Integrand size = 245, antiderivative size = 26 \[ \int \frac {e^4 \left (4 x^4-4 x^5+x^6\right )+e^{\frac {2 x^2}{e^4 (-2+x)}} \left (-512 x^3+128 x^4+e^4 \left (-512+384 x^2-128 x^3\right )\right )+\left (e^{\frac {2 x^2}{e^4 (-2+x)}} \left (-512 x^2+128 x^3\right )+e^4 \left (12 x^3-12 x^4+3 x^5\right )\right ) \log (x)+e^4 \left (12 x^2-12 x^3+3 x^4\right ) \log ^2(x)+e^4 \left (4 x-4 x^2+x^3\right ) \log ^3(x)}{e^4 \left (4 x^4-4 x^5+x^6\right )+e^4 \left (12 x^3-12 x^4+3 x^5\right ) \log (x)+e^4 \left (12 x^2-12 x^3+3 x^4\right ) \log ^2(x)+e^4 \left (4 x-4 x^2+x^3\right ) \log ^3(x)} \, dx=-3+x+\frac {64 e^{\frac {2 x^2}{e^4 (-2+x)}}}{(x+\log (x))^2} \] Output:
64*exp(x^2/(-2+x)/exp(2)^2)^2/(x+ln(x))^2-3+x
Time = 0.14 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {e^4 \left (4 x^4-4 x^5+x^6\right )+e^{\frac {2 x^2}{e^4 (-2+x)}} \left (-512 x^3+128 x^4+e^4 \left (-512+384 x^2-128 x^3\right )\right )+\left (e^{\frac {2 x^2}{e^4 (-2+x)}} \left (-512 x^2+128 x^3\right )+e^4 \left (12 x^3-12 x^4+3 x^5\right )\right ) \log (x)+e^4 \left (12 x^2-12 x^3+3 x^4\right ) \log ^2(x)+e^4 \left (4 x-4 x^2+x^3\right ) \log ^3(x)}{e^4 \left (4 x^4-4 x^5+x^6\right )+e^4 \left (12 x^3-12 x^4+3 x^5\right ) \log (x)+e^4 \left (12 x^2-12 x^3+3 x^4\right ) \log ^2(x)+e^4 \left (4 x-4 x^2+x^3\right ) \log ^3(x)} \, dx=x+\frac {64 e^{\frac {2 x^2}{e^4 (-2+x)}}}{(x+\log (x))^2} \] Input:
Integrate[(E^4*(4*x^4 - 4*x^5 + x^6) + E^((2*x^2)/(E^4*(-2 + x)))*(-512*x^ 3 + 128*x^4 + E^4*(-512 + 384*x^2 - 128*x^3)) + (E^((2*x^2)/(E^4*(-2 + x)) )*(-512*x^2 + 128*x^3) + E^4*(12*x^3 - 12*x^4 + 3*x^5))*Log[x] + E^4*(12*x ^2 - 12*x^3 + 3*x^4)*Log[x]^2 + E^4*(4*x - 4*x^2 + x^3)*Log[x]^3)/(E^4*(4* x^4 - 4*x^5 + x^6) + E^4*(12*x^3 - 12*x^4 + 3*x^5)*Log[x] + E^4*(12*x^2 - 12*x^3 + 3*x^4)*Log[x]^2 + E^4*(4*x - 4*x^2 + x^3)*Log[x]^3),x]
Output:
x + (64*E^((2*x^2)/(E^4*(-2 + x))))/(x + Log[x])^2
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.
