3.36.76 \(\int \frac {-4+13 x-11 x^2+2 x^3+(4-9 x+2 x^2) \log (4)+(-2-3 x+x^2+x^3+(-2-2 x+2 x^2) \log (4)) \log (1+x-x^2)}{(-4+x+8 x^2-6 x^3+x^4+(4+3 x-5 x^2+x^3) \log (4)) \log (1+x-x^2)} \, dx\)

Optimal. Leaf size=25 \[ \log \left (\frac {(4-x)^2 \log \left (1+x-x^2\right )}{-1+x+\log (4)}\right ) \]

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Rubi [A]  time = 0.81, antiderivative size = 40, normalized size of antiderivative = 1.60, number of steps used = 6, number of rules used = 4, integrand size = 108, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {6688, 6742, 72, 6684} \begin {gather*} \log \left (\log \left (-x^2+x+1\right )\right )+\frac {(6+\log (16)) \log (4-x)}{3+\log (4)}-\log (-x+1-\log (4)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-4 + 13*x - 11*x^2 + 2*x^3 + (4 - 9*x + 2*x^2)*Log[4] + (-2 - 3*x + x^2 + x^3 + (-2 - 2*x + 2*x^2)*Log[4]
)*Log[1 + x - x^2])/((-4 + x + 8*x^2 - 6*x^3 + x^4 + (4 + 3*x - 5*x^2 + x^3)*Log[4])*Log[1 + x - x^2]),x]

[Out]

((6 + Log[16])*Log[4 - x])/(3 + Log[4]) - Log[1 - x - Log[4]] + Log[Log[1 + x - x^2]]

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6688

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-\left (\left (4-9 x+2 x^2\right ) (-1+x+\log (4))\right )-\left (-1-x+x^2\right ) (2+x+\log (16)) \log \left (1+x-x^2\right )}{\left (4+3 x-5 x^2+x^3\right ) (1-x-\log (4)) \log \left (1+x-x^2\right )} \, dx\\ &=\int \left (\frac {2+x+\log (16)}{(-4+x) (-1+x+\log (4))}+\frac {-1+2 x}{\left (-1-x+x^2\right ) \log \left (1+x-x^2\right )}\right ) \, dx\\ &=\int \frac {2+x+\log (16)}{(-4+x) (-1+x+\log (4))} \, dx+\int \frac {-1+2 x}{\left (-1-x+x^2\right ) \log \left (1+x-x^2\right )} \, dx\\ &=\log \left (\log \left (1+x-x^2\right )\right )+\int \left (\frac {1}{1-x-\log (4)}+\frac {6+\log (16)}{(-4+x) (3+\log (4))}\right ) \, dx\\ &=\frac {(6+\log (16)) \log (4-x)}{3+\log (4)}-\log (1-x-\log (4))+\log \left (\log \left (1+x-x^2\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.34, size = 40, normalized size = 1.60 \begin {gather*} \frac {(6+\log (16)) \log (4-x)}{3+\log (4)}-\log (1-x-\log (4))+\log \left (\log \left (1+x-x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4 + 13*x - 11*x^2 + 2*x^3 + (4 - 9*x + 2*x^2)*Log[4] + (-2 - 3*x + x^2 + x^3 + (-2 - 2*x + 2*x^2)*
Log[4])*Log[1 + x - x^2])/((-4 + x + 8*x^2 - 6*x^3 + x^4 + (4 + 3*x - 5*x^2 + x^3)*Log[4])*Log[1 + x - x^2]),x
]

[Out]

((6 + Log[16])*Log[4 - x])/(3 + Log[4]) - Log[1 - x - Log[4]] + Log[Log[1 + x - x^2]]

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fricas [A]  time = 0.53, size = 27, normalized size = 1.08 \begin {gather*} -\log \left (x + 2 \, \log \relax (2) - 1\right ) + 2 \, \log \left (x - 4\right ) + \log \left (\log \left (-x^{2} + x + 1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*(2*x^2-2*x-2)*log(2)+x^3+x^2-3*x-2)*log(-x^2+x+1)+2*(2*x^2-9*x+4)*log(2)+2*x^3-11*x^2+13*x-4)/(2
*(x^3-5*x^2+3*x+4)*log(2)+x^4-6*x^3+8*x^2+x-4)/log(-x^2+x+1),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.18, size = 27, normalized size = 1.08 \begin {gather*} -\log \left (x + 2 \, \log \relax (2) - 1\right ) + 2 \, \log \left (x - 4\right ) + \log \left (\log \left (-x^{2} + x + 1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*(2*x^2-2*x-2)*log(2)+x^3+x^2-3*x-2)*log(-x^2+x+1)+2*(2*x^2-9*x+4)*log(2)+2*x^3-11*x^2+13*x-4)/(2
*(x^3-5*x^2+3*x+4)*log(2)+x^4-6*x^3+8*x^2+x-4)/log(-x^2+x+1),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.14, size = 28, normalized size = 1.12

