Integrand size = 19, antiderivative size = 97 \[ \int \frac {\sec (c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {\text {arctanh}(\sin (c+d x))}{a^3 d}-\frac {\sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {7 \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {22 \sin (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )} \]
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Time = 0.22 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {2845, 3057, 12, 3855} \[ \int \frac {\sec (c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {\text {arctanh}(\sin (c+d x))}{a^3 d}-\frac {22 \sin (c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}-\frac {7 \sin (c+d x)}{15 a d (a \cos (c+d x)+a)^2}-\frac {\sin (c+d x)}{5 d (a \cos (c+d x)+a)^3} \]
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Rule 12
Rule 2845
Rule 3057
Rule 3855
Rubi steps \begin{align*} \text {integral}& = -\frac {\sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {\int \frac {(5 a-2 a \cos (c+d x)) \sec (c+d x)}{(a+a \cos (c+d x))^2} \, dx}{5 a^2} \\ & = -\frac {\sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {7 \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {\int \frac {\left (15 a^2-7 a^2 \cos (c+d x)\right ) \sec (c+d x)}{a+a \cos (c+d x)} \, dx}{15 a^4} \\ & = -\frac {\sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {7 \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {22 \sin (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {\int 15 a^3 \sec (c+d x) \, dx}{15 a^6} \\ & = -\frac {\sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {7 \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {22 \sin (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {\int \sec (c+d x) \, dx}{a^3} \\ & = \frac {\text {arctanh}(\sin (c+d x))}{a^3 d}-\frac {\sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {7 \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {22 \sin (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(201\) vs. \(2(97)=194\).
Time = 0.55 (sec) , antiderivative size = 201, normalized size of antiderivative = 2.07 \[ \int \frac {\sec (c+d x)}{(a+a \cos (c+d x))^3} \, dx=-\frac {2 \cos \left (\frac {1}{2} (c+d x)\right ) \left (60 \cos ^5\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+3 \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+14 \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+88 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+3 \cos \left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {c}{2}\right )+14 \cos ^3\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {c}{2}\right )\right )}{15 a^3 d (1+\cos (c+d x))^3} \]
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Time = 0.89 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.77
method | result | size |
derivativedivides | \(\frac {-\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {4 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 d \,a^{3}}\) | \(75\) |
default | \(\frac {-\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {4 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 d \,a^{3}}\) | \(75\) |
parallelrisch | \(\frac {-3 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-20 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-60 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+60 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-105 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{60 a^{3} d}\) | \(75\) |
norman | \(\frac {-\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d a}-\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{3 d a}-\frac {\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )}{20 d a}}{a^{2}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{3} d}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{3} d}\) | \(101\) |
risch | \(-\frac {2 i \left (15 \,{\mathrm e}^{4 i \left (d x +c \right )}+75 \,{\mathrm e}^{3 i \left (d x +c \right )}+145 \,{\mathrm e}^{2 i \left (d x +c \right )}+95 \,{\mathrm e}^{i \left (d x +c \right )}+22\right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{3} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{3}}\) | \(111\) |
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Time = 0.25 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.63 \[ \int \frac {\sec (c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {15 \, {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (22 \, \cos \left (d x + c\right )^{2} + 51 \, \cos \left (d x + c\right ) + 32\right )} \sin \left (d x + c\right )}{30 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]
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\[ \int \frac {\sec (c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {\int \frac {\sec {\left (c + d x \right )}}{\cos ^{3}{\left (c + d x \right )} + 3 \cos ^{2}{\left (c + d x \right )} + 3 \cos {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]
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Time = 0.23 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.23 \[ \int \frac {\sec (c+d x)}{(a+a \cos (c+d x))^3} \, dx=-\frac {\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}}{60 \, d} \]
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Time = 0.33 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.97 \[ \int \frac {\sec (c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {\frac {60 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {60 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} - \frac {3 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 20 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]
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Time = 14.18 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.60 \[ \int \frac {\sec (c+d x)}{(a+a \cos (c+d x))^3} \, dx=-\frac {105\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-120\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )+20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{60\,a^3\,d} \]
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