Optimal. Leaf size=275 \[ -\frac {3 b^2 \text {Li}_3\left (e^{2 i \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right ) \left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )}{2 c}+\frac {3 i b \text {Li}_2\left (e^{2 i \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right ) \left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^2}{2 c}+\frac {i \left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^4}{4 b c}-\frac {\log \left (1-e^{2 i \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right ) \left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^3}{c}-\frac {3 i b^3 \text {Li}_4\left (e^{2 i \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right )}{4 c} \]
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Rubi [A] time = 0.22, antiderivative size = 275, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6681, 4625, 3717, 2190, 2531, 6609, 2282, 6589} \[ -\frac {3 b^2 \text {PolyLog}\left (3,e^{2 i \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right ) \left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )}{2 c}+\frac {3 i b \text {PolyLog}\left (2,e^{2 i \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right ) \left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^2}{2 c}-\frac {3 i b^3 \text {PolyLog}\left (4,e^{2 i \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right )}{4 c}+\frac {i \left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^4}{4 b c}-\frac {\log \left (1-e^{2 i \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right ) \left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^3}{c} \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2282
Rule 2531
Rule 3717
Rule 4625
Rule 6589
Rule 6609
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 )^3}{1-c^2 x^2} \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \sin ^{-1}(x)\right )^3}{x} \, dx,x,\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{c}\\ &=-\frac {\operatorname {Subst}\left (\int (a+b x)^3 \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 )^4}{4 b c}+\frac {(2 i) \operatorname {Subst}\left (\int \frac {e^{2 i x} (a+b x)^3}{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 )^4}{4 b c}-\frac {\left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3 \log \left (1-e^{2 i \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{c}+\frac {(3 b) \operatorname {Subst}\left (\int (a+b x)^2 \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 )^4}{4 b c}-\frac {\left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3 \log \left (1-e^{2 i \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{c}+\frac {3 i b \left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2 \text {Li}_2\left (e^{2 i \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{2 c}-\frac {\left (3 i b^2\right ) \operatorname {Subst}\left (\int (a+b x) \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 )^4}{4 b c}-\frac {\left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3 \log \left (1-e^{2 i \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{c}+\frac {3 i b \left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2 \text {Li}_2\left (e^{2 i \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{2 c}-\frac {3 b^2 \left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right ) \text {Li}_3\left (e^{2 i \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{2 c}+\frac {\left (3 b^3\right ) \operatorname {Subst}\left (\int \text {Li}_3\left (e^{2 i x}\right ) \, dx,x,\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 )^4}{4 b c}-\frac {\left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3 \log \left (1-e^{2 i \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{c}+\frac {3 i b \left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2 \text {Li}_2\left (e^{2 i \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{2 c}-\frac {3 b^2 \left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right ) \text {Li}_3\left (e^{2 i \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{2 c}-\frac {\left (3 i b^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3(x)}{x} \, dx,x,e^{2 i \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{4 c}\\ &=\frac {i \left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^4}{4 b c}-\frac {\left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3 \log \left (1-e^{2 i \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{c}+\frac {3 i b \left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2 \text {Li}_2\left (e^{2 i \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{2 c}-\frac {3 b^2 \left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right ) \text {Li}_3\left (e^{2 i \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{2 c}-\frac {3 i b^3 \text {Li}_4\left (e^{2 i \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{4 c}\\ \end {align*}
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Mathematica [F] time = 0.42, size = 0, normalized size = 0.00 \[ \int \frac {\left (a+b \sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3}{1-c^2 x^2} \, dx \]
Verification is Not applicable to the result.
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fricas [F] time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {b^{3} \arcsin \left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right )^{3} + 3 \, a b^{2} \arcsin \left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right )^{2} + 3 \, a^{2} b \arcsin \left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right ) + a^{3}}{c^{2} x^{2} - 1}, x\right ) \]
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 \[ \int -\frac {{\left (b \arcsin \left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right ) + a\right )}^{3}}{c^{2} x^{2} - 1}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.83, size = 1232, normalized size = 4.48 \[ \text {result too large to display} \]
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 \[ \frac {1}{2} \, a^{3} {\left (\frac {\log \left (c x + 1\right )}{c} - \frac {\log \left (c x - 1\right )}{c}\right )} - \int \frac {b^{3} \arctan \left (\sqrt {-c x + 1}, \sqrt {2} \sqrt {c} \sqrt {x}\right )^{3} + 3 \, a b^{2} \arctan \left (\sqrt {-c x + 1}, \sqrt {2} \sqrt {c} \sqrt {x}\right )^{2} + 3 \, a^{2} b \arctan \left (\sqrt {-c x + 1}, \sqrt {2} \sqrt {c} \sqrt {x}\right )}{c^{2} x^{2} - 1}\,{d x} \]
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 \[ \int -\frac {{\left (a+b\,\mathrm {asin}\left (\frac {\sqrt {1-c\,x}}{\sqrt {c\,x+1}}\right )\right )}^3}{c^2\,x^2-1} \,d x \]
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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