method	result	size
default	\(-\ln \left (x +2 \ln \relax (2)-1\right )+2 \ln \left (x -4\right )+\ln \left (\ln \left (-x^{2}+x +1\right )\right )\)	\(28\)
norman	\(-\ln \left (x +2 \ln \relax (2)-1\right )+2 \ln \left (x -4\right )+\ln \left (\ln \left (-x^{2}+x +1\right )\right )\)	\(28\)
risch	\(-\ln \left (x +2 \ln \relax (2)-1\right )+2 \ln \left (x -4\right )+\ln \left (\ln \left (-x^{2}+x +1\right )\right )\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*(2*x^2-2*x-2)*ln(2)+x^3+x^2-3*x-2)*ln(-x^2+x+1)+2*(2*x^2-9*x+4)*ln(2)+2*x^3-11*x^2+13*x-4)/(2*(x^3-5*x
^2+3*x+4)*ln(2)+x^4-6*x^3+8*x^2+x-4)/ln(-x^2+x+1),x,method=_RETURNVERBOSE)

[Out]

-ln(x+2*ln(2)-1)+2*ln(x-4)+ln(ln(-x^2+x+1))

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maxima [A]  time = 0.57, size = 27, normalized size = 1.08 \begin {gather*} -\log \left (x + 2 \, \log \relax (2) - 1\right ) + 2 \, \log \left (x - 4\right ) + \log \left (\log \left (-x^{2} + x + 1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*(2*x^2-2*x-2)*log(2)+x^3+x^2-3*x-2)*log(-x^2+x+1)+2*(2*x^2-9*x+4)*log(2)+2*x^3-11*x^2+13*x-4)/(2
*(x^3-5*x^2+3*x+4)*log(2)+x^4-6*x^3+8*x^2+x-4)/log(-x^2+x+1),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.40, size = 107, normalized size = 4.28 \begin {gather*} \ln \left (\ln \left (-x^2+x+1\right )\right )+\frac {\ln \left (x-4\right )\,\left (2\,\ln \left (16\right )-\ln \relax (4)+\sqrt {4\,\ln \left (256\right )-10\,\ln \relax (4)+{\ln \relax (4)}^2+9}+9\right )}{2\,\sqrt {4\,\ln \left (256\right )-10\,\ln \relax (4)+{\ln \relax (4)}^2+9}}+\frac {\ln \left (x+\ln \relax (4)-1\right )\,\left (\ln \relax (4)-2\,\ln \left (16\right )+\sqrt {4\,\ln \left (256\right )-10\,\ln \relax (4)+{\ln \relax (4)}^2+9}-9\right )}{2\,\sqrt {4\,\ln \left (256\right )-10\,\ln \relax (4)+{\ln \relax (4)}^2+9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((13*x + 2*log(2)*(2*x^2 - 9*x + 4) - log(x - x^2 + 1)*(3*x + 2*log(2)*(2*x - 2*x^2 + 2) - x^2 - x^3 + 2) -
11*x^2 + 2*x^3 - 4)/(log(x - x^2 + 1)*(x + 2*log(2)*(3*x - 5*x^2 + x^3 + 4) + 8*x^2 - 6*x^3 + x^4 - 4)),x)

[Out]

log(log(x - x^2 + 1)) + (log(x - 4)*(2*log(16) - log(4) + (4*log(256) - 10*log(4) + log(4)^2 + 9)^(1/2) + 9))/
(2*(4*log(256) - 10*log(4) + log(4)^2 + 9)^(1/2)) + (log(x + log(4) - 1)*(log(4) - 2*log(16) + (4*log(256) - 1
0*log(4) + log(4)^2 + 9)^(1/2) - 9))/(2*(4*log(256) - 10*log(4) + log(4)^2 + 9)^(1/2))

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sympy [A]  time = 0.68, size = 26, normalized size = 1.04 \begin {gather*} 2 \log {\left (x - 4 \right )} - \log {\left (x - 1 + 2 \log {\relax (2 )} \right )} + \log {\left (\log {\left (- x^{2} + x + 1 \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*(2*x**2-2*x-2)*ln(2)+x**3+x**2-3*x-2)*ln(-x**2+x+1)+2*(2*x**2-9*x+4)*ln(2)+2*x**3-11*x**2+13*x-4
)/(2*(x**3-5*x**2+3*x+4)*ln(2)+x**4-6*x**3+8*x**2+x-4)/ln(-x**2+x+1),x)

[Out]

2*log(x - 4) - log(x - 1 + 2*log(2)) + log(log(-x**2 + x + 1))

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