3.84.34 \(\int \frac {e^{-x^2} (-22 e^{x^2}+22 e^{x^2} \log (x)+x^x (2+(-2-2 x+4 x^2) \log (x)-2 x \log ^2(x)))}{\log ^2(x)} \, dx\)

Optimal. Leaf size=21 \[ \frac {2 x \left (11-e^{-x^2} x^x\right )}{\log (x)} \]

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Rubi [A]  time = 0.57, antiderivative size = 26, normalized size of antiderivative = 1.24, number of steps used = 9, number of rules used = 6, integrand size = 55, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.109, Rules used = {6688, 6742, 2360, 2297, 2298, 2288} \begin {gather*} \frac {22 x}{\log (x)}-\frac {2 e^{-x^2} x^{x+1}}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-22*E^x^2 + 22*E^x^2*Log[x] + x^x*(2 + (-2 - 2*x + 4*x^2)*Log[x] - 2*x*Log[x]^2))/(E^x^2*Log[x]^2),x]

[Out]

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

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 2297

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[(x*(a + b*Log[c*x^n])^(p + 1))/(b*n*(p + 1))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

Rule 2298

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, x]

Rule 2360

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(Log[(c_.)*(x_)^(n_.)]*(e_.) + (d_))^(q_.), x_Symbol] :> Int[E
xpandIntegrand[(a + b*Log[c*x^n])^p*(d + e*Log[c*x^n])^q, x], x] /; FreeQ[{a, b, c, d, e, n}, x] && IntegerQ[p
] && IntegerQ[q]

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 {-22+22 \log (x)-2 e^{-x^2} x^x \left (-1+\left (1+x-2 x^2\right ) \log (x)+x \log ^2(x)\right )}{\log ^2(x)} \, dx\\ &=\int \left (\frac {22 (-1+\log (x))}{\log ^2(x)}-\frac {2 e^{-x^2} x^x \left (-1+\log (x)+x \log (x)-2 x^2 \log (x)+x \log ^2(x)\right )}{\log ^2(x)}\right ) \, dx\\ &=-\left (2 \int \frac {e^{-x^2} x^x \left (-1+\log (x)+x \log (x)-2 x^2 \log (x)+x \log ^2(x)\right )}{\log ^2(x)} \, dx\right )+22 \int \frac {-1+\log (x)}{\log ^2(x)} \, dx\\ &=-\frac {2 e^{-x^2} x^{1+x}}{\log (x)}+22 \int \left (-\frac {1}{\log ^2(x)}+\frac {1}{\log (x)}\right ) \, dx\\ &=-\frac {2 e^{-x^2} x^{1+x}}{\log (x)}-22 \int \frac {1}{\log ^2(x)} \, dx+22 \int \frac {1}{\log (x)} \, dx\\ &=\frac {22 x}{\log (x)}-\frac {2 e^{-x^2} x^{1+x}}{\log (x)}+22 \text {li}(x)-22 \int \frac {1}{\log (x)} \, dx\\ &=\frac {22 x}{\log (x)}-\frac {2 e^{-x^2} x^{1+x}}{\log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.17, size = 21, normalized size = 1.00 \begin {gather*} \frac {2 x \left (11-e^{-x^2} x^x\right )}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-22*E^x^2 + 22*E^x^2*Log[x] + x^x*(2 + (-2 - 2*x + 4*x^2)*Log[x] - 2*x*Log[x]^2))/(E^x^2*Log[x]^2),
x]

[Out]

(2*x*(11 - x^x/E^x^2))/Log[x]

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fricas [A]  time = 0.87, size = 25, normalized size = 1.19 \begin {gather*} -\frac {2 \, {\left (x x^{x} - 11 \, x e^{\left (x^{2}\right )}\right )} e^{\left (-x^{2}\right )}}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*log(x)^2+(4*x^2-2*x-2)*log(x)+2)*exp(x*log(x))+22*exp(x^2)*log(x)-22*exp(x^2))/exp(x^2)/log(x
)^2,x, algorithm="fricas")

[Out]

-2*(x*x^x - 11*x*e^(x^2))*e^(-x^2)/log(x)

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giac [A]  time = 0.70, size = 26, normalized size = 1.24 \begin {gather*} -\frac {2 \, x e^{\left (-x^{2} + x \log \relax (x)\right )}}{\log \relax (x)} + \frac {22 \, x}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*log(x)^2+(4*x^2-2*x-2)*log(x)+2)*exp(x*log(x))+22*exp(x^2)*log(x)-22*exp(x^2))/exp(x^2)/log(x
)^2,x, algorithm="giac")

[Out]

-2*x*e^(-x^2 + x*log(x))/log(x) + 22*x/log(x)

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maple [A]  time = 0.06, size = 25, normalized size = 1.19

method	result	size
risch	\(\frac {22 x}{\ln \relax (x )}-\frac {2 x \,{\mathrm e}^{-x^{2}} x^{x}}{\ln \relax (x )}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x*ln(x)^2+(4*x^2-2*x-2)*ln(x)+2)*exp(x*ln(x))+22*exp(x^2)*ln(x)-22*exp(x^2))/exp(x^2)/ln(x)^2,x,metho
d=_RETURNVERBOSE)

[Out]

22*x/ln(x)-2*x*exp(-x^2)/ln(x)*x^x

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maxima [A]  time = 0.41, size = 25, normalized size = 1.19 \begin {gather*} -\frac {2 \, {\left (x x^{x} - 11 \, x e^{\left (x^{2}\right )}\right )} e^{\left (-x^{2}\right )}}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*log(x)^2+(4*x^2-2*x-2)*log(x)+2)*exp(x*log(x))+22*exp(x^2)*log(x)-22*exp(x^2))/exp(x^2)/log(x
)^2,x, algorithm="maxima")

[Out]

-2*(x*x^x - 11*x*e^(x^2))*e^(-x^2)/log(x)

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mupad [B]  time = 5.33, size = 25, normalized size = 1.19 \begin {gather*} \frac {2\,x\,{\mathrm {e}}^{-x^2}\,\left (11\,{\mathrm {e}}^{x^2}-x^x\right )}{\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-x^2)*(22*exp(x^2) - 22*exp(x^2)*log(x) + exp(x*log(x))*(2*x*log(x)^2 + log(x)*(2*x - 4*x^2 + 2) - 2
)))/log(x)^2,x)

[Out]

(2*x*exp(-x^2)*(11*exp(x^2) - x^x))/log(x)

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sympy [A]  time = 0.40, size = 24, normalized size = 1.14 \begin {gather*} \frac {22 x}{\log {\relax (x )}} - \frac {2 x e^{- x^{2}} e^{x \log {\relax (x )}}}{\log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*ln(x)**2+(4*x**2-2*x-2)*ln(x)+2)*exp(x*ln(x))+22*exp(x**2)*ln(x)-22*exp(x**2))/exp(x**2)/ln(x
)**2,x)

[Out]

22*x/log(x) - 2*x*exp(-x**2)*exp(x*log(x))/log(x)

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