Optimal. Leaf size=130 \[ \frac {2 \tan ^5(c+d x)}{5 a^2 d}+\frac {5 \tan ^3(c+d x)}{3 a^2 d}+\frac {4 \tan (c+d x)}{a^2 d}-\frac {\cot (c+d x)}{a^2 d}-\frac {2 \sec ^5(c+d x)}{5 a^2 d}-\frac {2 \sec ^3(c+d x)}{3 a^2 d}-\frac {2 \sec (c+d x)}{a^2 d}+\frac {2 \tanh ^{-1}(\cos (c+d x))}{a^2 d} \]
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Rubi [A] time = 0.31, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {2875, 2873, 3767, 2622, 302, 207, 2620, 270} \[ \frac {2 \tan ^5(c+d x)}{5 a^2 d}+\frac {5 \tan ^3(c+d x)}{3 a^2 d}+\frac {4 \tan (c+d x)}{a^2 d}-\frac {\cot (c+d x)}{a^2 d}-\frac {2 \sec ^5(c+d x)}{5 a^2 d}-\frac {2 \sec ^3(c+d x)}{3 a^2 d}-\frac {2 \sec (c+d x)}{a^2 d}+\frac {2 \tanh ^{-1}(\cos (c+d x))}{a^2 d} \]
Antiderivative was successfully verified.
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Rule 207
Rule 270
Rule 302
Rule 2620
Rule 2622
Rule 2873
Rule 2875
Rule 3767
Rubi steps
\begin {align*} \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac {\int \csc ^2(c+d x) \sec ^6(c+d x) (a-a \sin (c+d x))^2 \, dx}{a^4}\\ &=\frac {\int \left (a^2 \sec ^6(c+d x)-2 a^2 \csc (c+d x) \sec ^6(c+d x)+a^2 \csc ^2(c+d x) \sec ^6(c+d x)\right ) \, dx}{a^4}\\ &=\frac {\int \sec ^6(c+d x) \, dx}{a^2}+\frac {\int \csc ^2(c+d x) \sec ^6(c+d x) \, dx}{a^2}-\frac {2 \int \csc (c+d x) \sec ^6(c+d x) \, dx}{a^2}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\left (1+x^2\right )^3}{x^2} \, dx,x,\tan (c+d x)\right )}{a^2 d}-\frac {\operatorname {Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,-\tan (c+d x)\right )}{a^2 d}-\frac {2 \operatorname {Subst}\left (\int \frac {x^6}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{a^2 d}\\ &=\frac {\tan (c+d x)}{a^2 d}+\frac {2 \tan ^3(c+d x)}{3 a^2 d}+\frac {\tan ^5(c+d x)}{5 a^2 d}+\frac {\operatorname {Subst}\left (\int \left (3+\frac {1}{x^2}+3 x^2+x^4\right ) \, dx,x,\tan (c+d x)\right )}{a^2 d}-\frac {2 \operatorname {Subst}\left (\int \left (1+x^2+x^4+\frac {1}{-1+x^2}\right ) \, dx,x,\sec (c+d x)\right )}{a^2 d}\\ &=-\frac {\cot (c+d x)}{a^2 d}-\frac {2 \sec (c+d x)}{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 {4 \tan (c+d x)}{a^2 d}+\frac {5 \tan ^3(c+d x)}{3 a^2 d}+\frac {2 \tan ^5(c+d x)}{5 a^2 d}-\frac {2 \operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{a^2 d}\\ &=\frac {2 \tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac {\cot (c+d x)}{a^2 d}-\frac {2 \sec (c+d x)}{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 {4 \tan (c+d x)}{a^2 d}+\frac {5 \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 [B] time = 0.83, size = 289, normalized size = 2.22 \[ -\frac {\csc ^3(c+d x) \left (58 \sin (c+d x)-168 \sin (2 (c+d x))+82 \sin (3 (c+d x))+28 \sin (4 (c+d x))+48 \cos (2 (c+d x))+112 \cos (3 (c+d x))-28 \cos (4 (c+d x))+90 \sin (2 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-15 \sin (4 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+60 \cos (3 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-4 \cos (c+d x) \left (-15 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+15 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+28\right )-60 \cos (3 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-90 \sin (2 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+15 \sin (4 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+40\right )}{15 a^2 d (\sin (c+d x)+1)^2 \left (\csc ^2\left (\frac {1}{2} (c+d x)\right )-\sec ^2\left (\frac {1}{2} (c+d x)\right )\right )} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.46, size = 218, normalized size = 1.68 \[ -\frac {56 \, \cos \left (d x + c\right )^{4} - 80 \, \cos \left (d x + c\right )^{2} - 15 \, {\left (2 \, \cos \left (d x + c\right )^{3} + {\left (\cos \left (d x + c\right )^{3} - 2 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 15 \, {\left (2 \, \cos \left (d x + c\right )^{3} + {\left (\cos \left (d x + c\right )^{3} - 2 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 2 \, {\left (41 \, \cos \left (d x + c\right )^{2} - 3\right )} \sin \left (d x + c\right ) + 9}{15 \, {\left (2 \, a^{2} d \cos \left (d x + c\right )^{3} - 2 \, a^{2} d \cos \left (d x + c\right ) + {\left (a^{2} d \cos \left (d x + c\right )^{3} - 2 \, a^{2} d \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.23, size = 161, normalized size = 1.24 \[ -\frac {\frac {120 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} - \frac {30 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{2}} - \frac {15 \, {\left (4 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 7 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} a^{2}} + \frac {465 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 1590 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2240 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1450 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 383}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{5}}}{60 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.56, size = 182, normalized size = 1.40 \[ \frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}-\frac {1}{4 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {1}{2 d \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}}-\frac {4}{5 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {2}{a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {13}{3 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {9}{2 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {31}{4 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.34, size = 309, normalized size = 2.38 \[ -\frac {\frac {\frac {244 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {571 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {320 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {475 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {660 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {255 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + 15}{\frac {a^{2} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {4 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {5 \, a^{2} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {5 \, a^{2} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {4 \, a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {a^{2} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} - \frac {15 \, \sin \left (d x + c\right )}{a^{2} {\left (\cos \left (d x + c\right ) + 1\right )}}}{30 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 10.78, size = 216, normalized size = 1.66 \[ \frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a^2\,d}-\frac {2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^2\,d}-\frac {-17\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-44\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-\frac {95\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}+\frac {64\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+\frac {571\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{15}+\frac {244\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{15}+1}{d\,\left (-2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-8\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+10\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+8\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )} \]
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 \[ \frac {\int \frac {\csc ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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