Integrand size = 118, antiderivative size = 31 \[ \int \frac {e^{25/x} (-25-x)-x^2+e^{50/x} \left (x+8 x^2\right )+e^{25/x} \left (-2 x^2-16 x^3\right ) \log (x \log (3))+\left (x^3+8 x^4\right ) \log ^2(x \log (3))}{2 e^{50/x} x-4 e^{25/x} x^2 \log (x \log (3))+2 x^3 \log ^2(x \log (3))} \, dx=\frac {1}{2} \left (x+4 x^2+\frac {1}{-\frac {e^{25/x}}{x}+\log (x \log (3))}\right ) \] Output:
1/2*x+2*x^2+1/2/(ln(x*ln(3))-exp(25/x)/x)
Time = 0.07 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {e^{25/x} (-25-x)-x^2+e^{50/x} \left (x+8 x^2\right )+e^{25/x} \left (-2 x^2-16 x^3\right ) \log (x \log (3))+\left (x^3+8 x^4\right ) \log ^2(x \log (3))}{2 e^{50/x} x-4 e^{25/x} x^2 \log (x \log (3))+2 x^3 \log ^2(x \log (3))} \, dx=-\frac {1}{2} x \left (-1-4 x+\frac {1}{e^{25/x}-x \log (x \log (3))}\right ) \] Input:
Integrate[(E^(25/x)*(-25 - x) - x^2 + E^(50/x)*(x + 8*x^2) + E^(25/x)*(-2* x^2 - 16*x^3)*Log[x*Log[3]] + (x^3 + 8*x^4)*Log[x*Log[3]]^2)/(2*E^(50/x)*x - 4*E^(25/x)*x^2*Log[x*Log[3]] + 2*x^3*Log[x*Log[3]]^2),x]
Output:
-1/2*(x*(-1 - 4*x + (E^(25/x) - x*Log[x*Log[3]])^(-1)))
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.
\(\displaystyle \int \frac {-x^2+e^{50/x} \left (8 x^2+x\right )+\left (8 x^4+x^3\right ) \log ^2(x \log (3))+e^{25/x} \left (-16 x^3-2 x^2\right ) \log (x \log (3))+e^{25/x} (-x-25)}{2 x^3 \log ^2(x \log (3))-4 e^{25/x} x^2 \log (x \log (3))+2 e^{50/x} x} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-x^2+e^{50/x} \left (8 x^2+x\right )+\left (8 x^4+x^3\right ) \log ^2(x \log (3))+e^{25/x} \left (-16 x^3-2 x^2\right ) \log (x \log (3))+e^{25/x} (-x-25)}{2 x \left (e^{25/x}-x \log (x \log (3))\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \int -\frac {x^2-\left (8 x^4+x^3\right ) \log ^2(x \log (3))+e^{25/x} (x+25)-e^{50/x} \left (8 x^2+x\right )+2 e^{25/x} \left (8 x^3+x^2\right ) \log (x \log (3))}{x \left (e^{25/x}-x \log (x \log (3))\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{2} \int \frac {x^2-\left (8 x^4+x^3\right ) \log ^2(x \log (3))+e^{25/x} (x+25)-e^{50/x} \left (8 x^2+x\right )+2 e^{25/x} \left (8 x^3+x^2\right ) \log (x \log (3))}{x \left (e^{25/x}-x \log (x \log (3))\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {1}{2} \int \left (-8 x+\frac {\log (x \log (3)) x+x+25 \log (x \log (3))}{\left (e^{25/x}-x \log (x \log (3))\right )^2}-1-\frac {x+25}{\left (x \log (x \log (3))-e^{25/x}\right ) x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-\int \frac {1}{e^{25/x}-x \log (x \log (3))}dx-\int \frac {x}{\left (x \log (x \log (3))-e^{25/x}\right )^2}dx-25 \int \frac {\log (x \log (3))}{\left (x \log (x \log (3))-e^{25/x}\right )^2}dx-\int \frac {x \log (x \log (3))}{\left (x \log (x \log (3))-e^{25/x}\right )^2}dx+25 \int \frac {1}{x \left (x \log (x \log (3))-e^{25/x}\right )}dx+4 x^2+x\right )\) |
Input:
Int[(E^(25/x)*(-25 - x) - x^2 + E^(50/x)*(x + 8*x^2) + E^(25/x)*(-2*x^2 - 16*x^3)*Log[x*Log[3]] + (x^3 + 8*x^4)*Log[x*Log[3]]^2)/(2*E^(50/x)*x - 4*E ^(25/x)*x^2*Log[x*Log[3]] + 2*x^3*Log[x*Log[3]]^2),x]
Output:
$Aborted
Time = 1.08 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00
method | result | size |
risch | \(2 x^{2}+\frac {x}{2}+\frac {x}{2 \ln \left (x \ln \left (3\right )\right ) x -2 \,{\mathrm e}^{\frac {25}{x}}}\) | \(31\) |
