Integrand size = 87, antiderivative size = 33 \[ \int \frac {\left (-20-20 x^2-5 x^4\right ) \log (3)+e^{-4 e^x+4 x+4 x^2} \left (e^x \left (-32 x-16 x^3\right ) \log (3)+\left (8+32 x+60 x^2+16 x^3+32 x^4\right ) \log (3)\right )}{16+16 x^2+4 x^4} \, dx=\left (-\frac {5 x}{4}+\frac {e^{4 \left (-e^x+x+x^2\right )}}{\frac {2}{x}+x}\right ) \log (3) \] Output:
ln(3)*(exp(-4*exp(x)+4*x^2+4*x)/(x+2/x)-5/4*x)
Time = 0.36 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97 \[ \int \frac {\left (-20-20 x^2-5 x^4\right ) \log (3)+e^{-4 e^x+4 x+4 x^2} \left (e^x \left (-32 x-16 x^3\right ) \log (3)+\left (8+32 x+60 x^2+16 x^3+32 x^4\right ) \log (3)\right )}{16+16 x^2+4 x^4} \, dx=\frac {1}{4} x \left (-5+\frac {4 e^{-4 e^x+4 x (1+x)}}{2+x^2}\right ) \log (3) \] Input:
Integrate[((-20 - 20*x^2 - 5*x^4)*Log[3] + E^(-4*E^x + 4*x + 4*x^2)*(E^x*( -32*x - 16*x^3)*Log[3] + (8 + 32*x + 60*x^2 + 16*x^3 + 32*x^4)*Log[3]))/(1 6 + 16*x^2 + 4*x^4),x]
Output:
(x*(-5 + (4*E^(-4*E^x + 4*x*(1 + x)))/(2 + x^2))*Log[3])/4
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.
\(\displaystyle \int \frac {\left (-5 x^4-20 x^2-20\right ) \log (3)+e^{4 x^2+4 x-4 e^x} \left (e^x \left (-16 x^3-32 x\right ) \log (3)+\left (32 x^4+16 x^3+60 x^2+32 x+8\right ) \log (3)\right )}{4 x^4+16 x^2+16} \, dx\) |
\(\Big \downarrow \) 1380 |
\(\displaystyle 4 \int -\frac {5 \log (3) \left (x^4+4 x^2+4\right )+4 e^{4 x^2+4 x-4 e^x} \left (4 e^x \left (x^3+2 x\right ) \log (3)-\left (8 x^4+4 x^3+15 x^2+8 x+2\right ) \log (3)\right )}{16 \left (x^2+2\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {1}{4} \int \frac {5 \log (3) \left (x^4+4 x^2+4\right )+4 e^{4 x^2+4 x-4 e^x} \left (4 e^x \left (x^3+2 x\right ) \log (3)-\left (8 x^4+4 x^3+15 x^2+8 x+2\right ) \log (3)\right )}{\left (x^2+2\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {1}{4} \int \left (\frac {4 e^{-4 \left (-x^2-x+e^x\right )} \log (3) \left (-8 x^4+4 e^x x^3-4 x^3-15 x^2+8 e^x x-8 x-2\right )}{\left (x^2+2\right )^2}+\log (243)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \left (32 \log (3) \int e^{-4 \left (-x^2-x+e^x\right )}dx-2 \log (3) \int \frac {e^{-4 \left (-x^2-x+e^x\right )}}{\left (i \sqrt {2}-x\right )^2}dx-16 i \sqrt {2} \log (3) \int \frac {e^{-4 \left (-x^2-x+e^x\right )}}{i \sqrt {2}-x}dx-8 \log (3) \int \frac {e^{-4 \left (-x^2-x+e^x\right )}}{i \sqrt {2}-x}dx+8 \log (3) \int \frac {e^{x-4 \left (-x^2-x+e^x\right )}}{i \sqrt {2}-x}dx-2 \log (3) \int \frac {e^{-4 \left (-x^2-x+e^x\right )}}{\left (x+i \sqrt {2}\right )^2}dx-16 i \sqrt {2} \log (3) \int \frac {e^{-4 \left (-x^2-x+e^x\right )}}{x+i \sqrt {2}}dx+8 \log (3) \int \frac {e^{-4 \left (-x^2-x+e^x\right )}}{x+i \sqrt {2}}dx-8 \log (3) \int \frac {e^{x-4 \left (-x^2-x+e^x\right )}}{x+i \sqrt {2}}dx+x (-\log (243))\right )\) |
