\(\int -\frac {3 e \log (i \pi +\log (6))}{x^2} \, dx\) [7640]

Optimal result

Integrand size = 15, antiderivative size = 15 \[ \int -\frac {3 e \log (i \pi +\log (6))}{x^2} \, dx=\frac {3 e \log (i \pi +\log (6))}{x} \]

[Out]

3*exp(1)/x*ln(ln(6)+I*Pi)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {12, 30} \[ \int -\frac {3 e \log (i \pi +\log (6))}{x^2} \, dx=\frac {3 e \log (\log (6)+i \pi )}{x} \]

[In]

Int[(-3*E*Log[I*Pi + Log[6]])/x^2,x]

[Out]

(3*E*Log[I*Pi + Log[6]])/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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = -\left ((3 e \log (i \pi +\log (6))) \int \frac {1}{x^2} \, dx\right ) \\ & = \frac {3 e \log (i \pi +\log (6))}{x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int -\frac {3 e \log (i \pi +\log (6))}{x^2} \, dx=\frac {3 e \log (i \pi +\log (6))}{x} \]

[In]

Integrate[(-3*E*Log[I*Pi + Log[6]])/x^2,x]

[Out]

(3*E*Log[I*Pi + Log[6]])/x

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.07

method	result	size
gosper	\(\frac {3 \,{\mathrm e} \ln \left (\ln \left (6\right )+i \pi \right )}{x}\)	\(16\)
default	\(\frac {3 \,{\mathrm e} \ln \left (\ln \left (6\right )+i \pi \right )}{x}\)	\(16\)
norman	\(\frac {3 \,{\mathrm e} \ln \left (\ln \left (6\right )+i \pi \right )}{x}\)	\(16\)
parallelrisch	\(\frac {3 \,{\mathrm e} \ln \left (\ln \left (6\right )+i \pi \right )}{x}\)	\(16\)
risch	\(\frac {3 \,{\mathrm e} \ln \left (\ln \left (2\right )+\ln \left (3\right )+i \pi \right )}{x}\)	\(18\)

[In]

int(-3*exp(1)*ln(ln(6)+I*Pi)/x^2,x,method=_RETURNVERBOSE)

[Out]

3*exp(1)/x*ln(ln(6)+I*Pi)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int -\frac {3 e \log (i \pi +\log (6))}{x^2} \, dx=\frac {3 \, e \log \left (i \, \pi + \log \left (6\right )\right )}{x} \]

[In]

integrate(-3*exp(1)*log(log(6)+I*pi)/x^2,x, algorithm="fricas")

[Out]

3*e*log(I*pi + log(6))/x

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int -\frac {3 e \log (i \pi +\log (6))}{x^2} \, dx=\frac {3 e \log {\left (\log {\left (6 \right )} + i \pi \right )}}{x} \]

[In]

integrate(-3*exp(1)*ln(ln(6)+I*pi)/x**2,x)

[Out]

3*E*log(log(6) + I*pi)/x

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int -\frac {3 e \log (i \pi +\log (6))}{x^2} \, dx=\frac {3 \, e \log \left (i \, \pi + \log \left (6\right )\right )}{x} \]

[In]

integrate(-3*exp(1)*log(log(6)+I*pi)/x^2,x, algorithm="maxima")

[Out]

3*e*log(I*pi + log(6))/x

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int -\frac {3 e \log (i \pi +\log (6))}{x^2} \, dx=\frac {3 \, e \log \left (i \, \pi + \log \left (6\right )\right )}{x} \]

[In]

integrate(-3*exp(1)*log(log(6)+I*pi)/x^2,x, algorithm="giac")

[Out]

3*e*log(I*pi + log(6))/x

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int -\frac {3 e \log (i \pi +\log (6))}{x^2} \, dx=\frac {3\,\mathrm {e}\,\ln \left (\ln \left (6\right )+\Pi \,1{}\mathrm {i}\right )}{x} \]

[In]

int(-(3*exp(1)*log(Pi*1i + log(6)))/x^2,x)

[Out]

(3*exp(1)*log(Pi*1i + log(6)))/x