Integrand size = 29, antiderivative size = 47 \[ \int \frac {4 x^2-5 x^3+6 x^4}{1-x+2 x^2} \, dx=-\frac {x^2}{2}+x^3-\frac {\arctan \left (\frac {1-4 x}{\sqrt {7}}\right )}{2 \sqrt {7}}+\frac {1}{4} \log \left (1-x+2 x^2\right ) \]
[Out]
Time = 0.05 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {1608, 1642, 648, 632, 210, 642} \[ \int \frac {4 x^2-5 x^3+6 x^4}{1-x+2 x^2} \, dx=-\frac {\arctan \left (\frac {1-4 x}{\sqrt {7}}\right )}{2 \sqrt {7}}+x^3-\frac {x^2}{2}+\frac {1}{4} \log \left (2 x^2-x+1\right ) \]
[In]
[Out]
Rule 210
Rule 632
Rule 642
Rule 648
Rule 1608
Rule 1642
Rubi steps \begin{align*} \text {integral}& = \int \frac {x^2 \left (4-5 x+6 x^2\right )}{1-x+2 x^2} \, dx \\ & = \int \left (-x+3 x^2+\frac {x}{1-x+2 x^2}\right ) \, dx \\ & = -\frac {x^2}{2}+x^3+\int \frac {x}{1-x+2 x^2} \, dx \\ & = -\frac {x^2}{2}+x^3+\frac {1}{4} \int \frac {1}{1-x+2 x^2} \, dx+\frac {1}{4} \int \frac {-1+4 x}{1-x+2 x^2} \, dx \\ & = -\frac {x^2}{2}+x^3+\frac {1}{4} \log \left (1-x+2 x^2\right )-\frac {1}{2} \text {Subst}\left (\int \frac {1}{-7-x^2} \, dx,x,-1+4 x\right ) \\ & = -\frac {x^2}{2}+x^3-\frac {\arctan \left (\frac {1-4 x}{\sqrt {7}}\right )}{2 \sqrt {7}}+\frac {1}{4} \log \left (1-x+2 x^2\right ) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00 \[ \int \frac {4 x^2-5 x^3+6 x^4}{1-x+2 x^2} \, dx=-\frac {x^2}{2}+x^3+\frac {\arctan \left (\frac {-1+4 x}{\sqrt {7}}\right )}{2 \sqrt {7}}+\frac {1}{4} \log \left (1-x+2 x^2\right ) \]
[In]
[Out]
Time = 0.41 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.83
method | result | size |
default | \(x^{3}-\frac {x^{2}}{2}+\frac {\ln \left (2 x^{2}-x +1\right )}{4}+\frac {\sqrt {7}\, \arctan \left (\frac {\left (-1+4 x \right ) \sqrt {7}}{7}\right )}{14}\) | \(39\) |
risch | \(x^{3}-\frac {x^{2}}{2}+\frac {\ln \left (16 x^{2}-8 x +8\right )}{4}+\frac {\sqrt {7}\, \arctan \left (\frac {\left (-1+4 x \right ) \sqrt {7}}{7}\right )}{14}\) | \(39\) |
[In]
[Out]
none
Time = 0.26 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.81 \[ \int \frac {4 x^2-5 x^3+6 x^4}{1-x+2 x^2} \, dx=x^{3} - \frac {1}{2} \, x^{2} + \frac {1}{14} \, \sqrt {7} \arctan \left (\frac {1}{7} \, \sqrt {7} {\left (4 \, x - 1\right )}\right ) + \frac {1}{4} \, \log \left (2 \, x^{2} - x + 1\right ) \]
[In]
[Out]
Time = 0.05 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.98 \[ \int \frac {4 x^2-5 x^3+6 x^4}{1-x+2 x^2} \, dx=x^{3} - \frac {x^{2}}{2} + \frac {\log {\left (x^{2} - \frac {x}{2} + \frac {1}{2} \right )}}{4} + \frac {\sqrt {7} \operatorname {atan}{\left (\frac {4 \sqrt {7} x}{7} - \frac {\sqrt {7}}{7} \right )}}{14} \]
[In]
[Out]
none
Time = 0.29 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.81 \[ \int \frac {4 x^2-5 x^3+6 x^4}{1-x+2 x^2} \, dx=x^{3} - \frac {1}{2} \, x^{2} + \frac {1}{14} \, \sqrt {7} \arctan \left (\frac {1}{7} \, \sqrt {7} {\left (4 \, x - 1\right )}\right ) + \frac {1}{4} \, \log \left (2 \, x^{2} - x + 1\right ) \]
[In]
[Out]
none
Time = 0.27 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.81 \[ \int \frac {4 x^2-5 x^3+6 x^4}{1-x+2 x^2} \, dx=x^{3} - \frac {1}{2} \, x^{2} + \frac {1}{14} \, \sqrt {7} \arctan \left (\frac {1}{7} \, \sqrt {7} {\left (4 \, x - 1\right )}\right ) + \frac {1}{4} \, \log \left (2 \, x^{2} - x + 1\right ) \]
[In]
[Out]
Time = 0.28 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.85 \[ \int \frac {4 x^2-5 x^3+6 x^4}{1-x+2 x^2} \, dx=\frac {\ln \left (2\,x^2-x+1\right )}{4}+\frac {\sqrt {7}\,\mathrm {atan}\left (\frac {4\,\sqrt {7}\,x}{7}-\frac {\sqrt {7}}{7}\right )}{14}-\frac {x^2}{2}+x^3 \]
[In]
[Out]