Optimal. Leaf size=205 \[ \frac {i b \text {Li}_2\left (e^{2 i \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right ) \left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )}{c}+\frac {i \left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^3}{3 b c}-\frac {\log \left (1-e^{2 i \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right ) \left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^2}{c}-\frac {b^2 \text {Li}_3\left (e^{2 i \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right )}{2 c} \]
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Rubi [A] time = 0.18, antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.175, Rules used = {6681, 4625, 3717, 2190, 2531, 2282, 6589} \[ \frac {i b \text {PolyLog}\left (2,e^{2 i \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right ) \left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )}{c}-\frac {b^2 \text {PolyLog}\left (3,e^{2 i \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right )}{2 c}+\frac {i \left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^3}{3 b c}-\frac {\log \left (1-e^{2 i \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right ) \left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^2}{c} \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2282
Rule 2531
Rule 3717
Rule 4625
Rule 6589
Rule 6681
Rubi steps
\begin {align*} \int \frac {\left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2}{1-c^2 x^2} \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \sin ^{-1}(x)\right )^2}{x} \, dx,x,\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{c}\\ &=-\frac {\operatorname {Subst}\left (\int (a+b x)^2 \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 )^3}{3 b c}+\frac {(2 i) \operatorname {Subst}\left (\int \frac {e^{2 i x} (a+b x)^2}{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 )^3}{3 b c}-\frac {\left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2 \log \left (1-e^{2 i \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{c}+\frac {(2 b) \operatorname {Subst}\left (\int (a+b x) \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 )^3}{3 b c}-\frac {\left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2 \log \left (1-e^{2 i \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{c}+\frac {i b \left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right ) \text {Li}_2\left (e^{2 i \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{c}-\frac {\left (i b^2\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (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 )^3}{3 b c}-\frac {\left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2 \log \left (1-e^{2 i \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{c}+\frac {i b \left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right ) \text {Li}_2\left (e^{2 i \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{c}-\frac {b^2 \operatorname {Subst}\left (\int \frac {\text {Li}_2(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 )^3}{3 b c}-\frac {\left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2 \log \left (1-e^{2 i \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{c}+\frac {i b \left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right ) \text {Li}_2\left (e^{2 i \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{c}-\frac {b^2 \text {Li}_3\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 = 0.86, size = 0, normalized size = 0.00 \[ \int \frac {\left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2}{1-c^2 x^2} \, dx \]
Verification is Not applicable to the result.
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fricas [F] time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {b^{2} \arcsin \left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right )^{2} + 2 \, a b \arcsin \left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right ) + a^{2}}{c^{2} x^{2} - 1}, x\right ) \]
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 \[ \int -\frac {{\left (b \arcsin \left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right ) + a\right )}^{2}}{c^{2} x^{2} - 1}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.01, size = 681, normalized size = 3.32 \[ -\frac {a^{2} \ln \left (c x -1\right )}{2 c}+\frac {a^{2} \ln \left (c x +1\right )}{2 c}+\frac {i b^{2} \arcsin \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right )^{3}}{3 c}-\frac {b^{2} \arcsin \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right )^{2} \ln \left (1+\frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}+\sqrt {1-\frac {-c x +1}{c x +1}}\right )}{c}+\frac {2 i b^{2} \arcsin \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right ) \polylog \left (2, -\frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}-\sqrt {1-\frac {-c x +1}{c x +1}}\right )}{c}-\frac {2 b^{2} \polylog \left (3, -\frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}-\sqrt {1-\frac {-c x +1}{c x +1}}\right )}{c}-\frac {b^{2} \arcsin \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right )^{2} \ln \left (1-\frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}-\sqrt {1-\frac {-c x +1}{c x +1}}\right )}{c}+\frac {2 i b^{2} \arcsin \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right ) \polylog \left (2, \frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}+\sqrt {1-\frac {-c x +1}{c x +1}}\right )}{c}-\frac {2 b^{2} \polylog \left (3, \frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}+\sqrt {1-\frac {-c x +1}{c x +1}}\right )}{c}+\frac {i a b \arcsin \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right )^{2}}{c}-\frac {2 a b \arcsin \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right ) \ln \left (1+\frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}+\sqrt {1-\frac {-c x +1}{c x +1}}\right )}{c}+\frac {2 i a b \polylog \left (2, -\frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}-\sqrt {1-\frac {-c x +1}{c x +1}}\right )}{c}-\frac {2 a b \arcsin \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right ) \ln \left (1-\frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}-\sqrt {1-\frac {-c x +1}{c x +1}}\right )}{c}+\frac {2 i a b \polylog \left (2, \frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}+\sqrt {1-\frac {-c x +1}{c x +1}}\right )}{c} \]
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 \[ \frac {1}{2} \, a^{2} {\left (\frac {\log \left (c x + 1\right )}{c} - \frac {\log \left (c x - 1\right )}{c}\right )} - \int \frac {b^{2} \arctan \left (\sqrt {-c x + 1}, \sqrt {2} \sqrt {c} \sqrt {x}\right )^{2} + 2 \, a b \arctan \left (\sqrt {-c x + 1}, \sqrt {2} \sqrt {c} \sqrt {x}\right )}{c^{2} x^{2} - 1}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int -\frac {{\left (a+b\,\mathrm {asin}\left (\frac {\sqrt {1-c\,x}}{\sqrt {c\,x+1}}\right )\right )}^2}{c^2\,x^2-1} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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