\(\int \frac {-4+40 x \log (x)-2 x \log ^2(x)-x \log ^3(x)}{64 x \log (x)} \, dx\) [119]

Optimal result

Integrand size = 32, antiderivative size = 22 \[ \int \frac {-4+40 x \log (x)-2 x \log ^2(x)-x \log ^3(x)}{64 x \log (x)} \, dx=\frac {1}{16} \left (x \left (10-\frac {\log ^2(x)}{4}\right )-\log (\log (x))\right ) \]

[Out]

1/16*x*(10-1/4*ln(x)^2)-1/16*ln(ln(x))

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {12, 6820, 2339, 29, 2332, 2333} \[ \int \frac {-4+40 x \log (x)-2 x \log ^2(x)-x \log ^3(x)}{64 x \log (x)} \, dx=\frac {5 x}{8}-\frac {1}{64} x \log ^2(x)-\frac {1}{16} \log (\log (x)) \]

[In]

Int[(-4 + 40*x*Log[x] - 2*x*Log[x]^2 - x*Log[x]^3)/(64*x*Log[x]),x]

[Out]

(5*x)/8 - (x*Log[x]^2)/64 - Log[Log[x]]/16

Rule 12

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 2332

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2333

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*Log[c*x^n])^p, x] - Dist[b*n*p, In
t[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*p]

Rule 2339

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 6820

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{64} \int \frac {-4+40 x \log (x)-2 x \log ^2(x)-x \log ^3(x)}{x \log (x)} \, dx \\ & = \frac {1}{64} \int \left (40-\frac {4}{x \log (x)}-2 \log (x)-\log ^2(x)\right ) \, dx \\ & = \frac {5 x}{8}-\frac {1}{64} \int \log ^2(x) \, dx-\frac {1}{32} \int \log (x) \, dx-\frac {1}{16} \int \frac {1}{x \log (x)} \, dx \\ & = \frac {21 x}{32}-\frac {1}{32} x \log (x)-\frac {1}{64} x \log ^2(x)+\frac {1}{32} \int \log (x) \, dx-\frac {1}{16} \text {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right ) \\ & = \frac {5 x}{8}-\frac {1}{64} x \log ^2(x)-\frac {1}{16} \log (\log (x)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {-4+40 x \log (x)-2 x \log ^2(x)-x \log ^3(x)}{64 x \log (x)} \, dx=\frac {5 x}{8}-\frac {1}{64} x \log ^2(x)-\frac {1}{16} \log (\log (x)) \]

[In]

Integrate[(-4 + 40*x*Log[x] - 2*x*Log[x]^2 - x*Log[x]^3)/(64*x*Log[x]),x]

[Out]

(5*x)/8 - (x*Log[x]^2)/64 - Log[Log[x]]/16

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77

method	result	size
default	\(-\frac {x \ln \left (x \right )^{2}}{64}+\frac {5 x}{8}-\frac {\ln \left (\ln \left (x \right )\right )}{16}\)	\(17\)
norman	\(-\frac {x \ln \left (x \right )^{2}}{64}+\frac {5 x}{8}-\frac {\ln \left (\ln \left (x \right )\right )}{16}\)	\(17\)
risch	\(-\frac {x \ln \left (x \right )^{2}}{64}+\frac {5 x}{8}-\frac {\ln \left (\ln \left (x \right )\right )}{16}\)	\(17\)
parallelrisch	\(-\frac {x \ln \left (x \right )^{2}}{64}+\frac {5 x}{8}-\frac {\ln \left (\ln \left (x \right )\right )}{16}\)	\(17\)
parts	\(-\frac {x \ln \left (x \right )^{2}}{64}+\frac {5 x}{8}-\frac {\ln \left (\ln \left (x \right )\right )}{16}\)	\(17\)

[In]

int(1/64*(-x*ln(x)^3-2*x*ln(x)^2+40*x*ln(x)-4)/x/ln(x),x,method=_RETURNVERBOSE)

[Out]

-1/64*x*ln(x)^2+5/8*x-1/16*ln(ln(x))

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.73 \[ \int \frac {-4+40 x \log (x)-2 x \log ^2(x)-x \log ^3(x)}{64 x \log (x)} \, dx=-\frac {1}{64} \, x \log \left (x\right )^{2} + \frac {5}{8} \, x - \frac {1}{16} \, \log \left (\log \left (x\right )\right ) \]

[In]

integrate(1/64*(-x*log(x)^3-2*x*log(x)^2+40*x*log(x)-4)/x/log(x),x, algorithm="fricas")

[Out]

-1/64*x*log(x)^2 + 5/8*x - 1/16*log(log(x))

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {-4+40 x \log (x)-2 x \log ^2(x)-x \log ^3(x)}{64 x \log (x)} \, dx=- \frac {x \log {\left (x \right )}^{2}}{64} + \frac {5 x}{8} - \frac {\log {\left (\log {\left (x \right )} \right )}}{16} \]

[In]

integrate(1/64*(-x*ln(x)**3-2*x*ln(x)**2+40*x*ln(x)-4)/x/ln(x),x)

[Out]

-x*log(x)**2/64 + 5*x/8 - log(log(x))/16

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \frac {-4+40 x \log (x)-2 x \log ^2(x)-x \log ^3(x)}{64 x \log (x)} \, dx=-\frac {1}{64} \, {\left (\log \left (x\right )^{2} - 2 \, \log \left (x\right ) + 2\right )} x - \frac {1}{32} \, x \log \left (x\right ) + \frac {21}{32} \, x - \frac {1}{16} \, \log \left (\log \left (x\right )\right ) \]

[In]

integrate(1/64*(-x*log(x)^3-2*x*log(x)^2+40*x*log(x)-4)/x/log(x),x, algorithm="maxima")

[Out]

-1/64*(log(x)^2 - 2*log(x) + 2)*x - 1/32*x*log(x) + 21/32*x - 1/16*log(log(x))

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.73 \[ \int \frac {-4+40 x \log (x)-2 x \log ^2(x)-x \log ^3(x)}{64 x \log (x)} \, dx=-\frac {1}{64} \, x \log \left (x\right )^{2} + \frac {5}{8} \, x - \frac {1}{16} \, \log \left (\log \left (x\right )\right ) \]

[In]

integrate(1/64*(-x*log(x)^3-2*x*log(x)^2+40*x*log(x)-4)/x/log(x),x, algorithm="giac")

[Out]

-1/64*x*log(x)^2 + 5/8*x - 1/16*log(log(x))

Mupad [B] (verification not implemented)

Time = 7.68 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.73 \[ \int \frac {-4+40 x \log (x)-2 x \log ^2(x)-x \log ^3(x)}{64 x \log (x)} \, dx=\frac {5\,x}{8}-\frac {\ln \left (\ln \left (x\right )\right )}{16}-\frac {x\,{\ln \left (x\right )}^2}{64} \]

[In]

int(-((x*log(x)^2)/32 + (x*log(x)^3)/64 - (5*x*log(x))/8 + 1/16)/(x*log(x)),x)

[Out]

(5*x)/8 - log(log(x))/16 - (x*log(x)^2)/64