Optimal. Leaf size=244 \[ \frac{i d \text{PolyLog}\left (2,\frac{c (-a-b x+i)}{-a c+b d+i c}\right )}{2 c^2}-\frac{i d \text{PolyLog}\left (2,\frac{c (a+b x+i)}{-b d+(a+i) c}\right )}{2 c^2}+\frac{i d \log (i a+i b x+1) \log \left (\frac{b (c x+d)}{b d+(-a+i) c}\right )}{2 c^2}-\frac{i d \log (-i a-i b x+1) \log \left (-\frac{b (c x+d)}{-b d+(a+i) c}\right )}{2 c^2}-\frac{(i a+i b x+1) \log (i a+i b x+1)}{2 b c}-\frac{(-i a-i b x+1) \log (-i (a+b x+i))}{2 b c} \]
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Rubi [A] time = 0.239131, antiderivative size = 244, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 7, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.438, Rules used = {5051, 2409, 2389, 2295, 2394, 2393, 2391} \[ \frac{i d \text{PolyLog}\left (2,\frac{c (-a-b x+i)}{-a c+b d+i c}\right )}{2 c^2}-\frac{i d \text{PolyLog}\left (2,\frac{c (a+b x+i)}{-b d+(a+i) c}\right )}{2 c^2}+\frac{i d \log (i a+i b x+1) \log \left (\frac{b (c x+d)}{b d+(-a+i) c}\right )}{2 c^2}-\frac{i d \log (-i a-i b x+1) \log \left (-\frac{b (c x+d)}{-b d+(a+i) c}\right )}{2 c^2}-\frac{(i a+i b x+1) \log (i a+i b x+1)}{2 b c}-\frac{(-i a-i b x+1) \log (-i (a+b x+i))}{2 b c} \]
Antiderivative was successfully verified.
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Rule 5051
Rule 2409
Rule 2389
Rule 2295
Rule 2394
Rule 2393
Rule 2391
Rubi steps
\begin{align*} \int \frac{\tan ^{-1}(a+b x)}{c+\frac{d}{x}} \, dx &=\frac{1}{2} i \int \frac{\log (1-i a-i b x)}{c+\frac{d}{x}} \, dx-\frac{1}{2} i \int \frac{\log (1+i a+i b x)}{c+\frac{d}{x}} \, dx\\ &=\frac{1}{2} i \int \left (\frac{\log (1-i a-i b x)}{c}-\frac{d \log (1-i a-i b x)}{c (d+c x)}\right ) \, dx-\frac{1}{2} i \int \left (\frac{\log (1+i a+i b x)}{c}-\frac{d \log (1+i a+i b x)}{c (d+c x)}\right ) \, dx\\ &=\frac{i \int \log (1-i a-i b x) \, dx}{2 c}-\frac{i \int \log (1+i a+i b x) \, dx}{2 c}-\frac{(i d) \int \frac{\log (1-i a-i b x)}{d+c x} \, dx}{2 c}+\frac{(i d) \int \frac{\log (1+i a+i b x)}{d+c x} \, dx}{2 c}\\ &=-\frac{i d \log (1-i a-i b x) \log \left (-\frac{b (d+c x)}{(i+a) c-b d}\right )}{2 c^2}+\frac{i d \log (1+i a+i b x) \log \left (\frac{b (d+c x)}{(i-a) c+b d}\right )}{2 c^2}-\frac{\operatorname{Subst}(\int \log (x) \, dx,x,1-i a-i b x)}{2 b c}-\frac{\operatorname{Subst}(\int \log (x) \, dx,x,1+i a+i b x)}{2 b c}+\frac{(b d) \int \frac{\log \left (-\frac{i b (d+c x)}{-(1-i a) c-i b d}\right )}{1-i a-i b x} \, dx}{2 c^2}+\frac{(b d) \int \frac{\log \left (\frac{i b (d+c x)}{-(1+i a) c+i b d}\right )}{1+i a+i b x} \, dx}{2 c^2}\\ &=-\frac{(1+i a+i b x) \log (1+i a+i b x)}{2 b c}-\frac{(1-i a-i b x) \log (-i (i+a+b x))}{2 b c}-\frac{i d \log (1-i a-i b x) \log \left (-\frac{b (d+c x)}{(i+a) c-b d}\right )}{2 c^2}+\frac{i d \log (1+i a+i b x) \log \left (\frac{b (d+c x)}{(i-a) c+b d}\right )}{2 c^2}+\frac{(i d) \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{c x}{-(1-i a) c-i b d}\right )}{x} \, dx,x,1-i a-i b x\right )}{2 