3.86.22 \(\int e^{-e^{4 x}} (e^{2 e^{-e^{4 x}} (x+e^{e^{4 x}} \log (2))} (2-8 e^{4 x} x)+e^{e^{4 x}} (6 e^{6 x}+8 x+e^{3 x} (4+12 x))+e^{e^{-e^{4 x}} (x+e^{e^{4 x}} \log (2))} (-2 e^{3 x}+e^{e^{4 x}} (-4-6 e^{3 x})-4 x+e^{4 x} (8 e^{3 x} x+16 x^2))) \, dx\)

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

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Rubi [A]  time = 1.23, antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 150, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.020, Rules used = {6688, 12, 6686} \begin {gather*} \left (2 x+e^{3 x}-2 e^{e^{-e^{4 x}} x}\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((2*(x + E^E^(4*x)*Log[2]))/E^E^(4*x))*(2 - 8*E^(4*x)*x) + E^E^(4*x)*(6*E^(6*x) + 8*x + E^(3*x)*(4 + 12
*x)) + E^((x + E^E^(4*x)*Log[2])/E^E^(4*x))*(-2*E^(3*x) + E^E^(4*x)*(-4 - 6*E^(3*x)) - 4*x + E^(4*x)*(8*E^(3*x
)*x + 16*x^2)))/E^E^(4*x),x]

[Out]

(E^(3*x) - 2*E^(x/E^E^(4*x)) + 2*x)^2

Rule 12

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

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int 2 e^{-e^{4 x}} \left (e^{3 x}-2 e^{e^{-e^{4 x}} x}+2 x\right ) \left (2 e^{e^{4 x}}-2 e^{e^{-e^{4 x}} x}+3 e^{e^{4 x}+3 x}+8 e^{\left (4+e^{-e^{4 x}}\right ) x} x\right ) \, dx\\ &=2 \int e^{-e^{4 x}} \left (e^{3 x}-2 e^{e^{-e^{4 x}} x}+2 x\right ) \left (2 e^{e^{4 x}}-2 e^{e^{-e^{4 x}} x}+3 e^{e^{4 x}+3 x}+8 e^{\left (4+e^{-e^{4 x}}\right ) x} x\right ) \, dx\\ &=\left (e^{3 x}-2 e^{e^{-e^{4 x}} x}+2 x\right )^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 26, normalized size = 1.00 \begin {gather*} \left (e^{3 x}-2 e^{e^{-e^{4 x}} x}+2 x\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((2*(x + E^E^(4*x)*Log[2]))/E^E^(4*x))*(2 - 8*E^(4*x)*x) + E^E^(4*x)*(6*E^(6*x) + 8*x + E^(3*x)*(
4 + 12*x)) + E^((x + E^E^(4*x)*Log[2])/E^E^(4*x))*(-2*E^(3*x) + E^E^(4*x)*(-4 - 6*E^(3*x)) - 4*x + E^(4*x)*(8*
E^(3*x)*x + 16*x^2)))/E^E^(4*x),x]

[Out]

(E^(3*x) - 2*E^(x/E^E^(4*x)) + 2*x)^2

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fricas [B]  time = 0.49, size = 66, normalized size = 2.54 \begin {gather*} 4 \, x^{2} - 2 \, {\left (2 \, x + e^{\left (3 \, x\right )}\right )} e^{\left ({\left (e^{\left (e^{\left (4 \, x\right )}\right )} \log \relax (2) + x\right )} e^{\left (-e^{\left (4 \, x\right )}\right )}\right )} + 4 \, x e^{\left (3 \, x\right )} + e^{\left (2 \, {\left (e^{\left (e^{\left (4 \, x\right )}\right )} \log \relax (2) + x\right )} e^{\left (-e^{\left (4 \, x\right )}\right )}\right )} + e^{\left (6 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x*exp(4*x)+2)*exp((log(2)*exp(exp(4*x))+x)/exp(exp(4*x)))^2+((-6*exp(3*x)-4)*exp(exp(4*x))+(8*x
*exp(3*x)+16*x^2)*exp(4*x)-2*exp(3*x)-4*x)*exp((log(2)*exp(exp(4*x))+x)/exp(exp(4*x)))+(6*exp(3*x)^2+(12*x+4)*
exp(3*x)+8*x)*exp(exp(4*x)))/exp(exp(4*x)),x, algorithm="fricas")

[Out]

