Optimal. Leaf size=187 \[ -\frac {i \text {Li}_2\left (-\frac {i \sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}\right )}{2 b}+\frac {i \text {Li}_2\left (\frac {i \sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}\right )}{2 b}-\frac {\sqrt {(a+b x)^2+1}}{2 b}+\frac {i \tan ^{-1}\left (\frac {\sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right ) \tan ^{-1}(a+b x)}{b}+\frac {(a+b x) \sqrt {(a+b x)^2+1} \tan ^{-1}(a+b x)}{2 b} \]
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Rubi [A] time = 0.22, antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {5057, 4952, 261, 4886} \[ -\frac {i \text {PolyLog}\left (2,-\frac {i \sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{2 b}+\frac {i \text {PolyLog}\left (2,\frac {i \sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{2 b}-\frac {\sqrt {(a+b x)^2+1}}{2 b}+\frac {i \tan ^{-1}\left (\frac {\sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right ) \tan ^{-1}(a+b x)}{b}+\frac {(a+b x) \sqrt {(a+b x)^2+1} \tan ^{-1}(a+b x)}{2 b} \]
Antiderivative was successfully verified.
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Rule 261
Rule 4886
Rule 4952
Rule 5057
Rubi steps
\begin {align*} \int \frac {(a+b x)^2 \tan ^{-1}(a+b x)}{\sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^2 \tan ^{-1}(x)}{\sqrt {1+x^2}} \, dx,x,a+b x\right )}{b}\\ &=\frac {(a+b x) \sqrt {1+(a+b x)^2} \tan ^{-1}(a+b x)}{2 b}-\frac {\operatorname {Subst}\left (\int \frac {x}{\sqrt {1+x^2}} \, dx,x,a+b x\right )}{2 b}-\frac {\operatorname {Subst}\left (\int \frac {\tan ^{-1}(x)}{\sqrt {1+x^2}} \, dx,x,a+b x\right )}{2 b}\\ &=-\frac {\sqrt {1+(a+b x)^2}}{2 b}+\frac {(a+b x) \sqrt {1+(a+b x)^2} \tan ^{-1}(a+b x)}{2 b}+\frac {i \tan ^{-1}(a+b x) \tan ^{-1}\left (\frac {\sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{b}-\frac {i \text {Li}_2\left (-\frac {i \sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{2 b}+\frac {i \text {Li}_2\left (\frac {i \sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{2 b}\\ \end {align*}
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Mathematica [A] time = 0.72, size = 145, normalized size = 0.78 \[ \frac {-i \text {Li}_2\left (-i e^{i \tan ^{-1}(a+b x)}\right )+i \text {Li}_2\left (i e^{i \tan ^{-1}(a+b x)}\right )-\sqrt {(a+b x)^2+1}+(a+b x) \sqrt {(a+b x)^2+1} \tan ^{-1}(a+b x)+\tan ^{-1}(a+b x) \left (-\log \left (1-i e^{i \tan ^{-1}(a+b x)}\right )\right )+\tan ^{-1}(a+b x) \log \left (1+i e^{i \tan ^{-1}(a+b x)}\right )}{2 b} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \arctan \left (b x + a\right )}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 3.16, size = 187, normalized size = 1.00 \[ \frac {\left (\arctan \left (b x +a \right ) x b +\arctan \left (b x +a \right ) a -1\right ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{2 b}+\frac {\arctan \left (b x +a \right ) \ln \left (1+\frac {i \left (1+i \left (b x +a \right )\right )}{\sqrt {1+\left (b x +a \right )^{2}}}\right )}{2 b}-\frac {\arctan \left (b x +a \right ) \ln \left (1-\frac {i \left (1+i \left (b x +a \right )\right )}{\sqrt {1+\left (b x +a \right )^{2}}}\right )}{2 b}-\frac {i \dilog \left (1+\frac {i \left (1+i \left (b x +a \right )\right )}{\sqrt {1+\left (b x +a \right )^{2}}}\right )}{2 b}+\frac {i \dilog \left (1-\frac {i \left (1+i \left (b x +a \right )\right )}{\sqrt {1+\left (b x +a \right )^{2}}}\right )}{2 b} \]
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 \[ \int \frac {{\left (b x + a\right )}^{2} \arctan \left (b x + a\right )}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {atan}\left (a+b\,x\right )\,{\left (a+b\,x\right )}^2}{\sqrt {a^2+2\,a\,b\,x+b^2\,x^2+1}} \,d x \]
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 \[ \int \frac {\left (a + b x\right )^{2} \operatorname {atan}{\left (a + b x \right )}}{\sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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