Optimal. Leaf size=125 \[ -\frac {\cosh ^3(x) (4 b-3 a \sinh (x))}{12 a^2}+\frac {2 b \left (a^2+b^2\right )^{3/2} \tanh ^{-1}\left (\frac {a-b \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{a^5}-\frac {\cosh (x) \left (8 b \left (a^2+b^2\right )-a \left (3 a^2+4 b^2\right ) \sinh (x)\right )}{8 a^4}+\frac {x \left (3 a^4+12 a^2 b^2+8 b^4\right )}{8 a^5} \]
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Rubi [A] time = 0.38, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {3872, 2865, 2735, 2660, 618, 204} \[ \frac {x \left (12 a^2 b^2+3 a^4+8 b^4\right )}{8 a^5}+\frac {2 b \left (a^2+b^2\right )^{3/2} \tanh ^{-1}\left (\frac {a-b \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{a^5}-\frac {\cosh (x) \left (8 b \left (a^2+b^2\right )-a \left (3 a^2+4 b^2\right ) \sinh (x)\right )}{8 a^4}-\frac {\cosh ^3(x) (4 b-3 a \sinh (x))}{12 a^2} \]
Antiderivative was successfully verified.
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Rule 204
Rule 618
Rule 2660
Rule 2735
Rule 2865
Rule 3872
Rubi steps
\begin {align*} \int \frac {\cosh ^4(x)}{a+b \text {csch}(x)} \, dx &=i \int \frac {\cosh ^4(x) \sinh (x)}{i b+i a \sinh (x)} \, dx\\ &=-\frac {\cosh ^3(x) (4 b-3 a \sinh (x))}{12 a^2}+\frac {\int \frac {\cosh ^2(x) \left (-i a b+i \left (3 a^2+4 b^2\right ) \sinh (x)\right )}{i b+i a \sinh (x)} \, dx}{4 a^2}\\ &=-\frac {\cosh ^3(x) (4 b-3 a \sinh (x))}{12 a^2}-\frac {\cosh (x) \left (8 b \left (a^2+b^2\right )-a \left (3 a^2+4 b^2\right ) \sinh (x)\right )}{8 a^4}+\frac {\int \frac {-i a b \left (5 a^2+4 b^2\right )+i \left (3 a^4+12 a^2 b^2+8 b^4\right ) \sinh (x)}{i b+i a \sinh (x)} \, dx}{8 a^4}\\ &=\frac {\left (3 a^4+12 a^2 b^2+8 b^4\right ) x}{8 a^5}-\frac {\cosh ^3(x) (4 b-3 a \sinh (x))}{12 a^2}-\frac {\cosh (x) \left (8 b \left (a^2+b^2\right )-a \left (3 a^2+4 b^2\right ) \sinh (x)\right )}{8 a^4}-\frac {\left (i b \left (a^2+b^2\right )^2\right ) \int \frac {1}{i b+i a \sinh (x)} \, dx}{a^5}\\ &=\frac {\left (3 a^4+12 a^2 b^2+8 b^4\right ) x}{8 a^5}-\frac {\cosh ^3(x) (4 b-3 a \sinh (x))}{12 a^2}-\frac {\cosh (x) \left (8 b \left (a^2+b^2\right )-a \left (3 a^2+4 b^2\right ) \sinh (x)\right )}{8 a^4}-\frac {\left (2 i b \left (a^2+b^2\right )^2\right ) \operatorname {Subst}\left (\int \frac {1}{i b+2 i a x-i b x^2} \, dx,x,\tanh \left (\frac {x}{2}\right )\right )}{a^5}\\ &=\frac {\left (3 a^4+12 a^2 b^2+8 b^4\right ) x}{8 a^5}-\frac {\cosh ^3(x) (4 b-3 a \sinh (x))}{12 a^2}-\frac {\cosh (x) \left (8 b \left (a^2+b^2\right )-a \left (3 a^2+4 b^2\right ) \sinh (x)\right )}{8 a^4}+\frac {\left (4 i b \left (a^2+b^2\right )^2\right ) \operatorname {Subst}\left (\int \frac {1}{-4 \left (a^2+b^2\right )-x^2} \, dx,x,2 i a-2 i b \tanh \left (\frac {x}{2}\right )\right )}{a^5}\\ &=\frac {\left (3 a^4+12 a^2 b^2+8 b^4\right ) x}{8 a^5}+\frac {2 b \left (a^2+b^2\right )^{3/2} \tanh ^{-1}\left (\frac {a-b \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{a^5}-\frac {\cosh ^3(x) (4 b-3 a \sinh (x))}{12 a^2}-\frac {\cosh (x) \left (8 b \left (a^2+b^2\right )-a \left (3 a^2+4 b^2\right ) \sinh (x)\right )}{8 a^4}\\ \end {align*}
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Mathematica [A] time = 1.21, size = 180, normalized size = 1.44 \[ \frac {-8 a^3 b \cosh (3 x)-24 a b \left (5 a^2+4 b^2\right ) \cosh (x)+3 \left (12 a^4 x+a^4 \sinh (4 x)+48 a^2 b^2 x+8 a^2 \left (a^2+b^2\right ) \sinh (2 x)+64 a^2 b \sqrt {-a^2-b^2} \tan ^{-1}\left (\frac {a-b \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2-b^2}}\right )+64 b^3 \sqrt {-a^2-b^2} \tan ^{-1}\left (\frac {a-b \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2-b^2}}\right )+32 b^4 x\right )}{96 a^5} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.59, size = 924, normalized size = 7.39 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.14, size = 221, normalized size = 1.77 \[ \frac {3 \, a^{3} e^{\left (4 \, x\right )} - 8 \, a^{2} b e^{\left (3 \, x\right )} + 24 \, a^{3} e^{\left (2 \, x\right )} + 24 \, a b^{2} e^{\left (2 \, x\right )} - 120 \, a^{2} b e^{x} - 96 \, b^{3} e^{x}}{192 \, a^{4}} + \frac {{\left (3 \, a^{4} + 12 \, a^{2} b^{2} + 8 \, b^{4}\right )} x}{8 \, a^{5}} - \frac {{\left (8 \, a^{3} b e^{x} + 3 \, a^{4} + 24 \, {\left (5 \, a^{3} b + 4 \, a b^{3}\right )} e^{\left (3 \, x\right )} + 24 \, {\left (a^{4} + a^{2} b^{2}\right )} e^{\left (2 \, x\right )}\right )} e^{\left (-4 \, x\right )}}{192 \, a^{5}} - \frac {{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \log \left (\frac {{\left | 2 \, a e^{x} + 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a e^{x} + 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} a^{5}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.15, size = 486, normalized size = 3.89 \[ \frac {3 \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{8 a}+\frac {7}{8 a \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}+\frac {5}{8 a \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {3 \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{8 a}-\frac {7}{8 a \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {5}{8 a \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {2 b \arctanh \left (\frac {2 \tanh \left (\frac {x}{2}\right ) b -2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{a \sqrt {a^{2}+b^{2}}}-\frac {4 b^{3} \arctanh \left (\frac {2 \tanh \left (\frac {x}{2}\right ) b -2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{a^{3} \sqrt {a^{2}+b^{2}}}-\frac {2 b^{5} \arctanh \left (\frac {2 \tanh \left (\frac {x}{2}\right ) b -2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{a^{5} \sqrt {a^{2}+b^{2}}}-\frac {1}{4 a \left (\tanh \left (\frac {x}{2}\right )+1\right )^{4}}+\frac {1}{4 a \left (\tanh \left (\frac {x}{2}\right )-1\right )^{4}}-\frac {b^{3}}{a^{4} \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {b}{3 a^{2} \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}}{a^{4} \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {b}{3 a^{2} \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 {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right ) b^{4}}{a^{5}}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right ) b^{4}}{a^{5}}+\frac {1}{2 a \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3}}+\frac {1}{2 a \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}+\frac {b}{2 a^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}+\frac {3 b}{2 a^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {b^{2}}{2 a^{3} \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {b}{2 a^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {3 b}{2 a^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {b^{2}}{2 a^{3} \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {3 \ln \left (\tanh \left (\frac {x}{2}\right )-1\right ) b^{2}}{2 a^{3}}+\frac {3 \ln \left (\tanh \left (\frac {x}{2}\right )+1\right ) b^{2}}{2 a^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.41, size = 217, normalized size = 1.74 \[ -\frac {{\left (8 \, a^{2} b e^{\left (-x\right )} - 3 \, a^{3} - 24 \, {\left (a^{3} + a b^{2}\right )} e^{\left (-2 \, x\right )} + 24 \, {\left (5 \, a^{2} b + 4 \, b^{3}\right )} e^{\left (-3 \, x\right )}\right )} e^{\left (4 \, x\right )}}{192 \, a^{4}} - \frac {8 \, a^{2} b e^{\left (-3 \, x\right )} + 3 \, a^{3} e^{\left (-4 \, x\right )} + 24 \, {\left (5 \, a^{2} b + 4 \, b^{3}\right )} e^{\left (-x\right )} + 24 \, {\left (a^{3} + a b^{2}\right )} e^{\left (-2 \, x\right )}}{192 \, a^{4}} + \frac {{\left (3 \, a^{4} + 12 \, a^{2} b^{2} + 8 \, b^{4}\right )} x}{8 \, a^{5}} - \frac {{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \log \left (\frac {a e^{\left (-x\right )} - b - \sqrt {a^{2} + b^{2}}}{a e^{\left (-x\right )} - b + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} a^{5}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.02, size = 247, normalized size = 1.98 \[ \frac {{\mathrm {e}}^{4\,x}}{64\,a}-\frac {{\mathrm {e}}^{-4\,x}}{64\,a}+\frac {x\,\left (3\,a^4+12\,a^2\,b^2+8\,b^4\right )}{8\,a^5}-\frac {{\mathrm {e}}^{-2\,x}\,\left (a^2+b^2\right )}{8\,a^3}+\frac {{\mathrm {e}}^{2\,x}\,\left (a^2+b^2\right )}{8\,a^3}-\frac {{\mathrm {e}}^{-x}\,\left (5\,a^2\,b+4\,b^3\right )}{8\,a^4}-\frac {b\,{\mathrm {e}}^{-3\,x}}{24\,a^2}-\frac {b\,{\mathrm {e}}^{3\,x}}{24\,a^2}-\frac {{\mathrm {e}}^x\,\left (5\,a^2\,b+4\,b^3\right )}{8\,a^4}-\frac {b\,\ln \left (\frac {2\,b\,{\mathrm {e}}^x\,{\left (a^2+b^2\right )}^2}{a^6}-\frac {2\,b\,\left (a-b\,{\mathrm {e}}^x\right )\,{\left (a^2+b^2\right )}^{3/2}}{a^6}\right )\,{\left (a^2+b^2\right )}^{3/2}}{a^5}+\frac {b\,\ln \left (\frac {2\,b\,\left (a-b\,{\mathrm {e}}^x\right )\,{\left (a^2+b^2\right )}^{3/2}}{a^6}+\frac {2\,b\,{\mathrm {e}}^x\,{\left (a^2+b^2\right )}^2}{a^6}\right )\,{\left (a^2+b^2\right )}^{3/2}}{a^5} \]
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 \[ \int \frac {\cosh ^{4}{\relax (x )}}{a + b \operatorname {csch}{\relax (x )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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