Optimal. Leaf size=125 \[ -\frac{i b d \text{PolyLog}\left (2,-\frac{\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{2 f^2 \left (a^2+b^2\right )}+\frac{b (c+d x) \log \left (1+\frac{\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{f \left (a^2+b^2\right )}+\frac{(c+d x)^2}{2 d (a+i b)} \]
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Rubi [A] time = 0.159057, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {3732, 2190, 2279, 2391} \[ \frac{b (c+d x) \log \left (1+\frac{\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{f \left (a^2+b^2\right )}-\frac{i b d \text{Li}_2\left (-\frac{\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{2 f^2 \left (a^2+b^2\right )}+\frac{(c+d x)^2}{2 d (a+i b)} \]
Antiderivative was successfully verified.
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Rule 3732
Rule 2190
Rule 2279
Rule 2391
Rubi steps
\begin{align*} \int \frac{c+d x}{a+b \tan (e+f x)} \, dx &=\frac{(c+d x)^2}{2 (a+i b) d}+(2 i b) \int \frac{e^{2 i (e+f x)} (c+d x)}{(a+i b)^2+\left (a^2+b^2\right ) e^{2 i (e+f x)}} \, dx\\ &=\frac{(c+d x)^2}{2 (a+i b) d}+\frac{b (c+d x) \log \left (1+\frac{\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) f}-\frac{(b d) \int \log \left (1+\frac{\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right ) \, dx}{\left (a^2+b^2\right ) f}\\ &=\frac{(c+d x)^2}{2 (a+i b) d}+\frac{b (c+d x) \log \left (1+\frac{\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) f}+\frac{(i b d) \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{\left (a^2+b^2\right ) x}{(a+i b)^2}\right )}{x} \, dx,x,e^{2 i (e+f x)}\right )}{2 \left (a^2+b^2\right ) f^2}\\ &=\frac{(c+d x)^2}{2 (a+i b) d}+\frac{b (c+d x) \log \left (1+\frac{\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) f}-\frac{i b d \text{Li}_2\left (-\frac{\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) f^2}\\ \end{align*}
Mathematica [A] time = 1.46755, size = 177, normalized size = 1.42 \[ \frac{i b d \text{PolyLog}\left (2,\frac{(-a-i b) e^{-2 i (e+f x)}}{a-i b}\right )}{2 f^2 \left (a^2+b^2\right )}+\frac{b (c+d x) \log \left (1+\frac{(a+i b) e^{-2 i (e+f x)}}{a-i b}\right )}{f \left (a^2+b^2\right )}+\frac{b (c+d x)^2}{d (a-i b) \left (-i a \left (1+e^{2 i e}\right )+b \left (-e^{2 i e}\right )+b\right )}+\frac{x \cos (e) (2 c+d x)}{2 (a \cos (e)+b \sin (e))} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.204, size = 462, normalized size = 3.7 \begin{align*} -{\frac{d{x}^{2}}{2\,ib-2\,a}}-{\frac{cx}{ib-a}}+{\frac{bc\ln \left ( i{{\rm e}^{2\,i \left ( fx+e \right ) }}b-a{{\rm e}^{2\,i \left ( fx+e \right ) }}-ib-a \right ) }{ \left ( a-ib \right ) f \left ( a+ib \right ) }}-2\,{\frac{bc\ln \left ({{\rm e}^{i \left ( fx+e \right ) }} \right ) }{ \left ( a-ib \right ) f \left ( a+ib \right ) }}-{\frac{bdx}{ \left ( a-ib \right ) f \left ( -ib-a \right ) }\ln \left ( 1-{\frac{ \left ( a-ib \right ){{\rm e}^{2\,i \left ( fx+e \right ) }}}{-ib-a}} \right ) }-{\frac{bde}{ \left ( a-ib \right ){f}^{2} \left ( -ib-a \right ) }\ln \left ( 1-{\frac{ \left ( a-ib \right ){{\rm e}^{2\,i \left ( fx+e \right ) }}}{-ib-a}} \right ) }+{\frac{ibd{x}^{2}}{ \left ( a-ib \right ) \left ( -ib-a \right ) }}+{\frac{2\,ibdex}{ \left ( a-ib \right ) f \left ( -ib-a \right ) }}+{\frac{ibd{e}^{2}}{ \left ( a-ib \right ){f}^{2} \left ( -ib-a \right ) }}+{\frac{{\frac{i}{2}}bd}{ \left ( a-ib \right ){f}^{2} \left ( -ib-a \right ) }{\it polylog} \left ( 2,{\frac{ \left ( a-ib \right ){{\rm e}^{2\,i \left ( fx+e \right ) }}}{-ib-a}} \right ) }-{\frac{bde\ln \left ( i{{\rm e}^{2\,i \left ( fx+e \right ) }}b-a{{\rm e}^{2\,i \left ( fx+e \right ) }}-ib-a \right ) }{ \left ( a-ib \right ){f}^{2} \left ( a+ib \right ) }}+2\,{\frac{bde\ln \left ({{\rm e}^{i \left ( fx+e \right ) }} \right ) }{ \left ( a-ib \right ){f}^{2} \left ( a+ib \right ) }} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 