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

Optimal. Leaf size=22 \[ -5-\frac {x}{3}-4 e^x x+x \left (-\frac {2}{x}+x\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {12, 2176, 2194} \begin {gather*} x^2-\frac {x}{3}+4 e^x-4 e^x (x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

4*E^x - x/3 + x^2 - 4*E^x*(1 + x)

Rule 12

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

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

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Mathematica [A]  time = 0.01, size = 15, normalized size = 0.68 \begin {gather*} -\frac {x}{3}-4 e^x x+x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*x - 4*E^x*x + x^2

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fricas [A]  time = 0.76, size = 12, normalized size = 0.55 \begin {gather*} x^{2} - 4 \, x e^{x} - \frac {1}{3} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-12*x-12)*exp(x)+2*x-1/3,x, algorithm="fricas")

[Out]

x^2 - 4*x*e^x - 1/3*x

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giac [A]  time = 0.35, size = 12, normalized size = 0.55 \begin {gather*} x^{2} - 4 \, x e^{x} - \frac {1}{3} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-12*x-12)*exp(x)+2*x-1/3,x, algorithm="giac")

[Out]

x^2 - 4*x*e^x - 1/3*x

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maple [A]  time = 0.01, size = 13, normalized size = 0.59

method	result	size
default	\(-\frac {x}{3}+x^{2}-4 \,{\mathrm e}^{x} x\)	\(13\)
norman	\(-\frac {x}{3}+x^{2}-4 \,{\mathrm e}^{x} x\)	\(13\)
risch	\(-\frac {x}{3}+x^{2}-4 \,{\mathrm e}^{x} x\)	\(13\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*(-12*x-12)*exp(x)+2*x-1/3,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.49, size = 12, normalized size = 0.55 \begin {gather*} x^{2} - 4 \, x e^{x} - \frac {1}{3} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-12*x-12)*exp(x)+2*x-1/3,x, algorithm="maxima")

[Out]

x^2 - 4*x*e^x - 1/3*x

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mupad [B]  time = 0.04, size = 12, normalized size = 0.55 \begin {gather*} -\frac {x\,\left (12\,{\mathrm {e}}^x-3\,x+1\right )}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2*x - (exp(x)*(12*x + 12))/3 - 1/3,x)

[Out]

-(x*(12*exp(x) - 3*x + 1))/3

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sympy [A]  time = 0.09, size = 12, normalized size = 0.55 \begin {gather*} x^{2} - 4 x e^{x} - \frac {x}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-12*x-12)*exp(x)+2*x-1/3,x)

[Out]

x**2 - 4*x*exp(x) - x/3

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