Optimal. Leaf size=181 \[ \frac{2 b \left (-2 a^2 d+a b c+b^2 d\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{f \left (a^2-b^2\right )^{3/2} (b c-a d)^2}+\frac{b^2 \cos (e+f x)}{f \left (a^2-b^2\right ) (b c-a d) (a+b \sin (e+f x))}+\frac{2 d^2 \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{f \sqrt{c^2-d^2} (b c-a d)^2} \]
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Rubi [A] time = 0.437859, antiderivative size = 181, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2802, 3001, 2660, 618, 204} \[ \frac{2 b \left (-2 a^2 d+a b c+b^2 d\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{f \left (a^2-b^2\right )^{3/2} (b c-a d)^2}+\frac{b^2 \cos (e+f x)}{f \left (a^2-b^2\right ) (b c-a d) (a+b \sin (e+f x))}+\frac{2 d^2 \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{f \sqrt{c^2-d^2} (b c-a d)^2} \]
Antiderivative was successfully verified.
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Rule 2802
Rule 3001
Rule 2660
Rule 618
Rule 204
Rubi steps
\begin{align*} \int \frac{1}{(a+b \sin (e+f x))^2 (c+d \sin (e+f x))} \, dx &=\frac{b^2 \cos (e+f x)}{\left (a^2-b^2\right ) (b c-a d) f (a+b \sin (e+f x))}-\frac{\int \frac{-a b c+a^2 d-b^2 d-a b d \sin (e+f x)}{(a+b \sin (e+f x)) (c+d \sin (e+f x))} \, dx}{\left (a^2-b^2\right ) (b c-a d)}\\ &=\frac{b^2 \cos (e+f x)}{\left (a^2-b^2\right ) (b c-a d) f (a+b \sin (e+f x))}+\frac{d^2 \int \frac{1}{c+d \sin (e+f x)} \, dx}{(b c-a d)^2}+\frac{\left (b \left (a b c-2 a^2 d+b^2 d\right )\right ) \int \frac{1}{a+b \sin (e+f x)} \, dx}{\left (a^2-b^2\right ) (b c-a d)^2}\\ &=\frac{b^2 \cos (e+f x)}{\left (a^2-b^2\right ) (b c-a d) f (a+b \sin (e+f x))}+\frac{\left (2 d^2\right ) \operatorname{Subst}\left (\int \frac{1}{c+2 d x+c x^2} \, dx,x,\tan \left (\frac{1}{2} (e+f x)\right )\right )}{(b c-a d)^2 f}+\frac{\left (2 b \left (a b c-2 a^2 d+b^2 d\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac{1}{2} (e+f x)\right )\right )}{\left (a^2-b^2\right ) (b c-a d)^2 f}\\ &=\frac{b^2 \cos (e+f x)}{\left (a^2-b^2\right ) (b c-a d) f (a+b \sin (e+f x))}-\frac{\left (4 d^2\right ) \operatorname{Subst}\left (\int \frac{1}{-4 \left (c^2-d^2\right )-x^2} \, dx,x,2 d+2 c \tan \left (\frac{1}{2} (e+f x)\right )\right )}{(b c-a d)^2 f}-\frac{\left (4 b \left (a b c-2 a^2 d+b^2 d\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac{1}{2} (e+f x)\right )\right )}{\left (a^2-b^2\right ) (b c-a d)^2 f}\\ &=\frac{2 b \left (a b c-2 a^2 d+b^2 d\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} (b c-a d)^2 f}+\frac{2 d^2 \tan ^{-1}\left (\frac{d+c \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c^2-d^2}}\right )}{(b c-a d)^2 \sqrt{c^2-d^2} f}+\frac{b^2 \cos (e+f x)}{\left (a^2-b^2\right ) (b c-a d) f (a+b \sin (e+f x))}\\ \end{align*}
Mathematica [A] time = 0.851623, size = 178, normalized size = 0.98 \[ \frac{\frac{2 b \left (-2 a^2 d+a b c+b^2 d\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} (b c-a d)^2}-\frac{b^2 \cos (e+f x)}{(a-b) (a+b) (a d-b c) (a+b \sin (e+f x))}+\frac{2 d^2 \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{\sqrt{c^2-d^2} (b c-a d)^2}}{f} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.119, size = 514, normalized size = 2.8 \begin{align*} 2\,{\frac{{d}^{2}}{f \left ({a}^{2}{d}^{2}-2\,abcd+{c}^{2}{b}^{2} \right ) \sqrt{{c}^{2}-{d}^{2}}}\arctan \left ( 1/2\,{\frac{2\,c\tan \left ( 1/2\,fx+e/2 \right ) +2\,d}{\sqrt{{c}^{2}-{d}^{2}}}} \right ) }-2\,{\frac{{b}^{3}\tan \left ( 1/2\,fx+e/2 \right ) d}{f \left ( da-cb \right ) ^{2} \left ( \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}a+2\,\tan \left ( 1/2\,fx+e/2 \right ) b+a \right ) \left ({a}^{2}-{b}^{2} \right ) }}+2\,{\frac{{b}^{4}\tan \left ( 1/2\,fx+e/2 \right ) c}{f \left ( da-cb \right ) ^{2} \left ( \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}a+2\,\tan \left ( 1/2\,fx+e/2 \right ) b+a \right ) a \left ({a}^{2}-{b}^{2} \right ) }}-2\,{\frac{a{b}^{2}d}{f \left ( da-cb \right ) ^{2} \left ( \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}a+2\,\tan \left ( 1/2\,fx+e/2 \right ) b+a \right ) \left ({a}^{2}-{b}^{2} \right ) }}+2\,{\frac{c{b}^{3}}{f \left ( da-cb \right ) ^{2} \left ( \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}a+2\,\tan \left ( 1/2\,fx+e/2 \right ) b+a \right ) \left ({a}^{2}-{b}^{2} \right ) }}-4\,{\frac{{a}^{2}bd}{f \left ( da-cb \right ) ^{2} \left ({a}^{2}-{b}^{2} \right ) ^{3/2}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,fx+e/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }+2\,{\frac{ca{b}^{2}}{f \left ( da-cb \right ) ^{2} \left ({a}^{2}-{b}^{2} \right ) ^{3/2}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,fx+e/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }+2\,{\frac{{b}^{3}d}{f \left ( da-cb \right ) ^{2} \left ({a}^{2}-{b}^{2} \right ) ^{3/2}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,fx+e/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) } \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 [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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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 [A] time = 1.42915, size = 410, normalized size = 2.27 \begin{align*} \frac{2 \,{\left (\frac{{\left (\pi \left \lfloor \frac{f x + e}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (c\right ) + \arctan \left (\frac{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + d}{\sqrt{c^{2} - d^{2}}}\right )\right )} d^{2}}{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt{c^{2} - d^{2}}} + \frac{{\left (a b^{2} c - 2 \, a^{2} b d + b^{3} d\right )}{\left (\pi \left \lfloor \frac{f x + e}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (a\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + b}{\sqrt{a^{2} - b^{2}}}\right )\right )}}{{\left (a^{2} b^{2} c^{2} - b^{4} c^{2} - 2 \, a^{3} b c d + 2 \, a b^{3} c d + a^{4} d^{2} - a^{2} b^{2} d^{2}\right )} \sqrt{a^{2} - b^{2}}} + \frac{b^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + a b^{2}}{{\left (a^{3} b c - a b^{3} c - a^{4} d + a^{2} b^{2} d\right )}{\left (a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 2 \, b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + a\right )}}\right )}}{f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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