3.32.21 \(\int \frac {3^{-1/x} \sqrt [x]{\log (x^4)} (-48+12 \log (x^4) \log (\frac {\log (x^4)}{3}))}{(625 x^2+50 x^2 \log (4)+x^2 \log ^2(4)) \log (x^4)+3^{-1/x} (-50 x^2-2 x^2 \log (4)) \log ^{1+\frac {1}{x}}(x^4)+3^{-2/x} x^2 \log ^{1+\frac {2}{x}}(x^4)} \, dx\)

Optimal. Leaf size=26 \[ \frac {12}{-25-\log (4)+3^{-1/x} \sqrt [x]{\log \left (x^4\right )}} \]

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Rubi [A]  time = 1.58, antiderivative size = 33, normalized size of antiderivative = 1.27, number of steps used = 4, number of rules used = 4, integrand size = 116, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {6688, 12, 6711, 32} \begin {gather*} \frac {12}{(25+\log (4)) \left (1-3^{\frac {1}{x}} (25+\log (4)) \log ^{-\frac {1}{x}}\left (x^4\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Log[x^4]^x^(-1)*(-48 + 12*Log[x^4]*Log[Log[x^4]/3]))/(3^x^(-1)*((625*x^2 + 50*x^2*Log[4] + x^2*Log[4]^2)*
Log[x^4] + ((-50*x^2 - 2*x^2*Log[4])*Log[x^4]^(1 + x^(-1)))/3^x^(-1) + (x^2*Log[x^4]^(1 + 2/x))/3^(2/x))),x]

[Out]

12/((25 + Log[4])*(1 - (3^x^(-1)*(25 + Log[4]))/Log[x^4]^x^(-1)))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6711

Int[(u_)*((a_.)*(v_)^(p_.) + (b_.)*(w_)^(q_.))^(m_.), x_Symbol] :> With[{c = Simplify[u/(p*w*D[v, x] - q*v*D[w
, x])]}, Dist[c*p, Subst[Int[(b + a*x^p)^m, x], x, v*w^(m*q + 1)], x] /; FreeQ[c, x]] /; FreeQ[{a, b, m, p, q}
, x] && EqQ[p + q*(m*p + 1), 0] && IntegerQ[p] && IntegerQ[m]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4\ 3^{1+\frac {1}{x}} \log ^{-1+\frac {1}{x}}\left (x^4\right ) \left (-4+\log \left (x^4\right ) \log \left (\frac {\log \left (x^4\right )}{3}\right )\right )}{x^2 \left (3^{\frac {1}{x}} (25+\log (4))-\sqrt [x]{\log \left (x^4\right )}\right )^2} \, dx\\ &=4 \int \frac {3^{1+\frac {1}{x}} \log ^{-1+\frac {1}{x}}\left (x^4\right ) \left (-4+\log \left (x^4\right ) \log \left (\frac {\log \left (x^4\right )}{3}\right )\right )}{x^2 \left (3^{\frac {1}{x}} (25+\log (4))-\sqrt [x]{\log \left (x^4\right )}\right )^2} \, dx\\ &=12 \operatorname {Subst}\left (\int \frac {1}{(-1+x (25+\log (4)))^2} \, dx,x,3^{\frac {1}{x}} \log ^{-\frac {1}{x}}\left (x^4\right )\right )\\ &=\frac {12}{(25+\log (4)) \left (1-3^{\frac {1}{x}} (25+\log (4)) \log ^{-\frac {1}{x}}\left (x^4\right )\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.60, size = 32, normalized size = 1.23 \begin {gather*} -\frac {4\ 3^{1+\frac {1}{x}}}{3^{\frac {1}{x}} (25+\log (4))-\sqrt [x]{\log \left (x^4\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Log[x^4]^x^(-1)*(-48 + 12*Log[x^4]*Log[Log[x^4]/3]))/(3^x^(-1)*((625*x^2 + 50*x^2*Log[4] + x^2*Log[
4]^2)*Log[x^4] + ((-50*x^2 - 2*x^2*Log[4])*Log[x^4]^(1 + x^(-1)))/3^x^(-1) + (x^2*Log[x^4]^(1 + 2/x))/3^(2/x))
),x]

