Optimal. Leaf size=73 \[ \frac {2 \tan ^5(c+d x)}{5 a^2 d}+\frac {\tan ^3(c+d x)}{3 a^2 d}-\frac {2 \sec ^5(c+d x)}{5 a^2 d}+\frac {2 \sec ^3(c+d x)}{3 a^2 d} \]
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Rubi [A] time = 0.19, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2711, 2607, 14, 2606, 30} \[ \frac {2 \tan ^5(c+d x)}{5 a^2 d}+\frac {\tan ^3(c+d x)}{3 a^2 d}-\frac {2 \sec ^5(c+d x)}{5 a^2 d}+\frac {2 \sec ^3(c+d x)}{3 a^2 d} \]
Antiderivative was successfully verified.
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Rule 14
Rule 30
Rule 2606
Rule 2607
Rule 2711
Rubi steps
\begin {align*} \int \frac {\tan ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac {\int \left (a^2 \sec ^4(c+d x) \tan ^2(c+d x)-2 a^2 \sec ^3(c+d x) \tan ^3(c+d x)+a^2 \sec ^2(c+d x) \tan ^4(c+d x)\right ) \, dx}{a^4}\\ &=\frac {\int \sec ^4(c+d x) \tan ^2(c+d x) \, dx}{a^2}+\frac {\int \sec ^2(c+d x) \tan ^4(c+d x) \, dx}{a^2}-\frac {2 \int \sec ^3(c+d x) \tan ^3(c+d x) \, dx}{a^2}\\ &=\frac {\operatorname {Subst}\left (\int x^4 \, dx,x,\tan (c+d x)\right )}{a^2 d}+\frac {\operatorname {Subst}\left (\int x^2 \left (1+x^2\right ) \, dx,x,\tan (c+d x)\right )}{a^2 d}-\frac {2 \operatorname {Subst}\left (\int x^2 \left (-1+x^2\right ) \, dx,x,\sec (c+d x)\right )}{a^2 d}\\ &=\frac {\tan ^5(c+d x)}{5 a^2 d}+\frac {\operatorname {Subst}\left (\int \left (x^2+x^4\right ) \, dx,x,\tan (c+d x)\right )}{a^2 d}-\frac {2 \operatorname {Subst}\left (\int \left (-x^2+x^4\right ) \, dx,x,\sec (c+d x)\right )}{a^2 d}\\ &=\frac {2 \sec ^3(c+d x)}{3 a^2 d}-\frac {2 \sec ^5(c+d x)}{5 a^2 d}+\frac {\tan ^3(c+d x)}{3 a^2 d}+\frac {2 \tan ^5(c+d x)}{5 a^2 d}\\ \end {align*}
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Mathematica [A] time = 0.26, size = 86, normalized size = 1.18 \[ -\frac {\sec (c+d x) \left (-35 \sin (c+d x)+11 \sin (2 (c+d x))+\sin (3 (c+d x))+\frac {55}{4} \cos (c+d x)+4 \cos (2 (c+d x))-\frac {11}{4} \cos (3 (c+d x))-20\right )}{60 a^2 d (\sin (c+d x)+1)^2} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.43, size = 77, normalized size = 1.05 \[ \frac {2 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 9\right )} \sin \left (d x + c\right ) - 6}{15 \, {\left (a^{2} d \cos \left (d x + c\right )^{3} - 2 \, a^{2} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, a^{2} d \cos \left (d x + c\right )\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.21, size = 94, normalized size = 1.29 \[ -\frac {\frac {15}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}} - \frac {15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 90 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 80 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 70 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 17}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{5}}}{60 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.41, size = 100, normalized size = 1.37 \[ \frac {-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {4}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {2}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {5}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {8}{32 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+32}}{a^{2} d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.33, size = 184, normalized size = 2.52 \[ \frac {8 \, {\left (\frac {4 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {5 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + 1\right )}}{15 \, {\left (a^{2} + \frac {4 \, a^{2} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {5 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {5 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {4 \, a^{2} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}\right )} d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 9.31, size = 132, normalized size = 1.81 \[ \frac {8\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left ({\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+4\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+5\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\right )}{15\,a^2\,d\,\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,{\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}^5} \]
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 \[ \frac {\int \frac {\sin ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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