3.50.76 \(\int \frac {1}{3} (-6 x+e^{x/3} (6+x+e^4 (3+x)+e^2 (6+2 x))+e^{x/3} (3+x) \log (x)) \, dx\)

Optimal. Leaf size=24 \[ x \left (-x+e^{x/3} \left (\left (1+e^2\right )^2+\log (x)\right )\right ) \]

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Rubi [B]  time = 0.08, antiderivative size = 85, normalized size of antiderivative = 3.54, number of steps used = 8, number of rules used = 5, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.104, Rules used = {12, 2187, 2176, 2194, 2554} \begin {gather*} -x^2-3 e^{x/3}+e^{x/3} \left (\left (1+e^2\right )^2 x+3 \left (2+2 e^2+e^4\right )\right )-3 \left (1+e^2\right )^2 e^{x/3}-3 e^{x/3} \log (x)+e^{x/3} (x+3) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-6*x + E^(x/3)*(6 + x + E^4*(3 + x) + E^2*(6 + 2*x)) + E^(x/3)*(3 + x)*Log[x])/3,x]

[Out]

-3*E^(x/3) - 3*E^(x/3)*(1 + E^2)^2 - x^2 + E^(x/3)*(3*(2 + 2*E^2 + E^4) + (1 + E^2)^2*x) - 3*E^(x/3)*Log[x] +
E^(x/3)*(3 + x)*Log[x]

Rule 12

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

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 2187

Int[((a_.) + (b_.)*((F_)^((g_.)*(v_)))^(n_.))^(p_.)*(u_)^(m_.), x_Symbol] :> Int[NormalizePowerOfLinear[u, x]^
m*(a + b*(F^(g*ExpandToSum[v, x]))^n)^p, x] /; FreeQ[{F, a, b, g, n, p}, x] && LinearQ[v, x] && PowerOfLinearQ
[u, x] &&  !(LinearMatchQ[v, x] && PowerOfLinearMatchQ[u, x]) && IntegerQ[m]

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 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{3} \int \left (-6 x+e^{x/3} \left (6+x+e^4 (3+x)+e^2 (6+2 x)\right )+e^{x/3} (3+x) \log (x)\right ) \, dx\\ &=-x^2+\frac {1}{3} \int e^{x/3} \left (6+x+e^4 (3+x)+e^2 (6+2 x)\right ) \, dx+\frac {1}{3} \int e^{x/3} (3+x) \log (x) \, dx\\ &=-x^2-3 e^{x/3} \log (x)+e^{x/3} (3+x) \log (x)-\frac {1}{3} \int 3 e^{x/3} \, dx+\frac {1}{3} \int e^{x/3} \left (3 \left (2+2 e^2+e^4\right )+\left (1+e^2\right )^2 x\right ) \, dx\\ &=-x^2+e^{x/3} \left (3 \left (2+2 e^2+e^4\right )+\left (1+e^2\right )^2 x\right )-3 e^{x/3} \log (x)+e^{x/3} (3+x) \log (x)-\left (1+e^2\right )^2 \int e^{x/3} \, dx-\int e^{x/3} \, dx\\ &=-3 e^{x/3}-3 e^{x/3} \left (1+e^2\right )^2-x^2+e^{x/3} \left (3 \left (2+2 e^2+e^4\right )+\left (1+e^2\right )^2 x\right )-3 e^{x/3} \log (x)+e^{x/3} (3+x) \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.10, size = 51, normalized size = 2.12 \begin {gather*} \frac {1}{3} \left (-9 e^{x/3}-3 x^2+e^{x/3} \left (9+3 \left (1+e^2\right )^2 x\right )+3 e^{x/3} x \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-6*x + E^(x/3)*(6 + x + E^4*(3 + x) + E^2*(6 + 2*x)) + E^(x/3)*(3 + x)*Log[x])/3,x]

[Out]

(-9*E^(x/3) - 3*x^2 + E^(x/3)*(9 + 3*(1 + E^2)^2*x) + 3*E^(x/3)*x*Log[x])/3

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fricas [A]  time = 0.85, size = 30, normalized size = 1.25 \begin {gather*} x e^{\left (\frac {1}{3} \, x\right )} \log \relax (x) - x^{2} + {\left (x e^{4} + 2 \, x e^{2} + x\right )} e^{\left (\frac {1}{3} \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(3+x)*exp(1/3*x)*log(x)+1/3*((3+x)*exp(2)^2+(2*x+6)*exp(2)+x+6)*exp(1/3*x)-2*x,x, algorithm="fri
cas")

[Out]

