Optimal. Leaf size=186 \[ -\frac {6 b^3 \text {Li}_2\left (e^{-2 \sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^3}+\frac {6 b^2 \log \left (1-e^{-2 \sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^3}-\frac {2 b \sqrt {(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d e^3 (c+d x)}+\frac {2 b \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d e^3}-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^4}{2 d e^3 (c+d x)^2}-\frac {3 b^4 \text {Li}_3\left (e^{-2 \sinh ^{-1}(c+d x)}\right )}{d e^3} \]
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Rubi [A] time = 0.33, antiderivative size = 186, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {5865, 12, 5661, 5723, 5659, 3716, 2190, 2531, 2282, 6589} \[ \frac {6 b^3 \text {PolyLog}\left (2,e^{2 \sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^3}-\frac {3 b^4 \text {PolyLog}\left (3,e^{2 \sinh ^{-1}(c+d x)}\right )}{d e^3}+\frac {6 b^2 \log \left (1-e^{2 \sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^3}-\frac {2 b \sqrt {(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d e^3 (c+d x)}-\frac {2 b \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d e^3}-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^4}{2 d e^3 (c+d x)^2} \]
Warning: Unable to verify antiderivative.
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Rule 12
Rule 2190
Rule 2282
Rule 2531
Rule 3716
Rule 5659
Rule 5661
Rule 5723
Rule 5865
Rule 6589
Rubi steps
\begin {align*} \int \frac {\left (a+b \sinh ^{-1}(c+d x)\right )^4}{(c e+d e x)^3} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \sinh ^{-1}(x)\right )^4}{e^3 x^3} \, dx,x,c+d x\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \sinh ^{-1}(x)\right )^4}{x^3} \, dx,x,c+d x\right )}{d e^3}\\ &=-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^4}{2 d e^3 (c+d x)^2}+\frac {(2 b) \operatorname {Subst}\left (\int \frac {\left (a+b \sinh ^{-1}(x)\right )^3}{x^2 \sqrt {1+x^2}} \, dx,x,c+d x\right )}{d e^3}\\ &=-\frac {2 b \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d e^3 (c+d x)}-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^4}{2 d e^3 (c+d x)^2}+\frac {\left (6 b^2\right ) \operatorname {Subst}\left (\int \frac {\left (a+b \sinh ^{-1}(x)\right )^2}{x} \, dx,x,c+d x\right )}{d e^3}\\ &=-\frac {2 b \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d e^3 (c+d x)}-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^4}{2 d e^3 (c+d x)^2}+\frac {\left (6 b^2\right ) \operatorname {Subst}\left (\int (a+b x)^2 \coth (x) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e^3}\\ &=-\frac {2 b \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d e^3}-\frac {2 b \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d e^3 (c+d x)}-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^4}{2 d e^3 (c+d x)^2}-\frac {\left (12 b^2\right ) \operatorname {Subst}\left (\int \frac {e^{2 x} (a+b x)^2}{1-e^{2 x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e^3}\\ &=-\frac {2 b \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d e^3}-\frac {2 b \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d e^3 (c+d x)}-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^4}{2 d e^3 (c+d x)^2}+\frac {6 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2 \log \left (1-e^{2 \sinh ^{-1}(c+d x)}\right )}{d e^3}-\frac {\left (12 b^3\right ) \operatorname {Subst}\left (\int (a+b x) \log \left (1-e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e^3}\\ &=-\frac {2 b \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d e^3}-\frac {2 b \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d e^3 (c+d x)}-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^4}{2 d e^3 (c+d x)^2}+\frac {6 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2 \log \left (1-e^{2 \sinh ^{-1}(c+d x)}\right )}{d e^3}+\frac {6 b^3 \left (a+b \sinh ^{-1}(c+d x)\right ) \text {Li}_2\left (e^{2 \sinh ^{-1}(c+d x)}\right )}{d e^3}-\frac {\left (6 b^4\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e^3}\\ &=-\frac {2 b \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d e^3}-\frac {2 b \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d