Optimal. Leaf size=81 \[ \frac {\left (a^2+b^2\right )^2 \log (a+b \sinh (x))}{b^5}-\frac {a \left (a^2+2 b^2\right ) \sinh (x)}{b^4}+\frac {\left (a^2+2 b^2\right ) \sinh ^2(x)}{2 b^3}-\frac {a \sinh ^3(x)}{3 b^2}+\frac {\sinh ^4(x)}{4 b} \]
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Rubi [A] time = 0.09, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2668, 697} \[ \frac {\left (a^2+2 b^2\right ) \sinh ^2(x)}{2 b^3}-\frac {a \left (a^2+2 b^2\right ) \sinh (x)}{b^4}+\frac {\left (a^2+b^2\right )^2 \log (a+b \sinh (x))}{b^5}-\frac {a \sinh ^3(x)}{3 b^2}+\frac {\sinh ^4(x)}{4 b} \]
Antiderivative was successfully verified.
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Rule 697
Rule 2668
Rubi steps
\begin {align*} \int \frac {\cosh ^5(x)}{a+b \sinh (x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (-b^2-x^2\right )^2}{a+x} \, dx,x,b \sinh (x)\right )}{b^5}\\ &=\frac {\operatorname {Subst}\left (\int \left (-a \left (a^2+2 b^2\right )+\left (a^2+2 b^2\right ) x-a x^2+x^3+\frac {\left (a^2+b^2\right )^2}{a+x}\right ) \, dx,x,b \sinh (x)\right )}{b^5}\\ &=\frac {\left (a^2+b^2\right )^2 \log (a+b \sinh (x))}{b^5}-\frac {a \left (a^2+2 b^2\right ) \sinh (x)}{b^4}+\frac {\left (a^2+2 b^2\right ) \sinh ^2(x)}{2 b^3}-\frac {a \sinh ^3(x)}{3 b^2}+\frac {\sinh ^4(x)}{4 b}\\ \end {align*}
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Mathematica [A] time = 0.09, size = 76, normalized size = 0.94 \[ \frac {6 b^2 \left (a^2+b^2\right ) \sinh ^2(x)-12 a b \left (a^2+2 b^2\right ) \sinh (x)+12 \left (a^2+b^2\right )^2 \log (a+b \sinh (x))-4 a b^3 \sinh ^3(x)+3 b^4 \cosh ^4(x)}{12 b^5} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.70, size = 865, normalized size = 10.68 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.49, size = 139, normalized size = 1.72 \[ \frac {3 \, b^{3} {\left (e^{\left (-x\right )} - e^{x}\right )}^{4} + 8 \, a b^{2} {\left (e^{\left (-x\right )} - e^{x}\right )}^{3} + 24 \, a^{2} b {\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 48 \, b^{3} {\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 96 \, a^{3} {\left (e^{\left (-x\right )} - e^{x}\right )} + 192 \, a b^{2} {\left (e^{\left (-x\right )} - e^{x}\right )}}{192 \, b^{4}} + \frac {{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \log \left ({\left | -b {\left (e^{\left (-x\right )} - e^{x}\right )} + 2 \, a \right |}\right )}{b^{5}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.05, size = 447, normalized size = 5.52 \[ \frac {1}{4 b \left (\tanh \left (\frac {x}{2}\right )-1\right )^{4}}+\frac {1}{4 b \left (\tanh \left (\frac {x}{2}\right )+1\right )^{4}}+\frac {1}{2 b \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3}}+\frac {9}{8 b \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}+\frac {7}{8 b \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {1}{2 b \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}+\frac {9}{8 b \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {7}{8 b \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {\ln \left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-2 \tanh \left (\frac {x}{2}\right ) b -a \right )}{b}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{b}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{b}-\frac {2 \ln \left (\tanh \left (\frac {x}{2}\right )-1\right ) a^{2}}{b^{3}}-\frac {2 \ln \left (\tanh \left (\frac {x}{2}\right )+1\right ) a^{2}}{b^{3}}+\frac {2 \ln \left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-2 \tanh \left (\frac {x}{2}\right ) b -a \right ) a^{2}}{b^{3}}+\frac {\ln \left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-2 \tanh \left (\frac {x}{2}\right ) b -a \right ) a^{4}}{b^{5}}+\frac {a^{3}}{b^{4} \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {a}{3 b^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right ) a^{4}}{b^{5}}+\frac {a^{2}}{2 b^{3} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {a}{3 b^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3}}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right ) a^{4}}{b^{5}}+\frac {a^{2}}{2 b^{3} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}+\frac {a^{3}}{b^{4} \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {2 a}{b^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {a}{2 b^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}+\frac {a^{2}}{2 b^{3} \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {2 a}{b^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {a}{2 b^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {a^{2}}{2 b^{3} \left (\tanh \left (\frac {x}{2}\right )+1\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.32, size = 180, normalized size = 2.22 \[ -\frac {{\left (8 \, a b^{2} e^{\left (-x\right )} - 3 \, b^{3} - 12 \, {\left (2 \, a^{2} b + 3 \, b^{3}\right )} e^{\left (-2 \, x\right )} + 24 \, {\left (4 \, a^{3} + 7 \, a b^{2}\right )} e^{\left (-3 \, x\right )}\right )} e^{\left (4 \, x\right )}}{192 \, b^{4}} + \frac {8 \, a b^{2} e^{\left (-3 \, x\right )} + 3 \, b^{3} e^{\left (-4 \, x\right )} + 24 \, {\left (4 \, a^{3} + 7 \, a b^{2}\right )} e^{\left (-x\right )} + 12 \, {\left (2 \, a^{2} b + 3 \, b^{3}\right )} e^{\left (-2 \, x\right )}}{192 \, b^{4}} + \frac {{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} x}{b^{5}} + \frac {{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \log \left (-2 \, a e^{\left (-x\right )} + b e^{\left (-2 \, x\right )} - b\right )}{b^{5}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.88, size = 169, normalized size = 2.09 \[ \frac {{\mathrm {e}}^{-4\,x}}{64\,b}+\frac {{\mathrm {e}}^{4\,x}}{64\,b}+\frac {\ln \left (2\,a\,{\mathrm {e}}^x-b+b\,{\mathrm {e}}^{2\,x}\right )\,\left (a^4+2\,a^2\,b^2+b^4\right )}{b^5}+\frac {{\mathrm {e}}^{-x}\,\left (4\,a^3+7\,a\,b^2\right )}{8\,b^4}+\frac {a\,{\mathrm {e}}^{-3\,x}}{24\,b^2}-\frac {a\,{\mathrm {e}}^{3\,x}}{24\,b^2}-\frac {x\,{\left (a^2+b^2\right )}^2}{b^5}+\frac {{\mathrm {e}}^{-2\,x}\,\left (2\,a^2+3\,b^2\right )}{16\,b^3}+\frac {{\mathrm {e}}^{2\,x}\,\left (2\,a^2+3\,b^2\right )}{16\,b^3}-\frac {{\mathrm {e}}^x\,\left (4\,a^3+7\,a\,b^2\right )}{8\,b^4} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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