Optimal. Leaf size=84 \[ \frac{i b \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )}{c d}-\frac{i b \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )}{c d}-\frac{2 i \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c d} \]
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Rubi [A] time = 0.0672187, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {4657, 4181, 2279, 2391} \[ \frac{i b \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )}{c d}-\frac{i b \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )}{c d}-\frac{2 i \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c d} \]
Antiderivative was successfully verified.
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Rule 4657
Rule 4181
Rule 2279
Rule 2391
Rubi steps
\begin{align*} \int \frac{a+b \sin ^{-1}(c x)}{d-c^2 d x^2} \, dx &=\frac{\operatorname{Subst}\left (\int (a+b x) \sec (x) \, dx,x,\sin ^{-1}(c x)\right )}{c d}\\ &=-\frac{2 i \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c d}-\frac{b \operatorname{Subst}\left (\int \log \left (1-i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c d}+\frac{b \operatorname{Subst}\left (\int \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 ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c d}+\frac{(i b) \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{c d}-\frac{(i b) \operatorname{Subst}\left (\int \frac{\log (1+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 ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c d}+\frac{i b \text{Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{c d}-\frac{i b \text{Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{c d}\\ \end{align*}
Mathematica [B] time = 0.229741, size = 207, normalized size = 2.46 \[ \frac{2 i b \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )-2 i b \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )-a \log (1-c x)+a \log (c x+1)-i \pi b \sin ^{-1}(c x)+2 b \sin ^{-1}(c x) \log \left (1-i e^{i \sin ^{-1}(c x)}\right )-2 b \sin ^{-1}(c x) \log \left (1+i e^{i \sin ^{-1}(c x)}\right )+\pi b \log \left (1-i e^{i \sin ^{-1}(c x)}\right )+\pi b \log \left (1+i e^{i \sin ^{-1}(c x)}\right )-\pi b \log \left (\sin \left (\frac{1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )-\pi b \log \left (-\cos \left (\frac{1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )}{2 c d} \]
Warning: Unable to verify antiderivative.
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Maple [B] time = 0.102, size = 426, normalized size = 5.1 \begin{align*}{\frac{a{\it Artanh} \left ( cx \right ) }{dc}}+{\frac{b{\it Artanh} \left ( cx \right ) \arcsin \left ( cx \right ) }{dc}}-{\frac{ib}{dc}{\it dilog} \left ({-i{\frac{1}{\sqrt{-{c}^{2}{x}^{2}+1}}}}-{icx{\frac{1}{\sqrt{-{c}^{2}{x}^{2}+1}}}} \right ) }+{\frac{ib{\it Artanh} \left ( cx \right ) }{dc}\ln \left ( \left ( 1-i \right ) \cosh \left ({\frac{{\it Artanh} \left ( cx \right ) }{2}} \right ) + \left ( 1+i \right ) \sinh \left ({\frac{{\it Artanh} \left ( cx \right ) }{2}} \right ) \right ) }-{\frac{ib}{dc}\ln \left ( \left ( 1-i \right ) \cosh \left ({\frac{{\it Artanh} \left ( cx \right ) }{2}} \right ) + \left ( 1+i \right ) \sinh \left ({\frac{{\it Artanh} \left ( cx \right ) }{2}} \right ) \right ) \ln \left ({-i{\frac{1}{\sqrt{-{c}^{2}{x}^{2}+1}}}}-{icx{\frac{1}{\sqrt{-{c}^{2}{x}^{2}+1}}}} \right ) }+{\frac{ib}{dc}{\it dilog} \left ({i{\frac{1}{\sqrt{-{c}^{2}{x}^{2}+1}}}}+{icx{\frac{1}{\sqrt{-{c}^{2}{x}^{2}+1}}}} \right ) }-{\frac{ib{\it Artanh} \left ( cx \right ) }{dc}\ln \left ( \left ( 1+i \right ) \cosh \left ({\frac{{\it Artanh} \left ( cx \right ) }{2}} \right ) + \left ( 1-i \right ) \sinh \left ({\frac{{\it Artanh} \left ( cx \right ) }{2}} \right ) \right ) }+{\frac{ib}{dc}\ln \left ( \left ( 1+i \right ) \cosh \left ({\frac{{\it Artanh} \left ( cx \right ) }{2}} \right ) + \left ( 1-i \right ) \sinh \left ({\frac{{\it Artanh} \left ( cx \right ) }{2}} \right ) \right ) \ln \left ({i{\frac{1}{\sqrt{-{c}^{2}{x}^{2}+1}}}}+{icx{\frac{1}{\sqrt{-{c}^{2}{x}^{2}+1}}}} \right ) }+{\frac{{\frac{i}{2}}b{\it Artanh} \left ( cx \right ) }{dc}\ln \left ({i{\frac{1}{\sqrt{-{c}^{2}{x}^{2}+1}}}}+{icx{\frac{1}{\sqrt{-{c}^{2}{x}^{2}+1}}}} \right ) }-{\frac{{\frac{i}{2}}b{\it Artanh} \left ( cx \right ) }{dc}\ln \left ({-i{\frac{1}{\sqrt{-{c}^{2}{x}^{2}+1}}}}-{icx{\frac{1}{\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{\left (\frac{\log \left (c x + 1\right )}{c d} - \frac{\log \left (c x - 1\right )}{c d}\right )} + \frac{{\left (c d \int \frac{\sqrt{c x + 1} \sqrt{-c x + 1}{\left (\log \left (c x + 1\right ) - \log \left (-c x + 1\right )\right )}}{c^{2} d x^{2} - d}\,{d x} + \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right ) \log \left (c x + 1\right ) - \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right ) \log \left (-c x + 1\right )\right )} b}{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 \arcsin \left (c x\right ) + a}{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}{c^{2} x^{2} - 1}\, dx + \int \frac{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{b \arcsin \left (c x\right ) + a}{c^{2} d x^{2} - d}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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