left ( 2,i \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) \right ) }-{\frac{2\,i{a}^{2}}{dc}\arctan \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{2} \, a^{2}{\left (\frac{\log \left (c x + 1\right )}{c d} - \frac{\log \left (c x - 1\right )}{c d}\right )} + \frac{b^{2} \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right )^{2} \log \left (c x + 1\right ) - b^{2} \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right )^{2} \log \left (-c x + 1\right ) - 2 \, c d \int \frac{2 \, a b \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right ) -{\left (b^{2} \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right ) \log \left (c x + 1\right ) - b^{2} \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right ) \log \left (-c x + 1\right )\right )} \sqrt{c x + 1} \sqrt{-c x + 1}}{c^{2} d x^{2} - d}\,{d x}}{2 \, c d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{b^{2} \arcsin \left (c x\right )^{2} + 2 \, a b \arcsin \left (c x\right ) + a^{2}}{c^{2} d x^{2} - d}, x\right ) \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*} - \frac{\int \frac{a^{2}}{c^{2} x^{2} - 1}\, dx + \int \frac{b^{2} \operatorname{asin}^{2}{\left (c x \right )}}{c^{2} x^{2} - 1}\, dx + \int \frac{2 a b \operatorname{asin}{\left (c x \right )}}{c^{2} x^{2} - 1}\, dx}{d} \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 -\frac{{\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{c^{2} d x^{2} - d}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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