\(\int \frac {e^{2/x} (3 x+3 x^2)+e^{2/x} (-3 x-3 x^2+x^3) \log (x)+e^{2/x} (-2-2 x) \log (x) \log (\frac {5 \log ^3(x)}{x+x^2})}{(x^2+x^3) \log (x)} \, dx\) [1339]

Optimal result

Integrand size = 83, antiderivative size = 25 \[ \int \frac {e^{2/x} \left (3 x+3 x^2\right )+e^{2/x} \left (-3 x-3 x^2+x^3\right ) \log (x)+e^{2/x} (-2-2 x) \log (x) \log \left (\frac {5 \log ^3(x)}{x+x^2}\right )}{\left (x^2+x^3\right ) \log (x)} \, dx=e^{2/x} \left (x+\log \left (\frac {5 \log ^3(x)}{x (1+x)}\right )\right ) \] Output:

exp(2/x)*(x+ln(5*ln(x)^3/x/(1+x)))
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 1.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {e^{2/x} \left (3 x+3 x^2\right )+e^{2/x} \left (-3 x-3 x^2+x^3\right ) \log (x)+e^{2/x} (-2-2 x) \log (x) \log \left (\frac {5 \log ^3(x)}{x+x^2}\right )}{\left (x^2+x^3\right ) \log (x)} \, dx=e^{2/x} \left (x+\log \left (\frac {5 \log ^3(x)}{x+x^2}\right )\right ) \] Input:

Integrate[(E^(2/x)*(3*x + 3*x^2) + E^(2/x)*(-3*x - 3*x^2 + x^3)*Log[x] + E 
^(2/x)*(-2 - 2*x)*Log[x]*Log[(5*Log[x]^3)/(x + x^2)])/((x^2 + x^3)*Log[x]) 
,x]
 

Output:

E^(2/x)*(x + Log[(5*Log[x]^3)/(x + x^2)])
 

Rubi [A] (verified)

Time = 10.91 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {2026, 7293, 2009}

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.

\(\displaystyle \int \frac {e^{2/x} \left (3 x^2+3 x\right )+e^{2/x} (-2 x-2) \log (x) \log \left (\frac {5 \log ^3(x)}{x^2+x}\right )+e^{2/x} \left (x^3-3 x^2-3 x\right ) \log (x)}{\left (x^3+x^2\right ) \log (x)} \, dx\)

\(\Big \downarrow \) 2026

\(\displaystyle \int \frac {e^{2/x} \left (3 x^2+3 x\right )+e^{2/x} (-2 x-2) \log (x) \log \left (\frac {5 \log ^3(x)}{x^2+x}\right )+e^{2/x} \left (x^3-3 x^2-3 x\right ) \log (x)}{x^2 (x+1) \log (x)}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (-\frac {e^{2/x} \left (x^3 \log (x)+3 x^2-2 x \log (x) \log \left (\frac {5 \log ^3(x)}{x^2+x}\right )-2 \log (x) \log \left (\frac {5 \log ^3(x)}{x^2+x}\right )-3 x^2 \log (x)+3 x-3 x \log (x)\right )}{x \log (x)}+\frac {e^{2/x} \left (x^3 \log (x)+3 x^2-2 x \log (x) \log \left (\frac {5 \log ^3(x)}{x^2+x}\right )-2 \log (x) \log \left (\frac {5 \log ^3(x)}{x^2+x}\right )-3 x^2 \log (x)+3 x-3 x \log (x)\right )}{(x+1) \log (x)}+\frac {e^{2/x} \left (x^3 \log (x)+3 x^2-2 x \log (x) \log \left (\frac {5 \log ^3(x)}{x^2+x}\right )-2 \log (x) \log \left (\frac {5 \log ^3(x)}{x^2+x}\right )-3 x^2 \log (x)+3 x-3 x \log (x)\right )}{x^2 \log (x)}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle e^{2/x} x+e^{2/x} \log \left (\frac {5 \log ^3(x)}{x (x+1)}\right )\)

Input:

