Integrand size = 101, antiderivative size = 31 \[ \int \frac {e^{\frac {2 \left (e^4 x+3 \log (4)\right )}{3 e^4}} \left (3 e^{2 x}+e^x \left (-15 x+25 x^2\right )\right )-2 e^{2 x+\frac {2 \left (e^4 x+3 \log (4)\right )}{3 e^4}} x \log (x)}{75 x^3-30 e^x x^2 \log (x)+3 e^{2 x} x \log ^2(x)} \, dx=\frac {e^{\frac {2 x}{3}+\frac {2 \log (4)}{e^4}}}{5 e^{-x} x-\log (x)} \]
Time = 0.98 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {e^{\frac {2 \left (e^4 x+3 \log (4)\right )}{3 e^4}} \left (3 e^{2 x}+e^x \left (-15 x+25 x^2\right )\right )-2 e^{2 x+\frac {2 \left (e^4 x+3 \log (4)\right )}{3 e^4}} x \log (x)}{75 x^3-30 e^x x^2 \log (x)+3 e^{2 x} x \log ^2(x)} \, dx=-\frac {16^{\frac {1}{e^4}} e^{5 x/3}}{-5 x+e^x \log (x)} \]
Integrate[(E^((2*(E^4*x + 3*Log[4]))/(3*E^4))*(3*E^(2*x) + E^x*(-15*x + 25 *x^2)) - 2*E^(2*x + (2*(E^4*x + 3*Log[4]))/(3*E^4))*x*Log[x])/(75*x^3 - 30 *E^x*x^2*Log[x] + 3*E^(2*x)*x*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.
\(\displaystyle \int \frac {\left (e^x \left (25 x^2-15 x\right )+3 e^{2 x}\right ) e^{\frac {2 \left (e^4 x+3 \log (4)\right )}{3 e^4}}-2 x e^{2 x+\frac {2 \left (e^4 x+3 \log (4)\right )}{3 e^4}} \log (x)}{75 x^3-30 e^x x^2 \log (x)+3 e^{2 x} x \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {16^{\frac {1}{e^4}} e^{5 x/3} \left (5 x (5 x-3)+3 e^x-2 e^x x \log (x)\right )}{3 x \left (5 x-e^x \log (x)\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} 16^{\frac {1}{e^4}} \int \frac {e^{5 x/3} \left (-5 (3-5 x) x-2 e^x \log (x) x+3 e^x\right )}{x \left (5 x-e^x \log (x)\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{3} 16^{\frac {1}{e^4}} \int \left (\frac {15 e^{5 x/3} (x \log (x)-\log (x)+1)}{\log (x) \left (e^x \log (x)-5 x\right )^2}+\frac {e^{5 x/3} (2 x \log (x)-3)}{x \log (x) \left (5 x-e^x \log (x)\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} 16^{\frac {1}{e^4}} \left (-45 \text {Subst}\left (\int \frac {e^{5 x}}{\left (15 x-e^{3 x} \log (3 x)\right )^2}dx,x,\frac {x}{3}\right )+6 \text {Subst}\left (\int \frac {e^{5 x}}{15 x-e^{3 x} \log (3 x)}dx,x,\frac {x}{3}\right )+45 \text {Subst}\left (\int \frac {e^{5 x}}{\log (3 x) \left (e^{3 x} \log (3 x)-15 x\right )^2}dx,x,\frac {x}{3}\right )+15 \int \frac {e^{5 x/3} x}{\left (5 x-e^x \log (x)\right )^2}dx-3 \int \frac {e^{5 x/3}}{x \log (x) \left (5 x-e^x \log (x)\right )}dx\right )\) |
Int[(E^((2*(E^4*x + 3*Log[4]))/(3*E^4))*(3*E^(2*x) + E^x*(-15*x + 25*x^2)) - 2*E^(2*x + (2*(E^4*x + 3*Log[4]))/(3*E^4))*x*Log[x])/(75*x^3 - 30*E^x*x ^2*Log[x] + 3*E^(2*x)*x*Log[x]^2),x]
3.23.10.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.63 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77
method | result | size |
risch | \(\frac {2^{4 \,{\mathrm e}^{-4}} {\mathrm e}^{\frac {5 x}{3}}}{-{\mathrm e}^{x} \ln \left (x \right )+5 x}\) | \(24\) |
parallelrisch | \(\frac {{\mathrm e}^{\frac {2 \left (6 \ln \left (2\right )+x \,{\mathrm e}^{4}\right ) {\mathrm e}^{-4}}{3}} {\mathrm e}^{x}}{-{\mathrm e}^{x} \ln \left (x \right )+5 x}\) | \(34\) |
