Integrand size = 53, antiderivative size = 21 \[ \int \frac {-8+24 x-24 x^2+8 x^3+(-8+8 x) \log \left (5 e^x\right )-8 \log ^2\left (5 e^x\right )}{-1+3 x-3 x^2+x^3} \, dx=4 \left (4+2 x+\frac {\log ^2\left (5 e^x\right )}{(-1+x)^2}\right ) \]
[Out]
Time = 0.07 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {6820, 2199, 31} \[ \int \frac {-8+24 x-24 x^2+8 x^3+(-8+8 x) \log \left (5 e^x\right )-8 \log ^2\left (5 e^x\right )}{-1+3 x-3 x^2+x^3} \, dx=8 x+\frac {4 \log ^2\left (5 e^x\right )}{(1-x)^2} \]
[In]
[Out]
Rule 31
Rule 2199
Rule 6820
Rubi steps \begin{align*} \text {integral}& = \int \left (8+\frac {8 \log \left (5 e^x\right )}{(-1+x)^2}-\frac {8 \log ^2\left (5 e^x\right )}{(-1+x)^3}\right ) \, dx \\ & = 8 x+8 \int \frac {\log \left (5 e^x\right )}{(-1+x)^2} \, dx-8 \int \frac {\log ^2\left (5 e^x\right )}{(-1+x)^3} \, dx \\ & = 8 x+\frac {8 \log \left (5 e^x\right )}{1-x}+\frac {4 \log ^2\left (5 e^x\right )}{(1-x)^2}+8 \int \frac {1}{-1+x} \, dx-8 \int \frac {\log \left (5 e^x\right )}{(-1+x)^2} \, dx \\ & = 8 x+\frac {4 \log ^2\left (5 e^x\right )}{(1-x)^2}+8 \log (1-x)-8 \int \frac {1}{-1+x} \, dx \\ & = 8 x+\frac {4 \log ^2\left (5 e^x\right )}{(1-x)^2} \\ \end{align*}
Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {-8+24 x-24 x^2+8 x^3+(-8+8 x) \log \left (5 e^x\right )-8 \log ^2\left (5 e^x\right )}{-1+3 x-3 x^2+x^3} \, dx=-12+8 x+\frac {4 \log ^2\left (5 e^x\right )}{(-1+x)^2} \]
[In]
[Out]
Time = 0.09 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90
method | result | size |
default | \(8 x +\frac {4 \ln \left (5 \,{\mathrm e}^{x}\right )^{2}}{\left (-1+x \right )^{2}}\) | \(19\) |
parts | \(8 x +\frac {4 \ln \left (5 \,{\mathrm e}^{x}\right )^{2}}{\left (-1+x \right )^{2}}\) | \(19\) |
parallelrisch | \(\frac {16+8 x^{3}+4 \ln \left (5 \,{\mathrm e}^{x}\right )^{2}-24 x}{x^{2}-2 x +1}\) | \(31\) |
risch | \(\frac {4 \ln \left ({\mathrm e}^{x}\right )^{2}}{x^{2}-2 x +1}+\frac {8 \ln \left (5\right ) \ln \left ({\mathrm e}^{x}\right )}{x^{2}-2 x +1}+\frac {8 x^{3}+4 \ln \left (5\right )^{2}-16 x^{2}+8 x}{x^{2}-2 x +1}\) | \(67\) |
[In]
[Out]
none
Time = 0.26 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.71 \[ \int \frac {-8+24 x-24 x^2+8 x^3+(-8+8 x) \log \left (5 e^x\right )-8 \log ^2\left (5 e^x\right )}{-1+3 x-3 x^2+x^3} \, dx=\frac {4 \, {\left (2 \, x^{3} - 4 \, x^{2} + 2 \, x \log \left (5\right ) + \log \left (5\right )^{2} + 4 \, x - 1\right )}}{x^{2} - 2 \, x + 1} \]
[In]
[Out]
Time = 0.17 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.29 \[ \int \frac {-8+24 x-24 x^2+8 x^3+(-8+8 x) \log \left (5 e^x\right )-8 \log ^2\left (5 e^x\right )}{-1+3 x-3 x^2+x^3} \, dx=8 x + \frac {x \left (8 + 8 \log {\left (5 \right )}\right ) - 4 + 4 \log {\left (5 \right )}^{2}}{x^{2} - 2 x + 1} \]
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. 151 vs. \(2 (19) = 38\).
Time = 0.19 (sec) , antiderivative size = 151, normalized size of antiderivative = 7.19 \[ \int \frac {-8+24 x-24 x^2+8 x^3+(-8+8 x) \log \left (5 e^x\right )-8 \log ^2\left (5 e^x\right )}{-1+3 x-3 x^2+x^3} \, dx=8 \, x - \frac {4 \, {\left (2 \, x - 1\right )} \log \left (5 \, e^{x}\right )}{x^{2} - 2 \, x + 1} + \frac {4 \, \log \left (5 \, e^{x}\right )^{2}}{x^{2} - 2 \, x + 1} - \frac {4 \, {\left (6 \, x - 5\right )}}{x^{2} - 2 \, x + 1} + \frac {12 \, {\left (4 \, x - 3\right )}}{x^{2} - 2 \, x + 1} - \frac {12 \, {\left (2 \, x - 1\right )}}{x^{2} - 2 \, x + 1} + \frac {4 \, \log \left (5 \, e^{x}\right )}{x^{2} - 2 \, x + 1} + \frac {8 \, \log \left (5 \, e^{x}\right )}{x - 1} + \frac {4}{x^{2} - 2 \, x + 1} - 8 \, \log \left (2 \, x - 2\right ) + 8 \, \log \left (x - 1\right ) \]
[In]
[Out]
none
Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.19 \[ \int \frac {-8+24 x-24 x^2+8 x^3+(-8+8 x) \log \left (5 e^x\right )-8 \log ^2\left (5 e^x\right )}{-1+3 x-3 x^2+x^3} \, dx=8 \, x + \frac {4 \, {\left (2 \, x \log \left (5\right ) + \log \left (5\right )^{2} + 2 \, x - 1\right )}}{{\left (x - 1\right )}^{2}} \]
[In]
[Out]
Time = 11.08 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.38 \[ \int \frac {-8+24 x-24 x^2+8 x^3+(-8+8 x) \log \left (5 e^x\right )-8 \log ^2\left (5 e^x\right )}{-1+3 x-3 x^2+x^3} \, dx=8\,x+\frac {8\,\ln \left (5\right )+8}{x-1}+\frac {4\,{\left (\ln \left (5\right )+1\right )}^2}{{\left (x-1\right )}^2} \]
[In]
[Out]