Integrand size = 30, antiderivative size = 16 \[ \int \frac {32-16 x+16 x \log (x)+18 x^2 \log ^2(x)}{9 x \log ^2(x)} \, dx=3+x^2+\frac {16 (-2+x)}{9 \log (x)} \]
[Out]
Time = 0.17 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.31, number of steps used = 12, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {12, 6873, 6874, 2395, 2334, 2335, 2339, 30} \[ \int \frac {32-16 x+16 x \log (x)+18 x^2 \log ^2(x)}{9 x \log ^2(x)} \, dx=x^2+\frac {16 x}{9 \log (x)}-\frac {32}{9 \log (x)} \]
[In]
[Out]
Rule 12
Rule 30
Rule 2334
Rule 2335
Rule 2339
Rule 2395
Rule 6873
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \frac {1}{9} \int \frac {32-16 x+16 x \log (x)+18 x^2 \log ^2(x)}{x \log ^2(x)} \, dx \\ & = \frac {1}{9} \int \frac {2 \left (16-8 x+8 x \log (x)+9 x^2 \log ^2(x)\right )}{x \log ^2(x)} \, dx \\ & = \frac {2}{9} \int \frac {16-8 x+8 x \log (x)+9 x^2 \log ^2(x)}{x \log ^2(x)} \, dx \\ & = \frac {2}{9} \int \left (9 x-\frac {8 (-2+x)}{x \log ^2(x)}+\frac {8}{\log (x)}\right ) \, dx \\ & = x^2-\frac {16}{9} \int \frac {-2+x}{x \log ^2(x)} \, dx+\frac {16}{9} \int \frac {1}{\log (x)} \, dx \\ & = x^2+\frac {16 \text {li}(x)}{9}-\frac {16}{9} \int \left (\frac {1}{\log ^2(x)}-\frac {2}{x \log ^2(x)}\right ) \, dx \\ & = x^2+\frac {16 \text {li}(x)}{9}-\frac {16}{9} \int \frac {1}{\log ^2(x)} \, dx+\frac {32}{9} \int \frac {1}{x \log ^2(x)} \, dx \\ & = x^2+\frac {16 x}{9 \log (x)}+\frac {16 \text {li}(x)}{9}-\frac {16}{9} \int \frac {1}{\log (x)} \, dx+\frac {32}{9} \text {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (x)\right ) \\ & = x^2-\frac {32}{9 \log (x)}+\frac {16 x}{9 \log (x)} \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.31 \[ \int \frac {32-16 x+16 x \log (x)+18 x^2 \log ^2(x)}{9 x \log ^2(x)} \, dx=x^2-\frac {32}{9 \log (x)}+\frac {16 x}{9 \log (x)} \]
[In]
[Out]
Time = 0.04 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88
method | result | size |
risch | \(x^{2}+\frac {-\frac {32}{9}+\frac {16 x}{9}}{\ln \left (x \right )}\) | \(14\) |
norman | \(\frac {-\frac {32}{9}+x^{2} \ln \left (x \right )+\frac {16 x}{9}}{\ln \left (x \right )}\) | \(17\) |
default | \(x^{2}+\frac {16 x}{9 \ln \left (x \right )}-\frac {32}{9 \ln \left (x \right )}\) | \(18\) |
parts | \(x^{2}+\frac {16 x}{9 \ln \left (x \right )}-\frac {32}{9 \ln \left (x \right )}\) | \(18\) |
parallelrisch | \(\frac {9 x^{2} \ln \left (x \right )-32+16 x}{9 \ln \left (x \right )}\) | \(19\) |
[In]
[Out]
none
Time = 0.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.12 \[ \int \frac {32-16 x+16 x \log (x)+18 x^2 \log ^2(x)}{9 x \log ^2(x)} \, dx=\frac {9 \, x^{2} \log \left (x\right ) + 16 \, x - 32}{9 \, \log \left (x\right )} \]
[In]
[Out]
Time = 0.04 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {32-16 x+16 x \log (x)+18 x^2 \log ^2(x)}{9 x \log ^2(x)} \, dx=x^{2} + \frac {16 x - 32}{9 \log {\left (x \right )}} \]
[In]
[Out]
Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.22 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.44 \[ \int \frac {32-16 x+16 x \log (x)+18 x^2 \log ^2(x)}{9 x \log ^2(x)} \, dx=x^{2} - \frac {32}{9 \, \log \left (x\right )} + \frac {16}{9} \, {\rm Ei}\left (\log \left (x\right )\right ) - \frac {16}{9} \, \Gamma \left (-1, -\log \left (x\right )\right ) \]
[In]
[Out]
none
Time = 0.27 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.81 \[ \int \frac {32-16 x+16 x \log (x)+18 x^2 \log ^2(x)}{9 x \log ^2(x)} \, dx=x^{2} + \frac {16 \, {\left (x - 2\right )}}{9 \, \log \left (x\right )} \]
[In]
[Out]
Time = 11.93 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {32-16 x+16 x \log (x)+18 x^2 \log ^2(x)}{9 x \log ^2(x)} \, dx=x^2+\frac {\frac {16\,x}{9}-\frac {32}{9}}{\ln \left (x\right )} \]
[In]
[Out]