Optimal. Leaf size=299 \[ -\frac{a^2 e^{i \left (e-\frac{c f}{d}\right )} (c+d x)^m \left (-\frac{i f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,-\frac{i f (c+d x)}{d}\right )}{f}+\frac{i a^2 2^{-m-3} e^{2 i \left (e-\frac{c f}{d}\right )} (c+d x)^m \left (-\frac{i f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,-\frac{2 i f (c+d x)}{d}\right )}{f}-\frac{a^2 e^{-i \left (e-\frac{c f}{d}\right )} (c+d x)^m \left (\frac{i f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,\frac{i f (c+d x)}{d}\right )}{f}-\frac{i a^2 2^{-m-3} e^{-2 i \left (e-\frac{c f}{d}\right )} (c+d x)^m \left (\frac{i f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,\frac{2 i f (c+d x)}{d}\right )}{f}+\frac{3 a^2 (c+d x)^{m+1}}{2 d (m+1)} \]
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Rubi [A] time = 0.369239, antiderivative size = 299, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3318, 3312, 3307, 2181, 3308} \[ -\frac{a^2 e^{i \left (e-\frac{c f}{d}\right )} (c+d x)^m \left (-\frac{i f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,-\frac{i f (c+d x)}{d}\right )}{f}+\frac{i a^2 2^{-m-3} e^{2 i \left (e-\frac{c f}{d}\right )} (c+d x)^m \left (-\frac{i f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,-\frac{2 i f (c+d x)}{d}\right )}{f}-\frac{a^2 e^{-i \left (e-\frac{c f}{d}\right )} (c+d x)^m \left (\frac{i f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,\frac{i f (c+d x)}{d}\right )}{f}-\frac{i a^2 2^{-m-3} e^{-2 i \left (e-\frac{c f}{d}\right )} (c+d x)^m \left (\frac{i f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,\frac{2 i f (c+d x)}{d}\right )}{f}+\frac{3 a^2 (c+d x)^{m+1}}{2 d (m+1)} \]
Antiderivative was successfully verified.
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Rule 3318
Rule 3312
Rule 3307
Rule 2181
Rule 3308
Rubi steps
\begin{align*} \int (c+d x)^m (a+a \sin (e+f x))^2 \, dx &=\left (4 a^2\right ) \int (c+d x)^m \sin ^4\left (\frac{1}{2} \left (e+\frac{\pi }{2}\right )+\frac{f x}{2}\right ) \, dx\\ &=\left (4 a^2\right ) \int \left (\frac{3}{8} (c+d x)^m-\frac{1}{8} (c+d x)^m \cos (2 e+2 f x)+\frac{1}{2} (c+d x)^m \sin (e+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 \cos (2 e+2 f x) \, dx+\left (2 a^2\right ) \int (c+d x)^m \sin (e+f x) \, dx\\ &=\frac{3 a^2 (c+d x)^{1+m}}{2 d (1+m)}+\left (i a^2\right ) \int e^{-i (e+f x)} (c+d x)^m \, dx-\left (i a^2\right ) \int e^{i (e+f x)} (c+d x)^m \, dx-\frac{1}{4} a^2 \int e^{-i (2 e+2 f x)} (c+d x)^m \, dx-\frac{1}{4} a^2 \int e^{i (2 e+2 f x)} (c+d x)^m \, dx\\ &=\frac{3 a^2 (c+d x)^{1+m}}{2 d (1+m)}-\frac{a^2 e^{i \left (e-\frac{c f}{d}\right )} (c+d x)^m \left (-\frac{i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac{i f (c+d x)}{d}\right )}{f}-\frac{a^2 e^{-i \left (e-\frac{c f}{d}\right )} (c+d x)^m \left (\frac{i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac{i f (c+d x)}{d}\right )}{f}+\frac{i 2^{-3-m} a^2 e^{2 i \left (e-\frac{c f}{d}\right )} (c+d x)^m \left (-\frac{i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac{2 i f (c+d x)}{d}\right )}{f}-\frac{i 2^{-3-m} a^2 e^{-2 i \left (e-\frac{c f}{d}\right )} (c+d x)^m \left (\frac{i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac{2 i f (c+d x)}{d}\right )}{f}\\ \end{align*}
Mathematica [A] time = 0.284949, size = 260, normalized size = 0.87 \[ \frac{1}{8} a^2 (c+d x)^m \left (-\frac{8 e^{i \left (e-\frac{c f}{d}\right )} \left (-\frac{i f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,-\frac{i f (c+d x)}{d}\right )}{f}+\frac{i 2^{-m} e^{2 i \left (e-\frac{c f}{d}\right )} \left (-\frac{i f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,-\frac{2 i f (c+d x)}{d}\right )}{f}-\frac{8 e^{-i \left (e-\frac{c f}{d}\right )} \left (\frac{i f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,\frac{i f (c+d x)}{d}\right )}{f}-\frac{i 2^{-m} e^{-2 i \left (e-\frac{c f}{d}\right )} \left (\frac{i f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,\frac{2 i f (c+d x)}{d}\right )}{f}+\frac{12 (c+d x)}{d (m+1)}\right ) \]
Antiderivative was successfully verified.
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Maple [F] time = 0.193, size = 0, normalized size = 0. \begin{align*} \int \left ( dx+c \right ) ^{m} \left ( a+a\sin \left ( fx+e \right ) \right ) ^{2}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.97924, size = 636, normalized size = 2.13 \begin{align*} \frac{{\left (-i \, a^{2} d m - i \, a^{2} d\right )} e^{\left (-\frac{d m \log \left (\frac{2 i \, f}{d}\right ) + 2 i \, d e - 2 i \, c f}{d}\right )} \Gamma \left (m + 1, \frac{2 i \, d f x + 2 i \, c f}{d}\right ) - 8 \,{\left (a^{2} d m + a^{2} d\right )} e^{\left (-\frac{d m \log \left (\frac{i \, f}{d}\right ) + i \, d e - i \, c f}{d}\right )} \Gamma \left (m + 1, \frac{i \, d f x + i \, c f}{d}\right ) - 8 \,{\left (a^{2} d m + a^{2} d\right )} e^{\left (-\frac{d m \log \left (-\frac{i \, f}{d}\right ) - i \, d e + i \, c f}{d}\right )} \Gamma \left (m + 1, \frac{-i \, d f x - i \, c f}{d}\right ) +{\left (i \, a^{2} d m + i \, a^{2} d\right )} e^{\left (-\frac{d m \log \left (-\frac{2 i \, f}{d}\right ) - 2 i \, d e + 2 i \, c f}{d}\right )} \Gamma \left (m + 1, \frac{-2 i \, d f x - 2 i \, c f}{d}\right ) + 12 \,{\left (a^{2} d f x + a^{2} c f\right )}{\left (d x + c\right )}^{m}}{8 \,{\left (d f m + d f\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} a^{2} \left (\int 2 \left (c + d x\right )^{m} \sin{\left (e + f x \right )}\, dx + \int \left (c + d x\right )^{m} \sin ^{2}{\left (e + f x \right )}\, dx + \int \left (c + d x\right )^{m}\, dx\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sin \left (f x + e\right ) + a\right )}^{2}{\left (d x + c\right )}^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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