Input:
Int[((-20 - 20*x^2 - 5*x^4)*Log[3] + E^(-4*E^x + 4*x + 4*x^2)*(E^x*(-32*x - 16*x^3)*Log[3] + (8 + 32*x + 60*x^2 + 16*x^3 + 32*x^4)*Log[3]))/(16 + 16 *x^2 + 4*x^4),x]
Output:
$Aborted
Time = 0.40 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97
method | result | size |
risch | \(-\frac {5 x \ln \left (3\right )}{4}+\frac {x \ln \left (3\right ) {\mathrm e}^{-4 \,{\mathrm e}^{x}+4 x^{2}+4 x}}{x^{2}+2}\) | \(32\) |
norman | \(\frac {x \ln \left (3\right ) {\mathrm e}^{-4 \,{\mathrm e}^{x}+4 x^{2}+4 x}-\frac {5 x \ln \left (3\right )}{2}-\frac {5 x^{3} \ln \left (3\right )}{4}}{x^{2}+2}\) | \(40\) |
parallelrisch | \(-\frac {5 x^{3} \ln \left (3\right )-4 x \ln \left (3\right ) {\mathrm e}^{-4 \,{\mathrm e}^{x}+4 x^{2}+4 x}+10 x \ln \left (3\right )}{4 \left (x^{2}+2\right )}\) | \(42\) |
Input:
int((((-16*x^3-32*x)*ln(3)*exp(x)+(32*x^4+16*x^3+60*x^2+32*x+8)*ln(3))*exp (-4*exp(x)+4*x^2+4*x)+(-5*x^4-20*x^2-20)*ln(3))/(4*x^4+16*x^2+16),x,method =_RETURNVERBOSE)
Output:
-5/4*x*ln(3)+x*ln(3)/(x^2+2)*exp(-4*exp(x)+4*x^2+4*x)
Time = 0.08 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.21 \[ \int \frac {\left (-20-20 x^2-5 x^4\right ) \log (3)+e^{-4 e^x+4 x+4 x^2} \left (e^x \left (-32 x-16 x^3\right ) \log (3)+\left (8+32 x+60 x^2+16 x^3+32 x^4\right ) \log (3)\right )}{16+16 x^2+4 x^4} \, dx=\frac {4 \, x e^{\left (4 \, x^{2} + 4 \, x - 4 \, e^{x}\right )} \log \left (3\right ) - 5 \, {\left (x^{3} + 2 \, x\right )} \log \left (3\right )}{4 \, {\left (x^{2} + 2\right )}} \] Input:
integrate((((-16*x^3-32*x)*log(3)*exp(x)+(32*x^4+16*x^3+60*x^2+32*x+8)*log (3))*exp(-4*exp(x)+4*x^2+4*x)+(-5*x^4-20*x^2-20)*log(3))/(4*x^4+16*x^2+16) ,x, algorithm="fricas")
Output:
1/4*(4*x*e^(4*x^2 + 4*x - 4*e^x)*log(3) - 5*(x^3 + 2*x)*log(3))/(x^2 + 2)
Time = 0.11 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97 \[ \int \frac {\left (-20-20 x^2-5 x^4\right ) \log (3)+e^{-4 e^x+4 x+4 x^2} \left (e^x \left (-32 x-16 x^3\right ) \log (3)+\left (8+32 x+60 x^2+16 x^3+32 x^4\right ) \log (3)\right )}{16+16 x^2+4 x^4} \, dx=- \frac {5 x \log {\left (3 \right )}}{4} + \frac {x e^{4 x^{2} + 4 x - 4 e^{x}} \log {\left (3 \right )}}{x^{2} + 2} \] Input:
integrate((((-16*x**3-32*x)*ln(3)*exp(x)+(32*x**4+16*x**3+60*x**2+32*x+8)* ln(3))*exp(-4*exp(x)+4*x**2+4*x)+(-5*x**4-20*x**2-20)*ln(3))/(4*x**4+16*x* *2+16),x)
Output:
-5*x*log(3)/4 + x*exp(4*x**2 + 4*x - 4*exp(x))*log(3)/(x**2 + 2)
Leaf count of result is larger than twice the leaf count of optimal. 108 vs. \(2 (33) = 66\).
