3.23.83 \(\int \frac {e^{-1/x} (18 e^x x+18 x^2+216 e^{\frac {1}{x}} x^2+(9 x+9 x^2+108 e^{\frac {1}{x}} x^2+e^x (9+9 x^2)) \log (x^2))}{x^2} \, dx\)

Optimal. Leaf size=26 \[ 36 \left (3 x+\frac {1}{4} e^{-1/x} \left (e^x+x\right )\right ) \log \left (x^2\right ) \]

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Rubi [B]  time = 1.02, antiderivative size = 56, normalized size of antiderivative = 2.15, number of steps used = 8, number of rules used = 3, integrand size = 68, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.044, Rules used = {6742, 2295, 2288} \begin {gather*} 9 e^{-1/x} x \log \left (x^2\right )+108 x \log \left (x^2\right )+\frac {9 e^{x-\frac {1}{x}} \left (x^2 \log \left (x^2\right )+\log \left (x^2\right )\right )}{\left (\frac {1}{x^2}+1\right ) x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(18*E^x*x + 18*x^2 + 216*E^x^(-1)*x^2 + (9*x + 9*x^2 + 108*E^x^(-1)*x^2 + E^x*(9 + 9*x^2))*Log[x^2])/(E^x^
(-1)*x^2),x]

[Out]

108*x*Log[x^2] + (9*x*Log[x^2])/E^x^(-1) + (9*E^(-x^(-1) + x)*(Log[x^2] + x^2*Log[x^2]))/((1 + x^(-2))*x^2)

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 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{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 {9 e^{-1/x} \left (2 x+24 e^{\frac {1}{x}} x+\log \left (x^2\right )+x \log \left (x^2\right )+12 e^{\frac {1}{x}} x \log \left (x^2\right )\right )}{x}+\frac {9 e^{-\frac {1}{x}+x} \left (2 x+\log \left (x^2\right )+x^2 \log \left (x^2\right )\right )}{x^2}\right ) \, dx\\ &=9 \int \frac {e^{-1/x} \left (2 x+24 e^{\frac {1}{x}} x+\log \left (x^2\right )+x \log \left (x^2\right )+12 e^{\frac {1}{x}} x \log \left (x^2\right )\right )}{x} \, dx+9 \int \frac {e^{-\frac {1}{x}+x} \left (2 x+\log \left (x^2\right )+x^2 \log \left (x^2\right )\right )}{x^2} \, dx\\ &=\frac {9 e^{-\frac {1}{x}+x} \left (\log \left (x^2\right )+x^2 \log \left (x^2\right )\right )}{\left (1+\frac {1}{x^2}\right ) x^2}+9 \int \left (12 \left (2+\log \left (x^2\right )\right )+\frac {e^{-1/x} \left (2 x+\log \left (x^2\right )+x \log \left (x^2\right )\right )}{x}\right ) \, dx\\ &=\frac {9 e^{-\frac {1}{x}+x} \left (\log \left (x^2\right )+x^2 \log \left (x^2\right )\right )}{\left (1+\frac {1}{x^2}\right ) x^2}+9 \int \frac {e^{-1/x} \left (2 x+\log \left (x^2\right )+x \log \left (x^2\right )\right )}{x} \, dx+108 \int \left (2+\log \left (x^2\right )\right ) \, dx\\ &=216 x+9 e^{-1/x} x \log \left (x^2\right )+\frac {9 e^{-\frac {1}{x}+x} \left (\log \left (x^2\right )+x^2 \log \left (x^2\right )\right )}{\left (1+\frac {1}{x^2}\right ) x^2}+108 \int \log \left (x^2\right ) \, dx\\ &=108 x \log \left (x^2\right )+9 e^{-1/x} x \log \left (x^2\right )+\frac {9 e^{-\frac {1}{x}+x} \left (\log \left (x^2\right )+x^2 \log \left (x^2\right )\right )}{\left (1+\frac {1}{x^2}\right ) x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.34, size = 26, normalized size = 1.00 \begin {gather*} 9 e^{-1/x} \left (e^x+x+12 e^{\frac {1}{x}} x\right ) \log \left (x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(18*E^x*x + 18*x^2 + 216*E^x^(-1)*x^2 + (9*x + 9*x^2 + 108*E^x^(-1)*x^2 + E^x*(9 + 9*x^2))*Log[x^2])
/(E^x^(-1)*x^2),x]

[Out]

(9*(E^x + x + 12*E^x^(-1)*x)*Log[x^2])/E^x^(-1)

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fricas [A]  time = 0.66, size = 23, normalized size = 0.88 \begin {gather*} 9 \, {\left (12 \, x e^{\frac {1}{x}} + x + e^{x}\right )} e^{\left (-\frac {1}{x}\right )} \log \left (x^{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((9*x^2+9)*exp(x)+108*x^2*exp(1/x)+9*x^2+9*x)*log(x^2)+18*exp(x)*x+216*x^2*exp(1/x)+18*x^2)/x^2/exp
(1/x),x, algorithm="fricas")

