3.66.27 \(\int \frac {e^{-1-6 x-x^2} (-1+(1+x-6 x^2-2 x^3) \log (x)+(-6 x-2 x^2) \log (x) \log (\frac {x}{\log (x)}))}{x \log (x)} \, dx\)

Optimal. Leaf size=23 \[ e^{-4 x-(1+x)^2} \left (x+\log \left (\frac {x}{\log (x)}\right )\right ) \]

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Rubi [A]  time = 1.55, antiderivative size = 35, normalized size of antiderivative = 1.52, number of steps used = 21, number of rules used = 8, integrand size = 57, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {6742, 2234, 2205, 2240, 2241, 2236, 2555, 12} \begin {gather*} e^{-x^2-6 x-1} x+e^{-x^2-6 x-1} \log \left (\frac {x}{\log (x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-1 - 6*x - x^2)*(-1 + (1 + x - 6*x^2 - 2*x^3)*Log[x] + (-6*x - 2*x^2)*Log[x]*Log[x/Log[x]]))/(x*Log[x]
),x]

[Out]

E^(-1 - 6*x - x^2)*x + E^(-1 - 6*x - x^2)*Log[x/Log[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 2205

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erf[(c + d*x)*Rt[-(b*Log[F]),
2]])/(2*d*Rt[-(b*Log[F]), 2]), x] /; FreeQ[{F, a, b, c, d}, x] && NegQ[b]

Rule 2234

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[F^(a - b^2/(4*c)), Int[F^((b + 2*c*x)^2/(4*c))
, x], x] /; FreeQ[{F, a, b, c}, x]

Rule 2236

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

Rule 2240

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

Rule 2241

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_.) + (e_.)*(x_))^(m_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*F^(a + b*x + c*x^2))/(2*c*Log[F]), x] + (-Dist[(b*e - 2*c*d)/(2*c), Int[(d + e*x)^(m - 1)*F^(a + b*x + c*x^2)
, x], x] - Dist[((m - 1)*e^2)/(2*c*Log[F]), Int[(d + e*x)^(m - 2)*F^(a + b*x + c*x^2), x], x]) /; FreeQ[{F, a,
b, c, d, e}, x] && NeQ[b*e - 2*c*d, 0] && GtQ[m, 1]

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 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {e^{-1-6 x-x^2} \left (-1+\log (x)+x \log (x)-6 x^2 \log (x)-2 x^3 \log (x)\right )}{x \log (x)}-2 e^{-1-6 x-x^2} (3+x) \log \left (\frac {x}{\log (x)}\right )\right ) \, dx\\ &=-\left (2 \int e^{-1-6 x-x^2} (3+x) \log \left (\frac {x}{\log (x)}\right ) \, dx\right )+\int \frac {e^{-1-6 x-x^2} \left (-1+\log (x)+x \log (x)-6 x^2 \log (x)-2 x^3 \log (x)\right )}{x \log (x)} \, dx\\ &=e^{-1-6 x-x^2} \log \left (\frac {x}{\log (x)}\right )+2 \int -\frac {e^{-1-6 x-x^2} (-1+\log (x))}{2 x \log (x)} \, dx+\int \left (\frac {e^{-1-6 x-x^2} \left (1+x-6 x^2-2 x^3\right )}{x}-\frac {e^{-1-6 x-x^2}}{x \log (x)}\right ) \, dx\\ &=e^{-1-6 x-x^2} \log \left (\frac {x}{\log (x)}\right )+\int \frac {e^{-1-6 x-x^2} \left (1+x-6 x^2-2 x^3\right )}{x} \, dx-\int \frac {e^{-1-6 x-x^2}}{x \log (x)} \, dx-\int \frac {e^{-1-6 x-x^2} (-1+\log (x))}{x \log (x)} \, dx\\ &=e^{-1-6 x-x^2} \log \left (\frac {x}{\log (x)}\right )+\int \left (e^{-1-6 x-x^2}+\frac {e^{-1-6 x-x^2}}{x}-6 e^{-1-6 x-x^2} x-2 e^{-1-6 x-x^2} x^2\right ) \, dx-\int \left (\frac {e^{-1-6 x-x^2}}{x}-\frac {e^{-1-6 x-x^2}}{x \log (x)}\right ) \, dx-\int \frac {e^{-1-6 x-x^2}}{x \log (x)} \, dx\\ &=e^{-1-6 x-x^2} \log \left (\frac {x}{\log (x)}\right )-2 \int e^{-1-6 x-x^2} x^2 \, dx-6 \int e^{-1-6 x-x^2} x \, dx+\int e^{-1-6 x-x^2} \, dx\\ &=3 e^{-1-6 x-x^2}+e^{-1-6 x-x^2} x+e^{-1-6 x-x^2} \log \left (\frac {x}{\log (x)}\right )+6 \int e^{-1-6 x-x^2} x \, dx+18 \int e^{-1-6 x-x^2} \, dx+e^8 \int e^{-\frac {1}{4} (-6-2 x)^2} \, dx-\int e^{-1-6 x-x^2} \, dx\\ &=e^{-1-6 x-x^2} x+\frac {1}{2} e^8 \sqrt {\pi } \text {erf}(3+x)+e^{-1-6 x-x^2} \log \left (\frac {x}{\log (x)}\right )-18 \int e^{-1-6 x-x^2} \, dx-e^8 \int e^{-\frac {1}{4} (-6-2 x)^2} \, dx+\left (18 e^8\right ) \int e^{-\frac {1}{4} (-6-2 x)^2} \, dx\\ &=e^{-1-6 x-x^2} x+9 e^8 \sqrt {\pi } \text {erf}(3+x)+e^{-1-6 x-x^2} \log \left (\frac {x}{\log (x)}\right )-\left (18 e^8\right ) \int e^{-\frac {1}{4} (-6-2 x)^2} \, dx\\ &=e^{-1-6 x-x^2} x+e^{-1-6 x-x^2} \log \left (\frac {x}{\log (x)}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.24, size = 22, normalized size = 0.96 \begin {gather*} e^{-1-6 x-x^2} \left (x+\log \left (\frac {x}{\log (x)}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-1 - 6*x - x^2)*(-1 + (1 + x - 6*x^2 - 2*x^3)*Log[x] + (-6*x - 2*x^2)*Log[x]*Log[x/Log[x]]))/(x*
Log[x]),x]

