Integrand size = 13, antiderivative size = 155 \[ \int \frac {\cosh ^4(x)}{a+b \tanh (x)} \, dx=-\frac {\left (3 a^2+9 a b+8 b^2\right ) \log (1-\tanh (x))}{16 (a+b)^3}+\frac {\left (3 a^2-9 a b+8 b^2\right ) \log (1+\tanh (x))}{16 (a-b)^3}-\frac {b^5 \log (a+b \tanh (x))}{\left (a^2-b^2\right )^3}-\frac {\cosh ^4(x) (b-a \tanh (x))}{4 \left (a^2-b^2\right )}+\frac {\cosh ^2(x) \left (4 b^3-a \left (7-\frac {3 a^2}{b^2}\right ) b^2 \tanh (x)\right )}{8 \left (a^2-b^2\right )^2} \]
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Time = 0.18 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {3587, 755, 837, 815} \[ \int \frac {\cosh ^4(x)}{a+b \tanh (x)} \, dx=-\frac {\left (3 a^2+9 a b+8 b^2\right ) \log (1-\tanh (x))}{16 (a+b)^3}+\frac {\left (3 a^2-9 a b+8 b^2\right ) \log (\tanh (x)+1)}{16 (a-b)^3}-\frac {\cosh ^4(x) (b-a \tanh (x))}{4 \left (a^2-b^2\right )}-\frac {b^5 \log (a+b \tanh (x))}{\left (a^2-b^2\right )^3}+\frac {\cosh ^2(x) \left (4 b^3-a b^2 \left (7-\frac {3 a^2}{b^2}\right ) \tanh (x)\right )}{8 \left (a^2-b^2\right )^2} \]
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Rule 755
Rule 815
Rule 837
Rule 3587
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{(a+x) \left (1-\frac {x^2}{b^2}\right )^3} \, dx,x,b \tanh (x)\right )}{b} \\ & = -\frac {\cosh ^4(x) (b-a \tanh (x))}{4 \left (a^2-b^2\right )}+\frac {b \text {Subst}\left (\int \frac {-4+\frac {3 a^2}{b^2}+\frac {3 a x}{b^2}}{(a+x) \left (1-\frac {x^2}{b^2}\right )^2} \, dx,x,b \tanh (x)\right )}{4 \left (a^2-b^2\right )} \\ & = -\frac {\cosh ^4(x) (b-a \tanh (x))}{4 \left (a^2-b^2\right )}+\frac {\cosh ^2(x) \left (4 b^3-a \left (7-\frac {3 a^2}{b^2}\right ) b^2 \tanh (x)\right )}{8 \left (a^2-b^2\right )^2}-\frac {b^5 \text {Subst}\left (\int \frac {-\frac {3 a^4-7 a^2 b^2+8 b^4}{b^6}+\frac {a \left (7-\frac {3 a^2}{b^2}\right ) x}{b^4}}{(a+x) \left (1-\frac {x^2}{b^2}\right )} \, dx,x,b \tanh (x)\right )}{8 \left (a^2-b^2\right )^2} \\ & = -\frac {\cosh ^4(x) (b-a \tanh (x))}{4 \left (a^2-b^2\right )}+\frac {\cosh ^2(x) \left (4 b^3-a \left (7-\frac {3 a^2}{b^2}\right ) b^2 \tanh (x)\right )}{8 \left (a^2-b^2\right )^2}-\frac {b^5 \text {Subst}\left (\int \left (-\frac {(a-b)^2 \left (3 a^2+9 a b+8 b^2\right )}{2 b^5 (a+b) (b-x)}+\frac {8}{(a-b) (a+b) (a+x)}-\frac {(a+b)^2 \left (3 a^2-9 a b+8 b^2\right )}{2 (a-b) b^5 (b+x)}\right ) \, dx,x,b \tanh (x)\right )}{8 \left (a^2-b^2\right )^2} \\ & = -\frac {\left (3 a^2+9 a b+8 b^2\right ) \log (1-\tanh (x))}{16 (a+b)^3}+\frac {\left (3 a^2-9 a b+8 b^2\right ) \log (1+\tanh (x))}{16 (a-b)^3}-\frac {b^5 \log (a+b \tanh (x))}{\left (a^2-b^2\right )^3}-\frac {\cosh ^4(x) (b-a \tanh (x))}{4 \left (a^2-b^2\right )}+\frac {\cosh ^2(x) \left (4 b^3-a \left (7-\frac {3 a^2}{b^2}\right ) b^2 \tanh (x)\right )}{8 \left (a^2-b^2\right )^2} \\ \end{align*}
Time = 0.26 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.33 \[ \int \frac {\cosh ^4(x)}{a+b \tanh (x)} \, dx=\frac {8 b^3 \left (a^2-b^2\right ) \cosh ^2(x)-4 b \left (a^2-b^2\right )^2 \cosh ^4(x)-3 a^5 \log (1-\tanh (x))+10 a^3 b^2 \log (1-\tanh (x))-15 a b^4 \log (1-\tanh (x))+8 b^5 \log (1-\tanh (x))+3 a^5 \log (1+\tanh (x))-10 a^3 b^2 \log (1+\tanh (x))+15 a b^4 \log (1+\tanh (x))+8 b^5 \log (1+\tanh (x))-16 b^5 \log (a+b \tanh (x))+4 a \left (a^2-b^2\right )^2 \cosh ^3(x) \sinh (x)+a \left (3 a^4-10 a^2 b^2+7 b^4\right ) \sinh (2 x)}{16 (a-b)^3 (a+b)^3} \]
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Time = 6.09 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.23
method | result | size |
risch | \(\frac {3 a^{2} x}{8 \left (a +b \right )^{3}}+\frac {9 a x b}{8 \left (a +b \right )^{3}}+\frac {x \,b^{2}}{\left (a +b \right )^{3}}+\frac {{\mathrm e}^{4 x}}{64 a +64 b}+\frac {{\mathrm e}^{2 x} a}{8 \left (a +b \right )^{2}}+\frac {3 \,{\mathrm e}^{2 x} b}{16 \left (a +b \right )^{2}}-\frac {{\mathrm e}^{-2 x} a}{8 \left (a -b \right )^{2}}+\frac {3 \,{\mathrm e}^{-2 x} b}{16 \left (a -b \right )^{2}}-\frac {{\mathrm e}^{-4 x}}{64 \left (a -b \right )}+\frac {2 b^{5} x}{a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}}-\frac {b^{5} \ln \left ({\mathrm e}^{2 x}+\frac {a -b}{a +b}\right )}{a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}}\) | \(191\) |