int((-2*x*exp(x)^2*exp(1/3*(6*ln(2)+x*exp(4))/exp(4))^2*ln(x)+(3*exp(x)^2+ (25*x^2-15*x)*exp(x))*exp(1/3*(6*ln(2)+x*exp(4))/exp(4))^2)/(3*x*exp(x)^2* ln(x)^2-30*x^2*exp(x)*ln(x)+75*x^3),x,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.74 \[ \int \frac {e^{\frac {2 \left (e^4 x+3 \log (4)\right )}{3 e^4}} \left (3 e^{2 x}+e^x \left (-15 x+25 x^2\right )\right )-2 e^{2 x+\frac {2 \left (e^4 x+3 \log (4)\right )}{3 e^4}} x \log (x)}{75 x^3-30 e^x x^2 \log (x)+3 e^{2 x} x \log ^2(x)} \, dx=-\frac {2^{4 \, e^{\left (-4\right )}} e^{\left (\frac {5}{3} \, x\right )}}{e^{x} \log \left (x\right ) - 5 \, x} \]
integrate((-2*x*exp(x)^2*exp(1/3*(6*log(2)+x*exp(4))/exp(4))^2*log(x)+(3*e xp(x)^2+(25*x^2-15*x)*exp(x))*exp(1/3*(6*log(2)+x*exp(4))/exp(4))^2)/(3*x* exp(x)^2*log(x)^2-30*x^2*exp(x)*log(x)+75*x^3),x, algorithm=\
Time = 0.11 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77 \[ \int \frac {e^{\frac {2 \left (e^4 x+3 \log (4)\right )}{3 e^4}} \left (3 e^{2 x}+e^x \left (-15 x+25 x^2\right )\right )-2 e^{2 x+\frac {2 \left (e^4 x+3 \log (4)\right )}{3 e^4}} x \log (x)}{75 x^3-30 e^x x^2 \log (x)+3 e^{2 x} x \log ^2(x)} \, dx=- \frac {2^{\frac {4}{e^{4}}} e^{\frac {5 x}{3}}}{- 5 x + e^{x} \log {\left (x \right )}} \]
integrate((-2*x*exp(x)**2*exp(1/3*(6*ln(2)+x*exp(4))/exp(4))**2*ln(x)+(3*e xp(x)**2+(25*x**2-15*x)*exp(x))*exp(1/3*(6*ln(2)+x*exp(4))/exp(4))**2)/(3* x*exp(x)**2*ln(x)**2-30*x**2*exp(x)*ln(x)+75*x**3),x)
Time = 0.36 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.74 \[ \int \frac {e^{\frac {2 \left (e^4 x+3 \log (4)\right )}{3 e^4}} \left (3 e^{2 x}+e^x \left (-15 x+25 x^2\right )\right )-2 e^{2 x+\frac {2 \left (e^4 x+3 \log (4)\right )}{3 e^4}} x \log (x)}{75 x^3-30 e^x x^2 \log (x)+3 e^{2 x} x \log ^2(x)} \, dx=-\frac {2^{4 \, e^{\left (-4\right )}} e^{\left (\frac {5}{3} \, x\right )}}{e^{x} \log \left (x\right ) - 5 \, x} \]
integrate((-2*x*exp(x)^2*exp(1/3*(6*log(2)+x*exp(4))/exp(4))^2*log(x)+(3*e xp(x)^2+(25*x^2-15*x)*exp(x))*exp(1/3*(6*log(2)+x*exp(4))/exp(4))^2)/(3*x* exp(x)^2*log(x)^2-30*x^2*exp(x)*log(x)+75*x^3),x, algorithm=\
Time = 0.41 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.42 \[ \int \frac {e^{\frac {2 \left (e^4 x+3 \log (4)\right )}{3 e^4}} \left (3 e^{2 x}+e^x \left (-15 x+25 x^2\right )\right )-2 e^{2 x+\frac {2 \left (e^4 x+3 \log (4)\right )}{3 e^4}} x \log (x)}{75 x^3-30 e^x x^2 \log (x)+3 e^{2 x} x \log ^2(x)} \, dx=-\frac {5 \cdot 2^{4 \, e^{\left (-4\right )}} x e^{\left (\frac {2}{3} \, x\right )} + 2^{4 \, e^{\left (-4\right )}} e^{\left (\frac {5}{3} \, x\right )} \log \left (x\right )}{e^{x} \log \left (x\right )^{2} - 5 \, x \log \left (x\right )} \]
integrate((-2*x*exp(x)^2*exp(1/3*(6*log(2)+x*exp(4))/exp(4))^2*log(x)+(3*e xp(x)^2+(25*x^2-15*x)*exp(x))*exp(1/3*(6*log(2)+x*exp(4))/exp(4))^2)/(3*x* exp(x)^2*log(x)^2-30*x^2*exp(x)*log(x)+75*x^3),x, algorithm=\
Time = 11.60 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.68 \[ \int \frac {e^{\frac {2 \left (e^4 x+3 \log (4)\right )}{3 e^4}} \left (3 e^{2 x}+e^x \left (-15 x+25 x^2\right )\right )-2 e^{2 x+\frac {2 \left (e^4 x+3 \log (4)\right )}{3 e^4}} x \log (x)}{75 x^3-30 e^x x^2 \log (x)+3 e^{2 x} x \log ^2(x)} \, dx=\frac {{16}^{{\mathrm {e}}^{-4}}\,{\mathrm {e}}^{\frac {5\,x}{3}}}{5\,x-{\mathrm {e}}^x\,\ln \left (x\right )} \]