Optimal. Leaf size=298 \[ -\frac{2 d^3 (4 c+3 d) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{a^3 f (c-d)^4 (c+d) \sqrt{c^2-d^2}}-\frac{d \left (-12 c^2 d+2 c^3+43 c d^2+72 d^3\right ) \cos (e+f x)}{15 a^3 f (c-d)^4 (c+d) (c+d \sin (e+f x))}-\frac{\left (2 c^2-12 c d+45 d^2\right ) \cos (e+f x)}{15 f (c-d)^3 \left (a^3 \sin (e+f x)+a^3\right ) (c+d \sin (e+f x))}-\frac{(2 c-9 d) \cos (e+f x)}{15 a f (c-d)^2 (a \sin (e+f x)+a)^2 (c+d \sin (e+f x))}-\frac{\cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3 (c+d \sin (e+f x))} \]
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Rubi [A] time = 0.730008, antiderivative size = 298, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {2766, 2978, 2754, 12, 2660, 618, 204} \[ -\frac{2 d^3 (4 c+3 d) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{a^3 f (c-d)^4 (c+d) \sqrt{c^2-d^2}}-\frac{d \left (-12 c^2 d+2 c^3+43 c d^2+72 d^3\right ) \cos (e+f x)}{15 a^3 f (c-d)^4 (c+d) (c+d \sin (e+f x))}-\frac{\left (2 c^2-12 c d+45 d^2\right ) \cos (e+f x)}{15 f (c-d)^3 \left (a^3 \sin (e+f x)+a^3\right ) (c+d \sin (e+f x))}-\frac{(2 c-9 d) \cos (e+f x)}{15 a f (c-d)^2 (a \sin (e+f x)+a)^2 (c+d \sin (e+f x))}-\frac{\cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3 (c+d \sin (e+f x))} \]
Antiderivative was successfully verified.
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Rule 2766
Rule 2978
Rule 2754
Rule 12
Rule 2660
Rule 618
Rule 204
Rubi steps
\begin{align*} \int \frac{1}{(a+a \sin (e+f x))^3 (c+d \sin (e+f x))^2} \, dx &=-\frac{\cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3 (c+d \sin (e+f x))}-\frac{\int \frac{-2 a (c-3 d)-3 a d \sin (e+f x)}{(a+a \sin (e+f x))^2 (c+d \sin (e+f x))^2} \, dx}{5 a^2 (c-d)}\\ &=-\frac{\cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3 (c+d \sin (e+f x))}-\frac{(2 c-9 d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))}+\frac{\int \frac{a^2 \left (2 c^2-8 c d+27 d^2\right )+2 a^2 (2 c-9 d) d \sin (e+f x)}{(a+a \sin (e+f x)) (c+d \sin (e+f x))^2} \, dx}{15 a^4 (c-d)^2}\\ &=-\frac{\cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3 (c+d \sin (e+f x))}-\frac{(2 c-9 d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))}-\frac{\left (2 c^2-12 c d+45 d^2\right ) \cos (e+f x)}{15 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right ) (c+d \sin (e+f x))}-\frac{\int \frac{-2 a^3 (c-36 d) d^2-a^3 d \left (2 c^2-12 c d+45 d^2\right ) \sin (e+f x)}{(c+d \sin (e+f x))^2} \, dx}{15 a^6 (c-d)^3}\\ &=-\frac{d \left (2 c^3-12 c^2 d+43 c d^2+72 d^3\right ) \cos (e+f x)}{15 a^3 (c-d)^4 (c+d) f (c+d \sin (e+f x))}-\frac{\cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3 (c+d \sin (e+f x))}-\frac{(2 c-9 d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))}-\frac{\left (2 c^2-12 c d+45 d^2\right ) \cos (e+f x)}{15 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right ) (c+d \sin (e+f x))}+\frac{\int -\frac{15 a^3 d^3 (4 c+3 d)}{c+d \sin (e+f x)} \, dx}{15 a^6 (c-d)^4 (c+d)}\\ &=-\frac{d \left (2 c^3-12 c^2 d+43 c d^2+72 d^3\right ) \cos (e+f x)}{15 a^3 (c-d)^4 (c+d) f (c+d \sin (e+f x))}-\frac{\cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3 (c+d \sin (e+f x))}-\frac{(2 c-9 d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))}-\frac{\left (2 c^2-12 c d+45 d^2\right ) \cos (e+f x)}{15 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right ) (c+d \sin (e+f x))}-\frac{\left (d^3 (4 c+3 d)\right ) \int \frac{1}{c+d \sin (e+f x)} \, dx}{a^3 (c-d)^4 (c+d)}\\ &=-\frac{d \left (2 c^3-12 c^2 d+43 c d^2+72 d^3\right ) \cos (e+f x)}{15 a^3 (c-d)^4 (c+d) f (c+d \sin (e+f x))}-\frac{\cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3 (c+d \sin (e+f x))}-\frac{(2 c-9 d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))}-\frac{\left (2 c^2-12 c d+45 d^2\right ) \cos (e+f x)}{15 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right ) (c+d \sin (e+f x))}-\frac{\left (2 d^3 (4 c+3 d)\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 )}{a^3 (c-d)^4 (c+d) f}\\ &=-\frac{d \left (2 c^3-12 c^2 d+43 c d^2+72 d^3\right ) \cos (e+f x)}{15 a^3 (c-d)^4 (c+d) f (c+d \sin (e+f x))}-\frac{\cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3 (c+d \sin (e+f x))}-\frac{(2 c-9 d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))}-\frac{\left (2 c^2-12 c d+45 d^2\right ) \cos (e+f x)}{15 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right ) (c+d \sin (e+f x))}+\frac{\left (4 d^3 (4 c+3 d)\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 )}{a^3 (c-d)^4 (c+d) f}\\ &=-\frac{2 d^3 (4 c+3 d) \tan ^{-1}\left (\frac{d+c \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c^2-d^2}}\right )}{a^3 (c-d)^4 (c+d) \sqrt{c^2-d^2} f}-\frac{d \left (2 c^3-12 c^2 d+43 c d^2+72 d^3\right ) \cos (e+f x)}{15 a^3 (c-d)^4 (c+d) f (c+d \sin (e+f x))}-\frac{\cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3 (c+d \sin (e+f x))}-\frac{(2 c-9 d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))}-\frac{\left (2 c^2-12 c d+45 d^2\right ) \cos (e+f x)}{15 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right ) (c+d \sin (e+f x))}\\ \end{align*}
