Integrand size = 20, antiderivative size = 282 \[ \int (c+d x)^m (a+b \cosh (e+f x))^2 \, dx=\frac {a^2 (c+d x)^{1+m}}{d (1+m)}+\frac {b^2 (c+d x)^{1+m}}{2 d (1+m)}+\frac {2^{-3-m} b^2 e^{2 e-\frac {2 c f}{d}} (c+d x)^m \left (-\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {2 f (c+d x)}{d}\right )}{f}+\frac {a b e^{e-\frac {c f}{d}} (c+d x)^m \left (-\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {f (c+d x)}{d}\right )}{f}-\frac {a b e^{-e+\frac {c f}{d}} (c+d x)^m \left (\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {f (c+d x)}{d}\right )}{f}-\frac {2^{-3-m} b^2 e^{-2 e+\frac {2 c f}{d}} (c+d x)^m \left (\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {2 f (c+d x)}{d}\right )}{f} \]
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Time = 0.26 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3398, 3388, 2212, 3393} \[ \int (c+d x)^m (a+b \cosh (e+f x))^2 \, dx=\frac {a^2 (c+d x)^{m+1}}{d (m+1)}+\frac {a b e^{e-\frac {c f}{d}} (c+d x)^m \left (-\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {f (c+d x)}{d}\right )}{f}-\frac {a b e^{\frac {c f}{d}-e} (c+d x)^m \left (\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {f (c+d x)}{d}\right )}{f}+\frac {b^2 2^{-m-3} e^{2 e-\frac {2 c f}{d}} (c+d x)^m \left (-\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {2 f (c+d x)}{d}\right )}{f}-\frac {b^2 2^{-m-3} e^{\frac {2 c f}{d}-2 e} (c+d x)^m \left (\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {2 f (c+d x)}{d}\right )}{f}+\frac {b^2 (c+d x)^{m+1}}{2 d (m+1)} \]
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Rule 2212
Rule 3388
Rule 3393
Rule 3398
Rubi steps \begin{align*} \text {integral}& = \int \left (a^2 (c+d x)^m+2 a b (c+d x)^m \cosh (e+f x)+b^2 (c+d x)^m \cosh ^2(e+f x)\right ) \, dx \\ & = \frac {a^2 (c+d x)^{1+m}}{d (1+m)}+(2 a b) \int (c+d x)^m \cosh (e+f x) \, dx+b^2 \int (c+d x)^m \cosh ^2(e+f x) \, dx \\ & = \frac {a^2 (c+d x)^{1+m}}{d (1+m)}+(a b) \int e^{-i (i e+i f x)} (c+d x)^m \, dx+(a b) \int e^{i (i e+i f x)} (c+d x)^m \, dx+b^2 \int \left (\frac {1}{2} (c+d x)^m+\frac {1}{2} (c+d x)^m \cosh (2 e+2 f x)\right ) \, dx \\ & = \frac {a^2 (c+d x)^{1+m}}{d (1+m)}+\frac {b^2 (c+d x)^{1+m}}{2 d (1+m)}+\frac {a b 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 b 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 {1}{2} b^2 \int (c+d x)^m \cosh (2 e+2 f x) \, dx \\ & = \frac {a^2 (c+d x)^{1+m}}{d (1+m)}+\frac {b^2 (c+d x)^{1+m}}{2 d (1+m)}+\frac {a b 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 b 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 {1}{4} b^2 \int e^{-i (2 i e+2 i f x)} (c+d x)^m \, dx+\frac {1}{4} b^2 \int e^{i (2 i e+2 i f x)} (c+d x)^m \, dx \\ & = \frac {a^2 (c+d x)^{1+m}}{d (1+m)}+\frac {b^2 (c+d x)^{1+m}}{2 d (1+m)}+\frac {2^{-3-m} b^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 b 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 b 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} b^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 = 0.52 (sec) , antiderivative size = 254, normalized size of antiderivative = 0.90 \[ \int (c+d x)^m (a+b \cosh (e+f x))^2 \, dx=\frac {(c+d x)^m \left (8 a^2 f (c+d x)+4 b^2 f (c+d x)+2^{-m} b^2 d e^{2 e-\frac {2 c f}{d}} (1+m) \left (-\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {2 f (c+d x)}{d}\right )+8 a b d e^{e-\frac {c f}{d}} (1+m) \left (-\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {f (c+d x)}{d}\right )-8 a b d e^{-e+\frac {c f}{d}} (1+m) \left (\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {f (c+d x)}{d}\right )-2^{-m} b^2 d e^{-2 e+\frac {2 c f}{d}} (1+m) \left (\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {2 f (c+d x)}{d}\right )\right )}{8 d f (1+m)} \]
