3.53.95 \(\int (-25+e^{e^{-x+x^2}} (-5+e^{-x+x^2} (-10+25 x-10 x^2))) \, dx\)

Optimal. Leaf size=24 \[ \frac {5}{2}-5 \left (5+e^{e^{-x+x^2}}\right ) (-2+x)+\log (3) \]

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Rubi [A]  time = 0.04, antiderivative size = 34, normalized size of antiderivative = 1.42, number of steps used = 2, number of rules used = 1, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.028, Rules used = {2288} \begin {gather*} \frac {5 e^{e^{x^2-x}} \left (2 x^2-5 x+2\right )}{1-2 x}-25 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[-25 + E^E^(-x + x^2)*(-5 + E^(-x + x^2)*(-10 + 25*x - 10*x^2)),x]

[Out]

-25*x + (5*E^E^(-x + x^2)*(2 - 5*x + 2*x^2))/(1 - 2*x)

Rule 2288

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 {gather*} \begin {aligned} \text {integral} &=-25 x+\int e^{e^{-x+x^2}} \left (-5+e^{-x+x^2} \left (-10+25 x-10 x^2\right )\right ) \, dx\\ &=-25 x+\frac {5 e^{e^{-x+x^2}} \left (2-5 x+2 x^2\right )}{1-2 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 19, normalized size = 0.79 \begin {gather*} -5 \left (e^{e^{(-1+x) x}} (-2+x)+5 x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[-25 + E^E^(-x + x^2)*(-5 + E^(-x + x^2)*(-10 + 25*x - 10*x^2)),x]

[Out]

-5*(E^E^((-1 + x)*x)*(-2 + x) + 5*x)

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fricas [A]  time = 0.51, size = 18, normalized size = 0.75 \begin {gather*} -5 \, {\left (x - 2\right )} e^{\left (e^{\left (x^{2} - x\right )}\right )} - 25 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-10*x^2+25*x-10)*exp(x^2-x)-5)*exp(exp(x^2-x))-25,x, algorithm="fricas")

[Out]

-5*(x - 2)*e^(e^(x^2 - x)) - 25*x

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giac [B]  time = 0.20, size = 51, normalized size = 2.12 \begin {gather*} -5 \, {\left (x e^{\left (x^{2} - x + e^{\left (x^{2} - x\right )}\right )} - 2 \, e^{\left (x^{2} - x + e^{\left (x^{2} - x\right )}\right )}\right )} e^{\left (-x^{2} + x\right )} - 25 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-10*x^2+25*x-10)*exp(x^2-x)-5)*exp(exp(x^2-x))-25,x, algorithm="giac")

[Out]

-5*(x*e^(x^2 - x + e^(x^2 - x)) - 2*e^(x^2 - x + e^(x^2 - x)))*e^(-x^2 + x) - 25*x

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maple [A]  time = 0.06, size = 18, normalized size = 0.75




method result size



risch \(\left (-5 x +10\right ) {\mathrm e}^{{\mathrm e}^{x \left (x -1\right )}}-25 x\) \(18\)
default \(-25 x -5 x \,{\mathrm e}^{{\mathrm e}^{x^{2}-x}}+10 \,{\mathrm e}^{{\mathrm e}^{x^{2}-x}}\) \(28\)
norman \(-25 x -5 x \,{\mathrm e}^{{\mathrm e}^{x^{2}-x}}+10 \,{\mathrm e}^{{\mathrm e}^{x^{2}-x}}\) \(28\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-10*x^2+25*x-10)*exp(x^2-x)-5)*exp(exp(x^2-x))-25,x,method=_RETURNVERBOSE)

[Out]

(-5*x+10)*exp(exp(x*(x-1)))-25*x

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maxima [A]  time = 0.41, size = 18, normalized size = 0.75 \begin {gather*} -5 \, {\left (x - 2\right )} e^{\left (e^{\left (x^{2} - x\right )}\right )} - 25 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-10*x^2+25*x-10)*exp(x^2-x)-5)*exp(exp(x^2-x))-25,x, algorithm="maxima")

[Out]

-5*(x - 2)*e^(e^(x^2 - x)) - 25*x

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mupad [B]  time = 3.48, size = 29, normalized size = 1.21 \begin {gather*} 10\,{\mathrm {e}}^{{\mathrm {e}}^{-x}\,{\mathrm {e}}^{x^2}}-25\,x-5\,x\,{\mathrm {e}}^{{\mathrm {e}}^{-x}\,{\mathrm {e}}^{x^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(- exp(exp(x^2 - x))*(exp(x^2 - x)*(10*x^2 - 25*x + 10) + 5) - 25,x)

[Out]

10*exp(exp(-x)*exp(x^2)) - 25*x - 5*x*exp(exp(-x)*exp(x^2))

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sympy [A]  time = 8.38, size = 15, normalized size = 0.62 \begin {gather*} - 25 x + \left (10 - 5 x\right ) e^{e^{x^{2} - x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-10*x**2+25*x-10)*exp(x**2-x)-5)*exp(exp(x**2-x))-25,x)

[Out]

-25*x + (10 - 5*x)*exp(exp(x**2 - x))

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