Integrand size = 127, antiderivative size = 25 \[ \int \frac {-28+13 x^2+8 x^3-2 x^4+\left (28+16 x-15 x^2-4 x^3+2 x^4\right ) \log \left (\frac {7+4 x-2 x^2}{-4 x+x^3}\right ) \log \left (\log \left (\frac {7+4 x-2 x^2}{-4 x+x^3}\right )\right )}{\left (28+16 x-15 x^2-4 x^3+2 x^4\right ) \log \left (\frac {7+4 x-2 x^2}{-4 x+x^3}\right )} \, dx=x \log \left (\log \left (-\frac {2}{x}+\frac {4-\frac {1}{x}}{-4+x^2}\right )\right ) \]
Time = 0.07 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {-28+13 x^2+8 x^3-2 x^4+\left (28+16 x-15 x^2-4 x^3+2 x^4\right ) \log \left (\frac {7+4 x-2 x^2}{-4 x+x^3}\right ) \log \left (\log \left (\frac {7+4 x-2 x^2}{-4 x+x^3}\right )\right )}{\left (28+16 x-15 x^2-4 x^3+2 x^4\right ) \log \left (\frac {7+4 x-2 x^2}{-4 x+x^3}\right )} \, dx=x \log \left (\log \left (\frac {7+4 x-2 x^2}{-4 x+x^3}\right )\right ) \]
Integrate[(-28 + 13*x^2 + 8*x^3 - 2*x^4 + (28 + 16*x - 15*x^2 - 4*x^3 + 2* x^4)*Log[(7 + 4*x - 2*x^2)/(-4*x + x^3)]*Log[Log[(7 + 4*x - 2*x^2)/(-4*x + x^3)]])/((28 + 16*x - 15*x^2 - 4*x^3 + 2*x^4)*Log[(7 + 4*x - 2*x^2)/(-4*x + x^3)]),x]
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {-2 x^4+8 x^3+13 x^2+\left (2 x^4-4 x^3-15 x^2+16 x+28\right ) \log \left (\frac {-2 x^2+4 x+7}{x^3-4 x}\right ) \log \left (\log \left (\frac {-2 x^2+4 x+7}{x^3-4 x}\right )\right )-28}{\left (2 x^4-4 x^3-15 x^2+16 x+28\right ) \log \left (\frac {-2 x^2+4 x+7}{x^3-4 x}\right )} \, dx\) |
\(\Big \downarrow \) 2463 |
\(\displaystyle \int \left (-\frac {-2 x^4+8 x^3+13 x^2+\left (2 x^4-4 x^3-15 x^2+16 x+28\right ) \log \left (\frac {-2 x^2+4 x+7}{x^3-4 x}\right ) \log \left (\log \left (\frac {-2 x^2+4 x+7}{x^3-4 x}\right )\right )-28}{28 (x-2) \log \left (\frac {-2 x^2+4 x+7}{x^3-4 x}\right )}-\frac {-2 x^4+8 x^3+13 x^2+\left (2 x^4-4 x^3-15 x^2+16 x+28\right ) \log \left (\frac {-2 x^2+4 x+7}{x^3-4 x}\right ) \log \left (\log \left (\frac {-2 x^2+4 x+7}{x^3-4 x}\right )\right )-28}{36 (x+2) \log \left (\frac {-2 x^2+4 x+7}{x^3-4 x}\right )}+\frac {2 (4 x-7) \left (-2 x^4+8 x^3+13 x^2+\left (2 x^4-4 x^3-15 x^2+16 x+28\right ) \log \left (\frac {-2 x^2+4 x+7}{x^3-4 x}\right ) \log \left (\log \left (\frac {-2 x^2+4 x+7}{x^3-4 x}\right )\right )-28\right )}{63 \left (2 x^2-4 x-7\right ) \log \left (\frac {-2 x^2+4 x+7}{x^3-4 x}\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\int \frac {1}{\log \left (\frac {-2 x^2+4 x+7}{x \left (x^2-4\right )}\right )}dx-\frac {14}{3} \sqrt {2} \int \frac {1}{\left (-4 x+6 \sqrt {2}+4\right ) \log \left (\frac {-2 x^2+4 x+7}{x \left (x^2-4\right )}\right )}dx-2 \int \frac {1}{(x-2) \log \left (\frac {-2 x^2+4 x+7}{x \left (x^2-4\right )}\right )}dx+2 \int \frac {1}{(x+2) \log \left (\frac {-2 x^2+4 x+7}{x \left (x^2-4\right )}\right )}dx+\frac {4}{3} \left (3+\sqrt {2}\right ) \int \frac {1}{\left (4 x-6 \sqrt {2}-4\right ) \log \left (\frac {-2 x^2+4 x+7}{x \left (x^2-4\right )}\right )}dx+\frac {4}{3} \left (3-\sqrt {2}\right ) \int \frac {1}{\left (4 x+6 \sqrt {2}-4\right ) \log \left (\frac {-2 x^2+4 x+7}{x \left (x^2-4\right )}\right )}dx-\frac {14}{3} \sqrt {2} \int \frac {1}{\left (4 x+6 \sqrt {2}-4\right ) \log \left (\frac {-2 x^2+4 x+7}{x \left (x^2-4\right )}\right )}dx+\int \log \left (\log \left (\frac {-2 x^2+4 x+7}{x \left (x^2-4\right )}\right )\right )dx\) |
Int[(-28 + 13*x^2 + 8*x^3 - 2*x^4 + (28 + 16*x - 15*x^2 - 4*x^3 + 2*x^4)*L og[(7 + 4*x - 2*x^2)/(-4*x + x^3)]*Log[Log[(7 + 4*x - 2*x^2)/(-4*x + x^3)] ])/((28 + 16*x - 15*x^2 - 4*x^3 + 2*x^4)*Log[(7 + 4*x - 2*x^2)/(-4*x + x^3 )]),x]
3.8.97.3.1 Defintions of rubi rules used
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px]}, Int[ExpandIntegr and[u, Qx^p, x], x] /; !SumQ[NonfreeFactors[Qx, x]]] /; PolyQ[Px, x] && Gt Q[Expon[Px, x], 2] && !BinomialQ[Px, x] && !TrinomialQ[Px, x] && ILtQ[p, 0]
Time = 8.87 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08
method | result | size |
parallelrisch | \(x \ln \left (\ln \left (-\frac {2 x^{2}-4 x -7}{x \left (x^{2}-4\right )}\right )\right )\) | \(27\) |
int(((2*x^4-4*x^3-15*x^2+16*x+28)*ln((-2*x^2+4*x+7)/(x^3-4*x))*ln(ln((-2*x ^2+4*x+7)/(x^3-4*x)))-2*x^4+8*x^3+13*x^2-28)/(2*x^4-4*x^3-15*x^2+16*x+28)/ ln((-2*x^2+4*x+7)/(x^3-4*x)),x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {-28+13 x^2+8 x^3-2 x^4+\left (28+16 x-15 x^2-4 x^3+2 x^4\right ) \log \left (\frac {7+4 x-2 x^2}{-4 x+x^3}\right ) \log \left (\log \left (\frac {7+4 x-2 x^2}{-4 x+x^3}\right )\right )}{\left (28+16 x-15 x^2-4 x^3+2 x^4\right ) \log \left (\frac {7+4 x-2 x^2}{-4 x+x^3}\right )} \, dx=x \log \left (\log \left (-\frac {2 \, x^{2} - 4 \, x - 7}{x^{3} - 4 \, x}\right )\right ) \]
integrate(((2*x^4-4*x^3-15*x^2+16*x+28)*log((-2*x^2+4*x+7)/(x^3-4*x))*log( log((-2*x^2+4*x+7)/(x^3-4*x)))-2*x^4+8*x^3+13*x^2-28)/(2*x^4-4*x^3-15*x^2+ 16*x+28)/log((-2*x^2+4*x+7)/(x^3-4*x)),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (17) = 34\).