[Out]

(-4*3^(1 + x^(-1)))/(3^x^(-1)*(25 + Log[4]) - Log[x^4]^x^(-1))

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fricas [A]  time = 0.52, size = 20, normalized size = 0.77 \begin {gather*} \frac {12}{\left (\frac {1}{3} \, \log \left (x^{4}\right )\right )^{\left (\frac {1}{x}\right )} - 2 \, \log \relax (2) - 25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((12*log(x^4)*log(1/3*log(x^4))-48)*exp(log(1/3*log(x^4))/x)/(x^2*log(x^4)*exp(log(1/3*log(x^4))/x)^2
+(-4*x^2*log(2)-50*x^2)*log(x^4)*exp(log(1/3*log(x^4))/x)+(4*x^2*log(2)^2+100*x^2*log(2)+625*x^2)*log(x^4)),x,
algorithm="fricas")

[Out]

12/((1/3*log(x^4))^(1/x) - 2*log(2) - 25)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {12 \, {\left (\log \left (x^{4}\right ) \log \left (\frac {1}{3} \, \log \left (x^{4}\right )\right ) - 4\right )} \left (\frac {1}{3} \, \log \left (x^{4}\right )\right )^{\left (\frac {1}{x}\right )}}{x^{2} \left (\frac {1}{3} \, \log \left (x^{4}\right )\right )^{\frac {2}{x}} \log \left (x^{4}\right ) - 2 \, {\left (2 \, x^{2} \log \relax (2) + 25 \, x^{2}\right )} \left (\frac {1}{3} \, \log \left (x^{4}\right )\right )^{\left (\frac {1}{x}\right )} \log \left (x^{4}\right ) + {\left (4 \, x^{2} \log \relax (2)^{2} + 100 \, x^{2} \log \relax (2) + 625 \, x^{2}\right )} \log \left (x^{4}\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((12*log(x^4)*log(1/3*log(x^4))-48)*exp(log(1/3*log(x^4))/x)/(x^2*log(x^4)*exp(log(1/3*log(x^4))/x)^2
+(-4*x^2*log(2)-50*x^2)*log(x^4)*exp(log(1/3*log(x^4))/x)+(4*x^2*log(2)^2+100*x^2*log(2)+625*x^2)*log(x^4)),x,
algorithm="giac")

[Out]

integrate(12*(log(x^4)*log(1/3*log(x^4)) - 4)*(1/3*log(x^4))^(1/x)/(x^2*(1/3*log(x^4))^(2/x)*log(x^4) - 2*(2*x
^2*log(2) + 25*x^2)*(1/3*log(x^4))^(1/x)*log(x^4) + (4*x^2*log(2)^2 + 100*x^2*log(2) + 625*x^2)*log(x^4)), x)

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maple [F]  time = 0.23, size = 0, normalized size = 0.00 \[\int \frac {\left (12 \ln \left (x^{4}\right ) \ln \left (\frac {\ln \left (x^{4}\right )}{3}\right )-48\right ) {\mathrm e}^{\frac {\ln \left (\frac {\ln \left (x^{4}\right )}{3}\right )}{x}}}{x^{2} \ln \left (x^{4}\right ) {\mathrm e}^{\frac {2 \ln \left (\ln \left (\left (x^{4}\right )^{\frac {1}{3}}\right )\right )}{x}}+\left (-4 x^{2} \ln \relax (2)-50 x^{2}\right ) \ln \left (x^{4}\right ) {\mathrm e}^{\frac {\ln \left (\frac {\ln \left (x^{4}\right )}{3}\right )}{x}}+\left (4 x^{2} \ln \relax (2)^{2}+100 x^{2} \ln \relax (2)+625 x^{2}\right ) \ln \left (x^{4}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((12*ln(x^4)*ln(1/3*ln(x^4))-48)*exp(ln(1/3*ln(x^4))/x)/(x^2*ln(x^4)*exp(ln(1/3*ln(x^4))/x)^2+(-4*x^2*ln(2)
-50*x^2)*ln(x^4)*exp(ln(1/3*ln(x^4))/x)+(4*x^2*ln(2)^2+100*x^2*ln(2)+625*x^2)*ln(x^4)),x)

