\(\int e^{1-x+27 x^3} (2 x-x^2+81 x^4) \, dx\) [724]

Optimal result

Integrand size = 27, antiderivative size = 16 \[ \int e^{1-x+27 x^3} \left (2 x-x^2+81 x^4\right ) \, dx=e^{1-x+27 x^3} x^2 \]

[Out]

x^2/exp(-27*x^3+x-1)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.88, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {1608, 2326} \[ \int e^{1-x+27 x^3} \left (2 x-x^2+81 x^4\right ) \, dx=\frac {e^{27 x^3-x+1} x \left (x-81 x^3\right )}{1-81 x^2} \]

[In]

Int[E^(1 - x + 27*x^3)*(2*x - x^2 + 81*x^4),x]

[Out]

(E^(1 - x + 27*x^3)*x*(x - 81*x^3))/(1 - 81*x^2)

Rule 1608

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 2326

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{align*} \text {integral}& = \int e^{1-x+27 x^3} x \left (2-x+81 x^3\right ) \, dx \\ & = \frac {e^{1-x+27 x^3} x \left (x-81 x^3\right )}{1-81 x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int e^{1-x+27 x^3} \left (2 x-x^2+81 x^4\right ) \, dx=e^{1-x+27 x^3} x^2 \]

[In]

Integrate[E^(1 - x + 27*x^3)*(2*x - x^2 + 81*x^4),x]

[Out]

E^(1 - x + 27*x^3)*x^2

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00

method	result	size
gosper	\({\mathrm e}^{27 x^{3}-x +1} x^{2}\)	\(16\)
norman	\({\mathrm e}^{27 x^{3}-x +1} x^{2}\)	\(16\)
risch	\({\mathrm e}^{27 x^{3}-x +1} x^{2}\)	\(16\)
parallelrisch	\({\mathrm e}^{27 x^{3}-x +1} x^{2}\)	\(16\)

[In]

int((81*x^4-x^2+2*x)/exp(-27*x^3+x-1),x,method=_RETURNVERBOSE)

[Out]

x^2/exp(-27*x^3+x-1)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int e^{1-x+27 x^3} \left (2 x-x^2+81 x^4\right ) \, dx=x^{2} e^{\left (27 \, x^{3} - x + 1\right )} \]

[In]

integrate((81*x^4-x^2+2*x)/exp(-27*x^3+x-1),x, algorithm="fricas")

[Out]

x^2*e^(27*x^3 - x + 1)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int e^{1-x+27 x^3} \left (2 x-x^2+81 x^4\right ) \, dx=x^{2} e^{27 x^{3} - x + 1} \]

[In]

integrate((81*x**4-x**2+2*x)/exp(-27*x**3+x-1),x)

[Out]

x**2*exp(27*x**3 - x + 1)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int e^{1-x+27 x^3} \left (2 x-x^2+81 x^4\right ) \, dx=x^{2} e^{\left (27 \, x^{3} - x + 1\right )} \]

[In]

integrate((81*x^4-x^2+2*x)/exp(-27*x^3+x-1),x, algorithm="maxima")

[Out]

x^2*e^(27*x^3 - x + 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int e^{1-x+27 x^3} \left (2 x-x^2+81 x^4\right ) \, dx=x^{2} e^{\left (27 \, x^{3} - x + 1\right )} \]

[In]

integrate((81*x^4-x^2+2*x)/exp(-27*x^3+x-1),x, algorithm="giac")

[Out]

x^2*e^(27*x^3 - x + 1)

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int e^{1-x+27 x^3} \left (2 x-x^2+81 x^4\right ) \, dx=x^2\,{\mathrm {e}}^{-x}\,\mathrm {e}\,{\mathrm {e}}^{27\,x^3} \]

[In]

int(exp(27*x^3 - x + 1)*(2*x - x^2 + 81*x^4),x)

[Out]

x^2*exp(-x)*exp(1)*exp(27*x^3)