3.48.42 \(\int -\frac {80 \log (x^2)}{x+4 x \log ^2(x^2)+4 x \log ^4(x^2)} \, dx\)

Optimal. Leaf size=14 \[ \frac {5}{\frac {1}{2}+\log ^2\left (x^2\right )} \]

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Rubi [A]  time = 0.05, antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {12, 199, 203, 261} \begin {gather*} \frac {10}{2 \log ^2\left (x^2\right )+1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-80*Log[x^2])/(x + 4*x*Log[x^2]^2 + 4*x*Log[x^2]^4),x]

[Out]

10/(1 + 2*Log[x^2]^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 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
 FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

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} &=-\left (80 \int \frac {\log \left (x^2\right )}{x+4 x \log ^2\left (x^2\right )+4 x \log ^4\left (x^2\right )} \, dx\right )\\ &=-\left (40 \operatorname {Subst}\left (\int \frac {x}{\left (1+2 x^2\right )^2} \, dx,x,\log \left (x^2\right )\right )\right )\\ &=\frac {10}{1+2 \log ^2\left (x^2\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 14, normalized size = 1.00 \begin {gather*} \frac {10}{1+2 \log ^2\left (x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-80*Log[x^2])/(x + 4*x*Log[x^2]^2 + 4*x*Log[x^2]^4),x]

[Out]

10/(1 + 2*Log[x^2]^2)

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fricas [A]  time = 0.93, size = 14, normalized size = 1.00 \begin {gather*} \frac {10}{2 \, \log \left (x^{2}\right )^{2} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-80*log(x^2)/(4*x*log(x^2)^4+4*x*log(x^2)^2+x),x, algorithm="fricas")

[Out]

10/(2*log(x^2)^2 + 1)

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giac [A]  time = 0.21, size = 14, normalized size = 1.00 \begin {gather*} \frac {10}{2 \, \log \left (x^{2}\right )^{2} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-80*log(x^2)/(4*x*log(x^2)^4+4*x*log(x^2)^2+x),x, algorithm="giac")

[Out]

10/(2*log(x^2)^2 + 1)

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maple [A]  time = 0.03, size = 15, normalized size = 1.07




method result size



default \(\frac {10}{2 \ln \left (x^{2}\right )^{2}+1}\) \(15\)
norman \(\frac {10}{2 \ln \left (x^{2}\right )^{2}+1}\) \(15\)
risch \(\frac {10}{2 \ln \left (x^{2}\right )^{2}+1}\) \(15\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-80*ln(x^2)/(4*x*ln(x^2)^4+4*x*ln(x^2)^2+x),x,method=_RETURNVERBOSE)

[Out]

10/(2*ln(x^2)^2+1)

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maxima [A]  time = 0.38, size = 12, normalized size = 0.86 \begin {gather*} \frac {10}{8 \, \log \relax (x)^{2} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-80*log(x^2)/(4*x*log(x^2)^4+4*x*log(x^2)^2+x),x, algorithm="maxima")

[Out]

10/(8*log(x)^2 + 1)

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mupad [B]  time = 3.30, size = 14, normalized size = 1.00 \begin {gather*} \frac {10}{2\,{\ln \left (x^2\right )}^2+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(80*log(x^2))/(x + 4*x*log(x^2)^2 + 4*x*log(x^2)^4),x)

[Out]

10/(2*log(x^2)^2 + 1)

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sympy [A]  time = 0.09, size = 10, normalized size = 0.71 \begin {gather*} \frac {10}{2 \log {\left (x^{2} \right )}^{2} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-80*ln(x**2)/(4*x*ln(x**2)**4+4*x*ln(x**2)**2+x),x)

[Out]

10/(2*log(x**2)**2 + 1)

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