Integrand size = 16, antiderivative size = 57 \[ \int \frac {1}{a+b x+c (d+e x)^2} \, dx=-\frac {2 \text {arctanh}\left (\frac {b+2 c e (d+e x)}{\sqrt {b^2+4 b c d e-4 a c e^2}}\right )}{\sqrt {b^2+4 b c d e-4 a c e^2}} \]
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Time = 0.05 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {2006, 632, 212} \[ \int \frac {1}{a+b x+c (d+e x)^2} \, dx=-\frac {2 \text {arctanh}\left (\frac {b+2 c e (d+e x)}{\sqrt {-4 a c e^2+b^2+4 b c d e}}\right )}{\sqrt {-4 a c e^2+b^2+4 b c d e}} \]
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Rule 212
Rule 632
Rule 2006
Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{a+c d^2+(b+2 c d e) x+c e^2 x^2} \, dx \\ & = -\left (2 \text {Subst}\left (\int \frac {1}{b^2+4 b c d e-4 a c e^2-x^2} \, dx,x,b+2 c d e+2 c e^2 x\right )\right ) \\ & = -\frac {2 \tanh ^{-1}\left (\frac {b+2 c e (d+e x)}{\sqrt {b^2+4 b c d e-4 a c e^2}}\right )}{\sqrt {b^2+4 b c d e-4 a c e^2}} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.07 \[ \int \frac {1}{a+b x+c (d+e x)^2} \, dx=\frac {2 \arctan \left (\frac {b+2 c e (d+e x)}{\sqrt {-b^2-4 b c d e+4 a c e^2}}\right )}{\sqrt {-b^2-4 b c d e+4 a c e^2}} \]
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Time = 1.31 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.07
method | result | size |
default | \(\frac {2 \arctan \left (\frac {2 c \,e^{2} x +2 d c e +b}{\sqrt {4 e^{2} a c -4 b c d e -b^{2}}}\right )}{\sqrt {4 e^{2} a c -4 b c d e -b^{2}}}\) | \(61\) |
risch | \(-\frac {\ln \left (2 c \,e^{2} x +2 d c e +\sqrt {-4 e^{2} a c +4 b c d e +b^{2}}+b \right )}{\sqrt {-4 e^{2} a c +4 b c d e +b^{2}}}+\frac {\ln \left (-2 c \,e^{2} x -2 d c e +\sqrt {-4 e^{2} a c +4 b c d e +b^{2}}-b \right )}{\sqrt {-4 e^{2} a c +4 b c d e +b^{2}}}\) | \(113\) |
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Time = 0.26 (sec) , antiderivative size = 240, normalized size of antiderivative = 4.21 \[ \int \frac {1}{a+b x+c (d+e x)^2} \, dx=\left [\frac {\log \left (\frac {2 \, c^{2} e^{4} x^{2} + 4 \, b c d e + 2 \, {\left (c^{2} d^{2} - a c\right )} e^{2} + b^{2} + 2 \, {\left (2 \, c^{2} d e^{3} + b c e^{2}\right )} x - \sqrt {4 \, b c d e - 4 \, a c e^{2} + b^{2}} {\left (2 \, c e^{2} x + 2 \, c d e + b\right )}}{c e^{2} x^{2} + c d^{2} + {\left (2 \, c d e + b\right )} x + a}\right )}{\sqrt {4 \, b c d e - 4 \, a c e^{2} + b^{2}}}, -\frac {2 \, \sqrt {-4 \, b c d e + 4 \, a c e^{2} - b^{2}} \arctan \left (-\frac {\sqrt {-4 \, b c d e + 4 \, a c e^{2} - b^{2}} {\left (2 \, c e^{2} x + 2 \, c d e + b\right )}}{4 \, b c d e - 4 \, a c e^{2} + b^{2}}\right )}{4 \, b c d e - 4 \, a c e^{2} + b^{2}}\right ] \]
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Leaf count of result is larger than twice the leaf count of optimal. 294 vs. \(2 (63) = 126\).
Time = 0.18 (sec) , antiderivative size = 294, normalized size of antiderivative = 5.16 \[ \int \frac {1}{a+b x+c (d+e x)^2} \, dx=- \sqrt {- \frac {1}{4 a c e^{2} - b^{2} - 4 b c d e}} \log {\left (x + \frac {- 4 a c e^{2} \sqrt {- \frac {1}{4 a c e^{2} - b^{2} - 4 b c d e}} + b^{2} \sqrt {- \frac {1}{4 a c e^{2} - b^{2} - 4 b c d e}} + 4 b c d e \sqrt {- \frac {1}{4 a c e^{2} - b^{2} - 4 b c d e}} + b + 2 c d e}{2 c e^{2}} \right )} + \sqrt {- \frac {1}{4 a c e^{2} - b^{2} - 4 b c d e}} \log {\left (x + \frac {4 a c e^{2} \sqrt {- \frac {1}{4 a c e^{2} - b^{2} - 4 b c d e}} - b^{2} \sqrt {- \frac {1}{4 a c e^{2} - b^{2} - 4 b c d e}} - 4 b c d e \sqrt {- \frac {1}{4 a c e^{2} - b^{2} - 4 b c d e}} + b + 2 c d e}{2 c e^{2}} \right )} \]
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Exception generated. \[ \int \frac {1}{a+b x+c (d+e x)^2} \, dx=\text {Exception raised: ValueError} \]
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Time = 0.27 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.05 \[ \int \frac {1}{a+b x+c (d+e x)^2} \, dx=\frac {2 \, \arctan \left (\frac {2 \, c e^{2} x + 2 \, c d e + b}{\sqrt {-4 \, b c d e + 4 \, a c e^{2} - b^{2}}}\right )}{\sqrt {-4 \, b c d e + 4 \, a c e^{2} - b^{2}}} \]
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Time = 0.04 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.44 \[ \int \frac {1}{a+b x+c (d+e x)^2} \, dx=\frac {2\,\mathrm {atan}\left (\frac {b+2\,c\,d\,e}{\sqrt {-b^2-4\,c\,d\,b\,e+4\,a\,c\,e^2}}+\frac {2\,c\,e^2\,x}{\sqrt {-b^2-4\,c\,d\,b\,e+4\,a\,c\,e^2}}\right )}{\sqrt {-b^2-4\,c\,d\,b\,e+4\,a\,c\,e^2}} \]
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