3.27.12 \(\int \frac {e^{\frac {3 x-e^8 x+e^x x+\log (x^2 \log ^2(x))}{x}} (2+(2+e^x x^2) \log (x)-\log (x) \log (x^2 \log ^2(x)))}{x^2 \log (x)} \, dx\)

Optimal. Leaf size=25 \[ e^{3-e^8+e^x+\frac {\log \left (x^2 \log ^2(x)\right )}{x}} \]

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Rubi [A]  time = 0.83, antiderivative size = 34, normalized size of antiderivative = 1.36, number of steps used = 1, number of rules used = 1, integrand size = 65, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.015, Rules used = {6706} \begin {gather*} e^{\frac {e^x x-e^8 x+3 x}{x}} \left (x^2 \log ^2(x)\right )^{\frac {1}{x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((3*x - E^8*x + E^x*x + Log[x^2*Log[x]^2])/x)*(2 + (2 + E^x*x^2)*Log[x] - Log[x]*Log[x^2*Log[x]^2]))/(x
^2*Log[x]),x]

[Out]

E^((3*x - E^8*x + E^x*x)/x)*(x^2*Log[x]^2)^x^(-1)

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=e^{\frac {3 x-e^8 x+e^x x}{x}} \left (x^2 \log ^2(x)\right )^{\frac {1}{x}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.51, size = 25, normalized size = 1.00 \begin {gather*} e^{3-e^8+e^x} \left (x^2 \log ^2(x)\right )^{\frac {1}{x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((3*x - E^8*x + E^x*x + Log[x^2*Log[x]^2])/x)*(2 + (2 + E^x*x^2)*Log[x] - Log[x]*Log[x^2*Log[x]^2
]))/(x^2*Log[x]),x]

[Out]

E^(3 - E^8 + E^x)*(x^2*Log[x]^2)^x^(-1)

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fricas [A]  time = 0.59, size = 30, normalized size = 1.20 \begin {gather*} e^{\left (-\frac {x e^{8} - x e^{x} - 3 \, x - \log \left (x^{2} \log \relax (x)^{2}\right )}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(-(x*e^8 - x*e^x - 3*x - log(x^2*log(x)^2))/x)

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giac [A]  time = 0.51, size = 22, normalized size = 0.88 \begin {gather*} e^{\left (\frac {\log \left (x^{2} \log \relax (x)^{2}\right )}{x} - e^{8} + e^{x} + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(log(x^2*log(x)^2)/x - e^8 + e^x + 3)

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maple [C]  time = 0.40, size = 229, normalized size = 9.16

method	result	size
risch	\({\mathrm e}^{-\frac {i \pi \mathrm {csgn}\left (i \ln \relax (x )^{2}\right )^{3}-2 i \pi \mathrm {csgn}\left (i \ln \relax (x )^{2}\right )^{2} \mathrm {csgn}\left (i \ln \relax (x )\right )+i \pi \,\mathrm {csgn}\left (i \ln \relax (x )^{2}\right ) \mathrm {csgn}\left (i \ln \relax (x )\right )^{2}+i \pi \,\mathrm {csgn}\left (i \ln \relax (x )^{2}\right ) \mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{2} \ln \relax (x )^{2}\right )-i \pi \,\mathrm {csgn}\left (i \ln \relax (x )^{2}\right ) \mathrm {csgn}\left (i x^{2} \ln \relax (x )^{2}\right )^{2}+i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}-2 i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+i \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{2} \ln \relax (x )^{2}\right )^{2}+i \pi \mathrm {csgn}\left (i x^{2} \ln \relax (x )^{2}\right )^{3}-2 \,{\mathrm e}^{x} x +2 x \,{\mathrm e}^{8}-4 \ln \relax (x )-4 \ln \left (\ln \relax (x )\right )-6 x}{2 x}}\)	\(229\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-ln(x)*ln(x^2*ln(x)^2)+(exp(x)*x^2+2)*ln(x)+2)*exp((ln(x^2*ln(x)^2)+exp(x)*x-x*exp(4)^2+3*x)/x)/x^2/ln(x)
,x,method=_RETURNVERBOSE)

[Out]

exp(-1/2*(I*Pi*csgn(I*ln(x)^2)^3-2*I*Pi*csgn(I*ln(x)^2)^2*csgn(I*ln(x))+I*Pi*csgn(I*ln(x)^2)*csgn(I*ln(x))^2+I
*Pi*csgn(I*ln(x)^2)*csgn(I*x^2)*csgn(I*x^2*ln(x)^2)-I*Pi*csgn(I*ln(x)^2)*csgn(I*x^2*ln(x)^2)^2+I*Pi*csgn(I*x^2
)^3-2*I*Pi*csgn(I*x^2)^2*csgn(I*x)+I*Pi*csgn(I*x^2)*csgn(I*x)^2-I*Pi*csgn(I*x^2)*csgn(I*x^2*ln(x)^2)^2+I*Pi*cs
gn(I*x^2*ln(x)^2)^3-2*exp(x)*x+2*x*exp(8)-4*ln(x)-4*ln(ln(x))-6*x)/x)

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maxima [A]  time = 0.72, size = 24, normalized size = 0.96 \begin {gather*} e^{\left (\frac {2 \, \log \relax (x)}{x} + \frac {2 \, \log \left (\log \relax (x)\right )}{x} - e^{8} + e^{x} + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(2*log(x)/x + 2*log(log(x))/x - e^8 + e^x + 3)

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mupad [B]  time = 1.68, size = 26, normalized size = 1.04 \begin {gather*} {\mathrm {e}}^{-{\mathrm {e}}^8}\,{\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^3\,{\left ({\ln \relax (x)}^2\right )}^{1/x}\,{\left (x^2\right )}^{1/x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((3*x + log(x^2*log(x)^2) - x*exp(8) + x*exp(x))/x)*(log(x)*(x^2*exp(x) + 2) - log(x)*log(x^2*log(x)^2
) + 2))/(x^2*log(x)),x)

[Out]

exp(-exp(8))*exp(exp(x))*exp(3)*(log(x)^2)^(1/x)*(x^2)^(1/x)

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sympy [A]  time = 0.81, size = 26, normalized size = 1.04 \begin {gather*} e^{\frac {x e^{x} - x e^{8} + 3 x + \log {\left (x^{2} \log {\relax (x )}^{2} \right )}}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp((x*exp(x) - x*exp(8) + 3*x + log(x**2*log(x)**2))/x)

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