3.6.38 \(\int \frac {1}{4} (47 x+24 x^2+e^{5-x} (96 x^2-32 x^3)) \, dx\)

Optimal. Leaf size=20 \[ 2 x^2 \left (\frac {47}{16}+x+4 e^{5-x} x\right ) \]

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Rubi [A]  time = 0.12, antiderivative size = 25, normalized size of antiderivative = 1.25, number of steps used = 12, number of rules used = 5, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {12, 1593, 2196, 2176, 2194} \begin {gather*} 8 e^{5-x} x^3+2 x^3+\frac {47 x^2}{8} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(47*x + 24*x^2 + E^(5 - x)*(96*x^2 - 32*x^3))/4,x]

[Out]

(47*x^2)/8 + 2*x^3 + 8*E^(5 - x)*x^3

Rule 12

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

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 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} &=\frac {1}{4} \int \left (47 x+24 x^2+e^{5-x} \left (96 x^2-32 x^3\right )\right ) \, dx\\ &=\frac {47 x^2}{8}+2 x^3+\frac {1}{4} \int e^{5-x} \left (96 x^2-32 x^3\right ) \, dx\\ &=\frac {47 x^2}{8}+2 x^3+\frac {1}{4} \int e^{5-x} (96-32 x) x^2 \, dx\\ &=\frac {47 x^2}{8}+2 x^3+\frac {1}{4} \int \left (96 e^{5-x} x^2-32 e^{5-x} x^3\right ) \, dx\\ &=\frac {47 x^2}{8}+2 x^3-8 \int e^{5-x} x^3 \, dx+24 \int e^{5-x} x^2 \, dx\\ &=\frac {47 x^2}{8}-24 e^{5-x} x^2+2 x^3+8 e^{5-x} x^3-24 \int e^{5-x} x^2 \, dx+48 \int e^{5-x} x \, dx\\ &=-48 e^{5-x} x+\frac {47 x^2}{8}+2 x^3+8 e^{5-x} x^3+48 \int e^{5-x} \, dx-48 \int e^{5-x} x \, dx\\ &=-48 e^{5-x}+\frac {47 x^2}{8}+2 x^3+8 e^{5-x} x^3-48 \int e^{5-x} \, dx\\ &=\frac {47 x^2}{8}+2 x^3+8 e^{5-x} x^3\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 22, normalized size = 1.10 \begin {gather*} \frac {1}{8} x^2 \left (47+\left (16+64 e^{5-x}\right ) x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(47*x + 24*x^2 + E^(5 - x)*(96*x^2 - 32*x^3))/4,x]

[Out]

(x^2*(47 + (16 + 64*E^(5 - x))*x))/8

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fricas [A]  time = 0.62, size = 22, normalized size = 1.10 \begin {gather*} 8 \, x^{3} e^{\left (-x + 5\right )} + 2 \, x^{3} + \frac {47}{8} \, x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-32*x^3+96*x^2)*exp(5-x)+6*x^2+47/4*x,x, algorithm="fricas")

[Out]

8*x^3*e^(-x + 5) + 2*x^3 + 47/8*x^2

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giac [A]  time = 0.31, size = 22, normalized size = 1.10 \begin {gather*} 8 \, x^{3} e^{\left (-x + 5\right )} + 2 \, x^{3} + \frac {47}{8} \, x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-32*x^3+96*x^2)*exp(5-x)+6*x^2+47/4*x,x, algorithm="giac")

[Out]

8*x^3*e^(-x + 5) + 2*x^3 + 47/8*x^2

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maple [A]  time = 0.04, size = 23, normalized size = 1.15

method	result	size
norman	\(\frac {47 x^{2}}{8}+2 x^{3}+8 \,{\mathrm e}^{5-x} x^{3}\)	\(23\)
risch	\(\frac {47 x^{2}}{8}+2 x^{3}+8 \,{\mathrm e}^{5-x} x^{3}\)	\(23\)
default	\(\frac {47 x^{2}}{8}+2 x^{3}-8 \,{\mathrm e}^{5-x} \left (5-x \right )^{3}+120 \,{\mathrm e}^{5-x} \left (5-x \right )^{2}-600 \,{\mathrm e}^{5-x} \left (5-x \right )+1000 \,{\mathrm e}^{5-x}\)	\(63\)
derivativedivides	\(-\frac {1175}{4}+\frac {235 x}{4}+\frac {47 \left (5-x \right )^{2}}{8}+2 x^{3}-8 \,{\mathrm e}^{5-x} \left (5-x \right )^{3}+120 \,{\mathrm e}^{5-x} \left (5-x \right )^{2}-600 \,{\mathrm e}^{5-x} \left (5-x \right )+1000 \,{\mathrm e}^{5-x}\)	\(71\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*(-32*x^3+96*x^2)*exp(5-x)+6*x^2+47/4*x,x,method=_RETURNVERBOSE)

[Out]

47/8*x^2+2*x^3+8*exp(5-x)*x^3

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maxima [A]  time = 0.47, size = 22, normalized size = 1.10 \begin {gather*} 8 \, x^{3} e^{\left (-x + 5\right )} + 2 \, x^{3} + \frac {47}{8} \, x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-32*x^3+96*x^2)*exp(5-x)+6*x^2+47/4*x,x, algorithm="maxima")

[Out]

8*x^3*e^(-x + 5) + 2*x^3 + 47/8*x^2

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mupad [B]  time = 0.47, size = 18, normalized size = 0.90 \begin {gather*} x^2\,\left (2\,x+8\,x\,{\mathrm {e}}^{5-x}+\frac {47}{8}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((47*x)/4 + (exp(5 - x)*(96*x^2 - 32*x^3))/4 + 6*x^2,x)

[Out]

x^2*(2*x + 8*x*exp(5 - x) + 47/8)

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sympy [A]  time = 0.10, size = 20, normalized size = 1.00 \begin {gather*} 8 x^{3} e^{5 - x} + 2 x^{3} + \frac {47 x^{2}}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-32*x**3+96*x**2)*exp(5-x)+6*x**2+47/4*x,x)

[Out]

8*x**3*exp(5 - x) + 2*x**3 + 47*x**2/8

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