\(\int \frac {e^{\frac {4 x+4 x^2+x^3+(4+4 x+x^2) \log (x)}{x^2}} (4 x^2-2 x^3+3 x^4+x^5+(-8 x-6 x^2-8 x^3-2 x^4) \log (x)+(4+16 x+9 x^2+x^3) \log ^2(x)+(-8-4 x) \log ^3(x))}{3 x^3} \, dx\) [5584]

Optimal result

Integrand size = 105, antiderivative size = 27 \[ \int \frac {e^{\frac {4 x+4 x^2+x^3+\left (4+4 x+x^2\right ) \log (x)}{x^2}} \left (4 x^2-2 x^3+3 x^4+x^5+\left (-8 x-6 x^2-8 x^3-2 x^4\right ) \log (x)+\left (4+16 x+9 x^2+x^3\right ) \log ^2(x)+(-8-4 x) \log ^3(x)\right )}{3 x^3} \, dx=\frac {1}{3} e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} (x-\log (x))^2 \]

[Out]

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

Rubi [F]

\[ \int \frac {e^{\frac {4 x+4 x^2+x^3+\left (4+4 x+x^2\right ) \log (x)}{x^2}} \left (4 x^2-2 x^3+3 x^4+x^5+\left (-8 x-6 x^2-8 x^3-2 x^4\right ) \log (x)+\left (4+16 x+9 x^2+x^3\right ) \log ^2(x)+(-8-4 x) \log ^3(x)\right )}{3 x^3} \, dx=\int \frac {e^{\frac {4 x+4 x^2+x^3+\left (4+4 x+x^2\right ) \log (x)}{x^2}} \left (4 x^2-2 x^3+3 x^4+x^5+\left (-8 x-6 x^2-8 x^3-2 x^4\right ) \log (x)+\left (4+16 x+9 x^2+x^3\right ) \log ^2(x)+(-8-4 x) \log ^3(x)\right )}{3 x^3} \, dx \]

[In]

Int[(E^((4*x + 4*x^2 + x^3 + (4 + 4*x + x^2)*Log[x])/x^2)*(4*x^2 - 2*x^3 + 3*x^4 + x^5 + (-8*x - 6*x^2 - 8*x^3
- 2*x^4)*Log[x] + (4 + 16*x + 9*x^2 + x^3)*Log[x]^2 + (-8 - 4*x)*Log[x]^3))/(3*x^3),x]

[Out]

(-2*Defer[Int][E^(((2 + x)^2*(x + Log[x]))/x^2), x])/3 + (4*Defer[Int][E^(((2 + x)^2*(x + Log[x]))/x^2)/x, x])
/3 + Defer[Int][E^(((2 + x)^2*(x + Log[x]))/x^2)*x, x] + Defer[Int][E^(((2 + x)^2*(x + Log[x]))/x^2)*x^2, x]/3
- (8*Defer[Int][E^(((2 + x)^2*(x + Log[x]))/x^2)*Log[x], x])/3 - (8*Defer[Int][(E^(((2 + x)^2*(x + Log[x]))/x
^2)*Log[x])/x^2, x])/3 - 2*Defer[Int][(E^(((2 + x)^2*(x + Log[x]))/x^2)*Log[x])/x, x] - (2*Defer[Int][E^(((2 +
x)^2*(x + Log[x]))/x^2)*x*Log[x], x])/3 + Defer[Int][E^(((2 + x)^2*(x + Log[x]))/x^2)*Log[x]^2, x]/3 + (4*Def
er[Int][(E^(((2 + x)^2*(x + Log[x]))/x^2)*Log[x]^2)/x^3, x])/3 + (16*Defer[Int][(E^(((2 + x)^2*(x + Log[x]))/x
^2)*Log[x]^2)/x^2, x])/3 + 3*Defer[Int][(E^(((2 + x)^2*(x + Log[x]))/x^2)*Log[x]^2)/x, x] - (8*Defer[Int][(E^(
((2 + x)^2*(x + Log[x]))/x^2)*Log[x]^3)/x^3, x])/3 - (4*Defer[Int][(E^(((2 + x)^2*(x + Log[x]))/x^2)*Log[x]^3)
/x^2, x])/3