[Out]

int((12*ln(x^4)*ln(1/3*ln(x^4))-48)*exp(ln(1/3*ln(x^4))/x)/(x^2*ln(x^4)*exp(ln(1/3*ln(x^4))/x)^2+(-4*x^2*ln(2)
-50*x^2)*ln(x^4)*exp(ln(1/3*ln(x^4))/x)+(4*x^2*ln(2)^2+100*x^2*ln(2)+625*x^2)*ln(x^4)),x)

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maxima [A]  time = 0.92, size = 40, normalized size = 1.54 \begin {gather*} -\frac {12 \cdot 3^{\left (\frac {1}{x}\right )}}{3^{\left (\frac {1}{x}\right )} {\left (2 \, \log \relax (2) + 25\right )} - e^{\left (\frac {2 \, \log \relax (2)}{x} + \frac {\log \left (\log \relax (x)\right )}{x}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((12*log(x^4)*log(1/3*log(x^4))-48)*exp(log(1/3*log(x^4))/x)/(x^2*log(x^4)*exp(log(1/3*log(x^4))/x)^2
+(-4*x^2*log(2)-50*x^2)*log(x^4)*exp(log(1/3*log(x^4))/x)+(4*x^2*log(2)^2+100*x^2*log(2)+625*x^2)*log(x^4)),x,
algorithm="maxima")

[Out]

-12*3^(1/x)/(3^(1/x)*(2*log(2) + 25) - e^(2*log(2)/x + log(log(x))/x))

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mupad [B]  time = 3.33, size = 63, normalized size = 2.42 \begin {gather*} -\frac {24\,\mathrm {atanh}\left (\frac {\ln \left (16\right )-\frac {2\,{\ln \left (x^4\right )}^{1/x}}{3^{1/x}}+50}{\sqrt {\ln \left (16\right )-2\,\ln \relax (4)}\,\sqrt {2\,\ln \relax (4)+\ln \left (16\right )+100}}\right )}{\sqrt {\ln \left (16\right )-2\,\ln \relax (4)}\,\sqrt {2\,\ln \relax (4)+\ln \left (16\right )+100}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(log(log(x^4)/3)/x)*(12*log(log(x^4)/3)*log(x^4) - 48))/(log(x^4)*(4*x^2*log(2)^2 + 100*x^2*log(2) + 6
25*x^2) + x^2*log(x^4)*exp((2*log(log(x^4)/3))/x) - log(x^4)*exp(log(log(x^4)/3)/x)*(4*x^2*log(2) + 50*x^2)),x
)

[Out]

-(24*atanh((log(16) - (2*log(x^4)^(1/x))/3^(1/x) + 50)/((log(16) - 2*log(4))^(1/2)*(2*log(4) + log(16) + 100)^
(1/2))))/((log(16) - 2*log(4))^(1/2)*(2*log(4) + log(16) + 100)^(1/2))

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sympy [A]  time = 0.38, size = 19, normalized size = 0.73 \begin {gather*} \frac {12}{e^{\frac {\log {\left (\frac {\log {\left (x^{4} \right )}}{3} \right )}}{x}} - 25 - 2 \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((12*ln(x**4)*ln(1/3*ln(x**4))-48)*exp(ln(1/3*ln(x**4))/x)/(x**2*ln(x**4)*exp(ln(1/3*ln(x**4))/x)**2+
(-4*x**2*ln(2)-50*x**2)*ln(x**4)*exp(ln(1/3*ln(x**4))/x)+(4*x**2*ln(2)**2+100*x**2*ln(2)+625*x**2)*ln(x**4)),x
)

[Out]

12/(exp(log(log(x**4)/3)/x) - 25 - 2*log(2))

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