3.30.86 \(\int \frac {e^{-3+\frac {e^{-3+e^{512+2 e^2+64 x+2 x^2}}}{x}} (4 e^3 x+e^{e^{512+2 e^2+64 x+2 x^2}} (-4+e^{512+2 e^2+64 x+2 x^2} (256 x+16 x^2)))}{x} \, dx\)

Optimal. Leaf size=28 \[ 4 e^{\frac {e^{-3+e^{2 e^2+2 (16+x)^2}}}{x}} x \]

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Rubi [B]  time = 0.43, antiderivative size = 154, normalized size of antiderivative = 5.50, number of steps used = 1, number of rules used = 1, integrand size = 89, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.011, Rules used = {2288} \begin {gather*} \frac {4 \left (1-4 e^{2 x^2+64 x+2 \left (256+e^2\right )} \left (x^2+16 x\right )\right ) \exp \left (e^{2 x^2+64 x+2 \left (256+e^2\right )}+\frac {e^{e^{2 x^2+64 x+2 \left (256+e^2\right )}-3}}{x}-3\right )}{x \left (\frac {e^{e^{2 x^2+64 x+2 \left (256+e^2\right )}-3}}{x^2}-\frac {4 (x+16) \exp \left (2 x^2+e^{2 x^2+64 x+2 \left (256+e^2\right )}+64 x+2 e^2+509\right )}{x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-3 + E^(-3 + E^(512 + 2*E^2 + 64*x + 2*x^2))/x)*(4*E^3*x + E^E^(512 + 2*E^2 + 64*x + 2*x^2)*(-4 + E^(5
12 + 2*E^2 + 64*x + 2*x^2)*(256*x + 16*x^2))))/x,x]

[Out]

(4*E^(-3 + E^(2*(256 + E^2) + 64*x + 2*x^2) + E^(-3 + E^(2*(256 + E^2) + 64*x + 2*x^2))/x)*(1 - 4*E^(2*(256 +
E^2) + 64*x + 2*x^2)*(16*x + x^2)))/(x*(E^(-3 + E^(2*(256 + E^2) + 64*x + 2*x^2))/x^2 - (4*E^(509 + 2*E^2 + E^
(2*(256 + E^2) + 64*x + 2*x^2) + 64*x + 2*x^2)*(16 + x))/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 {4 \exp \left (-3+e^{2 \left (256+e^2\right )+64 x+2 x^2}+\frac {e^{-3+e^{2 \left (256+e^2\right )+64 x+2 x^2}}}{x}\right ) \left (1-4 e^{2 \left (256+e^2\right )+64 x+2 x^2} \left (16 x+x^2\right )\right )}{x \left (\frac {e^{-3+e^{2 \left (256+e^2\right )+64 x+2 x^2}}}{x^2}-\frac {4 \exp \left (509+2 e^2+e^{2 \left (256+e^2\right )+64 x+2 x^2}+64 x+2 x^2\right ) (16+x)}{x}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.25, size = 26, normalized size = 0.93 \begin {gather*} 4 e^{\frac {e^{-3+e^{2 \left (e^2+(16+x)^2\right )}}}{x}} x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-3 + E^(-3 + E^(512 + 2*E^2 + 64*x + 2*x^2))/x)*(4*E^3*x + E^E^(512 + 2*E^2 + 64*x + 2*x^2)*(-4
+ E^(512 + 2*E^2 + 64*x + 2*x^2)*(256*x + 16*x^2))))/x,x]

[Out]

4*E^(E^(-3 + E^(2*(E^2 + (16 + x)^2)))/x)*x

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fricas [A]  time = 1.72, size = 37, normalized size = 1.32 \begin {gather*} 4 \, x e^{\left (-\frac {{\left (3 \, x e^{3} - e^{\left (e^{\left (2 \, x^{2} + 64 \, x + 2 \, e^{2} + 512\right )}\right )}\right )} e^{\left (-3\right )}}{x} + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x^2+256*x)*exp(exp(2)+x^2+32*x+256)^2-4)*exp(exp(exp(2)+x^2+32*x+256)^2)+4*x*exp(3))*exp(exp(e
xp(exp(2)+x^2+32*x+256)^2)/x/exp(3))/x/exp(3),x, algorithm="fricas")

[Out]

