Optimal. Leaf size=141 \[ \frac{i b \text{PolyLog}\left (2,e^{2 i \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )}\right )}{2 c}+\frac{i \left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )\right )^2}{2 b c}-\frac{\log \left (1-e^{2 i \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )}\right ) \left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )\right )}{c} \]
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Rubi [A] time = 0.111446, antiderivative size = 141, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 7, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.184, Rules used = {206, 6681, 4625, 3717, 2190, 2279, 2391} \[ \frac{i b \text{PolyLog}\left (2,e^{2 i \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )}\right )}{2 c}+\frac{i \left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )\right )^2}{2 b c}-\frac{\log \left (1-e^{2 i \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )}\right ) \left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )\right )}{c} \]
Antiderivative was successfully verified.
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Rule 206
Rule 6681
Rule 4625
Rule 3717
Rule 2190
Rule 2279
Rule 2391
Rubi steps
\begin{align*} \int \frac{a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )}{1-c^2 x^2} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{a+b \sin ^{-1}(x)}{x} \, dx,x,\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )}{c}\\ &=-\frac{\operatorname{Subst}\left (\int (a+b x) \cot (x) \, dx,x,\sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )}{c}\\ &=\frac{i \left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )^2}{2 b c}+\frac{(2 i) \operatorname{Subst}\left (\int \frac{e^{2 i x} (a+b x)}{1-e^{2 i x}} \, dx,x,\sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )}{c}\\ &=\frac{i \left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )^2}{2 b c}-\frac{\left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right ) \log \left (1-e^{2 i \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )}\right )}{c}+\frac{b \operatorname{Subst}\left (\int \log \left (1-e^{2 i x}\right ) \, dx,x,\sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )}{c}\\ &=\frac{i \left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )^2}{2 b c}-\frac{\left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right ) \log \left (1-e^{2 i \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )}\right )}{c}-\frac{(i b) \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{2 i \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )}\right )}{2 c}\\ &=\frac{i \left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )^2}{2 b c}-\frac{\left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right ) \log \left (1-e^{2 i \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )}\right )}{c}+\frac{i b \text{Li}_2\left (e^{2 i \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )}\right )}{2 c}\\ \end{align*}
Mathematica [F] time = 1.07632, size = 0, normalized size = 0. \[ \int \frac{a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )}{1-c^2 x^2} \, dx \]
Verification is Not applicable to the result.
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Maple [A] time = 0.006, size = 276, normalized size = 2. \begin{align*}{\frac{a\ln \left ( cx+1 \right ) }{2\,c}}-{\frac{a\ln \left ( cx-1 \right ) }{2\,c}}+{\frac{{\frac{i}{2}}b}{c} \left ( \arcsin \left ({\sqrt{-cx+1}{\frac{1}{\sqrt{cx+1}}}} \right ) \right ) ^{2}}-{\frac{b}{c}\arcsin \left ({\sqrt{-cx+1}{\frac{1}{\sqrt{cx+1}}}} \right ) \ln \left ( 1+{i\sqrt{-cx+1}{\frac{1}{\sqrt{cx+1}}}}+\sqrt{1-{\frac{-cx+1}{cx+1}}} \right ) }+{\frac{ib}{c}{\it polylog} \left ( 2,{-i\sqrt{-cx+1}{\frac{1}{\sqrt{cx+1}}}}-\sqrt{1-{\frac{-cx+1}{cx+1}}} \right ) }-{\frac{b}{c}\arcsin \left ({\sqrt{-cx+1}{\frac{1}{\sqrt{cx+1}}}} \right ) \ln \left ( 1-{i\sqrt{-cx+1}{\frac{1}{\sqrt{cx+1}}}}-\sqrt{1-{\frac{-cx+1}{cx+1}}} \right ) }+{\frac{ib}{c}{\it polylog} \left ( 2,{i\sqrt{-cx+1}{\frac{1}{\sqrt{cx+1}}}}+\sqrt{1-{\frac{-cx+1}{cx+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} - \frac{\log \left (c x - 1\right )}{c}\right )} - b \int \frac{\arctan \left (\sqrt{-c x + 1}, \sqrt{2} \sqrt{c} \sqrt{x}\right )}{c^{2} x^{2} - 1}\,{d x} \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 (\frac{\sqrt{-c x + 1}}{\sqrt{c x + 1}}\right ) + a}{c^{2} x^{2} - 1}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [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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Giac [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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