3.94.1 \(\int \frac {-3 e+(24 x-8 e^{\frac {2}{3} (4+4 x)} x-6 x^2) \log ^2(x)}{3 x \log ^2(x)} \, dx\)

Optimal. Leaf size=30 \[ -e^{2+2 x+\frac {2 (1+x)}{3}}-(-4+x)^2+\frac {e}{\log (x)} \]

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Rubi [A]  time = 0.31, antiderivative size = 28, normalized size of antiderivative = 0.93, number of steps used = 6, number of rules used = 5, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.116, Rules used = {12, 6688, 2194, 2302, 30} \begin {gather*} -x^2+8 x-e^{\frac {8 x}{3}+\frac {8}{3}}+\frac {e}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-3*E + (24*x - 8*E^((2*(4 + 4*x))/3)*x - 6*x^2)*Log[x]^2)/(3*x*Log[x]^2),x]

[Out]

-E^(8/3 + (8*x)/3) + 8*x - x^2 + E/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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{3} \int \frac {-3 e+\left (24 x-8 e^{\frac {2}{3} (4+4 x)} x-6 x^2\right ) \log ^2(x)}{x \log ^2(x)} \, dx\\ &=\frac {1}{3} \int \left (24-8 e^{\frac {8}{3}+\frac {8 x}{3}}-6 x-\frac {3 e}{x \log ^2(x)}\right ) \, dx\\ &=8 x-x^2-\frac {8}{3} \int e^{\frac {8}{3}+\frac {8 x}{3}} \, dx-e \int \frac {1}{x \log ^2(x)} \, dx\\ &=-e^{\frac {8}{3}+\frac {8 x}{3}}+8 x-x^2-e \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (x)\right )\\ &=-e^{\frac {8}{3}+\frac {8 x}{3}}+8 x-x^2+\frac {e}{\log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 28, normalized size = 0.93 \begin {gather*} -e^{\frac {8}{3}+\frac {8 x}{3}}+8 x-x^2+\frac {e}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-3*E + (24*x - 8*E^((2*(4 + 4*x))/3)*x - 6*x^2)*Log[x]^2)/(3*x*Log[x]^2),x]

[Out]

-E^(8/3 + (8*x)/3) + 8*x - x^2 + E/Log[x]

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fricas [A]  time = 0.88, size = 27, normalized size = 0.90 \begin {gather*} -\frac {{\left (x^{2} - 8 \, x + e^{\left (\frac {8}{3} \, x + \frac {8}{3}\right )}\right )} \log \relax (x) - e}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-8*x*exp(4/3*x+4/3)^2-6*x^2+24*x)*log(x)^2-3*exp(1))/x/log(x)^2,x, algorithm="fricas")

[Out]

-((x^2 - 8*x + e^(8/3*x + 8/3))*log(x) - e)/log(x)

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giac [A]  time = 49.75, size = 31, normalized size = 1.03 \begin {gather*} -\frac {x^{2} \log \relax (x) - 8 \, x \log \relax (x) + e^{\left (\frac {8}{3} \, x + \frac {8}{3}\right )} \log \relax (x) - e}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-8*x*exp(4/3*x+4/3)^2-6*x^2+24*x)*log(x)^2-3*exp(1))/x/log(x)^2,x, algorithm="giac")

[Out]

-(x^2*log(x) - 8*x*log(x) + e^(8/3*x + 8/3)*log(x) - e)/log(x)

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maple [A]  time = 0.04, size = 25, normalized size = 0.83

method	result	size
risch	\(-x^{2}+8 x -{\mathrm e}^{\frac {8 x}{3}+\frac {8}{3}}+\frac {{\mathrm e}}{\ln \relax (x )}\)	\(25\)
default	\(-x^{2}+8 x -{\mathrm e}^{\frac {8 x}{3}+\frac {8}{3}}+\frac {{\mathrm e}}{\ln \relax (x )}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*((-8*x*exp(4/3*x+4/3)^2-6*x^2+24*x)*ln(x)^2-3*exp(1))/x/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

-x^2+8*x-exp(8/3*x+8/3)+exp(1)/ln(x)

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maxima [A]  time = 0.43, size = 24, normalized size = 0.80 \begin {gather*} -x^{2} + 8 \, x + \frac {e}{\log \relax (x)} - e^{\left (\frac {8}{3} \, x + \frac {8}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-8*x*exp(4/3*x+4/3)^2-6*x^2+24*x)*log(x)^2-3*exp(1))/x/log(x)^2,x, algorithm="maxima")

[Out]

-x^2 + 8*x + e/log(x) - e^(8/3*x + 8/3)

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mupad [B]  time = 7.38, size = 24, normalized size = 0.80 \begin {gather*} 8\,x-{\mathrm {e}}^{\frac {8\,x}{3}+\frac {8}{3}}+\frac {\mathrm {e}}{\ln \relax (x)}-x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(1) + (log(x)^2*(8*x*exp((8*x)/3 + 8/3) - 24*x + 6*x^2))/3)/(x*log(x)^2),x)

[Out]

8*x - exp((8*x)/3 + 8/3) + exp(1)/log(x) - x^2

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sympy [A]  time = 0.28, size = 22, normalized size = 0.73 \begin {gather*} - x^{2} + 8 x - e^{\frac {8 x}{3} + \frac {8}{3}} + \frac {e}{\log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-8*x*exp(4/3*x+4/3)**2-6*x**2+24*x)*ln(x)**2-3*exp(1))/x/ln(x)**2,x)

[Out]

-x**2 + 8*x - exp(8*x/3 + 8/3) + E/log(x)

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