3.49.43 \(\int \frac {-4+2 x-7 x^2-4 x^3+x^4+2 x^5-2 x \log (\log (4))}{1-2 x^2+x^4} \, dx\)

Optimal. Leaf size=26 \[ x+x^2-\frac {5 x+\log (\log (4))}{\left (\frac {1}{x}-x\right ) x} \]

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Rubi [A]  time = 0.04, antiderivative size = 23, normalized size of antiderivative = 0.88, number of steps used = 5, number of rules used = 4, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {6, 28, 1814, 1586} \begin {gather*} x^2-\frac {5 x+\log (\log (4))}{1-x^2}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-4 + 2*x - 7*x^2 - 4*x^3 + x^4 + 2*x^5 - 2*x*Log[Log[4]])/(1 - 2*x^2 + x^4),x]

[Out]

x + x^2 - (5*x + Log[Log[4]])/(1 - x^2)

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 1586

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

Rule 1814

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[((a
*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-4-7 x^2-4 x^3+x^4+2 x^5+x (2-2 \log (\log (4)))}{1-2 x^2+x^4} \, dx\\ &=\int \frac {-4-7 x^2-4 x^3+x^4+2 x^5+x (2-2 \log (\log (4)))}{\left (-1+x^2\right )^2} \, dx\\ &=-\frac {5 x+\log (\log (4))}{1-x^2}+\frac {1}{2} \int \frac {-2-4 x+2 x^2+4 x^3}{-1+x^2} \, dx\\ &=-\frac {5 x+\log (\log (4))}{1-x^2}+\frac {1}{2} \int (2+4 x) \, dx\\ &=x+x^2-\frac {5 x+\log (\log (4))}{1-x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 20, normalized size = 0.77 \begin {gather*} x+x^2+\frac {5 x+\log (\log (4))}{-1+x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4 + 2*x - 7*x^2 - 4*x^3 + x^4 + 2*x^5 - 2*x*Log[Log[4]])/(1 - 2*x^2 + x^4),x]

[Out]

x + x^2 + (5*x + Log[Log[4]])/(-1 + x^2)

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fricas [A]  time = 0.73, size = 28, normalized size = 1.08 \begin {gather*} \frac {x^{4} + x^{3} - x^{2} + 4 \, x + \log \left (2 \, \log \relax (2)\right )}{x^{2} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*log(2*log(2))+2*x^5+x^4-4*x^3-7*x^2+2*x-4)/(x^4-2*x^2+1),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.15, size = 22, normalized size = 0.85 \begin {gather*} x^{2} + x + \frac {5 \, x + \log \left (2 \, \log \relax (2)\right )}{x^{2} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*log(2*log(2))+2*x^5+x^4-4*x^3-7*x^2+2*x-4)/(x^4-2*x^2+1),x, algorithm="giac")

[Out]

x^2 + x + (5*x + log(2*log(2)))/(x^2 - 1)

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maple [A]  time = 0.06, size = 23, normalized size = 0.88

method	result	size
risch	\(x^{2}+x +\frac {5 x +\ln \relax (2)+\ln \left (\ln \relax (2)\right )}{x^{2}-1}\)	\(23\)
gosper	\(\frac {x^{4}+x^{3}+\ln \left (2 \ln \relax (2)\right )+4 x -1}{x^{2}-1}\)	\(25\)
norman	\(\frac {x^{3}+x^{4}+4 x -1+\ln \relax (2)+\ln \left (\ln \relax (2)\right )}{x^{2}-1}\)	\(25\)
default	\(x^{2}+x -\frac {-\frac {5}{2}-\frac {\ln \left (2 \ln \relax (2)\right )}{2}}{x -1}-\frac {-\frac {5}{2}+\frac {\ln \left (2 \ln \relax (2)\right )}{2}}{x +1}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*x*ln(2*ln(2))+2*x^5+x^4-4*x^3-7*x^2+2*x-4)/(x^4-2*x^2+1),x,method=_RETURNVERBOSE)

[Out]

x^2+x+(5*x+ln(2)+ln(ln(2)))/(x^2-1)

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maxima [A]  time = 0.36, size = 22, normalized size = 0.85 \begin {gather*} x^{2} + x + \frac {5 \, x + \log \left (2 \, \log \relax (2)\right )}{x^{2} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*log(2*log(2))+2*x^5+x^4-4*x^3-7*x^2+2*x-4)/(x^4-2*x^2+1),x, algorithm="maxima")

[Out]

x^2 + x + (5*x + log(2*log(2)))/(x^2 - 1)

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mupad [B]  time = 0.07, size = 20, normalized size = 0.77 \begin {gather*} x+\frac {5\,x+\ln \left (\ln \relax (4)\right )}{x^2-1}+x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x*log(2*log(2)) - 2*x + 7*x^2 + 4*x^3 - x^4 - 2*x^5 + 4)/(x^4 - 2*x^2 + 1),x)

[Out]

x + (5*x + log(log(4)))/(x^2 - 1) + x^2

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sympy [A]  time = 0.40, size = 20, normalized size = 0.77 \begin {gather*} x^{2} + x + \frac {5 x + \log {\left (\log {\relax (2 )} \right )} + \log {\relax (2 )}}{x^{2} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*ln(2*ln(2))+2*x**5+x**4-4*x**3-7*x**2+2*x-4)/(x**4-2*x**2+1),x)

[Out]

x**2 + x + (5*x + log(log(2)) + log(2))/(x**2 - 1)

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