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

Optimal. Leaf size=22 \[ \frac {1}{16} \left (x \left (10-\frac {\log ^2(x)}{4}\right )-\log (\log (x))\right ) \]

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Rubi [A]  time = 0.12, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {12, 6688, 2302, 29, 2295, 2296} \begin {gather*} \frac {5 x}{8}-\frac {1}{64} x \log ^2(x)-\frac {1}{16} \log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[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 2295

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

Rule 2296

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 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}{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} \operatorname {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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 22, normalized size = 1.00 \begin {gather*} \frac {5 x}{8}-\frac {1}{64} x \log ^2(x)-\frac {1}{16} \log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.96, size = 16, normalized size = 0.73 \begin {gather*} -\frac {1}{64} \, x \log \relax (x)^{2} + \frac {5}{8} \, x - \frac {1}{16} \, \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [A]  time = 0.31, size = 16, normalized size = 0.73 \begin {gather*} -\frac {1}{64} \, x \log \relax (x)^{2} + \frac {5}{8} \, x - \frac {1}{16} \, \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maple [A]  time = 0.02, size = 17, normalized size = 0.77

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

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maxima [A]  time = 0.40, size = 27, normalized size = 1.23 \begin {gather*} -\frac {1}{64} \, {\left (\log \relax (x)^{2} - 2 \, \log \relax (x) + 2\right )} x - \frac {1}{32} \, x \log \relax (x) + \frac {21}{32} \, x - \frac {1}{16} \, \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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mupad [B]  time = 0.28, size = 16, normalized size = 0.73 \begin {gather*} \frac {5\,x}{8}-\frac {\ln \left (\ln \relax (x)\right )}{16}-\frac {x\,{\ln \relax (x)}^2}{64} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 0.11, size = 19, normalized size = 0.86 \begin {gather*} - \frac {x \log {\relax (x )}^{2}}{64} + \frac {5 x}{8} - \frac {\log {\left (\log {\relax (x )} \right )}}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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