Integrand size = 144, antiderivative size = 33 \[ \int \frac {e^{10-6 e^{4/x}} \left (96 e^{4/x}+e^{-4+6 e^{4/x}-2 x+2 x^2+4 x \log (x)+2 \log ^2(x)} \left (2 x^2+4 x^3+\left (4 x+4 x^2\right ) \log (x)\right )+e^{-2+3 e^{4/x}-x+x^2+2 x \log (x)+\log ^2(x)} \left (48 e^{4/x}+4 x^2+8 x^3+\left (8 x+8 x^2\right ) \log (x)\right )\right )}{x^2} \, dx=3+\left (2 e^{5-3 e^{4/x}}+e^{3-x+(x+\log (x))^2}\right )^2 \]
Time = 0.21 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.58 \[ \int \frac {e^{10-6 e^{4/x}} \left (96 e^{4/x}+e^{-4+6 e^{4/x}-2 x+2 x^2+4 x \log (x)+2 \log ^2(x)} \left (2 x^2+4 x^3+\left (4 x+4 x^2\right ) \log (x)\right )+e^{-2+3 e^{4/x}-x+x^2+2 x \log (x)+\log ^2(x)} \left (48 e^{4/x}+4 x^2+8 x^3+\left (8 x+8 x^2\right ) \log (x)\right )\right )}{x^2} \, dx=e^{6-6 e^{4/x}-2 x} \left (2 e^{2+x}+e^{3 e^{4/x}+x^2+\log ^2(x)} x^{2 x}\right )^2 \]
Integrate[(E^(10 - 6*E^(4/x))*(96*E^(4/x) + E^(-4 + 6*E^(4/x) - 2*x + 2*x^ 2 + 4*x*Log[x] + 2*Log[x]^2)*(2*x^2 + 4*x^3 + (4*x + 4*x^2)*Log[x]) + E^(- 2 + 3*E^(4/x) - x + x^2 + 2*x*Log[x] + Log[x]^2)*(48*E^(4/x) + 4*x^2 + 8*x ^3 + (8*x + 8*x^2)*Log[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.
\(\displaystyle \int \frac {e^{10-6 e^{4/x}} \left (\left (4 x^3+2 x^2+\left (4 x^2+4 x\right ) \log (x)\right ) \exp \left (2 x^2-2 x+6 e^{4/x}+2 \log ^2(x)+4 x \log (x)-4\right )+e^{x^2-x+3 e^{4/x}+\log ^2(x)+2 x \log (x)-2} \left (8 x^3+4 x^2+\left (8 x^2+8 x\right ) \log (x)+48 e^{4/x}\right )+96 e^{4/x}\right )}{x^2} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{-2 \left (3 e^{4/x}-5\right )} \left (\left (4 x^3+2 x^2+\left (4 x^2+4 x\right ) \log (x)\right ) \exp \left (2 x^2-2 x+6 e^{4/x}+2 \log ^2(x)+4 x \log (x)-4\right )+e^{x^2-x+3 e^{4/x}+\log ^2(x)+2 x \log (x)-2} \left (8 x^3+4 x^2+\left (8 x^2+8 x\right ) \log (x)+48 e^{4/x}\right )+96 e^{4/x}\right )}{x^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (2 x^{4 x-1} \left (2 x^2+x+2 x \log (x)+2 \log (x)\right ) \exp \left (2 \left (x^2-x+3 e^{4/x}+\log ^2(x)-2\right )-2 \left (3 e^{4/x}-5\right )\right )+4 x^{2 x-2} \left (2 x^3+x^2+2 x^2 \log (x)+12 e^{4/x}+2 x \log (x)\right ) \exp \left (x^2-x+3 e^{4/x}-2 \left (3 e^{4/x}-5\right )+\log ^2(x)-2\right )+\frac {96 e^{\frac {4}{x}-2 \left (3 e^{4/x}-5\right )}}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 