3.87.47 \(\int \frac {\log (\log (5 (i \pi +\log (4))))}{4+4 x+x^2} \, dx\)

Optimal. Leaf size=19 \[ -\frac {\log (\log (5 (i \pi +\log (4))))}{2+x} \]

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Rubi [A]  time = 0.00, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {12, 27, 32} \begin {gather*} -\frac {\log (\log (5 (\log (4)+i \pi )))}{x+2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(Log[Log[5*(I*Pi + Log[4])]]/(2 + 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 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

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

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Mathematica [A]  time = 0.00, size = 19, normalized size = 1.00 \begin {gather*} -\frac {\log (\log (5 (i \pi +\log (4))))}{2+x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(Log[Log[5*(I*Pi + Log[4])]]/(2 + x))

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fricas [A]  time = 1.02, size = 17, normalized size = 0.89 \begin {gather*} -\frac {\log \left (\log \left (5 i \, \pi + 10 \, \log \relax (2)\right )\right )}{x + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(log(10*log(2)+5*I*pi))/(x^2+4*x+4),x, algorithm="fricas")

[Out]

-log(log(5*I*pi + 10*log(2)))/(x + 2)

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giac [A]  time = 0.19, size = 17, normalized size = 0.89 \begin {gather*} -\frac {\log \left (\log \left (5 i \, \pi + 10 \, \log \relax (2)\right )\right )}{x + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(log(10*log(2)+5*I*pi))/(x^2+4*x+4),x, algorithm="giac")

[Out]

-log(log(5*I*pi + 10*log(2)))/(x + 2)

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maple [A]  time = 0.46, size = 19, normalized size = 1.00

method	result	size
gosper	\(-\frac {\ln \left (\ln \left (10 \ln \relax (2)+5 i \pi \right )\right )}{2+x}\)	\(19\)
default	\(-\frac {\ln \left (\ln \left (10 \ln \relax (2)+5 i \pi \right )\right )}{2+x}\)	\(19\)
risch	\(-\frac {\ln \left (\ln \left (5 i \left (-2 i \ln \relax (2)+\pi \right )\right )\right )}{2+x}\)	\(20\)
norman	\(-\frac {\ln \left (\ln \relax (5)+\ln \left (2 \ln \relax (2)+i \pi \right )\right )}{2+x}\)	\(22\)
meijerg	\(\frac {\ln \left (\ln \left (10 \ln \relax (2)+5 i \pi \right )\right ) x}{2 x +4}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(ln(10*ln(2)+5*I*Pi))/(x^2+4*x+4),x,method=_RETURNVERBOSE)

[Out]

-ln(ln(10*ln(2)+5*I*Pi))/(2+x)

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maxima [A]  time = 0.37, size = 17, normalized size = 0.89 \begin {gather*} -\frac {\log \left (\log \left (5 i \, \pi + 10 \, \log \relax (2)\right )\right )}{x + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(log(10*log(2)+5*I*pi))/(x^2+4*x+4),x, algorithm="maxima")

[Out]

-log(log(5*I*pi + 10*log(2)))/(x + 2)

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mupad [B]  time = 0.08, size = 18, normalized size = 0.95 \begin {gather*} -\frac {\ln \left (\ln \left (10\,\ln \relax (2)+\Pi \,5{}\mathrm {i}\right )\right )}{x+2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(log(Pi*5i + 10*log(2)))/(4*x + x^2 + 4),x)

[Out]

-log(log(Pi*5i + 10*log(2)))/(x + 2)

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sympy [A]  time = 0.10, size = 19, normalized size = 1.00 \begin {gather*} - \frac {\log {\left (\log {\relax (5 )} + \log {\left (2 \log {\relax (2 )} + i \pi \right )} \right )}}{x + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(ln(10*ln(2)+5*I*pi))/(x**2+4*x+4),x)

[Out]

-log(log(5) + log(2*log(2) + I*pi))/(x + 2)

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