Integrand size = 25, antiderivative size = 49 \[ \int \frac {A+C \sec ^2(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {A x}{a}+\frac {C \text {arctanh}(\sin (c+d x))}{a d}-\frac {(A+C) \tan (c+d x)}{a d (1+\sec (c+d x))} \]
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Time = 0.13 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {4136, 3855, 4004, 3879} \[ \int \frac {A+C \sec ^2(c+d x)}{a+a \sec (c+d x)} \, dx=-\frac {(A+C) \tan (c+d x)}{a d (\sec (c+d x)+1)}+\frac {A x}{a}+\frac {C \text {arctanh}(\sin (c+d x))}{a d} \]
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Rule 3855
Rule 3879
Rule 4004
Rule 4136
Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {a A-a C \sec (c+d x)}{a+a \sec (c+d x)} \, dx}{a}+\frac {C \int \sec (c+d x) \, dx}{a} \\ & = \frac {A x}{a}+\frac {C \text {arctanh}(\sin (c+d x))}{a d}+(-A-C) \int \frac {\sec (c+d x)}{a+a \sec (c+d x)} \, dx \\ & = \frac {A x}{a}+\frac {C \text {arctanh}(\sin (c+d x))}{a d}-\frac {(A+C) \tan (c+d x)}{d (a+a \sec (c+d x))} \\ \end{align*}
Time = 0.64 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.73 \[ \int \frac {A+C \sec ^2(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {A d x+C \text {arctanh}(\sin (c+d x))-(A+C) \tan \left (\frac {1}{2} (c+d x)\right )}{a d} \]
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Time = 0.14 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.20
method | result | size |
parallelrisch | \(\frac {-C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (-A -C \right )+d x A}{a d}\) | \(59\) |
derivativedivides | \(\frac {-\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+2 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d a}\) | \(75\) |
default | \(\frac {-\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+2 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d a}\) | \(75\) |
risch | \(\frac {A x}{a}-\frac {2 i A}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}-\frac {2 i C}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{a d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{a d}\) | \(97\) |
norman | \(\frac {\frac {A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a}+\frac {\left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}-\frac {A x}{a}-\frac {\left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{a d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}+\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a d}-\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a d}\) | \(125\) |
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Time = 0.28 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.80 \[ \int \frac {A+C \sec ^2(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {2 \, A d x \cos \left (d x + c\right ) + 2 \, A d x + {\left (C \cos \left (d x + c\right ) + C\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (C \cos \left (d x + c\right ) + C\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (A + C\right )} \sin \left (d x + c\right )}{2 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \]
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\[ \int \frac {A+C \sec ^2(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {\int \frac {A}{\sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{2}{\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \]
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Leaf count of result is larger than twice the leaf count of optimal. 125 vs. \(2 (49) = 98\).
Time = 0.32 (sec) , antiderivative size = 125, normalized size of antiderivative = 2.55 \[ \int \frac {A+C \sec ^2(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {A {\left (\frac {2 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} + C {\left (\frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} - \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{d} \]
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Time = 0.30 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.63 \[ \int \frac {A+C \sec ^2(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {\frac {{\left (d x + c\right )} A}{a} + \frac {C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a} - \frac {C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a} - \frac {A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a}}{d} \]
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Time = 15.01 (sec) , antiderivative size = 101, normalized size of antiderivative = 2.06 \[ \int \frac {A+C \sec ^2(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {2\,A\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+2\,C\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{a\,d}-\frac {A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+C\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a\,d\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )} \]
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