4 \int \exp \left (x^2-x+3 e^{4/x}+\log ^2(x)-2 \left (-5+3 e^{4/x}\right )-2\right ) x^{2 x}dx+48 \int \exp \left (x^2-x+3 e^{4/x}+\log ^2(x)-2 \left (-5+3 e^{4/x}\right )-2+\frac {4}{x}\right ) x^{2 x-2}dx+8 \int \exp \left (x^2-x+3 e^{4/x}+\log ^2(x)-2 \left (-5+3 e^{4/x}\right )-2\right ) x^{2 x+1}dx+8 \int \exp \left (x^2-x+3 e^{4/x}+\log ^2(x)-2 \left (-5+3 e^{4/x}\right )-2\right ) x^{2 x} \log (x)dx+8 \int \exp \left (x^2-x+3 e^{4/x}+\log ^2(x)-2 \left (-5+3 e^{4/x}\right )-2\right ) x^{2 x-1} \log (x)dx+2 \int e^{2 \left (x^2-x+\log ^2(x)+3\right )} x^{4 x}dx+4 \int e^{2 \left (x^2-x+\log ^2(x)+3\right )} x^{4 x+1}dx+4 \int e^{2 \left (x^2-x+\log ^2(x)+3\right )} x^{4 x} \log (x)dx+4 \int e^{2 \left (x^2-x+\log ^2(x)+3\right )} x^{4 x-1} \log (x)dx+4 e^{10-6 e^{4/x}}\) |
Int[(E^(10 - 6*E^(4/x))*(96*E^(4/x) + E^(-4 + 6*E^(4/x) - 2*x + 2*x^2 + 4* x*Log[x] + 2*Log[x]^2)*(2*x^2 + 4*x^3 + (4*x + 4*x^2)*Log[x]) + E^(-2 + 3* E^(4/x) - x + x^2 + 2*x*Log[x] + Log[x]^2)*(48*E^(4/x) + 4*x^2 + 8*x^3 + ( 8*x + 8*x^2)*Log[x])))/x^2,x]
3.16.84.3.1 Defintions of rubi rules used
Leaf count of result is larger than twice the leaf count of optimal. \(67\) vs. \(2(32)=64\).
Time = 2.50 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.06
method | result | size |
risch | \(4 \,{\mathrm e}^{-6 \,{\mathrm e}^{\frac {4}{x}}+10}+x^{4 x} {\mathrm e}^{2 \ln \left (x \right )^{2}+6+2 x^{2}-2 x}+4 x^{2 x} {\mathrm e}^{-3 \,{\mathrm e}^{\frac {4}{x}}+8+\ln \left (x \right )^{2}+x^{2}-x}\) | \(68\) |
parallelrisch | \(-\left (-4-{\mathrm e}^{6 \,{\mathrm e}^{\frac {4}{x}}-10} {\mathrm e}^{2 \ln \left (x \right )^{2}+4 x \ln \left (x \right )+2 x^{2}-2 x +6}-4 \,{\mathrm e}^{3 \,{\mathrm e}^{\frac {4}{x}}-5} {\mathrm e}^{\ln \left (x \right )^{2}+2 x \ln \left (x \right )+x^{2}-x +3}\right ) {\mathrm e}^{-6 \,{\mathrm e}^{\frac {4}{x}}+10}\) | \(84\) |
int((((4*x^2+4*x)*ln(x)+4*x^3+2*x^2)*exp(3*exp(4/x)-5)^2*exp(ln(x)^2+2*x*l n(x)+x^2-x+3)^2+((8*x^2+8*x)*ln(x)+48*exp(4/x)+8*x^3+4*x^2)*exp(3*exp(4/x) -5)*exp(ln(x)^2+2*x*ln(x)+x^2-x+3)+96*exp(4/x))/x^2/exp(3*exp(4/x)-5)^2,x, method=_RETURNVERBOSE)
4*exp(-6*exp(4/x)+10)+(x^(2*x))^2*exp(2*ln(x)^2+6+2*x^2-2*x)+4*x^(2*x)*exp (-3*exp(4/x)+8+ln(x)^2+x^2-x)
Leaf count of result is larger than twice the leaf count of optimal. 64 vs. \(2 (30) = 60\).
