3.45.26 \(\int (2-2 e^{32+4 e^2-2 x}+2 x+2 e^{16+2 e^2-x} x) \, dx\)

Optimal. Leaf size=21 \[ 2+\left (1-e^{16+2 e^2-x}+x\right )^2 \]

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Rubi [B]  time = 0.03, antiderivative size = 51, normalized size of antiderivative = 2.43, number of steps used = 4, number of rules used = 2, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {2194, 2176} \begin {gather*} x^2-2 e^{2 \left (8+e^2\right )-x} x+2 x+e^{4 \left (8+e^2\right )-2 x}-2 e^{2 \left (8+e^2\right )-x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(4*(8 + E^2) - 2*x) - 2*E^(2*(8 + E^2) - x) + 2*x - 2*E^(2*(8 + E^2) - x)*x + x^2

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=2 x+x^2-2 \int e^{32+4 e^2-2 x} \, dx+2 \int e^{16+2 e^2-x} x \, dx\\ &=e^{4 \left (8+e^2\right )-2 x}+2 x-2 e^{2 \left (8+e^2\right )-x} x+x^2+2 \int e^{16+2 e^2-x} \, dx\\ &=e^{4 \left (8+e^2\right )-2 x}-2 e^{2 \left (8+e^2\right )-x}+2 x-2 e^{2 \left (8+e^2\right )-x} x+x^2\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.04, size = 57, normalized size = 2.71 \begin {gather*} 2 \left (\frac {1}{2} e^{32+4 e^2-2 x}+x+\frac {x^2}{2}+e^{-x} \left (-e^{16+2 e^2}-e^{16+2 e^2} x\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.55, size = 32, normalized size = 1.52 \begin {gather*} x^{2} - 2 \, {\left (x + 1\right )} e^{\left (-x + 2 \, e^{2} + 16\right )} + 2 \, x + e^{\left (-2 \, x + 4 \, e^{2} + 32\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-2*exp(16-x)^2*exp(exp(2))^4+2*x*exp(16-x)*exp(exp(2))^2+2*x+2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.12, size = 32, normalized size = 1.52 \begin {gather*} x^{2} - 2 \, {\left (x + 1\right )} e^{\left (-x + 2 \, e^{2} + 16\right )} + 2 \, x + e^{\left (-2 \, x + 4 \, e^{2} + 32\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-2*exp(16-x)^2*exp(exp(2))^4+2*x*exp(16-x)*exp(exp(2))^2+2*x+2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 35, normalized size = 1.67

method	result	size
risch	\({\mathrm e}^{-2 x +32+4 \,{\mathrm e}^{2}}+2 \left (-x -1\right ) {\mathrm e}^{2 \,{\mathrm e}^{2}+16-x}+x^{2}+2 x\)	\(35\)
norman	\(x^{2}+{\mathrm e}^{-2 x +32} {\mathrm e}^{4 \,{\mathrm e}^{2}}+2 x -2 \,{\mathrm e}^{2 \,{\mathrm e}^{2}} {\mathrm e}^{16-x}-2 x \,{\mathrm e}^{16-x} {\mathrm e}^{2 \,{\mathrm e}^{2}}\)	\(49\)
default	\(x^{2}+2 x +{\mathrm e}^{-2 x +32} {\mathrm e}^{4 \,{\mathrm e}^{2}}+2 \,{\mathrm e}^{2 \,{\mathrm e}^{2}} \left ({\mathrm e}^{16-x} \left (16-x \right )-17 \,{\mathrm e}^{16-x}\right )\)	\(50\)
derivativedivides	\(\left (16-x \right )^{2}-544+34 x +{\mathrm e}^{-2 x +32} {\mathrm e}^{4 \,{\mathrm e}^{2}}-2 \,{\mathrm e}^{2 \,{\mathrm e}^{2}} \left (-{\mathrm e}^{16-x} \left (16-x \right )+17 \,{\mathrm e}^{16-x}\right )\)	\(56\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-2*exp(16-x)^2*exp(exp(2))^4+2*x*exp(16-x)*exp(exp(2))^2+2*x+2,x,method=_RETURNVERBOSE)

[Out]

exp(-2*x+32+4*exp(2))+2*(-x-1)*exp(2*exp(2)+16-x)+x^2+2*x

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maxima [B]  time = 0.37, size = 40, normalized size = 1.90 \begin {gather*} x^{2} - 2 \, {\left (x e^{\left (2 \, e^{2} + 16\right )} + e^{\left (2 \, e^{2} + 16\right )}\right )} e^{\left (-x\right )} + 2 \, x + e^{\left (-2 \, x + 4 \, e^{2} + 32\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-2*exp(16-x)^2*exp(exp(2))^4+2*x*exp(16-x)*exp(exp(2))^2+2*x+2,x, algorithm="maxima")

[Out]

x^2 - 2*(x*e^(2*e^2 + 16) + e^(2*e^2 + 16))*e^(-x) + 2*x + e^(-2*x + 4*e^2 + 32)

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mupad [B]  time = 0.07, size = 42, normalized size = 2.00 \begin {gather*} 2\,x-2\,{\mathrm {e}}^{2\,{\mathrm {e}}^2-x+16}+{\mathrm {e}}^{4\,{\mathrm {e}}^2-2\,x+32}-2\,x\,{\mathrm {e}}^{2\,{\mathrm {e}}^2-x+16}+x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2*x - 2*exp(4*exp(2))*exp(32 - 2*x) + 2*x*exp(2*exp(2))*exp(16 - x) + 2,x)

[Out]

2*x - 2*exp(2*exp(2) - x + 16) + exp(4*exp(2) - 2*x + 32) - 2*x*exp(2*exp(2) - x + 16) + x^2

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sympy [B]  time = 0.12, size = 44, normalized size = 2.10 \begin {gather*} x^{2} + 2 x + \left (- 2 x e^{2 e^{2}} - 2 e^{2 e^{2}}\right ) e^{16 - x} + e^{32 - 2 x} e^{4 e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-2*exp(16-x)**2*exp(exp(2))**4+2*x*exp(16-x)*exp(exp(2))**2+2*x+2,x)

[Out]

x**2 + 2*x + (-2*x*exp(2*exp(2)) - 2*exp(2*exp(2)))*exp(16 - x) + exp(32 - 2*x)*exp(4*exp(2))

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