Integrand size = 70, antiderivative size = 25 \[ \int \frac {8 x+40 e^{2 x^2} x+\left (8 x+e^{2 x^2} \left (40 x-80 x^3\right )\right ) \log (x) \log \left (\log ^2(x)\right )}{\left (3+30 e^{2 x^2}+75 e^{4 x^2}\right ) \log (x)} \, dx=\frac {4 x^2 \log \left (\log ^2(x)\right )}{3 \left (1+5 e^{2 x^2}\right )} \] Output:
4/3*ln(ln(x)^2)/(1+5*exp(2*x^2))*x^2
Time = 0.19 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {8 x+40 e^{2 x^2} x+\left (8 x+e^{2 x^2} \left (40 x-80 x^3\right )\right ) \log (x) \log \left (\log ^2(x)\right )}{\left (3+30 e^{2 x^2}+75 e^{4 x^2}\right ) \log (x)} \, dx=\frac {4 x^2 \log \left (\log ^2(x)\right )}{3 \left (1+5 e^{2 x^2}\right )} \] Input:
Integrate[(8*x + 40*E^(2*x^2)*x + (8*x + E^(2*x^2)*(40*x - 80*x^3))*Log[x] *Log[Log[x]^2])/((3 + 30*E^(2*x^2) + 75*E^(4*x^2))*Log[x]),x]
Output:
(4*x^2*Log[Log[x]^2])/(3*(1 + 5*E^(2*x^2)))
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.
\(\displaystyle \int \frac {40 e^{2 x^2} x+\left (e^{2 x^2} \left (40 x-80 x^3\right )+8 x\right ) \log (x) \log \left (\log ^2(x)\right )+8 x}{\left (30 e^{2 x^2}+75 e^{4 x^2}+3\right ) \log (x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {40 e^{2 x^2} x+\left (e^{2 x^2} \left (40 x-80 x^3\right )+8 x\right ) \log (x) \log \left (\log ^2(x)\right )+8 x}{3 \left (5 e^{2 x^2}+1\right )^2 \log (x)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} \int \frac {8 \left (5 e^{2 x^2} x+x+\left (x+5 e^{2 x^2} \left (x-2 x^3\right )\right ) \log (x) \log \left (\log ^2(x)\right )\right )}{\left (1+5 e^{2 x^2}\right )^2 \log (x)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {8}{3} \int \frac {5 e^{2 x^2} x+x+\left (x+5 e^{2 x^2} \left (x-2 x^3\right )\right ) \log (x) \log \left (\log ^2(x)\right )}{\left (1+5 e^{2 x^2}\right )^2 \log (x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {8}{3} \int \left (\frac {2 x^3 \log \left (\log ^2(x)\right )}{\left (1+5 e^{2 x^2}\right )^2}-\frac {x \left (2 \log (x) \log \left (\log ^2(x)\right ) x^2-\log (x) \log \left (\log ^2(x)\right )-1\right )}{\left (1+5 e^{2 x^2}\right ) \log (x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {8}{3} \left (\int \frac {x \log \left (\log ^2(x)\right )}{1+5 e^{2 x^2}}dx+\int \frac {x}{\left (1+5 e^{2 x^2}\right ) \log (x)}dx+2 \int \frac {x^3 \log \left (\log ^2(x)\right )}{\left (1+5 e^{2 x^2}\right )^2}dx-2 \int \frac {x^3 \log \left (\log ^2(x)\right )}{1+5 e^{2 x^2}}dx\right )\) |
Input:
Int[(8*x + 40*E^(2*x^2)*x + (8*x + E^(2*x^2)*(40*x - 80*x^3))*Log[x]*Log[L og[x]^2])/((3 + 30*E^(2*x^2) + 75*E^(4*x^2))*Log[x]),x]
Output:
$Aborted
Time = 2.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92
method | result | size |
parallelrisch | \(\frac {4 \ln \left (\ln \left (x \right )^{2}\right ) x^{2}}{3 \left (1+5 \,{\mathrm e}^{2 x^{2}}\right )}\) | \(23\) |
risch | \(\frac {8 x^{2} \ln \left (\ln \left (x \right )\right )}{3 \left (1+5 \,{\mathrm e}^{2 x^{2}}\right )}-\frac {2 i x^{2} \pi \,\operatorname {csgn}\left (i \ln \left (x \right )^{2}\right ) \left (\operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )^{2}-2 \,\operatorname {csgn}\left (i \ln \left (x \right )^{2}\right ) \operatorname {csgn}\left (i \ln \left (x \right )\right )+\operatorname {csgn}\left (i \ln \left (x \right )\right )^{2}\right )}{3 \left (1+5 \,{\mathrm e}^{2 x^{2}}\right )}\) | \(84\) |
Input:
int((((-80*x^3+40*x)*exp(2*x^2)+8*x)*ln(x)*ln(ln(x)^2)+40*x*exp(2*x^2)+8*x )/(75*exp(2*x^2)^2+30*exp(2*x^2)+3)/ln(x),x,method=_RETURNVERBOSE)
Output:
4/3*ln(ln(x)^2)/(1+5*exp(2*x^2))*x^2
Time = 0.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {8 x+40 e^{2 x^2} x+\left (8 x+e^{2 x^2} \left (40 x-80 x^3\right )\right ) \log (x) \log \left (\log ^2(x)\right )}{\left (3+30 e^{2 x^2}+75 e^{4 x^2}\right ) \log (x)} \, dx=\frac {4 \, x^{2} \log \left (\log \left (x\right )^{2}\right )}{3 \, {\left (5 \, e^{\left (2 \, x^{2}\right )} + 1\right )}} \] Input:
integrate((((-80*x^3+40*x)*exp(2*x^2)+8*x)*log(x)*log(log(x)^2)+40*x*exp(2 *x^2)+8*x)/(75*exp(2*x^2)^2+30*exp(2*x^2)+3)/log(x),x, algorithm="fricas")
Output:
4/3*x^2*log(log(x)^2)/(5*e^(2*x^2) + 1)
Time = 0.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {8 x+40 e^{2 x^2} x+\left (8 x+e^{2 x^2} \left (40 x-80 x^3\right )\right ) \log (x) \log \left (\log ^2(x)\right )}{\left (3+30 e^{2 x^2}+75 e^{4 x^2}\right ) \log (x)} \, dx=\frac {4 x^{2} \log {\left (\log {\left (x \right )}^{2} \right )}}{15 e^{2 x^{2}} + 3} \] Input:
integrate((((-80*x**3+40*x)*exp(2*x**2)+8*x)*ln(x)*ln(ln(x)**2)+40*x*exp(2 *x**2)+8*x)/(75*exp(2*x**2)**2+30*exp(2*x**2)+3)/ln(x),x)
Output:
4*x**2*log(log(x)**2)/(15*exp(2*x**2) + 3)
Time = 0.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {8 x+40 e^{2 x^2} x+\left (8 x+e^{2 x^2} \left (40 x-80 x^3\right )\right ) \log (x) \log \left (\log ^2(x)\right )}{\left (3+30 e^{2 x^2}+75 e^{4 x^2}\right ) \log (x)} \, dx=\frac {8 \, x^{2} \log \left (\log \left (x\right )\right )}{3 \, {\left (5 \, e^{\left (2 \, x^{2}\right )} + 1\right )}} \] Input:
integrate((((-80*x^3+40*x)*exp(2*x^2)+8*x)*log(x)*log(log(x)^2)+40*x*exp(2 *x^2)+8*x)/(75*exp(2*x^2)^2+30*exp(2*x^2)+3)/log(x),x, algorithm="maxima")
Output:
8/3*x^2*log(log(x))/(5*e^(2*x^2) + 1)
Time = 0.16 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {8 x+40 e^{2 x^2} x+\left (8 x+e^{2 x^2} \left (40 x-80 x^3\right )\right ) \log (x) \log \left (\log ^2(x)\right )}{\left (3+30 e^{2 x^2}+75 e^{4 x^2}\right ) \log (x)} \, dx=\frac {4 \, x^{2} \log \left (\log \left (x\right )^{2}\right )}{3 \, {\left (5 \, e^{\left (2 \, x^{2}\right )} + 1\right )}} \] Input:
integrate((((-80*x^3+40*x)*exp(2*x^2)+8*x)*log(x)*log(log(x)^2)+40*x*exp(2 *x^2)+8*x)/(75*exp(2*x^2)^2+30*exp(2*x^2)+3)/log(x),x, algorithm="giac")
Output:
4/3*x^2*log(log(x)^2)/(5*e^(2*x^2) + 1)
Time = 0.71 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {8 x+40 e^{2 x^2} x+\left (8 x+e^{2 x^2} \left (40 x-80 x^3\right )\right ) \log (x) \log \left (\log ^2(x)\right )}{\left (3+30 e^{2 x^2}+75 e^{4 x^2}\right ) \log (x)} \, dx=\frac {4\,x^2\,\ln \left ({\ln \left (x\right )}^2\right )}{15\,\left ({\mathrm {e}}^{2\,x^2}+\frac {1}{5}\right )} \] Input:
int((8*x + 40*x*exp(2*x^2) + log(log(x)^2)*log(x)*(8*x + exp(2*x^2)*(40*x - 80*x^3)))/(log(x)*(30*exp(2*x^2) + 75*exp(4*x^2) + 3)),x)
Output:
(4*x^2*log(log(x)^2))/(15*(exp(2*x^2) + 1/5))
Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {8 x+40 e^{2 x^2} x+\left (8 x+e^{2 x^2} \left (40 x-80 x^3\right )\right ) \log (x) \log \left (\log ^2(x)\right )}{\left (3+30 e^{2 x^2}+75 e^{4 x^2}\right ) \log (x)} \, dx=\frac {4 \,\mathrm {log}\left (\mathrm {log}\left (x \right )^{2}\right ) x^{2}}{15 e^{2 x^{2}}+3} \] Input:
int((((-80*x^3+40*x)*exp(2*x^2)+8*x)*log(x)*log(log(x)^2)+40*x*exp(2*x^2)+ 8*x)/(75*exp(2*x^2)^2+30*exp(2*x^2)+3)/log(x),x)
Output:
(4*log(log(x)**2)*x**2)/(3*(5*e**(2*x**2) + 1))