Integrand size = 98, antiderivative size = 25 \[ \int \frac {e^{x \log (3) \log \left (x^2 \log \left (x^2\right )\right )} \left (\left (-4 x-2 x^2\right ) \log (3)+\left (4+3 x+\left (-4 x-2 x^2\right ) \log (3)\right ) \log \left (x^2\right )+\left (-2 x-x^2\right ) \log (3) \log \left (x^2\right ) \log \left (x^2 \log \left (x^2\right )\right )\right )}{\left (4 x^3+4 x^4+x^5\right ) \log \left (x^2\right )} \, dx=-\frac {e^{x \log (3) \log \left (x^2 \log \left (x^2\right )\right )}}{x^2 (2+x)} \]
\[ \int \frac {e^{x \log (3) \log \left (x^2 \log \left (x^2\right )\right )} \left (\left (-4 x-2 x^2\right ) \log (3)+\left (4+3 x+\left (-4 x-2 x^2\right ) \log (3)\right ) \log \left (x^2\right )+\left (-2 x-x^2\right ) \log (3) \log \left (x^2\right ) \log \left (x^2 \log \left (x^2\right )\right )\right )}{\left (4 x^3+4 x^4+x^5\right ) \log \left (x^2\right )} \, dx=\int \frac {e^{x \log (3) \log \left (x^2 \log \left (x^2\right )\right )} \left (\left (-4 x-2 x^2\right ) \log (3)+\left (4+3 x+\left (-4 x-2 x^2\right ) \log (3)\right ) \log \left (x^2\right )+\left (-2 x-x^2\right ) \log (3) \log \left (x^2\right ) \log \left (x^2 \log \left (x^2\right )\right )\right )}{\left (4 x^3+4 x^4+x^5\right ) \log \left (x^2\right )} \, dx \]
Integrate[(E^(x*Log[3]*Log[x^2*Log[x^2]])*((-4*x - 2*x^2)*Log[3] + (4 + 3* x + (-4*x - 2*x^2)*Log[3])*Log[x^2] + (-2*x - x^2)*Log[3]*Log[x^2]*Log[x^2 *Log[x^2]]))/((4*x^3 + 4*x^4 + x^5)*Log[x^2]),x]
Integrate[(E^(x*Log[3]*Log[x^2*Log[x^2]])*((-4*x - 2*x^2)*Log[3] + (4 + 3* x + (-4*x - 2*x^2)*Log[3])*Log[x^2] + (-2*x - x^2)*Log[3]*Log[x^2]*Log[x^2 *Log[x^2]]))/((4*x^3 + 4*x^4 + x^5)*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.
| \(\displaystyle \int \frac {e^{x \log (3) \log \left (x^2 \log \left (x^2\right )\right )} \left (\left (-2 x^2-4 x\right ) \log (3)+\left (\left (-2 x^2-4 x\right ) \log (3)+3 x+4\right ) \log \left (x^2\right )+\left (-x^2-2 x\right ) \log (3) \log \left (x^2\right ) \log \left (x^2 \log \left (x^2\right )\right )\right )}{\left (x^5+4 x^4+4 x^3\right ) \log \left (x^2\right )} \, dx\) |
| \(\Big \downarrow \) 2026 |
| \(\displaystyle \int \frac {e^{x \log (3) \log \left (x^2 \log \left (x^2\right )\right )} \left (\left (-2 x^2-4 x\right ) \log (3)+\left (\left (-2 x^2-4 x\right ) \log (3)+3 x+4\right ) \log \left (x^2\right )+\left (-x^2-2 x\right ) \log (3) \log \left (x^2\right ) \log \left (x^2 \log \left (x^2\right )\right )\right )}{x^3 \left (x^2+4 x+4\right ) \log \left (x^2\right )}dx\) |
| \(\Big \downarrow \) 2007 |