Time = 0.20 (sec) , antiderivative size = 108, normalized size of antiderivative = 3.27 \[ \int \frac {\left (-20-20 x^2-5 x^4\right ) \log (3)+e^{-4 e^x+4 x+4 x^2} \left (e^x \left (-32 x-16 x^3\right ) \log (3)+\left (8+32 x+60 x^2+16 x^3+32 x^4\right ) \log (3)\right )}{16+16 x^2+4 x^4} \, dx=\frac {5}{8} \, {\left (3 \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} x\right ) - 2 \, x - \frac {2 \, x}{x^{2} + 2}\right )} \log \left (3\right ) - \frac {5}{8} \, {\left (\sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} x\right ) + \frac {2 \, x}{x^{2} + 2}\right )} \log \left (3\right ) - \frac {5}{4} \, {\left (\sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} x\right ) - \frac {2 \, x}{x^{2} + 2}\right )} \log \left (3\right ) + \frac {x e^{\left (4 \, x^{2} + 4 \, x - 4 \, e^{x}\right )} \log \left (3\right )}{x^{2} + 2} \] Input:
integrate((((-16*x^3-32*x)*log(3)*exp(x)+(32*x^4+16*x^3+60*x^2+32*x+8)*log (3))*exp(-4*exp(x)+4*x^2+4*x)+(-5*x^4-20*x^2-20)*log(3))/(4*x^4+16*x^2+16) ,x, algorithm="maxima")
Output:
5/8*(3*sqrt(2)*arctan(1/2*sqrt(2)*x) - 2*x - 2*x/(x^2 + 2))*log(3) - 5/8*( sqrt(2)*arctan(1/2*sqrt(2)*x) + 2*x/(x^2 + 2))*log(3) - 5/4*(sqrt(2)*arcta n(1/2*sqrt(2)*x) - 2*x/(x^2 + 2))*log(3) + x*e^(4*x^2 + 4*x - 4*e^x)*log(3 )/(x^2 + 2)
\[ \int \frac {\left (-20-20 x^2-5 x^4\right ) \log (3)+e^{-4 e^x+4 x+4 x^2} \left (e^x \left (-32 x-16 x^3\right ) \log (3)+\left (8+32 x+60 x^2+16 x^3+32 x^4\right ) \log (3)\right )}{16+16 x^2+4 x^4} \, dx=\int { -\frac {4 \, {\left (4 \, {\left (x^{3} + 2 \, x\right )} e^{x} \log \left (3\right ) - {\left (8 \, x^{4} + 4 \, x^{3} + 15 \, x^{2} + 8 \, x + 2\right )} \log \left (3\right )\right )} e^{\left (4 \, x^{2} + 4 \, x - 4 \, e^{x}\right )} + 5 \, {\left (x^{4} + 4 \, x^{2} + 4\right )} \log \left (3\right )}{4 \, {\left (x^{4} + 4 \, x^{2} + 4\right )}} \,d x } \] Input:
integrate((((-16*x^3-32*x)*log(3)*exp(x)+(32*x^4+16*x^3+60*x^2+32*x+8)*log (3))*exp(-4*exp(x)+4*x^2+4*x)+(-5*x^4-20*x^2-20)*log(3))/(4*x^4+16*x^2+16) ,x, algorithm="giac")
Output:
integrate(-1/4*(4*(4*(x^3 + 2*x)*e^x*log(3) - (8*x^4 + 4*x^3 + 15*x^2 + 8* x + 2)*log(3))*e^(4*x^2 + 4*x - 4*e^x) + 5*(x^4 + 4*x^2 + 4)*log(3))/(x^4 + 4*x^2 + 4), x)
Time = 0.54 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97 \[ \int \frac {\left (-20-20 x^2-5 x^4\right ) \log (3)+e^{-4 e^x+4 x+4 x^2} \left (e^x \left (-32 x-16 x^3\right ) \log (3)+\left (8+32 x+60 x^2+16 x^3+32 x^4\right ) \log (3)\right )}{16+16 x^2+4 x^4} \, dx=\frac {x\,{\mathrm {e}}^{4\,x}\,{\mathrm {e}}^{4\,x^2}\,{\mathrm {e}}^{-4\,{\mathrm {e}}^x}\,\ln \left (3\right )}{x^2+2}-\frac {5\,x\,\ln \left (3\right )}{4} \] Input:
int((exp(4*x - 4*exp(x) + 4*x^2)*(log(3)*(32*x + 60*x^2 + 16*x^3 + 32*x^4 + 8) - exp(x)*log(3)*(32*x + 16*x^3)) - log(3)*(20*x^2 + 5*x^4 + 20))/(16* x^2 + 4*x^4 + 16),x)
Output:
(x*exp(4*x)*exp(4*x^2)*exp(-4*exp(x))*log(3))/(x^2 + 2) - (5*x*log(3))/4
Time = 0.18 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.70 \[ \int \frac {\left (-20-20 x^2-5 x^4\right ) \log (3)+e^{-4 e^x+4 x+4 x^2} \left (e^x \left (-32 x-16 x^3\right ) \log (3)+\left (8+32 x+60 x^2+16 x^3+32 x^4\right ) \log (3)\right )}{16+16 x^2+4 x^4} \, dx=\frac {\mathrm {log}\left (3\right ) x \left (-5 e^{4 e^{x}} x^{2}-10 e^{4 e^{x}}+4 e^{4 x^{2}+4 x}\right )}{4 e^{4 e^{x}} \left (x^{2}+2\right )} \] Input:
int((((-16*x^3-32*x)*log(3)*exp(x)+(32*x^4+16*x^3+60*x^2+32*x+8)*log(3))*e xp(-4*exp(x)+4*x^2+4*x)+(-5*x^4-20*x^2-20)*log(3))/(4*x^4+16*x^2+16),x)
Output:
(log(3)*x*( - 5*e**(4*e**x)*x**2 - 10*e**(4*e**x) + 4*e**(4*x**2 + 4*x)))/ (4*e**(4*e**x)*(x**2 + 2))