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

Optimal. Leaf size=16 \[ e+\frac {4-\log (\log (x))}{25 x} \]

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Rubi [C]  time = 0.18, antiderivative size = 53, normalized size of antiderivative = 3.31, number of steps used = 9, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {12, 6742, 2309, 2178, 2366, 6482, 2522} \begin {gather*} \frac {4}{25} \log (x) \text {Ei}(-\log (x))-\frac {1}{25} (4 \log (x)+1) \text {Ei}(-\log (x))+\frac {\text {Ei}(-\log (x))}{25}+\frac {4}{25 x}-\frac {\log (\log (x))}{25 x} \end {gather*}

Antiderivative was successfully verified.

[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 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 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 2366

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 2522

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 6482

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 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \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} \text {Ei}(-\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 {\text {Ei}(-\log (x))}{x} \, dx\\ &=-\frac {1}{25} \text {Ei}(-\log (x)) (1+4 \log (x))-\frac {\log (\log (x))}{25 x}+\frac {1}{25} \operatorname {Subst}\left (\int \frac {e^{-x}}{x} \, dx,x,\log (x)\right )+\frac {4}{25} \operatorname {Subst}(\int \text {Ei}(-x) \, dx,x,\log (x))\\ &=\frac {4}{25 x}+\frac {\text {Ei}(-\log (x))}{25}+\frac {4}{25} \text {Ei}(-\log (x)) \log (x)-\frac {1}{25} \text {Ei}(-\log (x)) (1+4 \log (x))-\frac {\log (\log (x))}{25 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 18, normalized size = 1.12 \begin {gather*} \frac {4}{25 x}-\frac {\log (\log (x))}{25 x} \end {gather*}

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.66, size = 10, normalized size = 0.62 \begin {gather*} -\frac {\log \left (\log \relax (x)\right ) - 4}{25 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [A]  time = 0.21, size = 14, normalized size = 0.88 \begin {gather*} -\frac {\log \left (\log \relax (x)\right )}{25 \, x} + \frac {4}{25 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.03, size = 12, normalized size = 0.75

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

Verification of antiderivative is not currently implemented for this CAS.

[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

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maxima [A]  time = 0.37, size = 14, normalized size = 0.88 \begin {gather*} -\frac {\log \left (\log \relax (x)\right )}{25 \, x} + \frac {4}{25 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 5.37, size = 10, normalized size = 0.62 \begin {gather*} -\frac {\ln \left (\ln \relax (x)\right )-4}{25\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.34, size = 12, normalized size = 0.75 \begin {gather*} - \frac {\log {\left (\log {\relax (x )} \right )}}{25 x} + \frac {4}{25 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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