3.85.68 \(\int \frac {-20 x^2+(40-10 x^2) \log (-4+x^2)+(8 x-2 x^3) \log (-4+x^2) \log ^2(x \log (-4+x^2))}{(-4 x+x^3) \log (-4+x^2) \log ^2(x \log (-4+x^2))} \, dx\)

Optimal. Leaf size=18 \[ 5-2 x+\frac {10}{\log \left (x \log \left (-4+x^2\right )\right )} \]

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Rubi [A]  time = 0.43, antiderivative size = 17, normalized size of antiderivative = 0.94, number of steps used = 4, number of rules used = 3, integrand size = 76, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.039, Rules used = {1593, 6688, 6686} \begin {gather*} \frac {10}{\log \left (x \log \left (x^2-4\right )\right )}-2 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-20*x^2 + (40 - 10*x^2)*Log[-4 + x^2] + (8*x - 2*x^3)*Log[-4 + x^2]*Log[x*Log[-4 + x^2]]^2)/((-4*x + x^3)
*Log[-4 + x^2]*Log[x*Log[-4 + x^2]]^2),x]

[Out]

-2*x + 10/Log[x*Log[-4 + x^2]]

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-20 x^2+\left (40-10 x^2\right ) \log \left (-4+x^2\right )+\left (8 x-2 x^3\right ) \log \left (-4+x^2\right ) \log ^2\left (x \log \left (-4+x^2\right )\right )}{x \left (-4+x^2\right ) \log \left (-4+x^2\right ) \log ^2\left (x \log \left (-4+x^2\right )\right )} \, dx\\ &=\int \left (-2+\frac {-\frac {10}{x}-\frac {20 x}{\left (-4+x^2\right ) \log \left (-4+x^2\right )}}{\log ^2\left (x \log \left (-4+x^2\right )\right )}\right ) \, dx\\ &=-2 x+\int \frac {-\frac {10}{x}-\frac {20 x}{\left (-4+x^2\right ) \log \left (-4+x^2\right )}}{\log ^2\left (x \log \left (-4+x^2\right )\right )} \, dx\\ &=-2 x+\frac {10}{\log \left (x \log \left (-4+x^2\right )\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 17, normalized size = 0.94 \begin {gather*} -2 x+\frac {10}{\log \left (x \log \left (-4+x^2\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-20*x^2 + (40 - 10*x^2)*Log[-4 + x^2] + (8*x - 2*x^3)*Log[-4 + x^2]*Log[x*Log[-4 + x^2]]^2)/((-4*x
+ x^3)*Log[-4 + x^2]*Log[x*Log[-4 + x^2]]^2),x]

[Out]

-2*x + 10/Log[x*Log[-4 + x^2]]

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fricas [A]  time = 1.01, size = 26, normalized size = 1.44 \begin {gather*} -\frac {2 \, {\left (x \log \left (x \log \left (x^{2} - 4\right )\right ) - 5\right )}}{\log \left (x \log \left (x^{2} - 4\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^3+8*x)*log(x^2-4)*log(x*log(x^2-4))^2+(-10*x^2+40)*log(x^2-4)-20*x^2)/(x^3-4*x)/log(x^2-4)/lo
g(x*log(x^2-4))^2,x, algorithm="fricas")

[Out]

-2*(x*log(x*log(x^2 - 4)) - 5)/log(x*log(x^2 - 4))

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giac [A]  time = 1.11, size = 17, normalized size = 0.94 \begin {gather*} -2 \, x + \frac {10}{\log \left (x \log \left (x^{2} - 4\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^3+8*x)*log(x^2-4)*log(x*log(x^2-4))^2+(-10*x^2+40)*log(x^2-4)-20*x^2)/(x^3-4*x)/log(x^2-4)/lo
g(x*log(x^2-4))^2,x, algorithm="giac")

[Out]

-2*x + 10/log(x*log(x^2 - 4))

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maple [C]  time = 0.45, size = 116, normalized size = 6.44

method	result	size
risch	\(-2 x +\frac {20 i}{\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \ln \left (x^{2}-4\right )\right ) \mathrm {csgn}\left (i x \ln \left (x^{2}-4\right )\right )-\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \ln \left (x^{2}-4\right )\right )^{2}-\pi \,\mathrm {csgn}\left (i \ln \left (x^{2}-4\right )\right ) \mathrm {csgn}\left (i x \ln \left (x^{2}-4\right )\right )^{2}+\pi \mathrm {csgn}\left (i x \ln \left (x^{2}-4\right )\right )^{3}+2 i \ln \relax (x )+2 i \ln \left (\ln \left (x^{2}-4\right )\right )}\)	\(116\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x^3+8*x)*ln(x^2-4)*ln(x*ln(x^2-4))^2+(-10*x^2+40)*ln(x^2-4)-20*x^2)/(x^3-4*x)/ln(x^2-4)/ln(x*ln(x^2-4
))^2,x,method=_RETURNVERBOSE)

[Out]

-2*x+20*I/(Pi*csgn(I*x)*csgn(I*ln(x^2-4))*csgn(I*x*ln(x^2-4))-Pi*csgn(I*x)*csgn(I*x*ln(x^2-4))^2-Pi*csgn(I*ln(
x^2-4))*csgn(I*x*ln(x^2-4))^2+Pi*csgn(I*x*ln(x^2-4))^3+2*I*ln(x)+2*I*ln(ln(x^2-4)))

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maxima [A]  time = 0.49, size = 35, normalized size = 1.94 \begin {gather*} -\frac {2 \, {\left (x \log \relax (x) + x \log \left (\log \left (x + 2\right ) + \log \left (x - 2\right )\right ) - 5\right )}}{\log \relax (x) + \log \left (\log \left (x + 2\right ) + \log \left (x - 2\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^3+8*x)*log(x^2-4)*log(x*log(x^2-4))^2+(-10*x^2+40)*log(x^2-4)-20*x^2)/(x^3-4*x)/log(x^2-4)/lo
g(x*log(x^2-4))^2,x, algorithm="maxima")

[Out]

-2*(x*log(x) + x*log(log(x + 2) + log(x - 2)) - 5)/(log(x) + log(log(x + 2) + log(x - 2)))

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mupad [B]  time = 5.47, size = 17, normalized size = 0.94 \begin {gather*} \frac {10}{\ln \left (x\,\ln \left (x^2-4\right )\right )}-2\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x^2 - 4)*(10*x^2 - 40) + 20*x^2 - log(x^2 - 4)*log(x*log(x^2 - 4))^2*(8*x - 2*x^3))/(log(x^2 - 4)*log
(x*log(x^2 - 4))^2*(4*x - x^3)),x)

[Out]

10/log(x*log(x^2 - 4)) - 2*x

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sympy [A]  time = 0.33, size = 14, normalized size = 0.78 \begin {gather*} - 2 x + \frac {10}{\log {\left (x \log {\left (x^{2} - 4 \right )} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x**3+8*x)*ln(x**2-4)*ln(x*ln(x**2-4))**2+(-10*x**2+40)*ln(x**2-4)-20*x**2)/(x**3-4*x)/ln(x**2-4
)/ln(x*ln(x**2-4))**2,x)

[Out]

-2*x + 10/log(x*log(x**2 - 4))

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