3.91.48 \(\int \frac {1}{5} (5+e^3+e^{4-2 x+x^2} (2-2 x)-10 x) \, dx\)

Optimal. Leaf size=28 \[ -5+x-x^2+\frac {1}{5} e^3 \left (-e^{(1-x)^2}+x\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 30, normalized size of antiderivative = 1.07, number of steps used = 3, number of rules used = 2, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {12, 2236} \begin {gather*} -x^2-\frac {1}{5} e^{x^2-2 x+4}+\frac {1}{5} \left (5+e^3\right ) x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5 + E^3 + E^(4 - 2*x + x^2)*(2 - 2*x) - 10*x)/5,x]

[Out]

-1/5*E^(4 - 2*x + x^2) + ((5 + E^3)*x)/5 - x^2

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \left (5+e^3+e^{4-2 x+x^2} (2-2 x)-10 x\right ) \, dx\\ &=\frac {1}{5} \left (5+e^3\right ) x-x^2+\frac {1}{5} \int e^{4-2 x+x^2} (2-2 x) \, dx\\ &=-\frac {1}{5} e^{4-2 x+x^2}+\frac {1}{5} \left (5+e^3\right ) x-x^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 30, normalized size = 1.07 \begin {gather*} \frac {1}{5} \left (-e^{4-2 x+x^2}+5 x+e^3 x-5 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5 + E^3 + E^(4 - 2*x + x^2)*(2 - 2*x) - 10*x)/5,x]

[Out]

(-E^(4 - 2*x + x^2) + 5*x + E^3*x - 5*x^2)/5

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fricas [A]  time = 0.55, size = 23, normalized size = 0.82 \begin {gather*} -x^{2} + \frac {1}{5} \, x e^{3} + x - \frac {1}{5} \, e^{\left (x^{2} - 2 \, x + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-2*x+2)*exp(3)*exp(x^2-2*x+1)+1/5*exp(3)-2*x+1,x, algorithm="fricas")

[Out]

-x^2 + 1/5*x*e^3 + x - 1/5*e^(x^2 - 2*x + 4)

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giac [A]  time = 0.23, size = 23, normalized size = 0.82 \begin {gather*} -x^{2} + \frac {1}{5} \, x e^{3} + x - \frac {1}{5} \, e^{\left (x^{2} - 2 \, x + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-2*x+2)*exp(3)*exp(x^2-2*x+1)+1/5*exp(3)-2*x+1,x, algorithm="giac")

[Out]

-x^2 + 1/5*x*e^3 + x - 1/5*e^(x^2 - 2*x + 4)

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maple [A]  time = 0.04, size = 24, normalized size = 0.86

method	result	size
risch	\(-x^{2}+x -\frac {{\mathrm e}^{x^{2}-2 x +4}}{5}+\frac {x \,{\mathrm e}^{3}}{5}\)	\(24\)
default	\(-x^{2}+x -\frac {{\mathrm e}^{3} {\mathrm e}^{x^{2}-2 x +1}}{5}+\frac {x \,{\mathrm e}^{3}}{5}\)	\(26\)
norman	\(\left (\frac {{\mathrm e}^{3}}{5}+1\right ) x -x^{2}-\frac {{\mathrm e}^{3} {\mathrm e}^{x^{2}-2 x +1}}{5}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(-2*x+2)*exp(3)*exp(x^2-2*x+1)+1/5*exp(3)-2*x+1,x,method=_RETURNVERBOSE)

[Out]

-x^2+x-1/5*exp(x^2-2*x+4)+1/5*x*exp(3)

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maxima [A]  time = 0.35, size = 23, normalized size = 0.82 \begin {gather*} -x^{2} + \frac {1}{5} \, x e^{3} + x - \frac {1}{5} \, e^{\left (x^{2} - 2 \, x + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-2*x+2)*exp(3)*exp(x^2-2*x+1)+1/5*exp(3)-2*x+1,x, algorithm="maxima")

[Out]

-x^2 + 1/5*x*e^3 + x - 1/5*e^(x^2 - 2*x + 4)

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mupad [B]  time = 8.80, size = 25, normalized size = 0.89 \begin {gather*} x\,\left (\frac {{\mathrm {e}}^3}{5}+1\right )-x^2-\frac {{\mathrm {e}}^{x^2-2\,x+4}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(3)/5 - 2*x - (exp(3)*exp(x^2 - 2*x + 1)*(2*x - 2))/5 + 1,x)

[Out]

x*(exp(3)/5 + 1) - x^2 - exp(x^2 - 2*x + 4)/5

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sympy [A]  time = 0.13, size = 26, normalized size = 0.93 \begin {gather*} - x^{2} + x \left (1 + \frac {e^{3}}{5}\right ) - \frac {e^{3} e^{x^{2} - 2 x + 1}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-2*x+2)*exp(3)*exp(x**2-2*x+1)+1/5*exp(3)-2*x+1,x)

[Out]

-x**2 + x*(1 + exp(3)/5) - exp(3)*exp(x**2 - 2*x + 1)/5

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