2.07047, size = 539, normalized size = 4.31 \begin{align*} \frac{{\left (a - i \, b\right )} d f^{2} x^{2} + 2 \,{\left (a - i \, b\right )} c f^{2} x - 2 i \, b d f x \arctan \left (\frac{2 \, a b \cos \left (2 \, f x + 2 \, e\right ) -{\left (a^{2} - b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )}{a^{2} + b^{2}}, \frac{2 \, a b \sin \left (2 \, f x + 2 \, e\right ) + a^{2} + b^{2} +{\left (a^{2} - b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right )}{a^{2} + b^{2}}\right ) + b d f x \log \left (\frac{{\left (a^{2} + b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right )^{2} + 4 \, a b \sin \left (2 \, f x + 2 \, e\right ) +{\left (a^{2} + b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )^{2} + a^{2} + b^{2} + 2 \,{\left (a^{2} - b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right )}{a^{2} + b^{2}}\right ) + 2 i \, b c f \arctan \left (-b \cos \left (2 \, f x + 2 \, e\right ) + a \sin \left (2 \, f x + 2 \, e\right ) + b, a \cos \left (2 \, f x + 2 \, e\right ) + b \sin \left (2 \, f x + 2 \, e\right ) + a\right ) + b c f \log \left ({\left (a^{2} + b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right )^{2} + 4 \, a b \sin \left (2 \, f x + 2 \, e\right ) +{\left (a^{2} + b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )^{2} + a^{2} + b^{2} + 2 \,{\left (a^{2} - b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right )\right ) - i \, b d{\rm Li}_2\left (\frac{{\left (i \, a + b\right )} e^{\left (2 i \, f x + 2 i \, e\right )}}{-i \, a + b}\right )}{2 \,{\left (a^{2} + b^{2}\right )} f^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 1.87714, size = 1273, normalized size = 10.18 \begin{align*} \frac{2 \, a d f^{2} x^{2} + 4 \, a c f^{2} x - i \, b d{\rm Li}_2\left (-\frac{{\left (2 i \, a b + 2 \, b^{2}\right )} \tan \left (f x + e\right )^{2} + 2 \, a^{2} - 2 i \, a b +{\left (2 i \, a^{2} + 4 \, a b - 2 i \, b^{2}\right )} \tan \left (f x + e\right )}{{\left (a^{2} + b^{2}\right )} \tan \left (f x + e\right )^{2} + a^{2} + b^{2}} + 1\right ) + i \, b d{\rm Li}_2\left (-\frac{{\left (-2 i \, a b + 2 \, b^{2}\right )} \tan \left (f x + e\right )^{2} + 2 \, a^{2} + 2 i \, a b +{\left (-2 i \, a^{2} + 4 \, a b + 2 i \, b^{2}\right )} \tan \left (f x + e\right )}{{\left (a^{2} + b^{2}\right )} \tan \left (f x + e\right )^{2} + a^{2} + b^{2}} + 1\right ) + 2 \,{\left (b d f x + b d e\right )} \log \left (\frac{{\left (2 i \, a b + 2 \, b^{2}\right )} \tan \left (f x + e\right )^{2} + 2 \, a^{2} - 2 i \, a b +{\left (2 i \, a^{2} + 4 \, a b - 2 i \, b^{2}\right )} \tan \left (f x + e\right )}{{\left (a^{2} + b^{2}\right )} \tan \left (f x + e\right )^{2} + a^{2} + b^{2}}\right ) + 2 \,{\left (b d f x + b d e\right )} \log \left (\frac{{\left (-2 i \, a b + 2 \, b^{2}\right )} \tan \left (f x + e\right )^{2} + 2 \, a^{2} + 2 i \, a b +{\left (-2 i \, a^{2} + 4 \, a b + 2 i \, b^{2}\right )} \tan \left (f x + e\right )}{{\left (a^{2} + b^{2}\right )} \tan \left (f x + e\right )^{2} + a^{2} + b^{2}}\right ) - 2 \,{\left (b d e - b c f\right )} \log \left (\frac{{\left (i \, a b + b^{2}\right )} \tan \left (f x + e\right )^{2} - a^{2} + i \, a b +{\left (i \, a^{2} + i \, b^{2}\right )} \tan \left (f x + e\right )}{\tan \left (f x + e\right )^{2} + 1}\right ) - 2 \,{\left (b d e - b c f\right )} \log \left (\frac{{\left (i \, a b - b^{2}\right )} \tan \left (f x + e\right )^{2} + a^{2} + i \, a b +{\left (i \, a^{2} + i \, b^{2}\right )} \tan \left (f x + e\right )}{\tan \left (f x + e\right )^{2} + 1}\right )}{4 \,{\left (a^{2} + b^{2}\right )} f^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{c + d x}{a + b \tan{\left (e + f x \right )}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{d x + c}{b \tan \left (f x + e\right ) + a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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