Integrand size = 65, antiderivative size = 23 \[ \int \frac {e^{-8 x-e^{\left .\frac {1}{4}\right /x} x+x \log \left (x^2\right )} \left (e^{3+\frac {1}{4 x}} (1-4 x)-24 e^3 x+4 e^3 x \log \left (x^2\right )\right )}{4 x} \, dx=e^{3+x \left (-8-e^{\left .\frac {1}{4}\right /x}+\log \left (x^2\right )\right )} \] Output:
exp(x*(ln(x^2)-8-exp(1/4/x)))*exp(3)
Time = 0.56 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {e^{-8 x-e^{\left .\frac {1}{4}\right /x} x+x \log \left (x^2\right )} \left (e^{3+\frac {1}{4 x}} (1-4 x)-24 e^3 x+4 e^3 x \log \left (x^2\right )\right )}{4 x} \, dx=e^{3-8 x-e^{\left .\frac {1}{4}\right /x} x} \left (x^2\right )^x \] Input:
Integrate[(E^(-8*x - E^(1/(4*x))*x + x*Log[x^2])*(E^(3 + 1/(4*x))*(1 - 4*x ) - 24*E^3*x + 4*E^3*x*Log[x^2]))/(4*x),x]
Output:
E^(3 - 8*x - E^(1/(4*x))*x)*(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.
\(\displaystyle \int \frac {e^{x \log \left (x^2\right )-e^{\left .\frac {1}{4}\right /x} x-8 x} \left (4 e^3 x \log \left (x^2\right )+e^{\frac {1}{4 x}+3} (1-4 x)-24 e^3 x\right )}{4 x} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \int \frac {e^{-e^{\left .\frac {1}{4}\right /x} x-8 x} \left (x^2\right )^x \left (e^{3+\frac {1}{4 x}} (1-4 x)-24 e^3 x+4 e^3 x \log \left (x^2\right )\right )}{x}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{4} \int \left (4 e^{-e^{\left .\frac {1}{4}\right /x} x-8 x+3} \left (x^2\right )^x \left (\log \left (x^2\right )-6\right )-\frac {e^{-e^{\left .\frac {1}{4}\right /x} x-8 x+3+\frac {1}{4 x}} \left (x^2\right )^x (4 x-1)}{x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \left (-24 \int e^{-e^{\left .\frac {1}{4}\right /x} x-8 x+3} \left (x^2\right )^xdx-4 \int e^{-e^{\left .\frac {1}{4}\right /x} x-8 x+3+\frac {1}{4 x}} \left (x^2\right )^xdx+\int \frac {e^{-e^{\left .\frac {1}{4}\right /x} x-8 x+3+\frac {1}{4 x}} \left (x^2\right )^x}{x}dx-8 \int \frac {\int e^{3-\left (8+e^{\left .\frac {1}{4}\right /x}\right ) x} \left (x^2\right )^xdx}{x}dx+4 \log \left (x^2\right ) \int e^{-e^{\left .\frac {1}{4}\right /x} x-8 x+3} \left (x^2\right )^xdx\right )\) |
Input:
Int[(E^(-8*x - E^(1/(4*x))*x + x*Log[x^2])*(E^(3 + 1/(4*x))*(1 - 4*x) - 24 *E^3*x + 4*E^3*x*Log[x^2]))/(4*x),x]
Output:
$Aborted
Time = 0.98 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96
method | result | size |
parallelrisch | \({\mathrm e}^{3} {\mathrm e}^{-x \left ({\mathrm e}^{\frac {1}{4 x}}-\ln \left (x^{2}\right )+8\right )}\) | \(22\) |
default | \({\mathrm e}^{3} {\mathrm e}^{x \ln \left (x^{2}\right )-x \,{\mathrm e}^{\frac {1}{4 x}}-8 x}\) | \(24\) |
norman | \({\mathrm e}^{3} {\mathrm e}^{x \ln \left (x^{2}\right )-x \,{\mathrm e}^{\frac {1}{4 x}}-8 x}\) | \(24\) |
risch | \(x^{2 x} {\mathrm e}^{3-\frac {i \pi \operatorname {csgn}\left (i x^{2}\right )^{3} x}{2}+i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2} x -\frac {i \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right ) x}{2}-x \,{\mathrm e}^{\frac {1}{4 x}}-8 x}\) | \(74\) |
Input:
int(1/4*(4*x*exp(3)*ln(x^2)+(-4*x+1)*exp(3)*exp(1/4/x)-24*x*exp(3))*exp(x* ln(x^2)-x*exp(1/4/x)-8*x)/x,x,method=_RETURNVERBOSE)
Output:
exp(3)*exp(-x*(exp(1/4/x)-ln(x^2)+8))