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\[\int \left (d x +c \right )^{m} \left (a +b \cosh \left (f x +e \right )\right )^{2}d x\]
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Time = 0.09 (sec) , antiderivative size = 509, normalized size of antiderivative = 1.80 \[ \int (c+d x)^m (a+b \cosh (e+f x))^2 \, dx=-\frac {{\left (b^{2} d m + b^{2} d\right )} \cosh \left (\frac {d m \log \left (\frac {2 \, f}{d}\right ) + 2 \, d e - 2 \, c f}{d}\right ) \Gamma \left (m + 1, \frac {2 \, {\left (d f x + c f\right )}}{d}\right ) + 8 \, {\left (a b d m + a b d\right )} \cosh \left (\frac {d m \log \left (\frac {f}{d}\right ) + d e - c f}{d}\right ) \Gamma \left (m + 1, \frac {d f x + c f}{d}\right ) - 8 \, {\left (a b d m + a b d\right )} \cosh \left (\frac {d m \log \left (-\frac {f}{d}\right ) - d e + c f}{d}\right ) \Gamma \left (m + 1, -\frac {d f x + c f}{d}\right ) - {\left (b^{2} d m + b^{2} d\right )} \cosh \left (\frac {d m \log \left (-\frac {2 \, f}{d}\right ) - 2 \, d e + 2 \, c f}{d}\right ) \Gamma \left (m + 1, -\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) - {\left (b^{2} d m + b^{2} d\right )} \Gamma \left (m + 1, \frac {2 \, {\left (d f x + c f\right )}}{d}\right ) \sinh \left (\frac {d m \log \left (\frac {2 \, f}{d}\right ) + 2 \, d e - 2 \, c f}{d}\right ) - 8 \, {\left (a b d m + a b d\right )} \Gamma \left (m + 1, \frac {d f x + c f}{d}\right ) \sinh \left (\frac {d m \log \left (\frac {f}{d}\right ) + d e - c f}{d}\right ) + 8 \, {\left (a b d m + a b d\right )} \Gamma \left (m + 1, -\frac {d f x + c f}{d}\right ) \sinh \left (\frac {d m \log \left (-\frac {f}{d}\right ) - d e + c f}{d}\right ) + {\left (b^{2} d m + b^{2} d\right )} \Gamma \left (m + 1, -\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) \sinh \left (\frac {d m \log \left (-\frac {2 \, f}{d}\right ) - 2 \, d e + 2 \, c f}{d}\right ) - 4 \, {\left ({\left (2 \, a^{2} + b^{2}\right )} d f x + {\left (2 \, a^{2} + b^{2}\right )} c f\right )} \cosh \left (m \log \left (d x + c\right )\right ) - 4 \, {\left ({\left (2 \, a^{2} + b^{2}\right )} d f x + {\left (2 \, a^{2} + b^{2}\right )} c f\right )} \sinh \left (m \log \left (d x + c\right )\right )}{8 \, {\left (d f m + d f\right )}} \]
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Exception generated. \[ \int (c+d x)^m (a+b \cosh (e+f x))^2 \, dx=\text {Exception raised: TypeError} \]
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Time = 0.10 (sec) , antiderivative size = 208, normalized size of antiderivative = 0.74 \[ \int (c+d x)^m (a+b \cosh (e+f x))^2 \, dx=-{\left (\frac {{\left (d x + c\right )}^{m + 1} e^{\left (-e + \frac {c f}{d}\right )} E_{-m}\left (\frac {{\left (d x + c\right )} f}{d}\right )}{d} + \frac {{\left (d x + c\right )}^{m + 1} e^{\left (e - \frac {c f}{d}\right )} E_{-m}\left (-\frac {{\left (d x + c\right )} f}{d}\right )}{d}\right )} a b - \frac {1}{4} \, {\left (\frac {{\left (d x + c\right )}^{m + 1} e^{\left (-2 \, e + \frac {2 \, c f}{d}\right )} E_{-m}\left (\frac {2 \, {\left (d x + c\right )} f}{d}\right )}{d} + \frac {{\left (d x + c\right )}^{m + 1} e^{\left (2 \, e - \frac {2 \, c f}{d}\right )} E_{-m}\left (-\frac {2 \, {\left (d x + c\right )} f}{d}\right )}{d} - \frac {2 \, {\left (d x + c\right )}^{m + 1}}{d {\left (m + 1\right )}}\right )} b^{2} + \frac {{\left (d x + c\right )}^{m + 1} a^{2}}{d {\left (m + 1\right )}} \]
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\[ \int (c+d x)^m (a+b \cosh (e+f x))^2 \, dx=\int { {\left (b \cosh \left (f x + e\right ) + a\right )}^{2} {\left (d x + c\right )}^{m} \,d x } \]
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Timed out. \[ \int (c+d x)^m (a+b \cosh (e+f x))^2 \, dx=\int {\left (a+b\,\mathrm {cosh}\left (e+f\,x\right )\right )}^2\,{\left (c+d\,x\right )}^m \,d x \]
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