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \int \frac {e^{\frac {4 x+4 x^2+x^3+\left (4+4 x+x^2\right ) \log (x)}{x^2}} \left (4 x^2-2 x^3+3 x^4+x^5+\left (-8 x-6 x^2-8 x^3-2 x^4\right ) \log (x)+\left (4+16 x+9 x^2+x^3\right ) \log ^2(x)+(-8-4 x) \log ^3(x)\right )}{x^3} \, dx \\ & = \frac {1}{3} \int \frac {e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} \left (4 x^2-2 x^3+3 x^4+x^5+\left (-8 x-6 x^2-8 x^3-2 x^4\right ) \log (x)+\left (4+16 x+9 x^2+x^3\right ) \log ^2(x)+(-8-4 x) \log ^3(x)\right )}{x^3} \, dx \\ & = \frac {1}{3} \int \left (\frac {e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} \left (4-2 x+3 x^2+x^3\right )}{x}-\frac {2 e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} \left (4+3 x+4 x^2+x^3\right ) \log (x)}{x^2}+\frac {e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} (2+x) \left (2+7 x+x^2\right ) \log ^2(x)}{x^3}-\frac {4 e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} (2+x) \log ^3(x)}{x^3}\right ) \, dx \\ & = \frac {1}{3} \int \frac {e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} \left (4-2 x+3 x^2+x^3\right )}{x} \, dx+\frac {1}{3} \int \frac {e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} (2+x) \left (2+7 x+x^2\right ) \log ^2(x)}{x^3} \, dx-\frac {2}{3} \int \frac {e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} \left (4+3 x+4 x^2+x^3\right ) \log (x)}{x^2} \, dx-\frac {4}{3} \int \frac {e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} (2+x) \log ^3(x)}{x^3} \, dx \\ & = \frac {1}{3} \int \left (-2 e^{\frac {(2+x)^2 (x+\log (x))}{x^2}}+\frac {4 e^{\frac {(2+x)^2 (x+\log (x))}{x^2}}}{x}+3 e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} x+e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} x^2\right ) \, dx+\frac {1}{3} \int \left (e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} \log ^2(x)+\frac {4 e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} \log ^2(x)}{x^3}+\frac {16 e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} \log ^2(x)}{x^2}+\frac {9 e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} \log ^2(x)}{x}\right ) \, dx-\frac {2}{3} \int \left (4 e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} \log (x)+\frac {4 e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} \log (x)}{x^2}+\frac {3 e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} \log (x)}{x}+e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} x \log (x)\right ) \, dx-\frac {4}{3} \int \left (\frac {2 e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} \log ^3(x)}{x^3}+\frac {e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} \log ^3(x)}{x^2}\right ) \, dx \\ & = \frac {1}{3} \int e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} x^2 \, dx+\frac {1}{3} \int e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} \log ^2(x) \, dx-\frac {2}{3} \int e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} \, dx-\frac {2}{3} \int e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} x \log (x) \, dx+\frac {4}{3} \int \frac {e^{\frac {(2+x)^2 (x+\log (x))}{x^2}}}{x} \, dx+\frac {4}{3} \int \frac {e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} \log ^2(x)}{x^3} \, dx-\frac {4}{3} \int \frac {e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} \log ^3(x)}{x^2} \, dx-2 \int \frac {e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} \log (x)}{x} \, dx-\frac {8}{3} \int e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} \log (x) \, dx-\frac {8}{3} \int \frac {e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} \log (x)}{x^2} \, dx-\frac {8}{3} \int \frac {e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} \log ^3(x)}{x^3} \, dx+3 \int \frac {e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} \log ^2(x)}{x} \, dx+\frac {16}{3} \int \frac {e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} \log ^2(x)}{x^2} \, dx+\int e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} x \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \frac {e^{\frac {4 x+4 x^2+x^3+\left (4+4 x+x^2\right ) \log (x)}{x^2}} \left (4 x^2-2 x^3+3 x^4+x^5+\left (-8 x-6 x^2-8 x^3-2 x^4\right ) \log (x)+\left (4+16 x+9 x^2+x^3\right ) \log ^2(x)+(-8-4 x) \log ^3(x)\right )}{3 x^3} \, dx=\frac {1}{3} e^{4+\frac {4}{x}+x} x^{\frac {(2+x)^2}{x^2}} (x-\log (x))^2 \]

[In]

Integrate[(E^((4*x + 4*x^2 + x^3 + (4 + 4*x + x^2)*Log[x])/x^2)*(4*x^2 - 2*x^3 + 3*x^4 + x^5 + (-8*x - 6*x^2 -
8*x^3 - 2*x^4)*Log[x] + (4 + 16*x + 9*x^2 + x^3)*Log[x]^2 + (-8 - 4*x)*Log[x]^3))/(3*x^3),x]

[Out]

