Integrand size = 32, antiderivative size = 85 \[ \int \frac {B \tan (c+d x)+C \tan ^2(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {(b B-a C) x}{a^2+b^2}-\frac {(a B+b C) \log (\cos (c+d x))}{\left (a^2+b^2\right ) d}-\frac {a (b B-a C) \log (a+b \tan (c+d x))}{b \left (a^2+b^2\right ) d} \]
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Time = 0.21 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1643, 649, 209, 266} \[ \int \frac {B \tan (c+d x)+C \tan ^2(c+d x)}{a+b \tan (c+d x)} \, dx=-\frac {a (b B-a C) \log (a+b \tan (c+d x))}{b d \left (a^2+b^2\right )}-\frac {(a B+b C) \log (\cos (c+d x))}{d \left (a^2+b^2\right )}+\frac {x (b B-a C)}{a^2+b^2} \]
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Rule 209
Rule 266
Rule 649
Rule 1643
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x (B+C x)}{(a+b x) \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{d} \\ & = \frac {\text {Subst}\left (\int \left (\frac {a (-b B+a C)}{\left (a^2+b^2\right ) (a+b x)}+\frac {b B-a C+(a B+b C) x}{\left (a^2+b^2\right ) \left (1+x^2\right )}\right ) \, dx,x,\tan (c+d x)\right )}{d} \\ & = -\frac {a (b B-a C) \log (a+b \tan (c+d x))}{b \left (a^2+b^2\right ) d}+\frac {\text {Subst}\left (\int \frac {b B-a C+(a B+b C) x}{1+x^2} \, dx,x,\tan (c+d x)\right )}{\left (a^2+b^2\right ) d} \\ & = -\frac {a (b B-a C) \log (a+b \tan (c+d x))}{b \left (a^2+b^2\right ) d}+\frac {(b B-a C) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{\left (a^2+b^2\right ) d}+\frac {(a B+b C) \text {Subst}\left (\int \frac {x}{1+x^2} \, dx,x,\tan (c+d x)\right )}{\left (a^2+b^2\right ) d} \\ & = \frac {(b B-a C) x}{a^2+b^2}-\frac {(a B+b C) \log (\cos (c+d x))}{\left (a^2+b^2\right ) d}-\frac {a (b B-a C) \log (a+b \tan (c+d x))}{b \left (a^2+b^2\right ) d} \\ \end{align*}
Result contains complex when optimal does not.
Time = 0.20 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.15 \[ \int \frac {B \tan (c+d x)+C \tan ^2(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {(a-i b) b (B+i C) \log (i-\tan (c+d x))+(a+i b) b (B-i C) \log (i+\tan (c+d x))+2 a (-b B+a C) \log (a+b \tan (c+d x))}{2 b \left (a^2+b^2\right ) d} \]
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Time = 0.08 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.02
method | result | size |
derivativedivides | \(\frac {\frac {\frac {\left (B a +C b \right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+\left (B b -C a \right ) \arctan \left (\tan \left (d x +c \right )\right )}{a^{2}+b^{2}}-\frac {a \left (B b -C a \right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right ) b}}{d}\) | \(87\) |
default | \(\frac {\frac {\frac {\left (B a +C b \right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+\left (B b -C a \right ) \arctan \left (\tan \left (d x +c \right )\right )}{a^{2}+b^{2}}-\frac {a \left (B b -C a \right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right ) b}}{d}\) | \(87\) |
norman | \(\frac {\left (B b -C a \right ) x}{a^{2}+b^{2}}+\frac {\left (B a +C b \right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d \left (a^{2}+b^{2}\right )}-\frac {a \left (B b -C a \right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{b \left (a^{2}+b^{2}\right ) d}\) | \(90\) |
parallelrisch | \(\frac {2 B \,b^{2} d x -2 C a b d x +B \ln \left (1+\tan \left (d x +c \right )^{2}\right ) a b -2 B \ln \left (a +b \tan \left (d x +c \right )\right ) a b +C \ln \left (1+\tan \left (d x +c \right )^{2}\right ) b^{2}+2 C \ln \left (a +b \tan \left (d x +c \right )\right ) a^{2}}{2 \left (a^{2}+b^{2}\right ) b d}\) | \(98\) |
risch | \(\frac {i x B}{i b -a}+\frac {x C}{i b -a}+\frac {2 i C x}{b}+\frac {2 i C c}{b d}+\frac {2 i a B x}{a^{2}+b^{2}}+\frac {2 i a B c}{\left (a^{2}+b^{2}\right ) d}-\frac {2 i a^{2} C x}{\left (a^{2}+b^{2}\right ) b}-\frac {2 i a^{2} C c}{\left (a^{2}+b^{2}\right ) b d}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) C}{b d}-\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right ) B}{\left (a^{2}+b^{2}\right ) d}+\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right ) C}{\left (a^{2}+b^{2}\right ) b d}\) | \(240\) |
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Time = 0.26 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.29 \[ \int \frac {B \tan (c+d x)+C \tan ^2(c+d x)}{a+b \tan (c+d x)} \, dx=-\frac {2 \, {\left (C a b - B b^{2}\right )} d x - {\left (C a^{2} - B a b\right )} \log \left (\frac {b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) + {\left (C a^{2} + C b^{2}\right )} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right )}{2 \, {\left (a^{2} b + b^{3}\right )} d} \]
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Result contains complex when optimal does not.
