Integrand size = 73, antiderivative size = 24 \[ \int \frac {-5+10 x+20 x^2+e^{4 x} (1-2 x) \log (5)+\left (10 x-2 e^{4 x} x \log (5)\right ) \log \left (\frac {e^{4 x} x^2}{-5+e^{4 x} \log (5)}\right )}{-5+e^{4 x} \log (5)} \, dx=x-x^2 \log \left (\frac {x^2}{-5 e^{-4 x}+\log (5)}\right ) \]
Time = 0.21 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {-5+10 x+20 x^2+e^{4 x} (1-2 x) \log (5)+\left (10 x-2 e^{4 x} x \log (5)\right ) \log \left (\frac {e^{4 x} x^2}{-5+e^{4 x} \log (5)}\right )}{-5+e^{4 x} \log (5)} \, dx=x-x^2 \log \left (-\frac {e^{4 x} x^2}{5-e^{4 x} \log (5)}\right ) \]
Integrate[(-5 + 10*x + 20*x^2 + E^(4*x)*(1 - 2*x)*Log[5] + (10*x - 2*E^(4* x)*x*Log[5])*Log[(E^(4*x)*x^2)/(-5 + E^(4*x)*Log[5])])/(-5 + E^(4*x)*Log[5 ]),x]
Time = 0.78 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.027, Rules used = {7293, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {20 x^2+\left (10 x-2 e^{4 x} x \log (5)\right ) \log \left (\frac {e^{4 x} x^2}{e^{4 x} \log (5)-5}\right )+10 x+e^{4 x} (1-2 x) \log (5)-5}{e^{4 x} \log (5)-5} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {20 x^2}{e^{4 x} \log (5)-5}-2 x \log \left (\frac {e^{4 x} x^2}{e^{4 x} \log (5)-5}\right )-2 x+1\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle x-x^2 \log \left (-\frac {e^{4 x} x^2}{5-e^{4 x} \log (5)}\right )\) |
Int[(-5 + 10*x + 20*x^2 + E^(4*x)*(1 - 2*x)*Log[5] + (10*x - 2*E^(4*x)*x*L og[5])*Log[(E^(4*x)*x^2)/(-5 + E^(4*x)*Log[5])])/(-5 + E^(4*x)*Log[5]),x]
3.22.57.3.1 Defintions of rubi rules used
Time = 0.49 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17
method | result | size |
norman | \(x -x^{2} \ln \left (\frac {x^{2} {\mathrm e}^{4 x}}{\ln \left (5\right ) {\mathrm e}^{4 x}-5}\right )\) | \(28\) |
parallelrisch | \(x -x^{2} \ln \left (\frac {x^{2} {\mathrm e}^{4 x}}{\ln \left (5\right ) {\mathrm e}^{4 x}-5}\right )\) | \(28\) |
risch | \(-x^{2} \ln \left ({\mathrm e}^{4 x}\right )+x^{2} \ln \left (\ln \left (5\right ) {\mathrm e}^{4 x}-5\right )-2 x^{2} \ln \left (x \right )+\frac {i \pi \,x^{2} \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )}{2}-i \pi \,x^{2} \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+\frac {i \pi \,x^{2} \operatorname {csgn}\left (i x^{2}\right )^{3}}{2}+\frac {i \pi \,x^{2} \operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{4 x}}{\ln \left (5\right ) {\mathrm e}^{4 x}-5}\right ) \operatorname {csgn}\left (\frac {i x^{2} {\mathrm e}^{4 x}}{\ln \left (5\right ) {\mathrm e}^{4 x}-5}\right )}{2}-\frac {i \pi \,x^{2} \operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (\frac {i x^{2} {\mathrm e}^{4 x}}{\ln \left (5\right ) {\mathrm e}^{4 x}-5}\right )^{2}}{2}+\frac {i \pi \,x^{2} \operatorname {csgn}\left (\frac {i}{\ln \left (5\right ) {\mathrm e}^{4 x}-5}\right ) \operatorname {csgn}\left (i {\mathrm e}^{4 x}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{4 x}}{\ln \left (5\right ) {\mathrm e}^{4 x}-5}\right )}{2}-\frac {i \pi \,x^{2} \operatorname {csgn}\left (\frac {i}{\ln \left (5\right ) {\mathrm e}^{4 x}-5}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{4 x}}{\ln \left (5\right ) {\mathrm e}^{4 x}-5}\right )^{2}}{2}-\frac {i \pi \,x^{2} \operatorname {csgn}\left (i {\mathrm e}^{4 x}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{4 x}}{\ln \left (5\right ) {\mathrm e}^{4 x}-5}\right )^{2}}{2}+\frac {i \pi \,x^{2} \operatorname {csgn}\left (\frac {i {\mathrm e}^{4 x}}{\ln \left (5\right ) {\mathrm e}^{4 x}-5}\right )^{3}}{2}-\frac {i \pi \,x^{2} \operatorname {csgn}\left (\frac {i {\mathrm e}^{4 x}}{\ln \left (5\right ) {\mathrm e}^{4 x}-5}\right ) \operatorname {csgn}\left (\frac {i x^{2} {\mathrm e}^{4 x}}{\ln \left (5\right ) {\mathrm e}^{4 x}-5}\right )^{2}}{2}+\frac {i \pi \,x^{2} \operatorname {csgn}\left (\frac {i x^{2} {\mathrm e}^{4 x}}{\ln \left (5\right ) {\mathrm e}^{4 x}-5}\right )^{3}}{2}+x\) | \(422\) |
