Optimal. Leaf size=77 \[ -\frac {2 c \tan ^{-1}\left (\frac {c+b \tan \left (\frac {a x}{2}\right )}{\sqrt {b^2-c^2}}\right )}{a \left (b^2-c^2\right )^{3/2}}-\frac {b \cos (a x)}{a \left (b^2-c^2\right ) (b+c \sin (a x))} \]
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Rubi [A]
time = 0.07, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2833, 12, 2739,
632, 210} \begin {gather*} -\frac {2 c \text {ArcTan}\left (\frac {b \tan \left (\frac {a x}{2}\right )+c}{\sqrt {b^2-c^2}}\right )}{a \left (b^2-c^2\right )^{3/2}}-\frac {b \cos (a x)}{a \left (b^2-c^2\right ) (c \sin (a x)+b)} \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 (a x)}{(b+c \sin (a x))^2} \, dx &=-\frac {b \cos (a x)}{a \left (b^2-c^2\right ) (b+c \sin (a x))}+\frac {\int \frac {c}{b+c \sin (a x)} \, dx}{-b^2+c^2}\\ &=-\frac {b \cos (a x)}{a \left (b^2-c^2\right ) (b+c \sin (a x))}-\frac {c \int \frac {1}{b+c \sin (a x)} \, dx}{b^2-c^2}\\ &=-\frac {b \cos (a x)}{a \left (b^2-c^2\right ) (b+c \sin (a x))}-\frac {(2 c) \text {Subst}\left (\int \frac {1}{b+2 c x+b x^2} \, dx,x,\tan \left (\frac {a x}{2}\right )\right )}{a \left (b^2-c^2\right )}\\ &=-\frac {b \cos (a x)}{a \left (b^2-c^2\right ) (b+c \sin (a x))}+\frac {(4 c) \text {Subst}\left (\int \frac {1}{-4 \left (b^2-c^2\right )-x^2} \, dx,x,2 c+2 b \tan \left (\frac {a x}{2}\right )\right )}{a \left (b^2-c^2\right )}\\ &=-\frac {2 c \tan ^{-1}\left (\frac {c+b \tan \left (\frac {a x}{2}\right )}{\sqrt {b^2-c^2}}\right )}{a \left (b^2-c^2\right )^{3/2}}-\frac {b \cos (a x)}{a \left (b^2-c^2\right ) (b+c \sin (a x))}\\ \end {align*}
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Mathematica [A]
time = 0.19, size = 76, normalized size = 0.99 \begin {gather*} -\frac {\frac {2 c \tan ^{-1}\left (\frac {c+b \tan \left (\frac {a x}{2}\right )}{\sqrt {b^2-c^2}}\right )}{\left (b^2-c^2\right )^{3/2}}+\frac {b \cos (a x)}{(b-c) (b+c) (b+c \sin (a x))}}{a} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.15, size = 107, normalized size = 1.39
method | result | size |
derivativedivides | \(\frac {\frac {-8 c \tan \left (\frac {a x}{2}\right )-8 b}{\left (4 b^{2}-4 c^{2}\right ) \left (b \left (\tan ^{2}\left (\frac {a x}{2}\right )\right )+2 c \tan \left (\frac {a x}{2}\right )+b \right )}-\frac {8 c \arctan \left (\frac {2 b \tan \left (\frac {a x}{2}\right )+2 c}{2 \sqrt {b^{2}-c^{2}}}\right )}{\left (4 b^{2}-4 c^{2}\right ) \sqrt {b^{2}-c^{2}}}}{a}\) | \(107\) |
default | \(\frac {\frac {-8 c \tan \left (\frac {a x}{2}\right )-8 b}{\left (4 b^{2}-4 c^{2}\right ) \left (b \left (\tan ^{2}\left (\frac {a x}{2}\right )\right )+2 c \tan \left (\frac {a x}{2}\right )+b \right )}-\frac {8 c \arctan \left (\frac {2 b \tan \left (\frac {a x}{2}\right )+2 c}{2 \sqrt {b^{2}-c^{2}}}\right )}{\left (4 b^{2}-4 c^{2}\right ) \sqrt {b^{2}-c^{2}}}}{a}\) | \(107\) |
