Integrand size = 92, antiderivative size = 31 \[ \int \frac {e^x \left (-6+11 x-3 e^3 x\right )+e^x \left (5 x+x^2\right ) \log (x)}{1008 x-504 e^3 x+63 e^6 x+\left (1008 x+168 x^2+e^3 \left (-252 x-42 x^2\right )\right ) \log (x)+\left (252 x+84 x^2+7 x^3\right ) \log ^2(x)} \, dx=\frac {e^x}{7 x \left (\log (x)+\frac {3 \left (4-e^3+2 \log (x)\right )}{x}\right )} \]
Time = 0.39 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77 \[ \int \frac {e^x \left (-6+11 x-3 e^3 x\right )+e^x \left (5 x+x^2\right ) \log (x)}{1008 x-504 e^3 x+63 e^6 x+\left (1008 x+168 x^2+e^3 \left (-252 x-42 x^2\right )\right ) \log (x)+\left (252 x+84 x^2+7 x^3\right ) \log ^2(x)} \, dx=\frac {e^x}{7 \left (12-3 e^3+6 \log (x)+x \log (x)\right )} \]
Integrate[(E^x*(-6 + 11*x - 3*E^3*x) + E^x*(5*x + x^2)*Log[x])/(1008*x - 5 04*E^3*x + 63*E^6*x + (1008*x + 168*x^2 + E^3*(-252*x - 42*x^2))*Log[x] + (252*x + 84*x^2 + 7*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.
\(\displaystyle \int \frac {e^x \left (x^2+5 x\right ) \log (x)+e^x \left (-3 e^3 x+11 x-6\right )}{\left (168 x^2+e^3 \left (-42 x^2-252 x\right )+1008 x\right ) \log (x)+\left (7 x^3+84 x^2+252 x\right ) \log ^2(x)+63 e^6 x-504 e^3 x+1008 x} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {e^x \left (x^2+5 x\right ) \log (x)+e^x \left (-3 e^3 x+11 x-6\right )}{\left (168 x^2+e^3 \left (-42 x^2-252 x\right )+1008 x\right ) \log (x)+\left (7 x^3+84 x^2+252 x\right ) \log ^2(x)+\left (1008-504 e^3\right ) x+63 e^6 x}dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {e^x \left (x^2+5 x\right ) \log (x)+e^x \left (-3 e^3 x+11 x-6\right )}{\left (168 x^2+e^3 \left (-42 x^2-252 x\right )+1008 x\right ) \log (x)+\left (7 x^3+84 x^2+252 x\right ) \log ^2(x)+\left (1008-504 e^3+63 e^6\right ) x}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {e^x \left (\left (11-3 e^3\right ) x+(x+5) x \log (x)-6\right )}{7 x \left (3 \left (e^3-4\right )-(x+6) \log (x)\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{7} \int -\frac {e^x \left (-((x+5) \log (x) x)-\left (11-3 e^3\right ) x+6\right )}{x \left ((x+6) \log (x)+3 \left (4-e^3\right )\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{7} \int \frac {e^x \left (-((x+5) \log (x) x)-\left (11-3 e^3\right ) x+6\right )}{x \left ((x+6) \log (x)+3 \left (4-e^3\right )\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {1}{7} \int \left (\frac {e^x (-x-5)}{(x+6) \left (x \log (x)+6 \log (x)+12 \left (1-\frac {e^3}{4}\right )\right )}+\frac {e^x \left (x^2+3 e^3 x+36\right )}{x (x+6) \left (x \log (x)+6 \log (x)+12 \left (1-\frac {e^3}{4}\right )\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{7} \left (-\int \frac {e^x}{-x \log (x)-6 \log (x)-12 \left (1-\frac {e^3}{4}\right )}dx-\int \frac {e^x}{\left (x \log (x)+6 \log (x)+12 \left (1-\frac {e^3}{4}\right )\right )^2}dx-6 \int \frac {e^x}{x \left (x \log (x)+6 \log (x)+12 \left (1-\frac {e^3}{4}\right )\right )^2}dx+3 \left (4-e^3\right ) \int \frac {e^x}{(x+6) \left (x \log (x)+6 \log (x)+12 \left (1-\frac {e^3}{4}\right )\right )^2}dx-\int \frac {e^x}{(x+6) \left (x \log (x)+6 \log (x)+12 \left (1-\frac {e^3}{4}\right )\right )}dx\right )\) |
Int[(E^x*(-6 + 11*x - 3*E^3*x) + E^x*(5*x + x^2)*Log[x])/(1008*x - 504*E^3 *x + 63*E^6*x + (1008*x + 168*x^2 + E^3*(-252*x - 42*x^2))*Log[x] + (252*x + 84*x^2 + 7*x^3)*Log[x]^2),x]
3.3.15.3.1 Defintions of rubi rules used
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
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 = 0.50 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.71
method | result | size |
