3.39.76 \(\int \frac {2+4 x-2 x^2-4 x^3+(-1-2 x+x^2+2 x^3) \log (4)}{-4 x^3+4 x^4+4 x^5+(4 x^2-4 x^3-4 x^4) \log (\frac {6-6 x-6 x^2}{x})+(-x+x^2+x^3) \log ^2(\frac {6-6 x-6 x^2}{x})} \, dx\)

Optimal. Leaf size=36 \[ \frac {-2+\log (4)}{x \left (-2+\frac {\log \left (3 \left (-3-x+\frac {2+x-x^2}{x}\right )\right )}{x}\right )} \]

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Rubi [A]  time = 0.28, antiderivative size = 26, normalized size of antiderivative = 0.72, number of steps used = 3, number of rules used = 3, integrand size = 110, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.027, Rules used = {6688, 12, 6686} \begin {gather*} \frac {2-\log (4)}{2 x-\log \left (-6 x+\frac {6}{x}-6\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 + 4*x - 2*x^2 - 4*x^3 + (-1 - 2*x + x^2 + 2*x^3)*Log[4])/(-4*x^3 + 4*x^4 + 4*x^5 + (4*x^2 - 4*x^3 - 4*x
^4)*Log[(6 - 6*x - 6*x^2)/x] + (-x + x^2 + x^3)*Log[(6 - 6*x - 6*x^2)/x]^2),x]

[Out]

(2 - Log[4])/(2*x - Log[-6 + 6/x - 6*x])

Rule 12

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

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 {\left (1+2 x-x^2-2 x^3\right ) (-2+\log (4))}{x \left (1-x-x^2\right ) \left (2 x-\log \left (-6+\frac {6}{x}-6 x\right )\right )^2} \, dx\\ &=(-2+\log (4)) \int \frac {1+2 x-x^2-2 x^3}{x \left (1-x-x^2\right ) \left (2 x-\log \left (-6+\frac {6}{x}-6 x\right )\right )^2} \, dx\\ &=\frac {2-\log (4)}{2 x-\log \left (-6+\frac {6}{x}-6 x\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 22, normalized size = 0.61 \begin {gather*} \frac {-2+\log (4)}{-2 x+\log \left (-6+\frac {6}{x}-6 x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 + 4*x - 2*x^2 - 4*x^3 + (-1 - 2*x + x^2 + 2*x^3)*Log[4])/(-4*x^3 + 4*x^4 + 4*x^5 + (4*x^2 - 4*x^3
- 4*x^4)*Log[(6 - 6*x - 6*x^2)/x] + (-x + x^2 + x^3)*Log[(6 - 6*x - 6*x^2)/x]^2),x]

[Out]

(-2 + Log[4])/(-2*x + Log[-6 + 6/x - 6*x])

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fricas [A]  time = 0.90, size = 26, normalized size = 0.72 \begin {gather*} -\frac {2 \, {\left (\log \relax (2) - 1\right )}}{2 \, x - \log \left (-\frac {6 \, {\left (x^{2} + x - 1\right )}}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(2*x^3+x^2-2*x-1)*log(2)-4*x^3-2*x^2+4*x+2)/((x^3+x^2-x)*log((-6*x^2-6*x+6)/x)^2+(-4*x^4-4*x^3+4*
x^2)*log((-6*x^2-6*x+6)/x)+4*x^5+4*x^4-4*x^3),x, algorithm="fricas")

[Out]

-2*(log(2) - 1)/(2*x - log(-6*(x^2 + x - 1)/x))

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giac [A]  time = 0.27, size = 31, normalized size = 0.86 \begin {gather*} -\frac {2 \, {\left (\log \relax (2) - 1\right )}}{2 \, x - \log \relax (2) - \log \left (-3 \, x^{2} - 3 \, x + 3\right ) + \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(2*x^3+x^2-2*x-1)*log(2)-4*x^3-2*x^2+4*x+2)/((x^3+x^2-x)*log((-6*x^2-6*x+6)/x)^2+(-4*x^4-4*x^3+4*
x^2)*log((-6*x^2-6*x+6)/x)+4*x^5+4*x^4-4*x^3),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.54, size = 31, normalized size = 0.86

method	result	size
norman	\(\frac {2-2 \ln \relax (2)}{2 x -\ln \left (\frac {-6 x^{2}-6 x +6}{x}\right )}\)	\(31\)
default	\(\frac {2 \ln \relax (2)}{\ln \relax (6)-2 x +\ln \left (-\frac {x^{2}+x -1}{x}\right )}-\frac {2}{\ln \relax (6)-2 x +\ln \left (-\frac {x^{2}+x -1}{x}\right )}\)	\(48\)
risch	\(-\frac {2 \ln \relax (2)}{2 x -\ln \left (\frac {-6 x^{2}-6 x +6}{x}\right )}+\frac {2}{2 x -\ln \left (\frac {-6 x^{2}-6 x +6}{x}\right )}\)	\(54\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*(2*x^3+x^2-2*x-1)*ln(2)-4*x^3-2*x^2+4*x+2)/((x^3+x^2-x)*ln((-6*x^2-6*x+6)/x)^2+(-4*x^4-4*x^3+4*x^2)*ln(
(-6*x^2-6*x+6)/x)+4*x^5+4*x^4-4*x^3),x,method=_RETURNVERBOSE)

[Out]

(2-2*ln(2))/(2*x-ln((-6*x^2-6*x+6)/x))

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maxima [C]  time = 0.47, size = 30, normalized size = 0.83 \begin {gather*} \frac {2 \, {\left (\log \relax (2) - 1\right )}}{i \, \pi - 2 \, x + \log \relax (3) + \log \relax (2) + \log \left (x^{2} + x - 1\right ) - \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(2*x^3+x^2-2*x-1)*log(2)-4*x^3-2*x^2+4*x+2)/((x^3+x^2-x)*log((-6*x^2-6*x+6)/x)^2+(-4*x^4-4*x^3+4*
x^2)*log((-6*x^2-6*x+6)/x)+4*x^5+4*x^4-4*x^3),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.47, size = 43, normalized size = 1.19 \begin {gather*} -\frac {\ln \relax (4)+x\,\left (\ln \left (16\right )-4\right )-2}{\left (2\,x+1\right )\,\left (2\,x-\ln \left (-\frac {6\,x^2+6\,x-6}{x}\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*log(2)*(2*x - x^2 - 2*x^3 + 1) - 4*x + 2*x^2 + 4*x^3 - 2)/(log(-(6*x + 6*x^2 - 6)/x)^2*(x^2 - x + x^3)
- log(-(6*x + 6*x^2 - 6)/x)*(4*x^3 - 4*x^2 + 4*x^4) - 4*x^3 + 4*x^4 + 4*x^5),x)

[Out]

-(log(4) + x*(log(16) - 4) - 2)/((2*x + 1)*(2*x - log(-(6*x + 6*x^2 - 6)/x)))

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sympy [A]  time = 0.18, size = 22, normalized size = 0.61 \begin {gather*} \frac {-2 + 2 \log {\relax (2 )}}{- 2 x + \log {\left (\frac {- 6 x^{2} - 6 x + 6}{x} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(2*x**3+x**2-2*x-1)*ln(2)-4*x**3-2*x**2+4*x+2)/((x**3+x**2-x)*ln((-6*x**2-6*x+6)/x)**2+(-4*x**4-4
*x**3+4*x**2)*ln((-6*x**2-6*x+6)/x)+4*x**5+4*x**4-4*x**3),x)

[Out]

(-2 + 2*log(2))/(-2*x + log((-6*x**2 - 6*x + 6)/x))

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