parallelrisch | \(-\frac {-4 x^{3} \ln \left (x \ln \left (3\right )\right )+4 \,{\mathrm e}^{\frac {25}{x}} x^{2}-x^{2} \ln \left (x \ln \left (3\right )\right )+x \,{\mathrm e}^{\frac {25}{x}}-x}{2 \left (\ln \left (x \ln \left (3\right )\right ) x -{\mathrm e}^{\frac {25}{x}}\right )}\) | \(64\) |
Input:
int(((8*x^4+x^3)*ln(x*ln(3))^2+(-16*x^3-2*x^2)*exp(25/x)*ln(x*ln(3))+(8*x^ 2+x)*exp(25/x)^2+(-x-25)*exp(25/x)-x^2)/(2*x^3*ln(x*ln(3))^2-4*x^2*exp(25/ x)*ln(x*ln(3))+2*x*exp(25/x)^2),x,method=_RETURNVERBOSE)
Output:
2*x^2+1/2*x+1/2*x/(ln(x*ln(3))*x-exp(25/x))
Time = 0.09 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.74 \[ \int \frac {e^{25/x} (-25-x)-x^2+e^{50/x} \left (x+8 x^2\right )+e^{25/x} \left (-2 x^2-16 x^3\right ) \log (x \log (3))+\left (x^3+8 x^4\right ) \log ^2(x \log (3))}{2 e^{50/x} x-4 e^{25/x} x^2 \log (x \log (3))+2 x^3 \log ^2(x \log (3))} \, dx=-\frac {{\left (4 \, x^{2} + x\right )} e^{\frac {25}{x}} - {\left (4 \, x^{3} + x^{2}\right )} \log \left (x \log \left (3\right )\right ) - x}{2 \, {\left (x \log \left (x \log \left (3\right )\right ) - e^{\frac {25}{x}}\right )}} \] Input:
integrate(((8*x^4+x^3)*log(x*log(3))^2+(-16*x^3-2*x^2)*exp(25/x)*log(x*log (3))+(8*x^2+x)*exp(25/x)^2+(-x-25)*exp(25/x)-x^2)/(2*x^3*log(x*log(3))^2-4 *x^2*exp(25/x)*log(x*log(3))+2*x*exp(25/x)^2),x, algorithm="fricas")
Output:
-1/2*((4*x^2 + x)*e^(25/x) - (4*x^3 + x^2)*log(x*log(3)) - x)/(x*log(x*log (3)) - e^(25/x))
Time = 0.11 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {e^{25/x} (-25-x)-x^2+e^{50/x} \left (x+8 x^2\right )+e^{25/x} \left (-2 x^2-16 x^3\right ) \log (x \log (3))+\left (x^3+8 x^4\right ) \log ^2(x \log (3))}{2 e^{50/x} x-4 e^{25/x} x^2 \log (x \log (3))+2 x^3 \log ^2(x \log (3))} \, dx=2 x^{2} + \frac {x}{2} - \frac {x}{- 2 x \log {\left (x \log {\left (3 \right )} \right )} + 2 e^{\frac {25}{x}}} \] Input:
integrate(((8*x**4+x**3)*ln(x*ln(3))**2+(-16*x**3-2*x**2)*exp(25/x)*ln(x*l n(3))+(8*x**2+x)*exp(25/x)**2+(-x-25)*exp(25/x)-x**2)/(2*x**3*ln(x*ln(3))* *2-4*x**2*exp(25/x)*ln(x*ln(3))+2*x*exp(25/x)**2),x)
Output:
2*x**2 + x/2 - x/(-2*x*log(x*log(3)) + 2*exp(25/x))
Leaf count of result is larger than twice the leaf count of optimal. 66 vs. \(2 (31) = 62\).
Time = 0.17 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.13 \[ \int \frac {e^{25/x} (-25-x)-x^2+e^{50/x} \left (x+8 x^2\right )+e^{25/x} \left (-2 x^2-16 x^3\right ) \log (x \log (3))+\left (x^3+8 x^4\right ) \log ^2(x \log (3))}{2 e^{50/x} x-4 e^{25/x} x^2 \log (x \log (3))+2 x^3 \log ^2(x \log (3))} \, dx=\frac {4 \, x^{3} \log \left (\log \left (3\right )\right ) + x^{2} \log \left (\log \left (3\right )\right ) - {\left (4 \, x^{2} + x\right )} e^{\frac {25}{x}} + {\left (4 \, x^{3} + x^{2}\right )} \log \left (x\right ) + x}{2 \, {\left (x \log \left (x\right ) + x \log \left (\log \left (3\right )\right ) - e^{\frac {25}{x}}\right )}} \] Input:
integrate(((8*x^4+x^3)*log(x*log(3))^2+(-16*x^3-2*x^2)*exp(25/x)*log(x*log (3))+(8*x^2+x)*exp(25/x)^2+(-x-25)*exp(25/x)-x^2)/(2*x^3*log(x*log(3))^2-4 *x^2*exp(25/x)*log(x*log(3))+2*x*exp(25/x)^2),x, algorithm="maxima")
Output:
1/2*(4*x^3*log(log(3)) + x^2*log(log(3)) - (4*x^2 + x)*e^(25/x) + (4*x^3 + x^2)*log(x) + x)/(x*log(x) + x*log(log(3)) - e^(25/x))
Leaf count of result is larger than twice the leaf count of optimal. 72 vs. \(2 (31) = 62\).
