Integrand size = 19, antiderivative size = 53 \[ \int \frac {1}{(d+e x) \left (b x+c x^2\right )} \, dx=\frac {\log (x)}{b d}-\frac {c \log (b+c x)}{b (c d-b e)}+\frac {e \log (d+e x)}{d (c d-b e)} \]
[Out]
Time = 0.03 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {712} \[ \int \frac {1}{(d+e x) \left (b x+c x^2\right )} \, dx=-\frac {c \log (b+c x)}{b (c d-b e)}+\frac {e \log (d+e x)}{d (c d-b e)}+\frac {\log (x)}{b d} \]
[In]
[Out]
Rule 712
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{b d x}+\frac {c^2}{b (-c d+b e) (b+c x)}+\frac {e^2}{d (c d-b e) (d+e x)}\right ) \, dx \\ & = \frac {\log (x)}{b d}-\frac {c \log (b+c x)}{b (c d-b e)}+\frac {e \log (d+e x)}{d (c d-b e)} \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.91 \[ \int \frac {1}{(d+e x) \left (b x+c x^2\right )} \, dx=\frac {c d \log (x)-b e \log (x)-c d \log (b+c x)+b e \log (d+e x)}{b c d^2-b^2 d e} \]
[In]
[Out]
Time = 1.91 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.92
method | result | size |
parallelrisch | \(\frac {\ln \left (x \right ) b e -d \ln \left (x \right ) c +\ln \left (c x +b \right ) c d -e \ln \left (e x +d \right ) b}{b d \left (b e -c d \right )}\) | \(49\) |
default | \(\frac {\ln \left (x \right )}{b d}+\frac {c \ln \left (c x +b \right )}{b \left (b e -c d \right )}-\frac {e \ln \left (e x +d \right )}{d \left (b e -c d \right )}\) | \(54\) |
norman | \(\frac {\ln \left (x \right )}{b d}+\frac {c \ln \left (c x +b \right )}{b \left (b e -c d \right )}-\frac {e \ln \left (e x +d \right )}{d \left (b e -c d \right )}\) | \(54\) |
risch | \(-\frac {e \ln \left (e x +d \right )}{d \left (b e -c d \right )}+\frac {\ln \left (-x \right )}{b d}+\frac {c \ln \left (c x +b \right )}{b \left (b e -c d \right )}\) | \(56\) |
[In]
[Out]
none
Time = 0.32 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.94 \[ \int \frac {1}{(d+e x) \left (b x+c x^2\right )} \, dx=-\frac {c d \log \left (c x + b\right ) - b e \log \left (e x + d\right ) - {\left (c d - b e\right )} \log \left (x\right )}{b c d^{2} - b^{2} d e} \]
[In]
[Out]
Timed out. \[ \int \frac {1}{(d+e x) \left (b x+c x^2\right )} \, dx=\text {Timed out} \]
[In]
[Out]
none
Time = 0.19 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(d+e x) \left (b x+c x^2\right )} \, dx=-\frac {c \log \left (c x + b\right )}{b c d - b^{2} e} + \frac {e \log \left (e x + d\right )}{c d^{2} - b d e} + \frac {\log \left (x\right )}{b d} \]
[In]
[Out]
none
Time = 0.27 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.25 \[ \int \frac {1}{(d+e x) \left (b x+c x^2\right )} \, dx=-\frac {c^{2} \log \left ({\left | c x + b \right |}\right )}{b c^{2} d - b^{2} c e} + \frac {e^{2} \log \left ({\left | e x + d \right |}\right )}{c d^{2} e - b d e^{2}} + \frac {\log \left ({\left | x \right |}\right )}{b d} \]
[In]
[Out]
Time = 9.78 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.75 \[ \int \frac {1}{(d+e x) \left (b x+c x^2\right )} \, dx=\frac {e\,\ln \left (\frac {{\left (d+e\,x\right )}^2}{x\,\left (b+c\,x\right )}\right )}{2\,c\,d^2-2\,b\,d\,e}-\frac {\ln \left (\frac {b-\sqrt {b^2}+2\,c\,x}{b+\sqrt {b^2}+2\,c\,x}\right )\,\left (b\,e-2\,c\,d\right )}{\left (2\,c\,d^2-2\,b\,d\,e\right )\,\sqrt {b^2}} \]
[In]
[Out]