Optimal. Leaf size=119 \[ -\frac{i b^2 \text{PolyLog}\left (2,-1+\frac{2}{1-i (c+d x)}\right )}{d e^2}-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^2}{d e^2 (c+d x)}-\frac{i \left (a+b \tan ^{-1}(c+d x)\right )^2}{d e^2}+\frac{2 b \log \left (2-\frac{2}{1-i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )}{d e^2} \]
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Rubi [A] time = 0.186854, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {5043, 12, 4852, 4924, 4868, 2447} \[ -\frac{i b^2 \text{PolyLog}\left (2,-1+\frac{2}{1-i (c+d x)}\right )}{d e^2}-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^2}{d e^2 (c+d x)}-\frac{i \left (a+b \tan ^{-1}(c+d x)\right )^2}{d e^2}+\frac{2 b \log \left (2-\frac{2}{1-i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )}{d e^2} \]
Antiderivative was successfully verified.
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Rule 5043
Rule 12
Rule 4852
Rule 4924
Rule 4868
Rule 2447
Rubi steps
\begin{align*} \int \frac{\left (a+b \tan ^{-1}(c+d x)\right )^2}{(c e+d e x)^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b \tan ^{-1}(x)\right )^2}{e^2 x^2} \, dx,x,c+d x\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b \tan ^{-1}(x)\right )^2}{x^2} \, dx,x,c+d x\right )}{d e^2}\\ &=-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^2}{d e^2 (c+d x)}+\frac{(2 b) \operatorname{Subst}\left (\int \frac{a+b \tan ^{-1}(x)}{x \left (1+x^2\right )} \, dx,x,c+d x\right )}{d e^2}\\ &=-\frac{i \left (a+b \tan ^{-1}(c+d x)\right )^2}{d e^2}-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^2}{d e^2 (c+d x)}+\frac{(2 i b) \operatorname{Subst}\left (\int \frac{a+b \tan ^{-1}(x)}{x (i+x)} \, dx,x,c+d x\right )}{d e^2}\\ &=-\frac{i \left (a+b \tan ^{-1}(c+d x)\right )^2}{d e^2}-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^2}{d e^2 (c+d x)}+\frac{2 b \left (a+b \tan ^{-1}(c+d x)\right ) \log \left (2-\frac{2}{1-i (c+d x)}\right )}{d e^2}-\frac{\left (2 b^2\right ) \operatorname{Subst}\left (\int \frac{\log \left (2-\frac{2}{1-i x}\right )}{1+x^2} \, dx,x,c+d x\right )}{d e^2}\\ &=-\frac{i \left (a+b \tan ^{-1}(c+d x)\right )^2}{d e^2}-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^2}{d e^2 (c+d x)}+\frac{2 b \left (a+b \tan ^{-1}(c+d x)\right ) \log \left (2-\frac{2}{1-i (c+d x)}\right )}{d e^2}-\frac{i b^2 \text{Li}_2\left (-1+\frac{2}{1-i (c+d x)}\right )}{d e^2}\\ \end{align*}
Mathematica [A] time = 0.190932, size = 135, normalized size = 1.13 \[ \frac{-i b^2 (c+d x) \text{PolyLog}\left (2,e^{2 i \tan ^{-1}(c+d x)}\right )+a \left (2 b (c+d x) \log \left (\frac{c+d x}{\sqrt{(c+d x)^2+1}}\right )-a\right )+2 b \tan ^{-1}(c+d x) \left (-a+b (c+d x) \log \left (1-e^{2 i \tan ^{-1}(c+d x)}\right )\right )-i b^2 (c+d x-i) \tan ^{-1}(c+d x)^2}{d e^2 (c+d x)} \]
Warning: Unable to verify antiderivative.
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Maple [B] time = 0.125, size = 471, normalized size = 4. \begin{align*} -{\frac{{a}^{2}}{d{e}^{2} \left ( dx+c \right ) }}-{\frac{{b}^{2} \left ( \arctan \left ( dx+c \right ) \right ) ^{2}}{d{e}^{2} \left ( dx+c \right ) }}-{\frac{{b}^{2}\arctan \left ( dx+c \right ) \ln \left ( 1+ \left ( dx+c \right ) ^{2} \right ) }{d{e}^{2}}}+2\,{\frac{{b}^{2}\ln \left ( dx+c \right ) \arctan \left ( dx+c \right ) }{d{e}^{2}}}-{\frac{i{b}^{2}\ln \left ( dx+c \right ) \ln \left ( 1-i \left ( dx+c \right ) \right ) }{d{e}^{2}}}-{\frac{i{b}^{2}{\it dilog} \left ( 1-i \left ( dx+c \right ) \right ) }{d{e}^{2}}}+{\frac{{\frac{i}{2}}{b}^{2}\ln \left ( dx+c-i \right ) \ln \left ( -{\frac{i}{2}} \left ( dx+c+i \right ) \right ) }{d{e}^{2}}}+{\frac{{\frac{i}{2}}{b}^{2}{\it dilog} \left ( -{\frac{i}{2}} \left ( dx+c+i \right ) \right ) }{d{e}^{2}}}+{\frac{{\frac{i}{2}}{b}^{2}\ln \left ( 1+ \left ( dx+c \right ) ^{2} \right ) \ln \left ( dx+c+i \right ) }{d{e}^{2}}}-{\frac{{\frac{i}{2}}{b}^{2}{\it dilog} \left ({\frac{i}{2}} \left ( dx+c-i \right ) \right ) }{d{e}^{2}}}+{\frac{{\frac{i}{4}}{b}^{2} \left ( \ln \left ( dx+c-i \right ) \right ) ^{2}}{d{e}^{2}}}-{\frac{{\frac{i}{2}}{b}^{2}\ln \left ( 1+ \left ( dx+c \right ) ^{2} \right ) \ln \left ( dx+c-i \right ) }{d{e}^{2}}}-{\frac{{\frac{i}{4}}{b}^{2} \left ( \ln \left ( dx+c+i \right ) \right ) ^{2}}{d{e}^{2}}}+{\frac{i{b}^{2}\ln \left ( dx+c \right ) \ln \left ( 1+i \left ( dx+c \right ) \right ) }{d{e}^{2}}}-{\frac{{\frac{i}{2}}{b}^{2}\ln \left ( dx+c+i \right ) \ln \left ({\frac{i}{2}} \left ( dx+c-i \right ) \right ) }{d{e}^{2}}}+{\frac{i{b}^{2}{\it dilog} \left ( 1+i \left ( dx+c \right ) \right ) }{d{e}^{2}}}-2\,{\frac{ab\arctan \left ( dx+c \right ) }{d{e}^{2} \left ( dx+c \right ) }}-{\frac{ab\ln \left ( 1+ \left ( dx+c \right ) ^{2} \right ) }{d{e}^{2}}}+2\,{\frac{ab\ln \left ( dx+c \right ) }{d{e}^{2}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [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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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b^{2} \arctan \left (d x + c\right )^{2} + 2 \, a b \arctan \left (d x + c\right ) + a^{2}}{d^{2} e^{2} x^{2} + 2 \, c d e^{2} x + c^{2} e^{2}}, x\right ) \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*} \frac{\int \frac{a^{2}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx + \int \frac{b^{2} \operatorname{atan}^{2}{\left (c + d x \right )}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx + \int \frac{2 a b \operatorname{atan}{\left (c + d x \right )}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx}{e^{2}} \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{{\left (b \arctan \left (d x + c\right ) + a\right )}^{2}}{{\left (d e x + c e\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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