3.48.2 \(\int (1+e^{e^{3-x+x \log ^2(x)}} (1+e^{3-x+x \log ^2(x)} (-x+2 x \log (x)+x \log ^2(x)))) \, dx\)

Optimal. Leaf size=22 \[ \frac {3}{4}+x+e^{e^{3-x+x \log ^2(x)}} x \]

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Rubi [B]  time = 0.09, antiderivative size = 46, normalized size of antiderivative = 2.09, number of steps used = 2, number of rules used = 1, integrand size = 49, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.020, Rules used = {2288} \begin {gather*} x+\frac {e^{e^{-x+x \log ^2(x)+3}} \left (x-x \log ^2(x)-2 x \log (x)\right )}{-\log ^2(x)-2 \log (x)+1} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 19, normalized size = 0.86 \begin {gather*} x+e^{e^{3-x+x \log ^2(x)}} x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x + E^E^(3 - x + x*Log[x]^2)*x

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fricas [A]  time = 0.49, size = 17, normalized size = 0.77 \begin {gather*} x e^{\left (e^{\left (x \log \relax (x)^{2} - x + 3\right )}\right )} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*log(x)^2+2*x*log(x)-x)*exp(x*log(x)^2+3-x)+1)*exp(exp(x*log(x)^2+3-x))+1,x, algorithm="fricas")

[Out]

x*e^(e^(x*log(x)^2 - x + 3)) + x

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int {\left ({\left (x \log \relax (x)^{2} + 2 \, x \log \relax (x) - x\right )} e^{\left (x \log \relax (x)^{2} - x + 3\right )} + 1\right )} e^{\left (e^{\left (x \log \relax (x)^{2} - x + 3\right )}\right )} + 1\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*log(x)^2+2*x*log(x)-x)*exp(x*log(x)^2+3-x)+1)*exp(exp(x*log(x)^2+3-x))+1,x, algorithm="giac")

[Out]

integrate(((x*log(x)^2 + 2*x*log(x) - x)*e^(x*log(x)^2 - x + 3) + 1)*e^(e^(x*log(x)^2 - x + 3)) + 1, x)

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maple [A]  time = 0.05, size = 18, normalized size = 0.82

method	result	size
risch	\(x \,{\mathrm e}^{{\mathrm e}^{x \ln \relax (x )^{2}+3-x}}+x\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x*ln(x)^2+2*x*ln(x)-x)*exp(x*ln(x)^2+3-x)+1)*exp(exp(x*ln(x)^2+3-x))+1,x,method=_RETURNVERBOSE)

[Out]

x*exp(exp(x*ln(x)^2+3-x))+x

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maxima [A]  time = 0.41, size = 17, normalized size = 0.77 \begin {gather*} x e^{\left (e^{\left (x \log \relax (x)^{2} - x + 3\right )}\right )} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*log(x)^2+2*x*log(x)-x)*exp(x*log(x)^2+3-x)+1)*exp(exp(x*log(x)^2+3-x))+1,x, algorithm="maxima")

[Out]

x*e^(e^(x*log(x)^2 - x + 3)) + x

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mupad [B]  time = 3.20, size = 19, normalized size = 0.86 \begin {gather*} x\,\left ({\mathrm {e}}^{{\mathrm {e}}^{-x}\,{\mathrm {e}}^3\,{\mathrm {e}}^{x\,{\ln \relax (x)}^2}}+1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(exp(x*log(x)^2 - x + 3))*(exp(x*log(x)^2 - x + 3)*(x*log(x)^2 - x + 2*x*log(x)) + 1) + 1,x)

[Out]

x*(exp(exp(-x)*exp(3)*exp(x*log(x)^2)) + 1)

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sympy [A]  time = 5.08, size = 15, normalized size = 0.68 \begin {gather*} x e^{e^{x \log {\relax (x )}^{2} - x + 3}} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*ln(x)**2+2*x*ln(x)-x)*exp(x*ln(x)**2+3-x)+1)*exp(exp(x*ln(x)**2+3-x))+1,x)

[Out]

x*exp(exp(x*log(x)**2 - x + 3)) + x

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