Integrand size = 63, antiderivative size = 23 \[ \int \frac {4 x^2+e^{4-16 x+16 x^2+(-4+8 x) \log (x)+\log ^2(x)} \left (160+272 x-992 x^2-128 x^3+\left (-64-264 x-32 x^2\right ) \log (x)\right )}{x^2} \, dx=4 \left (x-\frac {e^{(-2+4 x+\log (x))^2} (8+x)}{x}\right ) \] Output:
4*x-4*exp((4*x-2+ln(x))^2)*(x+8)/x
Time = 2.70 (sec) , antiderivative size = 46, normalized size of antiderivative = 2.00 \[ \int \frac {4 x^2+e^{4-16 x+16 x^2+(-4+8 x) \log (x)+\log ^2(x)} \left (160+272 x-992 x^2-128 x^3+\left (-64-264 x-32 x^2\right ) \log (x)\right )}{x^2} \, dx=4 x-\frac {4 e^{4-16 x+16 x^2+\log ^2(x)} x^{-5+8 x} \left (8+33 x+4 x^2\right )}{1+4 x} \] Input:
Integrate[(4*x^2 + E^(4 - 16*x + 16*x^2 + (-4 + 8*x)*Log[x] + Log[x]^2)*(1 60 + 272*x - 992*x^2 - 128*x^3 + (-64 - 264*x - 32*x^2)*Log[x]))/x^2,x]
Output:
4*x - (4*E^(4 - 16*x + 16*x^2 + Log[x]^2)*x^(-5 + 8*x)*(8 + 33*x + 4*x^2)) /(1 + 4*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.
\(\displaystyle \int \frac {4 x^2+e^{16 x^2-16 x+\log ^2(x)+(8 x-4) \log (x)+4} \left (-128 x^3-992 x^2+\left (-32 x^2-264 x-64\right ) \log (x)+272 x+160\right )}{x^2} \, dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \int \left (4-8 x^{8 x-6} e^{4 (1-2 x)^2+\log ^2(x)} \left (16 x^3+124 x^2+4 x^2 \log (x)-34 x+33 x \log (x)+8 \log (x)-20\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 160 \int e^{4 (1-2 x)^2+\log ^2(x)} x^{8 x-6}dx+272 \int e^{4 (1-2 x)^2+\log ^2(x)} x^{8 x-5}dx-992 \int e^{4 (1-2 x)^2+\log ^2(x)} x^{8 x-4}dx-128 \int e^{4 (1-2 x)^2+\log ^2(x)} x^{8 x-3}dx-64 \int e^{4 (1-2 x)^2+\log ^2(x)} x^{8 x-6} \log (x)dx-264 \int e^{4 (1-2 x)^2+\log ^2(x)} x^{8 x-5} \log (x)dx-32 \int e^{4 (1-2 x)^2+\log ^2(x)} x^{8 x-4} \log (x)dx+4 x\) |
Input:
Int[(4*x^2 + E^(4 - 16*x + 16*x^2 + (-4 + 8*x)*Log[x] + Log[x]^2)*(160 + 2 72*x - 992*x^2 - 128*x^3 + (-64 - 264*x - 32*x^2)*Log[x]))/x^2,x]
Output:
$Aborted
Time = 0.29 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.52
method | result | size |
risch | \(4 x -\frac {4 \left (x +8\right ) x^{8 x -4} {\mathrm e}^{\ln \left (x \right )^{2}+4+16 x^{2}-16 x}}{x}\) | \(35\) |
default | \(4 x +\frac {-4 x \,{\mathrm e}^{\ln \left (x \right )^{2}+\left (8 x -4\right ) \ln \left (x \right )+16 x^{2}-16 x +4}-32 \,{\mathrm e}^{\ln \left (x \right )^{2}+\left (8 x -4\right ) \ln \left (x \right )+16 x^{2}-16 x +4}}{x}\) | \(61\) |
parts | \(4 x +\frac {-4 x \,{\mathrm e}^{\ln \left (x \right )^{2}+\left (8 x -4\right ) \ln \left (x \right )+16 x^{2}-16 x +4}-32 \,{\mathrm e}^{\ln \left (x \right )^{2}+\left (8 x -4\right ) \ln \left (x \right )+16 x^{2}-16 x +4}}{x}\) | \(61\) |