4*x*e^(-(3*x*e^3 - e^(e^(2*x^2 + 64*x + 2*e^2 + 512)))*e^(-3)/x + 3)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {4 \, {\left (x e^{3} + {\left (4 \, {\left (x^{2} + 16 \, x\right )} e^{\left (2 \, x^{2} + 64 \, x + 2 \, e^{2} + 512\right )} - 1\right )} e^{\left (e^{\left (2 \, x^{2} + 64 \, x + 2 \, e^{2} + 512\right )}\right )}\right )} e^{\left (\frac {e^{\left (e^{\left (2 \, x^{2} + 64 \, x + 2 \, e^{2} + 512\right )} - 3\right )}}{x} - 3\right )}}{x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x^2+256*x)*exp(exp(2)+x^2+32*x+256)^2-4)*exp(exp(exp(2)+x^2+32*x+256)^2)+4*x*exp(3))*exp(exp(e
xp(exp(2)+x^2+32*x+256)^2)/x/exp(3))/x/exp(3),x, algorithm="giac")

[Out]

integrate(4*(x*e^3 + (4*(x^2 + 16*x)*e^(2*x^2 + 64*x + 2*e^2 + 512) - 1)*e^(e^(2*x^2 + 64*x + 2*e^2 + 512)))*e
^(e^(e^(2*x^2 + 64*x + 2*e^2 + 512) - 3)/x - 3)/x, x)

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maple [A]  time = 0.15, size = 27, normalized size = 0.96

method	result	size
risch	\(4 x \,{\mathrm e}^{\frac {{\mathrm e}^{{\mathrm e}^{2 \,{\mathrm e}^{2}+2 x^{2}+64 x +512}-3}}{x}}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((16*x^2+256*x)*exp(exp(2)+x^2+32*x+256)^2-4)*exp(exp(exp(2)+x^2+32*x+256)^2)+4*x*exp(3))*exp(exp(exp(exp
(2)+x^2+32*x+256)^2)/x/exp(3))/x/exp(3),x,method=_RETURNVERBOSE)

[Out]

4*x*exp(1/x*exp(exp(2*exp(2)+2*x^2+64*x+512)-3))

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maxima [A]  time = 0.52, size = 26, normalized size = 0.93 \begin {gather*} 4 \, x e^{\left (\frac {e^{\left (e^{\left (2 \, x^{2} + 64 \, x + 2 \, e^{2} + 512\right )} - 3\right )}}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x^2+256*x)*exp(exp(2)+x^2+32*x+256)^2-4)*exp(exp(exp(2)+x^2+32*x+256)^2)+4*x*exp(3))*exp(exp(e
xp(exp(2)+x^2+32*x+256)^2)/x/exp(3))/x/exp(3),x, algorithm="maxima")

[Out]

4*x*e^(e^(e^(2*x^2 + 64*x + 2*e^2 + 512) - 3)/x)

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mupad [B]  time = 2.11, size = 29, normalized size = 1.04 \begin {gather*} 4\,x\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{{\mathrm {e}}^{2\,{\mathrm {e}}^2}\,{\mathrm {e}}^{64\,x}\,{\mathrm {e}}^{512}\,{\mathrm {e}}^{2\,x^2}}\,{\mathrm {e}}^{-3}}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-3)*exp((exp(-3)*exp(exp(64*x + 2*exp(2) + 2*x^2 + 512)))/x)*(exp(exp(64*x + 2*exp(2) + 2*x^2 + 512))
*(exp(64*x + 2*exp(2) + 2*x^2 + 512)*(256*x + 16*x^2) - 4) + 4*x*exp(3)))/x,x)

[Out]

4*x*exp((exp(exp(2*exp(2))*exp(64*x)*exp(512)*exp(2*x^2))*exp(-3))/x)

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sympy [A]  time = 119.70, size = 27, normalized size = 0.96 \begin {gather*} 4 x e^{\frac {e^{e^{2 x^{2} + 64 x + 2 e^{2} + 512}}}{x e^{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x**2+256*x)*exp(exp(2)+x**2+32*x+256)**2-4)*exp(exp(exp(2)+x**2+32*x+256)**2)+4*x*exp(3))*exp(
exp(exp(exp(2)+x**2+32*x+256)**2)/x/exp(3))/x/exp(3),x)

[Out]

4*x*exp(exp(-3)*exp(exp(2*x**2 + 64*x + 2*exp(2) + 512))/x)

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