3.64.1 \(\int (2 e^{5+2 x}+e^x (-1-2 x-7 x^2-6 x^3-x^4)) \, dx\)

Optimal. Leaf size=22 \[ -6+e^x \left (-1+e^{5+x}-\left (x+x^2\right )^2\right ) \]

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Rubi [A]  time = 0.12, antiderivative size = 37, normalized size of antiderivative = 1.68, number of steps used = 19, number of rules used = 3, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {2194, 2196, 2176} \begin {gather*} -e^x x^4-2 e^x x^3-e^x x^2-e^x+e^{2 x+5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[2*E^(5 + 2*x) + E^x*(-1 - 2*x - 7*x^2 - 6*x^3 - x^4),x]

[Out]

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

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]

Rule 2196

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !$UseGamma === True

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=2 \int e^{5+2 x} \, dx+\int e^x \left (-1-2 x-7 x^2-6 x^3-x^4\right ) \, dx\\ &=e^{5+2 x}+\int \left (-e^x-2 e^x x-7 e^x x^2-6 e^x x^3-e^x x^4\right ) \, dx\\ &=e^{5+2 x}-2 \int e^x x \, dx-6 \int e^x x^3 \, dx-7 \int e^x x^2 \, dx-\int e^x \, dx-\int e^x x^4 \, dx\\ &=-e^x+e^{5+2 x}-2 e^x x-7 e^x x^2-6 e^x x^3-e^x x^4+2 \int e^x \, dx+4 \int e^x x^3 \, dx+14 \int e^x x \, dx+18 \int e^x x^2 \, dx\\ &=e^x+e^{5+2 x}+12 e^x x+11 e^x x^2-2 e^x x^3-e^x x^4-12 \int e^x x^2 \, dx-14 \int e^x \, dx-36 \int e^x x \, dx\\ &=-13 e^x+e^{5+2 x}-24 e^x x-e^x x^2-2 e^x x^3-e^x x^4+24 \int e^x x \, dx+36 \int e^x \, dx\\ &=23 e^x+e^{5+2 x}-e^x x^2-2 e^x x^3-e^x x^4-24 \int e^x \, dx\\ &=-e^x+e^{5+2 x}-e^x x^2-2 e^x x^3-e^x x^4\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 26, normalized size = 1.18 \begin {gather*} e^x \left (-1+e^{5+x}-x^2-2 x^3-x^4\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[2*E^(5 + 2*x) + E^x*(-1 - 2*x - 7*x^2 - 6*x^3 - x^4),x]

[Out]

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

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fricas [A]  time = 0.70, size = 24, normalized size = 1.09 \begin {gather*} -{\left (x^{4} + 2 \, x^{3} + x^{2} + 1\right )} e^{x} + e^{\left (2 \, x + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*exp(x)*exp(5+x)+(-x^4-6*x^3-7*x^2-2*x-1)*exp(x),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.15, size = 24, normalized size = 1.09 \begin {gather*} -{\left (x^{4} + 2 \, x^{3} + x^{2} + 1\right )} e^{x} + e^{\left (2 \, x + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*exp(x)*exp(5+x)+(-x^4-6*x^3-7*x^2-2*x-1)*exp(x),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 28, normalized size = 1.27

method	result	size
risch	\({\mathrm e}^{5+2 x}+\left (-x^{4}-2 x^{3}-x^{2}-1\right ) {\mathrm e}^{x}\)	\(28\)
default	\(-{\mathrm e}^{x} x^{2}-2 \,{\mathrm e}^{x} x^{3}-{\mathrm e}^{x} x^{4}-{\mathrm e}^{x}+{\mathrm e}^{5} {\mathrm e}^{2 x}\)	\(34\)
norman	\(-{\mathrm e}^{x} x^{2}-2 \,{\mathrm e}^{x} x^{3}-{\mathrm e}^{x} x^{4}-{\mathrm e}^{x}+{\mathrm e}^{5} {\mathrm e}^{2 x}\)	\(34\)
meijerg	\(-{\mathrm e}^{5} \left (1-{\mathrm e}^{2 x}\right )-\frac {\left (5 x^{4}-20 x^{3}+60 x^{2}-120 x +120\right ) {\mathrm e}^{x}}{5}+\frac {3 \left (-4 x^{3}+12 x^{2}-24 x +24\right ) {\mathrm e}^{x}}{2}-\frac {7 \left (3 x^{2}-6 x +6\right ) {\mathrm e}^{x}}{3}+\left (-2 x +2\right ) {\mathrm e}^{x}+1-{\mathrm e}^{x}\)	\(84\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2*exp(x)*exp(5+x)+(-x^4-6*x^3-7*x^2-2*x-1)*exp(x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.36, size = 69, normalized size = 3.14 \begin {gather*} -{\left (x^{4} - 4 \, x^{3} + 12 \, x^{2} - 24 \, x + 24\right )} e^{x} - 6 \, {\left (x^{3} - 3 \, x^{2} + 6 \, x - 6\right )} e^{x} - 7 \, {\left (x^{2} - 2 \, x + 2\right )} e^{x} - 2 \, {\left (x - 1\right )} e^{x} + e^{\left (2 \, x + 5\right )} - e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*exp(x)*exp(5+x)+(-x^4-6*x^3-7*x^2-2*x-1)*exp(x),x, algorithm="maxima")

[Out]

-(x^4 - 4*x^3 + 12*x^2 - 24*x + 24)*e^x - 6*(x^3 - 3*x^2 + 6*x - 6)*e^x - 7*(x^2 - 2*x + 2)*e^x - 2*(x - 1)*e^
x + e^(2*x + 5) - e^x

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mupad [B]  time = 0.08, size = 23, normalized size = 1.05 \begin {gather*} -{\mathrm {e}}^x\,\left (x^2-{\mathrm {e}}^{x+5}+2\,x^3+x^4+1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2*exp(x + 5)*exp(x) - exp(x)*(2*x + 7*x^2 + 6*x^3 + x^4 + 1),x)

[Out]

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

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sympy [A]  time = 0.12, size = 26, normalized size = 1.18 \begin {gather*} \left (- x^{4} - 2 x^{3} - x^{2} - 1\right ) e^{x} + e^{5} e^{2 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*exp(x)*exp(5+x)+(-x**4-6*x**3-7*x**2-2*x-1)*exp(x),x)

[Out]

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

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