Optimal. Leaf size=270 \[ -\frac{24 b^3 \text{PolyLog}\left (3,-e^{i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{d e^2}+\frac{24 b^3 \text{PolyLog}\left (3,e^{i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{d e^2}+\frac{12 i b^2 \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e^2}-\frac{12 i b^2 \text{PolyLog}\left (2,e^{i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e^2}-\frac{24 i b^4 \text{PolyLog}\left (4,-e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}+\frac{24 i b^4 \text{PolyLog}\left (4,e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}-\frac{\left (a+b \sin ^{-1}(c+d x)\right )^4}{d e^2 (c+d x)}-\frac{8 b \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )^3}{d e^2} \]
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Rubi [A] time = 0.312955, antiderivative size = 270, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.391, Rules used = {4805, 12, 4627, 4709, 4183, 2531, 6609, 2282, 6589} \[ -\frac{24 b^3 \text{PolyLog}\left (3,-e^{i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{d e^2}+\frac{24 b^3 \text{PolyLog}\left (3,e^{i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{d e^2}+\frac{12 i b^2 \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e^2}-\frac{12 i b^2 \text{PolyLog}\left (2,e^{i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e^2}-\frac{24 i b^4 \text{PolyLog}\left (4,-e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}+\frac{24 i b^4 \text{PolyLog}\left (4,e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}-\frac{\left (a+b \sin ^{-1}(c+d x)\right )^4}{d e^2 (c+d x)}-\frac{8 b \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )^3}{d e^2} \]
Antiderivative was successfully verified.
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Rule 4805
Rule 12
Rule 4627
Rule 4709
Rule 4183
Rule 2531
Rule 6609
Rule 2282
Rule 6589
Rubi steps
\begin{align*} \int \frac{\left (a+b \sin ^{-1}(c+d x)\right )^4}{(c e+d e x)^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b \sin ^{-1}(x)\right )^4}{e^2 x^2} \, dx,x,c+d x\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b \sin ^{-1}(x)\right )^4}{x^2} \, dx,x,c+d x\right )}{d e^2}\\ &=-\frac{\left (a+b \sin ^{-1}(c+d x)\right )^4}{d e^2 (c+d x)}+\frac{(4 b) \operatorname{Subst}\left (\int \frac{\left (a+b \sin ^{-1}(x)\right )^3}{x \sqrt{1-x^2}} \, dx,x,c+d x\right )}{d e^2}\\ &=-\frac{\left (a+b \sin ^{-1}(c+d x)\right )^4}{d e^2 (c+d x)}+\frac{(4 b) \operatorname{Subst}\left (\int (a+b x)^3 \csc (x) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e^2}\\ &=-\frac{\left (a+b \sin ^{-1}(c+d x)\right )^4}{d e^2 (c+d x)}-\frac{8 b \left (a+b \sin ^{-1}(c+d x)\right )^3 \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}-\frac{\left (12 b^2\right ) \operatorname{Subst}\left (\int (a+b x)^2 \log \left (1-e^{i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e^2}+\frac{\left (12 b^2\right ) \operatorname{Subst}\left (\int (a+b x)^2 \log \left (1+e^{i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e^2}\\ &=-\frac{\left (a+b \sin ^{-1}(c+d x)\right )^4}{d e^2 (c+d x)}-\frac{8 b \left (a+b \sin ^{-1}(c+d x)\right )^3 \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}+\frac{12 i b^2 \left (a+b \sin ^{-1}(c+d x)\right )^2 \text{Li}_2\left (-e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}-\frac{12 i b^2 \left (a+b \sin ^{-1}(c+d x)\right )^2 \text{Li}_2\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}-\frac{\left (24 i b^3\right ) \operatorname{Subst}\left (\int (a+b x) \text{Li}_2\left (-e^{i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e^2}+\frac{\left (24 i b^3\right ) \operatorname{Subst}\left (\int (a+b x) \text{Li}_2\left (e^{i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e^2}\\ &=-\frac{\left (a+b \sin ^{-1}(c+d x)\right )^4}{d e^2 (c+d x)}-\frac{8 b \left (a+b \sin ^{-1}(c+d x)\right )^3 \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}+\frac{12 i b^2 \left (a+b \sin ^{-1}(c+d x)\right )^2 \text{Li}_2\left (-e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}-\frac{12 i b^2 \left (a+b \sin ^{-1}(c+d x)\right )^2 \text{Li}_2\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}-\frac{24 b^3 \left (a+b \sin ^{-1}(c+d x)\right ) \text{Li}_3\left (-e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}+\frac{24 b^3 \left (a+b \sin ^{-1}(c+d x)\right ) \text{Li}_3\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}+\frac{\left (24 b^4\right ) \operatorname{Subst}\left (\int \text{Li}_3\left (-e^{i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e^2}-\frac{\left (24 b^4\right ) \operatorname{Subst}\left (\int \text{Li}_3\left (e^{i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e^2}\\ &=-\frac{\left (a+b \sin ^{-1}(c+d x)\right )^4}{d e^2 (c+d x)}-\frac{8 b \left (a+b \sin ^{-1}(c+d