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

Optimal. Leaf size=26 \[ \frac {4 \left (-4+i \pi +\log \left (5-e^{3-e^3}\right )\right )}{x^2} \]

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Rubi [A]  time = 0.02, antiderivative size = 28, normalized size of antiderivative = 1.08, number of steps used = 2, number of rules used = 2, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.054, Rules used = {12, 30} \begin {gather*} -\frac {4 \left (4-i \pi -\log \left (5-e^{3-e^3}\right )\right )}{x^2} \end {gather*}

Antiderivative was successfully verified.

[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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 30, normalized size = 1.15 \begin {gather*} -\frac {32-8 i \pi -8 \log \left (5-e^{3-e^3}\right )}{2 x^2} \end {gather*}

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.60, size = 26, normalized size = 1.00 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [A]  time = 0.13, size = 26, normalized size = 1.00 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.02, size = 23, normalized size = 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	\(-\frac {-8 \ln \left (\left (-5 \,{\mathrm e}^{{\mathrm e}^{3}-3}+1\right ) {\mathrm e}^{3-{\mathrm e}^{3}}\right )+32}{2 x^{2}}\)	\(28\)
risch	\(\frac {4 i \pi }{x^{2}}+\frac {4 \ln \left (5 \,{\mathrm e}^{{\mathrm e}^{3}}-{\mathrm e}^{3}\right )}{x^{2}}-\frac {4 \,{\mathrm e}^{3}}{x^{2}}-\frac {16}{x^{2}}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[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

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maxima [A]  time = 0.38, size = 26, normalized size = 1.00 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 4.63, size = 19, normalized size = 0.73 \begin {gather*} \frac {4\,\ln \left ({\mathrm {e}}^{-{\mathrm {e}}^3}\,{\mathrm {e}}^3-5\right )-16}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 0.10, size = 32, normalized size = 1.23 \begin {gather*} - \frac {- 8 \log {\left (-1 + \frac {5 e^{e^{3}}}{e^{3}} \right )} + 8 + 8 e^{3} - 8 i \pi }{2 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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