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

Optimal. Leaf size=25 \[ e^{2/x} \left (x+\log \left (\frac {5 \log ^3(x)}{x (1+x)}\right )\right ) \]

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Rubi [A]  time = 3.28, antiderivative size = 33, normalized size of antiderivative = 1.32, number of steps used = 26, number of rules used = 11, integrand size = 83, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {1593, 6688, 6742, 2206, 2210, 2222, 2228, 2178, 2209, 2555, 12} \begin {gather*} e^{2/x} x+e^{2/x} \log \left (\frac {5 \log ^3(x)}{x (x+1)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2206

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_)), x_Symbol] :> Simp[((c + d*x)*F^(a + b*(c + d*x)^n))/d, x]
- Dist[b*n*Log[F], Int[(c + d*x)^n*F^(a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[2/n]
&& ILtQ[n, 0]

Rule 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rule 2210

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> Simp[(F^a*ExpIntegralEi[
b*(c + d*x)^n*Log[F]])/(f*n), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[d*e - c*f, 0]

Rule 2222

Int[(F_)^((a_.) + (b_.)/((c_.) + (d_.)*(x_)))/((e_.) + (f_.)*(x_)), x_Symbol] :> Dist[d/f, Int[F^(a + b/(c + d
*x))/(c + d*x), x], x] - Dist[(d*e - c*f)/f, Int[F^(a + b/(c + d*x))/((c + d*x)*(e + f*x)), x], x] /; FreeQ[{F
, a, b, c, d, e, f}, x] && NeQ[d*e - c*f, 0]

Rule 2228

Int[(F_)^((a_.) + (b_.)/((c_.) + (d_.)*(x_)))/(((e_.) + (f_.)*(x_))*((g_.) + (h_.)*(x_))), x_Symbol] :> -Dist[
d/(f*(d*g - c*h)), Subst[Int[F^(a - (b*h)/(d*g - c*h) + (d*b*x)/(d*g - c*h))/x, x], x, (g + h*x)/(c + d*x)], x
] /; FreeQ[{F, a, b, c, d, e, f}, x] && EqQ[d*e - c*f, 0]

Rule 2555

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[w*Simplify
[D[u, x]/u], x], x] /; InverseFunctionFreeQ[w, x]] /; ProductQ[u]

Rule 6688

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\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 )}{x^2 (1+x) \log (x)} \, dx\\ &=\int \frac {e^{2/x} \left (3 x (1+x)+\log (x) \left (x \left (-3-3 x+x^2\right )-2 (1+x) \log \left (\frac {5 \log ^3(x)}{x+x^2}\right )\right )\right )}{x^2 (1+x) \log (x)} \, dx\\ &=\int \left (\frac {e^{2/x} \left (3+3 x-3 \log (x)-3 x \log (x)+x^2 \log (x)\right )}{x (1+x) \log (x)}-\frac {2 e^{2/x} \log \left (\frac {5 \log ^3(x)}{x (1+x)}\right )}{x^2}\right ) \, dx\\ &=-\left (2 \int \frac {e^{2/x} \log \left (\frac {5 \log ^3(x)}{x (1+x)}\right )}{x^2} \, dx\right )+\int \frac {e^{2/x} \left (3+3 x-3 \log (x)-3 x \log (x)+x^2 \log (x)\right )}{x (1+x) \log (x)} \, dx\\ &=e^{2/x} \log \left (\frac {5 \log ^3(x)}{x (1+x)}\right )+2 \int \frac {e^{2/x} (-3-3 x+\log (x)+2 x \log (x))}{2 x (1+x) \log (x)} \, dx+\int \left (\frac {e^{2/x} \left (-3-3 x+x^2\right )}{x (1+x)}+\frac {3 e^{2/x}}{x \log (x)}\right ) \, dx\\ &=e^{2/x} \log \left (\frac {5 \log ^3(x)}{x (1+x)}\right )+3 \int \frac {e^{2/x}}{x \log (x)} \, dx+\int \frac {e^{2/x} \left (-3-3 x+x^2\right )}{x (1+x)} \, dx+\int \frac {e^{2/x} (-3-3 x+\log (x)+2 x \log (x))}{x (1+x) \log (x)} \, dx\\ &=e^{2/x} \log \left (\frac {5 \log ^3(x)}{x (1+x)}\right )+3 \int \frac {e^{2/x}}{x \log (x)} \, dx+\int \left (e^{2/x}+\frac {e^{2/x}}{-1-x}-\frac {3 e^{2/x}}{x}\right ) \, dx+\int \left (\frac {e^{2/x} (1+2 x)}{x (1+x)}-\frac {3 e^{2/x}}{x \log (x)}\right ) \, dx\\ &=e^{2/x} \log \left (\frac {5 \log ^3(x)}{x (1+x)}\right )-3 \int \frac {e^{2/x}}{x} \, dx+\int e^{2/x} \, dx+\int \frac {e^{2/x}}{-1-x} \, dx+\int \frac {e^{2/x} (1+2 x)}{x (1+x)} \, dx\\ &=e^{2/x} x+3 \text {Ei}\left (\frac {2}{x}\right )+e^{2/x} \log \left (\frac {5 \log ^3(x)}{x (1+x)}\right )+2 \int \frac {e^{2/x}}{x} \, dx-\int \frac {e^{2/x}}{x} \, dx-\int \frac {e^{2/x}}{(-1-x) x} \, dx+\int \left (\frac {e^{2/x}}{x}+\frac {e^{2/x}}{1+x}\right ) \, dx\\ &=e^{2/x} x+2 \text {Ei}\left (\frac {2}{x}\right )+e^{2/x} \log \left (\frac {5 \log ^3(x)}{x (1+x)}\right )+\int \frac {e^{2/x}}{x} \, dx+\int \frac {e^{2/x}}{1+x} \, dx-\operatorname {Subst}\left (\int \frac {e^{-2-2 x}}{x} \, dx,x,\frac {-1-x}{x}\right )\\ &=e^{2/x} x-\frac {\text {Ei}\left (2+\frac {2}{x}\right )}{e^2}+\text {Ei}\left (\frac {2}{x}\right )+e^{2/x} \log \left (\frac {5 \log ^3(x)}{x (1+x)}\right )+\int \frac {e^{2/x}}{x} \, dx-\int \frac {e^{2/x}}{x (1+x)} \, dx\\ &=e^{2/x} x-\frac {\text {Ei}\left (2+\frac {2}{x}\right )}{e^2}+e^{2/x} \log \left (\frac {5 \log ^3(x)}{x (1+x)}\right )+\operatorname {Subst}\left (\int \frac {e^{-2+2 x}}{x} \, dx,x,\frac {1+x}{x}\right )\\ &=e^{2/x} x+e^{2/x} \log \left (\frac {5 \log ^3(x)}{x (1+x)}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.22, size = 24, normalized size = 0.96 \begin {gather*} e^{2/x} \left (x+\log \left (\frac {5 \log ^3(x)}{x+x^2}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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]