\(\displaystyle \int \frac {e^4 \left (x^3-4 x^2+4 x\right ) \log ^3(x)+e^4 \left (x^6-4 x^5+4 x^4\right )+e^{\frac {2 x^2}{e^4 (x-2)}} \left (128 x^4-512 x^3+e^4 \left (-128 x^3+384 x^2-512\right )\right )+e^4 \left (3 x^4-12 x^3+12 x^2\right ) \log ^2(x)+\left (e^{\frac {2 x^2}{e^4 (x-2)}} \left (128 x^3-512 x^2\right )+e^4 \left (3 x^5-12 x^4+12 x^3\right )\right ) \log (x)}{e^4 \left (x^3-4 x^2+4 x\right ) \log ^3(x)+e^4 \left (x^6-4 x^5+4 x^4\right )+e^4 \left (3 x^5-12 x^4+12 x^3\right ) \log (x)+e^4 \left (3 x^4-12 x^3+12 x^2\right ) \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^4 \left (x^3-4 x^2+4 x\right ) \log ^3(x)+e^4 \left (x^6-4 x^5+4 x^4\right )+e^{\frac {2 x^2}{e^4 (x-2)}} \left (128 x^4-512 x^3+e^4 \left (-128 x^3+384 x^2-512\right )\right )+e^4 \left (3 x^4-12 x^3+12 x^2\right ) \log ^2(x)+\left (e^{\frac {2 x^2}{e^4 (x-2)}} \left (128 x^3-512 x^2\right )+e^4 \left (3 x^5-12 x^4+12 x^3\right )\right ) \log (x)}{e^4 (2-x)^2 x (x+\log (x))^3}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {e^4 \left (x^3-4 x^2+4 x\right ) \log ^3(x)+3 e^4 \left (x^4-4 x^3+4 x^2\right ) \log ^2(x)-\left (128 e^{-\frac {2 x^2}{e^4 (2-x)}} \left (4 x^2-x^3\right )-3 e^4 \left (x^5-4 x^4+4 x^3\right )\right ) \log (x)+e^4 \left (x^6-4 x^5+4 x^4\right )-128 e^{-\frac {2 x^2}{e^4 (2-x)}} \left (-x^4+4 x^3+e^4 \left (x^3-3 x^2+4\right )\right )}{(2-x)^2 x (x+\log (x))^3}dx}{e^4}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {\int \left (\frac {e^4 x^3}{(x+\log (x))^3}+\frac {3 e^4 \log (x) x^2}{(x+\log (x))^3}+\frac {3 e^4 \log ^2(x) x}{(x+\log (x))^3}+\frac {e^4 \log ^3(x)}{(x+\log (x))^3}+\frac {128 e^{\frac {2 x^2}{e^4 (x-2)}} \left (x^4+\log (x) x^3-4 \left (1+\frac {e^4}{4}\right ) x^3-4 \log (x) x^2+3 e^4 x^2-4 e^4\right )}{(2-x)^2 (x+\log (x))^3 x}\right )dx}{e^4}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-128 \int \frac {e^{\frac {2 x^2}{e^4 (x-2)}+4}}{(x+\log (x))^3}dx-128 \int \frac {e^{\frac {2 x^2}{e^4 (x-2)}+4}}{x (x+\log (x))^3}dx+128 \int \frac {e^{\frac {2 x^2}{e^4 (x-2)}}}{(x+\log (x))^2}dx-512 \int \frac {e^{\frac {2 x^2}{e^4 (x-2)}}}{(x-2)^2 (x+\log (x))^2}dx+e^4 x}{e^4}\) |
Input:
Int[(E^4*(4*x^4 - 4*x^5 + x^6) + E^((2*x^2)/(E^4*(-2 + x)))*(-512*x^3 + 12 8*x^4 + E^4*(-512 + 384*x^2 - 128*x^3)) + (E^((2*x^2)/(E^4*(-2 + x)))*(-51 2*x^2 + 128*x^3) + E^4*(12*x^3 - 12*x^4 + 3*x^5))*Log[x] + E^4*(12*x^2 - 1 2*x^3 + 3*x^4)*Log[x]^2 + E^4*(4*x - 4*x^2 + x^3)*Log[x]^3)/(E^4*(4*x^4 - 4*x^5 + x^6) + E^4*(12*x^3 - 12*x^4 + 3*x^5)*Log[x] + E^4*(12*x^2 - 12*x^3 + 3*x^4)*Log[x]^2 + E^4*(4*x - 4*x^2 + x^3)*Log[x]^3),x]
