Optimal. Leaf size=66 \[ -\frac {2 b \tan ^{-1}\left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}-\frac {a \cos (x)}{\left (a^2-b^2\right ) (a+b \sin (x))} \]
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Rubi [A]
time = 0.05, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 5, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.454, Rules used = {2833, 12, 2739,
632, 210} \begin {gather*} -\frac {2 b \text {ArcTan}\left (\frac {a \tan \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}-\frac {a \cos (x)}{\left (a^2-b^2\right ) (a+b \sin (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 {\sin (x)}{(a+b \sin (x))^2} \, dx &=-\frac {a \cos (x)}{\left (a^2-b^2\right ) (a+b \sin (x))}+\frac {\int \frac {b}{a+b \sin (x)} \, dx}{-a^2+b^2}\\ &=-\frac {a \cos (x)}{\left (a^2-b^2\right ) (a+b \sin (x))}-\frac {b \int \frac {1}{a+b \sin (x)} \, dx}{a^2-b^2}\\ &=-\frac {a \cos (x)}{\left (a^2-b^2\right ) (a+b \sin (x))}-\frac {(2 b) \text {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )}{a^2-b^2}\\ &=-\frac {a \cos (x)}{\left (a^2-b^2\right ) (a+b \sin (x))}+\frac {(4 b) \text {Subst}\left (\int \frac {1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac {x}{2}\right )\right )}{a^2-b^2}\\ &=-\frac {2 b \tan ^{-1}\left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}-\frac {a \cos (x)}{\left (a^2-b^2\right ) (a+b \sin (x))}\\ \end {align*}
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Mathematica [A]
time = 0.08, size = 67, normalized size = 1.02 \begin {gather*} -\frac {2 b \tan ^{-1}\left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}-\frac {a \cos (x)}{(a-b) (a+b) (a+b \sin (x))} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.18, size = 99, normalized size = 1.50
method | result | size |
default | \(\frac {-8 b \tan \left (\frac {x}{2}\right )-8 a}{\left (4 a^{2}-4 b^{2}\right ) \left (a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )+2 b \tan \left (\frac {x}{2}\right )+a \right )}-\frac {8 b \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (4 a^{2}-4 b^{2}\right ) \sqrt {a^{2}-b^{2}}}\) | \(99\) |
risch | \(\frac {2 i a \left (-i a \,{\mathrm e}^{i x}+b \right )}{b \left (-a^{2}+b^{2}\right ) \left (b \,{\mathrm e}^{2 i x}-b +2 i a \,{\mathrm e}^{i x}\right )}-\frac {i b \ln \left ({\mathrm e}^{i x}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{\sqrt {a^{2}-b^{2}}\, b}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right )}+\frac {i b \ln \left ({\mathrm e}^{i x}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{\sqrt {a^{2}-b^{2}}\, b}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right )}\) | \(199\) |
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 = 0.38, size = 266, normalized size = 4.03 \begin {gather*} \left [\frac {{\left (b^{2} \sin \left (x\right ) + a b\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (x\right ) \sin \left (x\right ) + b \cos \left (x\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2}}\right ) - 2 \, {\left (a^{3} - a b^{2}\right )} \cos \left (x\right )}{2 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4} + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (x\right )\right )}}, \frac {{\left (b^{2} \sin \left (x\right ) + a b\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (x\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (x\right )}\right ) - {\left (a^{3} - a b^{2}\right )} \cos \left (x\right )}{a^{5} - 2 \, a^{3} b^{2} + a b^{4} + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (x\right )}\right ] \end {gather*}
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 \begin {gather*} \int \frac {\sin {\left (x \right )}}{\left (a + b \sin {\left (x \right )}\right )^{2}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.43, size = 90, normalized size = 1.36 \begin {gather*} -\frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, x\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} b}{{\left (a^{2} - b^{2}\right )}^{\frac {3}{2}}} - \frac {2 \, {\left (b \tan \left (\frac {1}{2} \, x\right ) + a\right )}}{{\left (a \tan \left (\frac {1}{2} \, x\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, x\right ) + a\right )} {\left (a^{2} - b^{2}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 6.49, size = 123, normalized size = 1.86 \begin {gather*} -\frac {\frac {2\,a}{a^2-b^2}+\frac {2\,b\,\mathrm {tan}\left (\frac {x}{2}\right )}{a^2-b^2}}{a\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {x}{2}\right )+a}-\frac {2\,b\,\mathrm {atan}\left (\frac {\left (a^2-b^2\right )\,\left (\frac {2\,b^2}{{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}+\frac {2\,a\,b\,\mathrm {tan}\left (\frac {x}{2}\right )}{{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}\right )}{2\,b}\right )}{{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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