Optimal. Leaf size=183 \[ \frac {a b^2 x}{\left (a^2+b^2\right )^2}+\frac {a x}{a^2+b^2}-\frac {a \tanh ^3(x)}{3 \left (a^2+b^2\right )}-\frac {a b^2 \tanh (x)}{\left (a^2+b^2\right )^2}-\frac {a \tanh (x)}{a^2+b^2}-\frac {b \text {sech}^3(x)}{3 \left (a^2+b^2\right )}+\frac {b \text {sech}(x)}{a^2+b^2}+\frac {2 b^5 \tanh ^{-1}\left (\frac {a-b \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{a \left (a^2+b^2\right )^{5/2}}+\frac {b^4 x}{a \left (a^2+b^2\right )^2}+\frac {b^3 \text {sech}(x)}{\left (a^2+b^2\right )^2} \]
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Rubi [A] time = 0.38, antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 9, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.692, Rules used = {3898, 2902, 2606, 3473, 8, 2735, 2660, 618, 204} \[ \frac {b^4 x}{a \left (a^2+b^2\right )^2}+\frac {a b^2 x}{\left (a^2+b^2\right )^2}+\frac {a x}{a^2+b^2}+\frac {2 b^5 \tanh ^{-1}\left (\frac {a-b \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{a \left (a^2+b^2\right )^{5/2}}-\frac {a \tanh ^3(x)}{3 \left (a^2+b^2\right )}-\frac {a b^2 \tanh (x)}{\left (a^2+b^2\right )^2}-\frac {a \tanh (x)}{a^2+b^2}-\frac {b \text {sech}^3(x)}{3 \left (a^2+b^2\right )}+\frac {b^3 \text {sech}(x)}{\left (a^2+b^2\right )^2}+\frac {b \text {sech}(x)}{a^2+b^2} \]
Antiderivative was successfully verified.
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Rule 8
Rule 204
Rule 618
Rule 2606
Rule 2660
Rule 2735
Rule 2902
Rule 3473
Rule 3898
Rubi steps
\begin {align*} \int \frac {\tanh ^4(x)}{a+b \text {csch}(x)} \, dx &=i \int \frac {\sinh (x) \tanh ^4(x)}{i b+i a \sinh (x)} \, dx\\ &=\frac {a \int \tanh ^4(x) \, dx}{a^2+b^2}-\frac {b \int \text {sech}(x) \tanh ^3(x) \, dx}{a^2+b^2}+\frac {\left (i b^2\right ) \int \frac {\sinh (x) \tanh ^2(x)}{i b+i a \sinh (x)} \, dx}{a^2+b^2}\\ &=-\frac {a \tanh ^3(x)}{3 \left (a^2+b^2\right )}+\frac {\left (a b^2\right ) \int \tanh ^2(x) \, dx}{\left (a^2+b^2\right )^2}-\frac {b^3 \int \text {sech}(x) \tanh (x) \, dx}{\left (a^2+b^2\right )^2}+\frac {\left (i b^4\right ) \int \frac {\sinh (x)}{i b+i a \sinh (x)} \, dx}{\left (a^2+b^2\right )^2}+\frac {a \int \tanh ^2(x) \, dx}{a^2+b^2}-\frac {b \operatorname {Subst}\left (\int \left (-1+x^2\right ) \, dx,x,\text {sech}(x)\right )}{a^2+b^2}\\ &=\frac {b^4 x}{a \left (a^2+b^2\right )^2}+\frac {b \text {sech}(x)}{a^2+b^2}-\frac {b \text {sech}^3(x)}{3 \left (a^2+b^2\right )}-\frac {a b^2 \tanh (x)}{\left (a^2+b^2\right )^2}-\frac {a \tanh (x)}{a^2+b^2}-\frac {a \tanh ^3(x)}{3 \left (a^2+b^2\right )}+\frac {\left (a b^2\right ) \int 1 \, dx}{\left (a^2+b^2\right )^2}+\frac {b^3 \operatorname {Subst}(\int 1 \, dx,x,\text {sech}(x))}{\left (a^2+b^2\right )^2}-\frac {\left (i b^5\right ) \int \frac {1}{i b+i a \sinh (x)} \, dx}{a \left (a^2+b^2\right )^2}+\frac {a \int 1 \, dx}{a^2+b^2}\\ &=\frac {a b^2 x}{\left (a^2+b^2\right )^2}+\frac {b^4 x}{a \left (a^2+b^2\right )^2}+\frac {a x}{a^2+b^2}+\frac {b^3 \text {sech}(x)}{\left (a^2+b^2\right )^2}+\frac {b \text {sech}(x)}{a^2+b^2}-\frac {b \text {sech}^3(x)}{3 \left (a^2+b^2\right )}-\frac {a b^2 \tanh (x)}{\left (a^2+b^2\right )^2}-\frac {a \tanh (x)}{a^2+b^2}-\frac {a \tanh ^3(x)}{3 \left (a^2+b^2\right )}-\frac {\left (2 i b^5\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 \left (a^2+b^2\right )^2}\\ &=\frac {a b^2 x}{\left (a^2+b^2\right )^2}+\frac {b^4 x}{a \left (a^2+b^2\right )^2}+\frac {a x}{a^2+b^2}+\frac {b^3 \text {sech}(x)}{\left (a^2+b^2\right )^2}+\frac {b \text {sech}(x)}{a^2+b^2}-\frac {b \text {sech}^3(x)}{3 \left (a^2+b^2\right )}-\frac {a b^2 \tanh (x)}{\left (a^2+b^2\right )^2}-\frac {a \tanh (x)}{a^2+b^2}-\frac {a \tanh ^3(x)}{3 \left (a^2+b^2\right )}+\frac {\left (4 i b^5\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 \left (a^2+b^2\right )^2}\\ &=\frac {a b^2 x}{\left (a^2+b^2\right )^2}+\frac {b^4 x}{a \left (a^2+b^2\right )^2}+\frac {a x}{a^2+b^2}+\frac {2 b^5 \tanh ^{-1}\left (\frac {a-b \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{a \left (a^2+b^2\right )^{5/2}}+\frac {b^3 \text {sech}(x)}{\left (a^2+b^2\right )^2}+\frac {b \text {sech}(x)}{a^2+b^2}-\frac {b \text {sech}^3(x)}{3 \left (a^2+b^2\right )}-\frac {a b^2 \tanh (x)}{\left (a^2+b^2\right )^2}-\frac {a \tanh (x)}{a^2+b^2}-\frac {a \tanh ^3(x)}{3 \left (a^2+b^2\right )}\\ \end {align*}
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Mathematica [A] time = 0.74, size = 141, normalized size = 0.77 \[ \frac {1}{3} \left (-\frac {a \left (4 a^2+7 b^2\right ) \tanh (x)}{\left (a^2+b^2\right )^2}-\frac {b \text {sech}^3(x)}{a^2+b^2}+\frac {3 b \left (a^2+2 b^2\right ) \text {sech}(x)}{\left (a^2+b^2\right )^2}+\frac {a \tanh (x) \text {sech}^2(x)}{a^2+b^2}+\frac {3 \left (x-\frac {2 b^5 \tan ^{-1}\left (\frac {a-b \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2-b^2}}\right )}{\left (-a^2-b^2\right )^{5/2}}\right )}{a}\right ) \]
Antiderivative was successfully verified.
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fricas [B] time = 0.64, size = 1746, normalized size = 9.54 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.15, size = 215, normalized size = 1.17 \[ -\frac {b^{5} \log \left (\frac {{\left | 2 \, a e^{x} + 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a e^{x} + 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \sqrt {a^{2} + b^{2}}} + \frac {x}{a} + \frac {2 \, {\left (3 \, a^{2} b e^{\left (5 \, x\right )} + 6 \, b^{3} e^{\left (5 \, x\right )} + 6 \, a^{3} e^{\left (4 \, x\right )} + 9 \, a b^{2} e^{\left (4 \, x\right )} + 2 \, a^{2} b e^{\left (3 \, x\right )} + 8 \, b^{3} e^{\left (3 \, x\right )} + 6 \, a^{3} e^{\left (2 \, x\right )} + 12 \, a b^{2} e^{\left (2 \, x\right )} + 3 \, a^{2} b e^{x} + 6 \, b^{3} e^{x} + 4 \, a^{3} + 7 \, a b^{2}\right )}}{3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} {\left (e^{\left (2 \, x\right )} + 1\right )}^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.20, size = 207, normalized size = 1.13 \[ -\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{a}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{a}-\frac {2 b^{5} \arctanh \left (\frac {2 \tanh \left (\frac {x}{2}\right ) b -2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{a \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \sqrt {a^{2}+b^{2}}}+\frac {2 \left (-a^{3}-2 a \,b^{2}\right ) \left (\tanh ^{5}\left (\frac {x}{2}\right )\right )+2 b^{3} \left (\tanh ^{4}\left (\frac {x}{2}\right )\right )+2 \left (-\frac {10}{3} a^{3}-\frac {16}{3} a \,b^{2}\right ) \left (\tanh ^{3}\left (\frac {x}{2}\right )\right )+2 \left (2 a^{2} b +4 b^{3}\right ) \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )+2 \left (-a^{3}-2 a \,b^{2}\right ) \tanh \left (\frac {x}{2}\right )+\frac {4 a^{2} b}{3}+\frac {10 b^{3}}{3}}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.41, size = 261, normalized size = 1.43 \[ -\frac {b^{5} \log \left (\frac {a e^{\left (-x\right )} - b - \sqrt {a^{2} + b^{2}}}{a e^{\left (-x\right )} - b + \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \sqrt {a^{2} + b^{2}}} - \frac {2 \, {\left (4 \, a^{3} + 7 \, a b^{2} - 3 \, {\left (a^{2} b + 2 \, b^{3}\right )} e^{\left (-x\right )} + 6 \, {\left (a^{3} + 2 \, a b^{2}\right )} e^{\left (-2 \, x\right )} - 2 \, {\left (a^{2} b + 4 \, b^{3}\right )} e^{\left (-3 \, x\right )} + 3 \, {\left (2 \, a^{3} + 3 \, a b^{2}\right )} e^{\left (-4 \, x\right )} - 3 \, {\left (a^{2} b + 2 \, b^{3}\right )} e^{\left (-5 \, x\right )}\right )}}{3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4} + 3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} e^{\left (-2 \, x\right )} + 3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} e^{\left (-4 \, x\right )} + {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} e^{\left (-6 \, x\right )}\right )}} + \frac {x}{a} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.43, size = 707, normalized size = 3.86 \[ \frac {x}{a}+\frac {\frac {8\,a}{3\,\left (a^2+b^2\right )}+\frac {8\,b\,{\mathrm {e}}^x}{3\,\left (a^2+b^2\right )}}{3\,{\mathrm {e}}^{2\,x}+3\,{\mathrm {e}}^{4\,x}+{\mathrm {e}}^{6\,x}+1}-\frac {\frac {8\,{\mathrm {e}}^x\,\left (a^2\,b+b^3\right )}{3\,{\left (a^2+b^2\right )}^2}+\frac {4\,\left (a^4+a^2\,b^2\right )}{a\,{\left (a^2+b^2\right )}^2}}{2\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^{4\,x}+1}+\frac {\frac {2\,{\mathrm {e}}^x\,\left (a^2\,b+2\,b^3\right )}{{\left (a^2+b^2\right )}^2}+\frac {2\,\left (2\,a^4+3\,a^2\,b^2\right )}{a\,{\left (a^2+b^2\right )}^2}}{{\mathrm {e}}^{2\,x}+1}+\frac {2\,\mathrm {atan}\left (\left ({\mathrm {e}}^x\,\left (\frac {2\,b^5}{a^3\,\sqrt {b^{10}}\,{\left (a^2+b^2\right )}^2\,\left (a^5+2\,a^3\,b^2+a\,b^4\right )}+\frac {2\,\left (2\,a^3\,b^3\,\sqrt {b^{10}}+a\,b^5\,\sqrt {b^{10}}+a^5\,b\,\sqrt {b^{10}}\right )}{a^2\,b^4\,\sqrt {-a^2\,{\left (a^2+b^2\right )}^5}\,\left (a^5+2\,a^3\,b^2+a\,b^4\right )\,\sqrt {-a^{12}-5\,a^{10}\,b^2-10\,a^8\,b^4-10\,a^6\,b^6-5\,a^4\,b^8-a^2\,b^{10}}}\right )-\frac {2\,\left (a^6\,\sqrt {b^{10}}+a^2\,b^4\,\sqrt {b^{10}}+2\,a^4\,b^2\,\sqrt {b^{10}}\right )}{a^2\,b^4\,\sqrt {-a^2\,{\left (a^2+b^2\right )}^5}\,\left (a^5+2\,a^3\,b^2+a\,b^4\right )\,\sqrt {-a^{12}-5\,a^{10}\,b^2-10\,a^8\,b^4-10\,a^6\,b^6-5\,a^4\,b^8-a^2\,b^{10}}}\right )\,\left (\frac {a^6\,\sqrt {-a^{12}-5\,a^{10}\,b^2-10\,a^8\,b^4-10\,a^6\,b^6-5\,a^4\,b^8-a^2\,b^{10}}}{2}+\frac {a^2\,b^4\,\sqrt {-a^{12}-5\,a^{10}\,b^2-10\,a^8\,b^4-10\,a^6\,b^6-5\,a^4\,b^8-a^2\,b^{10}}}{2}+a^4\,b^2\,\sqrt {-a^{12}-5\,a^{10}\,b^2-10\,a^8\,b^4-10\,a^6\,b^6-5\,a^4\,b^8-a^2\,b^{10}}\right )\right )\,\sqrt {b^{10}}}{\sqrt {-a^{12}-5\,a^{10}\,b^2-10\,a^8\,b^4-10\,a^6\,b^6-5\,a^4\,b^8-a^2\,b^{10}}} \]
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 \[ \int \frac {\tanh ^{4}{\relax (x )}}{a + b \operatorname {csch}{\relax (x )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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