c^2}-\frac{(i d) \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{c x}{-(1+i a) c+i b d}\right )}{x} \, dx,x,1+i a+i b x\right )}{2 c^2}\\ &=-\frac{(1+i a+i b x) \log (1+i a+i b x)}{2 b c}-\frac{(1-i a-i b x) \log (-i (i+a+b x))}{2 b c}-\frac{i d \log (1-i a-i b x) \log \left (-\frac{b (d+c x)}{(i+a) c-b d}\right )}{2 c^2}+\frac{i d \log (1+i a+i b x) \log \left (\frac{b (d+c x)}{(i-a) c+b d}\right )}{2 c^2}+\frac{i d \text{Li}_2\left (\frac{c (i-a-b x)}{(i-a) c+b d}\right )}{2 c^2}-\frac{i d \text{Li}_2\left (\frac{c (i+a+b x)}{(i+a) c-b d}\right )}{2 c^2}\\ \end{align*}
Mathematica [B] time = 11.0746, size = 771, normalized size = 3.16 \[ \frac{i b d (b d-a c) \text{PolyLog}\left (2,\exp \left (2 i \left (\tan ^{-1}(a+b x)-\tan ^{-1}\left (a-\frac{b d}{c}\right )\right )\right )\right )+i b d (a c-b d) \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(a+b x)}\right )+b c d \sqrt{a^2-\frac{2 a b d}{c}+\frac{b^2 d^2}{c^2}+1} \tan ^{-1}(a+b x)^2 e^{-i \tan ^{-1}\left (a-\frac{b d}{c}\right )}-2 a^2 c^2 \tan ^{-1}(a+b x)+2 b^2 d^2 \tan ^{-1}\left (a-\frac{b d}{c}\right ) \log \left (1-\exp \left (2 i \left (\tan ^{-1}(a+b x)-\tan ^{-1}\left (a-\frac{b d}{c}\right )\right )\right )\right )-2 b^2 d^2 \tan ^{-1}(a+b x) \log \left (1-\exp \left (2 i \left (\tan ^{-1}(a+b x)-\tan ^{-1}\left (a-\frac{b d}{c}\right )\right )\right )\right )-2 i b^2 d^2 \tan ^{-1}(a+b x) \tan ^{-1}\left (a-\frac{b d}{c}\right )-2 b^2 d^2 \tan ^{-1}\left (a-\frac{b d}{c}\right ) \log \left (-\sin \left (\tan ^{-1}\left (a-\frac{b d}{c}\right )-\tan ^{-1}(a+b x)\right )\right )+2 b^2 c d x \tan ^{-1}(a+b x)+\pi b^2 d^2 \log \left (\frac{1}{\sqrt{(a+b x)^2+1}}\right )-i b^2 d^2 \tan ^{-1}(a+b x)^2-i \pi b^2 d^2 \tan ^{-1}(a+b x)-\pi b^2 d^2 \log \left (1+e^{-2 i \tan ^{-1}(a+b x)}\right )+2 b^2 d^2 \tan ^{-1}(a+b x) \log \left (1+e^{2 i \tan ^{-1}(a+b x)}\right )-2 a c^2 \log \left (\frac{1}{\sqrt{(a+b x)^2+1}}\right )-2 a b c^2 x \tan ^{-1}(a+b x)-2 a b c d \tan ^{-1}\left (a-\frac{b d}{c}\right ) \log \left (1-\exp \left (2 i \left (\tan ^{-1}(a+b x)-\tan ^{-1}\left (a-\frac{b d}{c}\right )\right )\right )\right )+2 a b c d \tan ^{-1}(a+b x) \log \left (1-\exp \left (2 i \left (\tan ^{-1}(a+b x)-\tan ^{-1}\left (a-\frac{b d}{c}\right )\right )\right )\right )+2 b c d \log \left (\frac{1}{\sqrt{(a+b x)^2+1}}\right )-\pi a b c d \log \left (\frac{1}{\sqrt{(a+b x)^2+1}}\right )+i a b c d \tan ^{-1}(a+b x)^2-b c d \tan ^{-1}(a+b x)^2+2 a b c d \tan ^{-1}(a+b x)+2 i a b c d \tan ^{-1}(a+b x) \tan ^{-1}\left (a-\frac{b d}{c}\right )+i \pi a b c d \tan ^{-1}(a+b x)+\pi a b c d \log \left (1+e^{-2 i \tan ^{-1}(a+b x)}\right )-2 a b c d \tan ^{-1}(a+b x) \log \left (1+e^{2 i \tan ^{-1}(a+b x)}\right )+2 a b c d \tan ^{-1}\left (a-\frac{b d}{c}\right ) \log \left (-\sin \left (\tan ^{-1}\left (a-\frac{b d}{c}\right )-\tan ^{-1}(a+b x)\right )\right )}{b c^2 (2 b d-2 a c)} \]
Warning: Unable to verify antiderivative.
