Integrand size = 94, antiderivative size = 25 \[ \int \frac {e^{\frac {\frac {3 x}{e}+(100+25 x) \log (x)+(-4-x) \log ^3(x)}{(4+x) \log (x)}} \left (\frac {3 \left (-4 x-x^2\right )}{e}+\frac {12 x \log (x)}{e}+\left (-32-16 x-2 x^2\right ) \log ^3(x)\right )}{\left (16 x+8 x^2+x^3\right ) \log ^2(x)} \, dx=e^{25+\frac {3 x}{e (4+x) \log (x)}-\log ^2(x)} \]
Time = 3.24 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {\frac {3 x}{e}+(100+25 x) \log (x)+(-4-x) \log ^3(x)}{(4+x) \log (x)}} \left (\frac {3 \left (-4 x-x^2\right )}{e}+\frac {12 x \log (x)}{e}+\left (-32-16 x-2 x^2\right ) \log ^3(x)\right )}{\left (16 x+8 x^2+x^3\right ) \log ^2(x)} \, dx=e^{25+\frac {3 x}{e (4+x) \log (x)}-\log ^2(x)} \]
Integrate[(E^(((3*x)/E + (100 + 25*x)*Log[x] + (-4 - x)*Log[x]^3)/((4 + x) *Log[x]))*((3*(-4*x - x^2))/E + (12*x*Log[x])/E + (-32 - 16*x - 2*x^2)*Log [x]^3))/((16*x + 8*x^2 + x^3)*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.
\(\displaystyle \int \frac {\left (\frac {3 \left (-x^2-4 x\right )}{e}+\left (-2 x^2-16 x-32\right ) \log ^3(x)+\frac {12 x \log (x)}{e}\right ) \exp \left (\frac {\frac {3 x}{e}+(-x-4) \log ^3(x)+(25 x+100) \log (x)}{(x+4) \log (x)}\right )}{\left (x^3+8 x^2+16 x\right ) \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\left (\frac {3 \left (-x^2-4 x\right )}{e}+\left (-2 x^2-16 x-32\right ) \log ^3(x)+\frac {12 x \log (x)}{e}\right ) \exp \left (\frac {\frac {3 x}{e}+(-x-4) \log ^3(x)+(25 x+100) \log (x)}{(x+4) \log (x)}\right )}{x \left (x^2+8 x+16\right ) \log ^2(x)}dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {\left (\frac {3 \left (-x^2-4 x\right )}{e}+\left (-2 x^2-16 x-32\right ) \log ^3(x)+\frac {12 x \log (x)}{e}\right ) \exp \left (\frac {\frac {3 x}{e}+(-x-4) \log ^3(x)+(25 x+100) \log (x)}{(x+4) \log (x)}\right )}{x (x+4)^2 \log ^2(x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {2 \log (x) \exp \left (\frac {\frac {3 x}{e}+(-x-4) \log ^3(x)+(25 x+100) \log (x)}{(x+4) \log (x)}\right )}{x}+\frac {12 \exp \left (\frac {\frac {3 x}{e}+(-x-4) \log ^3(x)+(25 x+100) \log (x)}{(x+4) \log (x)}-1\right )}{(x+4)^2 \log (x)}-\frac {3 \exp \left (\frac {\frac {3 x}{e}+(-x-4) \log ^3(x)+(25 x+100) \log (x)}{(x+4) \log (x)}-1\right )}{(x+4) \log ^2(x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 12 \int \frac {\exp \left (\frac {(-x-4) \log ^3(x)+(25 x+100) \log (x)+\frac {3 x}{e}}{(x+4) \log (x)}-1\right )}{(x+4)^2 \log (x)}dx-2 \int \frac {\exp \left (\frac {(-x-4) \log ^3(x)+(25 x+100) \log (x)+\frac {3 x}{e}}{(x+4) \log (x)}\right ) \log (x)}{x}dx-3 \int \frac {\exp \left (\frac {(-x-4) \log ^3(x)+(25 x+100) \log (x)+\frac {3 x}{e}}{(x+4) \log (x)}-1\right )}{(x+4) \log ^2(x)}dx\) |
Int[(E^(((3*x)/E + (100 + 25*x)*Log[x] + (-4 - x)*Log[x]^3)/((4 + x)*Log[x ]))*((3*(-4*x - x^2))/E + (12*x*Log[x])/E + (-32 - 16*x - 2*x^2)*Log[x]^3) )/((16*x + 8*x^2 + x^3)*Log[x]^2),x]
