3.45.60 \(\int \frac {(-3+3 \log (x)) (i \pi +\log (\log (1+5 \log (4))))}{\log ^2(x)} \, dx\)

Optimal. Leaf size=27 \[ \frac {3 x \left (i \pi +\log \left (-\log \left (\frac {5}{5+25 \log (4)}\right )\right )\right )}{\log (x)} \]

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Rubi [A]  time = 0.02, antiderivative size = 21, normalized size of antiderivative = 0.78, number of steps used = 6, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {12, 2360, 2297, 2298} \begin {gather*} \frac {3 x (\log (\log (1+5 \log (4)))+i \pi )}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-3 + 3*Log[x])*(I*Pi + Log[Log[1 + 5*Log[4]]]))/Log[x]^2,x]

[Out]

(3*x*(I*Pi + Log[Log[1 + 5*Log[4]]]))/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 2297

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

Rule 2298

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, x]

Rule 2360

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(Log[(c_.)*(x_)^(n_.)]*(e_.) + (d_))^(q_.), x_Symbol] :> Int[E
xpandIntegrand[(a + b*Log[c*x^n])^p*(d + e*Log[c*x^n])^q, x], x] /; FreeQ[{a, b, c, d, e, n}, x] && IntegerQ[p
] && IntegerQ[q]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=(i \pi +\log (\log (1+5 \log (4)))) \int \frac {-3+3 \log (x)}{\log ^2(x)} \, dx\\ &=(i \pi +\log (\log (1+5 \log (4)))) \int \left (-\frac {3}{\log ^2(x)}+\frac {3}{\log (x)}\right ) \, dx\\ &=-\left ((3 (i \pi +\log (\log (1+5 \log (4))))) \int \frac {1}{\log ^2(x)} \, dx\right )+(3 (i \pi +\log (\log (1+5 \log (4))))) \int \frac {1}{\log (x)} \, dx\\ &=\frac {3 x (i \pi +\log (\log (1+5 \log (4))))}{\log (x)}+3 (i \pi +\log (\log (1+5 \log (4)))) \text {li}(x)-(3 (i \pi +\log (\log (1+5 \log (4))))) \int \frac {1}{\log (x)} \, dx\\ &=\frac {3 x (i \pi +\log (\log (1+5 \log (4))))}{\log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 21, normalized size = 0.78 \begin {gather*} \frac {3 x (i \pi +\log (\log (1+5 \log (4))))}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-3 + 3*Log[x])*(I*Pi + Log[Log[1 + 5*Log[4]]]))/Log[x]^2,x]

[Out]

(3*x*(I*Pi + Log[Log[1 + 5*Log[4]]]))/Log[x]

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fricas [A]  time = 0.42, size = 17, normalized size = 0.63 \begin {gather*} \frac {3 \, x \log \left (-\log \left (10 \, \log \relax (2) + 1\right )\right )}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*log(x)-3)*log(-log(10*log(2)+1))/log(x)^2,x, algorithm="fricas")

[Out]

3*x*log(-log(10*log(2) + 1))/log(x)

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giac [A]  time = 3.98, size = 17, normalized size = 0.63 \begin {gather*} \frac {3 \, x \log \left (-\log \left (10 \, \log \relax (2) + 1\right )\right )}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*log(x)-3)*log(-log(10*log(2)+1))/log(x)^2,x, algorithm="giac")

[Out]

3*x*log(-log(10*log(2) + 1))/log(x)

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maple [A]  time = 0.03, size = 18, normalized size = 0.67

method	result	size
default	\(\frac {3 \ln \left (-\ln \left (10 \ln \relax (2)+1\right )\right ) x}{\ln \relax (x )}\)	\(18\)
risch	\(\frac {3 \left (\ln \left (\ln \left (10 \ln \relax (2)+1\right )\right )+i \pi \right ) x}{\ln \relax (x )}\)	\(21\)
norman	\(\frac {\left (3 \ln \left (\ln \left (10 \ln \relax (2)+1\right )\right )+3 i \pi \right ) x}{\ln \relax (x )}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*ln(x)-3)*ln(-ln(10*ln(2)+1))/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

3*ln(-ln(10*ln(2)+1))*x/ln(x)

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maxima [C]  time = 0.40, size = 24, normalized size = 0.89 \begin {gather*} 3 \, {\left ({\rm Ei}\left (\log \relax (x)\right ) - \Gamma \left (-1, -\log \relax (x)\right )\right )} \log \left (-\log \left (10 \, \log \relax (2) + 1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*log(x)-3)*log(-log(10*log(2)+1))/log(x)^2,x, algorithm="maxima")

[Out]

3*(Ei(log(x)) - gamma(-1, -log(x)))*log(-log(10*log(2) + 1))

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mupad [B]  time = 3.37, size = 17, normalized size = 0.63 \begin {gather*} \frac {3\,x\,\ln \left (-\ln \left (10\,\ln \relax (2)+1\right )\right )}{\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(-log(10*log(2) + 1))*(3*log(x) - 3))/log(x)^2,x)

[Out]

(3*x*log(-log(10*log(2) + 1)))/log(x)

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sympy [A]  time = 0.14, size = 22, normalized size = 0.81 \begin {gather*} \frac {3 x \log {\left (\log {\left (1 + 10 \log {\relax (2 )} \right )} \right )} + 3 i \pi x}{\log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*ln(x)-3)*ln(-ln(10*ln(2)+1))/ln(x)**2,x)

[Out]

(3*x*log(log(1 + 10*log(2))) + 3*I*pi*x)/log(x)

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