Optimal. Leaf size=141 \[ \frac {i \left (a+b \text {ArcSin}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2}{2 b c}-\frac {\left (a+b \text {ArcSin}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right ) \log \left (1-e^{2 i \text {ArcSin}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{c}+\frac {i b \text {PolyLog}\left (2,e^{2 i \text {ArcSin}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{2 c} \]
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Rubi [A]
time = 0.08, antiderivative size = 141, normalized size of antiderivative = 1.00, 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 = {212, 6813,
4721, 3798, 2221, 2317, 2438} \begin {gather*} \frac {i \left (a+b \text {ArcSin}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^2}{2 b c}-\frac {\log \left (1-e^{2 i \text {ArcSin}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right ) \left (a+b \text {ArcSin}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )}{c}+\frac {i b \text {Li}_2\left (e^{2 i \text {ArcSin}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right )}{2 c} \end {gather*}
Antiderivative was successfully verified.
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Rule 212
Rule 2221
Rule 2317
Rule 2438
Rule 3798
Rule 4721
Rule 6813
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 {\text {Subst}\left (\int \frac {a+b \sin ^{-1}(x)}{x} \, dx,x,\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{c}\\ &=-\frac {\text {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) \text {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 \text {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) \text {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*}
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Mathematica [F]
time = 0.79, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a+b \text {ArcSin}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{1-c^2 x^2} \, dx \end {gather*}
Verification is not applicable to the result.
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Maple [A]
time = 0.15, size = 276, normalized size = 1.96
method | result | size |
default | \(-\frac {a \ln \left (c x -1\right )}{2 c}+\frac {a \ln \left (c x +1\right )}{2 c}+\frac {i b \arcsin \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right )^{2}}{2 c}-\frac {b \arcsin \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right ) \ln \left (\frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}+\sqrt {1-\frac {-c x +1}{c x +1}}+1\right )}{c}+\frac {i b \polylog \left (2, -\frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}-\sqrt {1-\frac {-c x +1}{c x +1}}\right )}{c}-\frac {b \arcsin \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right ) \ln \left (1-\frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}-\sqrt {1-\frac {-c x +1}{c x +1}}\right )}{c}+\frac {i b \polylog \left (2, \frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}+\sqrt {1-\frac {-c x +1}{c x +1}}\right )}{c}\) | \(276\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int -\frac {a+b\,\mathrm {asin}\left (\frac {\sqrt {1-c\,x}}{\sqrt {c\,x+1}}\right )}{c^2\,x^2-1} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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