Time = 0.09 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.48 \[ \int \frac {e^{-8 x-e^{\left .\frac {1}{4}\right /x} x+x \log \left (x^2\right )} \left (e^{3+\frac {1}{4 x}} (1-4 x)-24 e^3 x+4 e^3 x \log \left (x^2\right )\right )}{4 x} \, dx=e^{\left ({\left (x e^{3} \log \left (x^{2}\right ) - 8 \, x e^{3} - x e^{\left (\frac {12 \, x + 1}{4 \, x}\right )}\right )} e^{\left (-3\right )} + 3\right )} \] Input:
integrate(1/4*(4*x*exp(3)*log(x^2)+(-4*x+1)*exp(3)*exp(1/4/x)-24*x*exp(3)) *exp(x*log(x^2)-x*exp(1/4/x)-8*x)/x,x, algorithm="fricas")
Output:
e^((x*e^3*log(x^2) - 8*x*e^3 - x*e^(1/4*(12*x + 1)/x))*e^(-3) + 3)
Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {e^{-8 x-e^{\left .\frac {1}{4}\right /x} x+x \log \left (x^2\right )} \left (e^{3+\frac {1}{4 x}} (1-4 x)-24 e^3 x+4 e^3 x \log \left (x^2\right )\right )}{4 x} \, dx=e^{3} e^{- x e^{\frac {1}{4 x}} + x \log {\left (x^{2} \right )} - 8 x} \] Input:
integrate(1/4*(4*x*exp(3)*ln(x**2)+(-4*x+1)*exp(3)*exp(1/4/x)-24*x*exp(3)) *exp(x*ln(x**2)-x*exp(1/4/x)-8*x)/x,x)
Output:
exp(3)*exp(-x*exp(1/(4*x)) + x*log(x**2) - 8*x)
Time = 0.12 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {e^{-8 x-e^{\left .\frac {1}{4}\right /x} x+x \log \left (x^2\right )} \left (e^{3+\frac {1}{4 x}} (1-4 x)-24 e^3 x+4 e^3 x \log \left (x^2\right )\right )}{4 x} \, dx=e^{\left (-x e^{\left (\frac {1}{4 \, x}\right )} + 2 \, x \log \left (x\right ) - 8 \, x + 3\right )} \] Input:
integrate(1/4*(4*x*exp(3)*log(x^2)+(-4*x+1)*exp(3)*exp(1/4/x)-24*x*exp(3)) *exp(x*log(x^2)-x*exp(1/4/x)-8*x)/x,x, algorithm="maxima")
Output:
e^(-x*e^(1/4/x) + 2*x*log(x) - 8*x + 3)
Time = 0.19 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {e^{-8 x-e^{\left .\frac {1}{4}\right /x} x+x \log \left (x^2\right )} \left (e^{3+\frac {1}{4 x}} (1-4 x)-24 e^3 x+4 e^3 x \log \left (x^2\right )\right )}{4 x} \, dx=e^{\left (-x e^{\left (\frac {1}{4 \, x}\right )} + x \log \left (x^{2}\right ) - 8 \, x + 3\right )} \] Input:
integrate(1/4*(4*x*exp(3)*log(x^2)+(-4*x+1)*exp(3)*exp(1/4/x)-24*x*exp(3)) *exp(x*log(x^2)-x*exp(1/4/x)-8*x)/x,x, algorithm="giac")
Output:
e^(-x*e^(1/4/x) + x*log(x^2) - 8*x + 3)
Time = 7.50 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {e^{-8 x-e^{\left .\frac {1}{4}\right /x} x+x \log \left (x^2\right )} \left (e^{3+\frac {1}{4 x}} (1-4 x)-24 e^3 x+4 e^3 x \log \left (x^2\right )\right )}{4 x} \, dx={\mathrm {e}}^{-8\,x}\,{\mathrm {e}}^3\,{\mathrm {e}}^{-x\,{\mathrm {e}}^{\frac {1}{4\,x}}}\,{\left (x^2\right )}^x \] Input:
int(-(exp(x*log(x^2) - 8*x - x*exp(1/(4*x)))*(24*x*exp(3) - 4*x*log(x^2)*e xp(3) + exp(3)*exp(1/(4*x))*(4*x - 1)))/(4*x),x)
Output:
exp(-8*x)*exp(3)*exp(-x*exp(1/(4*x)))*(x^2)^x
\[ \int \frac {e^{-8 x-e^{\left .\frac {1}{4}\right /x} x+x \log \left (x^2\right )} \left (e^{3+\frac {1}{4 x}} (1-4 x)-24 e^3 x+4 e^3 x \log \left (x^2\right )\right )}{4 x} \, dx=\int \frac {\left (4 x \,{\mathrm e}^{3} \mathrm {log}\left (x^{2}\right )+\left (-4 x +1\right ) {\mathrm e}^{3} {\mathrm e}^{\frac {1}{4 x}}-24 x \,{\mathrm e}^{3}\right ) {\mathrm e}^{\mathrm {log}\left (x^{2}\right ) x -x \,{\mathrm e}^{\frac {1}{4 x}}-8 x}}{4 x}d x \] Input:
int(1/4*(4*x*exp(3)*log(x^2)+(-4*x+1)*exp(3)*exp(1/4/x)-24*x*exp(3))*exp(x *log(x^2)-x*exp(1/4/x)-8*x)/x,x)
Output:
int(1/4*(4*x*exp(3)*log(x^2)+(-4*x+1)*exp(3)*exp(1/4/x)-24*x*exp(3))*exp(x *log(x^2)-x*exp(1/4/x)-8*x)/x,x)