Optimal. Leaf size=106 \[ -\frac {2 \left (2 a^2-b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^3 d \sqrt {a^2-b^2}}+\frac {2 a x}{b^3}+\frac {\cos (c+d x) (2 a+b \sin (c+d x))}{b^2 d (a+b \sin (c+d x))} \]
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Rubi [A] time = 0.15, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2863, 2735, 2660, 618, 204} \[ -\frac {2 \left (2 a^2-b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^3 d \sqrt {a^2-b^2}}+\frac {\cos (c+d x) (2 a+b \sin (c+d x))}{b^2 d (a+b \sin (c+d x))}+\frac {2 a x}{b^3} \]
Antiderivative was successfully verified.
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Rule 204
Rule 618
Rule 2660
Rule 2735
Rule 2863
Rubi steps
\begin {align*} \int \frac {\cos ^2(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx &=\frac {\cos (c+d x) (2 a+b \sin (c+d x))}{b^2 d (a+b \sin (c+d x))}-\frac {\int \frac {-b-2 a \sin (c+d x)}{a+b \sin (c+d x)} \, dx}{b^2}\\ &=\frac {2 a x}{b^3}+\frac {\cos (c+d x) (2 a+b \sin (c+d x))}{b^2 d (a+b \sin (c+d x))}-\frac {\left (2 a^2-b^2\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{b^3}\\ &=\frac {2 a x}{b^3}+\frac {\cos (c+d x) (2 a+b \sin (c+d x))}{b^2 d (a+b \sin (c+d x))}-\frac {\left (2 \left (2 a^2-b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{b^3 d}\\ &=\frac {2 a x}{b^3}+\frac {\cos (c+d x) (2 a+b \sin (c+d x))}{b^2 d (a+b \sin (c+d x))}+\frac {\left (4 \left (2 a^2-b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{b^3 d}\\ &=\frac {2 a x}{b^3}-\frac {2 \left (2 a^2-b^2\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^3 \sqrt {a^2-b^2} d}+\frac {\cos (c+d x) (2 a+b \sin (c+d x))}{b^2 d (a+b \sin (c+d x))}\\ \end {align*}
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Mathematica [A] time = 1.07, size = 130, normalized size = 1.23 \[ \frac {\frac {4 a^2 c+4 a^2 d x+4 a b (c+d x) \sin (c+d x)+4 a b \cos (c+d x)+b^2 \sin (2 (c+d x))}{a+b \sin (c+d x)}-\frac {4 \left (2 a^2-b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}}{2 b^3 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.78, size = 479, normalized size = 4.52 \[ \left [\frac {4 \, {\left (a^{4} - a^{2} b^{2}\right )} d x + {\left (2 \, a^{3} - a b^{2} + {\left (2 \, a^{2} b - b^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) + 4 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right ) + 2 \, {\left (2 \, {\left (a^{3} b - a b^{3}\right )} d x + {\left (a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{2} b^{4} - b^{6}\right )} d \sin \left (d x + c\right ) + {\left (a^{3} b^{3} - a b^{5}\right )} d\right )}}, \frac {2 \, {\left (a^{4} - a^{2} b^{2}\right )} d x + {\left (2 \, a^{3} - a b^{2} + {\left (2 \, a^{2} b - b^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) + 2 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right ) + {\left (2 \, {\left (a^{3} b - a b^{3}\right )} d x + {\left (a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{{\left (a^{2} b^{4} - b^{6}\right )} d \sin \left (d x + c\right ) + {\left (a^{3} b^{3} - a b^{5}\right )} d}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.19, size = 191, normalized size = 1.80 \[ \frac {2 \, {\left (\frac {{\left (d x + c\right )} a}{b^{3}} - \frac {{\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} {\left (2 \, a^{2} - b^{2}\right )}}{\sqrt {a^{2} - b^{2}} b^{3}} + \frac {b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, a}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )} b^{2}}\right )}}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.45, size = 229, normalized size = 2.16 \[ \frac {2}{d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {4 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{d \,b^{3}}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d b \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )}+\frac {2 a}{d \,b^{2} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )}-\frac {4 a^{2} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \,b^{3} \sqrt {a^{2}-b^{2}}}+\frac {2 \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d b \sqrt {a^{2}-b^{2}}} \]
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 = 10.02, size = 269, normalized size = 2.54 \[ \frac {\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{b}+\frac {4\,a}{b^2}+\frac {6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{b}+\frac {4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{b^2}}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}+\frac {2\,a\,x}{b^3}-\frac {\ln \left (b+a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\sqrt {b^2-a^2}\right )\,\sqrt {-\left (a+b\right )\,\left (a-b\right )}\,\left (2\,a^2-b^2\right )}{d\,\left (b^5-a^2\,b^3\right )}-\frac {\ln \left (b+a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sqrt {b^2-a^2}\right )\,\left (2\,a^2\,\sqrt {b^2-a^2}-b^2\,\sqrt {b^2-a^2}\right )}{b^3\,d\,\left (a^2-b^2\right )} \]
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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