3.103.26 \(\int \frac {1+(-1-162 x^2+216 e^e x^2-108 e^{2 e} x^2+24 e^{3 e} x^2-2 e^{4 e} x^2) \log (\frac {1}{x})}{x \log (\frac {1}{x})} \, dx\)

Optimal. Leaf size=25 \[ -4-\left (3-e^e\right )^4 x^2-\log \left (x \log \left (\frac {1}{x}\right )\right ) \]

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Rubi [A]  time = 0.32, antiderivative size = 26, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 4, integrand size = 62, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {6688, 14, 2302, 29} \begin {gather*} -\left (3-e^e\right )^4 x^2-\log (x)-\log \left (\log \left (\frac {1}{x}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + (-1 - 162*x^2 + 216*E^E*x^2 - 108*E^(2*E)*x^2 + 24*E^(3*E)*x^2 - 2*E^(4*E)*x^2)*Log[x^(-1)])/(x*Log[x
^(-1)]),x]

[Out]

-((3 - E^E)^4*x^2) - Log[x] - Log[Log[x^(-1)]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-1-2 \left (-3+e^e\right )^4 x^2+\frac {1}{\log \left (\frac {1}{x}\right )}}{x} \, dx\\ &=\int \left (\frac {-1-2 \left (3-e^e\right )^4 x^2}{x}+\frac {1}{x \log \left (\frac {1}{x}\right )}\right ) \, dx\\ &=\int \frac {-1-2 \left (3-e^e\right )^4 x^2}{x} \, dx+\int \frac {1}{x \log \left (\frac {1}{x}\right )} \, dx\\ &=\int \left (-\frac {1}{x}-2 \left (3-e^e\right )^4 x\right ) \, dx-\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log \left (\frac {1}{x}\right )\right )\\ &=-\left (3-e^e\right )^4 x^2-\log (x)-\log \left (\log \left (\frac {1}{x}\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 24, normalized size = 0.96 \begin {gather*} -\left (-3+e^e\right )^4 x^2-\log (x)-\log \left (\log \left (\frac {1}{x}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + (-1 - 162*x^2 + 216*E^E*x^2 - 108*E^(2*E)*x^2 + 24*E^(3*E)*x^2 - 2*E^(4*E)*x^2)*Log[x^(-1)])/(x
*Log[x^(-1)]),x]

[Out]

-((-3 + E^E)^4*x^2) - Log[x] - Log[Log[x^(-1)]]

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fricas [B]  time = 0.70, size = 55, normalized size = 2.20 \begin {gather*} -x^{2} e^{\left (4 \, e\right )} + 12 \, x^{2} e^{\left (3 \, e\right )} - 54 \, x^{2} e^{\left (2 \, e\right )} + 108 \, x^{2} e^{e} - 81 \, x^{2} + \log \left (\frac {1}{x}\right ) - \log \left (\log \left (\frac {1}{x}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2*exp(exp(1))^4+24*x^2*exp(exp(1))^3-108*x^2*exp(exp(1))^2+216*x^2*exp(exp(1))-162*x^2-1)*log
(1/x)+1)/x/log(1/x),x, algorithm="fricas")

[Out]

-x^2*e^(4*e) + 12*x^2*e^(3*e) - 54*x^2*e^(2*e) + 108*x^2*e^e - 81*x^2 + log(1/x) - log(log(1/x))

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giac [B]  time = 0.15, size = 53, normalized size = 2.12 \begin {gather*} -x^{2} e^{\left (4 \, e\right )} + 12 \, x^{2} e^{\left (3 \, e\right )} - 54 \, x^{2} e^{\left (2 \, e\right )} + 108 \, x^{2} e^{e} - 81 \, x^{2} - \log \relax (x) - \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2*exp(exp(1))^4+24*x^2*exp(exp(1))^3-108*x^2*exp(exp(1))^2+216*x^2*exp(exp(1))-162*x^2-1)*log
(1/x)+1)/x/log(1/x),x, algorithm="giac")

[Out]

-x^2*e^(4*e) + 12*x^2*e^(3*e) - 54*x^2*e^(2*e) + 108*x^2*e^e - 81*x^2 - log(x) - log(log(x))

