Optimal. Leaf size=71 \[ \frac {4 \tan (c+d x)}{15 a^2 d}-\frac {2 \sec (c+d x)}{15 d \left (a^2 \sin (c+d x)+a^2\right )}+\frac {\sec (c+d x)}{5 d (a \sin (c+d x)+a)^2} \]
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Rubi [A] time = 0.13, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2859, 2672, 3767, 8} \[ \frac {4 \tan (c+d x)}{15 a^2 d}-\frac {2 \sec (c+d x)}{15 d \left (a^2 \sin (c+d x)+a^2\right )}+\frac {\sec (c+d x)}{5 d (a \sin (c+d x)+a)^2} \]
Antiderivative was successfully verified.
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Rule 8
Rule 2672
Rule 2859
Rule 3767
Rubi steps
\begin {align*} \int \frac {\sec (c+d x) \tan (c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac {\sec (c+d x)}{5 d (a+a \sin (c+d x))^2}+\frac {2 \int \frac {\sec ^2(c+d x)}{a+a \sin (c+d x)} \, dx}{5 a}\\ &=\frac {\sec (c+d x)}{5 d (a+a \sin (c+d x))^2}-\frac {2 \sec (c+d x)}{15 d \left (a^2+a^2 \sin (c+d x)\right )}+\frac {4 \int \sec ^2(c+d x) \, dx}{15 a^2}\\ &=\frac {\sec (c+d x)}{5 d (a+a \sin (c+d x))^2}-\frac {2 \sec (c+d x)}{15 d \left (a^2+a^2 \sin (c+d x)\right )}-\frac {4 \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 a^2 d}\\ &=\frac {\sec (c+d x)}{5 d (a+a \sin (c+d x))^2}-\frac {2 \sec (c+d x)}{15 d \left (a^2+a^2 \sin (c+d x)\right )}+\frac {4 \tan (c+d x)}{15 a^2 d}\\ \end {align*}
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Mathematica [A] time = 0.25, size = 82, normalized size = 1.15 \[ -\frac {\sec (c+d x) (-80 \sin (c+d x)-4 \sin (2 (c+d x))+16 \sin (3 (c+d x))-5 \cos (c+d x)+64 \cos (2 (c+d x))+\cos (3 (c+d x))-80)}{240 a^2 d (\sin (c+d x)+1)^2} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.48, size = 80, normalized size = 1.13 \[ \frac {8 \, \cos \left (d x + c\right )^{2} + 2 \, {\left (2 \, \cos \left (d x + c\right )^{2} - 3\right )} \sin \left (d x + c\right ) - 9}{15 \, {\left (a^{2} d \cos \left (d x + c\right )^{3} - 2 \, a^{2} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, a^{2} d \cos \left (d x + c\right )\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.19, size = 94, normalized size = 1.32 \[ -\frac {\frac {15}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}} - \frac {15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 30 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 50 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 7}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{5}}}{60 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.36, size = 100, normalized size = 1.41 \[ \frac {-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {4}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {2}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {7}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {3}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {4}{16 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+16}}{a^{2} d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.33, size = 204, normalized size = 2.87 \[ \frac {2 \, {\left (\frac {4 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {20 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 1\right )}}{15 \, {\left (a^{2} + \frac {4 \, a^{2} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {5 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {5 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {4 \, a^{2} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}\right )} d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 9.39, size = 159, normalized size = 2.24 \[ \frac {\frac {2\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{15}+\frac {8\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{15}+\frac {8\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+\frac {8\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+2\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{a^2\,d\,\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,{\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}^5} \]
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 \[ \frac {\int \frac {\sin {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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