Integrand size = 95, antiderivative size = 24 \[ \int \frac {5 e^x \left (1+x-2 x^2\right )+5 e^x \left (3 x-2 x^2\right ) \log \left (\frac {5}{x}\right )}{4 x-16 x^2+16 x^3+\left (8 x-32 x^2+32 x^3\right ) \log \left (\frac {5}{x}\right )+\left (4 x-16 x^2+16 x^3\right ) \log ^2\left (\frac {5}{x}\right )} \, dx=\frac {5 e^x}{(1-2 x) \left (4+4 \log \left (\frac {5}{x}\right )\right )} \]
Time = 0.38 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {5 e^x \left (1+x-2 x^2\right )+5 e^x \left (3 x-2 x^2\right ) \log \left (\frac {5}{x}\right )}{4 x-16 x^2+16 x^3+\left (8 x-32 x^2+32 x^3\right ) \log \left (\frac {5}{x}\right )+\left (4 x-16 x^2+16 x^3\right ) \log ^2\left (\frac {5}{x}\right )} \, dx=-\frac {5 e^x}{4 (-1+2 x) \left (1+\log \left (\frac {5}{x}\right )\right )} \]
Integrate[(5*E^x*(1 + x - 2*x^2) + 5*E^x*(3*x - 2*x^2)*Log[5/x])/(4*x - 16 *x^2 + 16*x^3 + (8*x - 32*x^2 + 32*x^3)*Log[5/x] + (4*x - 16*x^2 + 16*x^3) *Log[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.
\(\displaystyle \int \frac {5 e^x \left (-2 x^2+x+1\right )+5 e^x \left (3 x-2 x^2\right ) \log \left (\frac {5}{x}\right )}{16 x^3-16 x^2+\left (16 x^3-16 x^2+4 x\right ) \log ^2\left (\frac {5}{x}\right )+\left (32 x^3-32 x^2+8 x\right ) \log \left (\frac {5}{x}\right )+4 x} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {5 e^x \left (-2 x^2+x-(2 x-3) x \log \left (\frac {5}{x}\right )+1\right )}{4 (1-2 x)^2 x \left (\log \left (\frac {5}{x}\right )+1\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {5}{4} \int \frac {e^x \left (-2 x^2+(3-2 x) \log \left (\frac {5}{x}\right ) x+x+1\right )}{(1-2 x)^2 x \left (\log \left (\frac {5}{x}\right )+1\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {5}{4} \int \left (\frac {e^x (3-2 x)}{(2 x-1)^2 \left (\log \left (\frac {5}{x}\right )+1\right )}-\frac {e^x}{x (2 x-1) \left (\log \left (\frac {5}{x}\right )+1\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {5}{4} \left (\int \frac {e^x}{x \left (\log \left (\frac {5}{x}\right )+1\right )^2}dx-2 \int \frac {e^x}{(2 x-1) \left (\log \left (\frac {5}{x}\right )+1\right )^2}dx+2 \int \frac {e^x}{(2 x-1)^2 \left (\log \left (\frac {5}{x}\right )+1\right )}dx-\int \frac {e^x}{(2 x-1) \left (\log \left (\frac {5}{x}\right )+1\right )}dx\right )\) |
Int[(5*E^x*(1 + x - 2*x^2) + 5*E^x*(3*x - 2*x^2)*Log[5/x])/(4*x - 16*x^2 + 16*x^3 + (8*x - 32*x^2 + 32*x^3)*Log[5/x] + (4*x - 16*x^2 + 16*x^3)*Log[5 /x]^2),x]
3.13.82.3.1 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]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 1.60 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04
method | result | size |
parallelrisch | \(-\frac {{\mathrm e}^{\ln \left (5\right )+x}}{4 \left (\ln \left (\frac {5}{x}\right )+1\right ) \left (-1+2 x \right )}\) | \(25\) |
risch | \(\frac {5 i {\mathrm e}^{x}}{2 \left (-1+2 x \right ) \left (-2 i \ln \left (5\right )+2 i \ln \left (x \right )-2 i\right )}\) | \(28\) |
