Integrand size = 12, antiderivative size = 71 \[ \int \frac {\sinh \left (a+b x^n\right )}{x^2} \, dx=-\frac {e^a \left (-b x^n\right )^{\frac {1}{n}} \Gamma \left (-\frac {1}{n},-b x^n\right )}{2 n x}+\frac {e^{-a} \left (b x^n\right )^{\frac {1}{n}} \Gamma \left (-\frac {1}{n},b x^n\right )}{2 n x} \]
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Time = 0.04 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {5468, 2250} \[ \int \frac {\sinh \left (a+b x^n\right )}{x^2} \, dx=\frac {e^{-a} \left (b x^n\right )^{\frac {1}{n}} \Gamma \left (-\frac {1}{n},b x^n\right )}{2 n x}-\frac {e^a \left (-b x^n\right )^{\frac {1}{n}} \Gamma \left (-\frac {1}{n},-b x^n\right )}{2 n x} \]
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Rule 2250
Rule 5468
Rubi steps \begin{align*} \text {integral}& = -\left (\frac {1}{2} \int \frac {e^{-a-b x^n}}{x^2} \, dx\right )+\frac {1}{2} \int \frac {e^{a+b x^n}}{x^2} \, dx \\ & = -\frac {e^a \left (-b x^n\right )^{\frac {1}{n}} \Gamma \left (-\frac {1}{n},-b x^n\right )}{2 n x}+\frac {e^{-a} \left (b x^n\right )^{\frac {1}{n}} \Gamma \left (-\frac {1}{n},b x^n\right )}{2 n x} \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.90 \[ \int \frac {\sinh \left (a+b x^n\right )}{x^2} \, dx=-\frac {e^a \left (-b x^n\right )^{\frac {1}{n}} \Gamma \left (-\frac {1}{n},-b x^n\right )-e^{-a} \left (b x^n\right )^{\frac {1}{n}} \Gamma \left (-\frac {1}{n},b x^n\right )}{2 n x} \]
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Result contains higher order function than in optimal. Order 5 vs. order 4.
Time = 0.58 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.08
method | result | size |
meijerg | \(-\frac {\operatorname {hypergeom}\left (\left [-\frac {1}{2 n}\right ], \left [\frac {1}{2}, 1-\frac {1}{2 n}\right ], \frac {x^{2 n} b^{2}}{4}\right ) \sinh \left (a \right )}{x}+\frac {x^{-1+n} b \operatorname {hypergeom}\left (\left [\frac {1}{2}-\frac {1}{2 n}\right ], \left [\frac {3}{2}, \frac {3}{2}-\frac {1}{2 n}\right ], \frac {x^{2 n} b^{2}}{4}\right ) \cosh \left (a \right )}{-1+n}\) | \(77\) |
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\[ \int \frac {\sinh \left (a+b x^n\right )}{x^2} \, dx=\int { \frac {\sinh \left (b x^{n} + a\right )}{x^{2}} \,d x } \]
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\[ \int \frac {\sinh \left (a+b x^n\right )}{x^2} \, dx=\int \frac {\sinh {\left (a + b x^{n} \right )}}{x^{2}}\, dx \]
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none
Time = 0.10 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.92 \[ \int \frac {\sinh \left (a+b x^n\right )}{x^2} \, dx=\frac {\left (b x^{n}\right )^{\left (\frac {1}{n}\right )} e^{\left (-a\right )} \Gamma \left (-\frac {1}{n}, b x^{n}\right )}{2 \, n x} - \frac {\left (-b x^{n}\right )^{\left (\frac {1}{n}\right )} e^{a} \Gamma \left (-\frac {1}{n}, -b x^{n}\right )}{2 \, n x} \]
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\[ \int \frac {\sinh \left (a+b x^n\right )}{x^2} \, dx=\int { \frac {\sinh \left (b x^{n} + a\right )}{x^{2}} \,d x } \]
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Timed out. \[ \int \frac {\sinh \left (a+b x^n\right )}{x^2} \, dx=\int \frac {\mathrm {sinh}\left (a+b\,x^n\right )}{x^2} \,d x \]
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