Optimal. Leaf size=183 \[ -\frac{a^2}{d (c+d x)}+\frac{2 a b f \text{CosIntegral}\left (\frac{c f}{d}+f x\right ) \cos \left (e-\frac{c f}{d}\right )}{d^2}-\frac{2 a b f \sin \left (e-\frac{c f}{d}\right ) \text{Si}\left (x f+\frac{c f}{d}\right )}{d^2}-\frac{2 a b \sin (e+f x)}{d (c+d x)}+\frac{b^2 f \text{CosIntegral}\left (\frac{2 c f}{d}+2 f x\right ) \sin \left (2 e-\frac{2 c f}{d}\right )}{d^2}+\frac{b^2 f \cos \left (2 e-\frac{2 c f}{d}\right ) \text{Si}\left (2 x f+\frac{2 c f}{d}\right )}{d^2}-\frac{b^2 \sin ^2(e+f x)}{d (c+d x)} \]
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Rubi [A] time = 0.334011, antiderivative size = 183, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {3317, 3297, 3303, 3299, 3302, 3313, 12} \[ -\frac{a^2}{d (c+d x)}+\frac{2 a b f \text{CosIntegral}\left (\frac{c f}{d}+f x\right ) \cos \left (e-\frac{c f}{d}\right )}{d^2}-\frac{2 a b f \sin \left (e-\frac{c f}{d}\right ) \text{Si}\left (x f+\frac{c f}{d}\right )}{d^2}-\frac{2 a b \sin (e+f x)}{d (c+d x)}+\frac{b^2 f \text{CosIntegral}\left (\frac{2 c f}{d}+2 f x\right ) \sin \left (2 e-\frac{2 c f}{d}\right )}{d^2}+\frac{b^2 f \cos \left (2 e-\frac{2 c f}{d}\right ) \text{Si}\left (2 x f+\frac{2 c f}{d}\right )}{d^2}-\frac{b^2 \sin ^2(e+f x)}{d (c+d x)} \]
Antiderivative was successfully verified.
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Rule 3317
Rule 3297
Rule 3303
Rule 3299
Rule 3302
Rule 3313
Rule 12
Rubi steps
\begin{align*} \int \frac{(a+b \sin (e+f x))^2}{(c+d x)^2} \, dx &=\int \left (\frac{a^2}{(c+d x)^2}+\frac{2 a b \sin (e+f x)}{(c+d x)^2}+\frac{b^2 \sin ^2(e+f x)}{(c+d x)^2}\right ) \, dx\\ &=-\frac{a^2}{d (c+d x)}+(2 a b) \int \frac{\sin (e+f x)}{(c+d x)^2} \, dx+b^2 \int \frac{\sin ^2(e+f x)}{(c+d x)^2} \, dx\\ &=-\frac{a^2}{d (c+d x)}-\frac{2 a b \sin (e+f x)}{d (c+d x)}-\frac{b^2 \sin ^2(e+f x)}{d (c+d x)}+\frac{(2 a b f) \int \frac{\cos (e+f x)}{c+d x} \, dx}{d}+\frac{\left (2 b^2 f\right ) \int \frac{\sin (2 e+2 f x)}{2 (c+d x)} \, dx}{d}\\ &=-\frac{a^2}{d (c+d x)}-\frac{2 a b \sin (e+f x)}{d (c+d x)}-\frac{b^2 \sin ^2(e+f x)}{d (c+d x)}+\frac{\left (b^2 f\right ) \int \frac{\sin (2 e+2 f x)}{c+d x} \, dx}{d}+\frac{\left (2 a b f \cos \left (e-\frac{c f}{d}\right )\right ) \int \frac{\cos \left (\frac{c f}{d}+f x\right )}{c+d x} \, dx}{d}-\frac{\left (2 a b f \sin \left (e-\frac{c f}{d}\right )\right ) \int \frac{\sin \left (\frac{c f}{d}+f x\right )}{c+d x} \, dx}{d}\\ &=-\frac{a^2}{d (c+d x)}+\frac{2 a b f \cos \left (e-\frac{c f}{d}\right ) \text{Ci}\left (\frac{c f}{d}+f x\right )}{d^2}-\frac{2 a b \sin (e+f x)}{d (c+d x)}-\frac{b^2 \sin ^2(e+f x)}{d (c+d x)}-\frac{2 a b f \sin \left (e-\frac{c f}{d}\right ) \text{Si}\left (\frac{c f}{d}+f x\right )}{d^2}+\frac{\left (b^2 f \cos \left (2 e-\frac{2 c f}{d}\right )\right ) \int \frac{\sin \left (\frac{2 c f}{d}+2 f x\right )}{c+d x} \, dx}{d}+\frac{\left (b^2 f \sin \left (2 e-\frac{2 c f}{d}\right )\right ) \int \frac{\cos \left (\frac{2 c f}{d}+2 f x\right )}{c+d x} \, dx}{d}\\ &=-\frac{a^2}{d (c+d x)}+\frac{2 a b f \cos \left (e-\frac{c f}{d}\right ) \text{Ci}\left (\frac{c f}{d}+f x\right )}{d^2}+\frac{b^2 f \text{Ci}\left (\frac{2 c f}{d}+2 f x\right ) \sin \left (2 e-\frac{2 c f}{d}\right )}{d^2}-\frac{2 a b \sin (e+f x)}{d (c+d x)}-\frac{b^2 \sin ^2(e+f x)}{d (c+d x)}-\frac{2 a b f \sin \left (e-\frac{c f}{d}\right ) \text{Si}\left (\frac{c f}{d}+f x\right )}{d^2}+\frac{b^2 f \cos \left (2 e-\frac{2 c f}{d}\right ) \text{Si}\left (\frac{2 c f}{d}+2 f x\right )}{d^2}\\ \end{align*}
Mathematica [A] time = 0.596356, size = 232, normalized size = 1.27 \[ \frac{-2 a^2 d+4 a b f (c+d x) \text{CosIntegral}\left (f \left (\frac{c}{d}+x\right )\right ) \cos \left (e-\frac{c f}{d}\right )-4 a b c f \sin \left (e-\frac{c f}{d}\right ) \text{Si}\left (f \left (\frac{c}{d}+x\right )\right )-4 a b d f x \sin \left (e-\frac{c f}{d}\right ) \text{Si}\left (f \left (\frac{c}{d}+x\right )\right )-4 a b d \sin (e+f x)+2 b^2 f (c+d x) \text{CosIntegral}\left (\frac{2 f (c+d x)}{d}\right ) \sin \left (2 e-\frac{2 c f}{d}\right )+2 b^2 c f \cos \left (2 e-\frac{2 c f}{d}\right ) \text{Si}\left (\frac{2 f (c+d x)}{d}\right )+2 b^2 d f x \cos \left (2 e-\frac{2 c f}{d}\right ) \text{Si}\left (\frac{2 f (c+d x)}{d}\right )+b^2 d \cos (2 (e+f x))-b^2 d}{2 d^2 (c+d x)} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.018, size = 301, normalized size = 1.6 \begin{align*}{\frac{1}{f} \left ( -{\frac{{a}^{2}{f}^{2}}{ \left ( \left ( fx+e \right ) d+cf-de \right ) d}}+2\,{f}^{2}ab \left ( -{\frac{\sin \left ( fx+e \right ) }{ \left ( \left ( fx+e \right ) d+cf-de \right ) d}}+{\frac{1}{d} \left ({\frac{1}{d}{\it Si} \left ( fx+e+{\frac{cf-de}{d}} \right ) \sin \left ({\frac{cf-de}{d}} \right ) }+{\frac{1}{d}{\it Ci} \left ( fx+e+{\frac{cf-de}{d}} \right ) \cos \left ({\frac{cf-de}{d}} \right ) } \right ) } \right ) -{\frac{{f}^{2}{b}^{2}}{ \left ( 2\, \left ( fx+e \right ) d+2\,cf-2\,de \right ) d}}-{\frac{{f}^{2}{b}^{2}}{4} \left ( -2\,{\frac{\cos \left ( 2\,fx+2\,e \right ) }{ \left ( \left ( fx+e \right ) d+cf-de \right ) d}}-2\,{\frac{1}{d} \left ( 2\,{\frac{1}{d}{\it Si} \left ( 2\,fx+2\,e+2\,{\frac{cf-de}{d}} \right ) \cos \left ( 2\,{\frac{cf-de}{d}} \right ) }-2\,{\frac{1}{d}{\it Ci} \left ( 2\,fx+2\,e+2\,{\frac{cf-de}{d}} \right ) \sin \left ( 2\,{\frac{cf-de}{d}} \right ) } \right ) } \right ) } \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [C] time = 1.50062, size = 498, normalized size = 2.72 \begin{align*} -\frac{\frac{64 \, a^{2} f^{2}}{{\left (f x + e\right )} d^{2} - d^{2} e + c d f} - \frac{64 \,{\left (f^{2}{\left (-i \, E_{2}\left (\frac{i \,{\left (f x + e\right )} d - i \, d e + i \, c f}{d}\right ) + i \, E_{2}\left (-\frac{i \,{\left (f x + e\right )} d - i \, d e + i \, c f}{d}\right )\right )} \cos \left (-\frac{d e - c f}{d}\right ) + f^{2}{\left (E_{2}\left (\frac{i \,{\left (f x + e\right )} d - i \, d e + i \, c f}{d}\right ) + E_{2}\left (-\frac{i \,{\left (f x + e\right )} d - i \, d e + i \, c f}{d}\right )\right )} \sin \left (-\frac{d e - c f}{d}\right )\right )} a b}{{\left (f x + e\right )} d^{2} - d^{2} e + c d f} - \frac{{\left (16 \, f^{2}{\left (E_{2}\left (\frac{2 i \,{\left (f x + e\right )} d - 2 i \, d e + 2 i \, c f}{d}\right ) + E_{2}\left (-\frac{2 i \,{\left (f x + e\right )} d - 2 i \, d e + 2 i \, c f}{d}\right )\right )} \cos \left (-\frac{2 \,{\left (d e - c f\right )}}{d}\right ) + f^{2}{\left (16 i \, E_{2}\left (\frac{2 i \,{\left (f x + e\right )} d - 2 i \, d e + 2 i \, c f}{d}\right ) - 16 i \, E_{2}\left (-\frac{2 i \,{\left (f x + e\right )} d - 2 i \, d e + 2 i \, c f}{d}\right )\right )} \sin \left (-\frac{2 \,{\left (d e - c f\right )}}{d}\right ) - 32 \, f^{2}\right )} b^{2}}{{\left (f x + e\right )} d^{2} - d^{2} e + c d f}}{64 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.26381, size = 693, normalized size = 3.79 \begin{align*} \frac{2 \, b^{2} d \cos \left (f x + e\right )^{2} - 4 \, a b d \sin \left (f x + e\right ) + 2 \,{\left (b^{2} d f x + b^{2} c f\right )} \cos \left (-\frac{2 \,{\left (d e - c f\right )}}{d}\right ) \operatorname{Si}\left (\frac{2 \,{\left (d f x + c f\right )}}{d}\right ) + 4 \,{\left (a b d f x + a b c f\right )} \sin \left (-\frac{d e - c f}{d}\right ) \operatorname{Si}\left (\frac{d f x + c f}{d}\right ) - 2 \,{\left (a^{2} + b^{2}\right )} d + 2 \,{\left ({\left (a b d f x + a b c f\right )} \operatorname{Ci}\left (\frac{d f x + c f}{d}\right ) +{\left (a b d f x + a b c f\right )} \operatorname{Ci}\left (-\frac{d f x + c f}{d}\right )\right )} \cos \left (-\frac{d e - c f}{d}\right ) -{\left ({\left (b^{2} d f x + b^{2} c f\right )} \operatorname{Ci}\left (\frac{2 \,{\left (d f x + c f\right )}}{d}\right ) +{\left (b^{2} d f x + b^{2} c f\right )} \operatorname{Ci}\left (-\frac{2 \,{\left (d f x + c f\right )}}{d}\right )\right )} \sin \left (-\frac{2 \,{\left (d e - c f\right )}}{d}\right )}{2 \,{\left (d^{3} x + c d^{2}\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*} \int \frac{\left (a + b \sin{\left (e + f x \right )}\right )^{2}}{\left (c + d x\right )^{2}}\, dx \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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