Integrand size = 13, antiderivative size = 162 \[ \int \frac {\sinh ^4(x)}{(a+b \sinh (x))^2} \, dx=\frac {\left (6 a^2-b^2\right ) x}{2 b^4}+\frac {2 a^3 \left (3 a^2+4 b^2\right ) \text {arctanh}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{b^4 \left (a^2+b^2\right )^{3/2}}-\frac {a \left (3 a^2+2 b^2\right ) \cosh (x)}{b^3 \left (a^2+b^2\right )}+\frac {\left (3 a^2+b^2\right ) \cosh (x) \sinh (x)}{2 b^2 \left (a^2+b^2\right )}-\frac {a^2 \cosh (x) \sinh ^2(x)}{b \left (a^2+b^2\right ) (a+b \sinh (x))} \]
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Time = 0.28 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {2871, 3128, 3102, 2814, 2739, 632, 212} \[ \int \frac {\sinh ^4(x)}{(a+b \sinh (x))^2} \, dx=-\frac {a^2 \sinh ^2(x) \cosh (x)}{b \left (a^2+b^2\right ) (a+b \sinh (x))}+\frac {\left (3 a^2+b^2\right ) \sinh (x) \cosh (x)}{2 b^2 \left (a^2+b^2\right )}+\frac {x \left (6 a^2-b^2\right )}{2 b^4}-\frac {a \left (3 a^2+2 b^2\right ) \cosh (x)}{b^3 \left (a^2+b^2\right )}+\frac {2 a^3 \left (3 a^2+4 b^2\right ) \text {arctanh}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{b^4 \left (a^2+b^2\right )^{3/2}} \]
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Rule 212
Rule 632
Rule 2739
Rule 2814
Rule 2871
Rule 3102
Rule 3128
Rubi steps \begin{align*} \text {integral}& = -\frac {a^2 \cosh (x) \sinh ^2(x)}{b \left (a^2+b^2\right ) (a+b \sinh (x))}+\frac {\int \frac {\sinh (x) \left (2 a^2-a b \sinh (x)+\left (3 a^2+b^2\right ) \sinh ^2(x)\right )}{a+b \sinh (x)} \, dx}{b \left (a^2+b^2\right )} \\ & = \frac {\left (3 a^2+b^2\right ) \cosh (x) \sinh (x)}{2 b^2 \left (a^2+b^2\right )}-\frac {a^2 \cosh (x) \sinh ^2(x)}{b \left (a^2+b^2\right ) (a+b \sinh (x))}+\frac {\int \frac {-a \left (3 a^2+b^2\right )+b \left (a^2-b^2\right ) \sinh (x)-2 a \left (3 a^2+2 b^2\right ) \sinh ^2(x)}{a+b \sinh (x)} \, dx}{2 b^2 \left (a^2+b^2\right )} \\ & = -\frac {a \left (3 a^2+2 b^2\right ) \cosh (x)}{b^3 \left (a^2+b^2\right )}+\frac {\left (3 a^2+b^2\right ) \cosh (x) \sinh (x)}{2 b^2 \left (a^2+b^2\right )}-\frac {a^2 \cosh (x) \sinh ^2(x)}{b \left (a^2+b^2\right ) (a+b \sinh (x))}+\frac {i \int \frac {i a b \left (3 a^2+b^2\right )-i \left (6 a^4+5 a^2 b^2-b^4\right ) \sinh (x)}{a+b \sinh (x)} \, dx}{2 b^3 \left (a^2+b^2\right )} \\ & = \frac {\left (6 a^2-b^2\right ) x}{2 b^4}-\frac {a \left (3 a^2+2 b^2\right ) \cosh (x)}{b^3 \left (a^2+b^2\right )}+\frac {\left (3 a^2+b^2\right ) \cosh (x) \sinh (x)}{2 b^2 \left (a^2+b^2\right )}-\frac {a^2 \cosh (x) \sinh ^2(x)}{b \left (a^2+b^2\right ) (a+b \sinh (x))}-\frac {\left (a^3 \left (3 a^2+4 b^2\right )\right ) \int \frac {1}{a+b \sinh (x)} \, dx}{b^4 \left (a^2+b^2\right )} \\ & = \frac {\left (6 a^2-b^2\right ) x}{2 b^4}-\frac {a \left (3 a^2+2 b^2\right ) \cosh (x)}{b^3 \left (a^2+b^2\right )}+\frac {\left (3 a^2+b^2\right ) \cosh (x) \sinh (x)}{2 b^2 \left (a^2+b^2\right )}-\frac {a^2 \cosh (x) \sinh ^2(x)}{b \left (a^2+b^2\right ) (a+b \sinh (x))}-\frac {\left (2 a^3 \left (3 a^2+4 b^2\right )\right ) \text {Subst}\left (\int \frac {1}{a+2 b x-a x^2} \, dx,x,\tanh \left (\frac {x}{2}\right )\right )}{b^4 \left (a^2+b^2\right )} \\ & = \frac {\left (6 a^2-b^2\right ) x}{2 b^4}-\frac {a \left (3 a^2+2 b^2\right ) \cosh (x)}{b^3 \left (a^2+b^2\right )}+\frac {\left (3 a^2+b^2\right ) \cosh (x) \sinh (x)}{2 b^2 \left (a^2+b^2\right )}-\frac {a^2 \cosh (x) \sinh ^2(x)}{b \left (a^2+b^2\right ) (a+b \sinh (x))}+\frac {\left (4 a^3 \left (3 a^2+4 b^2\right )\right ) \text {Subst}\left (\int \frac {1}{4 \left (a^2+b^2\right )-x^2} \, dx,x,2 b-2 a \tanh \left (\frac {x}{2}\right )\right )}{b^4 \left (a^2+b^2\right )} \\ & = \frac {\left (6 a^2-b^2\right ) x}{2 b^4}+\frac {2 a^3 \left (3 a^2+4 b^2\right ) \text {arctanh}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{b^4 \left (a^2+b^2\right )^{3/2}}-\frac {a \left (3 a^2+2 b^2\right ) \cosh (x)}{b^3 \left (a^2+b^2\right )}+\frac {\left (3 a^2+b^2\right ) \cosh (x) \sinh (x)}{2 b^2 \left (a^2+b^2\right )}-\frac {a^2 \cosh (x) \sinh ^2(x)}{b \left (a^2+b^2\right ) (a+b \sinh (x))} \\ \end{align*}
Time = 1.02 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.73 \[ \int \frac {\sinh ^4(x)}{(a+b \sinh (x))^2} \, dx=\frac {-2 \left (-6 a^2+b^2\right ) x+\frac {8 a^3 \left (3 a^2+4 b^2\right ) \arctan \left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2-b^2}}\right )}{\left (-a^2-b^2\right )^{3/2}}-8 a b \cosh (x)-\frac {4 a^4 b \cosh (x)}{\left (a^2+b^2\right ) (a+b \sinh (x))}+b^2 \sinh (2 x)}{4 b^4} \]
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Time = 0.80 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.35
method | result | size |
default | \(\frac {2 a^{3} \left (\frac {\frac {b^{2} \tanh \left (\frac {x}{2}\right )}{a^{2}+b^{2}}+\frac {a b}{a^{2}+b^{2}}}{\tanh \left (\frac {x}{2}\right )^{2} a -2 b \tanh \left (\frac {x}{2}\right )-a}-\frac {\left (3 a^{2}+4 b^{2}\right ) \operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\right )}{b^{4}}-\frac {1}{2 b^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {-b +4 a}{2 b^{3} \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {\left (6 a^{2}-b^{2}\right ) \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{2 b^{4}}+\frac {1}{2 b^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {-b -4 a}{2 b^{3} \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {\left (-6 a^{2}+b^{2}\right ) \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2 b^{4}}\) | \(218\) |
risch | \(\frac {3 x \,a^{2}}{b^{4}}-\frac {x}{2 b^{2}}+\frac {{\mathrm e}^{2 x}}{8 b^{2}}-\frac {a \,{\mathrm e}^{x}}{b^{3}}-\frac {a \,{\mathrm e}^{-x}}{b^{3}}-\frac {{\mathrm e}^{-2 x}}{8 b^{2}}+\frac {2 a^{4} \left ({\mathrm e}^{x} a -b \right )}{b^{4} \left (a^{2}+b^{2}\right ) \left (b \,{\mathrm e}^{2 x}+2 \,{\mathrm e}^{x} a -b \right )}+\frac {3 a^{5} \ln \left ({\mathrm e}^{x}+\frac {a \left (a^{2}+b^{2}\right )^{\frac {3}{2}}+a^{4}+2 a^{2} b^{2}+b^{4}}{b \left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}} b^{4}}+\frac {4 a^{3} \ln \left ({\mathrm e}^{x}+\frac {a \left (a^{2}+b^{2}\right )^{\frac {3}{2}}+a^{4}+2 a^{2} b^{2}+b^{4}}{b \left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}} b^{2}}-\frac {3 a^{5} \ln \left ({\mathrm e}^{x}+\frac {a \left (a^{2}+b^{2}\right )^{\frac {3}{2}}-a^{4}-2 a^{2} b^{2}-b^{4}}{b \left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}} b^{4}}-\frac {4 a^{3} \ln \left ({\mathrm e}^{x}+\frac {a \left (a^{2}+b^{2}\right )^{\frac {3}{2}}-a^{4}-2 a^{2} b^{2}-b^{4}}{b \left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}} b^{2}}\) | \(343\) |
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Leaf count of result is larger than twice the leaf count of optimal. 1769 vs. \(2 (154) = 308\).
