Integrand size = 8, antiderivative size = 60 \[ \int x \cot ^{-1}(a+b x) \, dx=\frac {x}{2 b}+\frac {1}{2} x^2 \cot ^{-1}(a+b x)-\frac {\left (1-a^2\right ) \arctan (a+b x)}{2 b^2}-\frac {a \log \left (1+(a+b x)^2\right )}{2 b^2} \]
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Time = 0.04 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {5156, 4973, 716, 649, 209, 266} \[ \int x \cot ^{-1}(a+b x) \, dx=-\frac {\left (1-a^2\right ) \arctan (a+b x)}{2 b^2}-\frac {a \log \left ((a+b x)^2+1\right )}{2 b^2}+\frac {1}{2} x^2 \cot ^{-1}(a+b x)+\frac {x}{2 b} \]
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Rule 209
Rule 266
Rule 649
Rule 716
Rule 4973
Rule 5156
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \left (-\frac {a}{b}+\frac {x}{b}\right ) \cot ^{-1}(x) \, dx,x,a+b x\right )}{b} \\ & = \frac {1}{2} x^2 \cot ^{-1}(a+b x)+\frac {1}{2} \text {Subst}\left (\int \frac {\left (-\frac {a}{b}+\frac {x}{b}\right )^2}{1+x^2} \, dx,x,a+b x\right ) \\ & = \frac {1}{2} x^2 \cot ^{-1}(a+b x)+\frac {1}{2} \text {Subst}\left (\int \left (\frac {1}{b^2}-\frac {1-a^2+2 a x}{b^2 \left (1+x^2\right )}\right ) \, dx,x,a+b x\right ) \\ & = \frac {x}{2 b}+\frac {1}{2} x^2 \cot ^{-1}(a+b x)-\frac {\text {Subst}\left (\int \frac {1-a^2+2 a x}{1+x^2} \, dx,x,a+b x\right )}{2 b^2} \\ & = \frac {x}{2 b}+\frac {1}{2} x^2 \cot ^{-1}(a+b x)-\frac {a \text {Subst}\left (\int \frac {x}{1+x^2} \, dx,x,a+b x\right )}{b^2}-\frac {\left (1-a^2\right ) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,a+b x\right )}{2 b^2} \\ & = \frac {x}{2 b}+\frac {1}{2} x^2 \cot ^{-1}(a+b x)-\frac {\left (1-a^2\right ) \arctan (a+b x)}{2 b^2}-\frac {a \log \left (1+(a+b x)^2\right )}{2 b^2} \\ \end{align*}
Result contains complex when optimal does not.
Time = 0.03 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.50 \[ \int x \cot ^{-1}(a+b x) \, dx=\frac {2 b x+2 b^2 x^2 \cot ^{-1}(a+b x)-i (-i+a)^2 \log (i-a-b x)-i \log (i+a+b x)-2 a \log (i+a+b x)+i a^2 \log (i+a+b x)}{4 b^2} \]
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Time = 0.26 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.05
method | result | size |
derivativedivides | \(\frac {\frac {\left (b x +a \right )^{2} \operatorname {arccot}\left (b x +a \right )}{2}-\operatorname {arccot}\left (b x +a \right ) a \left (b x +a \right )+\frac {b x}{2}+\frac {a}{2}-\frac {a \ln \left (1+\left (b x +a \right )^{2}\right )}{2}-\frac {\arctan \left (b x +a \right )}{2}}{b^{2}}\) | \(63\) |
default | \(\frac {\frac {\left (b x +a \right )^{2} \operatorname {arccot}\left (b x +a \right )}{2}-\operatorname {arccot}\left (b x +a \right ) a \left (b x +a \right )+\frac {b x}{2}+\frac {a}{2}-\frac {a \ln \left (1+\left (b x +a \right )^{2}\right )}{2}-\frac {\arctan \left (b x +a \right )}{2}}{b^{2}}\) | \(63\) |
parallelrisch | \(-\frac {-\operatorname {arccot}\left (b x +a \right ) b^{2} x^{2}+\operatorname {arccot}\left (b x +a \right ) a^{2}+a \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )-b x -\operatorname {arccot}\left (b x +a \right )+2 a}{2 b^{2}}\) | \(66\) |
parts | \(\frac {x^{2} \operatorname {arccot}\left (b x +a \right )}{2}+\frac {b \left (\frac {x}{b^{2}}+\frac {-\frac {a \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{b}+\frac {\left (a^{2}-1\right ) \arctan \left (\frac {2 b^{2} x +2 a b}{2 b}\right )}{b}}{b^{2}}\right )}{2}\) | \(77\) |