e^3 (c+d x)}-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^4}{2 d e^3 (c+d x)^2}+\frac {6 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2 \log \left (1-e^{2 \sinh ^{-1}(c+d x)}\right )}{d e^3}+\frac {6 b^3 \left (a+b \sinh ^{-1}(c+d x)\right ) \text {Li}_2\left (e^{2 \sinh ^{-1}(c+d x)}\right )}{d e^3}-\frac {\left (3 b^4\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{2 \sinh ^{-1}(c+d x)}\right )}{d e^3}\\ &=-\frac {2 b \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d e^3}-\frac {2 b \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d e^3 (c+d x)}-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^4}{2 d e^3 (c+d x)^2}+\frac {6 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2 \log \left (1-e^{2 \sinh ^{-1}(c+d x)}\right )}{d e^3}+\frac {6 b^3 \left (a+b \sinh ^{-1}(c+d x)\right ) \text {Li}_2\left (e^{2 \sinh ^{-1}(c+d x)}\right )}{d e^3}-\frac {3 b^4 \text {Li}_3\left (e^{2 \sinh ^{-1}(c+d x)}\right )}{d e^3}\\ \end {align*}
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Mathematica [C] time = 1.27, size = 360, normalized size = 1.94 \[ \frac {-\frac {2 a^4}{(c+d x)^2}-\frac {8 a^3 b \sqrt {(c+d x)^2+1}}{c+d x}-\frac {8 a^3 b \sinh ^{-1}(c+d x)}{(c+d x)^2}+24 a^2 b^2 \left (\log (c+d x)-\frac {\sinh ^{-1}(c+d x)^2}{2 (c+d x)^2}-\frac {\sqrt {(c+d x)^2+1} \sinh ^{-1}(c+d x)}{c+d x}\right )+8 a b^3 \left (\sinh ^{-1}(c+d x) \left (-\frac {\sinh ^{-1}(c+d x)^2}{(c+d x)^2}-\frac {3 \sqrt {(c+d x)^2+1} \sinh ^{-1}(c+d x)}{c+d x}+3 \sinh ^{-1}(c+d x)+6 \log \left (1-e^{-2 \sinh ^{-1}(c+d x)}\right )\right )-3 \text {Li}_2\left (e^{-2 \sinh ^{-1}(c+d x)}\right )\right )+b^4 \left (24 \sinh ^{-1}(c+d x) \text {Li}_2\left (e^{2 \sinh ^{-1}(c+d x)}\right )-12 \text {Li}_3\left (e^{2 \sinh ^{-1}(c+d x)}\right )-\frac {8 \sqrt {(c+d x)^2+1} \sinh ^{-1}(c+d x)^3}{c+d x}-8 \sinh ^{-1}(c+d x)^3+24 \sinh ^{-1}(c+d x)^2 \log \left (1-e^{2 \sinh ^{-1}(c+d x)}\right )+i \pi ^3\right )-\frac {2 b^4 \sinh ^{-1}(c+d x)^4}{(c+d x)^2}}{4 d e^3} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.65, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{4} \operatorname {arsinh}\left (d x + c\right )^{4} + 4 \, a b^{3} \operatorname {arsinh}\left (d x + c\right )^{3} + 6 \, a^{2} b^{2} \operatorname {arsinh}\left (d x + c\right )^{2} + 4 \, a^{3} b \operatorname {arsinh}\left (d x + c\right ) + a^{4}}{d^{3} e^{3} x^{3} + 3 \, c d^{2} e^{3} x^{2} + 3 \, c^{2} d e^{3} x + c^{3} e^{3}}, x\right ) \]
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 \[ \int \frac {{\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}^{4}}{{\left (d e x + c e\right )}^{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.22, size = 723, normalized size = 3.89 \[ -\frac {a^{4}}{2 d \,e^{3} \left (d x +c \right )^{2}}-\frac {2 b^{4} \arcsinh \left (d x +c \right )^{3} \sqrt {1+\left (d x +c \right )^{2}}}{d \,e^{3} \left (d x +c \right )}-\frac {2 b^{4} \arcsinh \left (d x +c \right )^{3}}{d \,e^{3}}-\frac {b^{4} \arcsinh \left (d x +c \right )^{4}}{2 d \,e^{3} \left (d x +c \right )^{2}}+\frac {6 b^{4} \arcsinh \left (d x +c \right )^{2} \ln \left (1+d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )}{d \,e^{3}}+\frac {12 b^{4} \arcsinh \left (d x +c \right ) \polylog \left (2, -d x -c -\sqrt {1+\left (d x +c \right )^{2}}\right )}{d \,e^{3}}-\frac {12 b^{4} \polylog \left (3, -d x -c -\sqrt {1+\left (d x +c \right )^{2}}\right )}{d \,e^{3}}+\frac {6 b^{4} \arcsinh \left (d x +c \right )^{2} \ln \left (1-d x -c -\sqrt {1+\left (d x +c \right )^{2}}\right )}{d \,e^{3}}+\frac {12 b^{4} \arcsinh \left (d x +c \right ) \polylog \left (2, d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )}{d \,e^{3}}-\frac {12 b^{4} \polylog \left (3, d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )}{d \,e^{3}}-\frac {6 a \,b^{3} \arcsinh \left (d x +c \right )^{2} \sqrt {1+\left (d x +c \right )^{2}}}{d \,e^{3} \left (d x +c \right )}-\frac {6 a \,b^{3} \arcsinh \left (d x +c \right )^{2}}{d \,e^{3}}-\frac {2 a \,b^{3} \arcsinh \left (d x +c \right )^{3}}{d \,e^{3} \left (d x +c \right )^{2}}+\frac {12 a \,b^{3} \arcsinh \left (d x +c \right ) \ln \left (1+d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )}{d \,e^{3}}+\frac {12 a \,b^{3} \polylog \left (2, -d x -c -\sqrt {1+\left (d x +c \right )^{2}}\right )}{d \,e^{3}}+\frac {12 a \,b^{3} \arcsinh \left (d x +c \right ) \ln \left (1-d x -c -\sqrt {1+\left (d x +c \right )^{2}}\right )}{d \,e^{3}}+\frac {12 a \,b^{3} \polylog \left (2, d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )}{d \,e^{3}}-\frac {6 a^{2} b^{2} \arcsinh \left (d x +c \right )}{d \,e^{3}}-\frac {6 a^{2} b^{2} \arcsinh \left (d x +c \right ) \sqrt {1+\left (d x +c \right )^{2}}}{d \,e^{3} \left (d x +c \right )}-\frac {3 a^{2} b^{2} \arcsinh \left (d x +c \right )^{2}}{d \,e^{3} \left (d x +c \right )^{2}}+\frac {6 a^{2} b^{2} \ln \left (\left (d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )^{2}-1\right )}{d \,e^{3}}-\frac {2 a^{3} b \arcsinh \left (d x +c \right )}{d \,e^{3} \left (d x +c \right )^{2}}-\frac {2 a^{3} b \sqrt {1+\left (d x +c \right )^{2}}}{d \,e^{3} \left (d x +c \right )} \]
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 {b^{4} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )^{4}}{2 \, {\left (d^{3} e^{3} x^{2} + 2 \, c d^{2} e^{3} x + c^{2} d e^{3}\right )}} - 6 \, {\left (\frac {\sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} d \operatorname {arsinh}\left (d x + c\right )}{d^{3} e^{3} x + c d^{2} e^{3}} - \frac {\log \left (d x + c\right )}{d e^{3}}\right )} a^{2} b^{2} - 2 \, a^{3} b {\left (\frac {\sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} d}{d^{3} e^{3} x + c d^{2} e^{3}} + \frac {\operatorname {arsinh}\left (d x + c\right )}{d^{3} e^{3} x^{2} + 2 \, c d^{2} e^{3} x + c^{2} d e^{3}}\right )} - \frac {3 \, a^{2} b^{2} \operatorname {arsinh}\left (d x + c\right )^{2}}{d^{3} e^{3} x^{2} + 2 \, c d^{2} e^{3} x + c^{2} d e^{3}} - \frac {a^{4}}{2 \, {\left (d^{3} e^{3} x^{2} + 2 \, c d^{2} e^{3} x + c^{2} d e^{3}\right )}} + \int \frac {2 \, {\left (2 \, {\left (c^{3} + c\right )} a b^{3} + {\left (c^{3} + c\right )} b^{4} + {\left (2 \, a b^{3} d^{3} + b^{4} d^{3}\right )} x^{3} + 3 \, {\left (2 \, a b^{3} c d^{2} + b^{4} c d^{2}\right )} x^{2} + {\left (2 \, {\left (3 \, c^{2} d + d\right )} a b^{3} + {\left (3 \, c^{2} d + d\right )} b^{4}\right )} x + {\left (b^{4} c^{2} + 2 \, {\left (c^{2} + 1\right )} a b^{3} + {\left (2 \, a b^{3} d^{2} + b^{4} d^{2}\right )} x^{2} + 2 \, {\left (2 \, a b^{3} c d + b^{4} c d\right )} x\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )^{3}}{d^{6} e^{3} x^{6} + 6 \, c d^{5} e^{3} x^{5} + c^{6} e^{3} + c^{4} e^{3} + {\left (15 \, c^{2} d^{4} e^{3} + d^{4} e^{3}\right )} x^{4} + 4 \, {\left (5 \, c^{3} d^{3} e^{3} + c d^{3} e^{3}\right )} x^{3} + 3 \, {\left (5 \, c^{4} d^{2} e^{3} + 2 \, c^{2} d^{2} e^{3}\right )} x^{2} + 2 \, {\left (3 \, c^{5} d e^{3} + 2 \, c^{3} d e^{3}\right )} x + {\left (d^{5} e^{3} x^{5} + 5 \, c d^{4} e^{3} x^{4} + c^{5} e^{3} + c^{3} e^{3} + {\left (10 \, c^{2} d^{3} e^{3} + d^{3} e^{3}\right )} x^{3} + {\left (10 \, c^{3} d^{2} e^{3} + 3 \, c d^{2} e^{3}\right )} x^{2} + {\left (5 \, c^{4} d e^{3} + 3 \, c^{2} d e^{3}\right )} x\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}}\,{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 \[ \int \frac {{\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right )}^4}{{\left (c\,e+d\,e\,x\right )}^3} \,d x \]
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 \[ \frac {\int \frac {a^{4}}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}}\, dx + \int \frac {b^{4} \operatorname {asinh}^{4}{\left (c + d x \right )}}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}}\, dx + \int \frac {4 a b^{3} \operatorname {asinh}^{3}{\left (c + d x \right )}}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}}\, dx + \int \frac {6 a^{2} b^{2} \operatorname {asinh}^{2}{\left (c + d x \right )}}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}}\, dx + \int \frac {4 a^{3} b \operatorname {asinh}{\left (c + d x \right )}}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}}\, dx}{e^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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