Integrand size = 33, antiderivative size = 18 \[ \int \left (-2-2 x+10 x^2+4 x^3+\left (-2-4 x+30 x^2+16 x^3\right ) \log (x)\right ) \, dx=x (-1+2 x) (2+x (6+2 x)) \log (x) \]
[Out]
Time = 0.03 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.50, number of steps used = 7, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {2403, 2332, 2341} \[ \int \left (-2-2 x+10 x^2+4 x^3+\left (-2-4 x+30 x^2+16 x^3\right ) \log (x)\right ) \, dx=4 x^4 \log (x)+10 x^3 \log (x)-2 x^2 \log (x)-2 x \log (x) \]
[In]
[Out]
Rule 2332
Rule 2341
Rule 2403
Rubi steps \begin{align*} \text {integral}& = -2 x-x^2+\frac {10 x^3}{3}+x^4+\int \left (-2-4 x+30 x^2+16 x^3\right ) \log (x) \, dx \\ & = -2 x-x^2+\frac {10 x^3}{3}+x^4+\int \left (-2 \log (x)-4 x \log (x)+30 x^2 \log (x)+16 x^3 \log (x)\right ) \, dx \\ & = -2 x-x^2+\frac {10 x^3}{3}+x^4-2 \int \log (x) \, dx-4 \int x \log (x) \, dx+16 \int x^3 \log (x) \, dx+30 \int x^2 \log (x) \, dx \\ & = -2 x \log (x)-2 x^2 \log (x)+10 x^3 \log (x)+4 x^4 \log (x) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.50 \[ \int \left (-2-2 x+10 x^2+4 x^3+\left (-2-4 x+30 x^2+16 x^3\right ) \log (x)\right ) \, dx=-2 x \log (x)-2 x^2 \log (x)+10 x^3 \log (x)+4 x^4 \log (x) \]
[In]
[Out]
Time = 0.05 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.28
method | result | size |
risch | \(\left (4 x^{4}+10 x^{3}-2 x^{2}-2 x \right ) \ln \left (x \right )\) | \(23\) |
default | \(4 x^{4} \ln \left (x \right )+10 x^{3} \ln \left (x \right )-2 x^{2} \ln \left (x \right )-2 x \ln \left (x \right )\) | \(28\) |
norman | \(4 x^{4} \ln \left (x \right )+10 x^{3} \ln \left (x \right )-2 x^{2} \ln \left (x \right )-2 x \ln \left (x \right )\) | \(28\) |
parallelrisch | \(4 x^{4} \ln \left (x \right )+10 x^{3} \ln \left (x \right )-2 x^{2} \ln \left (x \right )-2 x \ln \left (x \right )\) | \(28\) |
parts | \(4 x^{4} \ln \left (x \right )+10 x^{3} \ln \left (x \right )-2 x^{2} \ln \left (x \right )-2 x \ln \left (x \right )\) | \(28\) |
[In]
[Out]
none
Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.28 \[ \int \left (-2-2 x+10 x^2+4 x^3+\left (-2-4 x+30 x^2+16 x^3\right ) \log (x)\right ) \, dx=2 \, {\left (2 \, x^{4} + 5 \, x^{3} - x^{2} - x\right )} \log \left (x\right ) \]
[In]
[Out]
Time = 0.05 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int \left (-2-2 x+10 x^2+4 x^3+\left (-2-4 x+30 x^2+16 x^3\right ) \log (x)\right ) \, dx=\left (4 x^{4} + 10 x^{3} - 2 x^{2} - 2 x\right ) \log {\left (x \right )} \]
[In]
[Out]
none
Time = 0.18 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.28 \[ \int \left (-2-2 x+10 x^2+4 x^3+\left (-2-4 x+30 x^2+16 x^3\right ) \log (x)\right ) \, dx=2 \, {\left (2 \, x^{4} + 5 \, x^{3} - x^{2} - x\right )} \log \left (x\right ) \]
[In]
[Out]
none
Time = 0.30 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.50 \[ \int \left (-2-2 x+10 x^2+4 x^3+\left (-2-4 x+30 x^2+16 x^3\right ) \log (x)\right ) \, dx=4 \, x^{4} \log \left (x\right ) + 10 \, x^{3} \log \left (x\right ) - 2 \, x^{2} \log \left (x\right ) - 2 \, x \log \left (x\right ) \]
[In]
[Out]
Time = 13.42 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \left (-2-2 x+10 x^2+4 x^3+\left (-2-4 x+30 x^2+16 x^3\right ) \log (x)\right ) \, dx=2\,x\,\ln \left (x\right )\,\left (2\,x-1\right )\,\left (x^2+3\,x+1\right ) \]
[In]
[Out]