Optimal. Leaf size=181 \[ -\frac {2 \left (A d (2 c+d)-B \left (c^2+c d+d^2\right )\right ) \tan ^{-1}\left (\frac {c \tan \left (\frac {1}{2} (e+f x)\right )+d}{\sqrt {c^2-d^2}}\right )}{a f (c-d) \left (c^2-d^2\right )^{3/2}}+\frac {d (B (2 c+d)-A (c+2 d)) \cos (e+f x)}{a f (c-d)^2 (c+d) (c+d \sin (e+f x))}-\frac {(A-B) \cos (e+f x)}{f (c-d) (a \sin (e+f x)+a) (c+d \sin (e+f x))} \]
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Rubi [A] time = 0.35, antiderivative size = 181, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {2978, 2754, 12, 2660, 618, 204} \[ -\frac {2 \left (A d (2 c+d)-B \left (c^2+c d+d^2\right )\right ) \tan ^{-1}\left (\frac {c \tan \left (\frac {1}{2} (e+f x)\right )+d}{\sqrt {c^2-d^2}}\right )}{a f (c-d) \left (c^2-d^2\right )^{3/2}}+\frac {d (B (2 c+d)-A (c+2 d)) \cos (e+f x)}{a f (c-d)^2 (c+d) (c+d \sin (e+f x))}-\frac {(A-B) \cos (e+f x)}{f (c-d) (a \sin (e+f x)+a) (c+d \sin (e+f x))} \]
Antiderivative was successfully verified.
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Rule 12
Rule 204
Rule 618
Rule 2660
Rule 2754
Rule 2978
Rubi steps
\begin {align*} \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c+d \sin (e+f x))^2} \, dx &=-\frac {(A-B) \cos (e+f x)}{(c-d) f (a+a \sin (e+f x)) (c+d \sin (e+f x))}-\frac {\int \frac {a (2 A d-B (c+d))-a (A-B) d \sin (e+f x)}{(c+d \sin (e+f x))^2} \, dx}{a^2 (c-d)}\\ &=\frac {d (B (2 c+d)-A (c+2 d)) \cos (e+f x)}{a (c-d)^2 (c+d) f (c+d \sin (e+f x))}-\frac {(A-B) \cos (e+f x)}{(c-d) f (a+a \sin (e+f x)) (c+d \sin (e+f x))}-\frac {\int \frac {a \left (A d (2 c+d)-B \left (c^2+c d+d^2\right )\right )}{c+d \sin (e+f x)} \, dx}{a^2 (c-d)^2 (c+d)}\\ &=\frac {d (B (2 c+d)-A (c+2 d)) \cos (e+f x)}{a (c-d)^2 (c+d) f (c+d \sin (e+f x))}-\frac {(A-B) \cos (e+f x)}{(c-d) f (a+a \sin (e+f x)) (c+d \sin (e+f x))}-\frac {\left (A d (2 c+d)-B \left (c^2+c d+d^2\right )\right ) \int \frac {1}{c+d \sin (e+f x)} \, dx}{a (c-d)^2 (c+d)}\\ &=\frac {d (B (2 c+d)-A (c+2 d)) \cos (e+f x)}{a (c-d)^2 (c+d) f (c+d \sin (e+f x))}-\frac {(A-B) \cos (e+f x)}{(c-d) f (a+a \sin (e+f x)) (c+d \sin (e+f x))}-\frac {\left (2 \left (A d (2 c+d)-B \left (c^2+c d+d^2\right )\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 )}{a (c-d)^2 (c+d) f}\\ &=\frac {d (B (2 c+d)-A (c+2 d)) \cos (e+f x)}{a (c-d)^2 (c+d) f (c+d \sin (e+f x))}-\frac {(A-B) \cos (e+f x)}{(c-d) f (a+a \sin (e+f x)) (c+d \sin (e+f x))}+\frac {\left (4 \left (A d (2 c+d)-B \left (c^2+c d+d^2\right )\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 )}{a (c-d)^2 (c+d) f}\\ &=-\frac {2 \left (A d (2 c+d)-B \left (c^2+c d+d^2\right )\right ) \tan ^{-1}\left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{a (c-d)^2 (c+d) \sqrt {c^2-d^2} f}+\frac {d (B (2 c+d)-A (c+2 d)) \cos (e+f x)}{a (c-d)^2 (c+d) f (c+d \sin (e+f x))}-\frac {(A-B) \cos (e+f x)}{(c-d) f (a+a \sin (e+f x)) (c+d \sin (e+f x))}\\ \end {align*}
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Mathematica [A] time = 1.22, size = 209, normalized size = 1.15 \[ \frac {\left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right ) \left (\frac {2 \left (B \left (c^2+c d+d^2\right )-A d (2 c+d)\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right ) \tan ^{-1}\left (\frac {c \tan \left (\frac {1}{2} (e+f x)\right )+d}{\sqrt {c^2-d^2}}\right )}{(c+d) \sqrt {c^2-d^2}}+\frac {d (B c-A d) \cos (e+f x) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )}{(c+d) (c+d \sin (e+f x))}+2 (A-B) \sin \left (\frac {1}{2} (e+f x)\right )\right )}{a f (c-d)^2 (\sin (e+f x)+1)} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.51, size = 1538, normalized size = 8.50 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.19, size = 443, normalized size = 2.45 \[ \frac {2 \, {\left (\frac {{\left (B c^{2} - 2 \, A c d + B c d - A d^{2} + B d^{2}\right )} {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (c) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + d}{\sqrt {c^{2} - d^{2}}}\right )\right )}}{{\left (a c^{3} - a c^{2} d - a c d^{2} + a d^{3}\right )} \sqrt {c^{2} - d^{2}}} - \frac {A c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - B c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + A c^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - B c^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - B c d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + A d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, A c^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 3 \, B c^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 3 \, A c d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 3 \, B c d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + A d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + A c^{3} - B c^{3} + A c^{2} d - 2 \, B c^{2} d + A c d^{2}}{{\left (a c^{4} - a c^{3} d - a c^{2} d^{2} + a c d^{3}\right )} {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + c\right )}}\right )}}{f} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.58, size = 615, normalized size = 3.40 \[ -\frac {2 d^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) A}{a f \left (c -d \right )^{2} \left (\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c +2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) d +c \right ) \left (c +d \right ) c}+\frac {2 d^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) B}{a f \left (c -d \right )^{2} \left (\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c +2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) d +c \right ) \left (c +d \right )}-\frac {2 d^{2} A}{a f \left (c -d \right )^{2} \left (\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c +2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) d +c \right ) \left (c +d \right )}+\frac {2 d B c}{a f \left (c -d \right )^{2} \left (\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c +2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) d +c \right ) \left (c +d \right )}-\frac {4 \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right ) A c d}{a f \left (c -d \right )^{2} \left (c +d \right ) \sqrt {c^{2}-d^{2}}}-\frac {2 \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right ) A \,d^{2}}{a f \left (c -d \right )^{2} \left (c +d \right ) \sqrt {c^{2}-d^{2}}}+\frac {2 \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right ) B \,c^{2}}{a f \left (c -d \right )^{2} \left (c +d \right ) \sqrt {c^{2}-d^{2}}}+\frac {2 \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right ) B c d}{a f \left (c -d \right )^{2} \left (c +d \right ) \sqrt {c^{2}-d^{2}}}+\frac {2 \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right ) B \,d^{2}}{a f \left (c -d \right )^{2} \left (c +d \right ) \sqrt {c^{2}-d^{2}}}-\frac {2 A}{a f \left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}+\frac {2 B}{a f \left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 15.53, size = 437, normalized size = 2.41 \[ \frac {2\,\mathrm {atan}\left (\frac {\frac {\left (2\,a\,c^3\,d-2\,a\,c^2\,d^2-2\,a\,c\,d^3+2\,a\,d^4\right )\,\left (B\,c^2-A\,d^2+B\,d^2-2\,A\,c\,d+B\,c\,d\right )}{a\,{\left (c+d\right )}^{3/2}\,{\left (c-d\right )}^{5/2}}+\frac {2\,c\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (a\,c^3-a\,c^2\,d-a\,c\,d^2+a\,d^3\right )\,\left (B\,c^2-A\,d^2+B\,d^2-2\,A\,c\,d+B\,c\,d\right )}{a\,{\left (c+d\right )}^{3/2}\,{\left (c-d\right )}^{5/2}}}{2\,B\,c^2-2\,A\,d^2+2\,B\,d^2-4\,A\,c\,d+2\,B\,c\,d}\right )\,\left (B\,c^2-A\,d^2+B\,d^2-2\,A\,c\,d+B\,c\,d\right )}{a\,f\,{\left (c+d\right )}^{3/2}\,{\left (c-d\right )}^{5/2}}-\frac {\frac {2\,\left (A\,c^2+A\,d^2-B\,c^2+A\,c\,d-2\,B\,c\,d\right )}{\left (c+d\right )\,{\left (c-d\right )}^2}+\frac {2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (A\,d^2+2\,A\,c\,d-3\,B\,c\,d\right )}{c\,{\left (c-d\right )}^2}+\frac {2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (A\,c^3+A\,d^3-B\,c^3+A\,c^2\,d-B\,c\,d^2-B\,c^2\,d\right )}{c\,\left (c+d\right )\,{\left (c-d\right )}^2}}{f\,\left (a\,c\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+\left (a\,c+2\,a\,d\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+\left (a\,c+2\,a\,d\right )\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+a\,c\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [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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