Optimal. Leaf size=134 \[ \frac {a (2 c-d) \tan ^{-1}\left (\frac {c \tan \left (\frac {1}{2} (e+f x)\right )+d}{\sqrt {c^2-d^2}}\right )}{f (c+d) \left (c^2-d^2\right )^{3/2}}-\frac {a (c-2 d) \cos (e+f x)}{2 f (c-d) (c+d)^2 (c+d \sin (e+f x))}-\frac {a \cos (e+f x)}{2 f (c+d) (c+d \sin (e+f x))^2} \]
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Rubi [A] time = 0.18, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {2754, 12, 2660, 618, 204} \[ \frac {a (2 c-d) \tan ^{-1}\left (\frac {c \tan \left (\frac {1}{2} (e+f x)\right )+d}{\sqrt {c^2-d^2}}\right )}{f (c+d) \left (c^2-d^2\right )^{3/2}}-\frac {a (c-2 d) \cos (e+f x)}{2 f (c-d) (c+d)^2 (c+d \sin (e+f x))}-\frac {a \cos (e+f x)}{2 f (c+d) (c+d \sin (e+f x))^2} \]
Antiderivative was successfully verified.
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Rule 12
Rule 204
Rule 618
Rule 2660
Rule 2754
Rubi steps
\begin {align*} \int \frac {a+a \sin (e+f x)}{(c+d \sin (e+f x))^3} \, dx &=-\frac {a \cos (e+f x)}{2 (c+d) f (c+d \sin (e+f x))^2}-\frac {\int \frac {-2 a (c-d)-a (c-d) \sin (e+f x)}{(c+d \sin (e+f x))^2} \, dx}{2 \left (c^2-d^2\right )}\\ &=-\frac {a \cos (e+f x)}{2 (c+d) f (c+d \sin (e+f x))^2}-\frac {a (c-2 d) \cos (e+f x)}{2 (c-d) (c+d)^2 f (c+d \sin (e+f x))}+\frac {\int \frac {a (c-d) (2 c-d)}{c+d \sin (e+f x)} \, dx}{2 \left (c^2-d^2\right )^2}\\ &=-\frac {a \cos (e+f x)}{2 (c+d) f (c+d \sin (e+f x))^2}-\frac {a (c-2 d) \cos (e+f x)}{2 (c-d) (c+d)^2 f (c+d \sin (e+f x))}+\frac {(a (2 c-d)) \int \frac {1}{c+d \sin (e+f x)} \, dx}{2 (c-d) (c+d)^2}\\ &=-\frac {a \cos (e+f x)}{2 (c+d) f (c+d \sin (e+f x))^2}-\frac {a (c-2 d) \cos (e+f x)}{2 (c-d) (c+d)^2 f (c+d \sin (e+f x))}+\frac {(a (2 c-d)) \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 )}{(c-d) (c+d)^2 f}\\ &=-\frac {a \cos (e+f x)}{2 (c+d) f (c+d \sin (e+f x))^2}-\frac {a (c-2 d) \cos (e+f x)}{2 (c-d) (c+d)^2 f (c+d \sin (e+f x))}-\frac {(2 a (2 c-d)) \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 )}{(c-d) (c+d)^2 f}\\ &=\frac {a (2 c-d) \tan ^{-1}\left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{(c-d) (c+d)^2 \sqrt {c^2-d^2} f}-\frac {a \cos (e+f x)}{2 (c+d) f (c+d \sin (e+f x))^2}-\frac {a (c-2 d) \cos (e+f x)}{2 (c-d) (c+d)^2 f (c+d \sin (e+f x))}\\ \end {align*}
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Mathematica [C] time = 1.20, size = 242, normalized size = 1.81 \[ \frac {a (\sin (e+f x)+1) \left (\frac {4 (2 c-d) (\cos (e)-i \sin (e)) \tan ^{-1}\left (\frac {(\cos (e)-i \sin (e)) \sec \left (\frac {f x}{2}\right ) \left (c \sin \left (\frac {f x}{2}\right )+d \cos \left (e+\frac {f x}{2}\right )\right )}{\sqrt {c^2-d^2} \sqrt {(\cos (e)-i \sin (e))^2}}\right )}{(c-d) \sqrt {c^2-d^2} \sqrt {(\cos (e)-i \sin (e))^2}}+\frac {2 (c+d) \csc (e) (c \cos (e)+d \sin (f x))}{d (c+d \sin (e+f x))^2}+\frac {2 (c-2 d) \csc (e) \sin (f x)+(2 d-4 c) \cot (e)}{(c-d) (c+d \sin (e+f x))}\right )}{4 f (c+d)^2 \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^2} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.48, size = 803, normalized size = 5.99 \[ \left [\frac {2 \, {\left (a c^{3} d - 2 \, a c^{2} d^{2} - a c d^{3} + 2 \, a d^{4}\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + {\left (2 \, a c^{3} - a c^{2} d + 2 \, a c d^{2} - a d^{3} - {\left (2 \, a c d^{2} - a d^{3}\right )} \cos \left (f x + e\right )^{2} + 2 \, {\left (2 \, a c^{2} d - a c d^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c^{2} + d^{2}} \log \left (\frac {{\left (2 \, c^{2} - d^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2} + 2 \, {\left (c \cos \left (f x + e\right ) \sin \left (f x + e\right ) + d \cos \left (f x + e\right )\right )} \sqrt {-c^{2} + d^{2}}}{d^{2} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2}}\right ) + 2 \, {\left (2 \, a c^{4} - 2 \, a c^{3} d - 3 \, a c^{2} d^{2} + 2 \, a c d^{3} + a d^{4}\right )} \cos \left (f x + e\right )}{4 \, {\left ({\left (c^{5} d^{2} + c^{4} d^{3} - 2 \, c^{3} d^{4} - 2 \, c^{2} d^{5} + c d^{6} + d^{7}\right )} f \cos \left (f x + e\right )^{2} - 2 \, {\left (c^{6} d + c^{5} d^{2} - 2 \, c^{4} d^{3} - 2 \, c^{3} d^{4} + c^{2} d^{5} + c d^{6}\right )} f \sin \left (f x + e\right ) - {\left (c^{7} + c^{6} d - c^{5} d^{2} - c^{4} d^{3} - c^{3} d^{4} - c^{2} d^{5} + c d^{6} + d^{7}\right )} f\right )}}, \frac {{\left (a c^{3} d - 2 \, a c^{2} d^{2} - a c d^{3} + 2 \, a d^{4}\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + {\left (2 \, a c^{3} - a c^{2} d + 2 \, a c d^{2} - a d^{3} - {\left (2 \, a c d^{2} - a d^{3}\right )} \cos \left (f x + e\right )^{2} + 2 \, {\left (2 \, a c^{2} d - a c d^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {c^{2} - d^{2}} \arctan \left (-\frac {c \sin \left (f x + e\right ) + d}{\sqrt {c^{2} - d^{2}} \cos \left (f x + e\right )}\right ) + {\left (2 \, a c^{4} - 2 \, a c^{3} d - 3 \, a c^{2} d^{2} + 2 \, a c d^{3} + a d^{4}\right )} \cos \left (f x + e\right )}{2 \, {\left ({\left (c^{5} d^{2} + c^{4} d^{3} - 2 \, c^{3} d^{4} - 2 \, c^{2} d^{5} + c d^{6} + d^{7}\right )} f \cos \left (f x + e\right )^{2} - 2 \, {\left (c^{6} d + c^{5} d^{2} - 2 \, c^{4} d^{3} - 2 \, c^{3} d^{4} + c^{2} d^{5} + c d^{6}\right )} f \sin \left (f x + e\right ) - {\left (c^{7} + c^{6} d - c^{5} d^{2} - c^{4} d^{3} - c^{3} d^{4} - c^{2} d^{5} + c d^{6} + d^{7}\right )} f\right )}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.36, size = 384, normalized size = 2.87 \[ \frac {\frac {{\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (c) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + d}{\sqrt {c^{2} - d^{2}}}\right )\right )} {\left (2 \, a c - a d\right )}}{{\left (c^{3} + c^{2} d - c d^{2} - d^{3}\right )} \sqrt {c^{2} - d^{2}}} - \frac {3 \, a c^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 2 \, a c^{2} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 2 \, a c d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 2 \, a c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 2 \, a c^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 3 \, a c^{2} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 4 \, a c d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 2 \, a d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 5 \, a c^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 6 \, a c^{2} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 2 \, a c d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2 \, a c^{4} - 2 \, a c^{3} d - a c^{2} d^{2}}{{\left (c^{5} + c^{4} d - c^{3} d^{2} - c^{2} d^{3}\right )} {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + c\right )}^{2}}}{f} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.28, size = 1104, normalized size = 8.24 \[ \text {result too large to display} \]
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 = 8.97, size = 445, normalized size = 3.32 \[ -\frac {\frac {-2\,a\,c^2+2\,a\,c\,d+a\,d^2}{-c^3-c^2\,d+c\,d^2+d^3}+\frac {a\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (c^2+2\,d^2\right )\,\left (-2\,c^2+2\,c\,d+d^2\right )}{c^2\,\left (-c^3-c^2\,d+c\,d^2+d^3\right )}+\frac {a\,d\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (-5\,c^2+6\,c\,d+2\,d^2\right )}{c\,\left (-c^3-c^2\,d+c\,d^2+d^3\right )}+\frac {a\,d\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (-3\,c^2+2\,c\,d+2\,d^2\right )}{c\,\left (-c^3-c^2\,d+c\,d^2+d^3\right )}}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (2\,c^2+4\,d^2\right )+c^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+c^2+4\,c\,d\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+4\,c\,d\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}-\frac {a\,\mathrm {atan}\left (\frac {\left (\frac {a\,\left (2\,c-d\right )\,\left (-2\,c^3\,d-2\,c^2\,d^2+2\,c\,d^3+2\,d^4\right )}{2\,{\left (c+d\right )}^{5/2}\,{\left (c-d\right )}^{3/2}\,\left (-c^3-c^2\,d+c\,d^2+d^3\right )}+\frac {a\,c\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (2\,c-d\right )}{{\left (c+d\right )}^{5/2}\,{\left (c-d\right )}^{3/2}}\right )\,\left (-c^3-c^2\,d+c\,d^2+d^3\right )}{2\,a\,c-a\,d}\right )\,\left (2\,c-d\right )}{f\,{\left (c+d\right )}^{5/2}\,{\left (c-d\right )}^{3/2}} \]
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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