3.18.92 \(\int \frac {324 x^3+e^{4/x} (-324+162 x)+e^{2/x} (324 x-486 x^2)}{\sqrt [256]{e}} \, dx\)

Optimal. Leaf size=28 \[ \frac {81 x^2 \left (-e^{2/x}+x\right )^2}{\sqrt [256]{e}}+4 \log (4) \]

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Rubi [A]  time = 0.22, antiderivative size = 43, normalized size of antiderivative = 1.54, number of steps used = 19, number of rules used = 6, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {12, 2226, 2206, 2210, 2214, 1593} \begin {gather*} \frac {81 x^4}{\sqrt [256]{e}}-162 e^{\frac {2}{x}-\frac {1}{256}} x^3+81 e^{\frac {4}{x}-\frac {1}{256}} x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(324*x^3 + E^(4/x)*(-324 + 162*x) + E^(2/x)*(324*x - 486*x^2))/E^(1/256),x]

[Out]

81*E^(-1/256 + 4/x)*x^2 - 162*E^(-1/256 + 2/x)*x^3 + (81*x^4)/E^(1/256)

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 2206

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_)), x_Symbol] :> Simp[((c + d*x)*F^(a + b*(c + d*x)^n))/d, x]
- Dist[b*n*Log[F], Int[(c + d*x)^n*F^(a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[2/n]
&& ILtQ[n, 0]

Rule 2210

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> Simp[(F^a*ExpIntegralEi[
b*(c + d*x)^n*Log[F]])/(f*n), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[d*e - c*f, 0]

Rule 2214

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*F^(a + b*(c + d*x)^n))/(d*(m + 1)), x] - Dist[(b*n*Log[F])/(m + 1), Int[(c + d*x)^(m + n)*F^(a + b*(c +
d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[(2*(m + 1))/n] && LtQ[-4, (m + 1)/n, 5] && IntegerQ[n
] && ((GtQ[n, 0] && LtQ[m, -1]) || (GtQ[-n, 0] && LeQ[-n, m + 1]))

Rule 2226

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*(u_), x_Symbol] :> Int[ExpandLinearProduct[F^(a + b*(c + d*
x)^n), u, c, d, x], x] /; FreeQ[{F, a, b, c, d, n}, x] && PolynomialQ[u, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \left (324 x^3+e^{4/x} (-324+162 x)+e^{2/x} \left (324 x-486 x^2\right )\right ) \, dx}{\sqrt [256]{e}}\\ &=\frac {81 x^4}{\sqrt [256]{e}}+\frac {\int e^{4/x} (-324+162 x) \, dx}{\sqrt [256]{e}}+\frac {\int e^{2/x} \left (324 x-486 x^2\right ) \, dx}{\sqrt [256]{e}}\\ &=\frac {81 x^4}{\sqrt [256]{e}}+\frac {\int e^{2/x} (324-486 x) x \, dx}{\sqrt [256]{e}}+\frac {\int \left (-324 e^{4/x}+162 e^{4/x} x\right ) \, dx}{\sqrt [256]{e}}\\ &=\frac {81 x^4}{\sqrt [256]{e}}+\frac {\int \left (324 e^{2/x} x-486 e^{2/x} x^2\right ) \, dx}{\sqrt [256]{e}}+\frac {162 \int e^{4/x} x \, dx}{\sqrt [256]{e}}-\frac {324 \int e^{4/x} \, dx}{\sqrt [256]{e}}\\ &=-324 e^{-\frac {1}{256}+\frac {4}{x}} x+81 e^{-\frac {1}{256}+\frac {4}{x}} x^2+\frac {81 x^4}{\sqrt [256]{e}}+\frac {324 \int e^{4/x} \, dx}{\sqrt [256]{e}}+\frac {324 \int e^{2/x} x \, dx}{\sqrt [256]{e}}-\frac {486 \int e^{2/x} x^2 \, dx}{\sqrt [256]{e}}-\frac {1296 \int \frac {e^{4/x}}{x} \, dx}{\sqrt [256]{e}}\\ &=162 e^{-\frac {1}{256}+\frac {2}{x}} x^2+81 e^{-\frac {1}{256}+\frac {4}{x}} x^2-162 e^{-\frac {1}{256}+\frac {2}{x}} x^3+\frac {81 x^4}{\sqrt [256]{e}}+\frac {1296 \text {Ei}\left (\frac {4}{x}\right )}{\sqrt [256]{e}}+\frac {324 \int e^{2/x} \, dx}{\sqrt [256]{e}}-\frac {324 \int e^{2/x} x \, dx}{\sqrt [256]{e}}+\frac {1296 \int \frac {e^{4/x}}{x} \, dx}{\sqrt [256]{e}}\\ &=324 e^{-\frac {1}{256}+\frac {2}{x}} x+81 e^{-\frac {1}{256}+\frac {4}{x}} x^2-162 e^{-\frac {1}{256}+\frac {2}{x}} x^3+\frac {81 x^4}{\sqrt [256]{e}}-\frac {324 \int e^{2/x} \, dx}{\sqrt [256]{e}}+\frac {648 \int \frac {e^{2/x}}{x} \, dx}{\sqrt [256]{e}}\\ &=81 e^{-\frac {1}{256}+\frac {4}{x}} x^2-162 e^{-\frac {1}{256}+\frac {2}{x}} x^3+\frac {81 x^4}{\sqrt [256]{e}}-\frac {648 \text {Ei}\left (\frac {2}{x}\right )}{\sqrt [256]{e}}-\frac {648 \int \frac {e^{2/x}}{x} \, dx}{\sqrt [256]{e}}\\ &=81 e^{-\frac {1}{256}+\frac {4}{x}} x^2-162 e^{-\frac {1}{256}+\frac {2}{x}} x^3+\frac {81 x^4}{\sqrt [256]{e}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 23, normalized size = 0.82 \begin {gather*} \frac {81 \left (e^{2/x}-x\right )^2 x^2}{\sqrt [256]{e}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(324*x^3 + E^(4/x)*(-324 + 162*x) + E^(2/x)*(324*x - 486*x^2))/E^(1/256),x]

