\(\int e^{3 x+4 e^{16 x-2 x \log (x)} x-8 e^{8 x-x \log (x)} x^2+4 x^3} (3+12 x^2+e^{16 x-2 x \log (x)} (4+56 x-8 x \log (x))+e^{8 x-x \log (x)} (-16 x-56 x^2+8 x^2 \log (x))) \, dx\) [7140]

Optimal result

Integrand size = 99, antiderivative size = 28 \[ \int e^{3 x+4 e^{16 x-2 x \log (x)} x-8 e^{8 x-x \log (x)} x^2+4 x^3} \left (3+12 x^2+e^{16 x-2 x \log (x)} (4+56 x-8 x \log (x))+e^{8 x-x \log (x)} \left (-16 x-56 x^2+8 x^2 \log (x)\right )\right ) \, dx=e^{x+2 \left (x+2 \left (e^{8 x-x \log (x)}-x\right )^2 x\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 1.26 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.010, Rules used = {6838} \[ \int e^{3 x+4 e^{16 x-2 x \log (x)} x-8 e^{8 x-x \log (x)} x^2+4 x^3} \left (3+12 x^2+e^{16 x-2 x \log (x)} (4+56 x-8 x \log (x))+e^{8 x-x \log (x)} \left (-16 x-56 x^2+8 x^2 \log (x)\right )\right ) \, dx=\exp \left (4 e^{16 x} x^{1-2 x}-8 e^{8 x} x^{2-x}+4 x^3+3 x\right ) \]

[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 6838

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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39 \[ \int e^{3 x+4 e^{16 x-2 x \log (x)} x-8 e^{8 x-x \log (x)} x^2+4 x^3} \left (3+12 x^2+e^{16 x-2 x \log (x)} (4+56 x-8 x \log (x))+e^{8 x-x \log (x)} \left (-16 x-56 x^2+8 x^2 \log (x)\right )\right ) \, dx=e^{3 x+4 x^3+4 e^{16 x} x^{1-2 x}-8 e^{8 x} x^{2-x}} \]

[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))

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29

method	result	size
risch	\({\mathrm e}^{x \left (-8 x^{-x} {\mathrm e}^{8 x} x +4 x^{-2 x} {\mathrm e}^{16 x}+4 x^{2}+3\right )}\)	\(36\)
parallelrisch	\({\mathrm e}^{4 x \,{\mathrm e}^{-2 x \ln \left (x \right )+16 x}-8 x^{2} {\mathrm e}^{-x \ln \left (x \right )+8 x}+4 x^{3}+3 x}\)	\(41\)

[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))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.36 \[ \int e^{3 x+4 e^{16 x-2 x \log (x)} x-8 e^{8 x-x \log (x)} x^2+4 x^3} \left (3+12 x^2+e^{16 x-2 x \log (x)} (4+56 x-8 x \log (x))+e^{8 x-x \log (x)} \left (-16 x-56 x^2+8 x^2 \log (x)\right )\right ) \, dx=e^{\left (4 \, x^{3} - 8 \, x^{2} e^{\left (-x \log \left (x\right ) + 8 \, x\right )} + 4 \, x e^{\left (-2 \, x \log \left (x\right ) + 16 \, x\right )} + 3 \, x\right )} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.43 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39 \[ \int e^{3 x+4 e^{16 x-2 x \log (x)} x-8 e^{8 x-x \log (x)} x^2+4 x^3} \left (3+12 x^2+e^{16 x-2 x \log (x)} (4+56 x-8 x \log (x))+e^{8 x-x \log (x)} \left (-16 x-56 x^2+8 x^2 \log (x)\right )\right ) \, dx=e^{4 x^{3} - 8 x^{2} e^{- x \log {\left (x \right )} + 8 x} + 4 x e^{- 2 x \log {\left (x \right )} + 16 x} + 3 x} \]

[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)

Maxima [F]

\[ \int e^{3 x+4 e^{16 x-2 x \log (x)} x-8 e^{8 x-x \log (x)} x^2+4 x^3} \left (3+12 x^2+e^{16 x-2 x \log (x)} (4+56 x-8 x \log (x))+e^{8 x-x \log (x)} \left (-16 x-56 x^2+8 x^2 \log (x)\right )\right ) \, dx=\int { {\left (12 \, x^{2} + 8 \, {\left (x^{2} \log \left (x\right ) - 7 \, x^{2} - 2 \, x\right )} e^{\left (-x \log \left (x\right ) + 8 \, x\right )} - 4 \, {\left (2 \, x \log \left (x\right ) - 14 \, x - 1\right )} e^{\left (-2 \, x \log \left (x\right ) + 16 \, x\right )} + 3\right )} e^{\left (4 \, x^{3} - 8 \, x^{2} e^{\left (-x \log \left (x\right ) + 8 \, x\right )} + 4 \, x e^{\left (-2 \, x \log \left (x\right ) + 16 \, x\right )} + 3 \, x\right )} \,d x } \]

[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)

Giac [A] (verification not implemented)

none

Time = 1.03 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.36 \[ \int e^{3 x+4 e^{16 x-2 x \log (x)} x-8 e^{8 x-x \log (x)} x^2+4 x^3} \left (3+12 x^2+e^{16 x-2 x \log (x)} (4+56 x-8 x \log (x))+e^{8 x-x \log (x)} \left (-16 x-56 x^2+8 x^2 \log (x)\right )\right ) \, dx=e^{\left (4 \, x^{3} - 8 \, x^{2} e^{\left (-x \log \left (x\right ) + 8 \, x\right )} + 4 \, x e^{\left (-2 \, x \log \left (x\right ) + 16 \, x\right )} + 3 \, x\right )} \]

[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)

Mupad [B] (verification not implemented)

Time = 12.64 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39 \[ \int e^{3 x+4 e^{16 x-2 x \log (x)} x-8 e^{8 x-x \log (x)} x^2+4 x^3} \left (3+12 x^2+e^{16 x-2 x \log (x)} (4+56 x-8 x \log (x))+e^{8 x-x \log (x)} \left (-16 x-56 x^2+8 x^2 \log (x)\right )\right ) \, dx={\mathrm {e}}^{-8\,x^{2-x}\,{\mathrm {e}}^{8\,x}}\,{\mathrm {e}}^{4\,x^{1-2\,x}\,{\mathrm {e}}^{16\,x}}\,{\mathrm {e}}^{3\,x}\,{\mathrm {e}}^{4\,x^3} \]

[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)