Optimal. Leaf size=326 \[ -\frac {d \left (c^2-5 c d-12 d^2\right ) \cos (e+f x)}{3 a^2 f (c-d)^3 (c+d) \sqrt {c+d \sin (e+f x)}}-\frac {\left (c^2-5 c d-12 d^2\right ) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{3 a^2 f (c-d)^3 (c+d) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}-\frac {(c-5 d) \cos (e+f x)}{3 a^2 f (c-d)^2 (\sin (e+f x)+1) \sqrt {c+d \sin (e+f x)}}+\frac {(c-5 d) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} F\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{3 a^2 f (c-d)^2 \sqrt {c+d \sin (e+f x)}}-\frac {\cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2 \sqrt {c+d \sin (e+f x)}} \]
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Rubi [A] time = 0.64, antiderivative size = 326, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {2766, 2978, 2754, 2752, 2663, 2661, 2655, 2653} \[ -\frac {d \left (c^2-5 c d-12 d^2\right ) \cos (e+f x)}{3 a^2 f (c-d)^3 (c+d) \sqrt {c+d \sin (e+f x)}}-\frac {\left (c^2-5 c d-12 d^2\right ) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{3 a^2 f (c-d)^3 (c+d) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}-\frac {(c-5 d) \cos (e+f x)}{3 a^2 f (c-d)^2 (\sin (e+f x)+1) \sqrt {c+d \sin (e+f x)}}+\frac {(c-5 d) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} F\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{3 a^2 f (c-d)^2 \sqrt {c+d \sin (e+f x)}}-\frac {\cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2 \sqrt {c+d \sin (e+f x)}} \]
Antiderivative was successfully verified.
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Rule 2653
Rule 2655
Rule 2661
Rule 2663
Rule 2752
Rule 2754
Rule 2766
Rule 2978
Rubi steps
\begin {align*} \int \frac {1}{(a+a \sin (e+f x))^2 (c+d \sin (e+f x))^{3/2}} \, dx &=-\frac {\cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2 \sqrt {c+d \sin (e+f x)}}-\frac {\int \frac {-\frac {1}{2} a (2 c-7 d)-\frac {3}{2} a d \sin (e+f x)}{(a+a \sin (e+f x)) (c+d \sin (e+f x))^{3/2}} \, dx}{3 a^2 (c-d)}\\ &=-\frac {(c-5 d) \cos (e+f x)}{3 a^2 (c-d)^2 f (1+\sin (e+f x)) \sqrt {c+d \sin (e+f x)}}-\frac {\cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2 \sqrt {c+d \sin (e+f x)}}+\frac {\int \frac {6 a^2 d^2+\frac {1}{2} a^2 (c-5 d) d \sin (e+f x)}{(c+d \sin (e+f x))^{3/2}} \, dx}{3 a^4 (c-d)^2}\\ &=-\frac {d \left (c^2-5 c d-12 d^2\right ) \cos (e+f x)}{3 a^2 (c-d)^3 (c+d) f \sqrt {c+d \sin (e+f x)}}-\frac {(c-5 d) \cos (e+f x)}{3 a^2 (c-d)^2 f (1+\sin (e+f x)) \sqrt {c+d \sin (e+f x)}}-\frac {\cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2 \sqrt {c+d \sin (e+f x)}}-\frac {2 \int \frac {-\frac {1}{4} a^2 d^2 (11 c+5 d)+\frac {1}{4} a^2 d \left (c^2-5 c d-12 d^2\right ) \sin (e+f x)}{\sqrt {c+d \sin (e+f x)}} \, dx}{3 a^4 (c-d)^3 (c+d)}\\ &=-\frac {d \left (c^2-5 c d-12 d^2\right ) \cos (e+f x)}{3 a^2 (c-d)^3 (c+d) f \sqrt {c+d \sin (e+f x)}}-\frac {(c-5 d) \cos (e+f x)}{3 a^2 (c-d)^2 f (1+\sin (e+f x)) \sqrt {c+d \sin (e+f x)}}-\frac {\cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2 \sqrt {c+d \sin (e+f x)}}+\frac {(c-5 d) \int \frac {1}{\sqrt {c+d \sin (e+f x)}} \, dx}{6 a^2 (c-d)^2}-\frac {\left (c^2-5 c d-12 d^2\right ) \int \sqrt {c+d \sin (e+f x)} \, dx}{6 a^2 (c-d)^3 (c+d)}\\ &=-\frac {d \left (c^2-5 c d-12 d^2\right ) \cos (e+f x)}{3 a^2 (c-d)^3 (c+d) f \sqrt {c+d \sin (e+f x)}}-\frac {(c-5 d) \cos (e+f x)}{3 a^2 (c-d)^2 f (1+\sin (e+f x)) \sqrt {c+d \sin (e+f x)}}-\frac {\cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2 \sqrt {c+d \sin (e+f x)}}-\frac {\left (\left (c^2-5 c d-12 d^2\right ) \sqrt {c+d \sin (e+f x)}\right ) \int \sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}} \, dx}{6 a^2 (c-d)^3 (c+d) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {\left ((c-5 d) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}\right ) \int \frac {1}{\sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}}} \, dx}{6 a^2 (c-d)^2 \sqrt {c+d \sin (e+f x)}}\\ &=-\frac {d \left (c^2-5 c d-12 d^2\right ) \cos (e+f x)}{3 a^2 (c-d)^3 (c+d) f \sqrt {c+d \sin (e+f x)}}-\frac {(c-5 d) \cos (e+f x)}{3 a^2 (c-d)^2 f (1+\sin (e+f x)) \sqrt {c+d \sin (e+f x)}}-\frac {\cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2 \sqrt {c+d \sin (e+f x)}}-\frac {\left (c^2-5 c d-12 d^2\right ) E\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {c+d \sin (e+f x)}}{3 a^2 (c-d)^3 (c+d) f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {(c-5 d) F\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}{3 a^2 (c-d)^2 f \sqrt {c+d \sin (e+f x)}}\\ \end {align*}
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Mathematica [A] time = 4.75, size = 405, normalized size = 1.24 \[ \frac {\left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^4 \left (\frac {\left (c^2-5 c d-12 d^2\right ) (c+d \sin (e+f x))+\left (c^2-5 c d-12 d^2\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} \left ((c+d) E\left (\frac {1}{4} (-2 e-2 f x+\pi )|\frac {2 d}{c+d}\right )-c F\left (\frac {1}{4} (-2 e-2 f x+\pi )|\frac {2 d}{c+d}\right )\right )-\left (d^2 (11 c+5 d) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} F\left (\frac {1}{4} (-2 e-2 f x+\pi )|\frac {2 d}{c+d}\right )\right )}{c+d}+(c+d \sin (e+f x)) \left (-\frac {2 \left (c^2-5 c d-9 d^2\right )}{c+d}+\frac {6 d^3 \cos (e+f x)}{(c+d) (c+d \sin (e+f x))}+\frac {2 (c-6 d) \sin \left (\frac {1}{2} (e+f x)\right )}{\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )}+\frac {d-c}{\left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^2}+\frac {2 (c-d) \sin \left (\frac {1}{2} (e+f x)\right )}{\left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^3}\right )\right )}{3 a^2 f (c-d)^3 (\sin (e+f x)+1)^2 \sqrt {c+d \sin (e+f x)}} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {d \sin \left (f x + e\right ) + c}}{a^{2} d^{2} \cos \left (f x + e\right )^{4} + 2 \, a^{2} c^{2} + 4 \, a^{2} c d + 2 \, a^{2} d^{2} - {\left (a^{2} c^{2} + 4 \, a^{2} c d + 3 \, a^{2} d^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, {\left (a^{2} c^{2} + 2 \, a^{2} c d + a^{2} d^{2} - {\left (a^{2} c d + a^{2} d^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (a \sin \left (f x + e\right ) + a\right )}^{2} {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 5.98, size = 1299, normalized size = 3.98 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^2\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {1}{c \sqrt {c + d \sin {\left (e + f x \right )}} \sin ^{2}{\left (e + f x \right )} + 2 c \sqrt {c + d \sin {\left (e + f x \right )}} \sin {\left (e + f x \right )} + c \sqrt {c + d \sin {\left (e + f x \right )}} + d \sqrt {c + d \sin {\left (e + f x \right )}} \sin ^{3}{\left (e + f x \right )} + 2 d \sqrt {c + d \sin {\left (e + f x \right )}} \sin ^{2}{\left (e + f x \right )} + d \sqrt {c + d \sin {\left (e + f x \right )}} \sin {\left (e + f x \right )}}\, dx}{a^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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