(E^(4 + 4/x + x)*x^((2 + x)^2/x^2)*(x - Log[x])^2)/3

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11

method	result	size
risch	\(\frac {\left (\ln \left (x \right )^{2}-2 x \ln \left (x \right )+x^{2}\right ) {\mathrm e}^{\frac {\left (2+x \right )^{2} \left (x +\ln \left (x \right )\right )}{x^{2}}}}{3}\)	\(30\)
parallelrisch	\(\frac {x^{2} {\mathrm e}^{\frac {\left (x^{2}+4 x +4\right ) \ln \left (x \right )+x^{3}+4 x^{2}+4 x}{x^{2}}}}{3}-\frac {2 \,{\mathrm e}^{\frac {\left (x^{2}+4 x +4\right ) \ln \left (x \right )+x^{3}+4 x^{2}+4 x}{x^{2}}} x \ln \left (x \right )}{3}+\frac {\ln \left (x \right )^{2} {\mathrm e}^{\frac {\left (x^{2}+4 x +4\right ) \ln \left (x \right )+x^{3}+4 x^{2}+4 x}{x^{2}}}}{3}\)	\(102\)

[In]

int(1/3*((-4*x-8)*ln(x)^3+(x^3+9*x^2+16*x+4)*ln(x)^2+(-2*x^4-8*x^3-6*x^2-8*x)*ln(x)+x^5+3*x^4-2*x^3+4*x^2)*exp
(((x^2+4*x+4)*ln(x)+x^3+4*x^2+4*x)/x^2)/x^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.59 \[ \int \frac {e^{\frac {4 x+4 x^2+x^3+\left (4+4 x+x^2\right ) \log (x)}{x^2}} \left (4 x^2-2 x^3+3 x^4+x^5+\left (-8 x-6 x^2-8 x^3-2 x^4\right ) \log (x)+\left (4+16 x+9 x^2+x^3\right ) \log ^2(x)+(-8-4 x) \log ^3(x)\right )}{3 x^3} \, dx=\frac {1}{3} \, {\left (x^{2} - 2 \, x \log \left (x\right ) + \log \left (x\right )^{2}\right )} e^{\left (\frac {x^{3} + 4 \, x^{2} + {\left (x^{2} + 4 \, x + 4\right )} \log \left (x\right ) + 4 \, x}{x^{2}}\right )} \]

[In]

integrate(1/3*((-4*x-8)*log(x)^3+(x^3+9*x^2+16*x+4)*log(x)^2+(-2*x^4-8*x^3-6*x^2-8*x)*log(x)+x^5+3*x^4-2*x^3+4
*x^2)*exp(((x^2+4*x+4)*log(x)+x^3+4*x^2+4*x)/x^2)/x^3,x, algorithm="fricas")

[Out]

1/3*(x^2 - 2*x*log(x) + log(x)^2)*e^((x^3 + 4*x^2 + (x^2 + 4*x + 4)*log(x) + 4*x)/x^2)

Sympy [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.63 \[ \int \frac {e^{\frac {4 x+4 x^2+x^3+\left (4+4 x+x^2\right ) \log (x)}{x^2}} \left (4 x^2-2 x^3+3 x^4+x^5+\left (-8 x-6 x^2-8 x^3-2 x^4\right ) \log (x)+\left (4+16 x+9 x^2+x^3\right ) \log ^2(x)+(-8-4 x) \log ^3(x)\right )}{3 x^3} \, dx=\frac {\left (x^{2} - 2 x \log {\left (x \right )} + \log {\left (x \right )}^{2}\right ) e^{\frac {x^{3} + 4 x^{2} + 4 x + \left (x^{2} + 4 x + 4\right ) \log {\left (x \right )}}{x^{2}}}}{3} \]

[In]

integrate(1/3*((-4*x-8)*ln(x)**3+(x**3+9*x**2+16*x+4)*ln(x)**2+(-2*x**4-8*x**3-6*x**2-8*x)*ln(x)+x**5+3*x**4-2
*x**3+4*x**2)*exp(((x**2+4*x+4)*ln(x)+x**3+4*x**2+4*x)/x**2)/x**3,x)

[Out]

(x**2 - 2*x*log(x) + log(x)**2)*exp((x**3 + 4*x**2 + 4*x + (x**2 + 4*x + 4)*log(x))/x**2)/3

