3.26.59 \(\int (1+e^{\frac {1}{4} (-4+e^6+12 x-4 e^3 x+4 x^2)} (-3+e^3-2 x)) \, dx\)

Optimal. Leaf size=24 \[ -e^{-1+\left (\frac {e^3}{2}-x\right )^2+3 x}+x \]

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Rubi [A]  time = 0.09, antiderivative size = 28, normalized size of antiderivative = 1.17, number of steps used = 3, number of rules used = 2, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {2244, 2236} \begin {gather*} x-e^{x^2+\left (3-e^3\right ) x+\frac {1}{4} \left (e^6-4\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1 + E^((-4 + E^6 + 12*x - 4*E^3*x + 4*x^2)/4)*(-3 + E^3 - 2*x),x]

[Out]

-E^((-4 + E^6)/4 + (3 - E^3)*x + x^2) + x

Rule 2236

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[(e*F^(a + b*x + c*x^2))/(
2*c*Log[F]), x] /; FreeQ[{F, a, b, c, d, e}, x] && EqQ[b*e - 2*c*d, 0]

Rule 2244

Int[(F_)^(v_)*(u_)^(m_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*F^ExpandToSum[v, x], x] /; FreeQ[{F, m}, x] &&
LinearQ[u, x] && QuadraticQ[v, x] &&  !(LinearMatchQ[u, x] && QuadraticMatchQ[v, x])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=x+\int e^{\frac {1}{4} \left (-4+e^6+12 x-4 e^3 x+4 x^2\right )} \left (-3+e^3-2 x\right ) \, dx\\ &=x+\int e^{\frac {1}{4} \left (-4+e^6\right )+\left (3-e^3\right ) x+x^2} \left (-3+e^3-2 x\right ) \, dx\\ &=-e^{\frac {1}{4} \left (-4+e^6\right )+\left (3-e^3\right ) x+x^2}+x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 26, normalized size = 1.08 \begin {gather*} -e^{-1+\frac {e^6}{4}-\left (-3+e^3\right ) x+x^2}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1 + E^((-4 + E^6 + 12*x - 4*E^3*x + 4*x^2)/4)*(-3 + E^3 - 2*x),x]

[Out]

-E^(-1 + E^6/4 - (-3 + E^3)*x + x^2) + x

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fricas [A]  time = 0.94, size = 22, normalized size = 0.92 \begin {gather*} x - e^{\left (x^{2} - x e^{3} + 3 \, x + \frac {1}{4} \, e^{6} - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x - e^(x^2 - x*e^3 + 3*x + 1/4*e^6 - 1)

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giac [A]  time = 0.19, size = 22, normalized size = 0.92 \begin {gather*} x - e^{\left (x^{2} - x e^{3} + 3 \, x + \frac {1}{4} \, e^{6} - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x - e^(x^2 - x*e^3 + 3*x + 1/4*e^6 - 1)

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maple [A]  time = 0.04, size = 23, normalized size = 0.96

method	result	size
risch	\(x -{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}-x \,{\mathrm e}^{3}+x^{2}+3 x -1}\)	\(23\)
norman	\(x -{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}-x \,{\mathrm e}^{3}+x^{2}+3 x -1}\)	\(25\)
default	\(x -\frac {i \sqrt {\pi }\, {\mathrm e}^{\frac {{\mathrm e}^{6}}{4}+2-\frac {\left (3-{\mathrm e}^{3}\right )^{2}}{4}} \erf \left (i x +\frac {i \left (3-{\mathrm e}^{3}\right )}{2}\right )}{2}+\frac {3 i \sqrt {\pi }\, {\mathrm e}^{\frac {{\mathrm e}^{6}}{4}-1-\frac {\left (3-{\mathrm e}^{3}\right )^{2}}{4}} \erf \left (i x +\frac {i \left (3-{\mathrm e}^{3}\right )}{2}\right )}{2}-{\mathrm e}^{x^{2}+\left (3-{\mathrm e}^{3}\right ) x +\frac {{\mathrm e}^{6}}{4}-1}-\frac {i \left (3-{\mathrm e}^{3}\right ) \sqrt {\pi }\, {\mathrm e}^{\frac {{\mathrm e}^{6}}{4}-1-\frac {\left (3-{\mathrm e}^{3}\right )^{2}}{4}} \erf \left (i x +\frac {i \left (3-{\mathrm e}^{3}\right )}{2}\right )}{2}\)	\(143\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x-exp(1/4*exp(6)-x*exp(3)+x^2+3*x-1)

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maxima [A]  time = 0.88, size = 22, normalized size = 0.92 \begin {gather*} x - e^{\left (x^{2} - x e^{3} + 3 \, x + \frac {1}{4} \, e^{6} - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x - e^(x^2 - x*e^3 + 3*x + 1/4*e^6 - 1)

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mupad [B]  time = 0.18, size = 25, normalized size = 1.04 \begin {gather*} x-{\mathrm {e}}^{3\,x}\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{-1}\,{\mathrm {e}}^{-x\,{\mathrm {e}}^3}\,{\left ({\mathrm {e}}^{{\mathrm {e}}^6}\right )}^{1/4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1 - exp(3*x + exp(6)/4 - x*exp(3) + x^2 - 1)*(2*x - exp(3) + 3),x)

[Out]

x - exp(3*x)*exp(x^2)*exp(-1)*exp(-x*exp(3))*exp(exp(6))^(1/4)

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sympy [A]  time = 0.12, size = 20, normalized size = 0.83 \begin {gather*} x - e^{x^{2} - x e^{3} + 3 x - 1 + \frac {e^{6}}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x - exp(x**2 - x*exp(3) + 3*x - 1 + exp(6)/4)

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