Integrand size = 124, antiderivative size = 29 \[ \int \frac {\left (\frac {3+3 x}{x}\right )^{\frac {e^{e^4 x} (5-x)}{x}} \left (\left (2 x+2 x^2\right ) \log \left (\frac {3+3 x}{x}\right )+e^{e^4 x} \log \left (\frac {3+3 x}{x}\right ) \left (-10+2 x+\left (-10-10 x+e^4 \left (10 x+8 x^2-2 x^3\right )\right ) \log \left (\frac {3+3 x}{x}\right )\right )\right )}{\left (x+x^2\right ) \log \left (\frac {3+3 x}{x}\right )} \, dx=4+2 \left (3+\frac {3}{x}\right )^{\frac {e^{e^4 x} (5-x)}{x}} x \]
Time = 0.16 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {\left (\frac {3+3 x}{x}\right )^{\frac {e^{e^4 x} (5-x)}{x}} \left (\left (2 x+2 x^2\right ) \log \left (\frac {3+3 x}{x}\right )+e^{e^4 x} \log \left (\frac {3+3 x}{x}\right ) \left (-10+2 x+\left (-10-10 x+e^4 \left (10 x+8 x^2-2 x^3\right )\right ) \log \left (\frac {3+3 x}{x}\right )\right )\right )}{\left (x+x^2\right ) \log \left (\frac {3+3 x}{x}\right )} \, dx=2 \left (3+\frac {3}{x}\right )^{-\frac {e^{e^4 x} (-5+x)}{x}} x \]
Integrate[(((3 + 3*x)/x)^((E^(E^4*x)*(5 - x))/x)*((2*x + 2*x^2)*Log[(3 + 3 *x)/x] + E^(E^4*x)*Log[(3 + 3*x)/x]*(-10 + 2*x + (-10 - 10*x + E^4*(10*x + 8*x^2 - 2*x^3))*Log[(3 + 3*x)/x])))/((x + x^2)*Log[(3 + 3*x)/x]),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.
| \(\displaystyle \int \frac {\left (\frac {3 x+3}{x}\right )^{\frac {e^{e^4 x} (5-x)}{x}} \left (\left (2 x^2+2 x\right ) \log \left (\frac {3 x+3}{x}\right )+e^{e^4 x} \left (\left (e^4 \left (-2 x^3+8 x^2+10 x\right )-10 x-10\right ) \log \left (\frac {3 x+3}{x}\right )+2 x-10\right ) \log \left (\frac {3 x+3}{x}\right )\right )}{\left (x^2+x\right ) \log \left (\frac {3 x+3}{x}\right )} \, dx\) |
| \(\Big \downarrow \) 2026 |
| \(\displaystyle \int \frac {\left (\frac {3 x+3}{x}\right )^{\frac {e^{e^4 x} (5-x)}{x}} \left (\left (2 x^2+2 x\right ) \log \left (\frac {3 x+3}{x}\right )+e^{e^4 x} \left (\left (e^4 \left (-2 x^3+8 x^2+10 x\right )-10 x-10\right ) \log \left (\frac {3 x+3}{x}\right )+2 x-10\right ) \log \left (\frac {3 x+3}{x}\right )\right )}{x (x+1) \log \left (\frac {3 x+3}{x}\right )}dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {\left (\frac {3}{x}+3\right )^{\frac {e^{e^4 x} (5-x)}{x}} \left (\left (2 x^2+2 x\right ) \log \left (\frac {3 x+3}{x}\right )+e^{e^4 x} \left (\left (e^4 \left (-2 x^3+8 x^2+10 x\right )-10 x-10\right ) \log \left (\frac {3 x+3}{x}\right )+2 x-10\right ) \log \left (\frac {3 x+3}{x}\right )\right )}{x (x+1) \log \left (\frac {3}{x}+3\right )}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {2 e^{e^4 x} \left (\frac {3}{x}+3\right )^{\frac {e^{e^4 x} (5-x)}{x}} \left (-e^4 x^3 \log \left (\frac {3}{x}+3\right )+4 e^4 x^2 \log \left (\frac {3}{x}+3\right )+x-5 \left (1-e^4\right ) x \log \left (\frac {3}{x}+3\right )-5 \log \left (\frac {3}{x}+3\right )-5\right )}{x (x+1)}+2 \left (\frac {3}{x}+3\right )^{\frac {e^{e^4 x} (5-x)}{x}}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 2 \int \left (3+\frac {3}{x}\right )^{\frac {e^{e^4 x} (5-x)}{x}}dx-10 \int \frac {e^{e^4 x} \left (3+\frac {3}{x}\right )^{-\frac {e^{e^4 x} (x-5)}{x}}}{x}dx+12 \int \frac {e^{e^4 x} \left (3+\frac {3}{x}\right )^{-\frac {e^{e^4 x} (x-5)}{x}}}{x+1}dx+10 \int \frac {\int e^{e^4 x+4} \left (3+\frac {3}{x}\right )^{-\frac {e^{e^4 x} (x-5)}{x}}dx}{x}dx-10 \int \frac {\int e^{e^4 x+4} \left (3+\frac {3}{x}\right )^{-\frac {e^{e^4 x} (x-5)}{x}}dx}{x+1}dx-10 \int \frac {\int \frac {e^{e^4 x} \left (3+\frac {3}{x}\right )^{-\frac {e^{e^4 x} (x-5)}{x}}}{x}dx}{x}dx+10 \int \frac {\int \frac {e^{e^4 x} \left (3+\frac {3}{x}\right )^{-\frac {e^{e^4 x} (x-5)}{x}}}{x}dx}{x+1}dx-2 \int \frac {\int e^{e^4 x+4} \left (3+\frac {3}{x}\right )^{-\frac {e^{e^4 x} (x-5)}{x}} xdx}{x}dx+2 \int \frac {\int e^{e^4 x+4} \left (3+\frac {3}{x}\right )^{-\frac {e^{e^4 x} (x-5)}{x}} xdx}{x+1}dx+10 \log \left (\frac {3}{x}+3\right ) \int e^{e^4 x+4} \left (3+\frac {3}{x}\right )^{-\frac {e^{e^4 x} (x-5)}{x}}dx-10 \log \left (\frac {3}{x}+3\right ) \int \frac {e^{e^4 x} \left (3+\frac {3}{x}\right )^{-\frac {e^{e^4 x} (x-5)}{x}}}{x}dx-2 \log \left (\frac {3}{x}+3\right ) \int e^{e^4 x+4} \left (3+\frac {3}{x}\right )^{-\frac {e^{e^4 x} (x-5)}{x}} xdx\) |
Int[(((3 + 3*x)/x)^((E^(E^4*x)*(5 - x))/x)*((2*x + 2*x^2)*Log[(3 + 3*x)/x] + E^(E^4*x)*Log[(3 + 3*x)/x]*(-10 + 2*x + (-10 - 10*x + E^4*(10*x + 8*x^2 - 2*x^3))*Log[(3 + 3*x)/x])))/((x + x^2)*Log[(3 + 3*x)/x]),x]
3.25.85.3.1 Defintions of rubi rules used
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])
