Integrand size = 14, antiderivative size = 128 \[ \int x^m \cosh ^2\left (a+b x^n\right ) \, dx=\frac {x^{1+m}}{2 (1+m)}-\frac {2^{-\frac {1+m+2 n}{n}} e^{2 a} x^{1+m} \left (-b x^n\right )^{-\frac {1+m}{n}} \Gamma \left (\frac {1+m}{n},-2 b x^n\right )}{n}-\frac {2^{-\frac {1+m+2 n}{n}} e^{-2 a} x^{1+m} \left (b x^n\right )^{-\frac {1+m}{n}} \Gamma \left (\frac {1+m}{n},2 b x^n\right )}{n} \]
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Time = 0.12 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {5471, 5469, 2250} \[ \int x^m \cosh ^2\left (a+b x^n\right ) \, dx=-\frac {e^{2 a} 2^{-\frac {m+2 n+1}{n}} x^{m+1} \left (-b x^n\right )^{-\frac {m+1}{n}} \Gamma \left (\frac {m+1}{n},-2 b x^n\right )}{n}-\frac {e^{-2 a} 2^{-\frac {m+2 n+1}{n}} x^{m+1} \left (b x^n\right )^{-\frac {m+1}{n}} \Gamma \left (\frac {m+1}{n},2 b x^n\right )}{n}+\frac {x^{m+1}}{2 (m+1)} \]
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Rule 2250
Rule 5469
Rule 5471
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {x^m}{2}+\frac {1}{2} x^m \cosh \left (2 a+2 b x^n\right )\right ) \, dx \\ & = \frac {x^{1+m}}{2 (1+m)}+\frac {1}{2} \int x^m \cosh \left (2 a+2 b x^n\right ) \, dx \\ & = \frac {x^{1+m}}{2 (1+m)}+\frac {1}{4} \int e^{-2 a-2 b x^n} x^m \, dx+\frac {1}{4} \int e^{2 a+2 b x^n} x^m \, dx \\ & = \frac {x^{1+m}}{2 (1+m)}-\frac {2^{-\frac {1+m+2 n}{n}} e^{2 a} x^{1+m} \left (-b x^n\right )^{-\frac {1+m}{n}} \Gamma \left (\frac {1+m}{n},-2 b x^n\right )}{n}-\frac {2^{-\frac {1+m+2 n}{n}} e^{-2 a} x^{1+m} \left (b x^n\right )^{-\frac {1+m}{n}} \Gamma \left (\frac {1+m}{n},2 b x^n\right )}{n} \\ \end{align*}
Time = 0.19 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.91 \[ \int x^m \cosh ^2\left (a+b x^n\right ) \, dx=-\frac {x^{1+m} \left (-2 n+2^{-\frac {1+m}{n}} e^{2 a} (1+m) \left (-b x^n\right )^{-\frac {1+m}{n}} \Gamma \left (\frac {1+m}{n},-2 b x^n\right )+2^{-\frac {1+m}{n}} e^{-2 a} (1+m) \left (b x^n\right )^{-\frac {1+m}{n}} \Gamma \left (\frac {1+m}{n},2 b x^n\right )\right )}{4 (1+m) n} \]
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\[\int x^{m} \cosh \left (a +b \,x^{n}\right )^{2}d x\]
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\[ \int x^m \cosh ^2\left (a+b x^n\right ) \, dx=\int { x^{m} \cosh \left (b x^{n} + a\right )^{2} \,d x } \]
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\[ \int x^m \cosh ^2\left (a+b x^n\right ) \, dx=\int x^{m} \cosh ^{2}{\left (a + b x^{n} \right )}\, dx \]
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none
Time = 0.11 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.79 \[ \int x^m \cosh ^2\left (a+b x^n\right ) \, dx=-\frac {x^{m + 1} e^{\left (-2 \, a\right )} \Gamma \left (\frac {m + 1}{n}, 2 \, b x^{n}\right )}{4 \, \left (2 \, b x^{n}\right )^{\frac {m + 1}{n}} n} - \frac {x^{m + 1} e^{\left (2 \, a\right )} \Gamma \left (\frac {m + 1}{n}, -2 \, b x^{n}\right )}{4 \, \left (-2 \, b x^{n}\right )^{\frac {m + 1}{n}} n} + \frac {x^{m + 1}}{2 \, {\left (m + 1\right )}} \]
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\[ \int x^m \cosh ^2\left (a+b x^n\right ) \, dx=\int { x^{m} \cosh \left (b x^{n} + a\right )^{2} \,d x } \]
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Timed out. \[ \int x^m \cosh ^2\left (a+b x^n\right ) \, dx=\int x^m\,{\mathrm {cosh}\left (a+b\,x^n\right )}^2 \,d x \]
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