Optimal. Leaf size=281 \[ -\frac{i \sqrt{(a+b x)^2+1} \text{PolyLog}\left (2,-\frac{i \sqrt{1+i (a+b x)}}{\sqrt{1-i (a+b x)}}\right )}{2 b \sqrt{c (a+b x)^2+c}}+\frac{i \sqrt{(a+b x)^2+1} \text{PolyLog}\left (2,\frac{i \sqrt{1+i (a+b x)}}{\sqrt{1-i (a+b x)}}\right )}{2 b \sqrt{c (a+b x)^2+c}}-\frac{\sqrt{c (a+b x)^2+c}}{2 b c}+\frac{i \sqrt{(a+b x)^2+1} \tan ^{-1}\left (\frac{\sqrt{1+i (a+b x)}}{\sqrt{1-i (a+b x)}}\right ) \tan ^{-1}(a+b x)}{b \sqrt{c (a+b x)^2+c}}+\frac{(a+b x) \sqrt{c (a+b x)^2+c} \tan ^{-1}(a+b x)}{2 b c} \]
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Rubi [A] time = 0.331005, antiderivative size = 281, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {5057, 4952, 261, 4890, 4886} \[ -\frac{i \sqrt{(a+b x)^2+1} \text{PolyLog}\left (2,-\frac{i \sqrt{1+i (a+b x)}}{\sqrt{1-i (a+b x)}}\right )}{2 b \sqrt{c (a+b x)^2+c}}+\frac{i \sqrt{(a+b x)^2+1} \text{PolyLog}\left (2,\frac{i \sqrt{1+i (a+b x)}}{\sqrt{1-i (a+b x)}}\right )}{2 b \sqrt{c (a+b x)^2+c}}-\frac{\sqrt{c (a+b x)^2+c}}{2 b c}+\frac{i \sqrt{(a+b x)^2+1} \tan ^{-1}\left (\frac{\sqrt{1+i (a+b x)}}{\sqrt{1-i (a+b x)}}\right ) \tan ^{-1}(a+b x)}{b \sqrt{c (a+b x)^2+c}}+\frac{(a+b x) \sqrt{c (a+b x)^2+c} \tan ^{-1}(a+b x)}{2 b c} \]
Antiderivative was successfully verified.
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Rule 5057
Rule 4952
Rule 261
Rule 4890
Rule 4886
Rubi steps
\begin{align*} \int \frac{(a+b x)^2 \tan ^{-1}(a+b x)}{\sqrt{\left (1+a^2\right ) c+2 a b c x+b^2 c x^2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^2 \tan ^{-1}(x)}{\sqrt{c+c x^2}} \, dx,x,a+b x\right )}{b}\\ &=\frac{(a+b x) \sqrt{c+c (a+b x)^2} \tan ^{-1}(a+b x)}{2 b c}-\frac{\operatorname{Subst}\left (\int \frac{x}{\sqrt{c+c x^2}} \, dx,x,a+b x\right )}{2 b}-\frac{\operatorname{Subst}\left (\int \frac{\tan ^{-1}(x)}{\sqrt{c+c x^2}} \, dx,x,a+b x\right )}{2 b}\\ &=-\frac{\sqrt{c+c (a+b x)^2}}{2 b c}+\frac{(a+b x) \sqrt{c+c (a+b x)^2} \tan ^{-1}(a+b x)}{2 b c}-\frac{\sqrt{1+(a+b x)^2} \operatorname{Subst}\left (\int \frac{\tan ^{-1}(x)}{\sqrt{1+x^2}} \, dx,x,a+b x\right )}{2 b \sqrt{c+c (a+b x)^2}}\\ &=-\frac{\sqrt{c+c (a+b x)^2}}{2 b c}+\frac{(a+b x) \sqrt{c+c (a+b x)^2} \tan ^{-1}(a+b x)}{2 b c}+\frac{i \sqrt{1+(a+b x)^2} \tan ^{-1}(a+b x) \tan ^{-1}\left (\frac{\sqrt{1+i (a+b x)}}{\sqrt{1-i (a+b x)}}\right )}{b \sqrt{c+c (a+b x)^2}}-\frac{i \sqrt{1+(a+b x)^2} \text{Li}_2\left (-\frac{i \sqrt{1+i (a+b x)}}{\sqrt{1-i (a+b x)}}\right )}{2 b \sqrt{c+c (a+b x)^2}}+\frac{i \sqrt{1+(a+b x)^2} \text{Li}_2\left (\frac{i \sqrt{1+i (a+b x)}}{\sqrt{1-i (a+b x)}}\right )}{2 b \sqrt{c+c (a+b x)^2}}\\ \end{align*}
Mathematica [A] time = 0.133035, size = 189, normalized size = 0.67 \[ \frac{\sqrt{a^2+2 a b x+b^2 x^2+1} \left (-i \text{PolyLog}\left (2,-i e^{i \tan ^{-1}(a+b x)}\right )+i \text{PolyLog}\left (2,i e^{i \tan ^{-1}(a+b x)}\right )-\sqrt{(a+b x)^2+1}+(a+b x) \sqrt{(a+b x)^2+1} \tan ^{-1}(a+b x)+\tan ^{-1}(a+b x) \left (-\log \left (1-i e^{i \tan ^{-1}(a+b x)}\right )\right )+\tan ^{-1}(a+b x) \log \left (1+i e^{i \tan ^{-1}(a+b x)}\right )\right )}{2 b \sqrt{c \left (a^2+2 a b x+b^2 x^2+1\right )}} \]
Warning: Unable to verify antiderivative.
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Maple [A] time = 1.125, size = 222, normalized size = 0.8 \begin{align*}{\frac{\arctan \left ( bx+a \right ) xb+\arctan \left ( bx+a \right ) a-1}{2\,bc}\sqrt{c \left ( -i+a+bx \right ) \left ( i+a+bx \right ) }}-{\frac{{\frac{i}{2}}}{bc} \left ( i\arctan \left ( bx+a \right ) \ln \left ( 1+{i \left ( 1+i \left ( bx+a \right ) \right ){\frac{1}{\sqrt{1+ \left ( bx+a \right ) ^{2}}}}} \right ) -i\arctan \left ( bx+a \right ) \ln \left ( 1-{i \left ( 1+i \left ( bx+a \right ) \right ){\frac{1}{\sqrt{1+ \left ( bx+a \right ) ^{2}}}}} \right ) +{\it dilog} \left ( 1+{i \left ( 1+i \left ( bx+a \right ) \right ){\frac{1}{\sqrt{1+ \left ( bx+a \right ) ^{2}}}}} \right ) -{\it dilog} \left ( 1-{i \left ( 1+i \left ( bx+a \right ) \right ){\frac{1}{\sqrt{1+ \left ( bx+a \right ) ^{2}}}}} \right ) \right ) \sqrt{c \left ( -i+a+bx \right ) \left ( i+a+bx \right ) }{\frac{1}{\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \arctan \left (b x + a\right )}{\sqrt{b^{2} c x^{2} + 2 \, a b c x +{\left (a^{2} + 1\right )} c}}, 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{{\left (b x + a\right )}^{2} \arctan \left (b x + a\right )}{\sqrt{b^{2} c x^{2} + 2 \, a b c x +{\left (a^{2} + 1\right )} c}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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