Optimal. Leaf size=186 \[ -\frac{3 b^2 \text{PolyLog}\left (3,e^{-2 \sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e}-\frac{3 b^3 \text{PolyLog}\left (4,e^{-2 \sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{d e}-\frac{2 b \text{PolyLog}\left (2,e^{-2 \sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d e}-\frac{3 b^4 \text{PolyLog}\left (5,e^{-2 \sinh ^{-1}(c+d x)}\right )}{2 d e}+\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^5}{5 b d e}+\frac{\log \left (1-e^{-2 \sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^4}{d e} \]
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Rubi [A] time = 0.258579, antiderivative size = 186, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.391, Rules used = {5865, 12, 5659, 3716, 2190, 2531, 6609, 2282, 6589} \[ -\frac{3 b^2 \text{PolyLog}\left (3,e^{2 \sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e}+\frac{3 b^3 \text{PolyLog}\left (4,e^{2 \sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{d e}+\frac{2 b \text{PolyLog}\left (2,e^{2 \sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d e}-\frac{3 b^4 \text{PolyLog}\left (5,e^{2 \sinh ^{-1}(c+d x)}\right )}{2 d e}-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^5}{5 b d e}+\frac{\log \left (1-e^{2 \sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^4}{d e} \]
Warning: Unable to verify antiderivative.
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Rule 5865
Rule 12
Rule 5659
Rule 3716
Rule 2190
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} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b \sinh ^{-1}(x)\right )^4}{e x} \, dx,x,c+d x\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b \sinh ^{-1}(x)\right )^4}{x} \, dx,x,c+d x\right )}{d e}\\ &=\frac{\operatorname{Subst}\left (\int (a+b x)^4 \coth (x) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e}\\ &=-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^5}{5 b d e}-\frac{2 \operatorname{Subst}\left (\int \frac{e^{2 x} (a+b x)^4}{1-e^{2 x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e}\\ &=-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^5}{5 b d e}+\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^4 \log \left (1-e^{2 \sinh ^{-1}(c+d x)}\right )}{d e}-\frac{(4 b) \operatorname{Subst}\left (\int (a+b x)^3 \log \left (1-e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e}\\ &=-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^5}{5 b d e}+\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^4 \log \left (1-e^{2 \sinh ^{-1}(c+d x)}\right )}{d e}+\frac{2 b \left (a+b \sinh ^{-1}(c+d x)\right )^3 \text{Li}_2\left (e^{2 \sinh ^{-1}(c+d x)}\right )}{d e}-\frac{\left (6 b^2\right ) \operatorname{Subst}\left (\int (a+b x)^2 \text{Li}_2\left (e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e}\\ &=-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^5}{5 b d e}+\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^4 \log \left (1-e^{2 \sinh ^{-1}(c+d x)}\right )}{d e}+\frac{2 b \left (a+b \sinh ^{-1}(c+d x)\right )^3 \text{Li}_2\left (e^{2 \sinh ^{-1}(c+d x)}\right )}{d e}-\frac{3 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2 \text{Li}_3\left (e^{2 \sinh ^{-1}(c+d x)}\right )}{d e}+\frac{\left (6 b^3\right ) \operatorname{Subst}\left (\int (a+b x) \text{Li}_3\left (e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e}\\ &=-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^5}{5 b d e}+\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^4 \log \left (1-e^{2 \sinh ^{-1}(c+d x)}\right )}{d e}+\frac{2 b \left (a+b \sinh ^{-1}(c+d x)\right )^3 \text{Li}_2\left (e^{2 \sinh ^{-1}(c+d x)}\right )}{d e}-\frac{3 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2 \text{Li}_3\left (e^{2 \sinh ^{-1}(c+d x)}\right )}{d e}+\frac{3 b^3 \left (a+b \sinh ^{-1}(c+d x)\right ) \text{Li}_4\left (e^{2 \sinh ^{-1}(c+d x)}\right )}{d e}-\frac{\left (3 b^4\right ) \operatorname{Subst}\left (\int \text{Li}_4\left (e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e}\\ &=-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^5}{5 b d e}+\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^4 \log \left (1-e^{2 \sinh ^{-1}(c+d x)}\right )}{d e}+\frac{2 b \left (a+b \sinh ^{-1}(c+d x)\right )^3 \text{Li}_2\left (e^{2 \sinh ^{-1}(c+d x)}\right )}{d e}-\frac{3 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2 \text{Li}_3\left (e^{2 \sinh ^{-1}(c+d x)}\right )}{d e}+\frac{3 b^3 \left (a+b \sinh ^{-1}(c+d x)\right ) \text{Li}_4\left (e^{2 \sinh ^{-1}(c+d x)}\right )}{d e}-\frac{\left (3 b^4\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_4(x)}{x} \, dx,x,e^{2 \sinh ^{-1}(c+d x)}\right )}{2 d e}\\ &=-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^5}{5 b d e}+\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^4 \log \left (1-e^{2 \sinh ^{-1}(c+d x)}\right )}{d e}+\frac{2 b \left (a+b \sinh ^{-1}(c+d x)\right )^3 \text{Li}_2\left (e^{2 \sinh ^{-1}(c+d x)}\right )}{d e}-\frac{3 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2 \text{Li}_3\left (e^{2 \sinh ^{-1}(c+d x)}\right )}{d e}+\frac{3 b^3 \left (a+b \sinh ^{-1}(c+d x)\right ) \text{Li}_4\left (e^{2 \sinh ^{-1}(c+d x)}\right )}{d e}-\frac{3 b^4 \text{Li}_5\left (e^{2 \sinh ^{-1}(c+d x)}\right )}{2 d e}\\ \end{align*}
Mathematica [A] time = 0.0641496, size = 157, normalized size = 0.84 \[ \frac{-3 b^2 \text{PolyLog}\left (3,e^{2 \sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^2+3 b^3 \text{PolyLog}\left (4,e^{2 \sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )+2 b \text{PolyLog}\left (2,e^{2 \sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^3-\frac{3}{2} b^4 \text{PolyLog}\left (5,e^{2 \sinh ^{-1}(c+d x)}\right )-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^5}{5 b}+\log \left (1-e^{2 \sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^4}{d e} \]
Warning: Unable to verify antiderivative.
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Maple [B] time = 0.06, size = 1153, normalized size = 6.2 \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 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^{4}}{c + d x}\, dx + \int \frac{b^{4} \operatorname{asinh}^{4}{\left (c + d x \right )}}{c + d x}\, dx + \int \frac{4 a b^{3} \operatorname{asinh}^{3}{\left (c + d x \right )}}{c + d x}\, dx + \int \frac{6 a^{2} b^{2} \operatorname{asinh}^{2}{\left (c + d x \right )}}{c + d x}\, dx + \int \frac{4 a^{3} b \operatorname{asinh}{\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 \operatorname{arsinh}\left (d x + c\right ) + a\right )}^{4}}{d e x + c e}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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