Integrand size = 56, antiderivative size = 23 \[ \int \frac {e^{2 x^2-4 x \log \left (\frac {1}{x^2}\right )+2 \log ^2\left (\frac {1}{x^2}\right )} \left (-4+\left (-8+32 x+16 x^2+(-32-16 x) \log \left (\frac {1}{x^2}\right )\right ) \log (x)\right )}{x^3 \log ^2(x)} \, dx=\frac {4 e^{2 \left (x-\log \left (\frac {1}{x^2}\right )\right )^2}}{x^2 \log (x)} \] Output:
4*exp((x-ln(1/x^2))^2)^2/x^2/ln(x)
Time = 0.43 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \[ \int \frac {e^{2 x^2-4 x \log \left (\frac {1}{x^2}\right )+2 \log ^2\left (\frac {1}{x^2}\right )} \left (-4+\left (-8+32 x+16 x^2+(-32-16 x) \log \left (\frac {1}{x^2}\right )\right ) \log (x)\right )}{x^3 \log ^2(x)} \, dx=\frac {4 e^{2 \left (x^2+\log ^2\left (\frac {1}{x^2}\right )\right )} \left (\frac {1}{x^2}\right )^{1-4 x}}{\log (x)} \] Input:
Integrate[(E^(2*x^2 - 4*x*Log[x^(-2)] + 2*Log[x^(-2)]^2)*(-4 + (-8 + 32*x + 16*x^2 + (-32 - 16*x)*Log[x^(-2)])*Log[x]))/(x^3*Log[x]^2),x]
Output:
(4*E^(2*(x^2 + Log[x^(-2)]^2))*(x^(-2))^(1 - 4*x))/Log[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.
\(\displaystyle \int \frac {e^{2 x^2+2 \log ^2\left (\frac {1}{x^2}\right )-4 x \log \left (\frac {1}{x^2}\right )} \left (\left (16 x^2+(-16 x-32) \log \left (\frac {1}{x^2}\right )+32 x-8\right ) \log (x)-4\right )}{x^3 \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{2 \left (x-\log \left (\frac {1}{x^2}\right )\right )^2} \left (\left (16 x^2+(-16 x-32) \log \left (\frac {1}{x^2}\right )+32 x-8\right ) \log (x)-4\right )}{x^3 \log ^2(x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {8 e^{2 \left (x-\log \left (\frac {1}{x^2}\right )\right )^2} \left (2 x^2-2 x \log \left (\frac {1}{x^2}\right )-4 \log \left (\frac {1}{x^2}\right )+4 x-1\right )}{x^3 \log (x)}-\frac {4 e^{2 \left (x-\log \left (\frac {1}{x^2}\right )\right )^2}}{x^3 \log ^2(x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 32 \int \frac {e^{2 \left (x-\log \left (\frac {1}{x^2}\right )\right )^2}}{x^2 \log (x)}dx+16 \int \frac {e^{2 \left (x-\log \left (\frac {1}{x^2}\right )\right )^2}}{x \log (x)}dx-16 \int \frac {e^{2 \left (x-\log \left (\frac {1}{x^2}\right )\right )^2} \log \left (\frac {1}{x^2}\right )}{x^2 \log (x)}dx-4 \int \frac {e^{2 \left (x-\log \left (\frac {1}{x^2}\right )\right )^2}}{x^3 \log ^2(x)}dx-8 \int \frac {e^{2 \left (x-\log \left (\frac {1}{x^2}\right )\right )^2}}{x^3 \log (x)}dx-32 \int \frac {e^{2 \left (x-\log \left (\frac {1}{x^2}\right )\right )^2} \log \left (\frac {1}{x^2}\right )}{x^3 \log (x)}dx\) |
Input:
Int[(E^(2*x^2 - 4*x*Log[x^(-2)] + 2*Log[x^(-2)]^2)*(-4 + (-8 + 32*x + 16*x ^2 + (-32 - 16*x)*Log[x^(-2)])*Log[x]))/(x^3*Log[x]^2),x]
Output:
$Aborted
Time = 0.83 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30
