Integrand size = 40, antiderivative size = 281 \[ \int \frac {(a+b x)^2 \cot ^{-1}(a+b x)}{\sqrt {\left (1+a^2\right ) c+2 a b c x+b^2 c x^2}} \, dx=\frac {\sqrt {c+c (a+b x)^2}}{2 b c}+\frac {(a+b x) \sqrt {c+c (a+b x)^2} \cot ^{-1}(a+b x)}{2 b c}+\frac {i \sqrt {1+(a+b x)^2} \cot ^{-1}(a+b x) \arctan \left (\frac {\sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{b \sqrt {c+c (a+b x)^2}}+\frac {i \sqrt {1+(a+b x)^2} \operatorname {PolyLog}\left (2,-\frac {i \sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{2 b \sqrt {c+c (a+b x)^2}}-\frac {i \sqrt {1+(a+b x)^2} \operatorname {PolyLog}\left (2,\frac {i \sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{2 b \sqrt {c+c (a+b x)^2}} \]
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Time = 0.24 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5166, 5073, 267, 5011, 5007} \[ \int \frac {(a+b x)^2 \cot ^{-1}(a+b x)}{\sqrt {\left (1+a^2\right ) c+2 a b c x+b^2 c x^2}} \, dx=\frac {i \sqrt {(a+b x)^2+1} \arctan \left (\frac {\sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right ) \cot ^{-1}(a+b x)}{b \sqrt {c (a+b x)^2+c}}+\frac {i \sqrt {(a+b x)^2+1} \operatorname {PolyLog}\left (2,-\frac {i \sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}\right )}{2 b \sqrt {c (a+b x)^2+c}}-\frac {i \sqrt {(a+b x)^2+1} \operatorname {PolyLog}\left (2,\frac {i \sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}\right )}{2 b \sqrt {c (a+b x)^2+c}}+\frac {\sqrt {c (a+b x)^2+c}}{2 b c}+\frac {(a+b x) \sqrt {c (a+b x)^2+c} \cot ^{-1}(a+b x)}{2 b c} \]
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Rule 267
Rule 5007
Rule 5011
Rule 5073
Rule 5166
Rubi steps \begin{align*} \text {integral}& = \frac {\text {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 {\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 {\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} \text {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) \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} \operatorname {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} \operatorname {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}} \\ \end{align*}
Time = 1.03 (sec) , antiderivative size = 207, normalized size of antiderivative = 0.74 \[ \int \frac {(a+b x)^2 \cot ^{-1}(a+b x)}{\sqrt {\left (1+a^2\right ) c+2 a b c x+b^2 c x^2}} \, dx=-\frac {\sqrt {c \left (1+a^2+2 a b x+b^2 x^2\right )} \left (-2 \cot \left (\frac {1}{2} \cot ^{-1}(a+b x)\right )-\cot ^{-1}(a+b x) \csc ^2\left (\frac {1}{2} \cot ^{-1}(a+b x)\right )-4 \cot ^{-1}(a+b x) \log \left (1-e^{i \cot ^{-1}(a+b x)}\right )+4 \cot ^{-1}(a+b x) \log \left (1+e^{i \cot ^{-1}(a+b x)}\right )-4 i \operatorname {PolyLog}\left (2,-e^{i \cot ^{-1}(a+b x)}\right )+4 i \operatorname {PolyLog}\left (2,e^{i \cot ^{-1}(a+b x)}\right )+\cot ^{-1}(a+b x) \sec ^2\left (\frac {1}{2} \cot ^{-1}(a+b x)\right )-2 \tan \left (\frac {1}{2} \cot ^{-1}(a+b x)\right )\right )}{8 b c (a+b x) \sqrt {1+\frac {1}{(a+b x)^2}}} \]
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Time = 1.43 (sec) , antiderivative size = 202, normalized size of antiderivative = 0.72
method | result | size |
default | \(\frac {\left (\operatorname {arccot}\left (b x +a \right ) b x +a \,\operatorname {arccot}\left (b x +a \right )+1\right ) \sqrt {c \left (b x +a -i\right ) \left (b x +a +i\right )}}{2 b c}-\frac {i \left (i \operatorname {arccot}\left (b x +a \right ) \ln \left (1-\frac {b x +a +i}{\sqrt {1+\left (b x +a \right )^{2}}}\right )-i \operatorname {arccot}\left (b x +a \right ) \ln \left (\frac {b x +a +i}{\sqrt {1+\left (b x +a \right )^{2}}}+1\right )+\operatorname {polylog}\left (2, \frac {b x +a +i}{\sqrt {1+\left (b x +a \right )^{2}}}\right )-\operatorname {polylog}\left (2, -\frac {b x +a +i}{\sqrt {1+\left (b x +a \right )^{2}}}\right )\right ) \sqrt {c \left (b x +a -i\right ) \left (b x +a +i\right )}}{2 \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, b c}\) | \(202\) |
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\[ \int \frac {(a+b x)^2 \cot ^{-1}(a+b x)}{\sqrt {\left (1+a^2\right ) c+2 a b c x+b^2 c x^2}} \, dx=\int { \frac {{\left (b x + a\right )}^{2} \operatorname {arccot}\left (b x + a\right )}{\sqrt {b^{2} c x^{2} + 2 \, a b c x + {\left (a^{2} + 1\right )} c}} \,d x } \]
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Timed out. \[ \int \frac {(a+b x)^2 \cot ^{-1}(a+b x)}{\sqrt {\left (1+a^2\right ) c+2 a b c x+b^2 c x^2}} \, dx=\text {Timed out} \]
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\[ \int \frac {(a+b x)^2 \cot ^{-1}(a+b x)}{\sqrt {\left (1+a^2\right ) c+2 a b c x+b^2 c x^2}} \, dx=\int { \frac {{\left (b x + a\right )}^{2} \operatorname {arccot}\left (b x + a\right )}{\sqrt {b^{2} c x^{2} + 2 \, a b c x + {\left (a^{2} + 1\right )} c}} \,d x } \]
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\[ \int \frac {(a+b x)^2 \cot ^{-1}(a+b x)}{\sqrt {\left (1+a^2\right ) c+2 a b c x+b^2 c x^2}} \, dx=\int { \frac {{\left (b x + a\right )}^{2} \operatorname {arccot}\left (b x + a\right )}{\sqrt {b^{2} c x^{2} + 2 \, a b c x + {\left (a^{2} + 1\right )} c}} \,d x } \]
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Timed out. \[ \int \frac {(a+b x)^2 \cot ^{-1}(a+b x)}{\sqrt {\left (1+a^2\right ) c+2 a b c x+b^2 c x^2}} \, dx=\int \frac {\mathrm {acot}\left (a+b\,x\right )\,{\left (a+b\,x\right )}^2}{\sqrt {c\,b^2\,x^2+2\,a\,c\,b\,x+c\,\left (a^2+1\right )}} \,d x \]
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