[Out]

E^(-1 - 6*x - x^2)*(x + Log[x/Log[x]])

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fricas [A]  time = 0.69, size = 33, normalized size = 1.43 \begin {gather*} x e^{\left (-x^{2} - 6 \, x - 1\right )} + e^{\left (-x^{2} - 6 \, x - 1\right )} \log \left (\frac {x}{\log \relax (x)}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*e^(-x^2 - 6*x - 1) + e^(-x^2 - 6*x - 1)*log(x/log(x))

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giac [B]  time = 0.25, size = 60, normalized size = 2.61 \begin {gather*} -{\left (x e^{\left (-x^{2} - 6 \, x\right )} \log \relax (x) - x e^{\left (-x^{2} - 6 \, x\right )} - e^{\left (-x^{2} - 6 \, x\right )} \log \relax (x) + e^{\left (-x^{2} - 6 \, x\right )} \log \left (\log \relax (x)\right )\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(x*e^(-x^2 - 6*x)*log(x) - x*e^(-x^2 - 6*x) - e^(-x^2 - 6*x)*log(x) + e^(-x^2 - 6*x)*log(log(x)))*e^(-1)

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maple [C]  time = 0.09, size = 123, normalized size = 5.35

method	result	size
risch	\(-{\mathrm e}^{-x^{2}-6 x -1} \ln \left (\ln \relax (x )\right )+\frac {\left (-i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i}{\ln \relax (x )}\right ) \mathrm {csgn}\left (\frac {i x}{\ln \relax (x )}\right )+i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i x}{\ln \relax (x )}\right )^{2}+i \pi \,\mathrm {csgn}\left (\frac {i}{\ln \relax (x )}\right ) \mathrm {csgn}\left (\frac {i x}{\ln \relax (x )}\right )^{2}-i \pi \mathrm {csgn}\left (\frac {i x}{\ln \relax (x )}\right )^{3}+2 x +2 \ln \relax (x )\right ) {\mathrm e}^{-x^{2}-6 x -1}}{2}\)	\(123\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x^2-6*x)*ln(x)*ln(x/ln(x))+(-2*x^3-6*x^2+x+1)*ln(x)-1)/x/exp(1/4*x^2+3/2*x+1/4)^4/ln(x),x,method=_RET
URNVERBOSE)

[Out]

-exp(-x^2-6*x-1)*ln(ln(x))+1/2*(-I*Pi*csgn(I*x)*csgn(I/ln(x))*csgn(I*x/ln(x))+I*Pi*csgn(I*x)*csgn(I*x/ln(x))^2
+I*Pi*csgn(I/ln(x))*csgn(I*x/ln(x))^2-I*Pi*csgn(I*x/ln(x))^3+2*x+2*ln(x))*exp(-x^2-6*x-1)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {1}{2} \, \sqrt {\pi } \operatorname {erf}\left (x + 3\right ) e^{8} - e^{\left (-x^{2} - 6 \, x - 1\right )} \log \left (\log \relax (x)\right ) - \int \frac {{\left (2 \, x^{3} + 6 \, x^{2} + 2 \, {\left (x^{2} + 3 \, x\right )} \log \relax (x) - 1\right )} e^{\left (-x^{2} - 6 \, x - 1\right )}}{x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(pi)*erf(x + 3)*e^8 - e^(-x^2 - 6*x - 1)*log(log(x)) - integrate((2*x^3 + 6*x^2 + 2*(x^2 + 3*x)*log(x)
- 1)*e^(-x^2 - 6*x - 1)/x, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {{\mathrm {e}}^{-x^2-6\,x-1}\,\left (\ln \left (\frac {x}{\ln \relax (x)}\right )\,\ln \relax (x)\,\left (2\,x^2+6\,x\right )-\ln \relax (x)\,\left (-2\,x^3-6\,x^2+x+1\right )+1\right )}{x\,\ln \relax (x)} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(- 6*x - x^2 - 1)*(log(x/log(x))*log(x)*(6*x + 2*x^2) - log(x)*(x - 6*x^2 - 2*x^3 + 1) + 1))/(x*log(x
)),x)

[Out]

int(-(exp(- 6*x - x^2 - 1)*(log(x/log(x))*log(x)*(6*x + 2*x^2) - log(x)*(x - 6*x^2 - 2*x^3 + 1) + 1))/(x*log(x
)), x)

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sympy [A]  time = 10.79, size = 19, normalized size = 0.83 \begin {gather*} \left (x + \log {\left (\frac {x}{\log {\relax (x )}} \right )}\right ) e^{- x^{2} - 6 x - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x + log(x/log(x)))*exp(-x**2 - 6*x - 1)

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