Int[(E^(2/x)*(3*x + 3*x^2) + E^(2/x)*(-3*x - 3*x^2 + x^3)*Log[x] + E^(2/x) 
*(-2 - 2*x)*Log[x]*Log[(5*Log[x]^3)/(x + x^2)])/((x^2 + x^3)*Log[x]),x]
 

Output:

E^(2/x)*x + E^(2/x)*Log[(5*Log[x]^3)/(x*(1 + x))]
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2026

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])
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

Maple [A] (verified)

Time = 1.23 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28

method	result	size
parallelrisch	\(x \,{\mathrm e}^{\frac {2}{x}}+{\mathrm e}^{\frac {2}{x}} \ln \left (\frac {5 \ln \left (x \right )^{3}}{x \left (1+x \right )}\right )\)	\(32\)
risch	\(-{\mathrm e}^{\frac {2}{x}} \ln \left (1+x \right )+3 \,{\mathrm e}^{\frac {2}{x}} \ln \left (\ln \left (x \right )\right )-\ln \left (x \right ) {\mathrm e}^{\frac {2}{x}}-\frac {i \pi \,\operatorname {csgn}\left (\frac {i}{1+x}\right ) \operatorname {csgn}\left (i \ln \left (x \right )^{3}\right ) \operatorname {csgn}\left (\frac {i \ln \left (x \right )^{3}}{1+x}\right ) {\mathrm e}^{\frac {2}{x}}}{2}+\frac {i \pi \,\operatorname {csgn}\left (\frac {i}{1+x}\right ) \operatorname {csgn}\left (\frac {i \ln \left (x \right )^{3}}{1+x}\right )^{2} {\mathrm e}^{\frac {2}{x}}}{2}-\frac {i \pi \operatorname {csgn}\left (i \ln \left (x \right )\right )^{2} \operatorname {csgn}\left (i \ln \left (x \right )^{2}\right ) {\mathrm e}^{\frac {2}{x}}}{2}-\frac {i \pi \operatorname {csgn}\left (\frac {i \ln \left (x \right )^{3}}{1+x}\right )^{3} {\mathrm e}^{\frac {2}{x}}}{2}+\frac {i \pi \,\operatorname {csgn}\left (i \ln \left (x \right )\right ) \operatorname {csgn}\left (i \ln \left (x \right )^{3}\right )^{2} {\mathrm e}^{\frac {2}{x}}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i \ln \left (x \right )\right ) \operatorname {csgn}\left (i \ln \left (x \right )^{2}\right ) \operatorname {csgn}\left (i \ln \left (x \right )^{3}\right ) {\mathrm e}^{\frac {2}{x}}}{2}+\frac {i \pi \,\operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (\frac {i \ln \left (x \right )^{3}}{x \left (1+x \right )}\right )^{2} {\mathrm e}^{\frac {2}{x}}}{2}-\frac {i \pi \operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )^{3} {\mathrm e}^{\frac {2}{x}}}{2}+\frac {i \pi \,\operatorname {csgn}\left (i \ln \left (x \right )^{3}\right ) \operatorname {csgn}\left (\frac {i \ln \left (x \right )^{3}}{1+x}\right )^{2} {\mathrm e}^{\frac {2}{x}}}{2}+\frac {i \pi \,\operatorname {csgn}\left (i \ln \left (x \right )^{2}\right ) \operatorname {csgn}\left (i \ln \left (x \right )^{3}\right )^{2} {\mathrm e}^{\frac {2}{x}}}{2}-\frac {i \pi \operatorname {csgn}\left (\frac {i \ln \left (x \right )^{3}}{x \left (1+x \right )}\right )^{3} {\mathrm e}^{\frac {2}{x}}}{2}+i \pi \,\operatorname {csgn}\left (i \ln \left (x \right )\right ) \operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )^{2} {\mathrm e}^{\frac {2}{x}}-\frac {i \pi \,\operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (\frac {i \ln \left (x \right )^{3}}{1+x}\right ) \operatorname {csgn}\left (\frac {i \ln \left (x \right )^{3}}{x \left (1+x \right )}\right ) {\mathrm e}^{\frac {2}{x}}}{2}-\frac {i \pi \operatorname {csgn}\left (i \ln \left (x \right )^{3}\right )^{3} {\mathrm e}^{\frac {2}{x}}}{2}+\frac {i \pi \,\operatorname {csgn}\left (\frac {i \ln \left (x \right )^{3}}{1+x}\right ) \operatorname {csgn}\left (\frac {i \ln \left (x \right )^{3}}{x \left (1+x \right )}\right )^{2} {\mathrm e}^{\frac {2}{x}}}{2}+\ln \left (5\right ) {\mathrm e}^{\frac {2}{x}}+x \,{\mathrm e}^{\frac {2}{x}}\)	\(512\)