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.78 \[ \int \frac {e^{\frac {4 x+4 x^2+x^3+\left (4+4 x+x^2\right ) \log (x)}{x^2}} \left (4 x^2-2 x^3+3 x^4+x^5+\left (-8 x-6 x^2-8 x^3-2 x^4\right ) \log (x)+\left (4+16 x+9 x^2+x^3\right ) \log ^2(x)+(-8-4 x) \log ^3(x)\right )}{3 x^3} \, dx=\frac {1}{3} \, {\left (x^{3} e^{4} - 2 \, x^{2} e^{4} \log \left (x\right ) + x e^{4} \log \left (x\right )^{2}\right )} e^{\left (x + \frac {4 \, \log \left (x\right )}{x} + \frac {4}{x} + \frac {4 \, \log \left (x\right )}{x^{2}}\right )} \]

[In]

integrate(1/3*((-4*x-8)*log(x)^3+(x^3+9*x^2+16*x+4)*log(x)^2+(-2*x^4-8*x^3-6*x^2-8*x)*log(x)+x^5+3*x^4-2*x^3+4
*x^2)*exp(((x^2+4*x+4)*log(x)+x^3+4*x^2+4*x)/x^2)/x^3,x, algorithm="maxima")

[Out]

1/3*(x^3*e^4 - 2*x^2*e^4*log(x) + x*e^4*log(x)^2)*e^(x + 4*log(x)/x + 4/x + 4*log(x)/x^2)

Giac [F]

\[ \int \frac {e^{\frac {4 x+4 x^2+x^3+\left (4+4 x+x^2\right ) \log (x)}{x^2}} \left (4 x^2-2 x^3+3 x^4+x^5+\left (-8 x-6 x^2-8 x^3-2 x^4\right ) \log (x)+\left (4+16 x+9 x^2+x^3\right ) \log ^2(x)+(-8-4 x) \log ^3(x)\right )}{3 x^3} \, dx=\int { \frac {{\left (x^{5} + 3 \, x^{4} - 4 \, {\left (x + 2\right )} \log \left (x\right )^{3} - 2 \, x^{3} + {\left (x^{3} + 9 \, x^{2} + 16 \, x + 4\right )} \log \left (x\right )^{2} + 4 \, x^{2} - 2 \, {\left (x^{4} + 4 \, x^{3} + 3 \, x^{2} + 4 \, x\right )} \log \left (x\right )\right )} e^{\left (\frac {x^{3} + 4 \, x^{2} + {\left (x^{2} + 4 \, x + 4\right )} \log \left (x\right ) + 4 \, x}{x^{2}}\right )}}{3 \, x^{3}} \,d x } \]

[In]

integrate(1/3*((-4*x-8)*log(x)^3+(x^3+9*x^2+16*x+4)*log(x)^2+(-2*x^4-8*x^3-6*x^2-8*x)*log(x)+x^5+3*x^4-2*x^3+4
*x^2)*exp(((x^2+4*x+4)*log(x)+x^3+4*x^2+4*x)/x^2)/x^3,x, algorithm="giac")

[Out]

integrate(1/3*(x^5 + 3*x^4 - 4*(x + 2)*log(x)^3 - 2*x^3 + (x^3 + 9*x^2 + 16*x + 4)*log(x)^2 + 4*x^2 - 2*(x^4 +
4*x^3 + 3*x^2 + 4*x)*log(x))*e^((x^3 + 4*x^2 + (x^2 + 4*x + 4)*log(x) + 4*x)/x^2)/x^3, x)

Mupad [B] (verification not implemented)

Time = 11.07 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.56 \[ \int \frac {e^{\frac {4 x+4 x^2+x^3+\left (4+4 x+x^2\right ) \log (x)}{x^2}} \left (4 x^2-2 x^3+3 x^4+x^5+\left (-8 x-6 x^2-8 x^3-2 x^4\right ) \log (x)+\left (4+16 x+9 x^2+x^3\right ) \log ^2(x)+(-8-4 x) \log ^3(x)\right )}{3 x^3} \, dx=x\,x^{4/x}\,x^{\frac {4}{x^2}}\,{\mathrm {e}}^{x+\frac {4}{x}+4}\,\left (\frac {x^2}{3}-\frac {2\,x\,\ln \left (x\right )}{3}+\frac {{\ln \left (x\right )}^2}{3}\right ) \]

[In]

int((exp((4*x + log(x)*(4*x + x^2 + 4) + 4*x^2 + x^3)/x^2)*(4*x^2 - log(x)*(8*x + 6*x^2 + 8*x^3 + 2*x^4) - 2*x
^3 + 3*x^4 + x^5 - log(x)^3*(4*x + 8) + log(x)^2*(16*x + 9*x^2 + x^3 + 4)))/(3*x^3),x)

[Out]

x*x^(4/x)*x^(4/x^2)*exp(x + 4/x + 4)*(log(x)^2/3 - (2*x*log(x))/3 + x^2/3)