Optimal. Leaf size=281 \[ \frac {i \sqrt {(a+b x)^2+1} \text {Li}_2\left (-\frac {i \sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}\right )}{2 b \sqrt {c (a+b x)^2+c}}-\frac {i \sqrt {(a+b x)^2+1} \text {Li}_2\left (\frac {i \sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}\right )}{2 b \sqrt {c (a+b x)^2+c}}+\frac {\sqrt {c (a+b x)^2+c}}{2 b c}+\frac {(a+b x) \sqrt {c (a+b x)^2+c} \cot ^{-1}(a+b x)}{2 b c}+\frac {i \sqrt {(a+b x)^2+1} \tan ^{-1}\left (\frac {\sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right ) \cot ^{-1}(a+b x)}{b \sqrt {c (a+b x)^2+c}} \]
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Rubi [A] time = 0.35, 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 = {5058, 4953, 261, 4891, 4887} \[ \frac {i \sqrt {(a+b x)^2+1} \text {PolyLog}\left (2,-\frac {i \sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{2 b \sqrt {c (a+b x)^2+c}}-\frac {i \sqrt {(a+b x)^2+1} \text {PolyLog}\left (2,\frac {i \sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{2 b \sqrt {c (a+b x)^2+c}}+\frac {\sqrt {c (a+b x)^2+c}}{2 b c}+\frac {(a+b x) \sqrt {c (a+b x)^2+c} \cot ^{-1}(a+b x)}{2 b c}+\frac {i \sqrt {(a+b x)^2+1} \tan ^{-1}\left (\frac {\sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right ) \cot ^{-1}(a+b x)}{b \sqrt {c (a+b x)^2+c}} \]
Antiderivative was successfully verified.
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Rule 261
Rule 4887
Rule 4891
Rule 4953
Rule 5058
Rubi steps
\begin {align*} \int \frac {(a+b x)^2 \cot ^{-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 \cot ^{-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} \cot ^{-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 {\cot ^{-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} \cot ^{-1}(a+b x)}{2 b c}-\frac {\sqrt {1+(a+b x)^2} \operatorname {Subst}\left (\int \frac {\cot ^{-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} \cot ^{-1}(a+b x)}{2 b c}+\frac {i \sqrt {1+(a+b x)^2} \cot ^{-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 = 0.93, size = 207, normalized size = 0.74 \[ -\frac {\sqrt {c \left (a^2+2 a b x+b^2 x^2+1\right )} \left (-4 i \text {Li}_2\left (-e^{i \cot ^{-1}(a+b x)}\right )+4 i \text {Li}_2\left (e^{i \cot ^{-1}(a+b x)}\right )-2 \cot \left (\frac {1}{2} \cot ^{-1}(a+b x)\right )-4 \cot ^{-1}(a+b x) \log \left (1-e^{i \cot ^{-1}(a+b x)}\right )+4 \cot ^{-1}(a+b x) \log \left (1+e^{i \cot ^{-1}(a+b x)}\right )-2 \tan \left (\frac {1}{2} \cot ^{-1}(a+b x)\right )-\cot ^{-1}(a+b x) \csc ^2\left (\frac {1}{2} \cot ^{-1}(a+b x)\right )+\cot ^{-1}(a+b x) \sec ^2\left (\frac {1}{2} \cot ^{-1}(a+b x)\right )\right )}{8 b c (a+b x) \sqrt {\frac {1}{(a+b x)^2}+1}} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 1.09, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \operatorname {arccot}\left (b x + a\right )}{\sqrt {b^{2} c x^{2} + 2 \, a b c x + {\left (a^{2} + 1\right )} c}}, x\right ) \]
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 \[ \int \frac {{\left (b x + a\right )}^{2} \operatorname {arccot}\left (b x + a\right )}{\sqrt {b^{2} c x^{2} + 2 \, a b c x + {\left (a^{2} + 1\right )} c}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 3.99, size = 202, normalized size = 0.72 \[ \frac {\left (\mathrm {arccot}\left (b x +a \right ) x b +\mathrm {arccot}\left (b x +a \right ) a +1\right ) \sqrt {c \left (b x +a -i\right ) \left (b x +a +i\right )}}{2 b c}-\frac {i \left (i \ln \left (1-\frac {b x +a +i}{\sqrt {1+\left (b x +a \right )^{2}}}\right ) \mathrm {arccot}\left (b x +a \right )-i \ln \left (1+\frac {b x +a +i}{\sqrt {1+\left (b x +a \right )^{2}}}\right ) \mathrm {arccot}\left (b x +a \right )+\polylog \left (2, \frac {b x +a +i}{\sqrt {1+\left (b x +a \right )^{2}}}\right )-\polylog \left (2, -\frac {b x +a +i}{\sqrt {1+\left (b x +a \right )^{2}}}\right )\right ) \sqrt {c \left (b x +a -i\right ) \left (b x +a +i\right )}}{2 \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, b c} \]
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 \[ \int \frac {{\left (b x + a\right )}^{2} \operatorname {arccot}\left (b x + a\right )}{\sqrt {b^{2} c x^{2} + 2 \, a b c x + {\left (a^{2} + 1\right )} c}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\mathrm {acot}\left (a+b\,x\right )\,{\left (a+b\,x\right )}^2}{\sqrt {c\,b^2\,x^2+2\,a\,c\,b\,x+c\,\left (a^2+1\right )}} \,d x \]
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 \[ \int \frac {\left (a + b x\right )^{2} \operatorname {acot}{\left (a + b x \right )}}{\sqrt {c \left (a^{2} + 2 a b x + b^{2} x^{2} + 1\right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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