3.7.70.3.1 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 = 3.86 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.52
method | result | size |
parallelrisch | \({\mathrm e}^{\frac {\left (-4-x \right ) \ln \left (x \right )^{3}+\left (25 x +100\right ) \ln \left (x \right )+x \,{\mathrm e}^{\ln \left (3\right )-1}}{\left (4+x \right ) \ln \left (x \right )}}\) | \(38\) |
risch | \({\mathrm e}^{\frac {-x \ln \left (x \right )^{3}-4 \ln \left (x \right )^{3}+25 x \ln \left (x \right )+3 \,{\mathrm e}^{-1} x +100 \ln \left (x \right )}{\left (4+x \right ) \ln \left (x \right )}}\) | \(40\) |
int(((-2*x^2-16*x-32)*ln(x)^3+4*x*exp(ln(3)-1)*ln(x)+(-x^2-4*x)*exp(ln(3)- 1))*exp(((-4-x)*ln(x)^3+(25*x+100)*ln(x)+x*exp(ln(3)-1))/(4+x)/ln(x))/(x^3 +8*x^2+16*x)/ln(x)^2,x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.44 \[ \int \frac {e^{\frac {\frac {3 x}{e}+(100+25 x) \log (x)+(-4-x) \log ^3(x)}{(4+x) \log (x)}} \left (\frac {3 \left (-4 x-x^2\right )}{e}+\frac {12 x \log (x)}{e}+\left (-32-16 x-2 x^2\right ) \log ^3(x)\right )}{\left (16 x+8 x^2+x^3\right ) \log ^2(x)} \, dx=e^{\left (-\frac {{\left (x + 4\right )} \log \left (x\right )^{3} - x e^{\left (\log \left (3\right ) - 1\right )} - 25 \, {\left (x + 4\right )} \log \left (x\right )}{{\left (x + 4\right )} \log \left (x\right )}\right )} \]
integrate(((-2*x^2-16*x-32)*log(x)^3+4*x*exp(log(3)-1)*log(x)+(-x^2-4*x)*e xp(log(3)-1))*exp(((-4-x)*log(x)^3+(25*x+100)*log(x)+x*exp(log(3)-1))/(4+x )/log(x))/(x^3+8*x^2+16*x)/log(x)^2,x, algorithm=\
Time = 0.29 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \[ \int \frac {e^{\frac {\frac {3 x}{e}+(100+25 x) \log (x)+(-4-x) \log ^3(x)}{(4+x) \log (x)}} \left (\frac {3 \left (-4 x-x^2\right )}{e}+\frac {12 x \log (x)}{e}+\left (-32-16 x-2 x^2\right ) \log ^3(x)\right )}{\left (16 x+8 x^2+x^3\right ) \log ^2(x)} \, dx=e^{\frac {\frac {3 x}{e} + \left (- x - 4\right ) \log {\left (x \right )}^{3} + \left (25 x + 100\right ) \log {\left (x \right )}}{\left (x + 4\right ) \log {\left (x \right )}}} \]
integrate(((-2*x**2-16*x-32)*ln(x)**3+4*x*exp(ln(3)-1)*ln(x)+(-x**2-4*x)*e xp(ln(3)-1))*exp(((-4-x)*ln(x)**3+(25*x+100)*ln(x)+x*exp(ln(3)-1))/(4+x)/l n(x))/(x**3+8*x**2+16*x)/ln(x)**2,x)
Time = 0.36 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.36 \[ \int \frac {e^{\frac {\frac {3 x}{e}+(100+25 x) \log (x)+(-4-x) \log ^3(x)}{(4+x) \log (x)}} \left (\frac {3 \left (-4 x-x^2\right )}{e}+\frac {12 x \log (x)}{e}+\left (-32-16 x-2 x^2\right ) \log ^3(x)\right )}{\left (16 x+8 x^2+x^3\right ) \log ^2(x)} \, dx=e^{\left (-\log \left (x\right )^{2} + \frac {3 \, e^{\left (-1\right )}}{\log \left (x\right )} - \frac {12}{{\left (x e + 4 \, e\right )} \log \left (x\right )} + 25\right )} \]
integrate(((-2*x^2-16*x-32)*log(x)^3+4*x*exp(log(3)-1)*log(x)+(-x^2-4*x)*e xp(log(3)-1))*exp(((-4-x)*log(x)^3+(25*x+100)*log(x)+x*exp(log(3)-1))/(4+x )/log(x))/(x^3+8*x^2+16*x)/log(x)^2,x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 86 vs. \(2 (25) = 50\).
