Integrand size = 62, antiderivative size = 29 \[ \int \frac {e^{-\frac {3-4 e^3}{4 x}} \left (e^x \left (-3+4 e^3-4 x^2\right )+\left (3 x-4 e^3 x+4 x^2\right ) \log (4)\right )}{4 x^2 \log (4)} \, dx=e^{-\frac {3}{4 x}+\frac {e^3}{x}} \left (x-\frac {e^x}{\log (4)}\right ) \] Output:
(x-1/2*exp(x)/ln(2))/exp(3/4/x-exp(3)/x)
Time = 0.45 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.21 \[ \int \frac {e^{-\frac {3-4 e^3}{4 x}} \left (e^x \left (-3+4 e^3-4 x^2\right )+\left (3 x-4 e^3 x+4 x^2\right ) \log (4)\right )}{4 x^2 \log (4)} \, dx=\frac {e^{\frac {-3+4 e^3}{4 x}} \left (-4 e^x+4 x \log (4)\right )}{4 \log (4)} \] Input:
Integrate[(E^x*(-3 + 4*E^3 - 4*x^2) + (3*x - 4*E^3*x + 4*x^2)*Log[4])/(4*E ^((3 - 4*E^3)/(4*x))*x^2*Log[4]),x]
Output:
(E^((-3 + 4*E^3)/(4*x))*(-4*E^x + 4*x*Log[4]))/(4*Log[4])
Time = 0.69 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.72, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {27, 25, 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.
\(\displaystyle \int \frac {e^{-\frac {3-4 e^3}{4 x}} \left (e^x \left (-4 x^2+4 e^3-3\right )+\left (4 x^2-4 e^3 x+3 x\right ) \log (4)\right )}{4 x^2 \log (4)} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int -\frac {e^{-\frac {3-4 e^3}{4 x}} \left (e^x \left (4 x^2-4 e^3+3\right )-\left (4 x^2-4 e^3 x+3 x\right ) \log (4)\right )}{x^2}dx}{4 \log (4)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {e^{-\frac {3-4 e^3}{4 x}} \left (e^x \left (4 x^2-4 e^3+3\right )-\left (4 x^2-4 e^3 x+3 x\right ) \log (4)\right )}{x^2}dx}{4 \log (4)}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {\int \left (\frac {e^{-\frac {3-4 e^3}{4 x}} \log (4) \left (-4 x+4 e^3-3\right )}{x}+\frac {e^{x-\frac {3-4 e^3}{4 x}} \left (4 x^2-4 e^3+3\right )}{x^2}\right )dx}{4 \log (4)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {4 e^{x-\frac {3-4 e^3}{4 x}}-4 e^{-\frac {3-4 e^3}{4 x}} x \log (4)}{4 \log (4)}\) |
Input:
Int[(E^x*(-3 + 4*E^3 - 4*x^2) + (3*x - 4*E^3*x + 4*x^2)*Log[4])/(4*E^((3 - 4*E^3)/(4*x))*x^2*Log[4]),x]
Output:
-1/4*(4*E^(-1/4*(3 - 4*E^3)/x + x) - (4*x*Log[4])/E^((3 - 4*E^3)/(4*x)))/L og[4]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 0.60 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00
method | result | size |
risch | \(\frac {\left (8 x \ln \left (2\right )-4 \,{\mathrm e}^{x}\right ) {\mathrm e}^{\frac {4 \,{\mathrm e}^{3}-3}{4 x}}}{8 \ln \left (2\right )}\) | \(29\) |
norman | \(\frac {\left (x^{2}-\frac {x \,{\mathrm e}^{x}}{2 \ln \left (2\right )}\right ) {\mathrm e}^{-\frac {-4 \,{\mathrm e}^{3}+3}{4 x}}}{x}\) | \(32\) |
parallelrisch | \(\frac {\left (32 x^{2} \ln \left (2\right )-16 \,{\mathrm e}^{x} x \right ) {\mathrm e}^{\frac {4 \,{\mathrm e}^{3}-3}{4 x}}}{32 \ln \left (2\right ) x}\) | \(37\) |
parts | \(\frac {3 x \,{\mathrm e}^{-\frac {-{\mathrm e}^{3}+\frac {3}{4}}{x}}}{4 \left (-{\mathrm e}^{3}+\frac {3}{4}\right )}+{\mathrm e}^{3} \left (-\frac {x \,{\mathrm e}^{-\frac {-{\mathrm e}^{3}+\frac {3}{4}}{x}}}{-{\mathrm e}^{3}+\frac {3}{4}}+\operatorname {expIntegral}_{1}\left (\frac {-{\mathrm e}^{3}+\frac {3}{4}}{x}\right )\right )-{\mathrm e}^{3} \operatorname {expIntegral}_{1}\left (\frac {-{\mathrm e}^{3}+\frac {3}{4}}{x}\right )-\frac {{\mathrm e}^{x} {\mathrm e}^{-\frac {-4 \,{\mathrm e}^{3}+3}{4 x}}}{2 \ln \left (2\right )}\) | \(104\) |
Input:
int(1/8*((4*exp(3)-4*x^2-3)*exp(x)+2*(-4*x*exp(3)+4*x^2+3*x)*ln(2))/x^2/ln (2)/exp(1/4*(-4*exp(3)+3)/x),x,method=_RETURNVERBOSE)
Output:
1/8/ln(2)*(8*x*ln(2)-4*exp(x))*exp(1/4*(4*exp(3)-3)/x)
Time = 0.08 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {e^{-\frac {3-4 e^3}{4 x}} \left (e^x \left (-3+4 e^3-4 x^2\right )+\left (3 x-4 e^3 x+4 x^2\right ) \log (4)\right )}{4 x^2 \log (4)} \, dx=\frac {{\left (2 \, x \log \left (2\right ) - e^{x}\right )} e^{\left (\frac {4 \, e^{3} - 3}{4 \, x}\right )}}{2 \, \log \left (2\right )} \] Input:
integrate(1/8*((4*exp(3)-4*x^2-3)*exp(x)+2*(-4*x*exp(3)+4*x^2+3*x)*log(2)) /x^2/log(2)/exp(1/4*(-4*exp(3)+3)/x),x, algorithm="fricas")
Output:
1/2*(2*x*log(2) - e^x)*e^(1/4*(4*e^3 - 3)/x)/log(2)
Time = 0.85 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {e^{-\frac {3-4 e^3}{4 x}} \left (e^x \left (-3+4 e^3-4 x^2\right )+\left (3 x-4 e^3 x+4 x^2\right ) \log (4)\right )}{4 x^2 \log (4)} \, dx=\frac {\left (2 x \log {\left (2 \right )} - e^{x}\right ) e^{- \frac {\frac {3}{4} - e^{3}}{x}}}{2 \log {\left (2 \right )}} \] Input:
integrate(1/8*((4*exp(3)-4*x**2-3)*exp(x)+2*(-4*x*exp(3)+4*x**2+3*x)*ln(2) )/x**2/ln(2)/exp(1/4*(-4*exp(3)+3)/x),x)
Output:
(2*x*log(2) - exp(x))*exp(-(3/4 - exp(3))/x)/(2*log(2))
Time = 0.18 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-\frac {3-4 e^3}{4 x}} \left (e^x \left (-3+4 e^3-4 x^2\right )+\left (3 x-4 e^3 x+4 x^2\right ) \log (4)\right )}{4 x^2 \log (4)} \, dx=\frac {{\left (2 \, x \log \left (2\right ) - e^{x}\right )} e^{\left (\frac {e^{3}}{x} - \frac {3}{4 \, x}\right )}}{2 \, \log \left (2\right )} \] Input:
integrate(1/8*((4*exp(3)-4*x^2-3)*exp(x)+2*(-4*x*exp(3)+4*x^2+3*x)*log(2)) /x^2/log(2)/exp(1/4*(-4*exp(3)+3)/x),x, algorithm="maxima")
Output:
1/2*(2*x*log(2) - e^x)*e^(e^3/x - 3/4/x)/log(2)
Time = 0.13 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.48 \[ \int \frac {e^{-\frac {3-4 e^3}{4 x}} \left (e^x \left (-3+4 e^3-4 x^2\right )+\left (3 x-4 e^3 x+4 x^2\right ) \log (4)\right )}{4 x^2 \log (4)} \, dx=\frac {2 \, x e^{\left (\frac {4 \, e^{3} - 3}{4 \, x}\right )} \log \left (2\right ) - e^{\left (\frac {4 \, x^{2} + 4 \, e^{3} - 3}{4 \, x}\right )}}{2 \, \log \left (2\right )} \] Input:
integrate(1/8*((4*exp(3)-4*x^2-3)*exp(x)+2*(-4*x*exp(3)+4*x^2+3*x)*log(2)) /x^2/log(2)/exp(1/4*(-4*exp(3)+3)/x),x, algorithm="giac")
Output:
1/2*(2*x*e^(1/4*(4*e^3 - 3)/x)*log(2) - e^(1/4*(4*x^2 + 4*e^3 - 3)/x))/log (2)
Time = 4.09 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-\frac {3-4 e^3}{4 x}} \left (e^x \left (-3+4 e^3-4 x^2\right )+\left (3 x-4 e^3 x+4 x^2\right ) \log (4)\right )}{4 x^2 \log (4)} \, dx=-\frac {{\mathrm {e}}^{\frac {{\mathrm {e}}^3}{x}-\frac {3}{4\,x}}\,\left (4\,{\mathrm {e}}^x-x\,\ln \left (256\right )\right )}{8\,\ln \left (2\right )} \] Input:
int((exp((exp(3) - 3/4)/x)*((log(2)*(3*x - 4*x*exp(3) + 4*x^2))/4 - (exp(x )*(4*x^2 - 4*exp(3) + 3))/8))/(x^2*log(2)),x)
Output:
-(exp(exp(3)/x - 3/(4*x))*(4*exp(x) - x*log(256)))/(8*log(2))
\[ \int \frac {e^{-\frac {3-4 e^3}{4 x}} \left (e^x \left (-3+4 e^3-4 x^2\right )+\left (3 x-4 e^3 x+4 x^2\right ) \log (4)\right )}{4 x^2 \log (4)} \, dx=\int \frac {\left (4 \,{\mathrm e}^{3}-4 x^{2}-3\right ) {\mathrm e}^{x}+2 \left (-4 x \,{\mathrm e}^{3}+4 x^{2}+3 x \right ) \mathrm {log}\left (2\right )}{8 x^{2} \mathrm {log}\left (2\right ) {\mathrm e}^{\frac {-4 \,{\mathrm e}^{3}+3}{4 x}}}d x \] Input:
int(1/8*((4*exp(3)-4*x^2-3)*exp(x)+2*(-4*x*exp(3)+4*x^2+3*x)*log(2))/x^2/l og(2)/exp(1/4*(-4*exp(3)+3)/x),x)
Output:
int(1/8*((4*exp(3)-4*x^2-3)*exp(x)+2*(-4*x*exp(3)+4*x^2+3*x)*log(2))/x^2/l og(2)/exp(1/4*(-4*exp(3)+3)/x),x)