[Out]

(81*(E^(2/x) - x)^2*x^2)/E^(1/256)

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fricas [A]  time = 0.78, size = 29, normalized size = 1.04 \begin {gather*} 81 \, {\left (x^{4} - 2 \, x^{3} e^{\frac {2}{x}} + x^{2} e^{\frac {4}{x}}\right )} e^{\left (-\frac {1}{256}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((162*x-324)*exp(2/x)^2+(-486*x^2+324*x)*exp(2/x)+324*x^3)/exp(1/256),x, algorithm="fricas")

[Out]

81*(x^4 - 2*x^3*e^(2/x) + x^2*e^(4/x))*e^(-1/256)

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giac [A]  time = 0.21, size = 29, normalized size = 1.04 \begin {gather*} 81 \, {\left (x^{4} - 2 \, x^{3} e^{\frac {2}{x}} + x^{2} e^{\frac {4}{x}}\right )} e^{\left (-\frac {1}{256}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((162*x-324)*exp(2/x)^2+(-486*x^2+324*x)*exp(2/x)+324*x^3)/exp(1/256),x, algorithm="giac")

[Out]

81*(x^4 - 2*x^3*e^(2/x) + x^2*e^(4/x))*e^(-1/256)

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maple [A]  time = 0.22, size = 36, normalized size = 1.29

method	result	size
default	\({\mathrm e}^{-\frac {1}{256}} \left (81 x^{2} {\mathrm e}^{\frac {4}{x}}-162 x^{3} {\mathrm e}^{\frac {2}{x}}+81 x^{4}\right )\)	\(36\)
derivativedivides	\(-{\mathrm e}^{-\frac {1}{256}} \left (-81 x^{4}+162 x^{3} {\mathrm e}^{\frac {2}{x}}-81 x^{2} {\mathrm e}^{\frac {4}{x}}\right )\)	\(37\)
risch	\(81 \,{\mathrm e}^{-\frac {1}{256}} x^{4}+81 x^{2} {\mathrm e}^{-\frac {x -1024}{256 x}}-162 x^{3} {\mathrm e}^{-\frac {x -512}{256 x}}\)	\(37\)
norman	\(81 \,{\mathrm e}^{-\frac {1}{256}} x^{4}+81 \,{\mathrm e}^{-\frac {1}{256}} x^{2} {\mathrm e}^{\frac {4}{x}}-162 \,{\mathrm e}^{-\frac {1}{256}} x^{3} {\mathrm e}^{\frac {2}{x}}\)	\(43\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((162*x-324)*exp(2/x)^2+(-486*x^2+324*x)*exp(2/x)+324*x^3)/exp(1/256),x,method=_RETURNVERBOSE)

[Out]

1/exp(1/256)*(81*x^2*exp(2/x)^2-162*x^3*exp(2/x)+81*x^4)

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maxima [C]  time = 0.51, size = 36, normalized size = 1.29 \begin {gather*} 81 \, {\left (x^{4} + x^{2} e^{\frac {4}{x}} + 16 \, \Gamma \left (-2, -\frac {2}{x}\right ) + 48 \, \Gamma \left (-3, -\frac {2}{x}\right )\right )} e^{\left (-\frac {1}{256}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((162*x-324)*exp(2/x)^2+(-486*x^2+324*x)*exp(2/x)+324*x^3)/exp(1/256),x, algorithm="maxima")

[Out]

81*(x^4 + x^2*e^(4/x) + 16*gamma(-2, -2/x) + 48*gamma(-3, -2/x))*e^(-1/256)

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mupad [B]  time = 1.16, size = 19, normalized size = 0.68 \begin {gather*} 81\,x^2\,{\mathrm {e}}^{-\frac {1}{256}}\,{\left (x-{\mathrm {e}}^{2/x}\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(-1/256)*(exp(4/x)*(162*x - 324) + exp(2/x)*(324*x - 486*x^2) + 324*x^3),x)

[Out]

81*x^2*exp(-1/256)*(x - exp(2/x))^2

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sympy [A]  time = 0.16, size = 44, normalized size = 1.57 \begin {gather*} \frac {81 x^{4}}{e^{\frac {1}{256}}} + \frac {- 162 x^{3} e^{\frac {1}{256}} e^{\frac {2}{x}} + 81 x^{2} e^{\frac {1}{256}} e^{\frac {4}{x}}}{e^{\frac {1}{128}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((162*x-324)*exp(2/x)**2+(-486*x**2+324*x)*exp(2/x)+324*x**3)/exp(1/256),x)

[Out]

81*x**4*exp(-1/256) + (-162*x**3*exp(1/256)*exp(2/x) + 81*x**2*exp(1/256)*exp(4/x))*exp(-1/128)

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