Optimal. Leaf size=55 \[ \frac {3 e^{3 x}}{4 \left (1-e^{4 x}\right )}-\frac {e^{3 x}}{\left (1-e^{4 x}\right )^2}+\frac {5}{8} \tan ^{-1}\left (e^x\right )-\frac {5}{8} \tanh ^{-1}\left (e^x\right ) \]
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Rubi [A] time = 0.05, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {2282, 12, 463, 457, 298, 203, 206} \[ \frac {3 e^{3 x}}{4 \left (1-e^{4 x}\right )}-\frac {e^{3 x}}{\left (1-e^{4 x}\right )^2}+\frac {5}{8} \tan ^{-1}\left (e^x\right )-\frac {5}{8} \tanh ^{-1}\left (e^x\right ) \]
Antiderivative was successfully verified.
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Rule 12
Rule 203
Rule 206
Rule 298
Rule 457
Rule 463
Rule 2282
Rubi steps
\begin {align*} \int e^x \coth ^2(2 x) \text {csch}(2 x) \, dx &=\operatorname {Subst}\left (\int \frac {2 x^2 \left (1+x^4\right )^2}{\left (-1+x^4\right )^3} \, dx,x,e^x\right )\\ &=2 \operatorname {Subst}\left (\int \frac {x^2 \left (1+x^4\right )^2}{\left (-1+x^4\right )^3} \, dx,x,e^x\right )\\ &=-\frac {e^{3 x}}{\left (1-e^{4 x}\right )^2}+\frac {1}{4} \operatorname {Subst}\left (\int \frac {x^2 \left (4+8 x^4\right )}{\left (-1+x^4\right )^2} \, dx,x,e^x\right )\\ &=-\frac {e^{3 x}}{\left (1-e^{4 x}\right )^2}+\frac {3 e^{3 x}}{4 \left (1-e^{4 x}\right )}+\frac {5}{4} \operatorname {Subst}\left (\int \frac {x^2}{-1+x^4} \, dx,x,e^x\right )\\ &=-\frac {e^{3 x}}{\left (1-e^{4 x}\right )^2}+\frac {3 e^{3 x}}{4 \left (1-e^{4 x}\right )}-\frac {5}{8} \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,e^x\right )+\frac {5}{8} \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,e^x\right )\\ &=-\frac {e^{3 x}}{\left (1-e^{4 x}\right )^2}+\frac {3 e^{3 x}}{4 \left (1-e^{4 x}\right )}+\frac {5}{8} \tan ^{-1}\left (e^x\right )-\frac {5}{8} \tanh ^{-1}\left (e^x\right )\\ \end {align*}
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Mathematica [C] time = 3.34, size = 161, normalized size = 2.93 \[ -\frac {16 e^{7 x} \left (e^{4 x}+1\right )^2 \, _5F_4\left (\frac {7}{4},2,2,2,2;1,1,1,\frac {19}{4};e^{4 x}\right )}{1155}-\frac {8 e^{7 x} \left (26 e^{4 x}+11 e^{8 x}+15\right ) \, _4F_3\left (\frac {7}{4},2,2,2;1,1,\frac {19}{4};e^{4 x}\right )}{1155}+\frac {e^{-5 x} \left (-7 \left (24152 e^{4 x}-10058 e^{8 x}-9048 e^{12 x}+513 e^{16 x}+25289\right ) \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};e^{4 x}\right )+244931 e^{4 x}+43161 e^{8 x}-26091 e^{12 x}+177023\right )}{10752} \]
Warning: Unable to verify antiderivative.
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fricas [B] time = 0.47, size = 557, normalized size = 10.13 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.12, size = 42, normalized size = 0.76 \[ -\frac {3 \, e^{\left (7 \, x\right )} + e^{\left (3 \, x\right )}}{4 \, {\left (e^{\left (4 \, x\right )} - 1\right )}^{2}} + \frac {5}{8} \, \arctan \left (e^{x}\right ) - \frac {5}{16} \, \log \left (e^{x} + 1\right ) + \frac {5}{16} \, \log \left ({\left | e^{x} - 1 \right |}\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.45, size = 56, normalized size = 1.02 \[ -\frac {{\mathrm e}^{3 x} \left (3 \,{\mathrm e}^{4 x}+1\right )}{4 \left ({\mathrm e}^{4 x}-1\right )^{2}}+\frac {5 \ln \left ({\mathrm e}^{x}-1\right )}{16}-\frac {5 \ln \left ({\mathrm e}^{x}+1\right )}{16}+\frac {5 i \ln \left ({\mathrm e}^{x}+i\right )}{16}-\frac {5 i \ln \left ({\mathrm e}^{x}-i\right )}{16} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.64, size = 47, normalized size = 0.85 \[ -\frac {3 \, e^{\left (7 \, x\right )} + e^{\left (3 \, x\right )}}{4 \, {\left (e^{\left (8 \, x\right )} - 2 \, e^{\left (4 \, x\right )} + 1\right )}} + \frac {5}{8} \, \arctan \left (e^{x}\right ) - \frac {5}{16} \, \log \left (e^{x} + 1\right ) + \frac {5}{16} \, \log \left (e^{x} - 1\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.87, size = 62, normalized size = 1.13 \[ \frac {5\,\ln \left (\frac {25\,{\mathrm {e}}^x}{16}-\frac {25}{16}\right )}{16}-\frac {5\,\ln \left (\frac {25\,{\mathrm {e}}^x}{16}+\frac {25}{16}\right )}{16}-\frac {5\,\mathrm {atan}\left ({\mathrm {e}}^{-x}\right )}{8}-\frac {{\mathrm {e}}^{3\,x}}{{\mathrm {e}}^{8\,x}-2\,{\mathrm {e}}^{4\,x}+1}-\frac {3\,{\mathrm {e}}^{3\,x}}{4\,\left ({\mathrm {e}}^{4\,x}-1\right )} \]
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 \[ \int e^{x} \coth ^{2}{\left (2 x \right )} \operatorname {csch}{\left (2 x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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