Integrand size = 154, antiderivative size = 25 \[ \int \frac {-4-9 x+8 x^2+(-7+8 x) \log \left (2 x^2\right )+\left (-2 x-2 \log \left (2 x^2\right )\right ) \log \left (x+\log \left (2 x^2\right )\right )}{49 x^3-56 x^4+16 x^5+\left (49 x^2-56 x^3+16 x^4\right ) \log \left (2 x^2\right )+\left (28 x^3-16 x^4+\left (28 x^2-16 x^3\right ) \log \left (2 x^2\right )\right ) \log \left (x+\log \left (2 x^2\right )\right )+\left (4 x^3+4 x^2 \log \left (2 x^2\right )\right ) \log ^2\left (x+\log \left (2 x^2\right )\right )} \, dx=\frac {1}{2 x \left (\frac {7}{2}-2 x+\log \left (x+\log \left (2 x^2\right )\right )\right )} \] Output:
1/2/x/(ln(ln(2*x^2)+x)-2*x+7/2)
Time = 0.51 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {-4-9 x+8 x^2+(-7+8 x) \log \left (2 x^2\right )+\left (-2 x-2 \log \left (2 x^2\right )\right ) \log \left (x+\log \left (2 x^2\right )\right )}{49 x^3-56 x^4+16 x^5+\left (49 x^2-56 x^3+16 x^4\right ) \log \left (2 x^2\right )+\left (28 x^3-16 x^4+\left (28 x^2-16 x^3\right ) \log \left (2 x^2\right )\right ) \log \left (x+\log \left (2 x^2\right )\right )+\left (4 x^3+4 x^2 \log \left (2 x^2\right )\right ) \log ^2\left (x+\log \left (2 x^2\right )\right )} \, dx=\frac {1}{x \left (7-4 x+2 \log \left (x+\log \left (2 x^2\right )\right )\right )} \] Input:
Integrate[(-4 - 9*x + 8*x^2 + (-7 + 8*x)*Log[2*x^2] + (-2*x - 2*Log[2*x^2] )*Log[x + Log[2*x^2]])/(49*x^3 - 56*x^4 + 16*x^5 + (49*x^2 - 56*x^3 + 16*x ^4)*Log[2*x^2] + (28*x^3 - 16*x^4 + (28*x^2 - 16*x^3)*Log[2*x^2])*Log[x + Log[2*x^2]] + (4*x^3 + 4*x^2*Log[2*x^2])*Log[x + Log[2*x^2]]^2),x]
Output:
1/(x*(7 - 4*x + 2*Log[x + Log[2*x^2]]))
Time = 0.62 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.013, Rules used = {7239, 7238}
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.
\(\displaystyle \int \frac {8 x^2+(8 x-7) \log \left (2 x^2\right )+\left (-2 \log \left (2 x^2\right )-2 x\right ) \log \left (\log \left (2 x^2\right )+x\right )-9 x-4}{16 x^5-56 x^4+49 x^3+\left (4 x^3+4 x^2 \log \left (2 x^2\right )\right ) \log ^2\left (\log \left (2 x^2\right )+x\right )+\left (16 x^4-56 x^3+49 x^2\right ) \log \left (2 x^2\right )+\left (-16 x^4+28 x^3+\left (28 x^2-16 x^3\right ) \log \left (2 x^2\right )\right ) \log \left (\log \left (2 x^2\right )+x\right )} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {8 x^2-2 x \log \left (\log \left (2 x^2\right )+x\right )+\log \left (2 x^2\right ) \left (-2 \log \left (\log \left (2 x^2\right )+x\right )+8 x-7\right )-9 x-4}{x^2 \left (\log \left (2 x^2\right )+x\right ) \left (2 \log \left (\log \left (2 x^2\right )+x\right )-4 x+7\right )^2}dx\) |
\(\Big \downarrow \) 7238 |