int(((-2*x^2+3*x)*exp(ln(5)+x)*ln(5/x)+(-2*x^2+x+1)*exp(ln(5)+x))/((16*x^3 -16*x^2+4*x)*ln(5/x)^2+(32*x^3-32*x^2+8*x)*ln(5/x)+16*x^3-16*x^2+4*x),x,me thod=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {5 e^x \left (1+x-2 x^2\right )+5 e^x \left (3 x-2 x^2\right ) \log \left (\frac {5}{x}\right )}{4 x-16 x^2+16 x^3+\left (8 x-32 x^2+32 x^3\right ) \log \left (\frac {5}{x}\right )+\left (4 x-16 x^2+16 x^3\right ) \log ^2\left (\frac {5}{x}\right )} \, dx=-\frac {e^{\left (x + \log \left (5\right )\right )}}{4 \, {\left ({\left (2 \, x - 1\right )} \log \left (\frac {5}{x}\right ) + 2 \, x - 1\right )}} \]
integrate(((-2*x^2+3*x)*exp(log(5)+x)*log(5/x)+(-2*x^2+x+1)*exp(log(5)+x)) /((16*x^3-16*x^2+4*x)*log(5/x)^2+(32*x^3-32*x^2+8*x)*log(5/x)+16*x^3-16*x^ 2+4*x),x, algorithm=\
Time = 0.13 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {5 e^x \left (1+x-2 x^2\right )+5 e^x \left (3 x-2 x^2\right ) \log \left (\frac {5}{x}\right )}{4 x-16 x^2+16 x^3+\left (8 x-32 x^2+32 x^3\right ) \log \left (\frac {5}{x}\right )+\left (4 x-16 x^2+16 x^3\right ) \log ^2\left (\frac {5}{x}\right )} \, dx=- \frac {5 e^{x}}{8 x \log {\left (\frac {5}{x} \right )} + 8 x - 4 \log {\left (\frac {5}{x} \right )} - 4} \]
integrate(((-2*x**2+3*x)*exp(ln(5)+x)*ln(5/x)+(-2*x**2+x+1)*exp(ln(5)+x))/ ((16*x**3-16*x**2+4*x)*ln(5/x)**2+(32*x**3-32*x**2+8*x)*ln(5/x)+16*x**3-16 *x**2+4*x),x)
Time = 0.32 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {5 e^x \left (1+x-2 x^2\right )+5 e^x \left (3 x-2 x^2\right ) \log \left (\frac {5}{x}\right )}{4 x-16 x^2+16 x^3+\left (8 x-32 x^2+32 x^3\right ) \log \left (\frac {5}{x}\right )+\left (4 x-16 x^2+16 x^3\right ) \log ^2\left (\frac {5}{x}\right )} \, dx=-\frac {5 \, e^{x}}{4 \, {\left (2 \, x {\left (\log \left (5\right ) + 1\right )} - {\left (2 \, x - 1\right )} \log \left (x\right ) - \log \left (5\right ) - 1\right )}} \]
integrate(((-2*x^2+3*x)*exp(log(5)+x)*log(5/x)+(-2*x^2+x+1)*exp(log(5)+x)) /((16*x^3-16*x^2+4*x)*log(5/x)^2+(32*x^3-32*x^2+8*x)*log(5/x)+16*x^3-16*x^ 2+4*x),x, algorithm=\
Time = 0.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {5 e^x \left (1+x-2 x^2\right )+5 e^x \left (3 x-2 x^2\right ) \log \left (\frac {5}{x}\right )}{4 x-16 x^2+16 x^3+\left (8 x-32 x^2+32 x^3\right ) \log \left (\frac {5}{x}\right )+\left (4 x-16 x^2+16 x^3\right ) \log ^2\left (\frac {5}{x}\right )} \, dx=-\frac {5 \, e^{x}}{4 \, {\left (2 \, x \log \left (\frac {5}{x}\right ) + 2 \, x - \log \left (\frac {5}{x}\right ) - 1\right )}} \]
integrate(((-2*x^2+3*x)*exp(log(5)+x)*log(5/x)+(-2*x^2+x+1)*exp(log(5)+x)) /((16*x^3-16*x^2+4*x)*log(5/x)^2+(32*x^3-32*x^2+8*x)*log(5/x)+16*x^3-16*x^ 2+4*x),x, algorithm=\
Time = 13.97 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {5 e^x \left (1+x-2 x^2\right )+5 e^x \left (3 x-2 x^2\right ) \log \left (\frac {5}{x}\right )}{4 x-16 x^2+16 x^3+\left (8 x-32 x^2+32 x^3\right ) \log \left (\frac {5}{x}\right )+\left (4 x-16 x^2+16 x^3\right ) \log ^2\left (\frac {5}{x}\right )} \, dx=-\frac {5\,{\mathrm {e}}^x}{4\,\left (\ln \left (\frac {5}{x}\right )+1\right )\,\left (2\,x-1\right )} \]