\(\int \frac {32-8 (i \pi +\log (-e^{3-e^3} (1-5 e^{-3+e^3})))}{x^3} \, dx\) [7472]

Optimal result

Integrand size = 37, antiderivative size = 26 \[ \int \frac {32-8 \left (i \pi +\log \left (-e^{3-e^3} \left (1-5 e^{-3+e^3}\right )\right )\right )}{x^3} \, dx=\frac {4 \left (-4+i \pi +\log \left (5-e^{3-e^3}\right )\right )}{x^2} \]

[Out]

4*(ln(exp(3-exp(3))-5)-4)/x^2

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.054, Rules used = {12, 30} \[ \int \frac {32-8 \left (i \pi +\log \left (-e^{3-e^3} \left (1-5 e^{-3+e^3}\right )\right )\right )}{x^3} \, dx=-\frac {4 \left (4-i \pi -\log \left (5-e^{3-e^3}\right )\right )}{x^2} \]

[In]

Int[(32 - 8*(I*Pi + Log[-(E^(3 - E^3)*(1 - 5*E^(-3 + E^3)))]))/x^3,x]

[Out]

(-4*(4 - I*Pi - Log[5 - E^(3 - E^3)]))/x^2

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 (8 \left (4-i \pi -\log \left (5-e^{3-e^3}\right )\right )\right ) \int \frac {1}{x^3} \, dx \\ & = -\frac {4 \left (4-i \pi -\log \left (5-e^{3-e^3}\right )\right )}{x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int \frac {32-8 \left (i \pi +\log \left (-e^{3-e^3} \left (1-5 e^{-3+e^3}\right )\right )\right )}{x^3} \, dx=-\frac {32-8 i \pi -8 \log \left (5-e^{3-e^3}\right )}{2 x^2} \]

[In]

Integrate[(32 - 8*(I*Pi + Log[-(E^(3 - E^3)*(1 - 5*E^(-3 + E^3)))]))/x^3,x]

[Out]

-1/2*(32 - (8*I)*Pi - 8*Log[5 - E^(3 - E^3)])/x^2

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88

method	result	size
norman	\(\frac {4 \ln \left (-5 \,{\mathrm e}^{{\mathrm e}^{3}-3}+1\right )-4-4 \,{\mathrm e}^{3}}{x^{2}}\)	\(23\)
gosper	\(\frac {4 \ln \left (\left (-5 \,{\mathrm e}^{{\mathrm e}^{3}-3}+1\right ) {\mathrm e}^{3-{\mathrm e}^{3}}\right )-16}{x^{2}}\)	\(26\)
default	\(-\frac {-8 \ln \left (\left (-5 \,{\mathrm e}^{{\mathrm e}^{3}-3}+1\right ) {\mathrm e}^{3-{\mathrm e}^{3}}\right )+32}{2 x^{2}}\)	\(28\)
parallelrisch	\(-\frac {-8 \ln \left (\left (-5 \,{\mathrm e}^{{\mathrm e}^{3}-3}+1\right ) {\mathrm e}^{3-{\mathrm e}^{3}}\right )+32}{2 x^{2}}\)	\(28\)
risch	\(\frac {4 i \pi }{x^{2}}-\frac {4 \,{\mathrm e}^{3}}{x^{2}}+\frac {4 \ln \left (5 \,{\mathrm e}^{{\mathrm e}^{3}}-{\mathrm e}^{3}\right )}{x^{2}}-\frac {16}{x^{2}}\)	\(37\)

[In]

int((-8*ln((-5*exp(exp(3)-3)+1)/exp(exp(3)-3))+32)/x^3,x,method=_RETURNVERBOSE)

[Out]

(4*ln(-5*exp(exp(3)-3)+1)-4-4*exp(3))/x^2

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {32-8 \left (i \pi +\log \left (-e^{3-e^3} \left (1-5 e^{-3+e^3}\right )\right )\right )}{x^3} \, dx=\frac {4 \, {\left (\log \left (-{\left (5 \, e^{\left (e^{3} - 3\right )} - 1\right )} e^{\left (-e^{3} + 3\right )}\right ) - 4\right )}}{x^{2}} \]

[In]

integrate((-8*log((-5*exp(exp(3)-3)+1)/exp(exp(3)-3))+32)/x^3,x, algorithm="fricas")

[Out]

4*(log(-(5*e^(e^3 - 3) - 1)*e^(-e^3 + 3)) - 4)/x^2

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {32-8 \left (i \pi +\log \left (-e^{3-e^3} \left (1-5 e^{-3+e^3}\right )\right )\right )}{x^3} \, dx=- \frac {- 8 \log {\left (-1 + \frac {5 e^{e^{3}}}{e^{3}} \right )} + 8 + 8 e^{3} - 8 i \pi }{2 x^{2}} \]

[In]

integrate((-8*ln((-5*exp(exp(3)-3)+1)/exp(exp(3)-3))+32)/x**3,x)

[Out]

-(-8*log(-1 + 5*exp(-3)*exp(exp(3))) + 8 + 8*exp(3) - 8*I*pi)/(2*x**2)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {32-8 \left (i \pi +\log \left (-e^{3-e^3} \left (1-5 e^{-3+e^3}\right )\right )\right )}{x^3} \, dx=\frac {4 \, {\left (\log \left (-{\left (5 \, e^{\left (e^{3} - 3\right )} - 1\right )} e^{\left (-e^{3} + 3\right )}\right ) - 4\right )}}{x^{2}} \]

[In]

integrate((-8*log((-5*exp(exp(3)-3)+1)/exp(exp(3)-3))+32)/x^3,x, algorithm="maxima")

[Out]

4*(log(-(5*e^(e^3 - 3) - 1)*e^(-e^3 + 3)) - 4)/x^2

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {32-8 \left (i \pi +\log \left (-e^{3-e^3} \left (1-5 e^{-3+e^3}\right )\right )\right )}{x^3} \, dx=\frac {4 \, {\left (\log \left (-{\left (5 \, e^{\left (e^{3} - 3\right )} - 1\right )} e^{\left (-e^{3} + 3\right )}\right ) - 4\right )}}{x^{2}} \]

[In]

integrate((-8*log((-5*exp(exp(3)-3)+1)/exp(exp(3)-3))+32)/x^3,x, algorithm="giac")

[Out]

4*(log(-(5*e^(e^3 - 3) - 1)*e^(-e^3 + 3)) - 4)/x^2

Mupad [B] (verification not implemented)

Time = 12.49 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.73 \[ \int \frac {32-8 \left (i \pi +\log \left (-e^{3-e^3} \left (1-5 e^{-3+e^3}\right )\right )\right )}{x^3} \, dx=\frac {4\,\ln \left ({\mathrm {e}}^{-{\mathrm {e}}^3}\,{\mathrm {e}}^3-5\right )-16}{x^2} \]

[In]

int(-(8*log(-exp(3 - exp(3))*(5*exp(exp(3) - 3) - 1)) - 32)/x^3,x)

[Out]

(4*log(exp(-exp(3))*exp(3) - 5) - 16)/x^2