| \(\displaystyle \int \frac {e^{x \log (3) \log \left (x^2 \log \left (x^2\right )\right )} \left (\left (-2 x^2-4 x\right ) \log (3)+\left (\left (-2 x^2-4 x\right ) \log (3)+3 x+4\right ) \log \left (x^2\right )+\left (-x^2-2 x\right ) \log (3) \log \left (x^2\right ) \log \left (x^2 \log \left (x^2\right )\right )\right )}{x^3 (x+2)^2 \log \left (x^2\right )}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {e^{x \log (3) \log \left (x^2 \log \left (x^2\right )\right )} \left (x^2 (-\log (9)) \log \left (x^2\right )-2 x^2 \log (3)+3 x \left (1-\frac {4 \log (3)}{3}\right ) \log \left (x^2\right )+4 \log \left (x^2\right )-4 x \log (3)\right )}{x^3 (x+2)^2 \log \left (x^2\right )}-\frac {\log (3) e^{x \log (3) \log \left (x^2 \log \left (x^2\right )\right )} \log \left (x^2 \log \left (x^2\right )\right )}{x^2 (x+2)}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {1}{4} (1+\log (81)) \int \frac {e^{x \log (3) \log \left (x^2 \log \left (x^2\right )\right )}}{x^2}dx+\frac {1}{4} \log (9) \int \frac {e^{x \log (3) \log \left (x^2 \log \left (x^2\right )\right )}}{x}dx+\frac {1}{4} \int \frac {e^{x \log (3) \log \left (x^2 \log \left (x^2\right )\right )}}{(x+2)^2}dx-\frac {1}{4} \log (9) \int \frac {e^{x \log (3) \log \left (x^2 \log \left (x^2\right )\right )}}{x+2}dx-\log (3) \int \frac {e^{x \log (3) \log \left (x^2 \log \left (x^2\right )\right )}}{x^2 \log \left (x^2\right )}dx+\frac {1}{2} \log (3) \int \frac {e^{x \log (3) \log \left (x^2 \log \left (x^2\right )\right )}}{x \log \left (x^2\right )}dx-\frac {1}{2} \log (3) \int \frac {e^{x \log (3) \log \left (x^2 \log \left (x^2\right )\right )}}{(x+2) \log \left (x^2\right )}dx-\frac {1}{2} \log (3) \int \frac {e^{x \log (3) \log \left (x^2 \log \left (x^2\right )\right )} \log \left (x^2 \log \left (x^2\right )\right )}{x^2}dx+\frac {1}{4} \log (3) \int \frac {e^{x \log (3) \log \left (x^2 \log \left (x^2\right )\right )} \log \left (x^2 \log \left (x^2\right )\right )}{x}dx-\frac {1}{4} \log (3) \int \frac {e^{x \log (3) \log \left (x^2 \log \left (x^2\right )\right )} \log \left (x^2 \log \left (x^2\right )\right )}{x+2}dx+\int \frac {e^{x \log (3) \log \left (x^2 \log \left (x^2\right )\right )}}{x^3}dx\) |
Int[(E^(x*Log[3]*Log[x^2*Log[x^2]])*((-4*x - 2*x^2)*Log[3] + (4 + 3*x + (- 4*x - 2*x^2)*Log[3])*Log[x^2] + (-2*x - x^2)*Log[3]*Log[x^2]*Log[x^2*Log[x ^2]]))/((4*x^3 + 4*x^4 + x^5)*Log[x^2]),x]
3.20.98.3.1 Defintions of rubi rules used
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{a = Rt[Coeff[Px, x, 0], Expon[Px, x]], b = Rt[Coeff[Px, x, Expon[Px, x]], Expon[Px, x]]}, Int[u*(a + b*x)^(Ex pon[Px, x]*p), x] /; EqQ[Px, (a + b*x)^Expon[Px, x]]] /; IntegerQ[p] && Pol yQ[Px, x] && GtQ[Expon[Px, x], 1] && NeQ[Coeff[Px, x, 0], 0]
