Integrand size = 10, antiderivative size = 62 \[ \int \frac {\arctan (a+b x)}{x^2} \, dx=-\frac {a b \arctan (a+b x)}{1+a^2}-\frac {\arctan (a+b x)}{x}+\frac {b \log (x)}{1+a^2}-\frac {b \log \left (1+(a+b x)^2\right )}{2 \left (1+a^2\right )} \]
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Time = 0.03 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {5153, 378, 720, 31, 649, 209, 266} \[ \int \frac {\arctan (a+b x)}{x^2} \, dx=-\frac {a b \arctan (a+b x)}{a^2+1}+\frac {b \log (x)}{a^2+1}-\frac {b \log \left ((a+b x)^2+1\right )}{2 \left (a^2+1\right )}-\frac {\arctan (a+b x)}{x} \]
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Rule 31
Rule 209
Rule 266
Rule 378
Rule 649
Rule 720
Rule 5153
Rubi steps \begin{align*} \text {integral}& = -\frac {\arctan (a+b x)}{x}+b \int \frac {1}{x \left (1+(a+b x)^2\right )} \, dx \\ & = -\frac {\arctan (a+b x)}{x}+b \text {Subst}\left (\int \frac {1}{(-a+x) \left (1+x^2\right )} \, dx,x,a+b x\right ) \\ & = -\frac {\arctan (a+b x)}{x}+\frac {b \text {Subst}\left (\int \frac {1}{-a+x} \, dx,x,a+b x\right )}{1+a^2}+\frac {b \text {Subst}\left (\int \frac {-a-x}{1+x^2} \, dx,x,a+b x\right )}{1+a^2} \\ & = -\frac {\arctan (a+b x)}{x}+\frac {b \log (x)}{1+a^2}-\frac {b \text {Subst}\left (\int \frac {x}{1+x^2} \, dx,x,a+b x\right )}{1+a^2}-\frac {(a b) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,a+b x\right )}{1+a^2} \\ & = -\frac {a b \arctan (a+b x)}{1+a^2}-\frac {\arctan (a+b x)}{x}+\frac {b \log (x)}{1+a^2}-\frac {b \log \left (1+(a+b x)^2\right )}{2 \left (1+a^2\right )} \\ \end{align*}
Result contains complex when optimal does not.
Time = 0.05 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.08 \[ \int \frac {\arctan (a+b x)}{x^2} \, dx=-\frac {\arctan (a+b x)}{x}+\frac {b (2 \log (x)+i (i+a) \log (i-a-b x)+(-1-i a) \log (i+a+b x))}{2 \left (1+a^2\right )} \]
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Time = 0.10 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.98
method | result | size |
derivativedivides | \(b \left (-\frac {\arctan \left (b x +a \right )}{b x}-\frac {\frac {\ln \left (1+\left (b x +a \right )^{2}\right )}{2}+a \arctan \left (b x +a \right )}{a^{2}+1}+\frac {\ln \left (-b x \right )}{a^{2}+1}\right )\) | \(61\) |
default | \(b \left (-\frac {\arctan \left (b x +a \right )}{b x}-\frac {\frac {\ln \left (1+\left (b x +a \right )^{2}\right )}{2}+a \arctan \left (b x +a \right )}{a^{2}+1}+\frac {\ln \left (-b x \right )}{a^{2}+1}\right )\) | \(61\) |
parts | \(-\frac {\arctan \left (b x +a \right )}{x}+b \left (\frac {\ln \left (x \right )}{a^{2}+1}-\frac {b \left (\frac {\ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{2 b}+\frac {a \arctan \left (\frac {2 b^{2} x +2 a b}{2 b}\right )}{b}\right )}{a^{2}+1}\right )\) | \(82\) |
parallelrisch | \(\frac {-2 x \arctan \left (b x +a \right ) a^{2} b^{2}+2 b^{2} \ln \left (x \right ) a x -b^{2} \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a x -2 \arctan \left (b x +a \right ) a^{3} b -2 \arctan \left (b x +a \right ) a b}{2 x a b \left (a^{2}+1\right )}\) | \(91\) |
