\(\int \frac {-1-4 \log (x)+\log (x) \log (\log (x))}{25 x^2 \log (x)} \, dx\) [7718]

Optimal result

Integrand size = 23, antiderivative size = 16 \[ \int \frac {-1-4 \log (x)+\log (x) \log (\log (x))}{25 x^2 \log (x)} \, dx=e+\frac {4-\log (\log (x))}{25 x} \]

[Out]

1/25/x*(4-ln(ln(x)))+exp(1)

Rubi [C] (verified)

Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.

Time = 0.11 (sec) , antiderivative size = 53, normalized size of antiderivative = 3.31, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {12, 6874, 2346, 2209, 2413, 6617, 2602} \[ \int \frac {-1-4 \log (x)+\log (x) \log (\log (x))}{25 x^2 \log (x)} \, dx=\frac {4}{25} \log (x) \operatorname {ExpIntegralEi}(-\log (x))-\frac {1}{25} (4 \log (x)+1) \operatorname {ExpIntegralEi}(-\log (x))+\frac {\operatorname {ExpIntegralEi}(-\log (x))}{25}+\frac {4}{25 x}-\frac {\log (\log (x))}{25 x} \]

[In]

Int[(-1 - 4*Log[x] + Log[x]*Log[Log[x]])/(25*x^2*Log[x]),x]

[Out]

4/(25*x) + ExpIntegralEi[-Log[x]]/25 + (4*ExpIntegralEi[-Log[x]]*Log[x])/25 - (ExpIntegralEi[-Log[x]]*(1 + 4*L
og[x]))/25 - Log[Log[x]]/(25*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 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 2413

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.) + Log[(f_.)*(x_)^(r_.)]*(e_.))*((g_.)*(x_))^(m_.), x_Sy
mbol] :> With[{u = IntHide[(g*x)^m*(a + b*Log[c*x^n])^p, x]}, Dist[d + e*Log[f*x^r], u, x] - Dist[e*r, Int[Sim
plifyIntegrand[u/x, x], x], x]] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, r}, x] &&  !(EqQ[p, 1] && EqQ[a, 0] &&
NeQ[d, 0])

Rule 2602

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

Rule 6617

Int[ExpIntegralEi[(a_.) + (b_.)*(x_)], x_Symbol] :> Simp[(a + b*x)*(ExpIntegralEi[a + b*x]/b), x] - Simp[E^(a
+ b*x)/b, x] /; FreeQ[{a, b}, x]

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{25} \int \frac {-1-4 \log (x)+\log (x) \log (\log (x))}{x^2 \log (x)} \, dx \\ & = \frac {1}{25} \int \left (\frac {-1-4 \log (x)}{x^2 \log (x)}+\frac {\log (\log (x))}{x^2}\right ) \, dx \\ & = \frac {1}{25} \int \frac {-1-4 \log (x)}{x^2 \log (x)} \, dx+\frac {1}{25} \int \frac {\log (\log (x))}{x^2} \, dx \\ & = -\frac {1}{25} \operatorname {ExpIntegralEi}(-\log (x)) (1+4 \log (x))-\frac {\log (\log (x))}{25 x}+\frac {1}{25} \int \frac {1}{x^2 \log (x)} \, dx+\frac {4}{25} \int \frac {\operatorname {ExpIntegralEi}(-\log (x))}{x} \, dx \\ & = -\frac {1}{25} \operatorname {ExpIntegralEi}(-\log (x)) (1+4 \log (x))-\frac {\log (\log (x))}{25 x}+\frac {1}{25} \text {Subst}\left (\int \frac {e^{-x}}{x} \, dx,x,\log (x)\right )+\frac {4}{25} \text {Subst}(\int \operatorname {ExpIntegralEi}(-x) \, dx,x,\log (x)) \\ & = \frac {4}{25 x}+\frac {\operatorname {ExpIntegralEi}(-\log (x))}{25}+\frac {4}{25} \operatorname {ExpIntegralEi}(-\log (x)) \log (x)-\frac {1}{25} \operatorname {ExpIntegralEi}(-\log (x)) (1+4 \log (x))-\frac {\log (\log (x))}{25 x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.12 \[ \int \frac {-1-4 \log (x)+\log (x) \log (\log (x))}{25 x^2 \log (x)} \, dx=\frac {4}{25 x}-\frac {\log (\log (x))}{25 x} \]

