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

Optimal. Leaf size=20 \[ e^{e^3}+\frac {e^x}{\log ^2(x)}-\log ^2(x) \]

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Rubi [A]  time = 0.28, antiderivative size = 15, normalized size of antiderivative = 0.75, number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {6742, 2301, 2202} \begin {gather*} \frac {e^x}{\log ^2(x)}-\log ^2(x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^x/Log[x]^2 - Log[x]^2

Rule 2202

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

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, 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 \left (-\frac {2 \log (x)}{x}+\frac {e^x (-2+x \log (x))}{x \log ^3(x)}\right ) \, dx\\ &=-\left (2 \int \frac {\log (x)}{x} \, dx\right )+\int \frac {e^x (-2+x \log (x))}{x \log ^3(x)} \, dx\\ &=\frac {e^x}{\log ^2(x)}-\log ^2(x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 15, normalized size = 0.75 \begin {gather*} \frac {e^x}{\log ^2(x)}-\log ^2(x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^x/Log[x]^2 - Log[x]^2

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fricas [A]  time = 0.70, size = 17, normalized size = 0.85 \begin {gather*} -\log \relax (x)^{2} + e^{\left (x - \log \left (\log \relax (x)^{2}\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(x)^2 + e^(x - log(log(x)^2))

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giac [A]  time = 0.13, size = 15, normalized size = 0.75 \begin {gather*} -\frac {\log \relax (x)^{4} - e^{x}}{\log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(log(x)^4 - e^x)/log(x)^2

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maple [A]  time = 0.13, size = 18, normalized size = 0.90

method	result	size
default	\({\mathrm e}^{-\ln \left (\ln \relax (x )^{2}\right )+x}-\ln \relax (x )^{2}\)	\(18\)
risch	\(-\ln \relax (x )^{2}-\frac {{\mathrm e}^{x} {\mathrm e}^{\frac {i \pi \mathrm {csgn}\left (i \ln \relax (x )^{2}\right )^{3}}{2}} {\mathrm e}^{\frac {i \pi \,\mathrm {csgn}\left (i \ln \relax (x )^{2}\right ) \mathrm {csgn}\left (i \ln \relax (x )\right )^{2}}{2}}}{\ln \relax (x )^{2}}\)	\(52\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(-ln(ln(x)^2)+x)-ln(x)^2

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maxima [A]  time = 0.43, size = 15, normalized size = 0.75 \begin {gather*} -\frac {\log \relax (x)^{4} - e^{x}}{\log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(log(x)^4 - e^x)/log(x)^2

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mupad [B]  time = 4.23, size = 14, normalized size = 0.70 \begin {gather*} \frac {{\mathrm {e}}^x-{\ln \relax (x)}^4}{{\ln \relax (x)}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(exp(x) - log(x)^4)/log(x)^2

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sympy [A]  time = 0.26, size = 12, normalized size = 0.60 \begin {gather*} \frac {e^{x}}{\log {\relax (x )}^{2}} - \log {\relax (x )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(x)/log(x)**2 - log(x)**2

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