Input:

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

Output:

x*exp(2/x)+exp(2/x)*ln(5*ln(x)^3/x/(1+x))
 

Fricas [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int \frac {e^{2/x} \left (3 x+3 x^2\right )+e^{2/x} \left (-3 x-3 x^2+x^3\right ) \log (x)+e^{2/x} (-2-2 x) \log (x) \log \left (\frac {5 \log ^3(x)}{x+x^2}\right )}{\left (x^2+x^3\right ) \log (x)} \, dx=x e^{\frac {2}{x}} + e^{\frac {2}{x}} \log \left (\frac {5 \, \log \left (x\right )^{3}}{x^{2} + x}\right ) \] Input:

integrate(((-2-2*x)*exp(2/x)*log(x)*log(5*log(x)^3/(x^2+x))+(x^3-3*x^2-3*x 
)*exp(2/x)*log(x)+(3*x^2+3*x)*exp(2/x))/(x^3+x^2)/log(x),x, algorithm="fri 
cas")
 

Output:

x*e^(2/x) + e^(2/x)*log(5*log(x)^3/(x^2 + x))
 

Sympy [A] (verification not implemented)

Time = 10.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int \frac {e^{2/x} \left (3 x+3 x^2\right )+e^{2/x} \left (-3 x-3 x^2+x^3\right ) \log (x)+e^{2/x} (-2-2 x) \log (x) \log \left (\frac {5 \log ^3(x)}{x+x^2}\right )}{\left (x^2+x^3\right ) \log (x)} \, dx=\left (x + \log {\left (\frac {5 \log {\left (x \right )}^{3}}{x^{2} + x} \right )}\right ) e^{\frac {2}{x}} \] Input:

integrate(((-2-2*x)*exp(2/x)*ln(x)*ln(5*ln(x)**3/(x**2+x))+(x**3-3*x**2-3* 
x)*exp(2/x)*ln(x)+(3*x**2+3*x)*exp(2/x))/(x**3+x**2)/ln(x),x)
 

Output:

(x + log(5*log(x)**3/(x**2 + x)))*exp(2/x)
 

Maxima [F]

\[ \int \frac {e^{2/x} \left (3 x+3 x^2\right )+e^{2/x} \left (-3 x-3 x^2+x^3\right ) \log (x)+e^{2/x} (-2-2 x) \log (x) \log \left (\frac {5 \log ^3(x)}{x+x^2}\right )}{\left (x^2+x^3\right ) \log (x)} \, dx=\int { -\frac {2 \, {\left (x + 1\right )} e^{\frac {2}{x}} \log \left (\frac {5 \, \log \left (x\right )^{3}}{x^{2} + x}\right ) \log \left (x\right ) - {\left (x^{3} - 3 \, x^{2} - 3 \, x\right )} e^{\frac {2}{x}} \log \left (x\right ) - 3 \, {\left (x^{2} + x\right )} e^{\frac {2}{x}}}{{\left (x^{3} + x^{2}\right )} \log \left (x\right )} \,d x } \] Input:

integrate(((-2-2*x)*exp(2/x)*log(x)*log(5*log(x)^3/(x^2+x))+(x^3-3*x^2-3*x 
)*exp(2/x)*log(x)+(3*x^2+3*x)*exp(2/x))/(x^3+x^2)/log(x),x, algorithm="max 
ima")
 

