Optimal. Leaf size=175 \[ -\frac{d \sin \left (c-\frac{d e}{f}\right ) \text{CosIntegral}\left (\frac{d e}{f}+d x\right )}{a f^2}-\frac{d \cos \left (2 c-\frac{2 d e}{f}\right ) \text{CosIntegral}\left (\frac{2 d e}{f}+2 d x\right )}{a f^2}+\frac{d \sin \left (2 c-\frac{2 d e}{f}\right ) \text{Si}\left (\frac{2 d e}{f}+2 d x\right )}{a f^2}-\frac{d \cos \left (c-\frac{d e}{f}\right ) \text{Si}\left (\frac{d e}{f}+d x\right )}{a f^2}+\frac{\sin (2 c+2 d x)}{2 a f (e+f x)}-\frac{\cos (c+d x)}{a f (e+f x)} \]
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Rubi [A] time = 0.334109, antiderivative size = 175, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 7, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {4523, 3297, 3303, 3299, 3302, 4406, 12} \[ -\frac{d \sin \left (c-\frac{d e}{f}\right ) \text{CosIntegral}\left (\frac{d e}{f}+d x\right )}{a f^2}-\frac{d \cos \left (2 c-\frac{2 d e}{f}\right ) \text{CosIntegral}\left (\frac{2 d e}{f}+2 d x\right )}{a f^2}+\frac{d \sin \left (2 c-\frac{2 d e}{f}\right ) \text{Si}\left (\frac{2 d e}{f}+2 d x\right )}{a f^2}-\frac{d \cos \left (c-\frac{d e}{f}\right ) \text{Si}\left (\frac{d e}{f}+d x\right )}{a f^2}+\frac{\sin (2 c+2 d x)}{2 a f (e+f x)}-\frac{\cos (c+d x)}{a f (e+f x)} \]
Antiderivative was successfully verified.
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Rule 4523
Rule 3297
Rule 3303
Rule 3299
Rule 3302
Rule 4406
Rule 12
Rubi steps
\begin{align*} \int \frac{\cos ^3(c+d x)}{(e+f x)^2 (a+a \sin (c+d x))} \, dx &=\frac{\int \frac{\cos (c+d x)}{(e+f x)^2} \, dx}{a}-\frac{\int \frac{\cos (c+d x) \sin (c+d x)}{(e+f x)^2} \, dx}{a}\\ &=-\frac{\cos (c+d x)}{a f (e+f x)}-\frac{\int \frac{\sin (2 c+2 d x)}{2 (e+f x)^2} \, dx}{a}-\frac{d \int \frac{\sin (c+d x)}{e+f x} \, dx}{a f}\\ &=-\frac{\cos (c+d x)}{a f (e+f x)}-\frac{\int \frac{\sin (2 c+2 d x)}{(e+f x)^2} \, dx}{2 a}-\frac{\left (d \cos \left (c-\frac{d e}{f}\right )\right ) \int \frac{\sin \left (\frac{d e}{f}+d x\right )}{e+f x} \, dx}{a f}-\frac{\left (d \sin \left (c-\frac{d e}{f}\right )\right ) \int \frac{\cos \left (\frac{d e}{f}+d x\right )}{e+f x} \, dx}{a f}\\ &=-\frac{\cos (c+d x)}{a f (e+f x)}-\frac{d \text{Ci}\left (\frac{d e}{f}+d x\right ) \sin \left (c-\frac{d e}{f}\right )}{a f^2}+\frac{\sin (2 c+2 d x)}{2 a f (e+f x)}-\frac{d \cos \left (c-\frac{d e}{f}\right ) \text{Si}\left (\frac{d e}{f}+d x\right )}{a f^2}-\frac{d \int \frac{\cos (2 c+2 d x)}{e+f x} \, dx}{a f}\\ &=-\frac{\cos (c+d x)}{a f (e+f x)}-\frac{d \text{Ci}\left (\frac{d e}{f}+d x\right ) \sin \left (c-\frac{d e}{f}\right )}{a f^2}+\frac{\sin (2 c+2 d x)}{2 a f (e+f x)}-\frac{d \cos \left (c-\frac{d e}{f}\right ) \text{Si}\left (\frac{d e}{f}+d x\right )}{a f^2}-\frac{\left (d \cos \left (2 c-\frac{2 d e}{f}\right )\right ) \int \frac{\cos \left (\frac{2 d e}{f}+2 d x\right )}{e+f x} \, dx}{a f}+\frac{\left (d \sin \left (2 c-\frac{2 d e}{f}\right )\right ) \int \frac{\sin \left (\frac{2 d e}{f}+2 d x\right )}{e+f x} \, dx}{a f}\\ &=-\frac{\cos (c+d x)}{a f (e+f x)}-\frac{d \cos \left (2 c-\frac{2 d e}{f}\right ) \text{Ci}\left (\frac{2 d e}{f}+2 d x\right )}{a f^2}-\frac{d \text{Ci}\left (\frac{d e}{f}+d x\right ) \sin \left (c-\frac{d e}{f}\right )}{a f^2}+\frac{\sin (2 c+2 d x)}{2 a f (e+f x)}-\frac{d \cos \left (c-\frac{d e}{f}\right ) \text{Si}\left (\frac{d e}{f}+d x\right )}{a f^2}+\frac{d \sin \left (2 c-\frac{2 d e}{f}\right ) \text{Si}\left (\frac{2 d e}{f}+2 d x\right )}{a f^2}\\ \end{align*}