Time = 0.26 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.94 \[ \int \frac {e^{10-6 e^{4/x}} \left (96 e^{4/x}+e^{-4+6 e^{4/x}-2 x+2 x^2+4 x \log (x)+2 \log ^2(x)} \left (2 x^2+4 x^3+\left (4 x+4 x^2\right ) \log (x)\right )+e^{-2+3 e^{4/x}-x+x^2+2 x \log (x)+\log ^2(x)} \left (48 e^{4/x}+4 x^2+8 x^3+\left (8 x+8 x^2\right ) \log (x)\right )\right )}{x^2} \, dx=e^{\left (2 \, x^{2} + 4 \, x \log \left (x\right ) + 2 \, \log \left (x\right )^{2} - 2 \, x + 6\right )} + 4 \, e^{\left (x^{2} + 2 \, x \log \left (x\right ) + \log \left (x\right )^{2} - x - 3 \, e^{\frac {4}{x}} + 8\right )} + 4 \, e^{\left (-6 \, e^{\frac {4}{x}} + 10\right )} \]
integrate((((4*x^2+4*x)*log(x)+4*x^3+2*x^2)*exp(3*exp(4/x)-5)^2*exp(log(x) ^2+2*x*log(x)+x^2-x+3)^2+((8*x^2+8*x)*log(x)+48*exp(4/x)+8*x^3+4*x^2)*exp( 3*exp(4/x)-5)*exp(log(x)^2+2*x*log(x)+x^2-x+3)+96*exp(4/x))/x^2/exp(3*exp( 4/x)-5)^2,x, algorithm=\
e^(2*x^2 + 4*x*log(x) + 2*log(x)^2 - 2*x + 6) + 4*e^(x^2 + 2*x*log(x) + lo g(x)^2 - x - 3*e^(4/x) + 8) + 4*e^(-6*e^(4/x) + 10)
Timed out. \[ \int \frac {e^{10-6 e^{4/x}} \left (96 e^{4/x}+e^{-4+6 e^{4/x}-2 x+2 x^2+4 x \log (x)+2 \log ^2(x)} \left (2 x^2+4 x^3+\left (4 x+4 x^2\right ) \log (x)\right )+e^{-2+3 e^{4/x}-x+x^2+2 x \log (x)+\log ^2(x)} \left (48 e^{4/x}+4 x^2+8 x^3+\left (8 x+8 x^2\right ) \log (x)\right )\right )}{x^2} \, dx=\text {Timed out} \]
integrate((((4*x**2+4*x)*ln(x)+4*x**3+2*x**2)*exp(3*exp(4/x)-5)**2*exp(ln( x)**2+2*x*ln(x)+x**2-x+3)**2+((8*x**2+8*x)*ln(x)+48*exp(4/x)+8*x**3+4*x**2 )*exp(3*exp(4/x)-5)*exp(ln(x)**2+2*x*ln(x)+x**2-x+3)+96*exp(4/x))/x**2/exp (3*exp(4/x)-5)**2,x)
Leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (30) = 60\).
Time = 0.31 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.97 \[ \int \frac {e^{10-6 e^{4/x}} \left (96 e^{4/x}+e^{-4+6 e^{4/x}-2 x+2 x^2+4 x \log (x)+2 \log ^2(x)} \left (2 x^2+4 x^3+\left (4 x+4 x^2\right ) \log (x)\right )+e^{-2+3 e^{4/x}-x+x^2+2 x \log (x)+\log ^2(x)} \left (48 e^{4/x}+4 x^2+8 x^3+\left (8 x+8 x^2\right ) \log (x)\right )\right )}{x^2} \, dx={\left (e^{\left (2 \, x^{2} + 4 \, x \log \left (x\right ) + 2 \, \log \left (x\right )^{2} + 6\right )} + 4 \, e^{\left (x^{2} + 2 \, x \log \left (x\right ) + \log \left (x\right )^{2} + x - 3 \, e^{\frac {4}{x}} + 8\right )}\right )} e^{\left (-2 \, x\right )} + 4 \, e^{\left (-6 \, e^{\frac {4}{x}} + 10\right )} \]
integrate((((4*x^2+4*x)*log(x)+4*x^3+2*x^2)*exp(3*exp(4/x)-5)^2*exp(log(x) ^2+2*x*log(x)+x^2-x+3)^2+((8*x^2+8*x)*log(x)+48*exp(4/x)+8*x^3+4*x^2)*exp( 3*exp(4/x)-5)*exp(log(x)^2+2*x*log(x)+x^2-x+3)+96*exp(4/x))/x^2/exp(3*exp( 4/x)-5)^2,x, algorithm=\
(e^(2*x^2 + 4*x*log(x) + 2*log(x)^2 + 6) + 4*e^(x^2 + 2*x*log(x) + log(x)^ 2 + x - 3*e^(4/x) + 8))*e^(-2*x) + 4*e^(-6*e^(4/x) + 10)
Leaf count of result is larger than twice the leaf count of optimal. 99 vs. \(2 (30) = 60\).