default | \(-\frac {1}{2 \left (2 a -2 b \right ) \left (\tanh \left (\frac {x}{2}\right )+1\right )^{4}}+\frac {2}{\left (4 a -4 b \right ) \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}-\frac {-5 a +7 b}{8 \left (a -b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {7 a -9 b}{8 \left (a -b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {\left (3 a^{2}-9 a b +8 b^{2}\right ) \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{8 \left (a -b \right )^{3}}-\frac {b^{5} \ln \left (\tanh \left (\frac {x}{2}\right )^{2} a +2 b \tanh \left (\frac {x}{2}\right )+a \right )}{\left (a +b \right )^{3} \left (a -b \right )^{3}}+\frac {1}{2 \left (2 a +2 b \right ) \left (\tanh \left (\frac {x}{2}\right )-1\right )^{4}}+\frac {2}{\left (4 a +4 b \right ) \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3}}-\frac {-7 a -9 b}{8 \left (a +b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {-5 a -7 b}{8 \left (a +b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {\left (-3 a^{2}-9 a b -8 b^{2}\right ) \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{8 \left (a +b \right )^{3}}\) | \(265\) |
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Leaf count of result is larger than twice the leaf count of optimal. 1281 vs. \(2 (147) = 294\).
Time = 0.28 (sec) , antiderivative size = 1281, normalized size of antiderivative = 8.26 \[ \int \frac {\cosh ^4(x)}{a+b \tanh (x)} \, dx=\text {Too large to display} \]
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\[ \int \frac {\cosh ^4(x)}{a+b \tanh (x)} \, dx=\int \frac {\cosh ^{4}{\left (x \right )}}{a + b \tanh {\left (x \right )}}\, dx \]
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Time = 0.20 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.06 \[ \int \frac {\cosh ^4(x)}{a+b \tanh (x)} \, dx=-\frac {b^{5} \log \left (-{\left (a - b\right )} e^{\left (-2 \, x\right )} - a - b\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} + \frac {{\left (3 \, a^{2} + 9 \, a b + 8 \, b^{2}\right )} x}{8 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}} + \frac {{\left (4 \, {\left (2 \, a + 3 \, b\right )} e^{\left (-2 \, x\right )} + a + b\right )} e^{\left (4 \, x\right )}}{64 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} - \frac {4 \, {\left (2 \, a - 3 \, b\right )} e^{\left (-2 \, x\right )} + {\left (a - b\right )} e^{\left (-4 \, x\right )}}{64 \, {\left (a^{2} - 2 \, a b + b^{2}\right )}} \]
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Time = 0.25 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.46 \[ \int \frac {\cosh ^4(x)}{a+b \tanh (x)} \, dx=-\frac {b^{5} \log \left ({\left | a e^{\left (2 \, x\right )} + b e^{\left (2 \, x\right )} + a - b \right |}\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} + \frac {{\left (3 \, a^{2} - 9 \, a b + 8 \, b^{2}\right )} x}{8 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )}} - \frac {{\left (18 \, a^{2} e^{\left (4 \, x\right )} - 54 \, a b e^{\left (4 \, x\right )} + 48 \, b^{2} e^{\left (4 \, x\right )} + 8 \, a^{2} e^{\left (2 \, x\right )} - 20 \, a b e^{\left (2 \, x\right )} + 12 \, b^{2} e^{\left (2 \, x\right )} + a^{2} - 2 \, a b + b^{2}\right )} e^{\left (-4 \, x\right )}}{64 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )}} + \frac {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 8 \, a e^{\left (2 \, x\right )} + 12 \, b e^{\left (2 \, x\right )}}{64 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} \]
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Time = 2.12 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.92 \[ \int \frac {\cosh ^4(x)}{a+b \tanh (x)} \, dx=\frac {{\mathrm {e}}^{4\,x}}{64\,a+64\,b}-\frac {{\mathrm {e}}^{-4\,x}}{64\,a-64\,b}-\frac {{\mathrm {e}}^{-2\,x}\,\left (2\,a-3\,b\right )}{16\,{\left (a-b\right )}^2}-\frac {b^5\,\ln \left (a-b+a\,{\mathrm {e}}^{2\,x}+b\,{\mathrm {e}}^{2\,x}\right )}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}+\frac {x\,\left (3\,a^2-9\,a\,b+8\,b^2\right )}{8\,{\left (a-b\right )}^3}+\frac {{\mathrm {e}}^{2\,x}\,\left (2\,a+3\,b\right )}{16\,{\left (a+b\right )}^2} \]
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