method | result | size |
parallelrisch | \(\frac {4 \,{\mathrm e}^{2 \ln \left (\frac {1}{x^{2}}\right )^{2}-4 x \ln \left (\frac {1}{x^{2}}\right )+2 x^{2}}}{x^{2} \ln \left (x \right )}\) | \(30\) |
risch | \(\frac {4 \,{\mathrm e}^{\frac {{\left (-i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}+2 i \pi \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )-i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2}+4 \ln \left (x \right )+2 x \right )}^{2}}{2}}}{x^{2} \ln \left (x \right )}\) | \(72\) |
Input:
int((((-16*x-32)*ln(1/x^2)+16*x^2+32*x-8)*ln(x)-4)*exp(ln(1/x^2)^2-2*x*ln( 1/x^2)+x^2)^2/x^3/ln(x)^2,x,method=_RETURNVERBOSE)
Output:
4/x^2*exp(ln(1/x^2)^2-2*x*ln(1/x^2)+x^2)^2/ln(x)
Time = 0.10 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.43 \[ \int \frac {e^{2 x^2-4 x \log \left (\frac {1}{x^2}\right )+2 \log ^2\left (\frac {1}{x^2}\right )} \left (-4+\left (-8+32 x+16 x^2+(-32-16 x) \log \left (\frac {1}{x^2}\right )\right ) \log (x)\right )}{x^3 \log ^2(x)} \, dx=-\frac {8 \, e^{\left (2 \, x^{2} - 4 \, x \log \left (\frac {1}{x^{2}}\right ) + 2 \, \log \left (\frac {1}{x^{2}}\right )^{2}\right )}}{x^{2} \log \left (\frac {1}{x^{2}}\right )} \] Input:
integrate((((-16*x-32)*log(1/x^2)+16*x^2+32*x-8)*log(x)-4)*exp(log(1/x^2)^ 2-2*x*log(1/x^2)+x^2)^2/x^3/log(x)^2,x, algorithm="fricas")
Output:
-8*e^(2*x^2 - 4*x*log(x^(-2)) + 2*log(x^(-2))^2)/(x^2*log(x^(-2)))
Time = 0.10 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {e^{2 x^2-4 x \log \left (\frac {1}{x^2}\right )+2 \log ^2\left (\frac {1}{x^2}\right )} \left (-4+\left (-8+32 x+16 x^2+(-32-16 x) \log \left (\frac {1}{x^2}\right )\right ) \log (x)\right )}{x^3 \log ^2(x)} \, dx=\frac {4 e^{2 x^{2} + 8 x \log {\left (x \right )} + 8 \log {\left (x \right )}^{2}}}{x^{2} \log {\left (x \right )}} \] Input:
integrate((((-16*x-32)*ln(1/x**2)+16*x**2+32*x-8)*ln(x)-4)*exp(ln(1/x**2)* *2-2*x*ln(1/x**2)+x**2)**2/x**3/ln(x)**2,x)
Output:
4*exp(2*x**2 + 8*x*log(x) + 8*log(x)**2)/(x**2*log(x))
Time = 0.14 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {e^{2 x^2-4 x \log \left (\frac {1}{x^2}\right )+2 \log ^2\left (\frac {1}{x^2}\right )} \left (-4+\left (-8+32 x+16 x^2+(-32-16 x) \log \left (\frac {1}{x^2}\right )\right ) \log (x)\right )}{x^3 \log ^2(x)} \, dx=\frac {4 \, e^{\left (2 \, x^{2} + 8 \, x \log \left (x\right ) + 8 \, \log \left (x\right )^{2}\right )}}{x^{2} \log \left (x\right )} \] Input:
integrate((((-16*x-32)*log(1/x^2)+16*x^2+32*x-8)*log(x)-4)*exp(log(1/x^2)^ 2-2*x*log(1/x^2)+x^2)^2/x^3/log(x)^2,x, algorithm="maxima")
Output:
4*e^(2*x^2 + 8*x*log(x) + 8*log(x)^2)/(x^2*log(x))
