Integrand size = 100, antiderivative size = 30 \[ \int \frac {e^{20+4 x} (-1-4 x)+2 x+e^{x^2} \left (e^{20+4 x} (-4-24 x)+8 x+8 x^2\right )+\left (1-4 e^{20+4 x}+e^{x^2} \left (4-16 e^{20+4 x}\right )\right ) \log \left (\frac {1}{5} \left (3+12 e^{x^2}\right )\right )}{1+4 e^{x^2}} \, dx=2+\left (-e^{4 (5+x)}+x\right ) \left (x+\log \left (\frac {1}{5} \left (3+12 e^{x^2}\right )\right )\right ) \] Output:
(x-exp(20+4*x))*(x+ln(12/5*exp(x^2)+3/5))+2
Time = 0.08 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {e^{20+4 x} (-1-4 x)+2 x+e^{x^2} \left (e^{20+4 x} (-4-24 x)+8 x+8 x^2\right )+\left (1-4 e^{20+4 x}+e^{x^2} \left (4-16 e^{20+4 x}\right )\right ) \log \left (\frac {1}{5} \left (3+12 e^{x^2}\right )\right )}{1+4 e^{x^2}} \, dx=-\left (\left (e^{4 (5+x)}-x\right ) \left (x+\log \left (\frac {3}{5} \left (1+4 e^{x^2}\right )\right )\right )\right ) \] Input:
Integrate[(E^(20 + 4*x)*(-1 - 4*x) + 2*x + E^x^2*(E^(20 + 4*x)*(-4 - 24*x) + 8*x + 8*x^2) + (1 - 4*E^(20 + 4*x) + E^x^2*(4 - 16*E^(20 + 4*x)))*Log[( 3 + 12*E^x^2)/5])/(1 + 4*E^x^2),x]
Output:
-((E^(4*(5 + x)) - x)*(x + Log[(3*(1 + 4*E^x^2))/5]))
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.
\(\displaystyle \int \frac {e^{x^2} \left (8 x^2+8 x+e^{4 x+20} (-24 x-4)\right )+\left (e^{x^2} \left (4-16 e^{4 x+20}\right )-4 e^{4 x+20}+1\right ) \log \left (\frac {1}{5} \left (12 e^{x^2}+3\right )\right )+e^{4 x+20} (-4 x-1)+2 x}{4 e^{x^2}+1} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (2 x^2+\frac {2 \left (e^{4 x+20}-x\right ) x}{4 e^{x^2}+1}-4 e^{4 x+20} \log \left (\frac {3}{5} \left (4 e^{x^2}+1\right )\right )+\log \left (\frac {3}{5} \left (4 e^{x^2}+1\right )\right )-6 e^{4 x+20} x+2 x-e^{4 x+20}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \int \frac {e^{4 x+20} x}{1+4 e^{x^2}}dx+8 \int \frac {e^{x^2+4 x+20} x}{1+4 e^{x^2}}dx-2 \int \frac {x^2}{1+4 e^{x^2}}dx-8 \int \frac {e^{x^2} x^2}{1+4 e^{x^2}}dx+\frac {2 x^3}{3}+x^2+x \log \left (\frac {3}{5} \left (4 e^{x^2}+1\right )\right )-e^{4 x+20} \log \left (\frac {3}{5} \left (4 e^{x^2}+1\right )\right )-\frac {3}{2} e^{4 x+20} x+\frac {1}{8} e^{4 x+20}\) |
Input:
Int[(E^(20 + 4*x)*(-1 - 4*x) + 2*x + E^x^2*(E^(20 + 4*x)*(-4 - 24*x) + 8*x + 8*x^2) + (1 - 4*E^(20 + 4*x) + E^x^2*(4 - 16*E^(20 + 4*x)))*Log[(3 + 12 *E^x^2)/5])/(1 + 4*E^x^2),x]
Output:
$Aborted
Time = 0.34 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13
method | result | size |
risch | \(\left (x -{\mathrm e}^{20+4 x}\right ) \ln \left (\frac {12 \,{\mathrm e}^{x^{2}}}{5}+\frac {3}{5}\right )+x^{2}-{\mathrm e}^{20+4 x} x\) | \(34\) |
parallelrisch | \(-{\mathrm e}^{20+4 x} \ln \left (\frac {12 \,{\mathrm e}^{x^{2}}}{5}+\frac {3}{5}\right )-{\mathrm e}^{20+4 x} x +\ln \left (\frac {12 \,{\mathrm e}^{x^{2}}}{5}+\frac {3}{5}\right ) x +x^{2}\) | \(42\) |
Input:
int((((-16*exp(20+4*x)+4)*exp(x^2)-4*exp(20+4*x)+1)*ln(12/5*exp(x^2)+3/5)+ ((-24*x-4)*exp(20+4*x)+8*x^2+8*x)*exp(x^2)+(-4*x-1)*exp(20+4*x)+2*x)/(4*ex p(x^2)+1),x,method=_RETURNVERBOSE)
Output:
(x-exp(20+4*x))*ln(12/5*exp(x^2)+3/5)+x^2-exp(20+4*x)*x
Time = 0.11 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10 \[ \int \frac {e^{20+4 x} (-1-4 x)+2 x+e^{x^2} \left (e^{20+4 x} (-4-24 x)+8 x+8 x^2\right )+\left (1-4 e^{20+4 x}+e^{x^2} \left (4-16 e^{20+4 x}\right )\right ) \log \left (\frac {1}{5} \left (3+12 e^{x^2}\right )\right )}{1+4 e^{x^2}} \, dx=x^{2} - x e^{\left (4 \, x + 20\right )} + {\left (x - e^{\left (4 \, x + 20\right )}\right )} \log \left (\frac {12}{5} \, e^{\left (x^{2}\right )} + \frac {3}{5}\right ) \] Input:
integrate((((-16*exp(20+4*x)+4)*exp(x^2)-4*exp(20+4*x)+1)*log(12/5*exp(x^2 )+3/5)+((-24*x-4)*exp(20+4*x)+8*x^2+8*x)*exp(x^2)+(-4*x-1)*exp(20+4*x)+2*x )/(4*exp(x^2)+1),x, algorithm="fricas")
Output:
x^2 - x*e^(4*x + 20) + (x - e^(4*x + 20))*log(12/5*e^(x^2) + 3/5)
Timed out. \[ \int \frac {e^{20+4 x} (-1-4 x)+2 x+e^{x^2} \left (e^{20+4 x} (-4-24 x)+8 x+8 x^2\right )+\left (1-4 e^{20+4 x}+e^{x^2} \left (4-16 e^{20+4 x}\right )\right ) \log \left (\frac {1}{5} \left (3+12 e^{x^2}\right )\right )}{1+4 e^{x^2}} \, dx=\text {Timed out} \] Input:
integrate((((-16*exp(20+4*x)+4)*exp(x**2)-4*exp(20+4*x)+1)*ln(12/5*exp(x** 2)+3/5)+((-24*x-4)*exp(20+4*x)+8*x**2+8*x)*exp(x**2)+(-4*x-1)*exp(20+4*x)+ 2*x)/(4*exp(x**2)+1),x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (24) = 48\).
