Integrand size = 113, antiderivative size = 21 \[ \int \frac {x+e^{2 x} x+e^{-8 \log ^2(x)} x+x^3+e^x \left (-x-2 x^2\right )+e^{-4 \log ^2(x)} \left (-2 e^x x+2 x^2-8 \log (x)\right )}{e^{2 x} x+e^{-8 \log ^2(x)} x-2 e^x x^2+x^3+e^{-4 \log ^2(x)} \left (-2 e^x x+2 x^2\right )} \, dx=\frac {1}{e^x-e^{-4 \log ^2(x)}-x}+x \] Output:
x+1/(exp(x)-exp(-ln(x)^2)^4-x)
Time = 0.12 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.71 \[ \int \frac {x+e^{2 x} x+e^{-8 \log ^2(x)} x+x^3+e^x \left (-x-2 x^2\right )+e^{-4 \log ^2(x)} \left (-2 e^x x+2 x^2-8 \log (x)\right )}{e^{2 x} x+e^{-8 \log ^2(x)} x-2 e^x x^2+x^3+e^{-4 \log ^2(x)} \left (-2 e^x x+2 x^2\right )} \, dx=x+\frac {e^{4 \log ^2(x)}}{-1+e^{x+4 \log ^2(x)}-e^{4 \log ^2(x)} x} \] Input:
Integrate[(x + E^(2*x)*x + x/E^(8*Log[x]^2) + x^3 + E^x*(-x - 2*x^2) + (-2 *E^x*x + 2*x^2 - 8*Log[x])/E^(4*Log[x]^2))/(E^(2*x)*x + x/E^(8*Log[x]^2) - 2*E^x*x^2 + x^3 + (-2*E^x*x + 2*x^2)/E^(4*Log[x]^2)),x]
Output:
x + E^(4*Log[x]^2)/(-1 + E^(x + 4*Log[x]^2) - E^(4*Log[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.
\(\displaystyle \int \frac {x^3+e^x \left (-2 x^2-x\right )+e^{-4 \log ^2(x)} \left (2 x^2-2 e^x x-8 \log (x)\right )+e^{2 x} x+x+x e^{-8 \log ^2(x)}}{x^3-2 e^x x^2+\left (2 x^2-2 e^x x\right ) e^{-4 \log ^2(x)}+e^{2 x} x+x e^{-8 \log ^2(x)}} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{8 \log ^2(x)} \left (x^3+e^x \left (-2 x^2-x\right )+e^{-4 \log ^2(x)} \left (2 x^2-2 e^x x-8 \log (x)\right )+e^{2 x} x+x+x e^{-8 \log ^2(x)}\right )}{x \left (x e^{4 \log ^2(x)}-e^{x+4 \log ^2(x)}+1\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {8 e^{4 \log ^2(x)} \log (x)}{x}+\frac {8 \left (x-e^x\right ) e^{8 \log ^2(x)} \log (x)}{x \left (x e^{4 \log ^2(x)}-e^{x+4 \log ^2(x)}+1\right )}-\frac {e^{8 \log ^2(x)} \left (e^x x-x-8 x \log (x)+8 e^x \log (x)\right )}{x \left (x e^{4 \log ^2(x)}-e^{x+4 \log ^2(x)}+1\right )^2}+1\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int \frac {e^{8 \log ^2(x)}}{\left (-e^{4 \log ^2(x)} x+e^{4 \log ^2(x)+x}-1\right )^2}dx-\int \frac {e^{8 \log ^2(x)+x}}{\left (-e^{4 \log ^2(x)} x+e^{4 \log ^2(x)+x}-1\right )^2}dx+8 \int \frac {e^{8 \log ^2(x)} \log (x)}{\left (-e^{4 \log ^2(x)} x+e^{4 \log ^2(x)+x}-1\right )^2}dx-8 \int \frac {e^{8 \log ^2(x)+x} \log (x)}{x \left (-e^{4 \log ^2(x)} x+e^{4 \log ^2(x)+x}-1\right )^2}dx-8 \int \frac {e^{8 \log ^2(x)} \log (x)}{-e^{4 \log ^2(x)} x+e^{4 \log ^2(x)+x}-1}dx+8 \int \frac {e^{8 \log ^2(x)+x} \log (x)}{x \left (-e^{4 \log ^2(x)} x+e^{4 \log ^2(x)+x}-1\right )}dx+x-e^{4 \log ^2(x)}\) |
