3.691 \(\int e^{3 x} (-x+x^2) \, dx\)

Optimal. Leaf size=32 \[ \frac{1}{3} e^{3 x} x^2-\frac{5}{9} e^{3 x} x+\frac{5 e^{3 x}}{27} \]

[Out]

(5*E^(3*x))/27 - (5*E^(3*x)*x)/9 + (E^(3*x)*x^2)/3

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Rubi [A]  time = 0.0506393, antiderivative size = 32, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {1593, 2196, 2176, 2194} \[ \frac{1}{3} e^{3 x} x^2-\frac{5}{9} e^{3 x} x+\frac{5 e^{3 x}}{27} \]

Antiderivative was successfully verified.

[In]

Int[E^(3*x)*(-x + x^2),x]

[Out]

(5*E^(3*x))/27 - (5*E^(3*x)*x)/9 + (E^(3*x)*x^2)/3

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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

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{align*} \int e^{3 x} \left (-x+x^2\right ) \, dx &=\int e^{3 x} (-1+x) x \, dx\\ &=\int \left (-e^{3 x} x+e^{3 x} x^2\right ) \, dx\\ &=-\int e^{3 x} x \, dx+\int e^{3 x} x^2 \, dx\\ &=-\frac{1}{3} e^{3 x} x+\frac{1}{3} e^{3 x} x^2+\frac{1}{3} \int e^{3 x} \, dx-\frac{2}{3} \int e^{3 x} x \, dx\\ &=\frac{e^{3 x}}{9}-\frac{5}{9} e^{3 x} x+\frac{1}{3} e^{3 x} x^2+\frac{2}{9} \int e^{3 x} \, dx\\ &=\frac{5 e^{3 x}}{27}-\frac{5}{9} e^{3 x} x+\frac{1}{3} e^{3 x} x^2\\ \end{align*}

Mathematica [A]  time = 0.0284713, size = 19, normalized size = 0.59 \[ \frac{1}{27} e^{3 x} \left (9 x^2-15 x+5\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[E^(3*x)*(-x + x^2),x]

[Out]

(E^(3*x)*(5 - 15*x + 9*x^2))/27

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Maple [A]  time = 0.022, size = 17, normalized size = 0.5 \begin{align*}{\frac{{{\rm e}^{3\,x}} \left ( 9\,{x}^{2}-15\,x+5 \right ) }{27}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(3*x)*(x^2-x),x)

[Out]

1/27*exp(3*x)*(9*x^2-15*x+5)

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Maxima [A]  time = 0.975857, size = 38, normalized size = 1.19 \begin{align*} \frac{1}{27} \,{\left (9 \, x^{2} - 6 \, x + 2\right )} e^{\left (3 \, x\right )} - \frac{1}{9} \,{\left (3 \, x - 1\right )} e^{\left (3 \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(3*x)*(x^2-x),x, algorithm="maxima")

[Out]

1/27*(9*x^2 - 6*x + 2)*e^(3*x) - 1/9*(3*x - 1)*e^(3*x)

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Fricas [A]  time = 0.808133, size = 45, normalized size = 1.41 \begin{align*} \frac{1}{27} \,{\left (9 \, x^{2} - 15 \, x + 5\right )} e^{\left (3 \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(3*x)*(x^2-x),x, algorithm="fricas")

[Out]

1/27*(9*x^2 - 15*x + 5)*e^(3*x)

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Sympy [A]  time = 0.096386, size = 15, normalized size = 0.47 \begin{align*} \frac{\left (9 x^{2} - 15 x + 5\right ) e^{3 x}}{27} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(3*x)*(x**2-x),x)

[Out]

(9*x**2 - 15*x + 5)*exp(3*x)/27

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Giac [A]  time = 1.2162, size = 22, normalized size = 0.69 \begin{align*} \frac{1}{27} \,{\left (9 \, x^{2} - 15 \, x + 5\right )} e^{\left (3 \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(3*x)*(x^2-x),x, algorithm="giac")

[Out]

1/27*(9*x^2 - 15*x + 5)*e^(3*x)