3.72.40 \(\int e^{3 x+4 e^{16 x-2 x \log (x)} x-8 e^{8 x-x \log (x)} x^2+4 x^3} (3+12 x^2+e^{16 x-2 x \log (x)} (4+56 x-8 x \log (x))+e^{8 x-x \log (x)} (-16 x-56 x^2+8 x^2 \log (x))) \, dx\)

Optimal. Leaf size=28 \[ e^{x+2 \left (x+2 \left (e^{8 x-x \log (x)}-x\right )^2 x\right )} \]

________________________________________________________________________________________

Rubi [A]  time = 2.14, antiderivative size = 39, normalized size of antiderivative = 1.39, number of steps used = 1, number of rules used = 1, integrand size = 99, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.010, Rules used = {6706} \begin {gather*} \exp \left (4 e^{16 x} x^{1-2 x}-8 e^{8 x} x^{2-x}+4 x^3+3 x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(3*x + 4*E^(16*x - 2*x*Log[x])*x - 8*E^(8*x - x*Log[x])*x^2 + 4*x^3)*(3 + 12*x^2 + E^(16*x - 2*x*Log[x])
*(4 + 56*x - 8*x*Log[x]) + E^(8*x - x*Log[x])*(-16*x - 56*x^2 + 8*x^2*Log[x])),x]

[Out]

E^(3*x + 4*x^3 + 4*E^(16*x)*x^(1 - 2*x) - 8*E^(8*x)*x^(2 - x))

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\exp \left (3 x+4 x^3+4 e^{16 x} x^{1-2 x}-8 e^{8 x} x^{2-x}\right )\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.20, size = 39, normalized size = 1.39 \begin {gather*} e^{3 x+4 x^3+4 e^{16 x} x^{1-2 x}-8 e^{8 x} x^{2-x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(3*x + 4*E^(16*x - 2*x*Log[x])*x - 8*E^(8*x - x*Log[x])*x^2 + 4*x^3)*(3 + 12*x^2 + E^(16*x - 2*x*L
og[x])*(4 + 56*x - 8*x*Log[x]) + E^(8*x - x*Log[x])*(-16*x - 56*x^2 + 8*x^2*Log[x])),x]

[Out]

E^(3*x + 4*x^3 + 4*E^(16*x)*x^(1 - 2*x) - 8*E^(8*x)*x^(2 - x))

________________________________________________________________________________________

fricas [A]  time = 1.00, size = 38, normalized size = 1.36 \begin {gather*} e^{\left (4 \, x^{3} - 8 \, x^{2} e^{\left (-x \log \relax (x) + 8 \, x\right )} + 4 \, x e^{\left (-2 \, x \log \relax (x) + 16 \, x\right )} + 3 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x*log(x)+56*x+4)*exp(-x*log(x)+8*x)^2+(8*x^2*log(x)-56*x^2-16*x)*exp(-x*log(x)+8*x)+12*x^2+3)*e
xp(4*x*exp(-x*log(x)+8*x)^2-8*x^2*exp(-x*log(x)+8*x)+4*x^3+3*x),x, algorithm="fricas")

[Out]

e^(4*x^3 - 8*x^2*e^(-x*log(x) + 8*x) + 4*x*e^(-2*x*log(x) + 16*x) + 3*x)

________________________________________________________________________________________

giac [A]  time = 1.73, size = 38, normalized size = 1.36 \begin {gather*} e^{\left (4 \, x^{3} - 8 \, x^{2} e^{\left (-x \log \relax (x) + 8 \, x\right )} + 4 \, x e^{\left (-2 \, x \log \relax (x) + 16 \, x\right )} + 3 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x*log(x)+56*x+4)*exp(-x*log(x)+8*x)^2+(8*x^2*log(x)-56*x^2-16*x)*exp(-x*log(x)+8*x)+12*x^2+3)*e
xp(4*x*exp(-x*log(x)+8*x)^2-8*x^2*exp(-x*log(x)+8*x)+4*x^3+3*x),x, algorithm="giac")

[Out]

e^(4*x^3 - 8*x^2*e^(-x*log(x) + 8*x) + 4*x*e^(-2*x*log(x) + 16*x) + 3*x)