Input:
Int[(x + E^(2*x)*x + x/E^(8*Log[x]^2) + x^3 + E^x*(-x - 2*x^2) + (-2*E^x*x + 2*x^2 - 8*Log[x])/E^(4*Log[x]^2))/(E^(2*x)*x + x/E^(8*Log[x]^2) - 2*E^x *x^2 + x^3 + (-2*E^x*x + 2*x^2)/E^(4*Log[x]^2)),x]
Output:
$Aborted
Time = 8.03 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95
method | result | size |
risch | \(x -\frac {1}{{\mathrm e}^{-4 \ln \left (x \right )^{2}}-{\mathrm e}^{x}+x}\) | \(20\) |
parallelrisch | \(\frac {{\mathrm e}^{-4 \ln \left (x \right )^{2}} x -1+x^{2}-{\mathrm e}^{x} x}{{\mathrm e}^{-4 \ln \left (x \right )^{2}}-{\mathrm e}^{x}+x}\) | \(40\) |
Input:
int((x*exp(-ln(x)^2)^8+(-8*ln(x)-2*exp(x)*x+2*x^2)*exp(-ln(x)^2)^4+x*exp(x )^2+(-2*x^2-x)*exp(x)+x^3+x)/(x*exp(-ln(x)^2)^8+(-2*exp(x)*x+2*x^2)*exp(-l n(x)^2)^4+x*exp(x)^2-2*exp(x)*x^2+x^3),x,method=_RETURNVERBOSE)
Output:
x-1/(exp(-4*ln(x)^2)-exp(x)+x)
Time = 0.10 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.67 \[ \int \frac {x+e^{2 x} x+e^{-8 \log ^2(x)} x+x^3+e^x \left (-x-2 x^2\right )+e^{-4 \log ^2(x)} \left (-2 e^x x+2 x^2-8 \log (x)\right )}{e^{2 x} x+e^{-8 \log ^2(x)} x-2 e^x x^2+x^3+e^{-4 \log ^2(x)} \left (-2 e^x x+2 x^2\right )} \, dx=\frac {x^{2} + x e^{\left (-4 \, \log \left (x\right )^{2}\right )} - x e^{x} - 1}{x + e^{\left (-4 \, \log \left (x\right )^{2}\right )} - e^{x}} \] Input:
integrate((x*exp(-log(x)^2)^8+(-8*log(x)-2*exp(x)*x+2*x^2)*exp(-log(x)^2)^ 4+x*exp(x)^2+(-2*x^2-x)*exp(x)+x^3+x)/(x*exp(-log(x)^2)^8+(-2*exp(x)*x+2*x ^2)*exp(-log(x)^2)^4+x*exp(x)^2-2*exp(x)*x^2+x^3),x, algorithm="fricas")
Output:
(x^2 + x*e^(-4*log(x)^2) - x*e^x - 1)/(x + e^(-4*log(x)^2) - e^x)
Time = 0.12 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {x+e^{2 x} x+e^{-8 \log ^2(x)} x+x^3+e^x \left (-x-2 x^2\right )+e^{-4 \log ^2(x)} \left (-2 e^x x+2 x^2-8 \log (x)\right )}{e^{2 x} x+e^{-8 \log ^2(x)} x-2 e^x x^2+x^3+e^{-4 \log ^2(x)} \left (-2 e^x x+2 x^2\right )} \, dx=x + \frac {1}{- x + e^{x} - e^{- 4 \log {\left (x \right )}^{2}}} \] Input:
integrate((x*exp(-ln(x)**2)**8+(-8*ln(x)-2*exp(x)*x+2*x**2)*exp(-ln(x)**2) **4+x*exp(x)**2+(-2*x**2-x)*exp(x)+x**3+x)/(x*exp(-ln(x)**2)**8+(-2*exp(x) *x+2*x**2)*exp(-ln(x)**2)**4+x*exp(x)**2-2*exp(x)*x**2+x**3),x)
Output:
x + 1/(-x + exp(x) - exp(-4*log(x)**2))
Leaf count of result is larger than twice the leaf count of optimal. 39 vs. \(2 (19) = 38\).