risch | \(\frac {i x^{2} \ln \left (1+i \left (b x +a \right )\right )}{4}-\frac {i x^{2} \ln \left (1-i \left (b x +a \right )\right )}{4}+\frac {\pi \,x^{2}}{4}+\frac {a^{2} \arctan \left (b x +a \right )}{2 b^{2}}-\frac {a \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{2 b^{2}}+\frac {x}{2 b}-\frac {\arctan \left (b x +a \right )}{2 b^{2}}\) | \(97\) |
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Time = 0.28 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.92 \[ \int x \cot ^{-1}(a+b x) \, dx=\frac {b^{2} x^{2} \operatorname {arccot}\left (b x + a\right ) + b x + {\left (a^{2} - 1\right )} \arctan \left (b x + a\right ) - a \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{2 \, b^{2}} \]
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Time = 0.26 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.30 \[ \int x \cot ^{-1}(a+b x) \, dx=\begin {cases} - \frac {a^{2} \operatorname {acot}{\left (a + b x \right )}}{2 b^{2}} - \frac {a \log {\left (a^{2} + 2 a b x + b^{2} x^{2} + 1 \right )}}{2 b^{2}} + \frac {x^{2} \operatorname {acot}{\left (a + b x \right )}}{2} + \frac {x}{2 b} + \frac {\operatorname {acot}{\left (a + b x \right )}}{2 b^{2}} & \text {for}\: b \neq 0 \\\frac {x^{2} \operatorname {acot}{\left (a \right )}}{2} & \text {otherwise} \end {cases} \]
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Time = 0.27 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.13 \[ \int x \cot ^{-1}(a+b x) \, dx=\frac {1}{2} \, x^{2} \operatorname {arccot}\left (b x + a\right ) + \frac {1}{2} \, b {\left (\frac {x}{b^{2}} + \frac {{\left (a^{2} - 1\right )} \arctan \left (\frac {b^{2} x + a b}{b}\right )}{b^{3}} - \frac {a \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{b^{3}}\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 210 vs. \(2 (50) = 100\).
Time = 0.34 (sec) , antiderivative size = 210, normalized size of antiderivative = 3.50 \[ \int x \cot ^{-1}(a+b x) \, dx=\frac {4 \, a \arctan \left (\frac {1}{b x + a}\right ) \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{3} + \arctan \left (\frac {1}{b x + a}\right ) \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{4} + 4 \, a \log \left (\frac {16 \, \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2}}{\tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{4} + 2 \, \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2} + 1}\right ) \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2} - 4 \, a \arctan \left (\frac {1}{b x + a}\right ) \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right ) + 2 \, \arctan \left (\frac {1}{b x + a}\right ) \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2} - 2 \, \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{3} + \arctan \left (\frac {1}{b x + a}\right ) + 2 \, \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )}{8 \, b^{2} \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2}} \]
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Time = 1.03 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.02 \[ \int x \cot ^{-1}(a+b x) \, dx=\frac {x^2\,\mathrm {acot}\left (a+b\,x\right )}{2}+\frac {\frac {\mathrm {acot}\left (a+b\,x\right )}{2}+\frac {b\,x}{2}-\frac {a^2\,\mathrm {acot}\left (a+b\,x\right )}{2}-\frac {a\,\ln \left (a^2+2\,a\,b\,x+b^2\,x^2+1\right )}{2}}{b^2} \]
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