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Maple [A] time = 0.06, size = 317, normalized size = 1.3 \begin{align*}{\frac{x\arctan \left ( bx+a \right ) }{c}}+{\frac{\arctan \left ( bx+a \right ) a}{bc}}-{\frac{\arctan \left ( bx+a \right ) d\ln \left ( c \left ( bx+a \right ) -ac+bd \right ) }{{c}^{2}}}-{\frac{\ln \left ({a}^{2}{c}^{2}-2\,abcd+{b}^{2}{d}^{2}+2\, \left ( c \left ( bx+a \right ) -ac+bd \right ) ac-2\, \left ( c \left ( bx+a \right ) -ac+bd \right ) bd+ \left ( c \left ( bx+a \right ) -ac+bd \right ) ^{2}+{c}^{2} \right ) }{2\,bc}}-{\frac{{\frac{i}{2}}d\ln \left ( c \left ( bx+a \right ) -ac+bd \right ) }{{c}^{2}}\ln \left ({\frac{ic-c \left ( bx+a \right ) }{ic-ac+bd}} \right ) }+{\frac{{\frac{i}{2}}d\ln \left ( c \left ( bx+a \right ) -ac+bd \right ) }{{c}^{2}}\ln \left ({\frac{ic+c \left ( bx+a \right ) }{ic+ac-bd}} \right ) }-{\frac{{\frac{i}{2}}d}{{c}^{2}}{\it dilog} \left ({\frac{ic-c \left ( bx+a \right ) }{ic-ac+bd}} \right ) }+{\frac{{\frac{i}{2}}d}{{c}^{2}}{\it dilog} \left ({\frac{ic+c \left ( bx+a \right ) }{ic+ac-bd}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 2.01758, size = 383, normalized size = 1.57 \begin{align*} -\frac{b d \arctan \left (b x + a\right ) \log \left (-\frac{b^{2} c^{2} x^{2} + 2 \, b^{2} c d x + b^{2} d^{2}}{2 \, a b c d - b^{2} d^{2} -{\left (a^{2} + 1\right )} c^{2}}\right ) + i \, b d{\rm Li}_2\left (-\frac{i \, b c x +{\left (i \, a - 1\right )} c}{{\left (-i \, a + 1\right )} c + i \, b d}\right ) - i \, b d{\rm Li}_2\left (-\frac{i \, b c x +{\left (i \, a + 1\right )} c}{{\left (-i \, a - 1\right )} c + i \, b d}\right ) - 2 \,{\left (b c x + a c\right )} \arctan \left (b x + a\right ) -{\left (b d \arctan \left (-\frac{b c^{2} x + b c d}{2 \, a b c d - b^{2} d^{2} -{\left (a^{2} + 1\right )} c^{2}}, \frac{a b c d - b^{2} d^{2} +{\left (a b c^{2} - b^{2} c d\right )} x}{2 \, a b c d - b^{2} d^{2} -{\left (a^{2} + 1\right )} c^{2}}\right ) - c\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{2 \, b c^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x \arctan \left (b x + a\right )}{c x + d}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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{\arctan \left (b x + a\right )}{c + \frac{d}{x}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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