risch | \(\frac {2 i b \left (c -i b \,{\mathrm e}^{i a x}\right )}{c \left (-b^{2}+c^{2}\right ) a \left (c \,{\mathrm e}^{2 i a x}-c +2 i b \,{\mathrm e}^{i a x}\right )}-\frac {i c \ln \left ({\mathrm e}^{i a x}+\frac {i \left (b \sqrt {b^{2}-c^{2}}+b^{2}-c^{2}\right )}{\sqrt {b^{2}-c^{2}}\, c}\right )}{\sqrt {b^{2}-c^{2}}\, \left (b +c \right ) \left (b -c \right ) a}+\frac {i c \ln \left ({\mathrm e}^{i a x}+\frac {i \left (b \sqrt {b^{2}-c^{2}}-b^{2}+c^{2}\right )}{\sqrt {b^{2}-c^{2}}\, c}\right )}{\sqrt {b^{2}-c^{2}}\, \left (b +c \right ) \left (b -c \right ) a}\) | \(213\) |
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 = 0.52, size = 312, normalized size = 4.05 \begin {gather*} \left [\frac {{\left (c^{2} \sin \left (a x\right ) + b c\right )} \sqrt {-b^{2} + c^{2}} \log \left (\frac {{\left (2 \, b^{2} - c^{2}\right )} \cos \left (a x\right )^{2} - 2 \, b c \sin \left (a x\right ) - b^{2} - c^{2} + 2 \, {\left (b \cos \left (a x\right ) \sin \left (a x\right ) + c \cos \left (a x\right )\right )} \sqrt {-b^{2} + c^{2}}}{c^{2} \cos \left (a x\right )^{2} - 2 \, b c \sin \left (a x\right ) - b^{2} - c^{2}}\right ) - 2 \, {\left (b^{3} - b c^{2}\right )} \cos \left (a x\right )}{2 \, {\left (a b^{5} - 2 \, a b^{3} c^{2} + a b c^{4} + {\left (a b^{4} c - 2 \, a b^{2} c^{3} + a c^{5}\right )} \sin \left (a x\right )\right )}}, \frac {{\left (c^{2} \sin \left (a x\right ) + b c\right )} \sqrt {b^{2} - c^{2}} \arctan \left (-\frac {b \sin \left (a x\right ) + c}{\sqrt {b^{2} - c^{2}} \cos \left (a x\right )}\right ) - {\left (b^{3} - b c^{2}\right )} \cos \left (a x\right )}{a b^{5} - 2 \, a b^{3} c^{2} + a b c^{4} + {\left (a b^{4} c - 2 \, a b^{2} c^{3} + a c^{5}\right )} \sin \left (a x\right )}\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 = 0.75, size = 98, normalized size = 1.27 \begin {gather*} -\frac {2 \, {\left (\frac {{\left (\pi \left \lfloor \frac {a x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (\frac {1}{2} \, a x\right ) + c}{\sqrt {b^{2} - c^{2}}}\right )\right )} c}{{\left (b^{2} - c^{2}\right )}^{\frac {3}{2}}} + \frac {c \tan \left (\frac {1}{2} \, a x\right ) + b}{{\left (b \tan \left (\frac {1}{2} \, a x\right )^{2} + 2 \, c \tan \left (\frac {1}{2} \, a x\right ) + b\right )} {\left (b^{2} - c^{2}\right )}}\right )}}{a} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 0.49, size = 133, normalized size = 1.73 \begin {gather*} -\frac {\frac {2\,b}{b^2-c^2}+\frac {2\,c\,\mathrm {tan}\left (\frac {a\,x}{2}\right )}{b^2-c^2}}{a\,\left (b\,{\mathrm {tan}\left (\frac {a\,x}{2}\right )}^2+2\,c\,\mathrm {tan}\left (\frac {a\,x}{2}\right )+b\right )}-\frac {2\,c\,\mathrm {atan}\left (\frac {\left (\frac {2\,c^2}{{\left (b+c\right )}^{3/2}\,{\left (b-c\right )}^{3/2}}+\frac {2\,b\,c\,\mathrm {tan}\left (\frac {a\,x}{2}\right )}{{\left (b+c\right )}^{3/2}\,{\left (b-c\right )}^{3/2}}\right )\,\left (b^2-c^2\right )}{2\,c}\right )}{a\,{\left (b+c\right )}^{3/2}\,{\left (b-c\right )}^{3/2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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