Optimal. Leaf size=83 \[ \frac {2 a \tan ^{-1}\left (\frac {c \tan \left (\frac {1}{2} (e+f x)\right )+d}{\sqrt {c^2-d^2}}\right )}{f (c+d) \sqrt {c^2-d^2}}-\frac {a \cos (e+f x)}{f (c+d) (c+d \sin (e+f x))} \]
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Rubi [A] time = 0.09, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 5, 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} \[ \frac {2 a \tan ^{-1}\left (\frac {c \tan \left (\frac {1}{2} (e+f x)\right )+d}{\sqrt {c^2-d^2}}\right )}{f (c+d) \sqrt {c^2-d^2}}-\frac {a \cos (e+f x)}{f (c+d) (c+d \sin (e+f x))} \]
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))^2} \, dx &=-\frac {a \cos (e+f x)}{(c+d) f (c+d \sin (e+f x))}-\frac {\int \frac {a (c-d)}{c+d \sin (e+f x)} \, dx}{-c^2+d^2}\\ &=-\frac {a \cos (e+f x)}{(c+d) f (c+d \sin (e+f x))}+\frac {a \int \frac {1}{c+d \sin (e+f x)} \, dx}{c+d}\\ &=-\frac {a \cos (e+f x)}{(c+d) f (c+d \sin (e+f x))}+\frac {(2 a) \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) f}\\ &=-\frac {a \cos (e+f x)}{(c+d) f (c+d \sin (e+f x))}-\frac {(4 a) \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) f}\\ &=\frac {2 a \tan ^{-1}\left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{(c+d) \sqrt {c^2-d^2} f}-\frac {a \cos (e+f x)}{(c+d) f (c+d \sin (e+f x))}\\ \end {align*}
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Mathematica [C] time = 0.59, size = 220, normalized size = 2.65 \[ \frac {a (\sin (e+f x)+1) \left (2 \sqrt {c^2-d^2} \csc (e) \sqrt {(\cos (e)-i \sin (e))^2} (c \cos (e)+d \sin (f x))+4 d (\cos (e)-i \sin (e)) (c+d \sin (e+f x)) \tan ^{-1}\left (\frac {(\cos (e)-i \sin (e)) \sec \left (\frac {f x}{2}\right ) \left (c \sin \left (\frac {f x}{2}\right )+d \cos \left (e+\frac {f x}{2}\right )\right )}{\sqrt {c^2-d^2} \sqrt {(\cos (e)-i \sin (e))^2}}\right )\right )}{2 d f (c+d) \sqrt {c^2-d^2} \sqrt {(\cos (e)-i \sin (e))^2} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^2 (c+d \sin (e+f x))} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.49, size = 362, normalized size = 4.36 \[ \left [-\frac {{\left (a d \sin \left (f x + e\right ) + a c\right )} \sqrt {-c^{2} + d^{2}} \log \left (\frac {{\left (2 \, c^{2} - d^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2} + 2 \, {\left (c \cos \left (f x + e\right ) \sin \left (f x + e\right ) + d \cos \left (f x + e\right )\right )} \sqrt {-c^{2} + d^{2}}}{d^{2} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2}}\right ) + 2 \, {\left (a c^{2} - a d^{2}\right )} \cos \left (f x + e\right )}{2 \, {\left ({\left (c^{3} d + c^{2} d^{2} - c d^{3} - d^{4}\right )} f \sin \left (f x + e\right ) + {\left (c^{4} + c^{3} d - c^{2} d^{2} - c d^{3}\right )} f\right )}}, -\frac {{\left (a d \sin \left (f x + e\right ) + a c\right )} \sqrt {c^{2} - d^{2}} \arctan \left (-\frac {c \sin \left (f x + e\right ) + d}{\sqrt {c^{2} - d^{2}} \cos \left (f x + e\right )}\right ) + {\left (a c^{2} - a d^{2}\right )} \cos \left (f x + e\right )}{{\left (c^{3} d + c^{2} d^{2} - c d^{3} - d^{4}\right )} f \sin \left (f x + e\right ) + {\left (c^{4} + c^{3} d - c^{2} d^{2} - c d^{3}\right )} f}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.34, size = 129, normalized size = 1.55 \[ \frac {2 \, {\left (\frac {{\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (c) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + d}{\sqrt {c^{2} - d^{2}}}\right )\right )} a}{\sqrt {c^{2} - d^{2}} {\left (c + d\right )}} - \frac {a d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + a c}{{\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + c\right )} {\left (c^{2} + c d\right )}}\right )}}{f} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.24, size = 147, normalized size = 1.77 \[ -\frac {2 a d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f \left (\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c +2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) d +c \right ) \left (c +d \right ) c}-\frac {2 a}{f \left (\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c +2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) d +c \right ) \left (c +d \right )}+\frac {2 a \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{f \left (c +d \right ) \sqrt {c^{2}-d^{2}}} \]
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 = 6.95, size = 140, normalized size = 1.69 \[ \frac {2\,a\,\mathrm {atan}\left (\frac {\left (c+d\right )\,\left (\frac {2\,a\,\left (d^2+c\,d\right )}{{\left (c+d\right )}^{5/2}\,\sqrt {c-d}}+\frac {2\,a\,c\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{{\left (c+d\right )}^{3/2}\,\sqrt {c-d}}\right )}{2\,a}\right )}{f\,{\left (c+d\right )}^{3/2}\,\sqrt {c-d}}-\frac {\frac {2\,a}{c+d}+\frac {2\,a\,d\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{c\,\left (c+d\right )}}{f\,\left (c\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+2\,d\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+c\right )} \]
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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