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maple [A]  time = 0.05, size = 45, normalized size = 1.80

method	result	size
norman	\(\ln \left (\frac {1}{x}\right )+\left (-{\mathrm e}^{4 \,{\mathrm e}}+12 \,{\mathrm e}^{3 \,{\mathrm e}}-54 \,{\mathrm e}^{2 \,{\mathrm e}}+108 \,{\mathrm e}^{{\mathrm e}}-81\right ) x^{2}-\ln \left (\ln \left (\frac {1}{x}\right )\right )\)	\(45\)
derivativedivides	\(-x^{2} {\mathrm e}^{4 \,{\mathrm e}}+12 x^{2} {\mathrm e}^{3 \,{\mathrm e}}-54 x^{2} {\mathrm e}^{2 \,{\mathrm e}}+\ln \left (\frac {1}{x}\right )+108 x^{2} {\mathrm e}^{{\mathrm e}}-\ln \left (\ln \left (\frac {1}{x}\right )\right )-81 x^{2}\)	\(56\)
default	\(-x^{2} {\mathrm e}^{4 \,{\mathrm e}}+12 x^{2} {\mathrm e}^{3 \,{\mathrm e}}-54 x^{2} {\mathrm e}^{2 \,{\mathrm e}}+\ln \left (\frac {1}{x}\right )+108 x^{2} {\mathrm e}^{{\mathrm e}}-\ln \left (\ln \left (\frac {1}{x}\right )\right )-81 x^{2}\)	\(56\)
risch	\(-x^{2} {\mathrm e}^{4 \,{\mathrm e}}+12 x^{2} {\mathrm e}^{3 \,{\mathrm e}}-54 x^{2} {\mathrm e}^{2 \,{\mathrm e}}+108 x^{2} {\mathrm e}^{{\mathrm e}}-81 x^{2}-\ln \relax (x )-\ln \left (\ln \left (\frac {1}{x}\right )\right )\)	\(56\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x^2*exp(exp(1))^4+24*x^2*exp(exp(1))^3-108*x^2*exp(exp(1))^2+216*x^2*exp(exp(1))-162*x^2-1)*ln(1/x)+1
)/x/ln(1/x),x,method=_RETURNVERBOSE)

[Out]

ln(1/x)+(-exp(exp(1))^4+12*exp(exp(1))^3-54*exp(exp(1))^2+108*exp(exp(1))-81)*x^2-ln(ln(1/x))

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maxima [B]  time = 0.39, size = 53, normalized size = 2.12 \begin {gather*} -x^{2} e^{\left (4 \, e\right )} + 12 \, x^{2} e^{\left (3 \, e\right )} - 54 \, x^{2} e^{\left (2 \, e\right )} + 108 \, x^{2} e^{e} - 81 \, x^{2} - \log \relax (x) - \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2*exp(exp(1))^4+24*x^2*exp(exp(1))^3-108*x^2*exp(exp(1))^2+216*x^2*exp(exp(1))-162*x^2-1)*log
(1/x)+1)/x/log(1/x),x, algorithm="maxima")

[Out]

-x^2*e^(4*e) + 12*x^2*e^(3*e) - 54*x^2*e^(2*e) + 108*x^2*e^e - 81*x^2 - log(x) - log(log(x))

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mupad [B]  time = 7.09, size = 33, normalized size = 1.32 \begin {gather*} \frac {x^2\,\ln \left (\frac {1}{x}\right )-x^4\,{\left ({\mathrm {e}}^{\mathrm {e}}-3\right )}^4}{x^2}-\ln \left (\ln \left (\frac {1}{x}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(1/x)*(108*x^2*exp(2*exp(1)) - 216*x^2*exp(exp(1)) - 24*x^2*exp(3*exp(1)) + 2*x^2*exp(4*exp(1)) + 162
*x^2 + 1) - 1)/(x*log(1/x)),x)

[Out]

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

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sympy [B]  time = 0.19, size = 46, normalized size = 1.84 \begin {gather*} - x^{2} \left (- 12 e^{3 e} - 108 e^{e} + 81 + 54 e^{2 e} + e^{4 e}\right ) - \log {\relax (x )} - \log {\left (\log {\left (\frac {1}{x} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x**2*exp(exp(1))**4+24*x**2*exp(exp(1))**3-108*x**2*exp(exp(1))**2+216*x**2*exp(exp(1))-162*x**
2-1)*ln(1/x)+1)/x/ln(1/x),x)

[Out]

-x**2*(-12*exp(3*E) - 108*exp(E) + 81 + 54*exp(2*E) + exp(4*E)) - log(x) - log(log(1/x))

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