3.15.47 \(\int e^x (-30 x-3 x^2+4 x^3) \, dx\)

Optimal. Leaf size=12 \[ e^x x^2 (-15+4 x) \]

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Rubi [A]  time = 0.08, antiderivative size = 17, normalized size of antiderivative = 1.42, number of steps used = 12, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {1594, 2196, 2176, 2194} \begin {gather*} 4 e^x x^3-15 e^x x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^x*(-30*x - 3*x^2 + 4*x^3),x]

[Out]

-15*E^x*x^2 + 4*E^x*x^3

Rule 1594

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

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} &=\int e^x x \left (-30-3 x+4 x^2\right ) \, dx\\ &=\int \left (-30 e^x x-3 e^x x^2+4 e^x x^3\right ) \, dx\\ &=-\left (3 \int e^x x^2 \, dx\right )+4 \int e^x x^3 \, dx-30 \int e^x x \, dx\\ &=-30 e^x x-3 e^x x^2+4 e^x x^3+6 \int e^x x \, dx-12 \int e^x x^2 \, dx+30 \int e^x \, dx\\ &=30 e^x-24 e^x x-15 e^x x^2+4 e^x x^3-6 \int e^x \, dx+24 \int e^x x \, dx\\ &=24 e^x-15 e^x x^2+4 e^x x^3-24 \int e^x \, dx\\ &=-15 e^x x^2+4 e^x x^3\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 12, normalized size = 1.00 \begin {gather*} e^x x^2 (-15+4 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^x*(-30*x - 3*x^2 + 4*x^3),x]

[Out]

E^x*x^2*(-15 + 4*x)

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fricas [A]  time = 0.66, size = 14, normalized size = 1.17 \begin {gather*} {\left (4 \, x^{3} - 15 \, x^{2}\right )} e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^3-3*x^2-30*x)*exp(x),x, algorithm="fricas")

[Out]

(4*x^3 - 15*x^2)*e^x

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giac [A]  time = 0.34, size = 14, normalized size = 1.17 \begin {gather*} {\left (4 \, x^{3} - 15 \, x^{2}\right )} e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^3-3*x^2-30*x)*exp(x),x, algorithm="giac")

[Out]

(4*x^3 - 15*x^2)*e^x

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maple [A]  time = 0.03, size = 12, normalized size = 1.00

method	result	size
gosper	\(x^{2} \left (4 x -15\right ) {\mathrm e}^{x}\)	\(12\)
risch	\(\left (4 x^{3}-15 x^{2}\right ) {\mathrm e}^{x}\)	\(15\)
default	\(-15 \,{\mathrm e}^{x} x^{2}+4 \,{\mathrm e}^{x} x^{3}\)	\(16\)
norman	\(-15 \,{\mathrm e}^{x} x^{2}+4 \,{\mathrm e}^{x} x^{3}\)	\(16\)
meijerg	\(-\left (-4 x^{3}+12 x^{2}-24 x +24\right ) {\mathrm e}^{x}-\left (3 x^{2}-6 x +6\right ) {\mathrm e}^{x}+15 \left (-2 x +2\right ) {\mathrm e}^{x}\)	\(44\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x^3-3*x^2-30*x)*exp(x),x,method=_RETURNVERBOSE)

[Out]

x^2*(4*x-15)*exp(x)

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maxima [B]  time = 0.40, size = 37, normalized size = 3.08 \begin {gather*} 4 \, {\left (x^{3} - 3 \, x^{2} + 6 \, x - 6\right )} e^{x} - 3 \, {\left (x^{2} - 2 \, x + 2\right )} e^{x} - 30 \, {\left (x - 1\right )} e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^3-3*x^2-30*x)*exp(x),x, algorithm="maxima")

[Out]

4*(x^3 - 3*x^2 + 6*x - 6)*e^x - 3*(x^2 - 2*x + 2)*e^x - 30*(x - 1)*e^x

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mupad [B]  time = 0.03, size = 11, normalized size = 0.92 \begin {gather*} x^2\,{\mathrm {e}}^x\,\left (4\,x-15\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(x)*(30*x + 3*x^2 - 4*x^3),x)

[Out]

x^2*exp(x)*(4*x - 15)

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sympy [A]  time = 0.09, size = 12, normalized size = 1.00 \begin {gather*} \left (4 x^{3} - 15 x^{2}\right ) e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x**3-3*x**2-30*x)*exp(x),x)

[Out]

(4*x**3 - 15*x**2)*exp(x)

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