Time = 0.51 (sec) , antiderivative size = 86, normalized size of antiderivative = 3.44 \[ \int \frac {e^{\frac {\frac {3 x}{e}+(100+25 x) \log (x)+(-4-x) \log ^3(x)}{(4+x) \log (x)}} \left (\frac {3 \left (-4 x-x^2\right )}{e}+\frac {12 x \log (x)}{e}+\left (-32-16 x-2 x^2\right ) \log ^3(x)\right )}{\left (16 x+8 x^2+x^3\right ) \log ^2(x)} \, dx=e^{\left (-\frac {x \log \left (x\right )^{3}}{x \log \left (x\right ) + 4 \, \log \left (x\right )} - \frac {4 \, \log \left (x\right )^{3}}{x \log \left (x\right ) + 4 \, \log \left (x\right )} + \frac {x e^{\left (\log \left (3\right ) - 1\right )}}{x \log \left (x\right ) + 4 \, \log \left (x\right )} + \frac {25 \, x \log \left (x\right )}{x \log \left (x\right ) + 4 \, \log \left (x\right )} + \frac {100 \, \log \left (x\right )}{x \log \left (x\right ) + 4 \, \log \left (x\right )}\right )} \]
integrate(((-2*x^2-16*x-32)*log(x)^3+4*x*exp(log(3)-1)*log(x)+(-x^2-4*x)*e xp(log(3)-1))*exp(((-4-x)*log(x)^3+(25*x+100)*log(x)+x*exp(log(3)-1))/(4+x )/log(x))/(x^3+8*x^2+16*x)/log(x)^2,x, algorithm=\
e^(-x*log(x)^3/(x*log(x) + 4*log(x)) - 4*log(x)^3/(x*log(x) + 4*log(x)) + x*e^(log(3) - 1)/(x*log(x) + 4*log(x)) + 25*x*log(x)/(x*log(x) + 4*log(x)) + 100*log(x)/(x*log(x) + 4*log(x)))
Time = 8.26 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.12 \[ \int \frac {e^{\frac {\frac {3 x}{e}+(100+25 x) \log (x)+(-4-x) \log ^3(x)}{(4+x) \log (x)}} \left (\frac {3 \left (-4 x-x^2\right )}{e}+\frac {12 x \log (x)}{e}+\left (-32-16 x-2 x^2\right ) \log ^3(x)\right )}{\left (16 x+8 x^2+x^3\right ) \log ^2(x)} \, dx=x^{\frac {25}{\ln \left (x\right )}}\,{\mathrm {e}}^{-\frac {4\,{\ln \left (x\right )}^2}{x+4}}\,{\mathrm {e}}^{-\frac {x\,{\ln \left (x\right )}^2}{x+4}}\,{\mathrm {e}}^{\frac {3\,x}{4\,\mathrm {e}\,\ln \left (x\right )+x\,\mathrm {e}\,\ln \left (x\right )}} \]
int(-(exp((x*exp(log(3) - 1) + log(x)*(25*x + 100) - log(x)^3*(x + 4))/(lo g(x)*(x + 4)))*(log(x)^3*(16*x + 2*x^2 + 32) + exp(log(3) - 1)*(4*x + x^2) - 4*x*exp(log(3) - 1)*log(x)))/(log(x)^2*(16*x + 8*x^2 + x^3)),x)