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

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giac [B]  time = 0.13, size = 45, normalized size = 1.88 \begin {gather*} x e^{\left (\frac {1}{3} \, x\right )} \log \relax (x) - x^{2} + {\left (x + 3\right )} e^{\left (\frac {1}{3} \, x\right )} + x e^{\left (\frac {1}{3} \, x + 4\right )} + 2 \, x e^{\left (\frac {1}{3} \, x + 2\right )} - 3 \, e^{\left (\frac {1}{3} \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(3+x)*exp(1/3*x)*log(x)+1/3*((3+x)*exp(2)^2+(2*x+6)*exp(2)+x+6)*exp(1/3*x)-2*x,x, algorithm="gia
c")

[Out]

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

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maple [A]  time = 0.05, size = 31, normalized size = 1.29

method	result	size
norman	\(\left ({\mathrm e}^{4}+2 \,{\mathrm e}^{2}+1\right ) x \,{\mathrm e}^{\frac {x}{3}}+{\mathrm e}^{\frac {x}{3}} \ln \relax (x ) x -x^{2}\)	\(31\)
risch	\({\mathrm e}^{\frac {x}{3}} \ln \relax (x ) x -3 \,{\mathrm e}^{\frac {x}{3}}+\frac {\left (3 x \,{\mathrm e}^{4}+6 \,{\mathrm e}^{2} x +3 x +9\right ) {\mathrm e}^{\frac {x}{3}}}{3}-x^{2}\)	\(42\)
default	\(-x^{2}+x \,{\mathrm e}^{\frac {x}{3}}+6 \,{\mathrm e}^{2} {\mathrm e}^{\frac {x}{3}}+3 \,{\mathrm e}^{4} {\mathrm e}^{\frac {x}{3}}+6 \,{\mathrm e}^{2} \left (\frac {x \,{\mathrm e}^{\frac {x}{3}}}{3}-{\mathrm e}^{\frac {x}{3}}\right )+3 \,{\mathrm e}^{4} \left (\frac {x \,{\mathrm e}^{\frac {x}{3}}}{3}-{\mathrm e}^{\frac {x}{3}}\right )+{\mathrm e}^{\frac {x}{3}} \ln \relax (x ) x\)	\(77\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*(3+x)*exp(1/3*x)*ln(x)+1/3*((3+x)*exp(2)^2+(2*x+6)*exp(2)+x+6)*exp(1/3*x)-2*x,x,method=_RETURNVERBOSE)

[Out]

(exp(2)^2+2*exp(2)+1)*x*exp(1/3*x)+exp(1/3*x)*ln(x)*x-x^2

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maxima [B]  time = 0.34, size = 73, normalized size = 3.04 \begin {gather*} x e^{\left (\frac {1}{3} \, x\right )} \log \relax (x) - x^{2} + {\left (x e^{4} - 3 \, e^{4}\right )} e^{\left (\frac {1}{3} \, x\right )} + 2 \, {\left (x e^{2} - 3 \, e^{2}\right )} e^{\left (\frac {1}{3} \, x\right )} + {\left (x - 3\right )} e^{\left (\frac {1}{3} \, x\right )} + 3 \, e^{\left (\frac {1}{3} \, x\right )} + 3 \, e^{\left (\frac {1}{3} \, x + 4\right )} + 6 \, e^{\left (\frac {1}{3} \, x + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(3+x)*exp(1/3*x)*log(x)+1/3*((3+x)*exp(2)^2+(2*x+6)*exp(2)+x+6)*exp(1/3*x)-2*x,x, algorithm="max
ima")

[Out]

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

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mupad [B]  time = 3.23, size = 27, normalized size = 1.12 \begin {gather*} x\,\left ({\mathrm {e}}^{x/3}\,\ln \relax (x)+{\mathrm {e}}^{x/3}\,{\left ({\mathrm {e}}^2+1\right )}^2\right )-x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x/3)*(x + exp(4)*(x + 3) + exp(2)*(2*x + 6) + 6))/3 - 2*x + (exp(x/3)*log(x)*(x + 3))/3,x)

[Out]

x*(exp(x/3)*log(x) + exp(x/3)*(exp(2) + 1)^2) - x^2

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sympy [A]  time = 0.32, size = 26, normalized size = 1.08 \begin {gather*} - x^{2} + \left (x \log {\relax (x )} + x + 2 x e^{2} + x e^{4}\right ) e^{\frac {x}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(3+x)*exp(1/3*x)*ln(x)+1/3*((3+x)*exp(2)**2+(2*x+6)*exp(2)+x+6)*exp(1/3*x)-2*x,x)

[Out]

-x**2 + (x*log(x) + x + 2*x*exp(2) + x*exp(4))*exp(x/3)

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