Optimal. Leaf size=281 \[ -\frac {\sqrt {c+c (a+b x)^2}}{2 b c}+\frac {(a+b x) \sqrt {c+c (a+b x)^2} \text {ArcTan}(a+b x)}{2 b c}+\frac {i \sqrt {1+(a+b x)^2} \text {ArcTan}(a+b x) \text {ArcTan}\left (\frac {\sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{b \sqrt {c+c (a+b x)^2}}-\frac {i \sqrt {1+(a+b x)^2} \text {PolyLog}\left (2,-\frac {i \sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{2 b \sqrt {c+c (a+b x)^2}}+\frac {i \sqrt {1+(a+b x)^2} \text {PolyLog}\left (2,\frac {i \sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{2 b \sqrt {c+c (a+b x)^2}} \]
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Rubi [A]
time = 0.24, antiderivative size = 281, normalized size of antiderivative = 1.00, 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 = {5165, 5072,
267, 5010, 5006} \begin {gather*} \frac {i \sqrt {(a+b x)^2+1} \text {ArcTan}\left (\frac {\sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right ) \text {ArcTan}(a+b x)}{b \sqrt {c (a+b x)^2+c}}+\frac {(a+b x) \text {ArcTan}(a+b x) \sqrt {c (a+b x)^2+c}}{2 b c}-\frac {i \sqrt {(a+b x)^2+1} \text {Li}_2\left (-\frac {i \sqrt {i (a+b x)+1}}{\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 {Li}_2\left (\frac {i \sqrt {i (a+b x)+1}}{\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} \end {gather*}
Antiderivative was successfully verified.
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Rule 267
Rule 5006
Rule 5010
Rule 5072
Rule 5165
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 {\text {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 {\text {Subst}\left (\int \frac {x}{\sqrt {c+c x^2}} \, dx,x,a+b x\right )}{2 b}-\frac {\text {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} \text {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*}
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Mathematica [A]
time = 0.10, size = 189, normalized size = 0.67 \begin {gather*} \frac {\sqrt {1+a^2+2 a b x+b^2 x^2} \left (-\sqrt {1+(a+b x)^2}+(a+b x) \sqrt {1+(a+b x)^2} \text {ArcTan}(a+b x)-\text {ArcTan}(a+b x) \log \left (1-i e^{i \text {ArcTan}(a+b x)}\right )+\text {ArcTan}(a+b x) \log \left (1+i e^{i \text {ArcTan}(a+b x)}\right )-i \text {PolyLog}\left (2,-i e^{i \text {ArcTan}(a+b x)}\right )+i \text {PolyLog}\left (2,i e^{i \text {ArcTan}(a+b x)}\right )\right )}{2 b \sqrt {c \left (1+a^2+2 a b x+b^2 x^2\right )}} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.48, size = 222, normalized size = 0.79
method | result | size |
default | \(\frac {\left (\arctan \left (b x +a \right ) b x +\arctan \left (b x +a \right ) a -1\right ) \sqrt {c \left (b x +a -i\right ) \left (b x +a +i\right )}}{2 b c}-\frac {i \left (i \arctan \left (b x +a \right ) \ln \left (1+\frac {i \left (1+i \left (b x +a \right )\right )}{\sqrt {1+\left (b x +a \right )^{2}}}\right )-i \arctan \left (b x +a \right ) \ln \left (1-\frac {i \left (1+i \left (b x +a \right )\right )}{\sqrt {1+\left (b x +a \right )^{2}}}\right )+\dilog \left (1+\frac {i \left (1+i \left (b x +a \right )\right )}{\sqrt {1+\left (b x +a \right )^{2}}}\right )-\dilog \left (1-\frac {i \left (1+i \left (b x +a \right )\right )}{\sqrt {1+\left (b x +a \right )^{2}}}\right )\right ) \sqrt {c \left (b x +a -i\right ) \left (b x +a +i\right )}}{2 \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, b c}\) | \(222\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b x\right )^{2} \operatorname {atan}{\left (a + b x \right )}}{\sqrt {c \left (a^{2} + 2 a b x + b^{2} x^{2} + 1\right )}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\mathrm {atan}\left (a+b\,x\right )\,{\left (a+b\,x\right )}^2}{\sqrt {c\,b^2\,x^2+2\,a\,c\,b\,x+c\,\left (a^2+1\right )}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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