Time = 0.11 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.86 \[ \int \frac {x+e^{2 x} x+e^{-8 \log ^2(x)} x+x^3+e^x \left (-x-2 x^2\right )+e^{-4 \log ^2(x)} \left (-2 e^x x+2 x^2-8 \log (x)\right )}{e^{2 x} x+e^{-8 \log ^2(x)} x-2 e^x x^2+x^3+e^{-4 \log ^2(x)} \left (-2 e^x x+2 x^2\right )} \, dx=\frac {{\left (x^{2} - x e^{x} - 1\right )} e^{\left (4 \, \log \left (x\right )^{2}\right )} + x}{{\left (x - e^{x}\right )} e^{\left (4 \, \log \left (x\right )^{2}\right )} + 1} \] Input:
integrate((x*exp(-log(x)^2)^8+(-8*log(x)-2*exp(x)*x+2*x^2)*exp(-log(x)^2)^ 4+x*exp(x)^2+(-2*x^2-x)*exp(x)+x^3+x)/(x*exp(-log(x)^2)^8+(-2*exp(x)*x+2*x ^2)*exp(-log(x)^2)^4+x*exp(x)^2-2*exp(x)*x^2+x^3),x, algorithm="maxima")
Output:
((x^2 - x*e^x - 1)*e^(4*log(x)^2) + x)/((x - e^x)*e^(4*log(x)^2) + 1)
Leaf count of result is larger than twice the leaf count of optimal. 607 vs. \(2 (19) = 38\).
Time = 0.21 (sec) , antiderivative size = 607, normalized size of antiderivative = 28.90 \[ \int \frac {x+e^{2 x} x+e^{-8 \log ^2(x)} x+x^3+e^x \left (-x-2 x^2\right )+e^{-4 \log ^2(x)} \left (-2 e^x x+2 x^2-8 \log (x)\right )}{e^{2 x} x+e^{-8 \log ^2(x)} x-2 e^x x^2+x^3+e^{-4 \log ^2(x)} \left (-2 e^x x+2 x^2\right )} \, dx =\text {Too large to display} \] Input:
integrate((x*exp(-log(x)^2)^8+(-8*log(x)-2*exp(x)*x+2*x^2)*exp(-log(x)^2)^ 4+x*exp(x)^2+(-2*x^2-x)*exp(x)+x^3+x)/(x*exp(-log(x)^2)^8+(-2*exp(x)*x+2*x ^2)*exp(-log(x)^2)^4+x*exp(x)^2-2*exp(x)*x^2+x^3),x, algorithm="giac")
Output:
(8*x^4*e^(4*log(x)^2)*log(x) + x^4*e^(4*log(x)^2) - x^4*e^(4*log(x)^2 + x) - 24*x^3*e^(4*log(x)^2 + x)*log(x) + 2*x^3*e^(4*log(x)^2 + 2*x) - 2*x^3*e ^(4*log(x)^2 + x) - 2*x^3*e^x + 16*x^3*log(x) - 8*x^2*e^(4*log(x)^2)*log(x ) + 8*x^2*e^(-4*log(x)^2)*log(x) + 24*x^2*e^(4*log(x)^2 + 2*x)*log(x) - 32 *x^2*e^x*log(x) + 2*x^3 - x^2*e^(4*log(x)^2) + x^2*e^(-4*log(x)^2) - x^2*e ^(4*log(x)^2 + 3*x) + x^2*e^(4*log(x)^2 + 2*x) + x^2*e^(4*log(x)^2 + x) - x^2*e^(-4*log(x)^2 + x) + 2*x^2*e^(2*x) - 2*x^2*e^x - 8*x*e^(4*log(x)^2 + 3*x)*log(x) + 16*x*e^(4*log(x)^2 + x)*log(x) - 8*x*e^(-4*log(x)^2 + x)*log (x) + 16*x*e^(2*x)*log(x) - x*e^(4*log(x)^2 + 2*x) + x*e^(4*log(x)^2 + x) + x*e^x - 8*x*log(x) - 8*e^(4*log(x)^2 + 2*x)*log(x) + 8*e^x*log(x) - x)/( 8*x^3*e^(4*log(x)^2)*log(x) + x^3*e^(4*log(x)^2) - x^3*e^(4*log(x)^2 + x) - 24*x^2*e^(4*log(x)^2 + x)*log(x) + 2*x^2*e^(4*log(x)^2 + 2*x) - 2*x^2*e^ (4*log(x)^2 + x) - 2*x^2*e^x + 16*x^2*log(x) + 8*x*e^(-4*log(x)^2)*log(x) + 24*x*e^(4*log(x)^2 + 2*x)*log(x) - 32*x*e^x*log(x) + 2*x^2 + x*e^(-4*log (x)^2) - x*e^(4*log(x)^2 + 3*x) + x*e^(4*log(x)^2 + 2*x) - x*e^(-4*log(x)^ 2 + x) + 2*x*e^(2*x) - 2*x*e^x - 8*e^(4*log(x)^2 + 3*x)*log(x) - 8*e^(-4*l og(x)^2 + x)*log(x) + 16*e^(2*x)*log(x))
Time = 2.75 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {x+e^{2 x} x+e^{-8 \log ^2(x)} x+x^3+e^x \left (-x-2 x^2\right )+e^{-4 \log ^2(x)} \left (-2 e^x x+2 x^2-8 \log (x)\right )}{e^{2 x} x+e^{-8 \log ^2(x)} x-2 e^x x^2+x^3+e^{-4 \log ^2(x)} \left (-2 e^x x+2 x^2\right )} \, dx=x-\frac {1}{x+{\mathrm {e}}^{-4\,{\ln \left (x\right )}^2}-{\mathrm {e}}^x} \] Input:
int((x + x*exp(2*x) - exp(x)*(x + 2*x^2) + x^3 + x*exp(-8*log(x)^2) - exp( -4*log(x)^2)*(8*log(x) + 2*x*exp(x) - 2*x^2))/(x*exp(2*x) - 2*x^2*exp(x) - exp(-4*log(x)^2)*(2*x*exp(x) - 2*x^2) + x^3 + x*exp(-8*log(x)^2)),x)
Output:
x - 1/(x + exp(-4*log(x)^2) - exp(x))
Time = 0.19 (sec) , antiderivative size = 63, normalized size of antiderivative = 3.00 \[ \int \frac {x+e^{2 x} x+e^{-8 \log ^2(x)} x+x^3+e^x \left (-x-2 x^2\right )+e^{-4 \log ^2(x)} \left (-2 e^x x+2 x^2-8 \log (x)\right )}{e^{2 x} x+e^{-8 \log ^2(x)} x-2 e^x x^2+x^3+e^{-4 \log ^2(x)} \left (-2 e^x x+2 x^2\right )} \, dx=\frac {e^{4 \mathrm {log}\left (x \right )^{2}+x} x -e^{4 \mathrm {log}\left (x \right )^{2}} x^{2}+e^{4 \mathrm {log}\left (x \right )^{2}}-x}{e^{4 \mathrm {log}\left (x \right )^{2}+x}-e^{4 \mathrm {log}\left (x \right )^{2}} x -1} \] Input:
int((x*exp(-log(x)^2)^8+(-8*log(x)-2*exp(x)*x+2*x^2)*exp(-log(x)^2)^4+x*ex p(x)^2+(-2*x^2-x)*exp(x)+x^3+x)/(x*exp(-log(x)^2)^8+(-2*exp(x)*x+2*x^2)*ex p(-log(x)^2)^4+x*exp(x)^2-2*exp(x)*x^2+x^3),x)
Output:
(e**(4*log(x)**2 + x)*x - e**(4*log(x)**2)*x**2 + e**(4*log(x)**2) - x)/(e **(4*log(x)**2 + x) - e**(4*log(x)**2)*x - 1)