Time = 0.21 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {e^{2 x^2-4 x \log \left (\frac {1}{x^2}\right )+2 \log ^2\left (\frac {1}{x^2}\right )} \left (-4+\left (-8+32 x+16 x^2+(-32-16 x) \log \left (\frac {1}{x^2}\right )\right ) \log (x)\right )}{x^3 \log ^2(x)} \, dx=\frac {4 \, e^{\left (2 \, x^{2} + 8 \, x \log \left (x\right ) + 8 \, \log \left (x\right )^{2}\right )}}{x^{2} \log \left (x\right )} \] Input:
integrate((((-16*x-32)*log(1/x^2)+16*x^2+32*x-8)*log(x)-4)*exp(log(1/x^2)^ 2-2*x*log(1/x^2)+x^2)^2/x^3/log(x)^2,x, algorithm="giac")
Output:
4*e^(2*x^2 + 8*x*log(x) + 8*log(x)^2)/(x^2*log(x))
Time = 3.29 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \[ \int \frac {e^{2 x^2-4 x \log \left (\frac {1}{x^2}\right )+2 \log ^2\left (\frac {1}{x^2}\right )} \left (-4+\left (-8+32 x+16 x^2+(-32-16 x) \log \left (\frac {1}{x^2}\right )\right ) \log (x)\right )}{x^3 \log ^2(x)} \, dx=\frac {4\,{\mathrm {e}}^{2\,{\ln \left (\frac {1}{x^2}\right )}^2}\,{\mathrm {e}}^{2\,x^2}\,{\left (x^8\right )}^x}{x^2\,\ln \left (x\right )} \] Input:
int((exp(2*log(1/x^2)^2 - 4*x*log(1/x^2) + 2*x^2)*(log(x)*(32*x + 16*x^2 - log(1/x^2)*(16*x + 32) - 8) - 4))/(x^3*log(x)^2),x)
Output:
(4*exp(2*log(1/x^2)^2)*exp(2*x^2)*(x^8)^x)/(x^2*log(x))
\[ \int \frac {e^{2 x^2-4 x \log \left (\frac {1}{x^2}\right )+2 \log ^2\left (\frac {1}{x^2}\right )} \left (-4+\left (-8+32 x+16 x^2+(-32-16 x) \log \left (\frac {1}{x^2}\right )\right ) \log (x)\right )}{x^3 \log ^2(x)} \, dx=32 \left (\int \frac {x^{8 x} e^{2 \mathrm {log}\left (x^{2}\right )^{2}+2 x^{2}} \mathrm {log}\left (x^{2}\right )}{\mathrm {log}\left (x \right ) x^{3}}d x \right )+16 \left (\int \frac {x^{8 x} e^{2 \mathrm {log}\left (x^{2}\right )^{2}+2 x^{2}} \mathrm {log}\left (x^{2}\right )}{\mathrm {log}\left (x \right ) x^{2}}d x \right )-4 \left (\int \frac {x^{8 x} e^{2 \mathrm {log}\left (x^{2}\right )^{2}+2 x^{2}}}{\mathrm {log}\left (x \right )^{2} x^{3}}d x \right )-8 \left (\int \frac {x^{8 x} e^{2 \mathrm {log}\left (x^{2}\right )^{2}+2 x^{2}}}{\mathrm {log}\left (x \right ) x^{3}}d x \right )+32 \left (\int \frac {x^{8 x} e^{2 \mathrm {log}\left (x^{2}\right )^{2}+2 x^{2}}}{\mathrm {log}\left (x \right ) x^{2}}d x \right )+16 \left (\int \frac {x^{8 x} e^{2 \mathrm {log}\left (x^{2}\right )^{2}+2 x^{2}}}{\mathrm {log}\left (x \right ) x}d x \right ) \] Input:
int((((-16*x-32)*log(1/x^2)+16*x^2+32*x-8)*log(x)-4)*exp(log(1/x^2)^2-2*x* log(1/x^2)+x^2)^2/x^3/log(x)^2,x)
Output:
4*(8*int((x**(8*x)*e**(2*log(x**2)**2 + 2*x**2)*log(x**2))/(log(x)*x**3),x ) + 4*int((x**(8*x)*e**(2*log(x**2)**2 + 2*x**2)*log(x**2))/(log(x)*x**2), x) - int((x**(8*x)*e**(2*log(x**2)**2 + 2*x**2))/(log(x)**2*x**3),x) - 2*i nt((x**(8*x)*e**(2*log(x**2)**2 + 2*x**2))/(log(x)*x**3),x) + 8*int((x**(8 *x)*e**(2*log(x**2)**2 + 2*x**2))/(log(x)*x**2),x) + 4*int((x**(8*x)*e**(2 *log(x**2)**2 + 2*x**2))/(log(x)*x),x))