Optimal. Leaf size=98 \[ \frac {2 (a c-b d) \tan ^{-1}\left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{\left (c^2-d^2\right )^{3/2} f}-\frac {(b c-a d) \cos (e+f x)}{\left (c^2-d^2\right ) f (c+d \sin (e+f x))} \]
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Rubi [A]
time = 0.08, antiderivative size = 98, 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 = {2833, 12, 2739,
632, 210} \begin {gather*} \frac {2 (a c-b d) \text {ArcTan}\left (\frac {c \tan \left (\frac {1}{2} (e+f x)\right )+d}{\sqrt {c^2-d^2}}\right )}{f \left (c^2-d^2\right )^{3/2}}-\frac {(b c-a d) \cos (e+f x)}{f \left (c^2-d^2\right ) (c+d \sin (e+f x))} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 210
Rule 632
Rule 2739
Rule 2833
Rubi steps
\begin {align*} \int \frac {a+b \sin (e+f x)}{(c+d \sin (e+f x))^2} \, dx &=-\frac {(b c-a d) \cos (e+f x)}{\left (c^2-d^2\right ) f (c+d \sin (e+f x))}+\frac {\int \frac {-a c+b d}{c+d \sin (e+f x)} \, dx}{-c^2+d^2}\\ &=-\frac {(b c-a d) \cos (e+f x)}{\left (c^2-d^2\right ) f (c+d \sin (e+f x))}+\frac {(a c-b d) \int \frac {1}{c+d \sin (e+f x)} \, dx}{c^2-d^2}\\ &=-\frac {(b c-a d) \cos (e+f x)}{\left (c^2-d^2\right ) f (c+d \sin (e+f x))}+\frac {(2 (a c-b d)) \text {Subst}\left (\int \frac {1}{c+2 d x+c x^2} \, dx,x,\tan \left (\frac {1}{2} (e+f x)\right )\right )}{\left (c^2-d^2\right ) f}\\ &=-\frac {(b c-a d) \cos (e+f x)}{\left (c^2-d^2\right ) f (c+d \sin (e+f x))}-\frac {(4 (a c-b d)) \text {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 )}{\left (c^2-d^2\right ) f}\\ &=\frac {2 (a c-b d) \tan ^{-1}\left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{\left (c^2-d^2\right )^{3/2} f}-\frac {(b c-a d) \cos (e+f x)}{\left (c^2-d^2\right ) f (c+d \sin (e+f x))}\\ \end {align*}
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Mathematica [A]
time = 0.44, size = 96, normalized size = 0.98 \begin {gather*} \frac {\frac {2 (a c-b d) \tan ^{-1}\left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{\left (c^2-d^2\right )^{3/2}}+\frac {(-b c+a d) \cos (e+f x)}{(c-d) (c+d) (c+d \sin (e+f x))}}{f} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.22, size = 142, normalized size = 1.45
method | result | size |
derivativedivides | \(\frac {\frac {\frac {2 d \left (a d -b c \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (c^{2}-d^{2}\right ) c}+\frac {2 \left (a d -b c \right )}{c^{2}-d^{2}}}{c \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c}+\frac {2 \left (a c -b d \right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c^{2}-d^{2}\right )^{\frac {3}{2}}}}{f}\) | \(142\) |
default | \(\frac {\frac {\frac {2 d \left (a d -b c \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (c^{2}-d^{2}\right ) c}+\frac {2 \left (a d -b c \right )}{c^{2}-d^{2}}}{c \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c}+\frac {2 \left (a c -b d \right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c^{2}-d^{2}\right )^{\frac {3}{2}}}}{f}\) | \(142\) |
risch | \(\frac {2 i \left (-a d +b c \right ) \left (i d +c \,{\mathrm e}^{i \left (f x +e \right )}\right )}{d \left (c^{2}-d^{2}\right ) f \left (-i {\mathrm e}^{2 i \left (f x +e \right )} d +i d +2 c \,{\mathrm e}^{i \left (f x +e \right )}\right )}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-c^{2}+d^{2}}\, c -c^{2}+d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) a c}{\sqrt {-c^{2}+d^{2}}\, \left (c +d \right ) \left (c -d \right ) f}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-c^{2}+d^{2}}\, c -c^{2}+d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) b d}{\sqrt {-c^{2}+d^{2}}\, \left (c +d \right ) \left (c -d \right ) f}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-c^{2}+d^{2}}\, c +c^{2}-d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) a c}{\sqrt {-c^{2}+d^{2}}\, \left (c +d \right ) \left (c -d \right ) f}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-c^{2}+d^{2}}\, c +c^{2}-d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) b d}{\sqrt {-c^{2}+d^{2}}\, \left (c +d \right ) \left (c -d \right ) f}\) | \(396\) |
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 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.36, size = 409, normalized size = 4.17 \begin {gather*} \left [-\frac {{\left (a c^{2} - b c d + {\left (a c d - b 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 (b c^{3} - a c^{2} d - b c d^{2} + a d^{3}\right )} \cos \left (f x + e\right )}{2 \, {\left ({\left (c^{4} d - 2 \, c^{2} d^{3} + d^{5}\right )} f \sin \left (f x + e\right ) + {\left (c^{5} - 2 \, c^{3} d^{2} + c d^{4}\right )} f\right )}}, -\frac {{\left (a c^{2} - b c d + {\left (a c d - b 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 (b c^{3} - a c^{2} d - b c d^{2} + a d^{3}\right )} \cos \left (f x + e\right )}{{\left (c^{4} d - 2 \, c^{2} d^{3} + d^{5}\right )} f \sin \left (f x + e\right ) + {\left (c^{5} - 2 \, c^{3} d^{2} + c d^{4}\right )} f}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.49, size = 158, normalized size = 1.61 \begin {gather*} \frac {2 \, {\left (\frac {{\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 c - b d\right )}}{{\left (c^{2} - d^{2}\right )}^{\frac {3}{2}}} - \frac {b c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - a d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + b c^{2} - a c d}{{\left (c^{3} - c d^{2}\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 )}}\right )}}{f} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 7.92, size = 214, normalized size = 2.18 \begin {gather*} \frac {\frac {2\,\left (a\,d-b\,c\right )}{c^2-d^2}+\frac {2\,d\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (a\,d-b\,c\right )}{c\,\left (c^2-d^2\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 )}+\frac {2\,\mathrm {atan}\left (\frac {\left (\frac {2\,\left (c^2\,d-d^3\right )\,\left (a\,c-b\,d\right )}{{\left (c+d\right )}^{3/2}\,\left (c^2-d^2\right )\,{\left (c-d\right )}^{3/2}}+\frac {2\,c\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (a\,c-b\,d\right )}{{\left (c+d\right )}^{3/2}\,{\left (c-d\right )}^{3/2}}\right )\,\left (c^2-d^2\right )}{2\,\left (a\,c-b\,d\right )}\right )\,\left (a\,c-b\,d\right )}{f\,{\left (c+d\right )}^{3/2}\,{\left (c-d\right )}^{3/2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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