\(\int \frac {-i \pi -e^2 \log (3)-\log (4 e^2-4 \log (4))}{x^2} \, dx\) [4168]

Optimal result

Integrand size = 30, antiderivative size = 30 \[ \int \frac {-i \pi -e^2 \log (3)-\log \left (4 e^2-4 \log (4)\right )}{x^2} \, dx=\frac {i \pi -x+e^2 \log (3)+\log \left (-4 \left (-e^2+\log (4)\right )\right )}{x} \]

[Out]

(ln(8*ln(2)-4*exp(2))+exp(2)*ln(3)-x)/x

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {12, 30} \[ \int \frac {-i \pi -e^2 \log (3)-\log \left (4 e^2-4 \log (4)\right )}{x^2} \, dx=\frac {i \pi +e^2 \log (3)+\log \left (4 \left (e^2-\log (4)\right )\right )}{x} \]

[In]

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

[Out]

(I*Pi + E^2*Log[3] + Log[4*(E^2 - Log[4])])/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 (-i \pi -e^2 \log (3)-\log \left (4 \left (e^2-\log (4)\right )\right )\right ) \int \frac {1}{x^2} \, dx \\ & = \frac {i \pi +e^2 \log (3)+\log \left (4 \left (e^2-\log (4)\right )\right )}{x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {-i \pi -e^2 \log (3)-\log \left (4 e^2-4 \log (4)\right )}{x^2} \, dx=\frac {i \pi +e^2 \log (3)+\log \left (4 \left (e^2-\log (4)\right )\right )}{x} \]

[In]

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

[Out]

(I*Pi + E^2*Log[3] + Log[4*(E^2 - Log[4])])/x

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.70

method	result	size
gosper	\(\frac {{\mathrm e}^{2} \ln \left (3\right )+\ln \left (8 \ln \left (2\right )-4 \,{\mathrm e}^{2}\right )}{x}\)	\(21\)
default	\(-\frac {-\ln \left (8 \ln \left (2\right )-4 \,{\mathrm e}^{2}\right )-{\mathrm e}^{2} \ln \left (3\right )}{x}\)	\(25\)
norman	\(\frac {{\mathrm e}^{2} \ln \left (3\right )+2 \ln \left (2\right )+\ln \left (2 \ln \left (2\right )-{\mathrm e}^{2}\right )}{x}\)	\(25\)
parallelrisch	\(-\frac {-\ln \left (8 \ln \left (2\right )-4 \,{\mathrm e}^{2}\right )-{\mathrm e}^{2} \ln \left (3\right )}{x}\)	\(25\)
risch	\(\frac {{\mathrm e}^{2} \ln \left (3\right )}{x}+\frac {2 \ln \left (2\right )}{x}+\frac {\ln \left (2 \ln \left (2\right )-{\mathrm e}^{2}\right )}{x}\)	\(31\)

[In]

int((-ln(8*ln(2)-4*exp(2))-exp(2)*ln(3))/x^2,x,method=_RETURNVERBOSE)

[Out]

(exp(2)*ln(3)+ln(8*ln(2)-4*exp(2)))/x

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.67 \[ \int \frac {-i \pi -e^2 \log (3)-\log \left (4 e^2-4 \log (4)\right )}{x^2} \, dx=\frac {e^{2} \log \left (3\right ) + \log \left (-4 \, e^{2} + 8 \, \log \left (2\right )\right )}{x} \]

[In]

integrate((-log(8*log(2)-4*exp(2))-exp(2)*log(3))/x^2,x, algorithm="fricas")

[Out]

(e^2*log(3) + log(-4*e^2 + 8*log(2)))/x

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {-i \pi -e^2 \log (3)-\log \left (4 e^2-4 \log (4)\right )}{x^2} \, dx=- \frac {- e^{2} \log {\left (3 \right )} - \log {\left (- 2 \log {\left (2 \right )} + e^{2} \right )} - 2 \log {\left (2 \right )} - i \pi }{x} \]

[In]

integrate((-ln(8*ln(2)-4*exp(2))-exp(2)*ln(3))/x**2,x)

[Out]

-(-exp(2)*log(3) - log(-2*log(2) + exp(2)) - 2*log(2) - I*pi)/x

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.67 \[ \int \frac {-i \pi -e^2 \log (3)-\log \left (4 e^2-4 \log (4)\right )}{x^2} \, dx=\frac {e^{2} \log \left (3\right ) + \log \left (-4 \, e^{2} + 8 \, \log \left (2\right )\right )}{x} \]

[In]

integrate((-log(8*log(2)-4*exp(2))-exp(2)*log(3))/x^2,x, algorithm="maxima")

[Out]

(e^2*log(3) + log(-4*e^2 + 8*log(2)))/x

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.67 \[ \int \frac {-i \pi -e^2 \log (3)-\log \left (4 e^2-4 \log (4)\right )}{x^2} \, dx=\frac {e^{2} \log \left (3\right ) + \log \left (-4 \, e^{2} + 8 \, \log \left (2\right )\right )}{x} \]

[In]

integrate((-log(8*log(2)-4*exp(2))-exp(2)*log(3))/x^2,x, algorithm="giac")

[Out]

(e^2*log(3) + log(-4*e^2 + 8*log(2)))/x

Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.67 \[ \int \frac {-i \pi -e^2 \log (3)-\log \left (4 e^2-4 \log (4)\right )}{x^2} \, dx=\frac {\ln \left (8\,\ln \left (2\right )-4\,{\mathrm {e}}^2\right )+{\mathrm {e}}^2\,\ln \left (3\right )}{x} \]

[In]

int(-(log(8*log(2) - 4*exp(2)) + exp(2)*log(3))/x^2,x)

[Out]

(log(8*log(2) - 4*exp(2)) + exp(2)*log(3))/x