Optimal. Leaf size=290 \[ \frac{d \left (a^2 d^2+b^2 \left (c^2-2 d^2\right )\right ) \cos (e+f x)}{f \left (a^2-b^2\right ) \left (c^2-d^2\right ) (b c-a d)^2 (c+d \sin (e+f x))}+\frac{2 b^2 \left (-3 a^2 d+a b c+2 b^2 d\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{f \left (a^2-b^2\right )^{3/2} (b c-a d)^3}+\frac{b^2 \cos (e+f x)}{f \left (a^2-b^2\right ) (b c-a d) (a+b \sin (e+f x)) (c+d \sin (e+f x))}+\frac{2 d^2 \left (-a c d+3 b c^2-2 b d^2\right ) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{f \left (c^2-d^2\right )^{3/2} (b c-a d)^3} \]
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Rubi [A] time = 1.19715, antiderivative size = 290, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {2802, 3055, 3001, 2660, 618, 204} \[ \frac{d \left (a^2 d^2+b^2 \left (c^2-2 d^2\right )\right ) \cos (e+f x)}{f \left (a^2-b^2\right ) \left (c^2-d^2\right ) (b c-a d)^2 (c+d \sin (e+f x))}+\frac{2 b^2 \left (-3 a^2 d+a b c+2 b^2 d\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{f \left (a^2-b^2\right )^{3/2} (b c-a d)^3}+\frac{b^2 \cos (e+f x)}{f \left (a^2-b^2\right ) (b c-a d) (a+b \sin (e+f x)) (c+d \sin (e+f x))}+\frac{2 d^2 \left (-a c d+3 b c^2-2 b d^2\right ) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{f \left (c^2-d^2\right )^{3/2} (b c-a d)^3} \]
Antiderivative was successfully verified.
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Rule 2802
Rule 3055
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))^2} \, 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)) (c+d \sin (e+f x))}-\frac{\int \frac{-a b c+a^2 d-2 b^2 d-a b d \sin (e+f x)+b^2 d \sin ^2(e+f x)}{(a+b \sin (e+f x)) (c+d \sin (e+f x))^2} \, dx}{\left (a^2-b^2\right ) (b c-a d)}\\ &=\frac{d \left (a^2 d^2+b^2 \left (c^2-2 d^2\right )\right ) \cos (e+f x)}{\left (a^2-b^2\right ) (b c-a d)^2 \left (c^2-d^2\right ) f (c+d \sin (e+f x))}+\frac{b^2 \cos (e+f x)}{\left (a^2-b^2\right ) (b c-a d) f (a+b \sin (e+f x)) (c+d \sin (e+f x))}-\frac{\int \frac{-a^3 c d^2-a b^2 c \left (c^2-2 d^2\right )+2 a^2 b d \left (c^2-d^2\right )-2 b^3 d \left (c^2-d^2\right )-b d (b c+a d) (a c-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)^2 \left (c^2-d^2\right )}\\ &=\frac{d \left (a^2 d^2+b^2 \left (c^2-2 d^2\right )\right ) \cos (e+f x)}{\left (a^2-b^2\right ) (b c-a d)^2 \left (c^2-d^2\right ) f (c+d \sin (e+f x))}+\frac{b^2 \cos (e+f x)}{\left (a^2-b^2\right ) (b c-a d) f (a+b \sin (e+f x)) (c+d \sin (e+f x))}+\frac{\left (b^2 \left (a b c-3 a^2 d+2 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)^3}+\frac{\left (d^2 \left (3 b c^2-a c d-2 b d^2\right )\right ) \int \frac{1}{c+d \sin (e+f x)} \, dx}{(b c-a d)^3 \left (c^2-d^2\right )}\\ &=\frac{d \left (a^2 d^2+b^2 \left (c^2-2 d^2\right )\right ) \cos (e+f x)}{\left (a^2-b^2\right ) (b c-a d)^2 \left (c^2-d^2\right ) f (c+d \sin (e+f x))}+\frac{b^2 \cos (e+f x)}{\left (a^2-b^2\right ) (b c-a d) f (a+b \sin (e+f x)) (c+d \sin (e+f x))}+\frac{\left (2 b^2 \left (a b c-3 a^2 d+2 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)^3 f}+\frac{\left (2 d^2 \left (3 b c^2-a c d-2 b d^2\right )\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)^3 \left (c^2-d^2\right ) f}\\ &=\frac{d \left (a^2 d^2+b^2 \left (c^2-2 d^2\right )\right ) \cos (e+f x)}{\left (a^2-b^2\right ) (b c-a d)^2 \left (c^2-d^2\right ) f (c+d \sin (e+f x))}+\frac{b^2 \cos (e+f x)}{\left (a^2-b^2\right ) (b c-a d) f (a+b \sin (e+f x)) (c+d \sin (e+f x))}-\frac{\left (4 b^2 \left (a b c-3 a^2 d+2 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)^3 f}-\frac{\left (4 d^2 \left (3 b c^2-a c d-2 b d^2\right )\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)^3 \left (c^2-d^2\right ) f}\\ &=\frac{2 b^2 \left (a b c-3 a^2 d+2 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)^3 f}+\frac{2 d^2 \left (3 b c^2-a c d-2 b d^2\right ) \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)^3 \left (c^2-d^2\right )^{3/2} f}+\frac{d \left (a^2 d^2+b^2 \left (c^2-2 d^2\right )\right ) \cos (e+f x)}{\left (a^2-b^2\right ) (b c-a d)^2 \left (c^2-d^2\right ) f (c+d \sin (e+f x))}+\frac{b^2 \cos (e+f x)}{\left (a^2-b^2\right ) (b c-a d) f (a+b \sin (e+f x)) (c+d \sin (e+f x))}\\ \end{align*}
Mathematica [A] time = 2.84357, size = 227, normalized size = 0.78 \[ \frac{\frac{2 b^2 \left (-3 a^2 d+a b c+2 b^2 d\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+\frac{b^3 (b c-a d) \cos (e+f x)}{(a-b) (a+b) (a+b \sin (e+f x))}-\frac{2 d^2 \left (a c d-3 b c^2+2 b d^2\right ) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{\left (c^2-d^2\right )^{3/2}}+\frac{d^3 (b c-a d) \cos (e+f x)}{(c-d) (c+d) (c+d \sin (e+f x))}}{f (b c-a d)^3} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.138, size = 886, normalized size = 3.1 \begin{align*} \text{result too large to display} \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 [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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