Integrand size = 123, antiderivative size = 23 \[ \int \frac {200 x^4+e^{5 x} \left (-200 x^4-1250 x^5\right )+\left (-750 x^4+750 e^{5 x} x^4\right ) \log \left (-1+e^{5 x}\right )+\left (-1250 x^4+1250 e^{5 x} x^4\right ) \log ^2\left (-1+e^{5 x}\right )}{-1+e^{5 x}+\left (10-10 e^{5 x}\right ) \log \left (-1+e^{5 x}\right )+\left (-25+25 e^{5 x}\right ) \log ^2\left (-1+e^{5 x}\right )} \, dx=10 x^4 \left (x+\frac {x}{-\frac {1}{5}+\log \left (-1+e^{5 x}\right )}\right ) \]
Time = 0.74 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \[ \int \frac {200 x^4+e^{5 x} \left (-200 x^4-1250 x^5\right )+\left (-750 x^4+750 e^{5 x} x^4\right ) \log \left (-1+e^{5 x}\right )+\left (-1250 x^4+1250 e^{5 x} x^4\right ) \log ^2\left (-1+e^{5 x}\right )}{-1+e^{5 x}+\left (10-10 e^{5 x}\right ) \log \left (-1+e^{5 x}\right )+\left (-25+25 e^{5 x}\right ) \log ^2\left (-1+e^{5 x}\right )} \, dx=-50 \left (-\frac {x^5}{5}-\frac {x^5}{-1+5 \log \left (-1+e^{5 x}\right )}\right ) \]
Integrate[(200*x^4 + E^(5*x)*(-200*x^4 - 1250*x^5) + (-750*x^4 + 750*E^(5* x)*x^4)*Log[-1 + E^(5*x)] + (-1250*x^4 + 1250*E^(5*x)*x^4)*Log[-1 + E^(5*x )]^2)/(-1 + E^(5*x) + (10 - 10*E^(5*x))*Log[-1 + E^(5*x)] + (-25 + 25*E^(5 *x))*Log[-1 + E^(5*x)]^2),x]
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.
\(\displaystyle \int \frac {200 x^4+\left (1250 e^{5 x} x^4-1250 x^4\right ) \log ^2\left (e^{5 x}-1\right )+\left (750 e^{5 x} x^4-750 x^4\right ) \log \left (e^{5 x}-1\right )+e^{5 x} \left (-1250 x^5-200 x^4\right )}{e^{5 x}+\left (25 e^{5 x}-25\right ) \log ^2\left (e^{5 x}-1\right )+\left (10-10 e^{5 x}\right ) \log \left (e^{5 x}-1\right )-1} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {50 x^4 \left (4 e^{5 x}+25 e^{5 x} x-25 e^{5 x} \log ^2\left (e^{5 x}-1\right )+25 \log ^2\left (e^{5 x}-1\right )-15 e^{5 x} \log \left (e^{5 x}-1\right )+15 \log \left (e^{5 x}-1\right )-4\right )}{\left (1-e^{5 x}\right ) \left (1-5 \log \left (e^{5 x}-1\right )\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 50 \int -\frac {x^4 \left (25 e^{5 x} \log ^2\left (-1+e^{5 x}\right )-25 \log ^2\left (-1+e^{5 x}\right )+15 e^{5 x} \log \left (-1+e^{5 x}\right )-15 \log \left (-1+e^{5 x}\right )-4 e^{5 x}-25 e^{5 x} x+4\right )}{\left (1-e^{5 x}\right ) \left (1-5 \log \left (-1+e^{5 x}\right )\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -50 \int \frac {x^4 \left (25 e^{5 x} \log ^2\left (-1+e^{5 x}\right )-25 \log ^2\left (-1+e^{5 x}\right )+15 e^{5 x} \log \left (-1+e^{5 x}\right )-15 \log \left (-1+e^{5 x}\right )-4 e^{5 x}-25 e^{5 x} x+4\right )}{\left (1-e^{5 x}\right ) \left (1-5 \log \left (-1+e^{5 x}\right )\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -50 \int \left (\frac {5 x^5}{\left (-1+e^x\right ) \left (5 \log \left (-1+e^{5 x}\right )-1\right )^2}-\frac {5 \left (4+3 e^x+2 e^{2 x}+e^{3 x}\right ) x^5}{\left (1+e^x+e^{2 x}+e^{3 x}+e^{4 x}\right ) \left (5 \log \left (-1+e^{5 x}\right )-1\right )^2}+\frac {\left (-25 \log ^2\left (-1+e^{5 x}\right )-15 \log \left (-1+e^{5 x}\right )+25 x+4\right ) x^4}{\left (5 \log \left (-1+e^{5 x}\right )-1\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -50 \left (25 \int \frac {x^5}{\left (5 \log \left (-1+e^{5 x}\right )-1\right )^2}dx+5 \int \frac {x^5}{\left (-1+e^x\right ) \left (5 \log \left (-1+e^{5 x}\right )-1\right )^2}dx-20 \int \frac {x^5}{\left (1+e^x+e^{2 x}+e^{3 x}+e^{4 x}\right ) \left (5 \log \left (-1+e^{5 x}\right )-1\right )^2}dx-15 \int \frac {e^x x^5}{\left (1+e^x+e^{2 x}+e^{3 x}+e^{4 x}\right ) \left (5 \log \left (-1+e^{5 x}\right )-1\right )^2}dx-10 \int \frac {e^{2 x} x^5}{\left (1+e^x+e^{2 x}+e^{3 x}+e^{4 x}\right ) \left (5 \log \left (-1+e^{5 x}\right )-1\right )^2}dx-5 \int \frac {e^{3 x} x^5}{\left (1+e^x+e^{2 x}+e^{3 x}+e^{4 x}\right ) \left (5 \log \left (-1+e^{5 x}\right )-1\right )^2}dx-5 \int \frac {x^4}{5 \log \left (-1+e^{5 x}\right )-1}dx-\frac {x^5}{5}\right )\) |
Int[(200*x^4 + E^(5*x)*(-200*x^4 - 1250*x^5) + (-750*x^4 + 750*E^(5*x)*x^4 )*Log[-1 + E^(5*x)] + (-1250*x^4 + 1250*E^(5*x)*x^4)*Log[-1 + E^(5*x)]^2)/ (-1 + E^(5*x) + (10 - 10*E^(5*x))*Log[-1 + E^(5*x)] + (-25 + 25*E^(5*x))*L og[-1 + E^(5*x)]^2),x]
3.8.12.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 0.71 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09
method | result | size |
risch | \(10 x^{5}+\frac {50 x^{5}}{-1+5 \ln \left ({\mathrm e}^{5 x}-1\right )}\) | \(25\) |
parallelrisch | \(\frac {2500 \ln \left ({\mathrm e}^{5 x}-1\right ) x^{5}+2000 x^{5}}{-50+250 \ln \left ({\mathrm e}^{5 x}-1\right )}\) | \(34\) |
int(((1250*x^4*exp(5*x)-1250*x^4)*ln(exp(5*x)-1)^2+(750*x^4*exp(5*x)-750*x ^4)*ln(exp(5*x)-1)+(-1250*x^5-200*x^4)*exp(5*x)+200*x^4)/((25*exp(5*x)-25) *ln(exp(5*x)-1)^2+(-10*exp(5*x)+10)*ln(exp(5*x)-1)+exp(5*x)-1),x,method=_R ETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.43 \[ \int \frac {200 x^4+e^{5 x} \left (-200 x^4-1250 x^5\right )+\left (-750 x^4+750 e^{5 x} x^4\right ) \log \left (-1+e^{5 x}\right )+\left (-1250 x^4+1250 e^{5 x} x^4\right ) \log ^2\left (-1+e^{5 x}\right )}{-1+e^{5 x}+\left (10-10 e^{5 x}\right ) \log \left (-1+e^{5 x}\right )+\left (-25+25 e^{5 x}\right ) \log ^2\left (-1+e^{5 x}\right )} \, dx=\frac {10 \, {\left (5 \, x^{5} \log \left (e^{\left (5 \, x\right )} - 1\right ) + 4 \, x^{5}\right )}}{5 \, \log \left (e^{\left (5 \, x\right )} - 1\right ) - 1} \]
integrate(((1250*x^4*exp(5*x)-1250*x^4)*log(exp(5*x)-1)^2+(750*x^4*exp(5*x )-750*x^4)*log(exp(5*x)-1)+(-1250*x^5-200*x^4)*exp(5*x)+200*x^4)/((25*exp( 5*x)-25)*log(exp(5*x)-1)^2+(-10*exp(5*x)+10)*log(exp(5*x)-1)+exp(5*x)-1),x , algorithm=\
Time = 0.09 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {200 x^4+e^{5 x} \left (-200 x^4-1250 x^5\right )+\left (-750 x^4+750 e^{5 x} x^4\right ) \log \left (-1+e^{5 x}\right )+\left (-1250 x^4+1250 e^{5 x} x^4\right ) \log ^2\left (-1+e^{5 x}\right )}{-1+e^{5 x}+\left (10-10 e^{5 x}\right ) \log \left (-1+e^{5 x}\right )+\left (-25+25 e^{5 x}\right ) \log ^2\left (-1+e^{5 x}\right )} \, dx=10 x^{5} + \frac {50 x^{5}}{5 \log {\left (e^{5 x} - 1 \right )} - 1} \]
integrate(((1250*x**4*exp(5*x)-1250*x**4)*ln(exp(5*x)-1)**2+(750*x**4*exp( 5*x)-750*x**4)*ln(exp(5*x)-1)+(-1250*x**5-200*x**4)*exp(5*x)+200*x**4)/((2 5*exp(5*x)-25)*ln(exp(5*x)-1)**2+(-10*exp(5*x)+10)*ln(exp(5*x)-1)+exp(5*x) -1),x)
Leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (23) = 46\).
