Optimal. Leaf size=102 \[ -\frac {b \sinh ^4(x)}{4 a^2}-\frac {b \left (a^2+b^2\right )^2 \log (a \sinh (x)+b)}{a^6}+\frac {\left (a^2+b^2\right )^2 \sinh (x)}{a^5}-\frac {b \left (2 a^2+b^2\right ) \sinh ^2(x)}{2 a^4}+\frac {\left (2 a^2+b^2\right ) \sinh ^3(x)}{3 a^3}+\frac {\sinh ^5(x)}{5 a} \]
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Rubi [A] time = 0.20, antiderivative size = 102, 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 = {3872, 2837, 12, 772} \[ \frac {\left (2 a^2+b^2\right ) \sinh ^3(x)}{3 a^3}-\frac {b \left (2 a^2+b^2\right ) \sinh ^2(x)}{2 a^4}+\frac {\left (a^2+b^2\right )^2 \sinh (x)}{a^5}-\frac {b \left (a^2+b^2\right )^2 \log (a \sinh (x)+b)}{a^6}-\frac {b \sinh ^4(x)}{4 a^2}+\frac {\sinh ^5(x)}{5 a} \]
Antiderivative was successfully verified.
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Rule 12
Rule 772
Rule 2837
Rule 3872
Rubi steps
\begin {align*} \int \frac {\cosh ^5(x)}{a+b \text {csch}(x)} \, dx &=i \int \frac {\cosh ^5(x) \sinh (x)}{i b+i a \sinh (x)} \, dx\\ &=-\frac {i \operatorname {Subst}\left (\int \frac {x \left (a^2-x^2\right )^2}{a (i b+x)} \, dx,x,i a \sinh (x)\right )}{a^5}\\ &=-\frac {i \operatorname {Subst}\left (\int \frac {x \left (a^2-x^2\right )^2}{i b+x} \, dx,x,i a \sinh (x)\right )}{a^6}\\ &=-\frac {i \operatorname {Subst}\left (\int \left (\left (a^2+b^2\right )^2-\frac {b \left (a^2+b^2\right )^2}{b-i x}+i b \left (2 a^2+b^2\right ) x-\left (2 a^2+b^2\right ) x^2-i b x^3+x^4\right ) \, dx,x,i a \sinh (x)\right )}{a^6}\\ &=-\frac {b \left (a^2+b^2\right )^2 \log (b+a \sinh (x))}{a^6}+\frac {\left (a^2+b^2\right )^2 \sinh (x)}{a^5}-\frac {b \left (2 a^2+b^2\right ) \sinh ^2(x)}{2 a^4}+\frac {\left (2 a^2+b^2\right ) \sinh ^3(x)}{3 a^3}-\frac {b \sinh ^4(x)}{4 a^2}+\frac {\sinh ^5(x)}{5 a}\\ \end {align*}
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Mathematica [A] time = 0.29, size = 97, normalized size = 0.95 \[ \frac {12 a^5 \sinh ^5(x)-15 a^4 b \sinh ^4(x)-30 a^2 b \left (2 a^2+b^2\right ) \sinh ^2(x)+60 a \left (a^2+b^2\right )^2 \sinh (x)-60 b \left (a^2+b^2\right )^2 \log (a \sinh (x)+b)+20 a^3 \left (2 a^2+b^2\right ) \sinh ^3(x)}{60 a^6} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.56, size = 1398, normalized size = 13.71 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.13, size = 194, normalized size = 1.90 \[ -\frac {6 \, a^{4} {\left (e^{\left (-x\right )} - e^{x}\right )}^{5} + 15 \, a^{3} b {\left (e^{\left (-x\right )} - e^{x}\right )}^{4} + 80 \, a^{4} {\left (e^{\left (-x\right )} - e^{x}\right )}^{3} + 40 \, a^{2} b^{2} {\left (e^{\left (-x\right )} - e^{x}\right )}^{3} + 240 \, a^{3} b {\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 120 \, a b^{3} {\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 480 \, a^{4} {\left (e^{\left (-x\right )} - e^{x}\right )} + 960 \, a^{2} b^{2} {\left (e^{\left (-x\right )} - e^{x}\right )} + 480 \, b^{4} {\left (e^{\left (-x\right )} - e^{x}\right )}}{960 \, a^{5}} - \frac {{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \log \left ({\left | -a {\left (e^{\left (-x\right )} - e^{x}\right )} + 2 \, b \right |}\right )}{a^{6}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.14, size = 600, normalized size = 5.88 \[ -\frac {7}{8 a \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {1}{a \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {7}{8 a \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {1}{a \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {1}{5 a \left (\tanh \left (\frac {x}{2}\right )-1\right )^{5}}-\frac {1}{5 a \left (\tanh \left (\frac {x}{2}\right )+1\right )^{5}}+\frac {1}{2 a \left (\tanh \left (\frac {x}{2}\right )+1\right )^{4}}-\frac {1}{2 a \left (\tanh \left (\frac {x}{2}\right )-1\right )^{4}}-\frac {b}{4 a^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{4}}+\frac {b^{3}}{2 a^{4} \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {b^{4}}{a^{5} \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {b^{5} \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{a^{6}}+\frac {b}{2 a^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}-\frac {b^{2}}{3 a^{3} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}+\frac {b^{2}}{2 a^{3} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {b^{3}}{2 a^{4} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {b^{5} \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{a^{6}}-\frac {b}{4 a^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{4}}-\frac {b^{3}}{2 a^{4} \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {b^{4}}{a^{5} \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {b}{2 a^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3}}-\frac {b^{2}}{3 a^{3} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3}}-\frac {b^{2}}{2 a^{3} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {b^{3}}{2 a^{4} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {b \ln \left (\left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) b -2 a \tanh \left (\frac {x}{2}\right )-b \right )}{a^{2}}-\frac {2 b^{3} \ln \left (\left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) b -2 a \tanh \left (\frac {x}{2}\right )-b \right )}{a^{4}}-\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 )}{a^{6}}-\frac {11}{12 a \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3}}-\frac {11}{12 a \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}-\frac {9 b}{8 a^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {7 b}{8 a^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {2 b^{2}}{a^{3} \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {b \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{a^{2}}+\frac {2 b^{3} \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{a^{4}}-\frac {9 b}{8 a^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {7 b}{8 a^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {2 b^{2}}{a^{3} \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {b \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{a^{2}}+\frac {2 b^{3} \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{a^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.32, size = 242, normalized size = 2.37 \[ -\frac {{\left (15 \, a^{3} b e^{\left (-x\right )} - 6 \, a^{4} - 10 \, {\left (5 \, a^{4} + 4 \, a^{2} b^{2}\right )} e^{\left (-2 \, x\right )} + 60 \, {\left (3 \, a^{3} b + 2 \, a b^{3}\right )} e^{\left (-3 \, x\right )} - 60 \, {\left (5 \, a^{4} + 14 \, a^{2} b^{2} + 8 \, b^{4}\right )} e^{\left (-4 \, x\right )}\right )} e^{\left (5 \, x\right )}}{960 \, a^{5}} - \frac {15 \, a^{3} b e^{\left (-4 \, x\right )} + 6 \, a^{4} e^{\left (-5 \, x\right )} + 60 \, {\left (5 \, a^{4} + 14 \, a^{2} b^{2} + 8 \, b^{4}\right )} e^{\left (-x\right )} + 60 \, {\left (3 \, a^{3} b + 2 \, a b^{3}\right )} e^{\left (-2 \, x\right )} + 10 \, {\left (5 \, a^{4} + 4 \, a^{2} b^{2}\right )} e^{\left (-3 \, x\right )}}{960 \, a^{5}} - \frac {{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} x}{a^{6}} - \frac {{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \log \left (-2 \, b e^{\left (-x\right )} + a e^{\left (-2 \, x\right )} - a\right )}{a^{6}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.07, size = 228, normalized size = 2.24 \[ \frac {{\mathrm {e}}^{5\,x}}{160\,a}-\frac {{\mathrm {e}}^{-5\,x}}{160\,a}-\frac {{\mathrm {e}}^{-2\,x}\,\left (3\,a^2\,b+2\,b^3\right )}{16\,a^4}-\frac {{\mathrm {e}}^{2\,x}\,\left (3\,a^2\,b+2\,b^3\right )}{16\,a^4}+\frac {{\mathrm {e}}^x\,\left (5\,a^4+14\,a^2\,b^2+8\,b^4\right )}{16\,a^5}-\frac {b\,{\mathrm {e}}^{-4\,x}}{64\,a^2}-\frac {b\,{\mathrm {e}}^{4\,x}}{64\,a^2}-\frac {\ln \left (2\,b\,{\mathrm {e}}^x-a+a\,{\mathrm {e}}^{2\,x}\right )\,\left (a^4\,b+2\,a^2\,b^3+b^5\right )}{a^6}-\frac {{\mathrm {e}}^{-x}\,\left (5\,a^4+14\,a^2\,b^2+8\,b^4\right )}{16\,a^5}-\frac {{\mathrm {e}}^{-3\,x}\,\left (5\,a^2+4\,b^2\right )}{96\,a^3}+\frac {{\mathrm {e}}^{3\,x}\,\left (5\,a^2+4\,b^2\right )}{96\,a^3}+\frac {b\,x\,{\left (a^2+b^2\right )}^2}{a^6} \]
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 \[ \int \frac {\cosh ^{5}{\relax (x )}}{a + b \operatorname {csch}{\relax (x )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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