Optimal. Leaf size=234 \[ \frac{24 b^3 \text{PolyLog}\left (3,-e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^2}-\frac{24 b^3 \text{PolyLog}\left (3,e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^2}-\frac{12 b^2 \text{PolyLog}\left (2,-e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^2}+\frac{12 b^2 \text{PolyLog}\left (2,e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^2}-\frac{24 b^4 \text{PolyLog}\left (4,-e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}+\frac{24 b^4 \text{PolyLog}\left (4,e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^4}{d e^2 (c+d x)}-\frac{8 b \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d e^2} \]
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Rubi [A] time = 0.317163, antiderivative size = 234, 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 = {5865, 12, 5661, 5760, 4182, 2531, 6609, 2282, 6589} \[ \frac{24 b^3 \text{PolyLog}\left (3,-e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^2}-\frac{24 b^3 \text{PolyLog}\left (3,e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^2}-\frac{12 b^2 \text{PolyLog}\left (2,-e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^2}+\frac{12 b^2 \text{PolyLog}\left (2,e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^2}-\frac{24 b^4 \text{PolyLog}\left (4,-e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}+\frac{24 b^4 \text{PolyLog}\left (4,e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^4}{d e^2 (c+d x)}-\frac{8 b \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d e^2} \]
Antiderivative was successfully verified.
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Rule 5865
Rule 12
Rule 5661
Rule 5760
Rule 4182
Rule 2531
Rule 6609
Rule 2282
Rule 6589
Rubi steps
\begin{align*} \int \frac{\left (a+b \sinh ^{-1}(c+d x)\right )^4}{(c e+d e x)^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b \sinh ^{-1}(x)\right )^4}{e^2 x^2} \, dx,x,c+d x\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b \sinh ^{-1}(x)\right )^4}{x^2} \, dx,x,c+d x\right )}{d e^2}\\ &=-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^4}{d e^2 (c+d x)}+\frac{(4 b) \operatorname{Subst}\left (\int \frac{\left (a+b \sinh ^{-1}(x)\right )^3}{x \sqrt{1+x^2}} \, dx,x,c+d x\right )}{d e^2}\\ &=-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^4}{d e^2 (c+d x)}+\frac{(4 b) \operatorname{Subst}\left (\int (a+b x)^3 \text{csch}(x) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e^2}\\ &=-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^4}{d e^2 (c+d x)}-\frac{8 b \left (a+b \sinh ^{-1}(c+d x)\right )^3 \tanh ^{-1}\left (e^{\sinh ^{-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^x\right ) \, dx,x,\sinh ^{-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^x\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e^2}\\ &=-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^4}{d e^2 (c+d x)}-\frac{8 b \left (a+b \sinh ^{-1}(c+d x)\right )^3 \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}-\frac{12 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2 \text{Li}_2\left (-e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}+\frac{12 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2 \text{Li}_2\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}+\frac{\left (24 b^3\right ) \operatorname{Subst}\left (\int (a+b x) \text{Li}_2\left (-e^x\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e^2}-\frac{\left (24 b^3\right ) \operatorname{Subst}\left (\int (a+b x) \text{Li}_2\left (e^x\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e^2}\\ &=-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^4}{d e^2 (c+d x)}-\frac{8 b \left (a+b \sinh ^{-1}(c+d x)\right )^3 \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}-\frac{12 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2 \text{Li}_2\left (-e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}+\frac{12 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2 \text{Li}_2\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}+\frac{24 b^3 \left (a+b \sinh ^{-1}(c+d x)\right ) \text{Li}_3\left (-e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}-\frac{24 b^3 \left (a+b \sinh ^{-1}(c+d x)\right ) \text{Li}_3\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}-\frac{\left (24 b^4\right ) \operatorname{Subst}\left (\int \text{Li}_3\left (-e^x\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e^2}+\frac{\left (24 b^4\right ) \operatorname{Subst}\left (\int \text{Li}_3\left (e^x\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e^2}\\ &=-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^4}{d e^2 (c+d x)}-\frac{8 b \left (a+b \sinh ^{-1}(c+d x)\right )^3 \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}-\frac{12 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2 \text{Li}_2\left (-e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}+\frac{12 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2 \text{Li}_2\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}+\frac{24 b^3 \left (a+b \sinh ^{-1}(c+d x)\right ) \text{Li}_3\left (-e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}-\frac{24 b^3 \left (a+b \sinh ^{-1}(c+d x)\right ) \text{Li}_3\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}-\frac{\left (24 b^4\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3(-x)}{x} \, dx,x,e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}+\frac{\left (24 b^4\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3(x)}{x} \, dx,x,e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}\\ &=-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^4}{d e^2 (c+d x)}-\frac{8 b \left (a+b \sinh ^{-1}(c+d x)\right )^3 \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}-\frac{12 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2 \text{Li}_2\left (-e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}+\frac{12 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2 \text{Li}_2\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}+\frac{24 b^3 \left (a+b \sinh ^{-1}(c+d x)\right ) \text{Li}_3\left (-e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}-\frac{24 b^3 \left (a+b \sinh ^{-1}(c+d x)\right ) \text{Li}_3\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}-\frac{24 b^4 \text{Li}_4\left (-e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}+\frac{24 b^4 \text{Li}_4\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}\\ \end{align*}
Mathematica [B] time = 1.76449, size = 501, normalized size = 2.14 \[ \frac{12 a^2 b^2 \left (2 \text{PolyLog}\left (2,-e^{-\sinh ^{-1}(c+d x)}\right )-2 \text{PolyLog}\left (2,e^{-\sinh ^{-1}(c+d x)}\right )+\sinh ^{-1}(c+d x) \left (-\frac{\sinh ^{-1}(c+d x)}{c+d x}+2 \log \left (1-e^{-\sinh ^{-1}(c+d x)}\right )-2 \log \left (e^{-\sinh ^{-1}(c+d x)}+1\right )\right )\right )+8 a b^3 \left (6 \sinh ^{-1}(c+d x) \text{PolyLog}\left (2,-e^{-\sinh ^{-1}(c+d x)}\right )-6 \sinh ^{-1}(c+d x) \text{PolyLog}\left (2,e^{-\sinh ^{-1}(c+d x)}\right )+6 \text{PolyLog}\left (3,-e^{-\sinh ^{-1}(c+d x)}\right )-6 \text{PolyLog}\left (3,e^{-\sinh ^{-1}(c+d x)}\right )-\frac{\sinh ^{-1}(c+d x)^3}{c+d x}+3 \sinh ^{-1}(c+d x)^2 \log \left (1-e^{-\sinh ^{-1}(c+d x)}\right )-3 \sinh ^{-1}(c+d x)^2 \log \left (e^{-\sinh ^{-1}(c+d x)}+1\right )\right )+b^4 \left (24 \sinh ^{-1}(c+d x)^2 \text{PolyLog}\left (2,-e^{-\sinh ^{-1}(c+d x)}\right )+24 \sinh ^{-1}(c+d x)^2 \text{PolyLog}\left (2,e^{\sinh ^{-1}(c+d x)}\right )+48 \sinh ^{-1}(c+d x) \text{PolyLog}\left (3,-e^{-\sinh ^{-1}(c+d x)}\right )-48 \sinh ^{-1}(c+d x) \text{PolyLog}\left (3,e^{\sinh ^{-1}(c+d x)}\right )+48 \text{PolyLog}\left (4,-e^{-\sinh ^{-1}(c+d x)}\right )+48 \text{PolyLog}\left (4,e^{\sinh ^{-1}(c+d x)}\right )-\frac{2 \sinh ^{-1}(c+d x)^4}{c+d x}-2 \sinh ^{-1}(c+d x)^4-8 \sinh ^{-1}(c+d x)^3 \log \left (e^{-\sinh ^{-1}(c+d x)}+1\right )+8 \sinh ^{-1}(c+d x)^3 \log \left (1-e^{\sinh ^{-1}(c+d x)}\right )+\pi ^4\right )+4 a^3 b \left (2 \log \left (\frac{2 \sinh ^2\left (\frac{1}{2} \sinh ^{-1}(c+d x)\right )}{c+d x}\right )-\frac{2 \sinh ^{-1}(c+d x)}{c+d x}\right )-\frac{2 a^4}{c+d x}}{2 d e^2} \]
Warning: Unable to verify antiderivative.
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Maple [B] time = 0.047, size = 820, normalized size = 3.5 \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} \operatorname{arsinh}\left (d x + c\right )^{4} + 4 \, a b^{3} \operatorname{arsinh}\left (d x + c\right )^{3} + 6 \, a^{2} b^{2} \operatorname{arsinh}\left (d x + c\right )^{2} + 4 \, a^{3} b \operatorname{arsinh}\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{asinh}^{4}{\left (c + d x \right )}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx + \int \frac{4 a b^{3} \operatorname{asinh}^{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{asinh}^{2}{\left (c + d x \right )}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx + \int \frac{4 a^{3} b \operatorname{asinh}{\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 \operatorname{arsinh}\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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