Integrand size = 14, antiderivative size = 86 \[ \int x^{3+m} \cosh ^2(a+b x) \, dx=\frac {x^{4+m}}{2 (4+m)}-\frac {2^{-6-m} e^{2 a} x^m (-b x)^{-m} \Gamma (4+m,-2 b x)}{b^4}-\frac {2^{-6-m} e^{-2 a} x^m (b x)^{-m} \Gamma (4+m,2 b x)}{b^4} \]
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Time = 0.11 (sec) , antiderivative size = 86, 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 = {3393, 3388, 2212} \[ \int x^{3+m} \cosh ^2(a+b x) \, dx=-\frac {e^{2 a} 2^{-m-6} x^m (-b x)^{-m} \Gamma (m+4,-2 b x)}{b^4}-\frac {e^{-2 a} 2^{-m-6} x^m (b x)^{-m} \Gamma (m+4,2 b x)}{b^4}+\frac {x^{m+4}}{2 (m+4)} \]
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Rule 2212
Rule 3388
Rule 3393
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {x^{3+m}}{2}+\frac {1}{2} x^{3+m} \cosh (2 a+2 b x)\right ) \, dx \\ & = \frac {x^{4+m}}{2 (4+m)}+\frac {1}{2} \int x^{3+m} \cosh (2 a+2 b x) \, dx \\ & = \frac {x^{4+m}}{2 (4+m)}+\frac {1}{4} \int e^{-i (2 i a+2 i b x)} x^{3+m} \, dx+\frac {1}{4} \int e^{i (2 i a+2 i b x)} x^{3+m} \, dx \\ & = \frac {x^{4+m}}{2 (4+m)}-\frac {2^{-6-m} e^{2 a} x^m (-b x)^{-m} \Gamma (4+m,-2 b x)}{b^4}-\frac {2^{-6-m} e^{-2 a} x^m (b x)^{-m} \Gamma (4+m,2 b x)}{b^4} \\ \end{align*}
Time = 0.08 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.92 \[ \int x^{3+m} \cosh ^2(a+b x) \, dx=\frac {1}{64} x^m \left (\frac {32 x^4}{4+m}-\frac {2^{-m} e^{2 a} (-b x)^{-m} \Gamma (4+m,-2 b x)}{b^4}-\frac {2^{-m} e^{-2 a} (b x)^{-m} \Gamma (4+m,2 b x)}{b^4}\right ) \]
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\[\int x^{3+m} \cosh \left (b x +a \right )^{2}d x\]
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none
Time = 0.08 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.58 \[ \int x^{3+m} \cosh ^2(a+b x) \, dx=\frac {4 \, b x \cosh \left ({\left (m + 3\right )} \log \left (x\right )\right ) - {\left (m + 4\right )} \cosh \left ({\left (m + 3\right )} \log \left (2 \, b\right ) + 2 \, a\right ) \Gamma \left (m + 4, 2 \, b x\right ) + {\left (m + 4\right )} \cosh \left ({\left (m + 3\right )} \log \left (-2 \, b\right ) - 2 \, a\right ) \Gamma \left (m + 4, -2 \, b x\right ) + {\left (m + 4\right )} \Gamma \left (m + 4, 2 \, b x\right ) \sinh \left ({\left (m + 3\right )} \log \left (2 \, b\right ) + 2 \, a\right ) - {\left (m + 4\right )} \Gamma \left (m + 4, -2 \, b x\right ) \sinh \left ({\left (m + 3\right )} \log \left (-2 \, b\right ) - 2 \, a\right ) + 4 \, b x \sinh \left ({\left (m + 3\right )} \log \left (x\right )\right )}{8 \, {\left (b m + 4 \, b\right )}} \]
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\[ \int x^{3+m} \cosh ^2(a+b x) \, dx=\int x^{m + 3} \cosh ^{2}{\left (a + b x \right )}\, dx \]
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Time = 0.08 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.83 \[ \int x^{3+m} \cosh ^2(a+b x) \, dx=-\frac {1}{4} \, \left (2 \, b x\right )^{-m - 4} x^{m + 4} e^{\left (-2 \, a\right )} \Gamma \left (m + 4, 2 \, b x\right ) - \frac {1}{4} \, \left (-2 \, b x\right )^{-m - 4} x^{m + 4} e^{\left (2 \, a\right )} \Gamma \left (m + 4, -2 \, b x\right ) + \frac {x^{m + 4}}{2 \, {\left (m + 4\right )}} \]
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\[ \int x^{3+m} \cosh ^2(a+b x) \, dx=\int { x^{m + 3} \cosh \left (b x + a\right )^{2} \,d x } \]
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Timed out. \[ \int x^{3+m} \cosh ^2(a+b x) \, dx=\int x^{m+3}\,{\mathrm {cosh}\left (a+b\,x\right )}^2 \,d x \]
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