Integrand size = 26, antiderivative size = 130 \[ \int \frac {b+2 c x}{(d+e x) \left (a+b x+c x^2\right )} \, dx=\frac {\sqrt {b^2-4 a c} e \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{c d^2-b d e+a e^2}-\frac {(2 c d-b e) \log (d+e x)}{c d^2-b d e+a e^2}+\frac {(2 c d-b e) \log \left (a+b x+c x^2\right )}{2 \left (c d^2-b d e+a e^2\right )} \]
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Time = 0.12 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {814, 648, 632, 212, 642} \[ \int \frac {b+2 c x}{(d+e x) \left (a+b x+c x^2\right )} \, dx=\frac {e \sqrt {b^2-4 a c} \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{a e^2-b d e+c d^2}+\frac {(2 c d-b e) \log \left (a+b x+c x^2\right )}{2 \left (a e^2-b d e+c d^2\right )}-\frac {(2 c d-b e) \log (d+e x)}{a e^2-b d e+c d^2} \]
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Rule 212
Rule 632
Rule 642
Rule 648
Rule 814
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {e (-2 c d+b e)}{\left (c d^2-b d e+a e^2\right ) (d+e x)}+\frac {b c d-b^2 e+2 a c e+c (2 c d-b e) x}{\left (c d^2-b d e+a e^2\right ) \left (a+b x+c x^2\right )}\right ) \, dx \\ & = -\frac {(2 c d-b e) \log (d+e x)}{c d^2-b d e+a e^2}+\frac {\int \frac {b c d-b^2 e+2 a c e+c (2 c d-b e) x}{a+b x+c x^2} \, dx}{c d^2-b d e+a e^2} \\ & = -\frac {(2 c d-b e) \log (d+e x)}{c d^2-b d e+a e^2}-\frac {\left (\left (b^2-4 a c\right ) e\right ) \int \frac {1}{a+b x+c x^2} \, dx}{2 \left (c d^2-b d e+a e^2\right )}+\frac {(2 c d-b e) \int \frac {b+2 c x}{a+b x+c x^2} \, dx}{2 \left (c d^2-b d e+a e^2\right )} \\ & = -\frac {(2 c d-b e) \log (d+e x)}{c d^2-b d e+a e^2}+\frac {(2 c d-b e) \log \left (a+b x+c x^2\right )}{2 \left (c d^2-b d e+a e^2\right )}+\frac {\left (\left (b^2-4 a c\right ) e\right ) \text {Subst}\left (\int \frac {1}{b^2-4 a c-x^2} \, dx,x,b+2 c x\right )}{c d^2-b d e+a e^2} \\ & = \frac {\sqrt {b^2-4 a c} e \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{c d^2-b d e+a e^2}-\frac {(2 c d-b e) \log (d+e x)}{c d^2-b d e+a e^2}+\frac {(2 c d-b e) \log \left (a+b x+c x^2\right )}{2 \left (c d^2-b d e+a e^2\right )} \\ \end{align*}
Time = 0.12 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.89 \[ \int \frac {b+2 c x}{(d+e x) \left (a+b x+c x^2\right )} \, dx=\frac {2 \left (b^2-4 a c\right ) e \arctan \left (\frac {b+2 c x}{\sqrt {-b^2+4 a c}}\right )+\sqrt {-b^2+4 a c} (2 c d-b e) (2 \log (d+e x)-\log (a+x (b+c x)))}{2 \sqrt {-b^2+4 a c} \left (-c d^2+e (b d-a e)\right )} \]
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Time = 0.66 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.16
method | result | size |
default | \(\frac {\left (b e -2 c d \right ) \ln \left (e x +d \right )}{e^{2} a -b d e +c \,d^{2}}+\frac {\frac {\left (-b c e +2 c^{2} d \right ) \ln \left (c \,x^{2}+b x +a \right )}{2 c}+\frac {2 \left (2 a c e -b^{2} e +b c d -\frac {\left (-b c e +2 c^{2} d \right ) b}{2 c}\right ) \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{e^{2} a -b d e +c \,d^{2}}\) | \(151\) |
risch | \(\text {Expression too large to display}\) | \(3112\) |
