3.83.84 \(\int e^{5 x} (-480 x^2+2400 x^3+4000 x^4) \, dx\)

Optimal. Leaf size=15 \[ 32 e^{5 x} x^3 (-5+25 x) \]

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Rubi [A]  time = 0.15, antiderivative size = 21, normalized size of antiderivative = 1.40, number of steps used = 15, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {1594, 2196, 2176, 2194} \begin {gather*} 800 e^{5 x} x^4-160 e^{5 x} x^3 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(5*x)*(-480*x^2 + 2400*x^3 + 4000*x^4),x]

[Out]

-160*E^(5*x)*x^3 + 800*E^(5*x)*x^4

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^{5 x} x^2 \left (-480+2400 x+4000 x^2\right ) \, dx\\ &=\int \left (-480 e^{5 x} x^2+2400 e^{5 x} x^3+4000 e^{5 x} x^4\right ) \, dx\\ &=-\left (480 \int e^{5 x} x^2 \, dx\right )+2400 \int e^{5 x} x^3 \, dx+4000 \int e^{5 x} x^4 \, dx\\ &=-96 e^{5 x} x^2+480 e^{5 x} x^3+800 e^{5 x} x^4+192 \int e^{5 x} x \, dx-1440 \int e^{5 x} x^2 \, dx-3200 \int e^{5 x} x^3 \, dx\\ &=\frac {192}{5} e^{5 x} x-384 e^{5 x} x^2-160 e^{5 x} x^3+800 e^{5 x} x^4-\frac {192}{5} \int e^{5 x} \, dx+576 \int e^{5 x} x \, dx+1920 \int e^{5 x} x^2 \, dx\\ &=-\frac {192 e^{5 x}}{25}+\frac {768}{5} e^{5 x} x-160 e^{5 x} x^3+800 e^{5 x} x^4-\frac {576}{5} \int e^{5 x} \, dx-768 \int e^{5 x} x \, dx\\ &=-\frac {768 e^{5 x}}{25}-160 e^{5 x} x^3+800 e^{5 x} x^4+\frac {768}{5} \int e^{5 x} \, dx\\ &=-160 e^{5 x} x^3+800 e^{5 x} x^4\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 15, normalized size = 1.00 \begin {gather*} 160 e^{5 x} x^3 (-1+5 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(5*x)*(-480*x^2 + 2400*x^3 + 4000*x^4),x]

[Out]

160*E^(5*x)*x^3*(-1 + 5*x)

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fricas [A]  time = 0.55, size = 17, normalized size = 1.13 \begin {gather*} 160 \, {\left (5 \, x^{4} - x^{3}\right )} e^{\left (5 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4000*x^4+2400*x^3-480*x^2)*exp(5*x),x, algorithm="fricas")

[Out]

160*(5*x^4 - x^3)*e^(5*x)

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giac [A]  time = 0.39, size = 17, normalized size = 1.13 \begin {gather*} 160 \, {\left (5 \, x^{4} - x^{3}\right )} e^{\left (5 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4000*x^4+2400*x^3-480*x^2)*exp(5*x),x, algorithm="giac")

[Out]

160*(5*x^4 - x^3)*e^(5*x)

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maple [A]  time = 0.04, size = 15, normalized size = 1.00

method	result	size
gosper	\(160 \,{\mathrm e}^{5 x} \left (5 x -1\right ) x^{3}\)	\(15\)
risch	\(\left (800 x^{4}-160 x^{3}\right ) {\mathrm e}^{5 x}\)	\(17\)
derivativedivides	\(800 x^{4} {\mathrm e}^{5 x}-160 \,{\mathrm e}^{5 x} x^{3}\)	\(20\)
default	\(800 x^{4} {\mathrm e}^{5 x}-160 \,{\mathrm e}^{5 x} x^{3}\)	\(20\)
norman	\(800 x^{4} {\mathrm e}^{5 x}-160 \,{\mathrm e}^{5 x} x^{3}\)	\(20\)
meijerg	\(\frac {32 \left (3125 x^{4}-2500 x^{3}+1500 x^{2}-600 x +120\right ) {\mathrm e}^{5 x}}{125}-\frac {24 \left (-500 x^{3}+300 x^{2}-120 x +24\right ) {\mathrm e}^{5 x}}{25}-\frac {32 \left (75 x^{2}-30 x +6\right ) {\mathrm e}^{5 x}}{25}\)	\(65\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4000*x^4+2400*x^3-480*x^2)*exp(5*x),x,method=_RETURNVERBOSE)

[Out]

160*exp(5*x)*(5*x-1)*x^3

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maxima [B]  time = 0.38, size = 64, normalized size = 4.27 \begin {gather*} \frac {32}{25} \, {\left (625 \, x^{4} - 500 \, x^{3} + 300 \, x^{2} - 120 \, x + 24\right )} e^{\left (5 \, x\right )} + \frac {96}{25} \, {\left (125 \, x^{3} - 75 \, x^{2} + 30 \, x - 6\right )} e^{\left (5 \, x\right )} - \frac {96}{25} \, {\left (25 \, x^{2} - 10 \, x + 2\right )} e^{\left (5 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4000*x^4+2400*x^3-480*x^2)*exp(5*x),x, algorithm="maxima")

[Out]

32/25*(625*x^4 - 500*x^3 + 300*x^2 - 120*x + 24)*e^(5*x) + 96/25*(125*x^3 - 75*x^2 + 30*x - 6)*e^(5*x) - 96/25
*(25*x^2 - 10*x + 2)*e^(5*x)

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mupad [B]  time = 5.03, size = 14, normalized size = 0.93 \begin {gather*} 160\,x^3\,{\mathrm {e}}^{5\,x}\,\left (5\,x-1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(5*x)*(2400*x^3 - 480*x^2 + 4000*x^4),x)

[Out]

160*x^3*exp(5*x)*(5*x - 1)

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sympy [A]  time = 0.10, size = 14, normalized size = 0.93 \begin {gather*} \left (800 x^{4} - 160 x^{3}\right ) e^{5 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4000*x**4+2400*x**3-480*x**2)*exp(5*x),x)

[Out]

(800*x**4 - 160*x**3)*exp(5*x)

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