Output:
$Aborted
Time = 1.04 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92
\[x +\frac {64 \,{\mathrm e}^{\frac {2 x^{2} {\mathrm e}^{-4}}{-2+x}}}{\left (x +\ln \left (x \right )\right )^{2}}\]
Input:
int(((x^3-4*x^2+4*x)*exp(2)^2*ln(x)^3+(3*x^4-12*x^3+12*x^2)*exp(2)^2*ln(x) ^2+((128*x^3-512*x^2)*exp(x^2/(-2+x)/exp(2)^2)^2+(3*x^5-12*x^4+12*x^3)*exp (2)^2)*ln(x)+((-128*x^3+384*x^2-512)*exp(2)^2+128*x^4-512*x^3)*exp(x^2/(-2 +x)/exp(2)^2)^2+(x^6-4*x^5+4*x^4)*exp(2)^2)/((x^3-4*x^2+4*x)*exp(2)^2*ln(x )^3+(3*x^4-12*x^3+12*x^2)*exp(2)^2*ln(x)^2+(3*x^5-12*x^4+12*x^3)*exp(2)^2* ln(x)+(x^6-4*x^5+4*x^4)*exp(2)^2),x)
Output:
x+64*exp(2*x^2/(-2+x)*exp(-4))/(x+ln(x))^2
Time = 0.12 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.85 \[ \int \frac {e^4 \left (4 x^4-4 x^5+x^6\right )+e^{\frac {2 x^2}{e^4 (-2+x)}} \left (-512 x^3+128 x^4+e^4 \left (-512+384 x^2-128 x^3\right )\right )+\left (e^{\frac {2 x^2}{e^4 (-2+x)}} \left (-512 x^2+128 x^3\right )+e^4 \left (12 x^3-12 x^4+3 x^5\right )\right ) \log (x)+e^4 \left (12 x^2-12 x^3+3 x^4\right ) \log ^2(x)+e^4 \left (4 x-4 x^2+x^3\right ) \log ^3(x)}{e^4 \left (4 x^4-4 x^5+x^6\right )+e^4 \left (12 x^3-12 x^4+3 x^5\right ) \log (x)+e^4 \left (12 x^2-12 x^3+3 x^4\right ) \log ^2(x)+e^4 \left (4 x-4 x^2+x^3\right ) \log ^3(x)} \, dx=\frac {x^{3} + 2 \, x^{2} \log \left (x\right ) + x \log \left (x\right )^{2} + 64 \, e^{\left (\frac {2 \, x^{2} e^{\left (-4\right )}}{x - 2}\right )}}{x^{2} + 2 \, x \log \left (x\right ) + \log \left (x\right )^{2}} \] Input:
integrate(((x^3-4*x^2+4*x)*exp(2)^2*log(x)^3+(3*x^4-12*x^3+12*x^2)*exp(2)^ 2*log(x)^2+((128*x^3-512*x^2)*exp(x^2/(-2+x)/exp(2)^2)^2+(3*x^5-12*x^4+12* x^3)*exp(2)^2)*log(x)+((-128*x^3+384*x^2-512)*exp(2)^2+128*x^4-512*x^3)*ex p(x^2/(-2+x)/exp(2)^2)^2+(x^6-4*x^5+4*x^4)*exp(2)^2)/((x^3-4*x^2+4*x)*exp( 2)^2*log(x)^3+(3*x^4-12*x^3+12*x^2)*exp(2)^2*log(x)^2+(3*x^5-12*x^4+12*x^3 )*exp(2)^2*log(x)+(x^6-4*x^5+4*x^4)*exp(2)^2),x, algorithm="fricas")
Output:
(x^3 + 2*x^2*log(x) + x*log(x)^2 + 64*e^(2*x^2*e^(-4)/(x - 2)))/(x^2 + 2*x *log(x) + log(x)^2)
Time = 0.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {e^4 \left (4 x^4-4 x^5+x^6\right )+e^{\frac {2 x^2}{e^4 (-2+x)}} \left (-512 x^3+128 x^4+e^4 \left (-512+384 x^2-128 x^3\right )\right )+\left (e^{\frac {2 x^2}{e^4 (-2+x)}} \left (-512 x^2+128 x^3\right )+e^4 \left (12 x^3-12 x^4+3 x^5\right )\right ) \log (x)+e^4 \left (12 x^2-12 x^3+3 x^4\right ) \log ^2(x)+e^4 \left (4 x-4 x^2+x^3\right ) \log ^3(x)}{e^4 \left (4 x^4-4 x^5+x^6\right )+e^4 \left (12 x^3-12 x^4+3 x^5\right ) \log (x)+e^4 \left (12 x^2-12 x^3+3 x^4\right ) \log ^2(x)+e^4 \left (4 x-4 x^2+x^3\right ) \log ^3(x)} \, dx=x + \frac {64 e^{\frac {2 x^{2}}{\left (x - 2\right ) e^{4}}}}{x^{2} + 2 x \log {\left (x \right )} + \log {\left (x \right )}^{2}} \] Input:
integrate(((x**3-4*x**2+4*x)*exp(2)**2*ln(x)**3+(3*x**4-12*x**3+12*x**2)*e xp(2)**2*ln(x)**2+((128*x**3-512*x**2)*exp(x**2/(-2+x)/exp(2)**2)**2+(3*x* *5-12*x**4+12*x**3)*exp(2)**2)*ln(x)+((-128*x**3+384*x**2-512)*exp(2)**2+1 28*x**4-512*x**3)*exp(x**2/(-2+x)/exp(2)**2)**2+(x**6-4*x**5+4*x**4)*exp(2 )**2)/((x**3-4*x**2+4*x)*exp(2)**2*ln(x)**3+(3*x**4-12*x**3+12*x**2)*exp(2 )**2*ln(x)**2+(3*x**5-12*x**4+12*x**3)*exp(2)**2*ln(x)+(x**6-4*x**5+4*x**4 )*exp(2)**2),x)
Output:
x + 64*exp(2*x**2*exp(-4)/(x - 2))/(x**2 + 2*x*log(x) + log(x)**2)
Leaf count of result is larger than twice the leaf count of optimal. 59 vs. \(2 (24) = 48\).
Time = 0.11 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.27 \[ \int \frac {e^4 \left (4 x^4-4 x^5+x^6\right )+e^{\frac {2 x^2}{e^4 (-2+x)}} \left (-512 x^3+128 x^4+e^4 \left (-512+384 x^2-128 x^3\right )\right )+\left (e^{\frac {2 x^2}{e^4 (-2+x)}} \left (-512 x^2+128 x^3\right )+e^4 \left (12 x^3-12 x^4+3 x^5\right )\right ) \log (x)+e^4 \left (12 x^2-12 x^3+3 x^4\right ) \log ^2(x)+e^4 \left (4 x-4 x^2+x^3\right ) \log ^3(x)}{e^4 \left (4 x^4-4 x^5+x^6\right )+e^4 \left (12 x^3-12 x^4+3 x^5\right ) \log (x)+e^4 \left (12 x^2-12 x^3+3 x^4\right ) \log ^2(x)+e^4 \left (4 x-4 x^2+x^3\right ) \log ^3(x)} \, dx=\frac {x^{3} + 2 \, x^{2} \log \left (x\right ) + x \log \left (x\right )^{2} + 64 \, e^{\left (2 \, x e^{\left (-4\right )} + \frac {8}{x e^{4} - 2 \, e^{4}} + 4 \, e^{\left (-4\right )}\right )}}{x^{2} + 2 \, x \log \left (x\right ) + \log \left (x\right )^{2}} \] Input:
integrate(((x^3-4*x^2+4*x)*exp(2)^2*log(x)^3+(3*x^4-12*x^3+12*x^2)*exp(2)^ 2*log(x)^2+((128*x^3-512*x^2)*exp(x^2/(-2+x)/exp(2)^2)^2+(3*x^5-12*x^4+12* x^3)*exp(2)^2)*log(x)+((-128*x^3+384*x^2-512)*exp(2)^2+128*x^4-512*x^3)*ex p(x^2/(-2+x)/exp(2)^2)^2+(x^6-4*x^5+4*x^4)*exp(2)^2)/((x^3-4*x^2+4*x)*exp( 2)^2*log(x)^3+(3*x^4-12*x^3+12*x^2)*exp(2)^2*log(x)^2+(3*x^5-12*x^4+12*x^3 )*exp(2)^2*log(x)+(x^6-4*x^5+4*x^4)*exp(2)^2),x, algorithm="maxima")
Output:
(x^3 + 2*x^2*log(x) + x*log(x)^2 + 64*e^(2*x*e^(-4) + 8/(x*e^4 - 2*e^4) + 4*e^(-4)))/(x^2 + 2*x*log(x) + log(x)^2)
Leaf count of result is larger than twice the leaf count of optimal. 52 vs. \(2 (24) = 48\).
