3.38.1 \(\int e^{2 x} (243 x^2+594 x^3+396 x^4+72 x^5) \, dx\)

Optimal. Leaf size=20 \[ 9 e^{2 x} x^3 \left (3+\log \left (e^{2 x}\right )\right )^2 \]

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Rubi [A]  time = 0.18, antiderivative size = 31, normalized size of antiderivative = 1.55, number of steps used = 20, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2196, 2176, 2194} \begin {gather*} 36 e^{2 x} x^5+108 e^{2 x} x^4+81 e^{2 x} x^3 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(2*x)*(243*x^2 + 594*x^3 + 396*x^4 + 72*x^5),x]

[Out]

81*E^(2*x)*x^3 + 108*E^(2*x)*x^4 + 36*E^(2*x)*x^5

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 \left (243 e^{2 x} x^2+594 e^{2 x} x^3+396 e^{2 x} x^4+72 e^{2 x} x^5\right ) \, dx\\ &=72 \int e^{2 x} x^5 \, dx+243 \int e^{2 x} x^2 \, dx+396 \int e^{2 x} x^4 \, dx+594 \int e^{2 x} x^3 \, dx\\ &=\frac {243}{2} e^{2 x} x^2+297 e^{2 x} x^3+198 e^{2 x} x^4+36 e^{2 x} x^5-180 \int e^{2 x} x^4 \, dx-243 \int e^{2 x} x \, dx-792 \int e^{2 x} x^3 \, dx-891 \int e^{2 x} x^2 \, dx\\ &=-\frac {243}{2} e^{2 x} x-324 e^{2 x} x^2-99 e^{2 x} x^3+108 e^{2 x} x^4+36 e^{2 x} x^5+\frac {243}{2} \int e^{2 x} \, dx+360 \int e^{2 x} x^3 \, dx+891 \int e^{2 x} x \, dx+1188 \int e^{2 x} x^2 \, dx\\ &=\frac {243 e^{2 x}}{4}+324 e^{2 x} x+270 e^{2 x} x^2+81 e^{2 x} x^3+108 e^{2 x} x^4+36 e^{2 x} x^5-\frac {891}{2} \int e^{2 x} \, dx-540 \int e^{2 x} x^2 \, dx-1188 \int e^{2 x} x \, dx\\ &=-162 e^{2 x}-270 e^{2 x} x+81 e^{2 x} x^3+108 e^{2 x} x^4+36 e^{2 x} x^5+540 \int e^{2 x} x \, dx+594 \int e^{2 x} \, dx\\ &=135 e^{2 x}+81 e^{2 x} x^3+108 e^{2 x} x^4+36 e^{2 x} x^5-270 \int e^{2 x} \, dx\\ &=81 e^{2 x} x^3+108 e^{2 x} x^4+36 e^{2 x} x^5\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 17, normalized size = 0.85 \begin {gather*} 9 e^{2 x} x^3 (3+2 x)^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(2*x)*(243*x^2 + 594*x^3 + 396*x^4 + 72*x^5),x]

[Out]

9*E^(2*x)*x^3*(3 + 2*x)^2

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fricas [A]  time = 0.68, size = 22, normalized size = 1.10 \begin {gather*} 9 \, {\left (4 \, x^{5} + 12 \, x^{4} + 9 \, x^{3}\right )} e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((72*x^5+396*x^4+594*x^3+243*x^2)*exp(x)^2,x, algorithm="fricas")

[Out]

9*(4*x^5 + 12*x^4 + 9*x^3)*e^(2*x)

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giac [A]  time = 0.14, size = 22, normalized size = 1.10 \begin {gather*} 9 \, {\left (4 \, x^{5} + 12 \, x^{4} + 9 \, x^{3}\right )} e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((72*x^5+396*x^4+594*x^3+243*x^2)*exp(x)^2,x, algorithm="giac")

[Out]

9*(4*x^5 + 12*x^4 + 9*x^3)*e^(2*x)

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maple [A]  time = 0.03, size = 17, normalized size = 0.85

method	result	size
gosper	\(9 x^{3} \left (2 x +3\right )^{2} {\mathrm e}^{2 x}\)	\(17\)
risch	\(\left (36 x^{5}+108 x^{4}+81 x^{3}\right ) {\mathrm e}^{2 x}\)	\(22\)
default	\(36 x^{5} {\mathrm e}^{2 x}+108 \,{\mathrm e}^{2 x} x^{4}+81 \,{\mathrm e}^{2 x} x^{3}\)	\(29\)
norman	\(36 x^{5} {\mathrm e}^{2 x}+108 \,{\mathrm e}^{2 x} x^{4}+81 \,{\mathrm e}^{2 x} x^{3}\)	\(29\)
meijerg	\(-\frac {3 \left (-192 x^{5}+480 x^{4}-960 x^{3}+1440 x^{2}-1440 x +720\right ) {\mathrm e}^{2 x}}{16}+\frac {99 \left (80 x^{4}-160 x^{3}+240 x^{2}-240 x +120\right ) {\mathrm e}^{2 x}}{40}-\frac {297 \left (-32 x^{3}+48 x^{2}-48 x +24\right ) {\mathrm e}^{2 x}}{32}+\frac {81 \left (12 x^{2}-12 x +6\right ) {\mathrm e}^{2 x}}{8}\)	\(96\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((72*x^5+396*x^4+594*x^3+243*x^2)*exp(x)^2,x,method=_RETURNVERBOSE)

[Out]

9*x^3*(2*x+3)^2*exp(x)^2

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maxima [B]  time = 0.37, size = 95, normalized size = 4.75 \begin {gather*} 9 \, {\left (4 \, x^{5} - 10 \, x^{4} + 20 \, x^{3} - 30 \, x^{2} + 30 \, x - 15\right )} e^{\left (2 \, x\right )} + 99 \, {\left (2 \, x^{4} - 4 \, x^{3} + 6 \, x^{2} - 6 \, x + 3\right )} e^{\left (2 \, x\right )} + \frac {297}{4} \, {\left (4 \, x^{3} - 6 \, x^{2} + 6 \, x - 3\right )} e^{\left (2 \, x\right )} + \frac {243}{4} \, {\left (2 \, x^{2} - 2 \, x + 1\right )} e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((72*x^5+396*x^4+594*x^3+243*x^2)*exp(x)^2,x, algorithm="maxima")

[Out]

9*(4*x^5 - 10*x^4 + 20*x^3 - 30*x^2 + 30*x - 15)*e^(2*x) + 99*(2*x^4 - 4*x^3 + 6*x^2 - 6*x + 3)*e^(2*x) + 297/
4*(4*x^3 - 6*x^2 + 6*x - 3)*e^(2*x) + 243/4*(2*x^2 - 2*x + 1)*e^(2*x)

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mupad [B]  time = 2.23, size = 16, normalized size = 0.80 \begin {gather*} 9\,x^3\,{\mathrm {e}}^{2\,x}\,{\left (2\,x+3\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(2*x)*(243*x^2 + 594*x^3 + 396*x^4 + 72*x^5),x)

[Out]

9*x^3*exp(2*x)*(2*x + 3)^2

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sympy [A]  time = 0.10, size = 19, normalized size = 0.95 \begin {gather*} \left (36 x^{5} + 108 x^{4} + 81 x^{3}\right ) e^{2 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((72*x**5+396*x**4+594*x**3+243*x**2)*exp(x)**2,x)

[Out]

(36*x**5 + 108*x**4 + 81*x**3)*exp(2*x)

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