Integrand size = 12, antiderivative size = 56 \[ \int \frac {1}{(3+5 \sec (c+d x))^2} \, dx=\frac {29 x}{576}+\frac {35 \arctan \left (\frac {\sin (c+d x)}{3+\cos (c+d x)}\right )}{288 d}-\frac {25 \tan (c+d x)}{48 d (3+5 \sec (c+d x))} \]
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Time = 0.10 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3870, 4004, 3916, 2736} \[ \int \frac {1}{(3+5 \sec (c+d x))^2} \, dx=\frac {35 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)+3}\right )}{288 d}-\frac {25 \tan (c+d x)}{48 d (5 \sec (c+d x)+3)}+\frac {29 x}{576} \]
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Rule 2736
Rule 3870
Rule 3916
Rule 4004
Rubi steps \begin{align*} \text {integral}& = -\frac {25 \tan (c+d x)}{48 d (3+5 \sec (c+d x))}+\frac {1}{48} \int \frac {16+15 \sec (c+d x)}{3+5 \sec (c+d x)} \, dx \\ & = \frac {x}{9}-\frac {25 \tan (c+d x)}{48 d (3+5 \sec (c+d x))}-\frac {35}{144} \int \frac {\sec (c+d x)}{3+5 \sec (c+d x)} \, dx \\ & = \frac {x}{9}-\frac {25 \tan (c+d x)}{48 d (3+5 \sec (c+d x))}-\frac {7}{144} \int \frac {1}{1+\frac {3}{5} \cos (c+d x)} \, dx \\ & = \frac {29 x}{576}+\frac {35 \arctan \left (\frac {\sin (c+d x)}{3+\cos (c+d x)}\right )}{288 d}-\frac {25 \tan (c+d x)}{48 d (3+5 \sec (c+d x))} \\ \end{align*}
Time = 0.27 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.30 \[ \int \frac {1}{(3+5 \sec (c+d x))^2} \, dx=\frac {160 (c+d x)+96 (c+d x) \cos (c+d x)+35 \arctan \left (2 \cot \left (\frac {1}{2} (c+d x)\right )\right ) (5+3 \cos (c+d x))-150 \sin (c+d x)}{288 d (5+3 \cos (c+d x))} \]
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Time = 0.42 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.04
method | result | size |
derivativedivides | \(\frac {-\frac {25 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{48 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+4\right )}-\frac {35 \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}\right )}{288}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{9}}{d}\) | \(58\) |
default | \(\frac {-\frac {25 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{48 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+4\right )}-\frac {35 \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}\right )}{288}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{9}}{d}\) | \(58\) |
risch | \(\frac {x}{9}-\frac {25 i \left (5 \,{\mathrm e}^{i \left (d x +c \right )}+3\right )}{72 d \left (3 \,{\mathrm e}^{2 i \left (d x +c \right )}+10 \,{\mathrm e}^{i \left (d x +c \right )}+3\right )}+\frac {35 i \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {1}{3}\right )}{576 d}-\frac {35 i \ln \left ({\mathrm e}^{i \left (d x +c \right )}+3\right )}{576 d}\) | \(86\) |
parallelrisch | \(\frac {\left (105 i \cos \left (d x +c \right )+175 i\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 i\right )+\left (-105 i \cos \left (d x +c \right )-175 i\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 i\right )+192 d x \cos \left (d x +c \right )+320 d x -300 \sin \left (d x +c \right )}{1728 d \cos \left (d x +c \right )+2880 d}\) | \(92\) |
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Time = 0.28 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.30 \[ \int \frac {1}{(3+5 \sec (c+d x))^2} \, dx=\frac {192 \, d x \cos \left (d x + c\right ) + 320 \, d x + 35 \, {\left (3 \, \cos \left (d x + c\right ) + 5\right )} \arctan \left (\frac {5 \, \cos \left (d x + c\right ) + 3}{4 \, \sin \left (d x + c\right )}\right ) - 300 \, \sin \left (d x + c\right )}{576 \, {\left (3 \, d \cos \left (d x + c\right ) + 5 \, d\right )}} \]
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\[ \int \frac {1}{(3+5 \sec (c+d x))^2} \, dx=\int \frac {1}{\left (5 \sec {\left (c + d x \right )} + 3\right )^{2}}\, dx \]
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Time = 0.28 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.57 \[ \int \frac {1}{(3+5 \sec (c+d x))^2} \, dx=-\frac {\frac {150 \, \sin \left (d x + c\right )}{{\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 4\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} - 64 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right ) + 35 \, \arctan \left (\frac {\sin \left (d x + c\right )}{2 \, {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}{288 \, d} \]
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Time = 0.28 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.05 \[ \int \frac {1}{(3+5 \sec (c+d x))^2} \, dx=\frac {29 \, d x + 29 \, c - \frac {300 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 4} + 70 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 3}\right )}{576 \, d} \]
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Time = 14.25 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.93 \[ \int \frac {1}{(3+5 \sec (c+d x))^2} \, dx=\frac {x}{9}-\frac {\frac {35\,\mathrm {atan}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}\right )}{288}+\frac {25\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{48\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+4\right )}}{d} \]
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