Time = 0.15 (sec) , antiderivative size = 72, normalized size of antiderivative = 2.32 \[ \int \frac {e^{25/x} (-25-x)-x^2+e^{50/x} \left (x+8 x^2\right )+e^{25/x} \left (-2 x^2-16 x^3\right ) \log (x \log (3))+\left (x^3+8 x^4\right ) \log ^2(x \log (3))}{2 e^{50/x} x-4 e^{25/x} x^2 \log (x \log (3))+2 x^3 \log ^2(x \log (3))} \, dx=\frac {4 \, x^{3} \log \left (x\right ) + 4 \, x^{3} \log \left (\log \left (3\right )\right ) - 4 \, x^{2} e^{\frac {25}{x}} + x^{2} \log \left (x\right ) + x^{2} \log \left (\log \left (3\right )\right ) - x e^{\frac {25}{x}} + x}{2 \, {\left (x \log \left (x\right ) + x \log \left (\log \left (3\right )\right ) - e^{\frac {25}{x}}\right )}} \] Input:
integrate(((8*x^4+x^3)*log(x*log(3))^2+(-16*x^3-2*x^2)*exp(25/x)*log(x*log (3))+(8*x^2+x)*exp(25/x)^2+(-x-25)*exp(25/x)-x^2)/(2*x^3*log(x*log(3))^2-4 *x^2*exp(25/x)*log(x*log(3))+2*x*exp(25/x)^2),x, algorithm="giac")
Output:
1/2*(4*x^3*log(x) + 4*x^3*log(log(3)) - 4*x^2*e^(25/x) + x^2*log(x) + x^2* log(log(3)) - x*e^(25/x) + x)/(x*log(x) + x*log(log(3)) - e^(25/x))
Time = 7.68 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.87 \[ \int \frac {e^{25/x} (-25-x)-x^2+e^{50/x} \left (x+8 x^2\right )+e^{25/x} \left (-2 x^2-16 x^3\right ) \log (x \log (3))+\left (x^3+8 x^4\right ) \log ^2(x \log (3))}{2 e^{50/x} x-4 e^{25/x} x^2 \log (x \log (3))+2 x^3 \log ^2(x \log (3))} \, dx=-\frac {x\,\left (x\,\ln \left (x\,\ln \left (3\right )\right )-4\,x\,{\mathrm {e}}^{25/x}-{\mathrm {e}}^{25/x}+4\,x^2\,\ln \left (x\,\ln \left (3\right )\right )+1\right )}{2\,\left ({\mathrm {e}}^{25/x}-x\,\ln \left (x\,\ln \left (3\right )\right )\right )} \] Input:
int(-(exp(25/x)*(x + 25) - exp(50/x)*(x + 8*x^2) + x^2 - log(x*log(3))^2*( x^3 + 8*x^4) + exp(25/x)*log(x*log(3))*(2*x^2 + 16*x^3))/(2*x^3*log(x*log( 3))^2 + 2*x*exp(50/x) - 4*x^2*exp(25/x)*log(x*log(3))),x)
Output:
-(x*(x*log(x*log(3)) - 4*x*exp(25/x) - exp(25/x) + 4*x^2*log(x*log(3)) + 1 ))/(2*(exp(25/x) - x*log(x*log(3))))
Time = 0.20 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.90 \[ \int \frac {e^{25/x} (-25-x)-x^2+e^{50/x} \left (x+8 x^2\right )+e^{25/x} \left (-2 x^2-16 x^3\right ) \log (x \log (3))+\left (x^3+8 x^4\right ) \log ^2(x \log (3))}{2 e^{50/x} x-4 e^{25/x} x^2 \log (x \log (3))+2 x^3 \log ^2(x \log (3))} \, dx=\frac {x \left (4 e^{\frac {25}{x}} x +e^{\frac {25}{x}}-4 \,\mathrm {log}\left (\mathrm {log}\left (3\right ) x \right ) x^{2}-\mathrm {log}\left (\mathrm {log}\left (3\right ) x \right ) x -1\right )}{2 e^{\frac {25}{x}}-2 \,\mathrm {log}\left (\mathrm {log}\left (3\right ) x \right ) x} \] Input:
int(((8*x^4+x^3)*log(x*log(3))^2+(-16*x^3-2*x^2)*exp(25/x)*log(x*log(3))+( 8*x^2+x)*exp(25/x)^2+(-x-25)*exp(25/x)-x^2)/(2*x^3*log(x*log(3))^2-4*x^2*e xp(25/x)*log(x*log(3))+2*x*exp(25/x)^2),x)
Output:
(x*(4*e**(25/x)*x + e**(25/x) - 4*log(log(3)*x)*x**2 - log(log(3)*x)*x - 1 ))/(2*(e**(25/x) - log(log(3)*x)*x))