risch | \(-\frac {{\mathrm e}^{x}}{7 \left (-x \ln \left (x \right )+3 \,{\mathrm e}^{3}-6 \ln \left (x \right )-12\right )}\) | \(22\) |
parallelrisch | \(-\frac {{\mathrm e}^{x}}{7 \left (-x \ln \left (x \right )+3 \,{\mathrm e}^{3}-6 \ln \left (x \right )-12\right )}\) | \(22\) |
int(((x^2+5*x)*exp(x)*ln(x)+(-3*x*exp(3)+11*x-6)*exp(x))/((7*x^3+84*x^2+25 2*x)*ln(x)^2+((-42*x^2-252*x)*exp(3)+168*x^2+1008*x)*ln(x)+63*x*exp(3)^2-5 04*x*exp(3)+1008*x),x,method=_RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.58 \[ \int \frac {e^x \left (-6+11 x-3 e^3 x\right )+e^x \left (5 x+x^2\right ) \log (x)}{1008 x-504 e^3 x+63 e^6 x+\left (1008 x+168 x^2+e^3 \left (-252 x-42 x^2\right )\right ) \log (x)+\left (252 x+84 x^2+7 x^3\right ) \log ^2(x)} \, dx=\frac {e^{x}}{7 \, {\left ({\left (x + 6\right )} \log \left (x\right ) - 3 \, e^{3} + 12\right )}} \]
integrate(((x^2+5*x)*exp(x)*log(x)+(-3*x*exp(3)+11*x-6)*exp(x))/((7*x^3+84 *x^2+252*x)*log(x)^2+((-42*x^2-252*x)*exp(3)+168*x^2+1008*x)*log(x)+63*x*e xp(3)^2-504*x*exp(3)+1008*x),x, algorithm=\
Time = 0.21 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.65 \[ \int \frac {e^x \left (-6+11 x-3 e^3 x\right )+e^x \left (5 x+x^2\right ) \log (x)}{1008 x-504 e^3 x+63 e^6 x+\left (1008 x+168 x^2+e^3 \left (-252 x-42 x^2\right )\right ) \log (x)+\left (252 x+84 x^2+7 x^3\right ) \log ^2(x)} \, dx=\frac {e^{x}}{7 x \log {\left (x \right )} + 42 \log {\left (x \right )} - 21 e^{3} + 84} \]
integrate(((x**2+5*x)*exp(x)*ln(x)+(-3*x*exp(3)+11*x-6)*exp(x))/((7*x**3+8 4*x**2+252*x)*ln(x)**2+((-42*x**2-252*x)*exp(3)+168*x**2+1008*x)*ln(x)+63* x*exp(3)**2-504*x*exp(3)+1008*x),x)
Time = 0.23 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.58 \[ \int \frac {e^x \left (-6+11 x-3 e^3 x\right )+e^x \left (5 x+x^2\right ) \log (x)}{1008 x-504 e^3 x+63 e^6 x+\left (1008 x+168 x^2+e^3 \left (-252 x-42 x^2\right )\right ) \log (x)+\left (252 x+84 x^2+7 x^3\right ) \log ^2(x)} \, dx=\frac {e^{x}}{7 \, {\left ({\left (x + 6\right )} \log \left (x\right ) - 3 \, e^{3} + 12\right )}} \]
integrate(((x^2+5*x)*exp(x)*log(x)+(-3*x*exp(3)+11*x-6)*exp(x))/((7*x^3+84 *x^2+252*x)*log(x)^2+((-42*x^2-252*x)*exp(3)+168*x^2+1008*x)*log(x)+63*x*e xp(3)^2-504*x*exp(3)+1008*x),x, algorithm=\
Time = 0.31 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.65 \[ \int \frac {e^x \left (-6+11 x-3 e^3 x\right )+e^x \left (5 x+x^2\right ) \log (x)}{1008 x-504 e^3 x+63 e^6 x+\left (1008 x+168 x^2+e^3 \left (-252 x-42 x^2\right )\right ) \log (x)+\left (252 x+84 x^2+7 x^3\right ) \log ^2(x)} \, dx=\frac {e^{x}}{7 \, {\left (x \log \left (x\right ) - 3 \, e^{3} + 6 \, \log \left (x\right ) + 12\right )}} \]
integrate(((x^2+5*x)*exp(x)*log(x)+(-3*x*exp(3)+11*x-6)*exp(x))/((7*x^3+84 *x^2+252*x)*log(x)^2+((-42*x^2-252*x)*exp(3)+168*x^2+1008*x)*log(x)+63*x*e xp(3)^2-504*x*exp(3)+1008*x),x, algorithm=\
Timed out. \[ \int \frac {e^x \left (-6+11 x-3 e^3 x\right )+e^x \left (5 x+x^2\right ) \log (x)}{1008 x-504 e^3 x+63 e^6 x+\left (1008 x+168 x^2+e^3 \left (-252 x-42 x^2\right )\right ) \log (x)+\left (252 x+84 x^2+7 x^3\right ) \log ^2(x)} \, dx=-\int \frac {{\mathrm {e}}^x\,\left (3\,x\,{\mathrm {e}}^3-11\,x+6\right )-{\mathrm {e}}^x\,\ln \left (x\right )\,\left (x^2+5\,x\right )}{\left (7\,x^3+84\,x^2+252\,x\right )\,{\ln \left (x\right )}^2+\left (1008\,x-{\mathrm {e}}^3\,\left (42\,x^2+252\,x\right )+168\,x^2\right )\,\ln \left (x\right )+1008\,x-504\,x\,{\mathrm {e}}^3+63\,x\,{\mathrm {e}}^6} \,d x \]
int(-(exp(x)*(3*x*exp(3) - 11*x + 6) - exp(x)*log(x)*(5*x + x^2))/(1008*x - 504*x*exp(3) + 63*x*exp(6) + log(x)^2*(252*x + 84*x^2 + 7*x^3) + log(x)* (1008*x - exp(3)*(252*x + 42*x^2) + 168*x^2)),x)