Integrand size = 23, antiderivative size = 58 \[ \int \frac {A+B \tan (c+d x)}{a+b \tan (c+d x)} \, dx=\frac {(a A+b B) x}{a^2+b^2}+\frac {(A b-a B) \log (a \cos (c+d x)+b \sin (c+d x))}{\left (a^2+b^2\right ) d} \]
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Time = 0.09 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {3612, 3611} \[ \int \frac {A+B \tan (c+d x)}{a+b \tan (c+d x)} \, dx=\frac {(A b-a B) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )}+\frac {x (a A+b B)}{a^2+b^2} \]
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Rule 3611
Rule 3612
Rubi steps \begin{align*} \text {integral}& = \frac {(a A+b B) x}{a^2+b^2}+\frac {(A b-a B) \int \frac {b-a \tan (c+d x)}{a+b \tan (c+d x)} \, dx}{a^2+b^2} \\ & = \frac {(a A+b B) x}{a^2+b^2}+\frac {(A b-a B) \log (a \cos (c+d x)+b \sin (c+d x))}{\left (a^2+b^2\right ) d} \\ \end{align*}
Time = 0.11 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.14 \[ \int \frac {A+B \tan (c+d x)}{a+b \tan (c+d x)} \, dx=\frac {2 (a A+b B) \arctan (\tan (c+d x))-(A b-a B) \left (\log \left (\sec ^2(c+d x)\right )-2 \log (a+b \tan (c+d x))\right )}{2 \left (a^2+b^2\right ) d} \]
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Time = 0.05 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.41
method | result | size |
derivativedivides | \(\frac {\frac {\frac {\left (-A b +B a \right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+\left (a A +B b \right ) \arctan \left (\tan \left (d x +c \right )\right )}{a^{2}+b^{2}}+\frac {\left (A b -B a \right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{a^{2}+b^{2}}}{d}\) | \(82\) |
default | \(\frac {\frac {\frac {\left (-A b +B a \right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+\left (a A +B b \right ) \arctan \left (\tan \left (d x +c \right )\right )}{a^{2}+b^{2}}+\frac {\left (A b -B a \right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{a^{2}+b^{2}}}{d}\) | \(82\) |
norman | \(\frac {\left (a A +B b \right ) x}{a^{2}+b^{2}}+\frac {\left (A b -B a \right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{d \left (a^{2}+b^{2}\right )}-\frac {\left (A b -B a \right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \left (a^{2}+b^{2}\right )}\) | \(85\) |
parallelrisch | \(-\frac {-2 A x a d -2 B b d x +A \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) b -2 A \ln \left (a +b \tan \left (d x +c \right )\right ) b -B \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a +2 B \ln \left (a +b \tan \left (d x +c \right )\right ) a}{2 d \left (a^{2}+b^{2}\right )}\) | \(87\) |
risch | \(\frac {i x B}{i b -a}-\frac {x A}{i b -a}-\frac {2 i A b x}{a^{2}+b^{2}}+\frac {2 i B x a}{a^{2}+b^{2}}-\frac {2 i A b c}{d \left (a^{2}+b^{2}\right )}+\frac {2 i B a c}{d \left (a^{2}+b^{2}\right )}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right ) A b}{d \left (a^{2}+b^{2}\right )}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right ) B a}{d \left (a^{2}+b^{2}\right )}\) | \(186\) |
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Time = 0.28 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.31 \[ \int \frac {A+B \tan (c+d x)}{a+b \tan (c+d x)} \, dx=\frac {2 \, {\left (A a + B b\right )} d x - {\left (B a - A b\right )} \log \left (\frac {b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2} + 1}\right )}{2 \, {\left (a^{2} + b^{2}\right )} d} \]
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Result contains complex when optimal does not.
Time = 0.47 (sec) , antiderivative size = 524, normalized size of antiderivative = 9.03 \[ \int \frac {A+B \tan (c+d x)}{a+b \tan (c+d x)} \, dx=\begin {cases} \frac {\tilde {\infty } x \left (A + B \tan {\left (c \right )}\right )}{\tan {\left (c \right )}} & \text {for}\: a = 0 \wedge b = 0 \wedge d = 0 \\\frac {A x + \frac {B \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d}}{a} & \text {for}\: b = 0 \\\frac {i A d x \tan {\left (c + d x \right )}}{2 b d \tan {\left (c + d x \right )} - 2 i b d} + \frac {A d x}{2 b d \tan {\left (c + d x \right )} - 2 i b d} + \frac {i A}{2 b d \tan {\left (c + d x \right )} - 2 i b d} + \frac {B d x \tan {\left (c + d x \right )}}{2 b d \tan {\left (c + d x \right )} - 2 i b d} - \frac {i B d x}{2 b d \tan {\left (c + d x \right )} - 2 i b d} - \frac {B}{2 b d \tan {\left (c + d x \right )} - 2 i b d} & \text {for}\: a = - i b \\- \frac {i A d x \tan {\left (c + d x \right )}}{2 b d \tan {\left (c + d x \right )} + 2 i b d} + \frac {A d x}{2 b d \tan {\left (c + d x \right )} + 2 i b d} - \frac {i A}{2 b d \tan {\left (c + d x \right )} + 2 i b d} + \frac {B d x \tan {\left (c + d x \right )}}{2 b d \tan {\left (c + d x \right )} + 2 i b d} + \frac {i B d x}{2 b d \tan {\left (c + d x \right )} + 2 i b d} - \frac {B}{2 b d \tan {\left (c + d x \right )} + 2 i b d} & \text {for}\: a = i b \\\frac {x \left (A + B \tan {\left (c \right )}\right )}{a + b \tan {\left (c \right )}} & \text {for}\: d = 0 \\\frac {2 A a d x}{2 a^{2} d + 2 b^{2} d} + \frac {2 A b \log {\left (\frac {a}{b} + \tan {\left (c + d x \right )} \right )}}{2 a^{2} d + 2 b^{2} d} - \frac {A b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 a^{2} d + 2 b^{2} d} - \frac {2 B a \log {\left (\frac {a}{b} + \tan {\left (c + d x \right )} \right )}}{2 a^{2} d + 2 b^{2} d} + \frac {B a \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 a^{2} d + 2 b^{2} d} + \frac {2 B b d x}{2 a^{2} d + 2 b^{2} d} & \text {otherwise} \end {cases} \]
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Time = 0.27 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.52 \[ \int \frac {A+B \tan (c+d x)}{a+b \tan (c+d x)} \, dx=\frac {\frac {2 \, {\left (A a + B b\right )} {\left (d x + c\right )}}{a^{2} + b^{2}} - \frac {2 \, {\left (B a - A b\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{2} + b^{2}} + \frac {{\left (B a - A b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}}}{2 \, d} \]
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Time = 0.35 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.62 \[ \int \frac {A+B \tan (c+d x)}{a+b \tan (c+d x)} \, dx=\frac {\frac {2 \, {\left (A a + B b\right )} {\left (d x + c\right )}}{a^{2} + b^{2}} + \frac {{\left (B a - A b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} - \frac {2 \, {\left (B a b - A b^{2}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{2} b + b^{3}}}{2 \, d} \]
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Time = 7.80 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.60 \[ \int \frac {A+B \tan (c+d x)}{a+b \tan (c+d x)} \, dx=\frac {\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (A\,b-B\,a\right )}{d\,\left (a^2+b^2\right )}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )}{2\,d\,\left (b+a\,1{}\mathrm {i}\right )}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (-B+A\,1{}\mathrm {i}\right )}{2\,d\,\left (a+b\,1{}\mathrm {i}\right )} \]
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