Integrand size = 20, antiderivative size = 30 \[ \int \frac {(1-2 x) (3+5 x)^2}{2+3 x} \, dx=\frac {118 x}{27}-\frac {5 x^2}{18}-\frac {50 x^3}{9}+\frac {7}{81} \log (2+3 x) \]
[Out]
Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {78} \[ \int \frac {(1-2 x) (3+5 x)^2}{2+3 x} \, dx=-\frac {50 x^3}{9}-\frac {5 x^2}{18}+\frac {118 x}{27}+\frac {7}{81} \log (3 x+2) \]
[In]
[Out]
Rule 78
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {118}{27}-\frac {5 x}{9}-\frac {50 x^2}{3}+\frac {7}{27 (2+3 x)}\right ) \, dx \\ & = \frac {118 x}{27}-\frac {5 x^2}{18}-\frac {50 x^3}{9}+\frac {7}{81} \log (2+3 x) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {(1-2 x) (3+5 x)^2}{2+3 x} \, dx=\frac {1}{486} \left (676+2124 x-135 x^2-2700 x^3+42 \log (2+3 x)\right ) \]
[In]
[Out]
Time = 1.78 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.70
method | result | size |
parallelrisch | \(-\frac {50 x^{3}}{9}-\frac {5 x^{2}}{18}+\frac {118 x}{27}+\frac {7 \ln \left (\frac {2}{3}+x \right )}{81}\) | \(21\) |
default | \(\frac {118 x}{27}-\frac {5 x^{2}}{18}-\frac {50 x^{3}}{9}+\frac {7 \ln \left (2+3 x \right )}{81}\) | \(23\) |
norman | \(\frac {118 x}{27}-\frac {5 x^{2}}{18}-\frac {50 x^{3}}{9}+\frac {7 \ln \left (2+3 x \right )}{81}\) | \(23\) |
risch | \(\frac {118 x}{27}-\frac {5 x^{2}}{18}-\frac {50 x^{3}}{9}+\frac {7 \ln \left (2+3 x \right )}{81}\) | \(23\) |
meijerg | \(\frac {7 \ln \left (1+\frac {3 x}{2}\right )}{81}+4 x +\frac {35 x \left (-\frac {9 x}{2}+6\right )}{27}-\frac {50 x \left (9 x^{2}-9 x +12\right )}{81}\) | \(34\) |
[In]
[Out]
none
Time = 0.23 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.73 \[ \int \frac {(1-2 x) (3+5 x)^2}{2+3 x} \, dx=-\frac {50}{9} \, x^{3} - \frac {5}{18} \, x^{2} + \frac {118}{27} \, x + \frac {7}{81} \, \log \left (3 \, x + 2\right ) \]
[In]
[Out]
Time = 0.04 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {(1-2 x) (3+5 x)^2}{2+3 x} \, dx=- \frac {50 x^{3}}{9} - \frac {5 x^{2}}{18} + \frac {118 x}{27} + \frac {7 \log {\left (3 x + 2 \right )}}{81} \]
[In]
[Out]
none
Time = 0.21 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.73 \[ \int \frac {(1-2 x) (3+5 x)^2}{2+3 x} \, dx=-\frac {50}{9} \, x^{3} - \frac {5}{18} \, x^{2} + \frac {118}{27} \, x + \frac {7}{81} \, \log \left (3 \, x + 2\right ) \]
[In]
[Out]
none
Time = 0.28 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.77 \[ \int \frac {(1-2 x) (3+5 x)^2}{2+3 x} \, dx=-\frac {50}{9} \, x^{3} - \frac {5}{18} \, x^{2} + \frac {118}{27} \, x + \frac {7}{81} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) \]
[In]
[Out]
Time = 0.05 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.67 \[ \int \frac {(1-2 x) (3+5 x)^2}{2+3 x} \, dx=\frac {118\,x}{27}+\frac {7\,\ln \left (x+\frac {2}{3}\right )}{81}-\frac {5\,x^2}{18}-\frac {50\,x^3}{9} \]
[In]
[Out]