4*x^2 - 2*(2*x + e^(3*x))*e^((e^(e^(4*x))*log(2) + x)*e^(-e^(4*x))) + 4*x*e^(3*x) + e^(2*(e^(e^(4*x))*log(2) +
x)*e^(-e^(4*x))) + e^(6*x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -2 \, {\left ({\left (4 \, x e^{\left (4 \, x\right )} - 1\right )} e^{\left (2 \, {\left (e^{\left (e^{\left (4 \, x\right )}\right )} \log \relax (2) + x\right )} e^{\left (-e^{\left (4 \, x\right )}\right )}\right )} - {\left (4 \, {\left (2 \, x^{2} + x e^{\left (3 \, x\right )}\right )} e^{\left (4 \, x\right )} - {\left (3 \, e^{\left (3 \, x\right )} + 2\right )} e^{\left (e^{\left (4 \, x\right )}\right )} - 2 \, x - e^{\left (3 \, x\right )}\right )} e^{\left ({\left (e^{\left (e^{\left (4 \, x\right )}\right )} \log \relax (2) + x\right )} e^{\left (-e^{\left (4 \, x\right )}\right )}\right )} - {\left (2 \, {\left (3 \, x + 1\right )} e^{\left (3 \, x\right )} + 4 \, x + 3 \, e^{\left (6 \, x\right )}\right )} e^{\left (e^{\left (4 \, x\right )}\right )}\right )} e^{\left (-e^{\left (4 \, x\right )}\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x*exp(4*x)+2)*exp((log(2)*exp(exp(4*x))+x)/exp(exp(4*x)))^2+((-6*exp(3*x)-4)*exp(exp(4*x))+(8*x
*exp(3*x)+16*x^2)*exp(4*x)-2*exp(3*x)-4*x)*exp((log(2)*exp(exp(4*x))+x)/exp(exp(4*x)))+(6*exp(3*x)^2+(12*x+4)*
exp(3*x)+8*x)*exp(exp(4*x)))/exp(exp(4*x)),x, algorithm="giac")

[Out]

integrate(-2*((4*x*e^(4*x) - 1)*e^(2*(e^(e^(4*x))*log(2) + x)*e^(-e^(4*x))) - (4*(2*x^2 + x*e^(3*x))*e^(4*x) -
(3*e^(3*x) + 2)*e^(e^(4*x)) - 2*x - e^(3*x))*e^((e^(e^(4*x))*log(2) + x)*e^(-e^(4*x))) - (2*(3*x + 1)*e^(3*x)
+ 4*x + 3*e^(6*x))*e^(e^(4*x)))*e^(-e^(4*x)), x)

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maple [B]  time = 0.08, size = 53, normalized size = 2.04

method	result	size
risch	\({\mathrm e}^{6 x}+4 x \,{\mathrm e}^{3 x}+4 x^{2}+4 \,{\mathrm e}^{2 \,{\mathrm e}^{-{\mathrm e}^{4 x}} x}+2 \left (-2 \,{\mathrm e}^{3 x}-4 x \right ) {\mathrm e}^{{\mathrm e}^{-{\mathrm e}^{4 x}} x}\)	\(53\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-8*x*exp(4*x)+2)*exp((ln(2)*exp(exp(4*x))+x)/exp(exp(4*x)))^2+((-6*exp(3*x)-4)*exp(exp(4*x))+(8*x*exp(3*
x)+16*x^2)*exp(4*x)-2*exp(3*x)-4*x)*exp((ln(2)*exp(exp(4*x))+x)/exp(exp(4*x)))+(6*exp(3*x)^2+(12*x+4)*exp(3*x)
+8*x)*exp(exp(4*x)))/exp(exp(4*x)),x,method=_RETURNVERBOSE)

[Out]

