3.64.93 \(\int -\frac {2 x \log (3) (i \pi +\log (4-\log (4)))}{e^2 (25+10 x^2+x^4)} \, dx\)

Optimal. Leaf size=30 \[ e^2+\frac {\log (3) (i \pi +\log (4-\log (4)))}{e^2 \left (5+x^2\right )} \]

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Rubi [A]  time = 0.01, antiderivative size = 26, normalized size of antiderivative = 0.87, number of steps used = 3, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {12, 28, 261} \begin {gather*} \frac {\log (3) (\log (4-\log (4))+i \pi )}{e^2 \left (x^2+5\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2*x*Log[3]*(I*Pi + Log[4 - Log[4]]))/(E^2*(25 + 10*x^2 + x^4)),x]

[Out]

(Log[3]*(I*Pi + Log[4 - Log[4]]))/(E^2*(5 + 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 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\frac {(2 \log (3) (i \pi +\log (4-\log (4)))) \int \frac {x}{25+10 x^2+x^4} \, dx}{e^2}\\ &=-\frac {(2 \log (3) (i \pi +\log (4-\log (4)))) \int \frac {x}{\left (5+x^2\right )^2} \, dx}{e^2}\\ &=\frac {\log (3) (i \pi +\log (4-\log (4)))}{e^2 \left (5+x^2\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 29, normalized size = 0.97 \begin {gather*} \frac {i \log (3) (\pi -i \log (4-\log (4)))}{e^2 \left (5+x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*x*Log[3]*(I*Pi + Log[4 - Log[4]]))/(E^2*(25 + 10*x^2 + x^4)),x]

[Out]

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

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fricas [A]  time = 0.71, size = 19, normalized size = 0.63 \begin {gather*} \frac {e^{\left (-2\right )} \log \relax (3) \log \left (2 \, \log \relax (2) - 4\right )}{x^{2} + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-2*x*log(3)*log(2*log(2)-4)/(x^4+10*x^2+25)/exp(2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.18, size = 19, normalized size = 0.63 \begin {gather*} \frac {e^{\left (-2\right )} \log \relax (3) \log \left (2 \, \log \relax (2) - 4\right )}{x^{2} + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-2*x*log(3)*log(2*log(2)-4)/(x^4+10*x^2+25)/exp(2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 22, normalized size = 0.73

method	result	size
gosper	\(\frac {\ln \relax (3) \ln \left (2 \ln \relax (2)-4\right ) {\mathrm e}^{-2}}{x^{2}+5}\)	\(22\)
default	\(\frac {\ln \relax (3) \ln \left (2 \ln \relax (2)-4\right ) {\mathrm e}^{-2}}{x^{2}+5}\)	\(22\)
norman	\(\frac {\ln \relax (3) \left (\ln \relax (2)+\ln \left (\ln \relax (2)-2\right )\right ) {\mathrm e}^{-2}}{x^{2}+5}\)	\(23\)
risch	\(\frac {\ln \relax (3) {\mathrm e}^{-2} \ln \relax (2)}{x^{2}+5}+\frac {\ln \relax (3) {\mathrm e}^{-2} \ln \left (\ln \relax (2)-2\right )}{x^{2}+5}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-2*x*ln(3)*ln(2*ln(2)-4)/(x^4+10*x^2+25)/exp(2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.37, size = 19, normalized size = 0.63 \begin {gather*} \frac {e^{\left (-2\right )} \log \relax (3) \log \left (2 \, \log \relax (2) - 4\right )}{x^{2} + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-2*x*log(3)*log(2*log(2)-4)/(x^4+10*x^2+25)/exp(2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.14, size = 17, normalized size = 0.57 \begin {gather*} \frac {\ln \left (\ln \relax (4)-4\right )\,{\mathrm {e}}^{-2}\,\ln \relax (3)}{x^2+5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x*exp(-2)*log(3)*log(2*log(2) - 4))/(10*x^2 + x^4 + 25),x)

[Out]

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

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sympy [A]  time = 0.17, size = 44, normalized size = 1.47 \begin {gather*} - \frac {- 2 \log {\relax (2 )} \log {\relax (3 )} - 2 \log {\relax (3 )} \log {\left (2 - \log {\relax (2 )} \right )} - 2 i \pi \log {\relax (3 )}}{2 x^{2} e^{2} + 10 e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-2*x*ln(3)*ln(2*ln(2)-4)/(x**4+10*x**2+25)/exp(2),x)

[Out]

-(-2*log(2)*log(3) - 2*log(3)*log(2 - log(2)) - 2*I*pi*log(3))/(2*x**2*exp(2) + 10*exp(2))

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