Integrand size = 20, antiderivative size = 263 \[ \int (c+d x)^m (a+a \cosh (e+f x))^2 \, dx=\frac {3 a^2 (c+d x)^{1+m}}{2 d (1+m)}+\frac {2^{-3-m} a^2 e^{2 e-\frac {2 c f}{d}} (c+d x)^m \left (-\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {2 f (c+d x)}{d}\right )}{f}+\frac {a^2 e^{e-\frac {c f}{d}} (c+d x)^m \left (-\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {f (c+d x)}{d}\right )}{f}-\frac {a^2 e^{-e+\frac {c f}{d}} (c+d x)^m \left (\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {f (c+d x)}{d}\right )}{f}-\frac {2^{-3-m} a^2 e^{-2 e+\frac {2 c f}{d}} (c+d x)^m \left (\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {2 f (c+d x)}{d}\right )}{f} \]
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Time = 0.29 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3399, 3393, 3388, 2212} \[ \int (c+d x)^m (a+a \cosh (e+f x))^2 \, dx=\frac {a^2 2^{-m-3} e^{2 e-\frac {2 c f}{d}} (c+d x)^m \left (-\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {2 f (c+d x)}{d}\right )}{f}+\frac {a^2 e^{e-\frac {c f}{d}} (c+d x)^m \left (-\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {f (c+d x)}{d}\right )}{f}-\frac {a^2 e^{\frac {c f}{d}-e} (c+d x)^m \left (\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {f (c+d x)}{d}\right )}{f}-\frac {a^2 2^{-m-3} e^{\frac {2 c f}{d}-2 e} (c+d x)^m \left (\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {2 f (c+d x)}{d}\right )}{f}+\frac {3 a^2 (c+d x)^{m+1}}{2 d (m+1)} \]
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Rule 2212
Rule 3388
Rule 3393
Rule 3399
Rubi steps \begin{align*} \text {integral}& = \left (4 a^2\right ) \int (c+d x)^m \sin ^4\left (\frac {1}{2} (i e+\pi )+\frac {i f x}{2}\right ) \, dx \\ & = \left (4 a^2\right ) \int \left (\frac {3}{8} (c+d x)^m+\frac {1}{2} (c+d x)^m \cosh (e+f x)+\frac {1}{8} (c+d x)^m \cosh (2 e+2 f x)\right ) \, dx \\ & = \frac {3 a^2 (c+d x)^{1+m}}{2 d (1+m)}+\frac {1}{2} a^2 \int (c+d x)^m \cosh (2 e+2 f x) \, dx+\left (2 a^2\right ) \int (c+d x)^m \cosh (e+f x) \, dx \\ & = \frac {3 a^2 (c+d x)^{1+m}}{2 d (1+m)}+\frac {1}{4} a^2 \int e^{-i (2 i e+2 i f x)} (c+d x)^m \, dx+\frac {1}{4} a^2 \int e^{i (2 i e+2 i f x)} (c+d x)^m \, dx+a^2 \int e^{-i (i e+i f x)} (c+d x)^m \, dx+a^2 \int e^{i (i e+i f x)} (c+d x)^m \, dx \\ & = \frac {3 a^2 (c+d x)^{1+m}}{2 d (1+m)}+\frac {2^{-3-m} a^2 e^{2 e-\frac {2 c f}{d}} (c+d x)^m \left (-\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {2 f (c+d x)}{d}\right )}{f}+\frac {a^2 e^{e-\frac {c f}{d}} (c+d x)^m \left (-\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {f (c+d x)}{d}\right )}{f}-\frac {a^2 e^{-e+\frac {c f}{d}} (c+d x)^m \left (\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {f (c+d x)}{d}\right )}{f}-\frac {2^{-3-m} a^2 e^{-2 e+\frac {2 c f}{d}} (c+d x)^m \left (\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {2 f (c+d x)}{d}\right )}{f} \\ \end{align*}
Time = 0.72 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.15 \[ \int (c+d x)^m (a+a \cosh (e+f x))^2 \, dx=-\frac {2^{-5-m} a^2 e^{-2 \left (e+\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {f^2 (c+d x)^2}{d^2}\right )^{-m} (1+\cosh (e+f x))^2 \left (-3 2^{2+m} e^{2 \left (e+\frac {c f}{d}\right )} f (c+d x) \left (-\frac {f^2 (c+d x)^2}{d^2}\right )^m-d e^{4 e} (1+m) \left (f \left (\frac {c}{d}+x\right )\right )^m \Gamma \left (1+m,-\frac {2 f (c+d x)}{d}\right )-2^{3+m} d e^{3 e+\frac {c f}{d}} (1+m) \left (f \left (\frac {c}{d}+x\right )\right )^m \Gamma \left (1+m,-\frac {f (c+d x)}{d}\right )+2^{3+m} d e^{e+\frac {3 c f}{d}} (1+m) \left (-\frac {f (c+d x)}{d}\right )^m \Gamma \left (1+m,\frac {f (c+d x)}{d}\right )+d e^{\frac {4 c f}{d}} (1+m) \left (-\frac {f (c+d x)}{d}\right )^m \Gamma \left (1+m,\frac {2 f (c+d x)}{d}\right )\right ) \text {sech}^4\left (\frac {1}{2} (e+f x)\right )}{d f (1+m)} \]