[Out]

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

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fricas [A]  time = 0.67, size = 30, normalized size = 1.20 \begin {gather*} x e^{\frac {2}{x}} + e^{\frac {2}{x}} \log \left (\frac {5 \, \log \relax (x)^{3}}{x^{2} + x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x-2)*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="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {2 \, {\left (x + 1\right )} e^{\frac {2}{x}} \log \left (\frac {5 \, \log \relax (x)^{3}}{x^{2} + x}\right ) \log \relax (x) - {\left (x^{3} - 3 \, x^{2} - 3 \, x\right )} e^{\frac {2}{x}} \log \relax (x) - 3 \, {\left (x^{2} + x\right )} e^{\frac {2}{x}}}{{\left (x^{3} + x^{2}\right )} \log \relax (x)}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x-2)*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="giac")

[Out]

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)

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maple [C]  time = 0.33, size = 512, normalized size = 20.48

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x-2)*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)

[Out]

-exp(2/x)*ln(x+1)+3*exp(2/x)*ln(ln(x))-exp(2/x)*ln(x)+I*Pi*csgn(I*ln(x))*csgn(I*ln(x)^2)^2*exp(2/x)-1/2*I*Pi*c
sgn(I*ln(x))*csgn(I*ln(x)^2)*csgn(I*ln(x)^3)*exp(2/x)+1/2*I*Pi*csgn(I*ln(x)^3)*csgn(I*ln(x)^3/(x+1))^2*exp(2/x
)-1/2*I*Pi*csgn(I/x/(x+1)*ln(x)^3)^3*exp(2/x)+1/2*I*Pi*csgn(I/(x+1))*csgn(I*ln(x)^3/(x+1))^2*exp(2/x)+1/2*I*Pi
*csgn(I*ln(x)^2)*csgn(I*ln(x)^3)^2*exp(2/x)+1/2*I*Pi*csgn(I*ln(x)^3/(x+1))*csgn(I/x/(x+1)*ln(x)^3)^2*exp(2/x)-
1/2*I*Pi*csgn(I/(x+1))*csgn(I*ln(x)^3)*csgn(I*ln(x)^3/(x+1))*exp(2/x)+1/2*I*Pi*csgn(I*ln(x))*csgn(I*ln(x)^3)^2
*exp(2/x)-1/2*I*Pi*csgn(I*ln(x)^3)^3*exp(2/x)+1/2*I*Pi*csgn(I/x)*csgn(I/x/(x+1)*ln(x)^3)^2*exp(2/x)-1/2*I*Pi*c
sgn(I*ln(x)^3/(x+1))^3*exp(2/x)-1/2*I*Pi*csgn(I/x)*csgn(I*ln(x)^3/(x+1))*csgn(I/x/(x+1)*ln(x)^3)*exp(2/x)-1/2*
I*Pi*csgn(I*ln(x)^2)^3*exp(2/x)-1/2*I*Pi*csgn(I*ln(x))^2*csgn(I*ln(x)^2)*exp(2/x)+ln(5)*exp(2/x)+x*exp(2/x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -e^{\frac {2}{x}} \log \left (x + 1\right ) - \int -\frac {{\left ({\left (x^{2} - 3 \, x - 2 \, \log \relax (5)\right )} \log \relax (x) + 2 \, \log \relax (x)^{2} - 6 \, \log \relax (x) \log \left (\log \relax (x)\right ) + 3 \, x\right )} e^{\frac {2}{x}}}{x^{2} \log \relax (x)}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x-2)*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="maxima")

[Out]

-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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {{\mathrm {e}}^{2/x}\,\ln \relax (x)\,\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 \relax (x)}^3}{x^2+x}\right )\,{\mathrm {e}}^{2/x}\,\ln \relax (x)\,\left (2\,x+2\right )}{\ln \relax (x)\,\left (x^3+x^2\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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)

[Out]

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)

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sympy [A]  time = 23.69, size = 19, normalized size = 0.76 \begin {gather*} \left (x + \log {\left (\frac {5 \log {\relax (x )}^{3}}{x^{2} + x} \right )}\right ) e^{\frac {2}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x-2)*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)

[Out]

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

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