norman | \(\frac {4 x^{2}-4 x \,{\mathrm e}^{\ln \left (x \right )^{2}+\left (8 x -4\right ) \ln \left (x \right )+16 x^{2}-16 x +4}-32 \,{\mathrm e}^{\ln \left (x \right )^{2}+\left (8 x -4\right ) \ln \left (x \right )+16 x^{2}-16 x +4}}{x}\) | \(62\) |
parallelrisch | \(\frac {4 x^{2}-4 x \,{\mathrm e}^{\ln \left (x \right )^{2}+\left (8 x -4\right ) \ln \left (x \right )+16 x^{2}-16 x +4}-32 \,{\mathrm e}^{\ln \left (x \right )^{2}+\left (8 x -4\right ) \ln \left (x \right )+16 x^{2}-16 x +4}}{x}\) | \(62\) |
Input:
int((((-32*x^2-264*x-64)*ln(x)-128*x^3-992*x^2+272*x+160)*exp(ln(x)^2+(8*x -4)*ln(x)+16*x^2-16*x+4)+4*x^2)/x^2,x,method=_RETURNVERBOSE)
Output:
4*x-4*(x+8)/x*x^(8*x-4)*exp(ln(x)^2+4+16*x^2-16*x)
Time = 0.09 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.65 \[ \int \frac {4 x^2+e^{4-16 x+16 x^2+(-4+8 x) \log (x)+\log ^2(x)} \left (160+272 x-992 x^2-128 x^3+\left (-64-264 x-32 x^2\right ) \log (x)\right )}{x^2} \, dx=\frac {4 \, {\left (x^{2} - {\left (x + 8\right )} e^{\left (16 \, x^{2} + 4 \, {\left (2 \, x - 1\right )} \log \left (x\right ) + \log \left (x\right )^{2} - 16 \, x + 4\right )}\right )}}{x} \] Input:
integrate((((-32*x^2-264*x-64)*log(x)-128*x^3-992*x^2+272*x+160)*exp(log(x )^2+(8*x-4)*log(x)+16*x^2-16*x+4)+4*x^2)/x^2,x, algorithm="fricas")
Output:
4*(x^2 - (x + 8)*e^(16*x^2 + 4*(2*x - 1)*log(x) + log(x)^2 - 16*x + 4))/x
Time = 0.12 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.57 \[ \int \frac {4 x^2+e^{4-16 x+16 x^2+(-4+8 x) \log (x)+\log ^2(x)} \left (160+272 x-992 x^2-128 x^3+\left (-64-264 x-32 x^2\right ) \log (x)\right )}{x^2} \, dx=4 x + \frac {\left (- 4 x - 32\right ) e^{16 x^{2} - 16 x + \left (8 x - 4\right ) \log {\left (x \right )} + \log {\left (x \right )}^{2} + 4}}{x} \] Input:
integrate((((-32*x**2-264*x-64)*ln(x)-128*x**3-992*x**2+272*x+160)*exp(ln( x)**2+(8*x-4)*ln(x)+16*x**2-16*x+4)+4*x**2)/x**2,x)
Output:
4*x + (-4*x - 32)*exp(16*x**2 - 16*x + (8*x - 4)*log(x) + log(x)**2 + 4)/x
Time = 0.13 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.61 \[ \int \frac {4 x^2+e^{4-16 x+16 x^2+(-4+8 x) \log (x)+\log ^2(x)} \left (160+272 x-992 x^2-128 x^3+\left (-64-264 x-32 x^2\right ) \log (x)\right )}{x^2} \, dx=4 \, x - \frac {4 \, {\left (x e^{4} + 8 \, e^{4}\right )} e^{\left (16 \, x^{2} + 8 \, x \log \left (x\right ) + \log \left (x\right )^{2} - 16 \, x\right )}}{x^{5}} \] Input:
integrate((((-32*x^2-264*x-64)*log(x)-128*x^3-992*x^2+272*x+160)*exp(log(x )^2+(8*x-4)*log(x)+16*x^2-16*x+4)+4*x^2)/x^2,x, algorithm="maxima")
Output:
4*x - 4*(x*e^4 + 8*e^4)*e^(16*x^2 + 8*x*log(x) + log(x)^2 - 16*x)/x^5
Leaf count of result is larger than twice the leaf count of optimal. 62 vs. \(2 (22) = 44\).