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Time = 10.48 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00
| method | result | size |
| parallelrisch | \(-\frac {{\mathrm e}^{x \ln \left (3\right ) \ln \left (x^{2} \ln \left (x^{2}\right )\right )}}{x^{2} \left (2+x \right )}\) | \(25\) |
int(((-x^2-2*x)*ln(3)*ln(x^2)*ln(x^2*ln(x^2))+((-2*x^2-4*x)*ln(3)+4+3*x)*l n(x^2)+(-2*x^2-4*x)*ln(3))*exp(x*ln(3)*ln(x^2*ln(x^2)))/(x^5+4*x^4+4*x^3)/ ln(x^2),x,method=_RETURNVERBOSE)
Time = 0.23 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {e^{x \log (3) \log \left (x^2 \log \left (x^2\right )\right )} \left (\left (-4 x-2 x^2\right ) \log (3)+\left (4+3 x+\left (-4 x-2 x^2\right ) \log (3)\right ) \log \left (x^2\right )+\left (-2 x-x^2\right ) \log (3) \log \left (x^2\right ) \log \left (x^2 \log \left (x^2\right )\right )\right )}{\left (4 x^3+4 x^4+x^5\right ) \log \left (x^2\right )} \, dx=-\frac {e^{\left (x \log \left (3\right ) \log \left (x^{2} \log \left (x^{2}\right )\right )\right )}}{x^{3} + 2 \, x^{2}} \]
integrate(((-x^2-2*x)*log(3)*log(x^2)*log(x^2*log(x^2))+((-2*x^2-4*x)*log( 3)+4+3*x)*log(x^2)+(-2*x^2-4*x)*log(3))*exp(x*log(3)*log(x^2*log(x^2)))/(x ^5+4*x^4+4*x^3)/log(x^2),x, algorithm=\
Time = 0.21 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {e^{x \log (3) \log \left (x^2 \log \left (x^2\right )\right )} \left (\left (-4 x-2 x^2\right ) \log (3)+\left (4+3 x+\left (-4 x-2 x^2\right ) \log (3)\right ) \log \left (x^2\right )+\left (-2 x-x^2\right ) \log (3) \log \left (x^2\right ) \log \left (x^2 \log \left (x^2\right )\right )\right )}{\left (4 x^3+4 x^4+x^5\right ) \log \left (x^2\right )} \, dx=- \frac {e^{x \log {\left (3 \right )} \log {\left (x^{2} \log {\left (x^{2} \right )} \right )}}}{x^{3} + 2 x^{2}} \]
integrate(((-x**2-2*x)*ln(3)*ln(x**2)*ln(x**2*ln(x**2))+((-2*x**2-4*x)*ln( 3)+4+3*x)*ln(x**2)+(-2*x**2-4*x)*ln(3))*exp(x*ln(3)*ln(x**2*ln(x**2)))/(x* *5+4*x**4+4*x**3)/ln(x**2),x)
Time = 0.34 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.40 \[ \int \frac {e^{x \log (3) \log \left (x^2 \log \left (x^2\right )\right )} \left (\left (-4 x-2 x^2\right ) \log (3)+\left (4+3 x+\left (-4 x-2 x^2\right ) \log (3)\right ) \log \left (x^2\right )+\left (-2 x-x^2\right ) \log (3) \log \left (x^2\right ) \log \left (x^2 \log \left (x^2\right )\right )\right )}{\left (4 x^3+4 x^4+x^5\right ) \log \left (x^2\right )} \, dx=-\frac {e^{\left (x \log \left (3\right ) \log \left (2\right ) + 2 \, x \log \left (3\right ) \log \left (x\right ) + x \log \left (3\right ) \log \left (\log \left (x\right )\right )\right )}}{x^{3} + 2 \, x^{2}} \]
