3.41.99 \(\int \frac {2 e^4+e^{\frac {16+32 x-9 x^2+4 x^3-x^4}{4 e^4}} (16-9 x+6 x^2-2 x^3)}{2 e^4} \, dx\)

Optimal. Leaf size=27 \[ e^{\frac {(4-x) x \left (9+\frac {4}{x}+x^2\right )}{4 e^4}}+x \]

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Rubi [A]  time = 0.11, antiderivative size = 31, normalized size of antiderivative = 1.15, number of steps used = 3, number of rules used = 2, integrand size = 58, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {12, 6706} \begin {gather*} e^{\frac {-x^4+4 x^3-9 x^2+32 x+16}{4 e^4}}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2*E^4 + E^((16 + 32*x - 9*x^2 + 4*x^3 - x^4)/(4*E^4))*(16 - 9*x + 6*x^2 - 2*x^3))/(2*E^4),x]

[Out]

E^((16 + 32*x - 9*x^2 + 4*x^3 - x^4)/(4*E^4)) + x

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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} &=\frac {\int \left (2 e^4+e^{\frac {16+32 x-9 x^2+4 x^3-x^4}{4 e^4}} \left (16-9 x+6 x^2-2 x^3\right )\right ) \, dx}{2 e^4}\\ &=x+\frac {\int e^{\frac {16+32 x-9 x^2+4 x^3-x^4}{4 e^4}} \left (16-9 x+6 x^2-2 x^3\right ) \, dx}{2 e^4}\\ &=e^{\frac {16+32 x-9 x^2+4 x^3-x^4}{4 e^4}}+x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.24, size = 43, normalized size = 1.59 \begin {gather*} e^{\frac {4}{e^4}+\frac {8 x}{e^4}-\frac {9 x^2}{4 e^4}+\frac {x^3}{e^4}-\frac {x^4}{4 e^4}}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*E^4 + E^((16 + 32*x - 9*x^2 + 4*x^3 - x^4)/(4*E^4))*(16 - 9*x + 6*x^2 - 2*x^3))/(2*E^4),x]

[Out]

E^(4/E^4 + (8*x)/E^4 - (9*x^2)/(4*E^4) + x^3/E^4 - x^4/(4*E^4)) + x

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fricas [A]  time = 0.87, size = 25, normalized size = 0.93 \begin {gather*} x + e^{\left (-\frac {1}{4} \, {\left (x^{4} - 4 \, x^{3} + 9 \, x^{2} - 32 \, x - 16\right )} e^{\left (-4\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-2*x^3+6*x^2-9*x+16)*exp(1/4*(-x^4+4*x^3-9*x^2+32*x+16)/exp(4))+2*exp(4))/exp(4),x, algorithm=
"fricas")

[Out]

x + e^(-1/4*(x^4 - 4*x^3 + 9*x^2 - 32*x - 16)*e^(-4))

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giac [A]  time = 0.12, size = 40, normalized size = 1.48 \begin {gather*} {\left (x e^{4} + e^{\left (-\frac {1}{4} \, x^{4} e^{\left (-4\right )} + x^{3} e^{\left (-4\right )} - \frac {9}{4} \, x^{2} e^{\left (-4\right )} + 8 \, x e^{\left (-4\right )} + 4 \, e^{\left (-4\right )} + 4\right )}\right )} e^{\left (-4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-2*x^3+6*x^2-9*x+16)*exp(1/4*(-x^4+4*x^3-9*x^2+32*x+16)/exp(4))+2*exp(4))/exp(4),x, algorithm=
"giac")

[Out]

(x*e^4 + e^(-1/4*x^4*e^(-4) + x^3*e^(-4) - 9/4*x^2*e^(-4) + 8*x*e^(-4) + 4*e^(-4) + 4))*e^(-4)

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maple [A]  time = 0.09, size = 19, normalized size = 0.70

method	result	size
risch	\(x +{\mathrm e}^{-\frac {\left (x -4\right ) \left (x^{3}+9 x +4\right ) {\mathrm e}^{-4}}{4}}\)	\(19\)
norman	\(x +{\mathrm e}^{\frac {\left (-x^{4}+4 x^{3}-9 x^{2}+32 x +16\right ) {\mathrm e}^{-4}}{4}}\)	\(30\)
default	\(\frac {{\mathrm e}^{-4} \left (2 \,{\mathrm e}^{4} {\mathrm e}^{\frac {\left (-x^{4}+4 x^{3}-9 x^{2}+32 x +16\right ) {\mathrm e}^{-4}}{4}}+2 x \,{\mathrm e}^{4}\right )}{2}\)	\(44\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*((-2*x^3+6*x^2-9*x+16)*exp(1/4*(-x^4+4*x^3-9*x^2+32*x+16)/exp(4))+2*exp(4))/exp(4),x,method=_RETURNVER
BOSE)

[Out]

x+exp(-1/4*(x-4)*(x^3+9*x+4)*exp(-4))

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maxima [A]  time = 0.53, size = 40, normalized size = 1.48 \begin {gather*} {\left (x e^{4} + e^{\left (-\frac {1}{4} \, x^{4} e^{\left (-4\right )} + x^{3} e^{\left (-4\right )} - \frac {9}{4} \, x^{2} e^{\left (-4\right )} + 8 \, x e^{\left (-4\right )} + 4 \, e^{\left (-4\right )} + 4\right )}\right )} e^{\left (-4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-2*x^3+6*x^2-9*x+16)*exp(1/4*(-x^4+4*x^3-9*x^2+32*x+16)/exp(4))+2*exp(4))/exp(4),x, algorithm=
"maxima")

[Out]

(x*e^4 + e^(-1/4*x^4*e^(-4) + x^3*e^(-4) - 9/4*x^2*e^(-4) + 8*x*e^(-4) + 4*e^(-4) + 4))*e^(-4)

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mupad [B]  time = 0.18, size = 33, normalized size = 1.22 \begin {gather*} x+{\mathrm {e}}^{-\frac {{\mathrm {e}}^{-4}\,x^4}{4}+{\mathrm {e}}^{-4}\,x^3-\frac {9\,{\mathrm {e}}^{-4}\,x^2}{4}+8\,{\mathrm {e}}^{-4}\,x+4\,{\mathrm {e}}^{-4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(-4)*(exp(4) - (exp(exp(-4)*(8*x - (9*x^2)/4 + x^3 - x^4/4 + 4))*(9*x - 6*x^2 + 2*x^3 - 16))/2),x)

[Out]

x + exp(4*exp(-4) + 8*x*exp(-4) - (9*x^2*exp(-4))/4 + x^3*exp(-4) - (x^4*exp(-4))/4)

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sympy [A]  time = 0.17, size = 26, normalized size = 0.96 \begin {gather*} x + e^{\frac {- \frac {x^{4}}{4} + x^{3} - \frac {9 x^{2}}{4} + 8 x + 4}{e^{4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-2*x**3+6*x**2-9*x+16)*exp(1/4*(-x**4+4*x**3-9*x**2+32*x+16)/exp(4))+2*exp(4))/exp(4),x)

[Out]

x + exp((-x**4/4 + x**3 - 9*x**2/4 + 8*x + 4)*exp(-4))

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