3.45.12 \(\int \frac {e^{-e^2 x+e^2 \log ^2(x)} (-3 e-3 e^3 x+6 e^3 \log (x))}{4 x^2} \, dx\)

Optimal. Leaf size=23 \[ \frac {3 e^{1+e^2 \left (-x+\log ^2(x)\right )}}{4 x} \]

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Rubi [B]  time = 0.10, antiderivative size = 53, normalized size of antiderivative = 2.30, number of steps used = 2, number of rules used = 2, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.049, Rules used = {12, 2288} \begin {gather*} \frac {3 e^{e^2 \log ^2(x)-e^2 x} \left (e^3 x-2 e^3 \log (x)\right )}{4 x^2 \left (e^2-\frac {2 e^2 \log (x)}{x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-(E^2*x) + E^2*Log[x]^2)*(-3*E - 3*E^3*x + 6*E^3*Log[x]))/(4*x^2),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{4} \int \frac {e^{-e^2 x+e^2 \log ^2(x)} \left (-3 e-3 e^3 x+6 e^3 \log (x)\right )}{x^2} \, dx\\ &=\frac {3 e^{-e^2 x+e^2 \log ^2(x)} \left (e^3 x-2 e^3 \log (x)\right )}{4 x^2 \left (e^2-\frac {2 e^2 \log (x)}{x}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 25, normalized size = 1.09 \begin {gather*} \frac {3 e^{1-e^2 x+e^2 \log ^2(x)}}{4 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-(E^2*x) + E^2*Log[x]^2)*(-3*E - 3*E^3*x + 6*E^3*Log[x]))/(4*x^2),x]

[Out]

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

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fricas [A]  time = 0.60, size = 20, normalized size = 0.87 \begin {gather*} \frac {3 \, e^{\left (e^{2} \log \relax (x)^{2} - x e^{2} + 1\right )}}{4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(6*exp(1)*exp(2)*log(x)-3*x*exp(1)*exp(2)-3*exp(1))*exp(exp(2)*log(x)^2-exp(2)*x)/x^2,x, algorit
hm="fricas")

[Out]

3/4*e^(e^2*log(x)^2 - x*e^2 + 1)/x

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {3 \, {\left (x e^{3} - 2 \, e^{3} \log \relax (x) + e\right )} e^{\left (e^{2} \log \relax (x)^{2} - x e^{2}\right )}}{4 \, x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(6*exp(1)*exp(2)*log(x)-3*x*exp(1)*exp(2)-3*exp(1))*exp(exp(2)*log(x)^2-exp(2)*x)/x^2,x, algorit
hm="giac")

[Out]

integrate(-3/4*(x*e^3 - 2*e^3*log(x) + e)*e^(e^2*log(x)^2 - x*e^2)/x^2, x)

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maple [A]  time = 0.06, size = 21, normalized size = 0.91

method	result	size
risch	\(\frac {3 \,{\mathrm e}^{{\mathrm e}^{2} \ln \relax (x )^{2}-{\mathrm e}^{2} x +1}}{4 x}\)	\(21\)
norman	\(\frac {3 \,{\mathrm e} \,{\mathrm e}^{{\mathrm e}^{2} \ln \relax (x )^{2}-{\mathrm e}^{2} x}}{4 x}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*(6*exp(1)*exp(2)*ln(x)-3*x*exp(1)*exp(2)-3*exp(1))*exp(exp(2)*ln(x)^2-exp(2)*x)/x^2,x,method=_RETURNVE
RBOSE)

[Out]

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

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maxima [A]  time = 0.46, size = 20, normalized size = 0.87 \begin {gather*} \frac {3 \, e^{\left (e^{2} \log \relax (x)^{2} - x e^{2} + 1\right )}}{4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(6*exp(1)*exp(2)*log(x)-3*x*exp(1)*exp(2)-3*exp(1))*exp(exp(2)*log(x)^2-exp(2)*x)/x^2,x, algorit
hm="maxima")

[Out]

3/4*e^(e^2*log(x)^2 - x*e^2 + 1)/x

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mupad [B]  time = 3.29, size = 21, normalized size = 0.91 \begin {gather*} \frac {3\,\mathrm {e}\,{\mathrm {e}}^{{\mathrm {e}}^2\,{\ln \relax (x)}^2}\,{\mathrm {e}}^{-x\,{\mathrm {e}}^2}}{4\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(exp(2)*log(x)^2 - x*exp(2))*(3*exp(1) + 3*x*exp(3) - 6*exp(3)*log(x)))/(4*x^2),x)

[Out]

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

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sympy [A]  time = 0.26, size = 22, normalized size = 0.96 \begin {gather*} \frac {3 e e^{- x e^{2} + e^{2} \log {\relax (x )}^{2}}}{4 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(6*exp(1)*exp(2)*ln(x)-3*x*exp(1)*exp(2)-3*exp(1))*exp(exp(2)*ln(x)**2-exp(2)*x)/x**2,x)

[Out]

3*E*exp(-x*exp(2) + exp(2)*log(x)**2)/(4*x)

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