\[\int \frac {\left (\left (\left (\left (-2 x^{3}+8 x^{2}+10 x \right ) {\mathrm e}^{4}-10 x -10\right ) \ln \left (\frac {3 x +3}{x}\right )+2 x -10\right ) {\mathrm e}^{\ln \left (\ln \left (\frac {3 x +3}{x}\right )\right )+x \,{\mathrm e}^{4}}+\left (2 x^{2}+2 x \right ) \ln \left (\frac {3 x +3}{x}\right )\right ) {\mathrm e}^{\frac {\left (5-x \right ) {\mathrm e}^{\ln \left (\ln \left (\frac {3 x +3}{x}\right )\right )+x \,{\mathrm e}^{4}}}{x}}}{\left (x^{2}+x \right ) \ln \left (\frac {3 x +3}{x}\right )}d x\]
int(((((-2*x^3+8*x^2+10*x)*exp(4)-10*x-10)*ln((3*x+3)/x)+2*x-10)*exp(ln(ln ((3*x+3)/x))+x*exp(4))+(2*x^2+2*x)*ln((3*x+3)/x))*exp((5-x)*exp(ln(ln((3*x +3)/x))+x*exp(4))/x)/(x^2+x)/ln((3*x+3)/x),x)
int(((((-2*x^3+8*x^2+10*x)*exp(4)-10*x-10)*ln((3*x+3)/x)+2*x-10)*exp(ln(ln ((3*x+3)/x))+x*exp(4))+(2*x^2+2*x)*ln((3*x+3)/x))*exp((5-x)*exp(ln(ln((3*x +3)/x))+x*exp(4))/x)/(x^2+x)/ln((3*x+3)/x),x)
Time = 0.28 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {\left (\frac {3+3 x}{x}\right )^{\frac {e^{e^4 x} (5-x)}{x}} \left (\left (2 x+2 x^2\right ) \log \left (\frac {3+3 x}{x}\right )+e^{e^4 x} \log \left (\frac {3+3 x}{x}\right ) \left (-10+2 x+\left (-10-10 x+e^4 \left (10 x+8 x^2-2 x^3\right )\right ) \log \left (\frac {3+3 x}{x}\right )\right )\right )}{\left (x+x^2\right ) \log \left (\frac {3+3 x}{x}\right )} \, dx=2 \, x e^{\left (-\frac {{\left (x - 5\right )} e^{\left (x e^{4} + \log \left (\log \left (\frac {3 \, {\left (x + 1\right )}}{x}\right )\right )\right )}}{x}\right )} \]
integrate(((((-2*x^3+8*x^2+10*x)*exp(4)-10*x-10)*log((3*x+3)/x)+2*x-10)*ex p(log(log((3*x+3)/x))+x*exp(4))+(2*x^2+2*x)*log((3*x+3)/x))*exp((5-x)*exp( log(log((3*x+3)/x))+x*exp(4))/x)/(x^2+x)/log((3*x+3)/x),x, algorithm=\
Time = 76.37 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {\left (\frac {3+3 x}{x}\right )^{\frac {e^{e^4 x} (5-x)}{x}} \left (\left (2 x+2 x^2\right ) \log \left (\frac {3+3 x}{x}\right )+e^{e^4 x} \log \left (\frac {3+3 x}{x}\right ) \left (-10+2 x+\left (-10-10 x+e^4 \left (10 x+8 x^2-2 x^3\right )\right ) \log \left (\frac {3+3 x}{x}\right )\right )\right )}{\left (x+x^2\right ) \log \left (\frac {3+3 x}{x}\right )} \, dx=2 x e^{\frac {\left (5 - x\right ) e^{x e^{4}} \log {\left (\frac {3 x + 3}{x} \right )}}{x}} \]
integrate(((((-2*x**3+8*x**2+10*x)*exp(4)-10*x-10)*ln((3*x+3)/x)+2*x-10)*e xp(ln(ln((3*x+3)/x))+x*exp(4))+(2*x**2+2*x)*ln((3*x+3)/x))*exp((5-x)*exp(l n(ln((3*x+3)/x))+x*exp(4))/x)/(x**2+x)/ln((3*x+3)/x),x)
Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (29) = 58\).