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Time = 0.34 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.76 \[ \int \frac {b+2 c x}{(d+e x) \left (a+b x+c x^2\right )} \, dx=\left [\frac {\sqrt {b^{2} - 4 \, a c} e \log \left (\frac {2 \, c^{2} x^{2} + 2 \, b c x + b^{2} - 2 \, a c + \sqrt {b^{2} - 4 \, a c} {\left (2 \, c x + b\right )}}{c x^{2} + b x + a}\right ) + {\left (2 \, c d - b e\right )} \log \left (c x^{2} + b x + a\right ) - 2 \, {\left (2 \, c d - b e\right )} \log \left (e x + d\right )}{2 \, {\left (c d^{2} - b d e + a e^{2}\right )}}, \frac {2 \, \sqrt {-b^{2} + 4 \, a c} e \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) + {\left (2 \, c d - b e\right )} \log \left (c x^{2} + b x + a\right ) - 2 \, {\left (2 \, c d - b e\right )} \log \left (e x + d\right )}{2 \, {\left (c d^{2} - b d e + a e^{2}\right )}}\right ] \]
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Timed out. \[ \int \frac {b+2 c x}{(d+e x) \left (a+b x+c x^2\right )} \, dx=\text {Timed out} \]
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Exception generated. \[ \int \frac {b+2 c x}{(d+e x) \left (a+b x+c x^2\right )} \, dx=\text {Exception raised: ValueError} \]
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Time = 0.35 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.12 \[ \int \frac {b+2 c x}{(d+e x) \left (a+b x+c x^2\right )} \, dx=\frac {{\left (2 \, c d - b e\right )} \log \left (c x^{2} + b x + a\right )}{2 \, {\left (c d^{2} - b d e + a e^{2}\right )}} - \frac {{\left (2 \, c d e - b e^{2}\right )} \log \left ({\left | e x + d \right |}\right )}{c d^{2} e - b d e^{2} + a e^{3}} - \frac {{\left (b^{2} e - 4 \, a c e\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (c d^{2} - b d e + a e^{2}\right )} \sqrt {-b^{2} + 4 \, a c}} \]
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Time = 13.28 (sec) , antiderivative size = 515, normalized size of antiderivative = 3.96 \[ \int \frac {b+2 c x}{(d+e x) \left (a+b x+c x^2\right )} \, dx=\frac {\ln \left (2\,b\,c^2\,e-\frac {\left (c\,d-\frac {b\,e}{2}+\frac {e\,\sqrt {b^2-4\,a\,c}}{2}\right )\,\left (2\,a\,c^2\,e^2-b^2\,c\,e^2+b\,c^2\,d\,e-c^2\,e\,x\,\left (b\,e-2\,c\,d\right )+\frac {c\,e\,\left (c\,d-\frac {b\,e}{2}+\frac {e\,\sqrt {b^2-4\,a\,c}}{2}\right )\,\left (b^2\,d\,e+2\,x\,b^2\,e^2+b\,c\,d^2-2\,x\,b\,c\,d\,e+a\,b\,e^2+2\,x\,c^2\,d^2-8\,a\,c\,d\,e-6\,a\,x\,c\,e^2\right )}{c\,d^2-b\,d\,e+a\,e^2}\right )}{c\,d^2-b\,d\,e+a\,e^2}+4\,c^3\,e\,x\right )\,\left (c\,d-e\,\left (\frac {b}{2}-\frac {\sqrt {b^2-4\,a\,c}}{2}\right )\right )}{c\,d^2-b\,d\,e+a\,e^2}+\frac {\ln \left (2\,b\,c^2\,e-\frac {\left (\frac {b\,e}{2}-c\,d+\frac {e\,\sqrt {b^2-4\,a\,c}}{2}\right )\,\left (b^2\,c\,e^2-2\,a\,c^2\,e^2-b\,c^2\,d\,e+c^2\,e\,x\,\left (b\,e-2\,c\,d\right )+\frac {c\,e\,\left (\frac {b\,e}{2}-c\,d+\frac {e\,\sqrt {b^2-4\,a\,c}}{2}\right )\,\left (b^2\,d\,e+2\,x\,b^2\,e^2+b\,c\,d^2-2\,x\,b\,c\,d\,e+a\,b\,e^2+2\,x\,c^2\,d^2-8\,a\,c\,d\,e-6\,a\,x\,c\,e^2\right )}{c\,d^2-b\,d\,e+a\,e^2}\right )}{c\,d^2-b\,d\,e+a\,e^2}+4\,c^3\,e\,x\right )\,\left (c\,d-e\,\left (\frac {b}{2}+\frac {\sqrt {b^2-4\,a\,c}}{2}\right )\right )}{c\,d^2-b\,d\,e+a\,e^2}+\frac {\ln \left (d+e\,x\right )\,\left (b\,e-2\,c\,d\right )}{c\,d^2-b\,d\,e+a\,e^2} \]
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