Integrand size = 33, antiderivative size = 123 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {C \text {arctanh}(\sin (c+d x))}{a^3 d}-\frac {(A+C) \sec ^2(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(3 A-7 C) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {(6 A-29 C) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )} \]
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Time = 0.40 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {4170, 4093, 4083, 3855, 3879} \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {(6 A-29 C) \tan (c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac {C \text {arctanh}(\sin (c+d x))}{a^3 d}-\frac {(A+C) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}-\frac {(3 A-7 C) \tan (c+d x)}{15 a d (a \sec (c+d x)+a)^2} \]
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Rule 3855
Rule 3879
Rule 4083
Rule 4093
Rule 4170
Rubi steps \begin{align*} \text {integral}& = -\frac {(A+C) \sec ^2(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {\int \frac {\sec ^2(c+d x) (-a (3 A-2 C)-5 a C \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{5 a^2} \\ & = -\frac {(A+C) \sec ^2(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(3 A-7 C) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {\int \frac {\sec (c+d x) \left (2 a^2 (3 A-7 C)+15 a^2 C \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4} \\ & = -\frac {(A+C) \sec ^2(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(3 A-7 C) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {(6 A-29 C) \int \frac {\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{15 a^2}+\frac {C \int \sec (c+d x) \, dx}{a^3} \\ & = \frac {C \text {arctanh}(\sin (c+d x))}{a^3 d}-\frac {(A+C) \sec ^2(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(3 A-7 C) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {(6 A-29 C) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )} \\ \end{align*}
Time = 3.76 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.92 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=-\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \left (A+C \sec ^2(c+d x)\right ) \left (240 C \cos ^5\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+\sec \left (\frac {c}{2}\right ) \left (-5 (3 A-29 C) \sin \left (\frac {d x}{2}\right )+15 (A-5 C) \sin \left (c+\frac {d x}{2}\right )-15 A \sin \left (c+\frac {3 d x}{2}\right )+95 C \sin \left (c+\frac {3 d x}{2}\right )-15 C \sin \left (2 c+\frac {3 d x}{2}\right )-3 A \sin \left (2 c+\frac {5 d x}{2}\right )+22 C \sin \left (2 c+\frac {5 d x}{2}\right )\right )\right )}{15 a^3 d (A+2 C+A \cos (2 (c+d x))) (1+\sec (c+d x))^3} \]
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Time = 0.22 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.71
method | result | size |
parallelrisch | \(\frac {-60 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+60 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\frac {20 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3}-5 A +35 C \right )}{60 a^{3} d}\) | \(87\) |
derivativedivides | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}-\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -4 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+4 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 d \,a^{3}}\) | \(105\) |
default | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}-\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -4 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+4 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 d \,a^{3}}\) | \(105\) |
risch | \(-\frac {2 i \left (15 C \,{\mathrm e}^{4 i \left (d x +c \right )}-15 A \,{\mathrm e}^{3 i \left (d x +c \right )}+75 C \,{\mathrm e}^{3 i \left (d x +c \right )}-15 A \,{\mathrm e}^{2 i \left (d x +c \right )}+145 C \,{\mathrm e}^{2 i \left (d x +c \right )}-15 A \,{\mathrm e}^{i \left (d x +c \right )}+95 C \,{\mathrm e}^{i \left (d x +c \right )}-3 A +22 C \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{a^{3} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{a^{3} d}\) | \(158\) |
norman | \(\frac {\frac {\left (A -9 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{10 a d}-\frac {\left (A -7 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}-\frac {\left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{20 a d}-\frac {\left (7 A -43 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{10 a d}+\frac {\left (9 A -11 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{60 a d}+\frac {\left (-59 C +9 A \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 a d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3} a^{2}}+\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{3} d}-\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{3} d}\) | \(209\) |
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Time = 0.28 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.50 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {15 \, {\left (C \cos \left (d x + c\right )^{3} + 3 \, C \cos \left (d x + c\right )^{2} + 3 \, C \cos \left (d x + c\right ) + C\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (C \cos \left (d x + c\right )^{3} + 3 \, C \cos \left (d x + c\right )^{2} + 3 \, C \cos \left (d x + c\right ) + C\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left ({\left (3 \, A - 22 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (3 \, A - 17 \, C\right )} \cos \left (d x + c\right ) + 3 \, A - 32 \, C\right )} \sin \left (d x + c\right )}{30 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]
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\[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {\int \frac {A \sec ^{2}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{4}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]
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Time = 0.22 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.36 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=-\frac {C {\left (\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} - \frac {3 \, A {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, d} \]
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Time = 0.34 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.07 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {\frac {60 \, C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {60 \, C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 20 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 105 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]
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Time = 15.32 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.89 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {2\,C\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^3\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {A+C}{4\,a^3}-\frac {A-3\,C}{2\,a^3}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A+C}{12\,a^3}-\frac {A-3\,C}{12\,a^3}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A+C\right )}{20\,a^3\,d} \]
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