Mathematica [A] time = 2.48324, size = 361, normalized size = 1.21 \[ \frac{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (2 \left (2 c^2-14 c d+57 d^2\right ) \sin \left (\frac{1}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^4-\frac{30 d^3 (4 c+3 d) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^5 \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{(c+d) \sqrt{c^2-d^2}}-\frac{15 d^4 \cos (e+f x) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^5}{(c+d) (c+d \sin (e+f x))}+6 (c-d)^2 \sin \left (\frac{1}{2} (e+f x)\right )-2 (c-6 d) (c-d) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3+4 (c-6 d) (c-d) \sin \left (\frac{1}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2-3 (c-d)^2 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )}{15 a^3 f (c-d)^4 (\sin (e+f x)+1)^3} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.118, size = 511, normalized size = 1.7 \begin{align*} -2\,{\frac{{d}^{5}\tan \left ( 1/2\,fx+e/2 \right ) }{f{a}^{3} \left ( c-d \right ) ^{4} \left ( c \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}+2\,\tan \left ( 1/2\,fx+e/2 \right ) d+c \right ) \left ( c+d \right ) c}}-2\,{\frac{{d}^{4}}{f{a}^{3} \left ( c-d \right ) ^{4} \left ( c \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}+2\,\tan \left ( 1/2\,fx+e/2 \right ) d+c \right ) \left ( c+d \right ) }}-8\,{\frac{c{d}^{3}}{f{a}^{3} \left ( c-d \right ) ^{4} \left ( c+d \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 ) }-6\,{\frac{{d}^{4}}{f{a}^{3} \left ( c-d \right ) ^{4} \left ( c+d \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 ) }-{\frac{16\,c}{3\,f{a}^{3} \left ( c-d \right ) ^{3}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-3}}+8\,{\frac{d}{f{a}^{3} \left ( c-d \right ) ^{3} \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{3}}}-8\,{\frac{d}{f{a}^{3} \left ( c-d \right ) ^{3} \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{2}}}+4\,{\frac{c}{f{a}^{3} \left ( c-d \right ) ^{3} \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{2}}}-2\,{\frac{{c}^{2}}{f{a}^{3} \left ( c-d \right ) ^{4} \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) }}+8\,{\frac{cd}{f{a}^{3} \left ( c-d \right ) ^{4} \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) }}-12\,{\frac{{d}^{2}}{f{a}^{3} \left ( c-d \right ) ^{4} \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) }}-{\frac{8}{5\,f{a}^{3} \left ( c-d \right ) ^{2}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-5}}+4\,{\frac{1}{f{a}^{3} \left ( c-d \right ) ^{2} \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{4}}} \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 [B] time = 2.63708, size = 6917, normalized size = 23.21 \begin{align*} \text{result too large to display} \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 = 1.43095, size = 698, normalized size = 2.34 \begin{align*} -\frac{2 \,{\left (\frac{15 \,{\left (4 \, c d^{3} + 3 \, d^{4}\right )}{\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 )}}{{\left (a^{3} c^{5} - 3 \, a^{3} c^{4} d + 2 \, a^{3} c^{3} d^{2} + 2 \, a^{3} c^{2} d^{3} - 3 \, a^{3} c d^{4} + a^{3} d^{5}\right )} \sqrt{c^{2} - d^{2}}} + \frac{15 \,{\left (d^{5} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + c d^{4}\right )}}{{\left (a^{3} c^{6} - 3 \, a^{3} c^{5} d + 2 \, a^{3} c^{4} d^{2} + 2 \, a^{3} c^{3} d^{3} - 3 \, a^{3} c^{2} d^{4} + a^{3} c d^{5}\right )}{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 2 \, d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + c\right )}} + \frac{15 \, c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 60 \, c d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 90 \, d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 30 \, c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 150 \, c d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 300 \, d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 40 \, c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 190 \, c d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 420 \, d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 20 \, c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 110 \, c d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 270 \, d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 7 \, c^{2} - 34 \, c d + 72 \, d^{2}}{{\left (a^{3} c^{4} - 4 \, a^{3} c^{3} d + 6 \, a^{3} c^{2} d^{2} - 4 \, a^{3} c d^{3} + a^{3} d^{4}\right )}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}^{5}}\right )}}{15 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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