Time = 0.23 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.00 \[ \int \frac {e^4 \left (4 x^4-4 x^5+x^6\right )+e^{\frac {2 x^2}{e^4 (-2+x)}} \left (-512 x^3+128 x^4+e^4 \left (-512+384 x^2-128 x^3\right )\right )+\left (e^{\frac {2 x^2}{e^4 (-2+x)}} \left (-512 x^2+128 x^3\right )+e^4 \left (12 x^3-12 x^4+3 x^5\right )\right ) \log (x)+e^4 \left (12 x^2-12 x^3+3 x^4\right ) \log ^2(x)+e^4 \left (4 x-4 x^2+x^3\right ) \log ^3(x)}{e^4 \left (4 x^4-4 x^5+x^6\right )+e^4 \left (12 x^3-12 x^4+3 x^5\right ) \log (x)+e^4 \left (12 x^2-12 x^3+3 x^4\right ) \log ^2(x)+e^4 \left (4 x-4 x^2+x^3\right ) \log ^3(x)} \, dx=\frac {x^{3} + 2 \, x^{2} \log \left (x\right ) + x \log \left (x\right )^{2} + 64 \, e^{\left (\frac {2 \, x^{2}}{x e^{4} - 2 \, e^{4}}\right )}}{x^{2} + 2 \, x \log \left (x\right ) + \log \left (x\right )^{2}} \] Input:
integrate(((x^3-4*x^2+4*x)*exp(2)^2*log(x)^3+(3*x^4-12*x^3+12*x^2)*exp(2)^ 2*log(x)^2+((128*x^3-512*x^2)*exp(x^2/(-2+x)/exp(2)^2)^2+(3*x^5-12*x^4+12* x^3)*exp(2)^2)*log(x)+((-128*x^3+384*x^2-512)*exp(2)^2+128*x^4-512*x^3)*ex p(x^2/(-2+x)/exp(2)^2)^2+(x^6-4*x^5+4*x^4)*exp(2)^2)/((x^3-4*x^2+4*x)*exp( 2)^2*log(x)^3+(3*x^4-12*x^3+12*x^2)*exp(2)^2*log(x)^2+(3*x^5-12*x^4+12*x^3 )*exp(2)^2*log(x)+(x^6-4*x^5+4*x^4)*exp(2)^2),x, algorithm="giac")
Output:
(x^3 + 2*x^2*log(x) + x*log(x)^2 + 64*e^(2*x^2/(x*e^4 - 2*e^4)))/(x^2 + 2* x*log(x) + log(x)^2)
Time = 3.20 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {e^4 \left (4 x^4-4 x^5+x^6\right )+e^{\frac {2 x^2}{e^4 (-2+x)}} \left (-512 x^3+128 x^4+e^4 \left (-512+384 x^2-128 x^3\right )\right )+\left (e^{\frac {2 x^2}{e^4 (-2+x)}} \left (-512 x^2+128 x^3\right )+e^4 \left (12 x^3-12 x^4+3 x^5\right )\right ) \log (x)+e^4 \left (12 x^2-12 x^3+3 x^4\right ) \log ^2(x)+e^4 \left (4 x-4 x^2+x^3\right ) \log ^3(x)}{e^4 \left (4 x^4-4 x^5+x^6\right )+e^4 \left (12 x^3-12 x^4+3 x^5\right ) \log (x)+e^4 \left (12 x^2-12 x^3+3 x^4\right ) \log ^2(x)+e^4 \left (4 x-4 x^2+x^3\right ) \log ^3(x)} \, dx=x+\frac {64\,{\mathrm {e}}^{\frac {2\,x^2\,{\mathrm {e}}^{-4}}{x-2}}}{{\left (x+\ln \left (x\right )\right )}^2} \] Input:
int((exp(4)*(4*x^4 - 4*x^5 + x^6) - log(x)*(exp((2*x^2*exp(-4))/(x - 2))*( 512*x^2 - 128*x^3) - exp(4)*(12*x^3 - 12*x^4 + 3*x^5)) - exp((2*x^2*exp(-4 ))/(x - 2))*(exp(4)*(128*x^3 - 384*x^2 + 512) + 512*x^3 - 128*x^4) + exp(4 )*log(x)^2*(12*x^2 - 12*x^3 + 3*x^4) + exp(4)*log(x)^3*(4*x - 4*x^2 + x^3) )/(exp(4)*(4*x^4 - 4*x^5 + x^6) + exp(4)*log(x)^2*(12*x^2 - 12*x^3 + 3*x^4 ) + exp(4)*log(x)^3*(4*x - 4*x^2 + x^3) + exp(4)*log(x)*(12*x^3 - 12*x^4 + 3*x^5)),x)
Output:
x + (64*exp((2*x^2*exp(-4))/(x - 2)))/(x + log(x))^2
Time = 0.22 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.12 \[ \int \frac {e^4 \left (4 x^4-4 x^5+x^6\right )+e^{\frac {2 x^2}{e^4 (-2+x)}} \left (-512 x^3+128 x^4+e^4 \left (-512+384 x^2-128 x^3\right )\right )+\left (e^{\frac {2 x^2}{e^4 (-2+x)}} \left (-512 x^2+128 x^3\right )+e^4 \left (12 x^3-12 x^4+3 x^5\right )\right ) \log (x)+e^4 \left (12 x^2-12 x^3+3 x^4\right ) \log ^2(x)+e^4 \left (4 x-4 x^2+x^3\right ) \log ^3(x)}{e^4 \left (4 x^4-4 x^5+x^6\right )+e^4 \left (12 x^3-12 x^4+3 x^5\right ) \log (x)+e^4 \left (12 x^2-12 x^3+3 x^4\right ) \log ^2(x)+e^4 \left (4 x-4 x^2+x^3\right ) \log ^3(x)} \, dx=\frac {64 e^{\frac {2 x^{2}}{e^{4} x -2 e^{4}}}+\mathrm {log}\left (x \right )^{2} x +2 \,\mathrm {log}\left (x \right ) x^{2}+x^{3}}{\mathrm {log}\left (x \right )^{2}+2 \,\mathrm {log}\left (x \right ) x +x^{2}} \] Input:
int(((x^3-4*x^2+4*x)*exp(2)^2*log(x)^3+(3*x^4-12*x^3+12*x^2)*exp(2)^2*log( x)^2+((128*x^3-512*x^2)*exp(x^2/(-2+x)/exp(2)^2)^2+(3*x^5-12*x^4+12*x^3)*e xp(2)^2)*log(x)+((-128*x^3+384*x^2-512)*exp(2)^2+128*x^4-512*x^3)*exp(x^2/ (-2+x)/exp(2)^2)^2+(x^6-4*x^5+4*x^4)*exp(2)^2)/((x^3-4*x^2+4*x)*exp(2)^2*l og(x)^3+(3*x^4-12*x^3+12*x^2)*exp(2)^2*log(x)^2+(3*x^5-12*x^4+12*x^3)*exp( 2)^2*log(x)+(x^6-4*x^5+4*x^4)*exp(2)^2),x)
Output:
(64*e**((2*x**2)/(e**4*x - 2*e**4)) + log(x)**2*x + 2*log(x)*x**2 + x**3)/ (log(x)**2 + 2*log(x)*x + x**2)