Integrand size = 15, antiderivative size = 38 \[ \int \frac {(1-2 x)^3}{(3+5 x)^3} \, dx=-\frac {8 x}{125}-\frac {1331}{1250 (3+5 x)^2}+\frac {726}{625 (3+5 x)}+\frac {132}{625} \log (3+5 x) \]
[Out]
Time = 0.01 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {45} \[ \int \frac {(1-2 x)^3}{(3+5 x)^3} \, dx=-\frac {8 x}{125}+\frac {726}{625 (5 x+3)}-\frac {1331}{1250 (5 x+3)^2}+\frac {132}{625} \log (5 x+3) \]
[In]
[Out]
Rule 45
Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {8}{125}+\frac {1331}{125 (3+5 x)^3}-\frac {726}{125 (3+5 x)^2}+\frac {132}{125 (3+5 x)}\right ) \, dx \\ & = -\frac {8 x}{125}-\frac {1331}{1250 (3+5 x)^2}+\frac {726}{625 (3+5 x)}+\frac {132}{625} \log (3+5 x) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.97 \[ \int \frac {(1-2 x)^3}{(3+5 x)^3} \, dx=\frac {\frac {5 \left (677+1548 x-280 x^2-400 x^3\right )}{(3+5 x)^2}+264 \log (6+10 x)}{1250} \]
[In]
[Out]
Time = 2.47 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.71
method | result | size |
risch | \(-\frac {8 x}{125}+\frac {\frac {726 x}{125}+\frac {121}{50}}{\left (3+5 x \right )^{2}}+\frac {132 \ln \left (3+5 x \right )}{625}\) | \(27\) |
default | \(-\frac {8 x}{125}-\frac {1331}{1250 \left (3+5 x \right )^{2}}+\frac {726}{625 \left (3+5 x \right )}+\frac {132 \ln \left (3+5 x \right )}{625}\) | \(31\) |
norman | \(\frac {-\frac {1063}{375} x -\frac {3889}{450} x^{2}-\frac {8}{5} x^{3}}{\left (3+5 x \right )^{2}}+\frac {132 \ln \left (3+5 x \right )}{625}\) | \(32\) |
parallelrisch | \(\frac {59400 \ln \left (x +\frac {3}{5}\right ) x^{2}-18000 x^{3}+71280 \ln \left (x +\frac {3}{5}\right ) x -97225 x^{2}+21384 \ln \left (x +\frac {3}{5}\right )-31890 x}{11250 \left (3+5 x \right )^{2}}\) | \(46\) |
meijerg | \(\frac {x \left (\frac {5 x}{3}+2\right )}{54 \left (1+\frac {5 x}{3}\right )^{2}}-\frac {x^{2}}{9 \left (1+\frac {5 x}{3}\right )^{2}}-\frac {2 x \left (15 x +6\right )}{75 \left (1+\frac {5 x}{3}\right )^{2}}+\frac {132 \ln \left (1+\frac {5 x}{3}\right )}{625}-\frac {2 x \left (\frac {100}{9} x^{2}+30 x +12\right )}{125 \left (1+\frac {5 x}{3}\right )^{2}}\) | \(72\) |
[In]
[Out]
none
Time = 0.22 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.24 \[ \int \frac {(1-2 x)^3}{(3+5 x)^3} \, dx=-\frac {2000 \, x^{3} + 2400 \, x^{2} - 264 \, {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (5 \, x + 3\right ) - 6540 \, x - 3025}{1250 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \]
[In]
[Out]
Time = 0.05 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.82 \[ \int \frac {(1-2 x)^3}{(3+5 x)^3} \, dx=- \frac {8 x}{125} - \frac {- 1452 x - 605}{6250 x^{2} + 7500 x + 2250} + \frac {132 \log {\left (5 x + 3 \right )}}{625} \]
[In]
[Out]
none
Time = 0.25 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.82 \[ \int \frac {(1-2 x)^3}{(3+5 x)^3} \, dx=-\frac {8}{125} \, x + \frac {121 \, {\left (12 \, x + 5\right )}}{250 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} + \frac {132}{625} \, \log \left (5 \, x + 3\right ) \]
[In]
[Out]
none
Time = 0.28 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.71 \[ \int \frac {(1-2 x)^3}{(3+5 x)^3} \, dx=-\frac {8}{125} \, x + \frac {121 \, {\left (12 \, x + 5\right )}}{250 \, {\left (5 \, x + 3\right )}^{2}} + \frac {132}{625} \, \log \left ({\left | 5 \, x + 3 \right |}\right ) \]
[In]
[Out]
Time = 1.36 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.68 \[ \int \frac {(1-2 x)^3}{(3+5 x)^3} \, dx=\frac {132\,\ln \left (x+\frac {3}{5}\right )}{625}-\frac {8\,x}{125}+\frac {\frac {726\,x}{3125}+\frac {121}{1250}}{x^2+\frac {6\,x}{5}+\frac {9}{25}} \]
[In]
[Out]