________________________________________________________________________________________

maple [A]  time = 0.10, size = 36, normalized size = 1.29

method	result	size
risch	\({\mathrm e}^{x \left (4 x^{-2 x} {\mathrm e}^{16 x}-8 x^{-x} {\mathrm e}^{8 x} x +4 x^{2}+3\right )}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-8*x*ln(x)+56*x+4)*exp(-x*ln(x)+8*x)^2+(8*x^2*ln(x)-56*x^2-16*x)*exp(-x*ln(x)+8*x)+12*x^2+3)*exp(4*x*exp
(-x*ln(x)+8*x)^2-8*x^2*exp(-x*ln(x)+8*x)+4*x^3+3*x),x,method=_RETURNVERBOSE)

[Out]

exp(x*(4*(x^(-x))^2*exp(16*x)-8*x^(-x)*exp(8*x)*x+4*x^2+3))

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int {\left (12 \, x^{2} + 8 \, {\left (x^{2} \log \relax (x) - 7 \, x^{2} - 2 \, x\right )} e^{\left (-x \log \relax (x) + 8 \, x\right )} - 4 \, {\left (2 \, x \log \relax (x) - 14 \, x - 1\right )} e^{\left (-2 \, x \log \relax (x) + 16 \, x\right )} + 3\right )} e^{\left (4 \, x^{3} - 8 \, x^{2} e^{\left (-x \log \relax (x) + 8 \, x\right )} + 4 \, x e^{\left (-2 \, x \log \relax (x) + 16 \, x\right )} + 3 \, x\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x*log(x)+56*x+4)*exp(-x*log(x)+8*x)^2+(8*x^2*log(x)-56*x^2-16*x)*exp(-x*log(x)+8*x)+12*x^2+3)*e
xp(4*x*exp(-x*log(x)+8*x)^2-8*x^2*exp(-x*log(x)+8*x)+4*x^3+3*x),x, algorithm="maxima")

[Out]

integrate((12*x^2 + 8*(x^2*log(x) - 7*x^2 - 2*x)*e^(-x*log(x) + 8*x) - 4*(2*x*log(x) - 14*x - 1)*e^(-2*x*log(x
) + 16*x) + 3)*e^(4*x^3 - 8*x^2*e^(-x*log(x) + 8*x) + 4*x*e^(-2*x*log(x) + 16*x) + 3*x), x)

________________________________________________________________________________________

mupad [B]  time = 4.22, size = 39, normalized size = 1.39 \begin {gather*} {\mathrm {e}}^{-8\,x^{2-x}\,{\mathrm {e}}^{8\,x}}\,{\mathrm {e}}^{4\,x^{1-2\,x}\,{\mathrm {e}}^{16\,x}}\,{\mathrm {e}}^{3\,x}\,{\mathrm {e}}^{4\,x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(3*x + 4*x*exp(16*x - 2*x*log(x)) - 8*x^2*exp(8*x - x*log(x)) + 4*x^3)*(exp(16*x - 2*x*log(x))*(56*x -
8*x*log(x) + 4) - exp(8*x - x*log(x))*(16*x - 8*x^2*log(x) + 56*x^2) + 12*x^2 + 3),x)

[Out]

exp(-8*x^(2 - x)*exp(8*x))*exp(4*x^(1 - 2*x)*exp(16*x))*exp(3*x)*exp(4*x^3)

________________________________________________________________________________________

sympy [A]  time = 0.91, size = 39, normalized size = 1.39 \begin {gather*} e^{4 x^{3} - 8 x^{2} e^{- x \log {\relax (x )} + 8 x} + 4 x e^{- 2 x \log {\relax (x )} + 16 x} + 3 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x*ln(x)+56*x+4)*exp(-x*ln(x)+8*x)**2+(8*x**2*ln(x)-56*x**2-16*x)*exp(-x*ln(x)+8*x)+12*x**2+3)*e
xp(4*x*exp(-x*ln(x)+8*x)**2-8*x**2*exp(-x*ln(x)+8*x)+4*x**3+3*x),x)

[Out]

exp(4*x**3 - 8*x**2*exp(-x*log(x) + 8*x) + 4*x*exp(-2*x*log(x) + 16*x) + 3*x)

________________________________________________________________________________________