Optimal. Leaf size=169 \[ \frac{3 b^2 \text{PolyLog}\left (3,e^{2 i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{2 d e}-\frac{3 i b \text{PolyLog}\left (2,e^{2 i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )^2}{2 d e}+\frac{3 i b^3 \text{PolyLog}\left (4,e^{2 i \sin ^{-1}(c+d x)}\right )}{4 d e}-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )^4}{4 b d e}+\frac{\log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )^3}{d e} \]
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Rubi [A] time = 0.203623, antiderivative size = 169, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.391, Rules used = {4805, 12, 4625, 3717, 2190, 2531, 6609, 2282, 6589} \[ \frac{3 b^2 \text{PolyLog}\left (3,e^{2 i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{2 d e}-\frac{3 i b \text{PolyLog}\left (2,e^{2 i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )^2}{2 d e}+\frac{3 i b^3 \text{PolyLog}\left (4,e^{2 i \sin ^{-1}(c+d x)}\right )}{4 d e}-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )^4}{4 b d e}+\frac{\log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )^3}{d e} \]
Antiderivative was successfully verified.
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Rule 4805
Rule 12
Rule 4625
Rule 3717
Rule 2190
Rule 2531
Rule 6609
Rule 2282
Rule 6589
Rubi steps
\begin{align*} \int \frac{\left (a+b \sin ^{-1}(c+d x)\right )^3}{c e+d e x} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b \sin ^{-1}(x)\right )^3}{e x} \, dx,x,c+d x\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b \sin ^{-1}(x)\right )^3}{x} \, dx,x,c+d x\right )}{d e}\\ &=\frac{\operatorname{Subst}\left (\int (a+b x)^3 \cot (x) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e}\\ &=-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )^4}{4 b d e}-\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}(c+d x)\right )}{d e}\\ &=-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )^4}{4 b d e}+\frac{\left (a+b \sin ^{-1}(c+d x)\right )^3 \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}-\frac{(3 b) \operatorname{Subst}\left (\int (a+b x)^2 \log \left (1-e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e}\\ &=-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )^4}{4 b d e}+\frac{\left (a+b \sin ^{-1}(c+d x)\right )^3 \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}-\frac{3 i b \left (a+b \sin ^{-1}(c+d x)\right )^2 \text{Li}_2\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{2 d e}+\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}(c+d x)\right )}{d e}\\ &=-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )^4}{4 b d e}+\frac{\left (a+b \sin ^{-1}(c+d x)\right )^3 \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}-\frac{3 i b \left (a+b \sin ^{-1}(c+d x)\right )^2 \text{Li}_2\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{2 d e}+\frac{3 b^2 \left (a+b \sin ^{-1}(c+d x)\right ) \text{Li}_3\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{2 d e}-\frac{\left (3 b^3\right ) \operatorname{Subst}\left (\int \text{Li}_3\left (e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{2 d e}\\ &=-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )^4}{4 b d e}+\frac{\left (a+b \sin ^{-1}(c+d x)\right )^3 \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}-\frac{3 i b \left (a+b \sin ^{-1}(c+d x)\right )^2 \text{Li}_2\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{2 d e}+\frac{3 b^2 \left (a+b \sin ^{-1}(c+d x)\right ) \text{Li}_3\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{2 d e}+\frac{\left (3 i b^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3(x)}{x} \, dx,x,e^{2 i \sin ^{-1}(c+d x)}\right )}{4 d e}\\ &=-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )^4}{4 b d e}+\frac{\left (a+b \sin ^{-1}(c+d x)\right )^3 \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}-\frac{3 i b \left (a+b \sin ^{-1}(c+d x)\right )^2 \text{Li}_2\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{2 d e}+\frac{3 b^2 \left (a+b \sin ^{-1}(c+d x)\right ) \text{Li}_3\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{2 d e}+\frac{3 i b^3 \text{Li}_4\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{4 d e}\\ \end{align*}
Mathematica [A] time = 0.197131, size = 304, normalized size = 1.8 \[ -\frac{i \left (96 a^2 b \text{PolyLog}\left (2,e^{2 i \sin ^{-1}(c+d x)}\right )-96 b^2 \sin ^{-1}(c+d x) \text{PolyLog}\left (2,e^{-2 i \sin ^{-1}(c+d x)}\right ) \left (2 a+b \sin ^{-1}(c+d x)\right )+96 i a b^2 \text{PolyLog}\left (3,e^{-2 i \sin ^{-1}(c+d x)}\right )+96 i b^3 \sin ^{-1}(c+d x) \text{PolyLog}\left (3,e^{-2 i \sin ^{-1}(c+d x)}\right )+48 b^3 \text{PolyLog}\left (4,e^{-2 i \sin ^{-1}(c+d x)}\right )+96 a^2 b \sin ^{-1}(c+d x)^2+192 i a^2 b \sin ^{-1}(c+d x) \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )+64 i a^3 \log (c+d x)-64 a b^2 \sin ^{-1}(c+d x)^3+192 i a b^2 \sin ^{-1}(c+d x)^2 \log \left (1-e^{-2 i \sin ^{-1}(c+d x)}\right )+8 \pi ^3 a b^2-16 b^3 \sin ^{-1}(c+d x)^4+64 i b^3 \sin ^{-1}(c+d x)^3 \log \left (1-e^{-2 i \sin ^{-1}(c+d x)}\right )+\pi ^4 b^3\right )}{64 d e} \]
Warning: Unable to verify antiderivative.
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Maple [B] time = 0.043, size = 828, normalized size = 4.9 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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^{3} \arcsin \left (d x + c\right )^{3} + 3 \, a b^{2} \arcsin \left (d x + c\right )^{2} + 3 \, a^{2} b \arcsin \left (d x + c\right ) + a^{3}}{d e x + c e}, 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^{3}}{c + d x}\, dx + \int \frac{b^{3} \operatorname{asin}^{3}{\left (c + d x \right )}}{c + d x}\, dx + \int \frac{3 a b^{2} \operatorname{asin}^{2}{\left (c + d x \right )}}{c + d x}\, dx + \int \frac{3 a^{2} b \operatorname{asin}{\left (c + d x \right )}}{c + d x}\, dx}{e} \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{{\left (b \arcsin \left (d x + c\right ) + a\right )}^{3}}{d e x + c e}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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