Time = 0.15 (sec) , antiderivative size = 62, normalized size of antiderivative = 2.70 \[ \int \frac {4 x^2+e^{4-16 x+16 x^2+(-4+8 x) \log (x)+\log ^2(x)} \left (160+272 x-992 x^2-128 x^3+\left (-64-264 x-32 x^2\right ) \log (x)\right )}{x^2} \, dx=\frac {4 \, {\left (x^{2} - x e^{\left (16 \, x^{2} + 8 \, x \log \left (x\right ) + \log \left (x\right )^{2} - 16 \, x - 4 \, \log \left (x\right ) + 4\right )} - 8 \, e^{\left (16 \, x^{2} + 8 \, x \log \left (x\right ) + \log \left (x\right )^{2} - 16 \, x - 4 \, \log \left (x\right ) + 4\right )}\right )}}{x} \] Input:
integrate((((-32*x^2-264*x-64)*log(x)-128*x^3-992*x^2+272*x+160)*exp(log(x )^2+(8*x-4)*log(x)+16*x^2-16*x+4)+4*x^2)/x^2,x, algorithm="giac")
Output:
4*(x^2 - x*e^(16*x^2 + 8*x*log(x) + log(x)^2 - 16*x - 4*log(x) + 4) - 8*e^ (16*x^2 + 8*x*log(x) + log(x)^2 - 16*x - 4*log(x) + 4))/x
Time = 2.93 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.52 \[ \int \frac {4 x^2+e^{4-16 x+16 x^2+(-4+8 x) \log (x)+\log ^2(x)} \left (160+272 x-992 x^2-128 x^3+\left (-64-264 x-32 x^2\right ) \log (x)\right )}{x^2} \, dx=4\,x-\frac {4\,x^{8\,x}\,{\mathrm {e}}^{-16\,x}\,{\mathrm {e}}^4\,{\mathrm {e}}^{16\,x^2}\,{\mathrm {e}}^{{\ln \left (x\right )}^2}}{x^4}-\frac {32\,x^{8\,x}\,{\mathrm {e}}^{-16\,x}\,{\mathrm {e}}^4\,{\mathrm {e}}^{16\,x^2}\,{\mathrm {e}}^{{\ln \left (x\right )}^2}}{x^5} \] Input:
int(-(exp(log(x)^2 - 16*x + log(x)*(8*x - 4) + 16*x^2 + 4)*(log(x)*(264*x + 32*x^2 + 64) - 272*x + 992*x^2 + 128*x^3 - 160) - 4*x^2)/x^2,x)
Output:
4*x - (4*x^(8*x)*exp(-16*x)*exp(4)*exp(16*x^2)*exp(log(x)^2))/x^4 - (32*x^ (8*x)*exp(-16*x)*exp(4)*exp(16*x^2)*exp(log(x)^2))/x^5
Time = 0.23 (sec) , antiderivative size = 67, normalized size of antiderivative = 2.91 \[ \int \frac {4 x^2+e^{4-16 x+16 x^2+(-4+8 x) \log (x)+\log ^2(x)} \left (160+272 x-992 x^2-128 x^3+\left (-64-264 x-32 x^2\right ) \log (x)\right )}{x^2} \, dx=\frac {-4 x^{8 x} e^{\mathrm {log}\left (x \right )^{2}+16 x^{2}} e^{4} x -32 x^{8 x} e^{\mathrm {log}\left (x \right )^{2}+16 x^{2}} e^{4}+4 e^{16 x} x^{6}}{e^{16 x} x^{5}} \] Input:
int((((-32*x^2-264*x-64)*log(x)-128*x^3-992*x^2+272*x+160)*exp(log(x)^2+(8 *x-4)*log(x)+16*x^2-16*x+4)+4*x^2)/x^2,x)
Output:
(4*( - x**(8*x)*e**(log(x)**2 + 16*x**2)*e**4*x - 8*x**(8*x)*e**(log(x)**2 + 16*x**2)*e**4 + e**(16*x)*x**6))/(e**(16*x)*x**5)