Output:

-e^(2/x)*log(x + 1) - integrate(-((x^2 - 3*x - 2*log(5))*log(x) + 2*log(x) 
^2 - 6*log(x)*log(log(x)) + 3*x)*e^(2/x)/(x^2*log(x)), x)
 

Giac [F]

\[ \int \frac {e^{2/x} \left (3 x+3 x^2\right )+e^{2/x} \left (-3 x-3 x^2+x^3\right ) \log (x)+e^{2/x} (-2-2 x) \log (x) \log \left (\frac {5 \log ^3(x)}{x+x^2}\right )}{\left (x^2+x^3\right ) \log (x)} \, dx=\int { -\frac {2 \, {\left (x + 1\right )} e^{\frac {2}{x}} \log \left (\frac {5 \, \log \left (x\right )^{3}}{x^{2} + x}\right ) \log \left (x\right ) - {\left (x^{3} - 3 \, x^{2} - 3 \, x\right )} e^{\frac {2}{x}} \log \left (x\right ) - 3 \, {\left (x^{2} + x\right )} e^{\frac {2}{x}}}{{\left (x^{3} + x^{2}\right )} \log \left (x\right )} \,d x } \] Input:

integrate(((-2-2*x)*exp(2/x)*log(x)*log(5*log(x)^3/(x^2+x))+(x^3-3*x^2-3*x 
)*exp(2/x)*log(x)+(3*x^2+3*x)*exp(2/x))/(x^3+x^2)/log(x),x, algorithm="gia 
c")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{2/x} \left (3 x+3 x^2\right )+e^{2/x} \left (-3 x-3 x^2+x^3\right ) \log (x)+e^{2/x} (-2-2 x) \log (x) \log \left (\frac {5 \log ^3(x)}{x+x^2}\right )}{\left (x^2+x^3\right ) \log (x)} \, dx=\int -\frac {{\mathrm {e}}^{2/x}\,\ln \left (x\right )\,\left (-x^3+3\,x^2+3\,x\right )-{\mathrm {e}}^{2/x}\,\left (3\,x^2+3\,x\right )+\ln \left (\frac {5\,{\ln \left (x\right )}^3}{x^2+x}\right )\,{\mathrm {e}}^{2/x}\,\ln \left (x\right )\,\left (2\,x+2\right )}{\ln \left (x\right )\,\left (x^3+x^2\right )} \,d x \] Input:

int(-(exp(2/x)*log(x)*(3*x + 3*x^2 - x^3) - exp(2/x)*(3*x + 3*x^2) + log(( 
5*log(x)^3)/(x + x^2))*exp(2/x)*log(x)*(2*x + 2))/(log(x)*(x^2 + x^3)),x)
 

Output:

int(-(exp(2/x)*log(x)*(3*x + 3*x^2 - x^3) - exp(2/x)*(3*x + 3*x^2) + log(( 
5*log(x)^3)/(x + x^2))*exp(2/x)*log(x)*(2*x + 2))/(log(x)*(x^2 + x^3)), x)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {e^{2/x} \left (3 x+3 x^2\right )+e^{2/x} \left (-3 x-3 x^2+x^3\right ) \log (x)+e^{2/x} (-2-2 x) \log (x) \log \left (\frac {5 \log ^3(x)}{x+x^2}\right )}{\left (x^2+x^3\right ) \log (x)} \, dx=e^{\frac {2}{x}} \left (\mathrm {log}\left (\frac {5 \mathrm {log}\left (x \right )^{3}}{x^{2}+x}\right )+x \right ) \] Input:

int(((-2-2*x)*exp(2/x)*log(x)*log(5*log(x)^3/(x^2+x))+(x^3-3*x^2-3*x)*exp( 
2/x)*log(x)+(3*x^2+3*x)*exp(2/x))/(x^3+x^2)/log(x),x)
 

Output:

e**(2/x)*(log((5*log(x)**3)/(x**2 + x)) + x)