\(\int \frac {e^x (-6+11 x-3 e^3 x)+e^x (5 x+x^2) \log (x)}{1008 x-504 e^3 x+63 e^6 x+(1008 x+168 x^2+e^3 (-252 x-42 x^2)) \log (x)+(252 x+84 x^2+7 x^3) \log ^2(x)} \, dx\) [6215]

Optimal result

Integrand size = 92, antiderivative size = 31 \[ \int \frac {e^x \left (-6+11 x-3 e^3 x\right )+e^x \left (5 x+x^2\right ) \log (x)}{1008 x-504 e^3 x+63 e^6 x+\left (1008 x+168 x^2+e^3 \left (-252 x-42 x^2\right )\right ) \log (x)+\left (252 x+84 x^2+7 x^3\right ) \log ^2(x)} \, dx=\frac {e^x}{7 x \left (\log (x)+\frac {3 \left (4-e^3+2 \log (x)\right )}{x}\right )} \]

[Out]

1/7*exp(x)/(3*(2*ln(x)+4-exp(3))/x+ln(x))/x

Rubi [F]

\[ \int \frac {e^x \left (-6+11 x-3 e^3 x\right )+e^x \left (5 x+x^2\right ) \log (x)}{1008 x-504 e^3 x+63 e^6 x+\left (1008 x+168 x^2+e^3 \left (-252 x-42 x^2\right )\right ) \log (x)+\left (252 x+84 x^2+7 x^3\right ) \log ^2(x)} \, dx=\int \frac {e^x \left (-6+11 x-3 e^3 x\right )+e^x \left (5 x+x^2\right ) \log (x)}{1008 x-504 e^3 x+63 e^6 x+\left (1008 x+168 x^2+e^3 \left (-252 x-42 x^2\right )\right ) \log (x)+\left (252 x+84 x^2+7 x^3\right ) \log ^2(x)} \, dx \]

[In]

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]

[Out]

-1/7*Defer[Int][E^x/(12*(1 - E^3/4) + 6*Log[x] + x*Log[x])^2, x] - (6*Defer[Int][E^x/(x*(12*(1 - E^3/4) + 6*Lo
g[x] + x*Log[x])^2), x])/7 + (3*(4 - E^3)*Defer[Int][E^x/((6 + x)*(12*(1 - E^3/4) + 6*Log[x] + x*Log[x])^2), x
])/7 + Defer[Int][E^x/((-6 - x)*(12*(1 - E^3/4) + 6*Log[x] + x*Log[x])), x]/7 + Defer[Int][E^x/(-3*(-4 + E^3)
+ (6 + x)*Log[x]), x]/7