Time = 0.56 (sec) , antiderivative size = 711, normalized size of antiderivative = 8.36 \[ \int \frac {B \tan (c+d x)+C \tan ^2(c+d x)}{a+b \tan (c+d x)} \, dx=\begin {cases} \frac {\tilde {\infty } x \left (B \tan {\left (c \right )} + C \tan ^{2}{\left (c \right )}\right )}{\tan {\left (c \right )}} & \text {for}\: a = 0 \wedge b = 0 \wedge d = 0 \\\frac {\frac {B \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - C x + \frac {C \tan {\left (c + d x \right )}}{d}}{a} & \text {for}\: b = 0 \\\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} + \frac {i C d x \tan {\left (c + d x \right )}}{2 b d \tan {\left (c + d x \right )} - 2 i b d} + \frac {C d x}{2 b d \tan {\left (c + d x \right )} - 2 i b d} + \frac {C \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )} \tan {\left (c + d x \right )}}{2 b d \tan {\left (c + d x \right )} - 2 i b d} - \frac {i C \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 b d \tan {\left (c + d x \right )} - 2 i b d} - \frac {i C}{2 b d \tan {\left (c + d x \right )} - 2 i b d} & \text {for}\: a = - i b \\\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} - \frac {i C d x \tan {\left (c + d x \right )}}{2 b d \tan {\left (c + d x \right )} + 2 i b d} + \frac {C d x}{2 b d \tan {\left (c + d x \right )} + 2 i b d} + \frac {C \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )} \tan {\left (c + d x \right )}}{2 b d \tan {\left (c + d x \right )} + 2 i b d} + \frac {i C \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 b d \tan {\left (c + d x \right )} + 2 i b d} + \frac {i C}{2 b d \tan {\left (c + d x \right )} + 2 i b d} & \text {for}\: a = i b \\\frac {x \left (B \tan {\left (c \right )} + C \tan ^{2}{\left (c \right )}\right )}{a + b \tan {\left (c \right )}} & \text {for}\: d = 0 \\- \frac {2 B a b \log {\left (\frac {a}{b} + \tan {\left (c + d x \right )} \right )}}{2 a^{2} b d + 2 b^{3} d} + \frac {B a b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 a^{2} b d + 2 b^{3} d} + \frac {2 B b^{2} d x}{2 a^{2} b d + 2 b^{3} d} + \frac {2 C a^{2} \log {\left (\frac {a}{b} + \tan {\left (c + d x \right )} \right )}}{2 a^{2} b d + 2 b^{3} d} - \frac {2 C a b d x}{2 a^{2} b d + 2 b^{3} d} + \frac {C b^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 a^{2} b d + 2 b^{3} d} & \text {otherwise} \end {cases} \]
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Time = 0.31 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.11 \[ \int \frac {B \tan (c+d x)+C \tan ^2(c+d x)}{a+b \tan (c+d x)} \, dx=-\frac {\frac {2 \, {\left (C a - B b\right )} {\left (d x + c\right )}}{a^{2} + b^{2}} - \frac {2 \, {\left (C a^{2} - B a b\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{2} b + b^{3}} - \frac {{\left (B a + C b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}}}{2 \, d} \]
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Time = 0.50 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.12 \[ \int \frac {B \tan (c+d x)+C \tan ^2(c+d x)}{a+b \tan (c+d x)} \, dx=-\frac {\frac {2 \, {\left (C a - B b\right )} {\left (d x + c\right )}}{a^{2} + b^{2}} - \frac {{\left (B a + C b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} - \frac {2 \, {\left (C a^{2} - B a b\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{2} b + b^{3}}}{2 \, d} \]
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Time = 8.91 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.18 \[ \int \frac {B \tan (c+d x)+C \tan ^2(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (-C+B\,1{}\mathrm {i}\right )}{2\,d\,\left (-b+a\,1{}\mathrm {i}\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B-C\,1{}\mathrm {i}\right )}{2\,d\,\left (a-b\,1{}\mathrm {i}\right )}-\frac {a\,\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (B\,b-C\,a\right )}{b\,d\,\left (a^2+b^2\right )} \]
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