Time = 0.55 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.45 \[ \int \frac {\left (\frac {3+3 x}{x}\right )^{\frac {e^{e^4 x} (5-x)}{x}} \left (\left (2 x+2 x^2\right ) \log \left (\frac {3+3 x}{x}\right )+e^{e^4 x} \log \left (\frac {3+3 x}{x}\right ) \left (-10+2 x+\left (-10-10 x+e^4 \left (10 x+8 x^2-2 x^3\right )\right ) \log \left (\frac {3+3 x}{x}\right )\right )\right )}{\left (x+x^2\right ) \log \left (\frac {3+3 x}{x}\right )} \, dx=2 \, x e^{\left (-e^{\left (x e^{4}\right )} \log \left (3\right ) - e^{\left (x e^{4}\right )} \log \left (x + 1\right ) + e^{\left (x e^{4}\right )} \log \left (x\right ) + \frac {5 \, e^{\left (x e^{4}\right )} \log \left (3\right )}{x} + \frac {5 \, e^{\left (x e^{4}\right )} \log \left (x + 1\right )}{x} - \frac {5 \, e^{\left (x e^{4}\right )} \log \left (x\right )}{x}\right )} \]
integrate(((((-2*x^3+8*x^2+10*x)*exp(4)-10*x-10)*log((3*x+3)/x)+2*x-10)*ex p(log(log((3*x+3)/x))+x*exp(4))+(2*x^2+2*x)*log((3*x+3)/x))*exp((5-x)*exp( log(log((3*x+3)/x))+x*exp(4))/x)/(x^2+x)/log((3*x+3)/x),x, algorithm=\
2*x*e^(-e^(x*e^4)*log(3) - e^(x*e^4)*log(x + 1) + e^(x*e^4)*log(x) + 5*e^( x*e^4)*log(3)/x + 5*e^(x*e^4)*log(x + 1)/x - 5*e^(x*e^4)*log(x)/x)
\[ \int \frac {\left (\frac {3+3 x}{x}\right )^{\frac {e^{e^4 x} (5-x)}{x}} \left (\left (2 x+2 x^2\right ) \log \left (\frac {3+3 x}{x}\right )+e^{e^4 x} \log \left (\frac {3+3 x}{x}\right ) \left (-10+2 x+\left (-10-10 x+e^4 \left (10 x+8 x^2-2 x^3\right )\right ) \log \left (\frac {3+3 x}{x}\right )\right )\right )}{\left (x+x^2\right ) \log \left (\frac {3+3 x}{x}\right )} \, dx=\int { -\frac {2 \, {\left ({\left ({\left ({\left (x^{3} - 4 \, x^{2} - 5 \, x\right )} e^{4} + 5 \, x + 5\right )} \log \left (\frac {3 \, {\left (x + 1\right )}}{x}\right ) - x + 5\right )} e^{\left (x e^{4} + \log \left (\log \left (\frac {3 \, {\left (x + 1\right )}}{x}\right )\right )\right )} - {\left (x^{2} + x\right )} \log \left (\frac {3 \, {\left (x + 1\right )}}{x}\right )\right )} e^{\left (-\frac {{\left (x - 5\right )} e^{\left (x e^{4} + \log \left (\log \left (\frac {3 \, {\left (x + 1\right )}}{x}\right )\right )\right )}}{x}\right )}}{{\left (x^{2} + x\right )} \log \left (\frac {3 \, {\left (x + 1\right )}}{x}\right )} \,d x } \]
integrate(((((-2*x^3+8*x^2+10*x)*exp(4)-10*x-10)*log((3*x+3)/x)+2*x-10)*ex p(log(log((3*x+3)/x))+x*exp(4))+(2*x^2+2*x)*log((3*x+3)/x))*exp((5-x)*exp( log(log((3*x+3)/x))+x*exp(4))/x)/(x^2+x)/log((3*x+3)/x),x, algorithm=\
integrate(-2*((((x^3 - 4*x^2 - 5*x)*e^4 + 5*x + 5)*log(3*(x + 1)/x) - x + 5)*e^(x*e^4 + log(log(3*(x + 1)/x))) - (x^2 + x)*log(3*(x + 1)/x))*e^(-(x - 5)*e^(x*e^4 + log(log(3*(x + 1)/x)))/x)/((x^2 + x)*log(3*(x + 1)/x)), x)
Time = 10.97 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {\left (\frac {3+3 x}{x}\right )^{\frac {e^{e^4 x} (5-x)}{x}} \left (\left (2 x+2 x^2\right ) \log \left (\frac {3+3 x}{x}\right )+e^{e^4 x} \log \left (\frac {3+3 x}{x}\right ) \left (-10+2 x+\left (-10-10 x+e^4 \left (10 x+8 x^2-2 x^3\right )\right ) \log \left (\frac {3+3 x}{x}\right )\right )\right )}{\left (x+x^2\right ) \log \left (\frac {3+3 x}{x}\right )} \, dx=2\,x\,{\left (\frac {3}{x}+3\right )}^{\frac {5\,{\mathrm {e}}^{x\,{\mathrm {e}}^4}}{x}-{\mathrm {e}}^{x\,{\mathrm {e}}^4}} \]