[Out]

9*(12*x*e^(1/x) + x + e^x)*e^(-1/x)*log(x^2)

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giac [A]  time = 0.34, size = 37, normalized size = 1.42 \begin {gather*} 9 \, x e^{\left (-\frac {1}{x}\right )} \log \left (x^{2}\right ) + 108 \, x \log \left (x^{2}\right ) + 9 \, e^{\left (\frac {x^{2} - 1}{x}\right )} \log \left (x^{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((9*x^2+9)*exp(x)+108*x^2*exp(1/x)+9*x^2+9*x)*log(x^2)+18*exp(x)*x+216*x^2*exp(1/x)+18*x^2)/x^2/exp
(1/x),x, algorithm="giac")

[Out]

9*x*e^(-1/x)*log(x^2) + 108*x*log(x^2) + 9*e^((x^2 - 1)/x)*log(x^2)

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maple [B]  time = 0.13, size = 84, normalized size = 3.23 \[-9 \left (-\frac {\ln \left (x^{2}\right )+2 \ln \left (\frac {1}{x}\right )}{x}+\frac {2 \ln \left (\frac {1}{x}\right )}{x}\right ) {\mathrm e}^{-\frac {1}{x}} x^{2}+\frac {9 \left (\ln \left (x^{2}\right )-2 \ln \relax (x )\right ) x \,{\mathrm e}^{x -\frac {1}{x}}+18 x \,{\mathrm e}^{x -\frac {1}{x}} \ln \relax (x )}{x}+108 x \ln \left (x^{2}\right )\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((9*x^2+9)*exp(x)+108*x^2*exp(1/x)+9*x^2+9*x)*ln(x^2)+18*exp(x)*x+216*x^2*exp(1/x)+18*x^2)/x^2/exp(1/x),x
)

[Out]

-9*(-1/x*(ln(x^2)+2*ln(1/x))+2*ln(1/x)/x)/exp(1/x)*x^2+9*((ln(x^2)-2*ln(x))*x*exp(x-1/x)+2*x*exp(x-1/x)*ln(x))
/x+108*x*ln(x^2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 216 \, x \log \relax (x) + 18 \, e^{\left (x - \frac {1}{x}\right )} \log \relax (x) + 18 \, \Gamma \left (-1, \frac {1}{x}\right ) + 9 \, \int \frac {2 \, {\left (x + 1\right )} e^{\left (-\frac {1}{x}\right )} \log \relax (x)}{x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((9*x^2+9)*exp(x)+108*x^2*exp(1/x)+9*x^2+9*x)*log(x^2)+18*exp(x)*x+216*x^2*exp(1/x)+18*x^2)/x^2/exp
(1/x),x, algorithm="maxima")

[Out]

216*x*log(x) + 18*e^(x - 1/x)*log(x) + 18*gamma(-1, 1/x) + 9*integrate(2*(x + 1)*e^(-1/x)*log(x)/x, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {{\mathrm {e}}^{-\frac {1}{x}}\,\left (\ln \left (x^2\right )\,\left (9\,x+108\,x^2\,{\mathrm {e}}^{1/x}+{\mathrm {e}}^x\,\left (9\,x^2+9\right )+9\,x^2\right )+216\,x^2\,{\mathrm {e}}^{1/x}+18\,x\,{\mathrm {e}}^x+18\,x^2\right )}{x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-1/x)*(log(x^2)*(9*x + 108*x^2*exp(1/x) + exp(x)*(9*x^2 + 9) + 9*x^2) + 216*x^2*exp(1/x) + 18*x*exp(x
) + 18*x^2))/x^2,x)

[Out]

int((exp(-1/x)*(log(x^2)*(9*x + 108*x^2*exp(1/x) + exp(x)*(9*x^2 + 9) + 9*x^2) + 216*x^2*exp(1/x) + 18*x*exp(x
) + 18*x^2))/x^2, x)

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sympy [A]  time = 0.45, size = 36, normalized size = 1.38 \begin {gather*} 108 x \log {\left (x^{2} \right )} + 9 x e^{- \frac {1}{x}} \log {\left (x^{2} \right )} + 9 e^{- \frac {1}{x}} e^{x} \log {\left (x^{2} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((9*x**2+9)*exp(x)+108*x**2*exp(1/x)+9*x**2+9*x)*ln(x**2)+18*exp(x)*x+216*x**2*exp(1/x)+18*x**2)/x*
*2/exp(1/x),x)

[Out]

108*x*log(x**2) + 9*x*exp(-1/x)*log(x**2) + 9*exp(-1/x)*exp(x)*log(x**2)

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