int(((-2*x*ln(5)*exp(4*x)+10*x)*ln(x^2*exp(4*x)/(ln(5)*exp(4*x)-5))+(1-2*x )*ln(5)*exp(4*x)+20*x^2+10*x-5)/(ln(5)*exp(4*x)-5),x,method=_RETURNVERBOSE )
Time = 0.29 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {-5+10 x+20 x^2+e^{4 x} (1-2 x) \log (5)+\left (10 x-2 e^{4 x} x \log (5)\right ) \log \left (\frac {e^{4 x} x^2}{-5+e^{4 x} \log (5)}\right )}{-5+e^{4 x} \log (5)} \, dx=-x^{2} \log \left (\frac {x^{2} e^{\left (4 \, x\right )}}{e^{\left (4 \, x\right )} \log \left (5\right ) - 5}\right ) + x \]
integrate(((-2*x*log(5)*exp(4*x)+10*x)*log(x^2*exp(4*x)/(log(5)*exp(4*x)-5 ))+(1-2*x)*log(5)*exp(4*x)+20*x^2+10*x-5)/(log(5)*exp(4*x)-5),x, algorithm =\
Time = 0.15 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {-5+10 x+20 x^2+e^{4 x} (1-2 x) \log (5)+\left (10 x-2 e^{4 x} x \log (5)\right ) \log \left (\frac {e^{4 x} x^2}{-5+e^{4 x} \log (5)}\right )}{-5+e^{4 x} \log (5)} \, dx=- x^{2} \log {\left (\frac {x^{2} e^{4 x}}{e^{4 x} \log {\left (5 \right )} - 5} \right )} + x \]
integrate(((-2*x*ln(5)*exp(4*x)+10*x)*ln(x**2*exp(4*x)/(ln(5)*exp(4*x)-5)) +(1-2*x)*ln(5)*exp(4*x)+20*x**2+10*x-5)/(ln(5)*exp(4*x)-5),x)
Time = 0.35 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.88 \[ \int \frac {-5+10 x+20 x^2+e^{4 x} (1-2 x) \log (5)+\left (10 x-2 e^{4 x} x \log (5)\right ) \log \left (\frac {e^{4 x} x^2}{-5+e^{4 x} \log (5)}\right )}{-5+e^{4 x} \log (5)} \, dx=-4 \, x^{3} - 2 \, x^{2} \log \left (x\right ) + \frac {1}{4} \, {\left (4 \, x^{2} + 1\right )} \log \left (e^{\left (4 \, x\right )} \log \left (5\right ) - 5\right ) + x - \frac {1}{4} \, \log \left (e^{\left (4 \, x\right )} \log \left (5\right ) - 5\right ) \]
integrate(((-2*x*log(5)*exp(4*x)+10*x)*log(x^2*exp(4*x)/(log(5)*exp(4*x)-5 ))+(1-2*x)*log(5)*exp(4*x)+20*x^2+10*x-5)/(log(5)*exp(4*x)-5),x, algorithm =\
-4*x^3 - 2*x^2*log(x) + 1/4*(4*x^2 + 1)*log(e^(4*x)*log(5) - 5) + x - 1/4* log(e^(4*x)*log(5) - 5)
Time = 0.29 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25 \[ \int \frac {-5+10 x+20 x^2+e^{4 x} (1-2 x) \log (5)+\left (10 x-2 e^{4 x} x \log (5)\right ) \log \left (\frac {e^{4 x} x^2}{-5+e^{4 x} \log (5)}\right )}{-5+e^{4 x} \log (5)} \, dx=-4 \, x^{3} - x^{2} \log \left (x^{2}\right ) + x^{2} \log \left (e^{\left (4 \, x\right )} \log \left (5\right ) - 5\right ) + x \]
integrate(((-2*x*log(5)*exp(4*x)+10*x)*log(x^2*exp(4*x)/(log(5)*exp(4*x)-5 ))+(1-2*x)*log(5)*exp(4*x)+20*x^2+10*x-5)/(log(5)*exp(4*x)-5),x, algorithm =\
Timed out. \[ \int \frac {-5+10 x+20 x^2+e^{4 x} (1-2 x) \log (5)+\left (10 x-2 e^{4 x} x \log (5)\right ) \log \left (\frac {e^{4 x} x^2}{-5+e^{4 x} \log (5)}\right )}{-5+e^{4 x} \log (5)} \, dx=\int \frac {10\,x+\ln \left (\frac {x^2\,{\mathrm {e}}^{4\,x}}{{\mathrm {e}}^{4\,x}\,\ln \left (5\right )-5}\right )\,\left (10\,x-2\,x\,{\mathrm {e}}^{4\,x}\,\ln \left (5\right )\right )+20\,x^2-{\mathrm {e}}^{4\,x}\,\ln \left (5\right )\,\left (2\,x-1\right )-5}{{\mathrm {e}}^{4\,x}\,\ln \left (5\right )-5} \,d x \]
int((10*x + log((x^2*exp(4*x))/(exp(4*x)*log(5) - 5))*(10*x - 2*x*exp(4*x) *log(5)) + 20*x^2 - exp(4*x)*log(5)*(2*x - 1) - 5)/(exp(4*x)*log(5) - 5),x )