Optimal. Leaf size=112 \[ \frac {24 \tan (c+d x)}{5 a^3 d}-\frac {3 \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac {3 \tan (c+d x)}{d \left (a^3 \cos (c+d x)+a^3\right )}-\frac {3 \tan (c+d x)}{5 a d (a \cos (c+d x)+a)^2}-\frac {\tan (c+d x)}{5 d (a \cos (c+d x)+a)^3} \]
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Rubi [A] time = 0.28, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2766, 2978, 2748, 3767, 8, 3770} \[ \frac {24 \tan (c+d x)}{5 a^3 d}-\frac {3 \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac {3 \tan (c+d x)}{d \left (a^3 \cos (c+d x)+a^3\right )}-\frac {3 \tan (c+d x)}{5 a d (a \cos (c+d x)+a)^2}-\frac {\tan (c+d x)}{5 d (a \cos (c+d x)+a)^3} \]
Antiderivative was successfully verified.
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Rule 8
Rule 2748
Rule 2766
Rule 2978
Rule 3767
Rule 3770
Rubi steps
\begin {align*} \int \frac {\sec ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx &=-\frac {\tan (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {\int \frac {(6 a-3 a \cos (c+d x)) \sec ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac {\tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {3 \tan (c+d x)}{5 a d (a+a \cos (c+d x))^2}+\frac {\int \frac {\left (27 a^2-18 a^2 \cos (c+d x)\right ) \sec ^2(c+d x)}{a+a \cos (c+d x)} \, dx}{15 a^4}\\ &=-\frac {\tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {3 \tan (c+d x)}{5 a d (a+a \cos (c+d x))^2}-\frac {3 \tan (c+d x)}{d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {\int \left (72 a^3-45 a^3 \cos (c+d x)\right ) \sec ^2(c+d x) \, dx}{15 a^6}\\ &=-\frac {\tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {3 \tan (c+d x)}{5 a d (a+a \cos (c+d x))^2}-\frac {3 \tan (c+d x)}{d \left (a^3+a^3 \cos (c+d x)\right )}-\frac {3 \int \sec (c+d x) \, dx}{a^3}+\frac {24 \int \sec ^2(c+d x) \, dx}{5 a^3}\\ &=-\frac {3 \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac {\tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {3 \tan (c+d x)}{5 a d (a+a \cos (c+d x))^2}-\frac {3 \tan (c+d x)}{d \left (a^3+a^3 \cos (c+d x)\right )}-\frac {24 \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{5 a^3 d}\\ &=-\frac {3 \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac {24 \tan (c+d x)}{5 a^3 d}-\frac {\tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {3 \tan (c+d x)}{5 a d (a+a \cos (c+d x))^2}-\frac {3 \tan (c+d x)}{d \left (a^3+a^3 \cos (c+d x)\right )}\\ \end {align*}
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Mathematica [B] time = 1.15, size = 286, normalized size = 2.55 \[ \frac {2 \cos \left (\frac {1}{2} (c+d x)\right ) \left (8 \tan \left (\frac {c}{2}\right ) \cos ^3\left (\frac {1}{2} (c+d x)\right )+\tan \left (\frac {c}{2}\right ) \cos \left (\frac {1}{2} (c+d x)\right )+\sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+20 \cos ^5\left (\frac {1}{2} (c+d x)\right ) \left (\frac {\sin (d x)}{\left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}+3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-3 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )+76 \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right )+8 \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right )\right )}{5 a^3 d (\cos (c+d x)+1)^3} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.89, size = 190, normalized size = 1.70 \[ -\frac {15 \, {\left (\cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (\cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (24 \, \cos \left (d x + c\right )^{3} + 57 \, \cos \left (d x + c\right )^{2} + 39 \, \cos \left (d x + c\right ) + 5\right )} \sin \left (d x + c\right )}{10 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right )\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.52, size = 122, normalized size = 1.09 \[ -\frac {\frac {60 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {60 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac {40 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{3}} - \frac {a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 10 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 85 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{20 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.08, size = 139, normalized size = 1.24 \[ \frac {\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )}{20 d \,a^{3}}+\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{3}}+\frac {17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}-\frac {1}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,a^{3}}-\frac {1}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 1.48, size = 165, normalized size = 1.47 \[ \frac {\frac {40 \, \sin \left (d x + c\right )}{{\left (a^{3} - \frac {a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}}{20 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.40, size = 111, normalized size = 0.99 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2\,a^3\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{20\,a^3\,d}-\frac {6\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^3\,d}-\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^3\right )}+\frac {17\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4\,a^3\,d} \]
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 \[ \frac {\int \frac {\sec ^{2}{\left (c + d x \right )}}{\cos ^{3}{\left (c + d x \right )} + 3 \cos ^{2}{\left (c + d x \right )} + 3 \cos {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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