3.26.28 \(\int \frac {-3 x^4+3 x^4 \log (x)+(-20 e^5+6 x^3) \log ^3(x)}{6 x^3 \log ^3(x)} \, dx\)

Optimal. Leaf size=23 \[ \frac {5 e^5}{3 x^2}+x+\frac {x^2}{4 \log ^2(x)} \]

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Rubi [A]  time = 0.20, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {12, 6688, 2306, 2309, 2178} \begin {gather*} \frac {5 e^5}{3 x^2}+\frac {x^2}{4 \log ^2(x)}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-3*x^4 + 3*x^4*Log[x] + (-20*E^5 + 6*x^3)*Log[x]^3)/(6*x^3*Log[x]^3),x]

[Out]

(5*E^5)/(3*x^2) + x + x^2/(4*Log[x]^2)

Rule 12

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

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2306

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log
[c*x^n])^(p + 1))/(b*d*n*(p + 1)), x] - Dist[(m + 1)/(b*n*(p + 1)), Int[(d*x)^m*(a + b*Log[c*x^n])^(p + 1), x]
, x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1] && LtQ[p, -1]

Rule 2309

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))^(p_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[E^((m + 1)*x)*(a
 + b*x)^p, x], x, Log[c*x]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[m]

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}{6} \int \frac {-3 x^4+3 x^4 \log (x)+\left (-20 e^5+6 x^3\right ) \log ^3(x)}{x^3 \log ^3(x)} \, dx\\ &=\frac {1}{6} \int \left (6-\frac {20 e^5}{x^3}-\frac {3 x}{\log ^3(x)}+\frac {3 x}{\log ^2(x)}\right ) \, dx\\ &=\frac {5 e^5}{3 x^2}+x-\frac {1}{2} \int \frac {x}{\log ^3(x)} \, dx+\frac {1}{2} \int \frac {x}{\log ^2(x)} \, dx\\ &=\frac {5 e^5}{3 x^2}+x+\frac {x^2}{4 \log ^2(x)}-\frac {x^2}{2 \log (x)}-\frac {1}{2} \int \frac {x}{\log ^2(x)} \, dx+\int \frac {x}{\log (x)} \, dx\\ &=\frac {5 e^5}{3 x^2}+x+\frac {x^2}{4 \log ^2(x)}-\int \frac {x}{\log (x)} \, dx+\operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )\\ &=\frac {5 e^5}{3 x^2}+x+\text {Ei}(2 \log (x))+\frac {x^2}{4 \log ^2(x)}-\operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )\\ &=\frac {5 e^5}{3 x^2}+x+\frac {x^2}{4 \log ^2(x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 23, normalized size = 1.00 \begin {gather*} \frac {5 e^5}{3 x^2}+x+\frac {x^2}{4 \log ^2(x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-3*x^4 + 3*x^4*Log[x] + (-20*E^5 + 6*x^3)*Log[x]^3)/(6*x^3*Log[x]^3),x]

[Out]

(5*E^5)/(3*x^2) + x + x^2/(4*Log[x]^2)

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fricas [A]  time = 0.85, size = 31, normalized size = 1.35 \begin {gather*} \frac {3 \, x^{4} + 4 \, {\left (3 \, x^{3} + 5 \, e^{5}\right )} \log \relax (x)^{2}}{12 \, x^{2} \log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*((-20*exp(5)+6*x^3)*log(x)^3+3*x^4*log(x)-3*x^4)/x^3/log(x)^3,x, algorithm="fricas")

[Out]

1/12*(3*x^4 + 4*(3*x^3 + 5*e^5)*log(x)^2)/(x^2*log(x)^2)

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giac [A]  time = 0.33, size = 32, normalized size = 1.39 \begin {gather*} \frac {12 \, x^{3} \log \relax (x)^{2} + 3 \, x^{4} + 20 \, e^{5} \log \relax (x)^{2}}{12 \, x^{2} \log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*((-20*exp(5)+6*x^3)*log(x)^3+3*x^4*log(x)-3*x^4)/x^3/log(x)^3,x, algorithm="giac")

[Out]

1/12*(12*x^3*log(x)^2 + 3*x^4 + 20*e^5*log(x)^2)/(x^2*log(x)^2)

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maple [A]  time = 0.04, size = 19, normalized size = 0.83




method result size



default \(\frac {5 \,{\mathrm e}^{5}}{3 x^{2}}+x +\frac {x^{2}}{4 \ln \relax (x )^{2}}\) \(19\)
risch \(\frac {3 x^{3}+5 \,{\mathrm e}^{5}}{3 x^{2}}+\frac {x^{2}}{4 \ln \relax (x )^{2}}\) \(26\)
norman \(\frac {x^{3} \ln \relax (x )^{2}+\frac {x^{4}}{4}+\frac {5 \,{\mathrm e}^{5} \ln \relax (x )^{2}}{3}}{x^{2} \ln \relax (x )^{2}}\) \(31\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/6*((-20*exp(5)+6*x^3)*ln(x)^3+3*x^4*ln(x)-3*x^4)/x^3/ln(x)^3,x,method=_RETURNVERBOSE)

[Out]

5/3*exp(5)/x^2+x+1/4*x^2/ln(x)^2

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maxima [C]  time = 0.47, size = 23, normalized size = 1.00 \begin {gather*} x + \frac {5 \, e^{5}}{3 \, x^{2}} + \Gamma \left (-1, -2 \, \log \relax (x)\right ) + 2 \, \Gamma \left (-2, -2 \, \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*((-20*exp(5)+6*x^3)*log(x)^3+3*x^4*log(x)-3*x^4)/x^3/log(x)^3,x, algorithm="maxima")

[Out]

x + 5/3*e^5/x^2 + gamma(-1, -2*log(x)) + 2*gamma(-2, -2*log(x))

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mupad [B]  time = 1.46, size = 22, normalized size = 0.96 \begin {gather*} \frac {x^3+\frac {5\,{\mathrm {e}}^5}{3}}{x^2}+\frac {x^2}{4\,{\ln \relax (x)}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x^4/2 - (x^4*log(x))/2 + (log(x)^3*(20*exp(5) - 6*x^3))/6)/(x^3*log(x)^3),x)

[Out]

((5*exp(5))/3 + x^3)/x^2 + x^2/(4*log(x)^2)

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sympy [A]  time = 0.13, size = 20, normalized size = 0.87 \begin {gather*} \frac {x^{2}}{4 \log {\relax (x )}^{2}} + x + \frac {5 e^{5}}{3 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*((-20*exp(5)+6*x**3)*ln(x)**3+3*x**4*ln(x)-3*x**4)/x**3/ln(x)**3,x)

[Out]

x**2/(4*log(x)**2) + x + 5*exp(5)/(3*x**2)

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