Integrand size = 18, antiderivative size = 98 \[ \int \frac {x \cosh (a+b x)}{\sinh ^{\frac {7}{2}}(a+b x)} \, dx=-\frac {2 x}{5 b \sinh ^{\frac {5}{2}}(a+b x)}-\frac {4 \cosh (a+b x)}{15 b^2 \sinh ^{\frac {3}{2}}(a+b x)}+\frac {4 i \operatorname {EllipticF}\left (\frac {1}{2} \left (i a-\frac {\pi }{2}+i b x\right ),2\right ) \sqrt {i \sinh (a+b x)}}{15 b^2 \sqrt {\sinh (a+b x)}} \]
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Time = 0.04 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {5480, 2716, 2721, 2720} \[ \int \frac {x \cosh (a+b x)}{\sinh ^{\frac {7}{2}}(a+b x)} \, dx=-\frac {4 \cosh (a+b x)}{15 b^2 \sinh ^{\frac {3}{2}}(a+b x)}+\frac {4 i \sqrt {i \sinh (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (i a+i b x-\frac {\pi }{2}\right ),2\right )}{15 b^2 \sqrt {\sinh (a+b x)}}-\frac {2 x}{5 b \sinh ^{\frac {5}{2}}(a+b x)} \]
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Rule 2716
Rule 2720
Rule 2721
Rule 5480
Rubi steps \begin{align*} \text {integral}& = -\frac {2 x}{5 b \sinh ^{\frac {5}{2}}(a+b x)}+\frac {2 \int \frac {1}{\sinh ^{\frac {5}{2}}(a+b x)} \, dx}{5 b} \\ & = -\frac {2 x}{5 b \sinh ^{\frac {5}{2}}(a+b x)}-\frac {4 \cosh (a+b x)}{15 b^2 \sinh ^{\frac {3}{2}}(a+b x)}-\frac {2 \int \frac {1}{\sqrt {\sinh (a+b x)}} \, dx}{15 b} \\ & = -\frac {2 x}{5 b \sinh ^{\frac {5}{2}}(a+b x)}-\frac {4 \cosh (a+b x)}{15 b^2 \sinh ^{\frac {3}{2}}(a+b x)}-\frac {\left (2 \sqrt {i \sinh (a+b x)}\right ) \int \frac {1}{\sqrt {i \sinh (a+b x)}} \, dx}{15 b \sqrt {\sinh (a+b x)}} \\ & = -\frac {2 x}{5 b \sinh ^{\frac {5}{2}}(a+b x)}-\frac {4 \cosh (a+b x)}{15 b^2 \sinh ^{\frac {3}{2}}(a+b x)}+\frac {4 i \operatorname {EllipticF}\left (\frac {1}{2} \left (i a-\frac {\pi }{2}+i b x\right ),2\right ) \sqrt {i \sinh (a+b x)}}{15 b^2 \sqrt {\sinh (a+b x)}} \\ \end{align*}
Time = 0.36 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.68 \[ \int \frac {x \cosh (a+b x)}{\sinh ^{\frac {7}{2}}(a+b x)} \, dx=-\frac {2 \left (3 b x-2 i \operatorname {EllipticF}\left (\frac {1}{4} (-2 i a+\pi -2 i b x),2\right ) (i \sinh (a+b x))^{5/2}+\sinh (2 (a+b x))\right )}{15 b^2 \sinh ^{\frac {5}{2}}(a+b x)} \]
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\[\int \frac {x \cosh \left (b x +a \right )}{\sinh \left (b x +a \right )^{\frac {7}{2}}}d x\]
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Exception generated. \[ \int \frac {x \cosh (a+b x)}{\sinh ^{\frac {7}{2}}(a+b x)} \, dx=\text {Exception raised: TypeError} \]
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Timed out. \[ \int \frac {x \cosh (a+b x)}{\sinh ^{\frac {7}{2}}(a+b x)} \, dx=\text {Timed out} \]
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\[ \int \frac {x \cosh (a+b x)}{\sinh ^{\frac {7}{2}}(a+b x)} \, dx=\int { \frac {x \cosh \left (b x + a\right )}{\sinh \left (b x + a\right )^{\frac {7}{2}}} \,d x } \]
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\[ \int \frac {x \cosh (a+b x)}{\sinh ^{\frac {7}{2}}(a+b x)} \, dx=\int { \frac {x \cosh \left (b x + a\right )}{\sinh \left (b x + a\right )^{\frac {7}{2}}} \,d x } \]
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Timed out. \[ \int \frac {x \cosh (a+b x)}{\sinh ^{\frac {7}{2}}(a+b x)} \, dx=\int \frac {x\,\mathrm {cosh}\left (a+b\,x\right )}{{\mathrm {sinh}\left (a+b\,x\right )}^{7/2}} \,d x \]
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