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^x \left (-6+11 x-3 e^3 x\right )+e^x \left (5 x+x^2\right ) \log (x)}{63 e^6 x+\left (1008-504 e^3\right ) 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 {e^x \left (-6+11 x-3 e^3 x\right )+e^x \left (5 x+x^2\right ) \log (x)}{\left (1008-504 e^3+63 e^6\right ) 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 {e^x \left (-6+\left (11-3 e^3\right ) x+x (5+x) \log (x)\right )}{7 x \left (3 \left (-4+e^3\right )-(6+x) \log (x)\right )^2} \, dx \\ & = \frac {1}{7} \int \frac {e^x \left (-6+\left (11-3 e^3\right ) x+x (5+x) \log (x)\right )}{x \left (3 \left (-4+e^3\right )-(6+x) \log (x)\right )^2} \, dx \\ & = \frac {1}{7} \int \left (\frac {e^x \left (-36-3 e^3 x-x^2\right )}{x (6+x) \left (12 \left (1-\frac {e^3}{4}\right )+6 \log (x)+x \log (x)\right )^2}+\frac {e^x (5+x)}{(6+x) \left (12 \left (1-\frac {e^3}{4}\right )+6 \log (x)+x \log (x)\right )}\right ) \, dx \\ & = \frac {1}{7} \int \frac {e^x \left (-36-3 e^3 x-x^2\right )}{x (6+x) \left (12 \left (1-\frac {e^3}{4}\right )+6 \log (x)+x \log (x)\right )^2} \, dx+\frac {1}{7} \int \frac {e^x (5+x)}{(6+x) \left (12 \left (1-\frac {e^3}{4}\right )+6 \log (x)+x \log (x)\right )} \, dx \\ & = \frac {1}{7} \int \left (-\frac {e^x}{\left (12 \left (1-\frac {e^3}{4}\right )+6 \log (x)+x \log (x)\right )^2}-\frac {6 e^x}{x \left (12 \left (1-\frac {e^3}{4}\right )+6 \log (x)+x \log (x)\right )^2}+\frac {3 e^x \left (4-e^3\right )}{(6+x) \left (12 \left (1-\frac {e^3}{4}\right )+6 \log (x)+x \log (x)\right )^2}\right ) \, dx+\frac {1}{7} \int \left (\frac {e^x}{12 \left (1-\frac {e^3}{4}\right )+6 \log (x)+x \log (x)}+\frac {e^x}{(-6-x) \left (12 \left (1-\frac {e^3}{4}\right )+6 \log (x)+x \log (x)\right )}\right ) \, dx \\ & = -\left (\frac {1}{7} \int \frac {e^x}{\left (12 \left (1-\frac {e^3}{4}\right )+6 \log (x)+x \log (x)\right )^2} \, dx\right )+\frac {1}{7} \int \frac {e^x}{12 \left (1-\frac {e^3}{4}\right )+6 \log (x)+x \log (x)} \, dx+\frac {1}{7} \int \frac {e^x}{(-6-x) \left (12 \left (1-\frac {e^3}{4}\right )+6 \log (x)+x \log (x)\right )} \, dx-\frac {6}{7} \int \frac {e^x}{x \left (12 \left (1-\frac {e^3}{4}\right )+6 \log (x)+x \log (x)\right )^2} \, dx+\frac {1}{7} \left (3 \left (4-e^3\right )\right ) \int \frac {e^x}{(6+x) \left (12 \left (1-\frac {e^3}{4}\right )+6 \log (x)+x \log (x)\right )^2} \, dx \\ & = -\left (\frac {1}{7} \int \frac {e^x}{\left (12 \left (1-\frac {e^3}{4}\right )+6 \log (x)+x \log (x)\right )^2} \, dx\right )+\frac {1}{7} \int \frac {e^x}{(-6-x) \left (12 \left (1-\frac {e^3}{4}\right )+6 \log (x)+x \log (x)\right )} \, dx+\frac {1}{7} \int \frac {e^x}{-3 \left (-4+e^3\right )+(6+x) \log (x)} \, dx-\frac {6}{7} \int \frac {e^x}{x \left (12 \left (1-\frac {e^3}{4}\right )+6 \log (x)+x \log (x)\right )^2} \, dx+\frac {1}{7} \left (3 \left (4-e^3\right )\right ) \int \frac {e^x}{(6+x) \left (12 \left (1-\frac {e^3}{4}\right )+6 \log (x)+x \log (x)\right )^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77 \[ \int \frac {e^x \left (-6+11 x-3 e^3 x\right )+e^x \left (5 x+x^2\right ) \log (x)}{1008 x-504 e^3 x+63 e^6 x+\left (1008 x+168 x^2+e^3 \left (-252 x-42 x^2\right )\right ) \log (x)+\left (252 x+84 x^2+7 x^3\right ) \log ^2(x)} \, dx=\frac {e^x}{7 \left (12-3 e^3+6 \log (x)+x \log (x)\right )} \]

[In]

Integrate[(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]

[Out]

E^x/(7*(12 - 3*E^3 + 6*Log[x] + x*Log[x]))

Maple [A] (verified)

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	\(-\frac {{\mathrm e}^{x}}{7 \left (-x \ln \left (x \right )+3 \,{\mathrm e}^{3}-6 \ln \left (x \right )-12\right )}\)	\(22\)

[In]

int(((x^2+5*x)*exp(x)*ln(x)+(-3*x*exp(3)+11*x-6)*exp(x))/((7*x^3+84*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,method=_RETURNVERBOSE)

[Out]

