Optimal. Leaf size=155 \[ -\frac {\left (3 a^2+9 a b+8 b^2\right ) \log (1-\coth (x))}{16 (a+b)^3}+\frac {\left (3 a^2-9 a b+8 b^2\right ) \log (\coth (x)+1)}{16 (a-b)^3}-\frac {\sinh ^4(x) (b-a \coth (x))}{4 \left (a^2-b^2\right )}-\frac {b^5 \log (a+b \coth (x))}{\left (a^2-b^2\right )^3}-\frac {\sinh ^2(x) \left (4 b^3-a b^2 \left (7-\frac {3 a^2}{b^2}\right ) \coth (x)\right )}{8 \left (a^2-b^2\right )^2} \]
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Rubi [A] time = 0.24, antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {3506, 741, 823, 801} \[ -\frac {b^5 \log (a+b \coth (x))}{\left (a^2-b^2\right )^3}-\frac {\left (3 a^2+9 a b+8 b^2\right ) \log (1-\coth (x))}{16 (a+b)^3}+\frac {\left (3 a^2-9 a b+8 b^2\right ) \log (\coth (x)+1)}{16 (a-b)^3}-\frac {\sinh ^4(x) (b-a \coth (x))}{4 \left (a^2-b^2\right )}-\frac {\sinh ^2(x) \left (4 b^3-a b^2 \left (7-\frac {3 a^2}{b^2}\right ) \coth (x)\right )}{8 \left (a^2-b^2\right )^2} \]
Antiderivative was successfully verified.
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Rule 741
Rule 801
Rule 823
Rule 3506
Rubi steps
\begin {align*} \int \frac {\sinh ^4(x)}{a+b \coth (x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{(a+x) \left (1-\frac {x^2}{b^2}\right )^3} \, dx,x,b \coth (x)\right )}{b}\\ &=-\frac {(b-a \coth (x)) \sinh ^4(x)}{4 \left (a^2-b^2\right )}+\frac {b \operatorname {Subst}\left (\int \frac {-4+\frac {3 a^2}{b^2}+\frac {3 a x}{b^2}}{(a+x) \left (1-\frac {x^2}{b^2}\right )^2} \, dx,x,b \coth (x)\right )}{4 \left (a^2-b^2\right )}\\ &=-\frac {\left (4 b^3-a \left (7-\frac {3 a^2}{b^2}\right ) b^2 \coth (x)\right ) \sinh ^2(x)}{8 \left (a^2-b^2\right )^2}-\frac {(b-a \coth (x)) \sinh ^4(x)}{4 \left (a^2-b^2\right )}-\frac {b^5 \operatorname {Subst}\left (\int \frac {-\frac {3 a^4-7 a^2 b^2+8 b^4}{b^6}+\frac {a \left (7-\frac {3 a^2}{b^2}\right ) x}{b^4}}{(a+x) \left (1-\frac {x^2}{b^2}\right )} \, dx,x,b \coth (x)\right )}{8 \left (a^2-b^2\right )^2}\\ &=-\frac {\left (4 b^3-a \left (7-\frac {3 a^2}{b^2}\right ) b^2 \coth (x)\right ) \sinh ^2(x)}{8 \left (a^2-b^2\right )^2}-\frac {(b-a \coth (x)) \sinh ^4(x)}{4 \left (a^2-b^2\right )}-\frac {b^5 \operatorname {Subst}\left (\int \left (-\frac {(a-b)^2 \left (3 a^2+9 a b+8 b^2\right )}{2 b^5 (a+b) (b-x)}+\frac {8}{(a-b) (a+b) (a+x)}-\frac {(a+b)^2 \left (3 a^2-9 a b+8 b^2\right )}{2 (a-b) b^5 (b+x)}\right ) \, dx,x,b \coth (x)\right )}{8 \left (a^2-b^2\right )^2}\\ &=-\frac {\left (3 a^2+9 a b+8 b^2\right ) \log (1-\coth (x))}{16 (a+b)^3}+\frac {\left (3 a^2-9 a b+8 b^2\right ) \log (1+\coth (x))}{16 (a-b)^3}-\frac {b^5 \log (a+b \coth (x))}{\left (a^2-b^2\right )^3}-\frac {\left (4 b^3-a \left (7-\frac {3 a^2}{b^2}\right ) b^2 \coth (x)\right ) \sinh ^2(x)}{8 \left (a^2-b^2\right )^2}-\frac {(b-a \coth (x)) \sinh ^4(x)}{4 \left (a^2-b^2\right )}\\ \end {align*}
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Mathematica [A] time = 0.29, size = 156, normalized size = 1.01 \[ \frac {12 a^5 x-8 a^5 \sinh (2 x)+a^5 \sinh (4 x)-40 a^3 b^2 x+24 a^3 b^2 \sinh (2 x)-2 a^3 b^2 \sinh (4 x)-b \left (a^2-b^2\right )^2 \cosh (4 x)+4 b \left (a^4-4 a^2 b^2+3 b^4\right ) \cosh (2 x)-32 b^5 \log (a \sinh (x)+b \cosh (x))+60 a b^4 x-16 a b^4 \sinh (2 x)+a b^4 \sinh (4 x)}{32 (a-b)^3 (a+b)^3} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.43, size = 1279, normalized size = 8.25 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.13, size = 229, normalized size = 1.48 \[ -\frac {b^{5} \log \left ({\left | -a e^{\left (2 \, x\right )} - b e^{\left (2 \, x\right )} + a - b \right |}\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} + \frac {{\left (3 \, a^{2} - 9 \, a b + 8 \, b^{2}\right )} x}{8 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )}} - \frac {{\left (18 \, a^{2} e^{\left (4 \, x\right )} - 54 \, a b e^{\left (4 \, x\right )} + 48 \, b^{2} e^{\left (4 \, x\right )} - 8 \, a^{2} e^{\left (2 \, x\right )} + 20 \, a b e^{\left (2 \, x\right )} - 12 \, b^{2} e^{\left (2 \, x\right )} + a^{2} - 2 \, a b + b^{2}\right )} e^{\left (-4 \, x\right )}}{64 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )}} + \frac {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} - 8 \, a e^{\left (2 \, x\right )} - 12 \, b e^{\left (2 \, x\right )}}{64 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.13, size = 354, normalized size = 2.28 \[ -\frac {b^{5} \ln \left (\left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) b +2 a \tanh \left (\frac {x}{2}\right )+b \right )}{\left (a -b \right )^{3} \left (a +b \right )^{3}}+\frac {16}{\left (64 a +64 b \right ) \left (\tanh \left (\frac {x}{2}\right )-1\right )^{4}}+\frac {64}{\left (128 a +128 b \right ) \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3}}-\frac {a}{8 \left (a +b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {3 b}{8 \left (a +b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {3 a}{8 \left (a +b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {5 b}{8 \left (a +b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {3 \ln \left (\tanh \left (\frac {x}{2}\right )-1\right ) a^{2}}{8 \left (a +b \right )^{3}}-\frac {9 \ln \left (\tanh \left (\frac {x}{2}\right )-1\right ) a b}{8 \left (a +b \right )^{3}}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right ) b^{2}}{\left (a +b \right )^{3}}-\frac {16}{\left (64 a -64 b \right ) \left (\tanh \left (\frac {x}{2}\right )+1\right )^{4}}+\frac {64}{\left (128 a -128 b \right ) \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}+\frac {a}{8 \left (a -b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {3 b}{8 \left (a -b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {3 a}{8 \left (a -b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {5 b}{8 \left (a -b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {3 \ln \left (\tanh \left (\frac {x}{2}\right )+1\right ) a^{2}}{8 \left (a -b \right )^{3}}-\frac {9 \ln \left (\tanh \left (\frac {x}{2}\right )+1\right ) a b}{8 \left (a -b \right )^{3}}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right ) b^{2}}{\left (a -b \right )^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.32, size = 166, normalized size = 1.07 \[ -\frac {b^{5} \log \left (-{\left (a - b\right )} e^{\left (-2 \, x\right )} + a + b\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} + \frac {{\left (3 \, a^{2} + 9 \, a b + 8 \, b^{2}\right )} x}{8 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}} - \frac {{\left (4 \, {\left (2 \, a + 3 \, b\right )} e^{\left (-2 \, x\right )} - a - b\right )} e^{\left (4 \, x\right )}}{64 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} + \frac {4 \, {\left (2 \, a - 3 \, b\right )} e^{\left (-2 \, x\right )} - {\left (a - b\right )} e^{\left (-4 \, x\right )}}{64 \, {\left (a^{2} - 2 \, a b + b^{2}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.63, size = 143, normalized size = 0.92 \[ \frac {{\mathrm {e}}^{4\,x}}{64\,a+64\,b}-\frac {{\mathrm {e}}^{-4\,x}}{64\,a-64\,b}+\frac {{\mathrm {e}}^{-2\,x}\,\left (2\,a-3\,b\right )}{16\,{\left (a-b\right )}^2}-\frac {b^5\,\ln \left (b-a+a\,{\mathrm {e}}^{2\,x}+b\,{\mathrm {e}}^{2\,x}\right )}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}+\frac {x\,\left (3\,a^2-9\,a\,b+8\,b^2\right )}{8\,{\left (a-b\right )}^3}-\frac {{\mathrm {e}}^{2\,x}\,\left (2\,a+3\,b\right )}{16\,{\left (a+b\right )}^2} \]
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 \[ \int \frac {\sinh ^{4}{\relax (x )}}{a + b \coth {\relax (x )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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