Optimal. Leaf size=161 \[ \frac{2 f (e+f x) \cos (c+d x)}{a d^2}-\frac{f (e+f x) \sin (c+d x) \cos (c+d x)}{2 a d^2}+\frac{f^2 \sin ^2(c+d x)}{4 a d^3}-\frac{2 f^2 \sin (c+d x)}{a d^3}-\frac{(e+f x)^2 \sin ^2(c+d x)}{2 a d}+\frac{(e+f x)^2 \sin (c+d x)}{a d}+\frac{e f x}{2 a d}+\frac{f^2 x^2}{4 a d} \]
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Rubi [A] time = 0.172781, antiderivative size = 161, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {4523, 3296, 2637, 4404, 3310} \[ \frac{2 f (e+f x) \cos (c+d x)}{a d^2}-\frac{f (e+f x) \sin (c+d x) \cos (c+d x)}{2 a d^2}+\frac{f^2 \sin ^2(c+d x)}{4 a d^3}-\frac{2 f^2 \sin (c+d x)}{a d^3}-\frac{(e+f x)^2 \sin ^2(c+d x)}{2 a d}+\frac{(e+f x)^2 \sin (c+d x)}{a d}+\frac{e f x}{2 a d}+\frac{f^2 x^2}{4 a d} \]
Antiderivative was successfully verified.
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Rule 4523
Rule 3296
Rule 2637
Rule 4404
Rule 3310
Rubi steps
\begin{align*} \int \frac{(e+f x)^2 \cos ^3(c+d x)}{a+a \sin (c+d x)} \, dx &=\frac{\int (e+f x)^2 \cos (c+d x) \, dx}{a}-\frac{\int (e+f x)^2 \cos (c+d x) \sin (c+d x) \, dx}{a}\\ &=\frac{(e+f x)^2 \sin (c+d x)}{a d}-\frac{(e+f x)^2 \sin ^2(c+d x)}{2 a d}+\frac{f \int (e+f x) \sin ^2(c+d x) \, dx}{a d}-\frac{(2 f) \int (e+f x) \sin (c+d x) \, dx}{a d}\\ &=\frac{2 f (e+f x) \cos (c+d x)}{a d^2}+\frac{(e+f x)^2 \sin (c+d x)}{a d}-\frac{f (e+f x) \cos (c+d x) \sin (c+d x)}{2 a d^2}+\frac{f^2 \sin ^2(c+d x)}{4 a d^3}-\frac{(e+f x)^2 \sin ^2(c+d x)}{2 a d}+\frac{f \int (e+f x) \, dx}{2 a d}-\frac{\left (2 f^2\right ) \int \cos (c+d x) \, dx}{a d^2}\\ &=\frac{e f x}{2 a d}+\frac{f^2 x^2}{4 a d}+\frac{2 f (e+f x) \cos (c+d x)}{a d^2}-\frac{2 f^2 \sin (c+d x)}{a d^3}+\frac{(e+f x)^2 \sin (c+d x)}{a d}-\frac{f (e+f x) \cos (c+d x) \sin (c+d x)}{2 a d^2}+\frac{f^2 \sin ^2(c+d x)}{4 a d^3}-\frac{(e+f x)^2 \sin ^2(c+d x)}{2 a d}\\ \end{align*}
Mathematica [A] time = 0.795724, size = 95, normalized size = 0.59 \[ \frac{\cos (2 (c+d x)) \left (2 d^2 (e+f x)^2-f^2\right )-4 \sin (c+d x) \left (d f (e+f x) \cos (c+d x)-2 \left (d^2 (e+f x)^2-2 f^2\right )\right )+16 d f (e+f x) \cos (c+d x)}{8 a d^3} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.062, size = 339, normalized size = 2.1 \begin{align*} -{\frac{1}{a{d}^{3}} \left ({f}^{2} \left ( -{\frac{ \left ( dx+c \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{2}}+ \left ( dx+c \right ) \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) -{\frac{ \left ( dx+c \right ) ^{2}}{4}}-{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{4}} \right ) -2\,c{f}^{2} \left ( -1/2\, \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}+1/4\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/4\,dx+c/4 \right ) +2\,def \left ( -1/2\, \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}+1/4\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/4\,dx+c/4 \right ) -{\frac{{c}^{2}{f}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{2}}+cdef \left ( \cos \left ( dx+c \right ) \right ) ^{2}-{\frac{{d}^{2}{e}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{2}}-{f}^{2} \left ( \left ( dx+c \right ) ^{2}\sin \left ( dx+c \right ) -2\,\sin \left ( dx+c \right ) +2\, \left ( dx+c \right ) \cos \left ( dx+c \right ) \right ) +2\,c{f}^{2} \left ( \cos \left ( dx+c \right ) + \left ( dx+c \right ) \sin \left ( dx+c \right ) \right ) -2\,def \left ( \cos \left ( dx+c \right ) + \left ( dx+c \right ) \sin \left ( dx+c \right ) \right ) -\sin \left ( dx+c \right ){c}^{2}{f}^{2}+2\,\sin \left ( dx+c \right ) cdef-\sin \left ( dx+c \right ){d}^{2}{e}^{2} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.09502, size = 390, normalized size = 2.42 \begin{align*} -\frac{\frac{4 \,{\left (\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right )\right )} e^{2}}{a} - \frac{8 \,{\left (\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right )\right )} c e f}{a d} + \frac{4 \,{\left (\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right )\right )} c^{2} f^{2}}{a d^{2}} - \frac{2 \,{\left (2 \,{\left (d x + c\right )} \cos \left (2 \, d x + 2 \, c\right ) + 8 \,{\left (d x + c\right )} \sin \left (d x + c\right ) + 8 \, \cos \left (d x + c\right ) - \sin \left (2 \, d x + 2 \, c\right )\right )} e f}{a d} + \frac{2 \,{\left (2 \,{\left (d x + c\right )} \cos \left (2 \, d x + 2 \, c\right ) + 8 \,{\left (d x + c\right )} \sin \left (d x + c\right ) + 8 \, \cos \left (d x + c\right ) - \sin \left (2 \, d x + 2 \, c\right )\right )} c f^{2}}{a d^{2}} - \frac{{\left ({\left (2 \,{\left (d x + c\right )}^{2} - 1\right )} \cos \left (2 \, d x + 2 \, c\right ) + 16 \,{\left (d x + c\right )} \cos \left (d x + c\right ) - 2 \,{\left (d x + c\right )} \sin \left (2 \, d x + 2 \, c\right ) + 8 \,{\left ({\left (d x + c\right )}^{2} - 2\right )} \sin \left (d x + c\right )\right )} f^{2}}{a d^{2}}}{8 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.72647, size = 327, normalized size = 2.03 \begin{align*} -\frac{d^{2} f^{2} x^{2} + 2 \, d^{2} e f x -{\left (2 \, d^{2} f^{2} x^{2} + 4 \, d^{2} e f x + 2 \, d^{2} e^{2} - f^{2}\right )} \cos \left (d x + c\right )^{2} - 8 \,{\left (d f^{2} x + d e f\right )} \cos \left (d x + c\right ) - 2 \,{\left (2 \, d^{2} f^{2} x^{2} + 4 \, d^{2} e f x + 2 \, d^{2} e^{2} - 4 \, f^{2} -{\left (d f^{2} x + d e f\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, a d^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 17.4589, size = 1705, normalized size = 10.59 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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