Time = 0.30 (sec) , antiderivative size = 1769, normalized size of antiderivative = 10.92 \[ \int \frac {\sinh ^4(x)}{(a+b \sinh (x))^2} \, dx=\text {Too large to display} \]
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Timed out. \[ \int \frac {\sinh ^4(x)}{(a+b \sinh (x))^2} \, dx=\text {Timed out} \]
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Time = 0.28 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.58 \[ \int \frac {\sinh ^4(x)}{(a+b \sinh (x))^2} \, dx=-\frac {{\left (3 \, a^{2} + 4 \, b^{2}\right )} a^{3} \log \left (\frac {b e^{\left (-x\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-x\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{2} b^{4} + b^{6}\right )} \sqrt {a^{2} + b^{2}}} + \frac {a^{2} b^{3} + b^{5} - 6 \, {\left (a^{3} b^{2} + a b^{4}\right )} e^{\left (-x\right )} - {\left (32 \, a^{4} b + 17 \, a^{2} b^{3} + b^{5}\right )} e^{\left (-2 \, x\right )} - 8 \, {\left (2 \, a^{5} - a^{3} b^{2} - a b^{4}\right )} e^{\left (-3 \, x\right )}}{8 \, {\left ({\left (a^{2} b^{5} + b^{7}\right )} e^{\left (-2 \, x\right )} + 2 \, {\left (a^{3} b^{4} + a b^{6}\right )} e^{\left (-3 \, x\right )} - {\left (a^{2} b^{5} + b^{7}\right )} e^{\left (-4 \, x\right )}\right )}} - \frac {8 \, a e^{\left (-x\right )} + b e^{\left (-2 \, x\right )}}{8 \, b^{3}} + \frac {{\left (6 \, a^{2} - b^{2}\right )} x}{2 \, b^{4}} \]
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Time = 0.28 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.45 \[ \int \frac {\sinh ^4(x)}{(a+b \sinh (x))^2} \, dx=-\frac {{\left (3 \, a^{5} + 4 \, a^{3} b^{2}\right )} \log \left (\frac {{\left | 2 \, b e^{x} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{x} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{2} b^{4} + b^{6}\right )} \sqrt {a^{2} + b^{2}}} + \frac {{\left (6 \, a^{2} - b^{2}\right )} x}{2 \, b^{4}} + \frac {b^{2} e^{\left (2 \, x\right )} - 8 \, a b e^{x}}{8 \, b^{4}} + \frac {{\left (a^{2} b^{3} + b^{5} + 8 \, {\left (2 \, a^{5} - a^{3} b^{2} - a b^{4}\right )} e^{\left (3 \, x\right )} - {\left (32 \, a^{4} b + 17 \, a^{2} b^{3} + b^{5}\right )} e^{\left (2 \, x\right )} + 6 \, {\left (a^{3} b^{2} + a b^{4}\right )} e^{x}\right )} e^{\left (-2 \, x\right )}}{8 \, {\left (a^{2} + b^{2}\right )} {\left (b e^{\left (2 \, x\right )} + 2 \, a e^{x} - b\right )} b^{4}} \]
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Time = 1.61 (sec) , antiderivative size = 305, normalized size of antiderivative = 1.88 \[ \int \frac {\sinh ^4(x)}{(a+b \sinh (x))^2} \, dx=\frac {{\mathrm {e}}^{2\,x}}{8\,b^2}-\frac {{\mathrm {e}}^{-2\,x}}{8\,b^2}-\frac {\frac {2\,a^4}{b^2\,\left (a^2\,b+b^3\right )}-\frac {2\,a^5\,{\mathrm {e}}^x}{b^3\,\left (a^2\,b+b^3\right )}}{2\,a\,{\mathrm {e}}^x-b+b\,{\mathrm {e}}^{2\,x}}+\frac {x\,\left (6\,a^2-b^2\right )}{2\,b^4}-\frac {a\,{\mathrm {e}}^x}{b^3}-\frac {a\,{\mathrm {e}}^{-x}}{b^3}-\frac {a^3\,\ln \left (\frac {2\,{\mathrm {e}}^x\,\left (3\,a^5+4\,a^3\,b^2\right )}{a^2\,b^5+b^7}-\frac {2\,a^3\,\left (b-a\,{\mathrm {e}}^x\right )\,\left (3\,a^2+4\,b^2\right )}{b^5\,{\left (a^2+b^2\right )}^{3/2}}\right )\,\left (3\,a^2+4\,b^2\right )}{b^4\,{\left (a^2+b^2\right )}^{3/2}}+\frac {a^3\,\ln \left (\frac {2\,{\mathrm {e}}^x\,\left (3\,a^5+4\,a^3\,b^2\right )}{a^2\,b^5+b^7}+\frac {2\,a^3\,\left (b-a\,{\mathrm {e}}^x\right )\,\left (3\,a^2+4\,b^2\right )}{b^5\,{\left (a^2+b^2\right )}^{3/2}}\right )\,\left (3\,a^2+4\,b^2\right )}{b^4\,{\left (a^2+b^2\right )}^{3/2}} \]
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