Time = 0.36 (sec) , antiderivative size = 70, normalized size of antiderivative = 3.04 \[ \int \frac {200 x^4+e^{5 x} \left (-200 x^4-1250 x^5\right )+\left (-750 x^4+750 e^{5 x} x^4\right ) \log \left (-1+e^{5 x}\right )+\left (-1250 x^4+1250 e^{5 x} x^4\right ) \log ^2\left (-1+e^{5 x}\right )}{-1+e^{5 x}+\left (10-10 e^{5 x}\right ) \log \left (-1+e^{5 x}\right )+\left (-25+25 e^{5 x}\right ) \log ^2\left (-1+e^{5 x}\right )} \, dx=\frac {10 \, {\left (5 \, x^{5} \log \left (e^{\left (4 \, x\right )} + e^{\left (3 \, x\right )} + e^{\left (2 \, x\right )} + e^{x} + 1\right ) + 5 \, x^{5} \log \left (e^{x} - 1\right ) + 4 \, x^{5}\right )}}{5 \, \log \left (e^{\left (4 \, x\right )} + e^{\left (3 \, x\right )} + e^{\left (2 \, x\right )} + e^{x} + 1\right ) + 5 \, \log \left (e^{x} - 1\right ) - 1} \]
integrate(((1250*x^4*exp(5*x)-1250*x^4)*log(exp(5*x)-1)^2+(750*x^4*exp(5*x )-750*x^4)*log(exp(5*x)-1)+(-1250*x^5-200*x^4)*exp(5*x)+200*x^4)/((25*exp( 5*x)-25)*log(exp(5*x)-1)^2+(-10*exp(5*x)+10)*log(exp(5*x)-1)+exp(5*x)-1),x , algorithm=\
10*(5*x^5*log(e^(4*x) + e^(3*x) + e^(2*x) + e^x + 1) + 5*x^5*log(e^x - 1) + 4*x^5)/(5*log(e^(4*x) + e^(3*x) + e^(2*x) + e^x + 1) + 5*log(e^x - 1) - 1)
Time = 0.36 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.43 \[ \int \frac {200 x^4+e^{5 x} \left (-200 x^4-1250 x^5\right )+\left (-750 x^4+750 e^{5 x} x^4\right ) \log \left (-1+e^{5 x}\right )+\left (-1250 x^4+1250 e^{5 x} x^4\right ) \log ^2\left (-1+e^{5 x}\right )}{-1+e^{5 x}+\left (10-10 e^{5 x}\right ) \log \left (-1+e^{5 x}\right )+\left (-25+25 e^{5 x}\right ) \log ^2\left (-1+e^{5 x}\right )} \, dx=\frac {10 \, {\left (5 \, x^{5} \log \left (e^{\left (5 \, x\right )} - 1\right ) + 4 \, x^{5}\right )}}{5 \, \log \left (e^{\left (5 \, x\right )} - 1\right ) - 1} \]
integrate(((1250*x^4*exp(5*x)-1250*x^4)*log(exp(5*x)-1)^2+(750*x^4*exp(5*x )-750*x^4)*log(exp(5*x)-1)+(-1250*x^5-200*x^4)*exp(5*x)+200*x^4)/((25*exp( 5*x)-25)*log(exp(5*x)-1)^2+(-10*exp(5*x)+10)*log(exp(5*x)-1)+exp(5*x)-1),x , algorithm=\
Time = 8.73 (sec) , antiderivative size = 79, normalized size of antiderivative = 3.43 \[ \int \frac {200 x^4+e^{5 x} \left (-200 x^4-1250 x^5\right )+\left (-750 x^4+750 e^{5 x} x^4\right ) \log \left (-1+e^{5 x}\right )+\left (-1250 x^4+1250 e^{5 x} x^4\right ) \log ^2\left (-1+e^{5 x}\right )}{-1+e^{5 x}+\left (10-10 e^{5 x}\right ) \log \left (-1+e^{5 x}\right )+\left (-25+25 e^{5 x}\right ) \log ^2\left (-1+e^{5 x}\right )} \, dx=\frac {10\,x^4\,{\mathrm {e}}^{-5\,x}\,\left ({\mathrm {e}}^{5\,x}+5\,x\,{\mathrm {e}}^{5\,x}-1\right )-50\,x^4\,\ln \left ({\mathrm {e}}^{5\,x}-1\right )\,{\mathrm {e}}^{-5\,x}\,\left ({\mathrm {e}}^{5\,x}-1\right )}{5\,\ln \left ({\mathrm {e}}^{5\,x}-1\right )-1}-10\,x^4\,{\mathrm {e}}^{-5\,x}+10\,x^4+10\,x^5 \]
int((log(exp(5*x) - 1)*(750*x^4*exp(5*x) - 750*x^4) - exp(5*x)*(200*x^4 + 1250*x^5) + log(exp(5*x) - 1)^2*(1250*x^4*exp(5*x) - 1250*x^4) + 200*x^4)/ (exp(5*x) - log(exp(5*x) - 1)*(10*exp(5*x) - 10) + log(exp(5*x) - 1)^2*(25 *exp(5*x) - 25) - 1),x)