integrate(((-x^2-2*x)*log(3)*log(x^2)*log(x^2*log(x^2))+((-2*x^2-4*x)*log( 3)+4+3*x)*log(x^2)+(-2*x^2-4*x)*log(3))*exp(x*log(3)*log(x^2*log(x^2)))/(x ^5+4*x^4+4*x^3)/log(x^2),x, algorithm=\
\[ \int \frac {e^{x \log (3) \log \left (x^2 \log \left (x^2\right )\right )} \left (\left (-4 x-2 x^2\right ) \log (3)+\left (4+3 x+\left (-4 x-2 x^2\right ) \log (3)\right ) \log \left (x^2\right )+\left (-2 x-x^2\right ) \log (3) \log \left (x^2\right ) \log \left (x^2 \log \left (x^2\right )\right )\right )}{\left (4 x^3+4 x^4+x^5\right ) \log \left (x^2\right )} \, dx=\int { -\frac {{\left ({\left (x^{2} + 2 \, x\right )} \log \left (3\right ) \log \left (x^{2} \log \left (x^{2}\right )\right ) \log \left (x^{2}\right ) + 2 \, {\left (x^{2} + 2 \, x\right )} \log \left (3\right ) + {\left (2 \, {\left (x^{2} + 2 \, x\right )} \log \left (3\right ) - 3 \, x - 4\right )} \log \left (x^{2}\right )\right )} e^{\left (x \log \left (3\right ) \log \left (x^{2} \log \left (x^{2}\right )\right )\right )}}{{\left (x^{5} + 4 \, x^{4} + 4 \, x^{3}\right )} \log \left (x^{2}\right )} \,d x } \]
integrate(((-x^2-2*x)*log(3)*log(x^2)*log(x^2*log(x^2))+((-2*x^2-4*x)*log( 3)+4+3*x)*log(x^2)+(-2*x^2-4*x)*log(3))*exp(x*log(3)*log(x^2*log(x^2)))/(x ^5+4*x^4+4*x^3)/log(x^2),x, algorithm=\
integrate(-((x^2 + 2*x)*log(3)*log(x^2*log(x^2))*log(x^2) + 2*(x^2 + 2*x)* log(3) + (2*(x^2 + 2*x)*log(3) - 3*x - 4)*log(x^2))*e^(x*log(3)*log(x^2*lo g(x^2)))/((x^5 + 4*x^4 + 4*x^3)*log(x^2)), x)
Timed out. \[ \int \frac {e^{x \log (3) \log \left (x^2 \log \left (x^2\right )\right )} \left (\left (-4 x-2 x^2\right ) \log (3)+\left (4+3 x+\left (-4 x-2 x^2\right ) \log (3)\right ) \log \left (x^2\right )+\left (-2 x-x^2\right ) \log (3) \log \left (x^2\right ) \log \left (x^2 \log \left (x^2\right )\right )\right )}{\left (4 x^3+4 x^4+x^5\right ) \log \left (x^2\right )} \, dx=\int -\frac {{\mathrm {e}}^{x\,\ln \left (3\right )\,\ln \left (x^2\,\ln \left (x^2\right )\right )}\,\left (\ln \left (3\right )\,\left (2\,x^2+4\,x\right )-\ln \left (x^2\right )\,\left (3\,x-\ln \left (3\right )\,\left (2\,x^2+4\,x\right )+4\right )+\ln \left (x^2\right )\,\ln \left (3\right )\,\ln \left (x^2\,\ln \left (x^2\right )\right )\,\left (x^2+2\,x\right )\right )}{\ln \left (x^2\right )\,\left (x^5+4\,x^4+4\,x^3\right )} \,d x \]
int(-(exp(x*log(3)*log(x^2*log(x^2)))*(log(3)*(4*x + 2*x^2) - log(x^2)*(3* x - log(3)*(4*x + 2*x^2) + 4) + log(x^2)*log(3)*log(x^2*log(x^2))*(2*x + x ^2)))/(log(x^2)*(4*x^3 + 4*x^4 + x^5)),x)