exp(6*x)+4*x*exp(3*x)+4*x^2+4*exp(2*exp(-exp(4*x))*x)+2*(-2*exp(3*x)-4*x)*exp(exp(-exp(4*x))*x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 4 \, x^{2} + \frac {4}{3} \, {\left (3 \, x - 1\right )} e^{\left (3 \, x\right )} + 4 \, e^{\left (2 \, x e^{\left (-e^{\left (4 \, x\right )}\right )}\right )} + e^{\left (6 \, x\right )} + \frac {4}{3} \, e^{\left (3 \, x\right )} - 2 \, \int -2 \, {\left (8 \, x^{2} e^{\left (4 \, x\right )} + 4 \, x e^{\left (7 \, x\right )} - {\left (3 \, e^{\left (3 \, x\right )} + 2\right )} e^{\left (e^{\left (4 \, x\right )}\right )} - 2 \, x - e^{\left (3 \, x\right )}\right )} e^{\left (x e^{\left (-e^{\left (4 \, x\right )}\right )} - e^{\left (4 \, x\right )}\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x*exp(4*x)+2)*exp((log(2)*exp(exp(4*x))+x)/exp(exp(4*x)))^2+((-6*exp(3*x)-4)*exp(exp(4*x))+(8*x
*exp(3*x)+16*x^2)*exp(4*x)-2*exp(3*x)-4*x)*exp((log(2)*exp(exp(4*x))+x)/exp(exp(4*x)))+(6*exp(3*x)^2+(12*x+4)*
exp(3*x)+8*x)*exp(exp(4*x)))/exp(exp(4*x)),x, algorithm="maxima")

[Out]

4*x^2 + 4/3*(3*x - 1)*e^(3*x) + 4*e^(2*x*e^(-e^(4*x))) + e^(6*x) + 4/3*e^(3*x) - 2*integrate(-2*(8*x^2*e^(4*x)
+ 4*x*e^(7*x) - (3*e^(3*x) + 2)*e^(e^(4*x)) - 2*x - e^(3*x))*e^(x*e^(-e^(4*x)) - e^(4*x)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} -\int {\mathrm {e}}^{-{\mathrm {e}}^{4\,x}}\,\left ({\mathrm {e}}^{{\mathrm {e}}^{-{\mathrm {e}}^{4\,x}}\,\left (x+{\mathrm {e}}^{{\mathrm {e}}^{4\,x}}\,\ln \relax (2)\right )}\,\left (4\,x+2\,{\mathrm {e}}^{3\,x}-{\mathrm {e}}^{4\,x}\,\left (8\,x\,{\mathrm {e}}^{3\,x}+16\,x^2\right )+{\mathrm {e}}^{{\mathrm {e}}^{4\,x}}\,\left (6\,{\mathrm {e}}^{3\,x}+4\right )\right )-{\mathrm {e}}^{{\mathrm {e}}^{4\,x}}\,\left (8\,x+6\,{\mathrm {e}}^{6\,x}+{\mathrm {e}}^{3\,x}\,\left (12\,x+4\right )\right )+{\mathrm {e}}^{2\,{\mathrm {e}}^{-{\mathrm {e}}^{4\,x}}\,\left (x+{\mathrm {e}}^{{\mathrm {e}}^{4\,x}}\,\ln \relax (2)\right )}\,\left (8\,x\,{\mathrm {e}}^{4\,x}-2\right )\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(-exp(4*x))*(exp(exp(-exp(4*x))*(x + exp(exp(4*x))*log(2)))*(4*x + 2*exp(3*x) - exp(4*x)*(8*x*exp(3*x)
+ 16*x^2) + exp(exp(4*x))*(6*exp(3*x) + 4)) - exp(exp(4*x))*(8*x + 6*exp(6*x) + exp(3*x)*(12*x + 4)) + exp(2*
exp(-exp(4*x))*(x + exp(exp(4*x))*log(2)))*(8*x*exp(4*x) - 2)),x)

[Out]

-int(exp(-exp(4*x))*(exp(exp(-exp(4*x))*(x + exp(exp(4*x))*log(2)))*(4*x + 2*exp(3*x) - exp(4*x)*(8*x*exp(3*x)
+ 16*x^2) + exp(exp(4*x))*(6*exp(3*x) + 4)) - exp(exp(4*x))*(8*x + 6*exp(6*x) + exp(3*x)*(12*x + 4)) + exp(2*
exp(-exp(4*x))*(x + exp(exp(4*x))*log(2)))*(8*x*exp(4*x) - 2)), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x*exp(4*x)+2)*exp((ln(2)*exp(exp(4*x))+x)/exp(exp(4*x)))**2+((-6*exp(3*x)-4)*exp(exp(4*x))+(8*x
*exp(3*x)+16*x**2)*exp(4*x)-2*exp(3*x)-4*x)*exp((ln(2)*exp(exp(4*x))+x)/exp(exp(4*x)))+(6*exp(3*x)**2+(12*x+4)
*exp(3*x)+8*x)*exp(exp(4*x)))/exp(exp(4*x)),x)

[Out]

Timed out

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