\(\int \frac {48-16 x-8 x^5+(-64+24 x+16 x^5) \log (x)-8 x^5 \log ^2(x)}{x^5-2 x^5 \log (x)+x^5 \log ^2(x)} \, dx\) [7757]

Optimal result

Integrand size = 54, antiderivative size = 24 \[ \int \frac {48-16 x-8 x^5+\left (-64+24 x+16 x^5\right ) \log (x)-8 x^5 \log ^2(x)}{x^5-2 x^5 \log (x)+x^5 \log ^2(x)} \, dx=2 (2-x) \left (4+\frac {4}{x^3 (-x+x \log (x))}\right ) \]

[Out]

2*(4/(x*ln(x)-x)/x^3+4)*(2-x)

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25, number of steps used = 18, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {6873, 12, 6874, 2395, 2343, 2346, 2209} \[ \int \frac {48-16 x-8 x^5+\left (-64+24 x+16 x^5\right ) \log (x)-8 x^5 \log ^2(x)}{x^5-2 x^5 \log (x)+x^5 \log ^2(x)} \, dx=-\frac {16}{x^4 (1-\log (x))}+\frac {8}{x^3 (1-\log (x))}-8 x \]

[In]

Int[(48 - 16*x - 8*x^5 + (-64 + 24*x + 16*x^5)*Log[x] - 8*x^5*Log[x]^2)/(x^5 - 2*x^5*Log[x] + x^5*Log[x]^2),x]

[Out]

-8*x - 16/(x^4*(1 - Log[x])) + 8/(x^3*(1 - 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 2209

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

Rule 2343

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 2346

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 2395

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol]
:> With[{u = ExpandIntegrand[(a + b*Log[c*x^n])^p, (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[
{a, b, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[m] && IntegerQ[r
]))

Rule 6873

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

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {8 \left (6-2 x-x^5-8 \log (x)+3 x \log (x)+2 x^5 \log (x)-x^5 \log ^2(x)\right )}{x^5 (1-\log (x))^2} \, dx \\ & = 8 \int \frac {6-2 x-x^5-8 \log (x)+3 x \log (x)+2 x^5 \log (x)-x^5 \log ^2(x)}{x^5 (1-\log (x))^2} \, dx \\ & = 8 \int \left (-1+\frac {-2+x}{x^5 (-1+\log (x))^2}+\frac {-8+3 x}{x^5 (-1+\log (x))}\right ) \, dx \\ & = -8 x+8 \int \frac {-2+x}{x^5 (-1+\log (x))^2} \, dx+8 \int \frac {-8+3 x}{x^5 (-1+\log (x))} \, dx \\ & = -8 x+8 \int \left (-\frac {2}{x^5 (-1+\log (x))^2}+\frac {1}{x^4 (-1+\log (x))^2}\right ) \, dx+8 \int \left (-\frac {8}{x^5 (-1+\log (x))}+\frac {3}{x^4 (-1+\log (x))}\right ) \, dx \\ & = -8 x+8 \int \frac {1}{x^4 (-1+\log (x))^2} \, dx-16 \int \frac {1}{x^5 (-1+\log (x))^2} \, dx+24 \int \frac {1}{x^4 (-1+\log (x))} \, dx-64 \int \frac {1}{x^5 (-1+\log (x))} \, dx \\ & = -8 x-\frac {16}{x^4 (1-\log (x))}+\frac {8}{x^3 (1-\log (x))}-24 \int \frac {1}{x^4 (-1+\log (x))} \, dx+24 \text {Subst}\left (\int \frac {e^{-3 x}}{-1+x} \, dx,x,\log (x)\right )+64 \int \frac {1}{x^5 (-1+\log (x))} \, dx-64 \text {Subst}\left (\int \frac {e^{-4 x}}{-1+x} \, dx,x,\log (x)\right ) \\ & = -8 x+\frac {24 \operatorname {ExpIntegralEi}(3 (1-\log (x)))}{e^3}-\frac {64 \operatorname {ExpIntegralEi}(4 (1-\log (x)))}{e^4}-\frac {16}{x^4 (1-\log (x))}+\frac {8}{x^3 (1-\log (x))}-24 \text {Subst}\left (\int \frac {e^{-3 x}}{-1+x} \, dx,x,\log (x)\right )+64 \text {Subst}\left (\int \frac {e^{-4 x}}{-1+x} \, dx,x,\log (x)\right ) \\ & = -8 x-\frac {16}{x^4 (1-\log (x))}+\frac {8}{x^3 (1-\log (x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71 \[ \int \frac {48-16 x-8 x^5+\left (-64+24 x+16 x^5\right ) \log (x)-8 x^5 \log ^2(x)}{x^5-2 x^5 \log (x)+x^5 \log ^2(x)} \, dx=-8 \left (x+\frac {-2+x}{x^4 (-1+\log (x))}\right ) \]

[In]