Mathematica [A] time = 0.571508, size = 203, normalized size = 1.16 \[ \frac{-2 d (e+f x) \sin \left (c-\frac{d e}{f}\right ) \text{CosIntegral}\left (d \left (\frac{e}{f}+x\right )\right )-2 d (e+f x) \cos \left (2 c-\frac{2 d e}{f}\right ) \text{CosIntegral}\left (\frac{2 d (e+f x)}{f}\right )+2 d e \sin \left (2 c-\frac{2 d e}{f}\right ) \text{Si}\left (\frac{2 d (e+f x)}{f}\right )+2 d f x \sin \left (2 c-\frac{2 d e}{f}\right ) \text{Si}\left (\frac{2 d (e+f x)}{f}\right )-2 d e \cos \left (c-\frac{d e}{f}\right ) \text{Si}\left (d \left (\frac{e}{f}+x\right )\right )-2 d f x \cos \left (c-\frac{d e}{f}\right ) \text{Si}\left (d \left (\frac{e}{f}+x\right )\right )+f \sin (2 (c+d x))-2 f \cos (c+d x)}{2 a f^2 (e+f x)} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.049, size = 230, normalized size = 1.3 \begin{align*} -{\frac{d}{a} \left ( -{\frac{\sin \left ( 2\,dx+2\,c \right ) }{ \left ( 2\, \left ( dx+c \right ) f-2\,cf+2\,de \right ) f}}+{\frac{1}{2\,f} \left ( 2\,{\frac{1}{f}{\it Si} \left ( 2\,dx+2\,c+2\,{\frac{-cf+de}{f}} \right ) \sin \left ( 2\,{\frac{-cf+de}{f}} \right ) }+2\,{\frac{1}{f}{\it Ci} \left ( 2\,dx+2\,c+2\,{\frac{-cf+de}{f}} \right ) \cos \left ( 2\,{\frac{-cf+de}{f}} \right ) } \right ) }+{\frac{\cos \left ( dx+c \right ) }{ \left ( \left ( dx+c \right ) f-cf+de \right ) f}}+{\frac{1}{f} \left ({\frac{1}{f}{\it Si} \left ( dx+c+{\frac{-cf+de}{f}} \right ) \cos \left ({\frac{-cf+de}{f}} \right ) }-{\frac{1}{f}{\it Ci} \left ( dx+c+{\frac{-cf+de}{f}} \right ) \sin \left ({\frac{-cf+de}{f}} \right ) } \right ) } \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [C] time = 1.61909, size = 414, normalized size = 2.37 \begin{align*} -\frac{2 \, d^{2}{\left (E_{2}\left (\frac{i \, d e + i \,{\left (d x + c\right )} f - i \, c f}{f}\right ) + E_{2}\left (-\frac{i \, d e + i \,{\left (d x + c\right )} f - i \, c f}{f}\right )\right )} \cos \left (-\frac{d e - c f}{f}\right ) - d^{2}{\left (i \, E_{2}\left (\frac{2 i \, d e + 2 i \,{\left (d x + c\right )} f - 2 i \, c f}{f}\right ) - i \, E_{2}\left (-\frac{2 i \, d e + 2 i \,{\left (d x + c\right )} f - 2 i \, c f}{f}\right )\right )} \cos \left (-\frac{2 \,{\left (d e - c f\right )}}{f}\right ) - d^{2}{\left (2 i \, E_{2}\left (\frac{i \, d e + i \,{\left (d x + c\right )} f - i \, c f}{f}\right ) - 2 i \, E_{2}\left (-\frac{i \, d e + i \,{\left (d x + c\right )} f - i \, c f}{f}\right )\right )} \sin \left (-\frac{d e - c f}{f}\right ) - d^{2}{\left (E_{2}\left (\frac{2 i \, d e + 2 i \,{\left (d x + c\right )} f - 2 i \, c f}{f}\right ) + E_{2}\left (-\frac{2 i \, d e + 2 i \,{\left (d x + c\right )} f - 2 i \, c f}{f}\right )\right )} \sin \left (-\frac{2 \,{\left (d e - c f\right )}}{f}\right )}{4 \,{\left (a d e f +{\left (d x + c\right )} a f^{2} - a c f^{2}\right )} d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.92626, size = 610, normalized size = 3.49 \begin{align*} \frac{2 \, f \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 2 \,{\left (d f x + d e\right )} \sin \left (-\frac{2 \,{\left (d e - c f\right )}}{f}\right ) \operatorname{Si}\left (\frac{2 \,{\left (d f x + d e\right )}}{f}\right ) - 2 \,{\left (d f x + d e\right )} \cos \left (-\frac{d e - c f}{f}\right ) \operatorname{Si}\left (\frac{d f x + d e}{f}\right ) - 2 \, f \cos \left (d x + c\right ) -{\left ({\left (d f x + d e\right )} \operatorname{Ci}\left (\frac{2 \,{\left (d f x + d e\right )}}{f}\right ) +{\left (d f x + d e\right )} \operatorname{Ci}\left (-\frac{2 \,{\left (d f x + d e\right )}}{f}\right )\right )} \cos \left (-\frac{2 \,{\left (d e - c f\right )}}{f}\right ) -{\left ({\left (d f x + d e\right )} \operatorname{Ci}\left (\frac{d f x + d e}{f}\right ) +{\left (d f x + d e\right )} \operatorname{Ci}\left (-\frac{d f x + d e}{f}\right )\right )} \sin \left (-\frac{d e - c f}{f}\right )}{2 \,{\left (a f^{3} x + a e f^{2}\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [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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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (d x + c\right )^{3}}{{\left (f x + e\right )}^{2}{\left (a \sin \left (d x + c\right ) + a\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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