x)\right )^3 \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}+\frac{12 i b^2 \left (a+b \sin ^{-1}(c+d x)\right )^2 \text{Li}_2\left (-e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}-\frac{12 i b^2 \left (a+b \sin ^{-1}(c+d x)\right )^2 \text{Li}_2\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}-\frac{24 b^3 \left (a+b \sin ^{-1}(c+d x)\right ) \text{Li}_3\left (-e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}+\frac{24 b^3 \left (a+b \sin ^{-1}(c+d x)\right ) \text{Li}_3\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}-\frac{\left (24 i b^4\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3(-x)}{x} \, dx,x,e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}+\frac{\left (24 i b^4\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3(x)}{x} \, dx,x,e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}\\ &=-\frac{\left (a+b \sin ^{-1}(c+d x)\right )^4}{d e^2 (c+d x)}-\frac{8 b \left (a+b \sin ^{-1}(c+d x)\right )^3 \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}+\frac{12 i b^2 \left (a+b \sin ^{-1}(c+d x)\right )^2 \text{Li}_2\left (-e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}-\frac{12 i b^2 \left (a+b \sin ^{-1}(c+d x)\right )^2 \text{Li}_2\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}-\frac{24 b^3 \left (a+b \sin ^{-1}(c+d x)\right ) \text{Li}_3\left (-e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}+\frac{24 b^3 \left (a+b \sin ^{-1}(c+d x)\right ) \text{Li}_3\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}-\frac{24 i b^4 \text{Li}_4\left (-e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}+\frac{24 i b^4 \text{Li}_4\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}\\ \end{align*}
Mathematica [B] time = 1.75632, size = 575, normalized size = 2.13 \[ \frac{6 a^2 b^2 \left (2 i \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c+d x)}\right )-2 i \text{PolyLog}\left (2,e^{i \sin ^{-1}(c+d x)}\right )+\sin ^{-1}(c+d x) \left (-\frac{\sin ^{-1}(c+d x)}{c+d x}+2 \log \left (1-e^{i \sin ^{-1}(c+d x)}\right )-2 \log \left (1+e^{i \sin ^{-1}(c+d x)}\right )\right )\right )+4 a b^3 \left (6 i \sin ^{-1}(c+d x) \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c+d x)}\right )-6 i \sin ^{-1}(c+d x) \text{PolyLog}\left (2,e^{i \sin ^{-1}(c+d x)}\right )-6 \text{PolyLog}\left (3,-e^{i \sin ^{-1}(c+d x)}\right )+6 \text{PolyLog}\left (3,e^{i \sin ^{-1}(c+d x)}\right )-\frac{\sin ^{-1}(c+d x)^3}{c+d x}+3 \sin ^{-1}(c+d x)^2 \log \left (1-e^{i \sin ^{-1}(c+d x)}\right )-3 \sin ^{-1}(c+d x)^2 \log \left (1+e^{i \sin ^{-1}(c+d x)}\right )\right )+b^4 \left (12 i \sin ^{-1}(c+d x)^2 \text{PolyLog}\left (2,e^{-i \sin ^{-1}(c+d x)}\right )+12 i \sin ^{-1}(c+d x)^2 \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c+d x)}\right )+24 \sin ^{-1}(c+d x) \text{PolyLog}\left (3,e^{-i \sin ^{-1}(c+d x)}\right )-24 \sin ^{-1}(c+d x) \text{PolyLog}\left (3,-e^{i \sin ^{-1}(c+d x)}\right )-24 i \text{PolyLog}\left (4,e^{-i \sin ^{-1}(c+d x)}\right )-24 i \text{PolyLog}\left (4,-e^{i \sin ^{-1}(c+d x)}\right )-\frac{\sin ^{-1}(c+d x)^4}{c+d x}+i \sin ^{-1}(c+d x)^4+4 \sin ^{-1}(c+d x)^3 \log \left (1-e^{-i \sin ^{-1}(c+d x)}\right )-4 \sin ^{-1}(c+d x)^3 \log \left (1+e^{i \sin ^{-1}(c+d x)}\right )-\frac{i \pi ^4}{2}\right )-4 a^3 b \left (\frac{\sin ^{-1}(c+d x)}{c+d x}-\log \left (\sin \left (\frac{1}{2} \sin ^{-1}(c+d x)\right )\right )+\log \left (\frac{1}{2} (c+d x) \csc \left (\frac{1}{2} \sin ^{-1}(c+d x)\right )\right )\right )-\frac{a^4}{c+d x}}{d e^2} \]
Warning: Unable to verify antiderivative.
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Maple [B] time = 0.076, size = 911, normalized size = 3.4 \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^{4} \arcsin \left (d x + c\right )^{4} + 4 \, a b^{3} \arcsin \left (d x + c\right )^{3} + 6 \, a^{2} b^{2} \arcsin \left (d x + c\right )^{2} + 4 \, a^{3} b \arcsin \left (d x + c\right ) + a^{4}}{d^{2} e^{2} x^{2} + 2 \, c d e^{2} x + c^{2} e^{2}}, 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^{4}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx + \int \frac{b^{4} \operatorname{asin}^{4}{\left (c + d x \right )}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx + \int \frac{4 a b^{3} \operatorname{asin}^{3}{\left (c + d x \right )}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx + \int \frac{6 a^{2} b^{2} \operatorname{asin}^{2}{\left (c + d x \right )}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx + \int \frac{4 a^{3} b \operatorname{asin}{\left (c + d x \right )}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx}{e^{2}} \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 )}^{4}}{{\left (d e x + c e\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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