[In]

Integrate[(-1 - 4*Log[x] + Log[x]*Log[Log[x]])/(25*x^2*Log[x]),x]

[Out]

4/(25*x) - Log[Log[x]]/(25*x)

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75

method	result	size
norman	\(\frac {\frac {4}{25}-\frac {\ln \left (\ln \left (x \right )\right )}{25}}{x}\)	\(12\)
parallelrisch	\(\frac {4-\ln \left (\ln \left (x \right )\right )}{25 x}\)	\(13\)
default	\(-\frac {\ln \left (\ln \left (x \right )\right )}{25 x}+\frac {4}{25 x}\)	\(15\)
risch	\(-\frac {\ln \left (\ln \left (x \right )\right )}{25 x}+\frac {4}{25 x}\)	\(15\)
parts	\(-\frac {\ln \left (\ln \left (x \right )\right )}{25 x}+\frac {4}{25 x}\)	\(15\)

[In]

int(1/25*(ln(x)*ln(ln(x))-4*ln(x)-1)/x^2/ln(x),x,method=_RETURNVERBOSE)

[Out]

(4/25-1/25*ln(ln(x)))/x

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.62 \[ \int \frac {-1-4 \log (x)+\log (x) \log (\log (x))}{25 x^2 \log (x)} \, dx=-\frac {\log \left (\log \left (x\right )\right ) - 4}{25 \, x} \]

[In]

integrate(1/25*(log(x)*log(log(x))-4*log(x)-1)/x^2/log(x),x, algorithm="fricas")

[Out]

-1/25*(log(log(x)) - 4)/x

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {-1-4 \log (x)+\log (x) \log (\log (x))}{25 x^2 \log (x)} \, dx=- \frac {\log {\left (\log {\left (x \right )} \right )}}{25 x} + \frac {4}{25 x} \]

[In]

integrate(1/25*(ln(x)*ln(ln(x))-4*ln(x)-1)/x**2/ln(x),x)

[Out]

-log(log(x))/(25*x) + 4/(25*x)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {-1-4 \log (x)+\log (x) \log (\log (x))}{25 x^2 \log (x)} \, dx=-\frac {\log \left (\log \left (x\right )\right )}{25 \, x} + \frac {4}{25 \, x} \]

[In]

integrate(1/25*(log(x)*log(log(x))-4*log(x)-1)/x^2/log(x),x, algorithm="maxima")

[Out]

-1/25*log(log(x))/x + 4/25/x

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {-1-4 \log (x)+\log (x) \log (\log (x))}{25 x^2 \log (x)} \, dx=-\frac {\log \left (\log \left (x\right )\right )}{25 \, x} + \frac {4}{25 \, x} \]

[In]

integrate(1/25*(log(x)*log(log(x))-4*log(x)-1)/x^2/log(x),x, algorithm="giac")

[Out]

-1/25*log(log(x))/x + 4/25/x

Mupad [B] (verification not implemented)

Time = 13.47 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.62 \[ \int \frac {-1-4 \log (x)+\log (x) \log (\log (x))}{25 x^2 \log (x)} \, dx=-\frac {\ln \left (\ln \left (x\right )\right )-4}{25\,x} \]

[In]

int(-((4*log(x))/25 - (log(log(x))*log(x))/25 + 1/25)/(x^2*log(x)),x)

[Out]

-(log(log(x)) - 4)/(25*x)