-1/7*exp(x)/(-x*ln(x)+3*exp(3)-6*ln(x)-12)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.58 \[ \int \frac {e^x \left (-6+11 x-3 e^3 x\right )+e^x \left (5 x+x^2\right ) \log (x)}{1008 x-504 e^3 x+63 e^6 x+\left (1008 x+168 x^2+e^3 \left (-252 x-42 x^2\right )\right ) \log (x)+\left (252 x+84 x^2+7 x^3\right ) \log ^2(x)} \, dx=\frac {e^{x}}{7 \, {\left ({\left (x + 6\right )} \log \left (x\right ) - 3 \, e^{3} + 12\right )}} \]

[In]

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*exp(3)^2-504*x*exp(3)+1008*x),x, algorithm="fricas")

[Out]

1/7*e^x/((x + 6)*log(x) - 3*e^3 + 12)

Sympy [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.65 \[ \int \frac {e^x \left (-6+11 x-3 e^3 x\right )+e^x \left (5 x+x^2\right ) \log (x)}{1008 x-504 e^3 x+63 e^6 x+\left (1008 x+168 x^2+e^3 \left (-252 x-42 x^2\right )\right ) \log (x)+\left (252 x+84 x^2+7 x^3\right ) \log ^2(x)} \, dx=\frac {e^{x}}{7 x \log {\left (x \right )} + 42 \log {\left (x \right )} - 21 e^{3} + 84} \]

[In]

integrate(((x**2+5*x)*exp(x)*ln(x)+(-3*x*exp(3)+11*x-6)*exp(x))/((7*x**3+84*x**2+252*x)*ln(x)**2+((-42*x**2-25
2*x)*exp(3)+168*x**2+1008*x)*ln(x)+63*x*exp(3)**2-504*x*exp(3)+1008*x),x)

[Out]

exp(x)/(7*x*log(x) + 42*log(x) - 21*exp(3) + 84)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.58 \[ \int \frac {e^x \left (-6+11 x-3 e^3 x\right )+e^x \left (5 x+x^2\right ) \log (x)}{1008 x-504 e^3 x+63 e^6 x+\left (1008 x+168 x^2+e^3 \left (-252 x-42 x^2\right )\right ) \log (x)+\left (252 x+84 x^2+7 x^3\right ) \log ^2(x)} \, dx=\frac {e^{x}}{7 \, {\left ({\left (x + 6\right )} \log \left (x\right ) - 3 \, e^{3} + 12\right )}} \]

[In]

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*exp(3)^2-504*x*exp(3)+1008*x),x, algorithm="maxima")

[Out]

1/7*e^x/((x + 6)*log(x) - 3*e^3 + 12)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.65 \[ \int \frac {e^x \left (-6+11 x-3 e^3 x\right )+e^x \left (5 x+x^2\right ) \log (x)}{1008 x-504 e^3 x+63 e^6 x+\left (1008 x+168 x^2+e^3 \left (-252 x-42 x^2\right )\right ) \log (x)+\left (252 x+84 x^2+7 x^3\right ) \log ^2(x)} \, dx=\frac {e^{x}}{7 \, {\left (x \log \left (x\right ) - 3 \, e^{3} + 6 \, \log \left (x\right ) + 12\right )}} \]

[In]

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*exp(3)^2-504*x*exp(3)+1008*x),x, algorithm="giac")

[Out]

1/7*e^x/(x*log(x) - 3*e^3 + 6*log(x) + 12)

Mupad [F(-1)]

Timed out. \[ \int \frac {e^x \left (-6+11 x-3 e^3 x\right )+e^x \left (5 x+x^2\right ) \log (x)}{1008 x-504 e^3 x+63 e^6 x+\left (1008 x+168 x^2+e^3 \left (-252 x-42 x^2\right )\right ) \log (x)+\left (252 x+84 x^2+7 x^3\right ) \log ^2(x)} \, dx=-\int \frac {{\mathrm {e}}^x\,\left (3\,x\,{\mathrm {e}}^3-11\,x+6\right )-{\mathrm {e}}^x\,\ln \left (x\right )\,\left (x^2+5\,x\right )}{\left (7\,x^3+84\,x^2+252\,x\right )\,{\ln \left (x\right )}^2+\left (1008\,x-{\mathrm {e}}^3\,\left (42\,x^2+252\,x\right )+168\,x^2\right )\,\ln \left (x\right )+1008\,x-504\,x\,{\mathrm {e}}^3+63\,x\,{\mathrm {e}}^6} \,d x \]

[In]

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)

[Out]

-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)