Integrate[(48 - 16*x - 8*x^5 + (-64 + 24*x + 16*x^5)*Log[x] - 8*x^5*Log[x]^2)/(x^5 - 2*x^5*Log[x] + x^5*Log[x]
^2),x]

[Out]

-8*(x + (-2 + x)/(x^4*(-1 + Log[x])))

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79

method	result	size
risch	\(-8 x -\frac {8 \left (-2+x \right )}{x^{4} \left (\ln \left (x \right )-1\right )}\)	\(19\)
default	\(-8 x +\frac {16}{x^{4} \left (\ln \left (x \right )-1\right )}-\frac {8}{x^{3} \left (\ln \left (x \right )-1\right )}\)	\(27\)
norman	\(\frac {16-8 x +8 x^{5}-8 x^{5} \ln \left (x \right )}{x^{4} \left (\ln \left (x \right )-1\right )}\)	\(28\)
parallelrisch	\(\frac {16-8 x +8 x^{5}-8 x^{5} \ln \left (x \right )}{x^{4} \left (\ln \left (x \right )-1\right )}\)	\(28\)

[In]

int((-8*x^5*ln(x)^2+(16*x^5+24*x-64)*ln(x)-8*x^5-16*x+48)/(x^5*ln(x)^2-2*x^5*ln(x)+x^5),x,method=_RETURNVERBOS
E)

[Out]

-8*x-8/x^4*(-2+x)/(ln(x)-1)

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25 \[ \int \frac {48-16 x-8 x^5+\left (-64+24 x+16 x^5\right ) \log (x)-8 x^5 \log ^2(x)}{x^5-2 x^5 \log (x)+x^5 \log ^2(x)} \, dx=-\frac {8 \, {\left (x^{5} \log \left (x\right ) - x^{5} + x - 2\right )}}{x^{4} \log \left (x\right ) - x^{4}} \]

[In]

integrate((-8*x^5*log(x)^2+(16*x^5+24*x-64)*log(x)-8*x^5-16*x+48)/(x^5*log(x)^2-2*x^5*log(x)+x^5),x, algorithm
="fricas")

[Out]

-8*(x^5*log(x) - x^5 + x - 2)/(x^4*log(x) - x^4)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71 \[ \int \frac {48-16 x-8 x^5+\left (-64+24 x+16 x^5\right ) \log (x)-8 x^5 \log ^2(x)}{x^5-2 x^5 \log (x)+x^5 \log ^2(x)} \, dx=- 8 x + \frac {16 - 8 x}{x^{4} \log {\left (x \right )} - x^{4}} \]

[In]

integrate((-8*x**5*ln(x)**2+(16*x**5+24*x-64)*ln(x)-8*x**5-16*x+48)/(x**5*ln(x)**2-2*x**5*ln(x)+x**5),x)

[Out]

-8*x + (16 - 8*x)/(x**4*log(x) - x**4)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25 \[ \int \frac {48-16 x-8 x^5+\left (-64+24 x+16 x^5\right ) \log (x)-8 x^5 \log ^2(x)}{x^5-2 x^5 \log (x)+x^5 \log ^2(x)} \, dx=-\frac {8 \, {\left (x^{5} \log \left (x\right ) - x^{5} + x - 2\right )}}{x^{4} \log \left (x\right ) - x^{4}} \]

[In]

integrate((-8*x^5*log(x)^2+(16*x^5+24*x-64)*log(x)-8*x^5-16*x+48)/(x^5*log(x)^2-2*x^5*log(x)+x^5),x, algorithm
="maxima")

[Out]

-8*(x^5*log(x) - x^5 + x - 2)/(x^4*log(x) - x^4)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {48-16 x-8 x^5+\left (-64+24 x+16 x^5\right ) \log (x)-8 x^5 \log ^2(x)}{x^5-2 x^5 \log (x)+x^5 \log ^2(x)} \, dx=-8 \, x - \frac {8 \, {\left (x - 2\right )}}{x^{4} \log \left (x\right ) - x^{4}} \]

[In]

integrate((-8*x^5*log(x)^2+(16*x^5+24*x-64)*log(x)-8*x^5-16*x+48)/(x^5*log(x)^2-2*x^5*log(x)+x^5),x, algorithm
="giac")

[Out]

-8*x - 8*(x - 2)/(x^4*log(x) - x^4)

Mupad [B] (verification not implemented)

Time = 13.76 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {48-16 x-8 x^5+\left (-64+24 x+16 x^5\right ) \log (x)-8 x^5 \log ^2(x)}{x^5-2 x^5 \log (x)+x^5 \log ^2(x)} \, dx=-8\,x-\frac {8\,x-16}{x^4\,\left (\ln \left (x\right )-1\right )} \]

[In]

int(-(16*x + 8*x^5*log(x)^2 - log(x)*(24*x + 16*x^5 - 64) + 8*x^5 - 48)/(x^5*log(x)^2 - 2*x^5*log(x) + x^5),x)

[Out]

- 8*x - (8*x - 16)/(x^4*(log(x) - 1))