Time = 0.68 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.84 \[ \int \frac {-28+13 x^2+8 x^3-2 x^4+\left (28+16 x-15 x^2-4 x^3+2 x^4\right ) \log \left (\frac {7+4 x-2 x^2}{-4 x+x^3}\right ) \log \left (\log \left (\frac {7+4 x-2 x^2}{-4 x+x^3}\right )\right )}{\left (28+16 x-15 x^2-4 x^3+2 x^4\right ) \log \left (\frac {7+4 x-2 x^2}{-4 x+x^3}\right )} \, dx=\left (x - \frac {1}{15}\right ) \log {\left (\log {\left (\frac {- 2 x^{2} + 4 x + 7}{x^{3} - 4 x} \right )} \right )} + \frac {\log {\left (\log {\left (\frac {- 2 x^{2} + 4 x + 7}{x^{3} - 4 x} \right )} \right )}}{15} \]
integrate(((2*x**4-4*x**3-15*x**2+16*x+28)*ln((-2*x**2+4*x+7)/(x**3-4*x))* ln(ln((-2*x**2+4*x+7)/(x**3-4*x)))-2*x**4+8*x**3+13*x**2-28)/(2*x**4-4*x** 3-15*x**2+16*x+28)/ln((-2*x**2+4*x+7)/(x**3-4*x)),x)
(x - 1/15)*log(log((-2*x**2 + 4*x + 7)/(x**3 - 4*x))) + log(log((-2*x**2 + 4*x + 7)/(x**3 - 4*x)))/15
Time = 0.24 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \frac {-28+13 x^2+8 x^3-2 x^4+\left (28+16 x-15 x^2-4 x^3+2 x^4\right ) \log \left (\frac {7+4 x-2 x^2}{-4 x+x^3}\right ) \log \left (\log \left (\frac {7+4 x-2 x^2}{-4 x+x^3}\right )\right )}{\left (28+16 x-15 x^2-4 x^3+2 x^4\right ) \log \left (\frac {7+4 x-2 x^2}{-4 x+x^3}\right )} \, dx=x \log \left (\log \left (-2 \, x^{2} + 4 \, x + 7\right ) - \log \left (x + 2\right ) - \log \left (x - 2\right ) - \log \left (x\right )\right ) \]
integrate(((2*x^4-4*x^3-15*x^2+16*x+28)*log((-2*x^2+4*x+7)/(x^3-4*x))*log( log((-2*x^2+4*x+7)/(x^3-4*x)))-2*x^4+8*x^3+13*x^2-28)/(2*x^4-4*x^3-15*x^2+ 16*x+28)/log((-2*x^2+4*x+7)/(x^3-4*x)),x, algorithm=\
Time = 0.63 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {-28+13 x^2+8 x^3-2 x^4+\left (28+16 x-15 x^2-4 x^3+2 x^4\right ) \log \left (\frac {7+4 x-2 x^2}{-4 x+x^3}\right ) \log \left (\log \left (\frac {7+4 x-2 x^2}{-4 x+x^3}\right )\right )}{\left (28+16 x-15 x^2-4 x^3+2 x^4\right ) \log \left (\frac {7+4 x-2 x^2}{-4 x+x^3}\right )} \, dx=x \log \left (\log \left (-\frac {2 \, x^{2} - 4 \, x - 7}{x^{3} - 4 \, x}\right )\right ) \]
integrate(((2*x^4-4*x^3-15*x^2+16*x+28)*log((-2*x^2+4*x+7)/(x^3-4*x))*log( log((-2*x^2+4*x+7)/(x^3-4*x)))-2*x^4+8*x^3+13*x^2-28)/(2*x^4-4*x^3-15*x^2+ 16*x+28)/log((-2*x^2+4*x+7)/(x^3-4*x)),x, algorithm=\
Time = 10.13 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {-28+13 x^2+8 x^3-2 x^4+\left (28+16 x-15 x^2-4 x^3+2 x^4\right ) \log \left (\frac {7+4 x-2 x^2}{-4 x+x^3}\right ) \log \left (\log \left (\frac {7+4 x-2 x^2}{-4 x+x^3}\right )\right )}{\left (28+16 x-15 x^2-4 x^3+2 x^4\right ) \log \left (\frac {7+4 x-2 x^2}{-4 x+x^3}\right )} \, dx=x\,\ln \left (\ln \left (-\frac {-2\,x^2+4\,x+7}{4\,x-x^3}\right )\right ) \]