Time = 0.34 (sec) , antiderivative size = 99, normalized size of antiderivative = 3.00 \[ \int \frac {e^{10-6 e^{4/x}} \left (96 e^{4/x}+e^{-4+6 e^{4/x}-2 x+2 x^2+4 x \log (x)+2 \log ^2(x)} \left (2 x^2+4 x^3+\left (4 x+4 x^2\right ) \log (x)\right )+e^{-2+3 e^{4/x}-x+x^2+2 x \log (x)+\log ^2(x)} \left (48 e^{4/x}+4 x^2+8 x^3+\left (8 x+8 x^2\right ) \log (x)\right )\right )}{x^2} \, dx={\left (e^{\left (2 \, x^{2} + 4 \, x \log \left (x\right ) + 2 \, \log \left (x\right )^{2} - 2 \, x + \frac {4}{x} + 6\right )} + 4 \, e^{\left (\frac {x^{3} + 2 \, x^{2} \log \left (x\right ) + x \log \left (x\right )^{2} - x^{2} - 3 \, x e^{\frac {4}{x}} + 8 \, x + 4}{x}\right )} + 4 \, e^{\left (-\frac {2 \, {\left (3 \, x e^{\frac {4}{x}} - 5 \, x - 2\right )}}{x}\right )}\right )} e^{\left (-\frac {4}{x}\right )} \]
integrate((((4*x^2+4*x)*log(x)+4*x^3+2*x^2)*exp(3*exp(4/x)-5)^2*exp(log(x) ^2+2*x*log(x)+x^2-x+3)^2+((8*x^2+8*x)*log(x)+48*exp(4/x)+8*x^3+4*x^2)*exp( 3*exp(4/x)-5)*exp(log(x)^2+2*x*log(x)+x^2-x+3)+96*exp(4/x))/x^2/exp(3*exp( 4/x)-5)^2,x, algorithm=\
(e^(2*x^2 + 4*x*log(x) + 2*log(x)^2 - 2*x + 4/x + 6) + 4*e^((x^3 + 2*x^2*l og(x) + x*log(x)^2 - x^2 - 3*x*e^(4/x) + 8*x + 4)/x) + 4*e^(-2*(3*x*e^(4/x ) - 5*x - 2)/x))*e^(-4/x)
Time = 8.53 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.12 \[ \int \frac {e^{10-6 e^{4/x}} \left (96 e^{4/x}+e^{-4+6 e^{4/x}-2 x+2 x^2+4 x \log (x)+2 \log ^2(x)} \left (2 x^2+4 x^3+\left (4 x+4 x^2\right ) \log (x)\right )+e^{-2+3 e^{4/x}-x+x^2+2 x \log (x)+\log ^2(x)} \left (48 e^{4/x}+4 x^2+8 x^3+\left (8 x+8 x^2\right ) \log (x)\right )\right )}{x^2} \, dx=4\,{\mathrm {e}}^{-6\,{\mathrm {e}}^{4/x}}\,{\mathrm {e}}^{10}+x^{4\,x}\,{\mathrm {e}}^{2\,{\ln \left (x\right )}^2}\,{\mathrm {e}}^{-2\,x}\,{\mathrm {e}}^6\,{\mathrm {e}}^{2\,x^2}+4\,x^{2\,x}\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{-3\,{\mathrm {e}}^{4/x}}\,{\mathrm {e}}^8\,{\mathrm {e}}^{{\ln \left (x\right )}^2} \]
int((exp(10 - 6*exp(4/x))*(96*exp(4/x) + exp(log(x)^2 - x + 2*x*log(x) + x ^2 + 3)*exp(3*exp(4/x) - 5)*(48*exp(4/x) + log(x)*(8*x + 8*x^2) + 4*x^2 + 8*x^3) + exp(2*log(x)^2 - 2*x + 4*x*log(x) + 2*x^2 + 6)*exp(6*exp(4/x) - 1 0)*(log(x)*(4*x + 4*x^2) + 2*x^2 + 4*x^3)))/x^2,x)