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\[\int \left (d x +c \right )^{m} \left (a +a \cosh \left (f x +e \right )\right )^{2}d x\]
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Time = 0.09 (sec) , antiderivative size = 493, normalized size of antiderivative = 1.87 \[ \int (c+d x)^m (a+a \cosh (e+f x))^2 \, dx=-\frac {{\left (a^{2} d m + a^{2} d\right )} \cosh \left (\frac {d m \log \left (\frac {2 \, f}{d}\right ) + 2 \, d e - 2 \, c f}{d}\right ) \Gamma \left (m + 1, \frac {2 \, {\left (d f x + c f\right )}}{d}\right ) + 8 \, {\left (a^{2} d m + a^{2} d\right )} \cosh \left (\frac {d m \log \left (\frac {f}{d}\right ) + d e - c f}{d}\right ) \Gamma \left (m + 1, \frac {d f x + c f}{d}\right ) - 8 \, {\left (a^{2} d m + a^{2} d\right )} \cosh \left (\frac {d m \log \left (-\frac {f}{d}\right ) - d e + c f}{d}\right ) \Gamma \left (m + 1, -\frac {d f x + c f}{d}\right ) - {\left (a^{2} d m + a^{2} d\right )} \cosh \left (\frac {d m \log \left (-\frac {2 \, f}{d}\right ) - 2 \, d e + 2 \, c f}{d}\right ) \Gamma \left (m + 1, -\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) - {\left (a^{2} d m + a^{2} d\right )} \Gamma \left (m + 1, \frac {2 \, {\left (d f x + c f\right )}}{d}\right ) \sinh \left (\frac {d m \log \left (\frac {2 \, f}{d}\right ) + 2 \, d e - 2 \, c f}{d}\right ) - 8 \, {\left (a^{2} d m + a^{2} d\right )} \Gamma \left (m + 1, \frac {d f x + c f}{d}\right ) \sinh \left (\frac {d m \log \left (\frac {f}{d}\right ) + d e - c f}{d}\right ) + 8 \, {\left (a^{2} d m + a^{2} d\right )} \Gamma \left (m + 1, -\frac {d f x + c f}{d}\right ) \sinh \left (\frac {d m \log \left (-\frac {f}{d}\right ) - d e + c f}{d}\right ) + {\left (a^{2} d m + a^{2} d\right )} \Gamma \left (m + 1, -\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) \sinh \left (\frac {d m \log \left (-\frac {2 \, f}{d}\right ) - 2 \, d e + 2 \, c f}{d}\right ) - 12 \, {\left (a^{2} d f x + a^{2} c f\right )} \cosh \left (m \log \left (d x + c\right )\right ) - 12 \, {\left (a^{2} d f x + a^{2} c f\right )} \sinh \left (m \log \left (d x + c\right )\right )}{8 \, {\left (d f m + d f\right )}} \]
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Exception generated. \[ \int (c+d x)^m (a+a \cosh (e+f x))^2 \, dx=\text {Exception raised: TypeError} \]
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Time = 0.09 (sec) , antiderivative size = 209, normalized size of antiderivative = 0.79 \[ \int (c+d x)^m (a+a \cosh (e+f x))^2 \, dx=-\frac {1}{4} \, {\left (\frac {{\left (d x + c\right )}^{m + 1} e^{\left (-2 \, e + \frac {2 \, c f}{d}\right )} E_{-m}\left (\frac {2 \, {\left (d x + c\right )} f}{d}\right )}{d} + \frac {{\left (d x + c\right )}^{m + 1} e^{\left (2 \, e - \frac {2 \, c f}{d}\right )} E_{-m}\left (-\frac {2 \, {\left (d x + c\right )} f}{d}\right )}{d} - \frac {2 \, {\left (d x + c\right )}^{m + 1}}{d {\left (m + 1\right )}}\right )} a^{2} - {\left (\frac {{\left (d x + c\right )}^{m + 1} e^{\left (-e + \frac {c f}{d}\right )} E_{-m}\left (\frac {{\left (d x + c\right )} f}{d}\right )}{d} + \frac {{\left (d x + c\right )}^{m + 1} e^{\left (e - \frac {c f}{d}\right )} E_{-m}\left (-\frac {{\left (d x + c\right )} f}{d}\right )}{d}\right )} a^{2} + \frac {{\left (d x + c\right )}^{m + 1} a^{2}}{d {\left (m + 1\right )}} \]
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\[ \int (c+d x)^m (a+a \cosh (e+f x))^2 \, dx=\int { {\left (a \cosh \left (f x + e\right ) + a\right )}^{2} {\left (d x + c\right )}^{m} \,d x } \]
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Timed out. \[ \int (c+d x)^m (a+a \cosh (e+f x))^2 \, dx=\int {\left (a+a\,\mathrm {cosh}\left (e+f\,x\right )\right )}^2\,{\left (c+d\,x\right )}^m \,d x \]
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