risch | \(\frac {i \ln \left (1+i \left (b x +a \right )\right )}{2 x}-\frac {i \left (a^{2} \ln \left (1-i \left (b x +a \right )\right )+\ln \left (1-i \left (b x +a \right )\right )-i \ln \left (\left (-a^{2} b +3 i a b \right ) x -3 a +2 i a^{2}-a^{3}\right ) b x +\ln \left (\left (-a^{2} b +3 i a b \right ) x -3 a +2 i a^{2}-a^{3}\right ) a b x -i \ln \left (\left (-a^{2} b -3 i a b \right ) x -3 a -2 i a^{2}-a^{3}\right ) b x +2 i \ln \left (x \right ) b x -\ln \left (\left (-a^{2} b -3 i a b \right ) x -3 a -2 i a^{2}-a^{3}\right ) a b x \right )}{2 x \left (i+a \right ) \left (a -i\right )}\) | \(210\) |
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Time = 0.28 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.92 \[ \int \frac {\arctan (a+b x)}{x^2} \, dx=-\frac {b x \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right ) - 2 \, b x \log \left (x\right ) + 2 \, {\left (a b x + a^{2} + 1\right )} \arctan \left (b x + a\right )}{2 \, {\left (a^{2} + 1\right )} x} \]
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Result contains complex when optimal does not.
Time = 0.64 (sec) , antiderivative size = 168, normalized size of antiderivative = 2.71 \[ \int \frac {\arctan (a+b x)}{x^2} \, dx=\begin {cases} - \frac {i b \operatorname {atan}{\left (b x - i \right )}}{2} - \frac {\operatorname {atan}{\left (b x - i \right )}}{x} - \frac {i}{2 x} & \text {for}\: a = - i \\\frac {i b \operatorname {atan}{\left (b x + i \right )}}{2} - \frac {\operatorname {atan}{\left (b x + i \right )}}{x} + \frac {i}{2 x} & \text {for}\: a = i \\- \frac {2 a^{2} \operatorname {atan}{\left (a + b x \right )}}{2 a^{2} x + 2 x} - \frac {2 a b x \operatorname {atan}{\left (a + b x \right )}}{2 a^{2} x + 2 x} + \frac {2 b x \log {\left (x \right )}}{2 a^{2} x + 2 x} - \frac {b x \log {\left (a^{2} + 2 a b x + b^{2} x^{2} + 1 \right )}}{2 a^{2} x + 2 x} - \frac {2 \operatorname {atan}{\left (a + b x \right )}}{2 a^{2} x + 2 x} & \text {otherwise} \end {cases} \]
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Time = 0.28 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.24 \[ \int \frac {\arctan (a+b x)}{x^2} \, dx=-\frac {1}{2} \, b {\left (\frac {2 \, a \arctan \left (\frac {b^{2} x + a b}{b}\right )}{a^{2} + 1} + \frac {\log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{a^{2} + 1} - \frac {2 \, \log \left (x\right )}{a^{2} + 1}\right )} - \frac {\arctan \left (b x + a\right )}{x} \]
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\[ \int \frac {\arctan (a+b x)}{x^2} \, dx=\int { \frac {\arctan \left (b x + a\right )}{x^{2}} \,d x } \]
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Time = 1.11 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.02 \[ \int \frac {\arctan (a+b x)}{x^2} \, dx=-\frac {\mathrm {atan}\left (a+b\,x\right )}{x}-\frac {\frac {b\,x\,\ln \left (a^2+2\,a\,b\,x+b^2\,x^2+1\right )}{2}-b\,x\,\ln \left (x\right )+a\,b\,x\,\mathrm {atan}\left (a+b\,x\right )}{x\,\left (a^2+1\right )} \]
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