Optimal. Leaf size=186 \[ -\frac {3 b^3 \text {Li}_4\left (e^{-2 \sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{d e}-\frac {3 b^2 \text {Li}_3\left (e^{-2 \sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e}-\frac {2 b \text {Li}_2\left (e^{-2 \sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d e}+\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^5}{5 b d e}+\frac {\log \left (1-e^{-2 \sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^4}{d e}-\frac {3 b^4 \text {Li}_5\left (e^{-2 \sinh ^{-1}(c+d x)}\right )}{2 d e} \]
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Rubi [A] time = 0.26, antiderivative size = 186, normalized size of antiderivative = 1.00, 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 12
Rule 2190
Rule 2282
Rule 2531
Rule 3716
Rule 5659
Rule 5865
Rule 6589
Rule 6609
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*}
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Mathematica [A] time = 0.07, size = 157, normalized size = 0.84 \[ \frac {3 b^3 \text {Li}_4\left (e^{2 \sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )-3 b^2 \text {Li}_3\left (e^{2 \sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^2+2 b \text {Li}_2\left (e^{2 \sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^3-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^5}{5 b}+\log \left (1-e^{2 \sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^4-\frac {3}{2} b^4 \text {Li}_5\left (e^{2 \sinh ^{-1}(c+d x)}\right )}{d e} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{4} \operatorname {arsinh}\left (d x + c\right )^{4} + 4 \, a b^{3} \operatorname {arsinh}\left (d x + c\right )^{3} + 6 \, a^{2} b^{2} \operatorname {arsinh}\left (d x + c\right )^{2} + 4 \, a^{3} b \operatorname {arsinh}\left (d x + c\right ) + a^{4}}{d e x + c e}, x\right ) \]
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 \[ \int \frac {{\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}^{4}}{d e x + c e}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.12, size = 1153, normalized size = 6.20 \[ \text {result too large to display} \]
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 \[ \frac {a^{4} \log \left (d e x + c e\right )}{d e} + \int \frac {b^{4} \log \left (d x + c + \sqrt {{\left (d x + c\right )}^{2} + 1}\right )^{4}}{d e x + c e} + \frac {4 \, a b^{3} \log \left (d x + c + \sqrt {{\left (d x + c\right )}^{2} + 1}\right )^{3}}{d e x + c e} + \frac {6 \, a^{2} b^{2} \log \left (d x + c + \sqrt {{\left (d x + c\right )}^{2} + 1}\right )^{2}}{d e x + c e} + \frac {4 \, a^{3} b \log \left (d x + c + \sqrt {{\left (d x + c\right )}^{2} + 1}\right )}{d e x + c e}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right )}^4}{c\,e+d\,e\,x} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a^{4}}{c + d x}\, dx + \int \frac {b^{4} \operatorname {asinh}^{4}{\left (c + d x \right )}}{c + d x}\, dx + \int \frac {4 a b^{3} \operatorname {asinh}^{3}{\left (c + d x \right )}}{c + d x}\, dx + \int \frac {6 a^{2} b^{2} \operatorname {asinh}^{2}{\left (c + d x \right )}}{c + d x}\, dx + \int \frac {4 a^{3} b \operatorname {asinh}{\left (c + d x \right )}}{c + d x}\, dx}{e} \]
Verification of antiderivative is not currently implemented for this CAS.
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