Time = 0.18 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.27 \[ \int \frac {e^{20+4 x} (-1-4 x)+2 x+e^{x^2} \left (e^{20+4 x} (-4-24 x)+8 x+8 x^2\right )+\left (1-4 e^{20+4 x}+e^{x^2} \left (4-16 e^{20+4 x}\right )\right ) \log \left (\frac {1}{5} \left (3+12 e^{x^2}\right )\right )}{1+4 e^{x^2}} \, dx=x^{2} - x {\left (\log \left (5\right ) - \log \left (3\right )\right )} - {\left (x e^{20} - {\left (\log \left (5\right ) - \log \left (3\right )\right )} e^{20}\right )} e^{\left (4 \, x\right )} + {\left (x - e^{\left (4 \, x + 20\right )} + 1\right )} \log \left (4 \, e^{\left (x^{2}\right )} + 1\right ) - \log \left (4 \, e^{\left (x^{2}\right )} + 1\right ) \] Input:
integrate((((-16*exp(20+4*x)+4)*exp(x^2)-4*exp(20+4*x)+1)*log(12/5*exp(x^2 )+3/5)+((-24*x-4)*exp(20+4*x)+8*x^2+8*x)*exp(x^2)+(-4*x-1)*exp(20+4*x)+2*x )/(4*exp(x^2)+1),x, algorithm="maxima")
Output:
x^2 - x*(log(5) - log(3)) - (x*e^20 - (log(5) - log(3))*e^20)*e^(4*x) + (x - e^(4*x + 20) + 1)*log(4*e^(x^2) + 1) - log(4*e^(x^2) + 1)
Time = 0.14 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.37 \[ \int \frac {e^{20+4 x} (-1-4 x)+2 x+e^{x^2} \left (e^{20+4 x} (-4-24 x)+8 x+8 x^2\right )+\left (1-4 e^{20+4 x}+e^{x^2} \left (4-16 e^{20+4 x}\right )\right ) \log \left (\frac {1}{5} \left (3+12 e^{x^2}\right )\right )}{1+4 e^{x^2}} \, dx=x^{2} - x e^{\left (4 \, x + 20\right )} + x \log \left (\frac {12}{5} \, e^{\left (x^{2}\right )} + \frac {3}{5}\right ) - e^{\left (4 \, x + 20\right )} \log \left (\frac {12}{5} \, e^{\left (x^{2}\right )} + \frac {3}{5}\right ) \] Input:
integrate((((-16*exp(20+4*x)+4)*exp(x^2)-4*exp(20+4*x)+1)*log(12/5*exp(x^2 )+3/5)+((-24*x-4)*exp(20+4*x)+8*x^2+8*x)*exp(x^2)+(-4*x-1)*exp(20+4*x)+2*x )/(4*exp(x^2)+1),x, algorithm="giac")
Output:
x^2 - x*e^(4*x + 20) + x*log(12/5*e^(x^2) + 3/5) - e^(4*x + 20)*log(12/5*e ^(x^2) + 3/5)
Time = 4.18 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.73 \[ \int \frac {e^{20+4 x} (-1-4 x)+2 x+e^{x^2} \left (e^{20+4 x} (-4-24 x)+8 x+8 x^2\right )+\left (1-4 e^{20+4 x}+e^{x^2} \left (4-16 e^{20+4 x}\right )\right ) \log \left (\frac {1}{5} \left (3+12 e^{x^2}\right )\right )}{1+4 e^{x^2}} \, dx=\left (x+\ln \left (\frac {12\,{\mathrm {e}}^{x^2}}{5}+\frac {3}{5}\right )\right )\,\left (x-{\mathrm {e}}^{4\,x+20}\right ) \] Input:
int((2*x - log((12*exp(x^2))/5 + 3/5)*(4*exp(4*x + 20) + exp(x^2)*(16*exp( 4*x + 20) - 4) - 1) - exp(4*x + 20)*(4*x + 1) + exp(x^2)*(8*x - exp(4*x + 20)*(24*x + 4) + 8*x^2))/(4*exp(x^2) + 1),x)
Output:
(x + log((12*exp(x^2))/5 + 3/5))*(x - exp(4*x + 20))
Time = 0.19 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.57 \[ \int \frac {e^{20+4 x} (-1-4 x)+2 x+e^{x^2} \left (e^{20+4 x} (-4-24 x)+8 x+8 x^2\right )+\left (1-4 e^{20+4 x}+e^{x^2} \left (4-16 e^{20+4 x}\right )\right ) \log \left (\frac {1}{5} \left (3+12 e^{x^2}\right )\right )}{1+4 e^{x^2}} \, dx=-e^{4 x} \mathrm {log}\left (\frac {12 e^{x^{2}}}{5}+\frac {3}{5}\right ) e^{20}-e^{4 x} e^{20} x +\mathrm {log}\left (\frac {12 e^{x^{2}}}{5}+\frac {3}{5}\right ) x +x^{2} \] Input:
int((((-16*exp(20+4*x)+4)*exp(x^2)-4*exp(20+4*x)+1)*log(12/5*exp(x^2)+3/5) +((-24*x-4)*exp(20+4*x)+8*x^2+8*x)*exp(x^2)+(-4*x-1)*exp(20+4*x)+2*x)/(4*e xp(x^2)+1),x)
Output:
- e**(4*x)*log((12*e**(x**2) + 3)/5)*e**20 - e**(4*x)*e**20*x + log((12*e **(x**2) + 3)/5)*x + x**2