\(\displaystyle \frac {1}{x \left (2 \log \left (\log \left (2 x^2\right )+x\right )-4 x+7\right )}\) |
Input:
Int[(-4 - 9*x + 8*x^2 + (-7 + 8*x)*Log[2*x^2] + (-2*x - 2*Log[2*x^2])*Log[ x + Log[2*x^2]])/(49*x^3 - 56*x^4 + 16*x^5 + (49*x^2 - 56*x^3 + 16*x^4)*Lo g[2*x^2] + (28*x^3 - 16*x^4 + (28*x^2 - 16*x^3)*Log[2*x^2])*Log[x + Log[2* x^2]] + (4*x^3 + 4*x^2*Log[2*x^2])*Log[x + Log[2*x^2]]^2),x]
Output:
1/(x*(7 - 4*x + 2*Log[x + Log[2*x^2]]))
Int[(u_)*(y_)^(m_.)*(z_)^(n_.), x_Symbol] :> With[{q = DerivativeDivides[y* z, u*z^(n - m), x]}, Simp[q*y^(m + 1)*(z^(m + 1)/(m + 1)), x] /; !FalseQ[q ]] /; FreeQ[{m, n}, x] && NeQ[m, -1]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.60 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96
method | result | size |
parallelrisch | \(-\frac {1}{x \left (-7+4 x -2 \ln \left (\ln \left (2 x^{2}\right )+x \right )\right )}\) | \(24\) |
Input:
int(((-2*ln(2*x^2)-2*x)*ln(ln(2*x^2)+x)+(8*x-7)*ln(2*x^2)+8*x^2-9*x-4)/((4 *x^2*ln(2*x^2)+4*x^3)*ln(ln(2*x^2)+x)^2+((-16*x^3+28*x^2)*ln(2*x^2)-16*x^4 +28*x^3)*ln(ln(2*x^2)+x)+(16*x^4-56*x^3+49*x^2)*ln(2*x^2)+16*x^5-56*x^4+49 *x^3),x,method=_RETURNVERBOSE)
Output:
-1/x/(-7+4*x-2*ln(ln(2*x^2)+x))
Time = 0.08 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {-4-9 x+8 x^2+(-7+8 x) \log \left (2 x^2\right )+\left (-2 x-2 \log \left (2 x^2\right )\right ) \log \left (x+\log \left (2 x^2\right )\right )}{49 x^3-56 x^4+16 x^5+\left (49 x^2-56 x^3+16 x^4\right ) \log \left (2 x^2\right )+\left (28 x^3-16 x^4+\left (28 x^2-16 x^3\right ) \log \left (2 x^2\right )\right ) \log \left (x+\log \left (2 x^2\right )\right )+\left (4 x^3+4 x^2 \log \left (2 x^2\right )\right ) \log ^2\left (x+\log \left (2 x^2\right )\right )} \, dx=-\frac {1}{4 \, x^{2} - 2 \, x \log \left (x + \log \left (2 \, x^{2}\right )\right ) - 7 \, x} \] Input:
integrate(((-2*log(2*x^2)-2*x)*log(log(2*x^2)+x)+(8*x-7)*log(2*x^2)+8*x^2- 9*x-4)/((4*x^2*log(2*x^2)+4*x^3)*log(log(2*x^2)+x)^2+((-16*x^3+28*x^2)*log (2*x^2)-16*x^4+28*x^3)*log(log(2*x^2)+x)+(16*x^4-56*x^3+49*x^2)*log(2*x^2) +16*x^5-56*x^4+49*x^3),x, algorithm="fricas")
Output:
-1/(4*x^2 - 2*x*log(x + log(2*x^2)) - 7*x)
Time = 0.14 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {-4-9 x+8 x^2+(-7+8 x) \log \left (2 x^2\right )+\left (-2 x-2 \log \left (2 x^2\right )\right ) \log \left (x+\log \left (2 x^2\right )\right )}{49 x^3-56 x^4+16 x^5+\left (49 x^2-56 x^3+16 x^4\right ) \log \left (2 x^2\right )+\left (28 x^3-16 x^4+\left (28 x^2-16 x^3\right ) \log \left (2 x^2\right )\right ) \log \left (x+\log \left (2 x^2\right )\right )+\left (4 x^3+4 x^2 \log \left (2 x^2\right )\right ) \log ^2\left (x+\log \left (2 x^2\right )\right )} \, dx=\frac {1}{- 4 x^{2} + 2 x \log {\left (x + \log {\left (2 x^{2} \right )} \right )} + 7 x} \] Input:
integrate(((-2*ln(2*x**2)-2*x)*ln(ln(2*x**2)+x)+(8*x-7)*ln(2*x**2)+8*x**2- 9*x-4)/((4*x**2*ln(2*x**2)+4*x**3)*ln(ln(2*x**2)+x)**2+((-16*x**3+28*x**2) *ln(2*x**2)-16*x**4+28*x**3)*ln(ln(2*x**2)+x)+(16*x**4-56*x**3+49*x**2)*ln (2*x**2)+16*x**5-56*x**4+49*x**3),x)
Output:
1/(-4*x**2 + 2*x*log(x + log(2*x**2)) + 7*x)
Time = 0.16 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {-4-9 x+8 x^2+(-7+8 x) \log \left (2 x^2\right )+\left (-2 x-2 \log \left (2 x^2\right )\right ) \log \left (x+\log \left (2 x^2\right )\right )}{49 x^3-56 x^4+16 x^5+\left (49 x^2-56 x^3+16 x^4\right ) \log \left (2 x^2\right )+\left (28 x^3-16 x^4+\left (28 x^2-16 x^3\right ) \log \left (2 x^2\right )\right ) \log \left (x+\log \left (2 x^2\right )\right )+\left (4 x^3+4 x^2 \log \left (2 x^2\right )\right ) \log ^2\left (x+\log \left (2 x^2\right )\right )} \, dx=-\frac {1}{4 \, x^{2} - 2 \, x \log \left (x + \log \left (2\right ) + 2 \, \log \left (x\right )\right ) - 7 \, x} \] Input:
integrate(((-2*log(2*x^2)-2*x)*log(log(2*x^2)+x)+(8*x-7)*log(2*x^2)+8*x^2- 9*x-4)/((4*x^2*log(2*x^2)+4*x^3)*log(log(2*x^2)+x)^2+((-16*x^3+28*x^2)*log (2*x^2)-16*x^4+28*x^3)*log(log(2*x^2)+x)+(16*x^4-56*x^3+49*x^2)*log(2*x^2) +16*x^5-56*x^4+49*x^3),x, algorithm="maxima")
Output:
-1/(4*x^2 - 2*x*log(x + log(2) + 2*log(x)) - 7*x)
Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {-4-9 x+8 x^2+(-7+8 x) \log \left (2 x^2\right )+\left (-2 x-2 \log \left (2 x^2\right )\right ) \log \left (x+\log \left (2 x^2\right )\right )}{49 x^3-56 x^4+16 x^5+\left (49 x^2-56 x^3+16 x^4\right ) \log \left (2 x^2\right )+\left (28 x^3-16 x^4+\left (28 x^2-16 x^3\right ) \log \left (2 x^2\right )\right ) \log \left (x+\log \left (2 x^2\right )\right )+\left (4 x^3+4 x^2 \log \left (2 x^2\right )\right ) \log ^2\left (x+\log \left (2 x^2\right )\right )} \, dx=-\frac {1}{4 \, x^{2} - 2 \, x \log \left (x + \log \left (2 \, x^{2}\right )\right ) - 7 \, x} \] Input:
integrate(((-2*log(2*x^2)-2*x)*log(log(2*x^2)+x)+(8*x-7)*log(2*x^2)+8*x^2- 9*x-4)/((4*x^2*log(2*x^2)+4*x^3)*log(log(2*x^2)+x)^2+((-16*x^3+28*x^2)*log (2*x^2)-16*x^4+28*x^3)*log(log(2*x^2)+x)+(16*x^4-56*x^3+49*x^2)*log(2*x^2) +16*x^5-56*x^4+49*x^3),x, algorithm="giac")
Output:
-1/(4*x^2 - 2*x*log(x + log(2*x^2)) - 7*x)
Time = 4.63 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {-4-9 x+8 x^2+(-7+8 x) \log \left (2 x^2\right )+\left (-2 x-2 \log \left (2 x^2\right )\right ) \log \left (x+\log \left (2 x^2\right )\right )}{49 x^3-56 x^4+16 x^5+\left (49 x^2-56 x^3+16 x^4\right ) \log \left (2 x^2\right )+\left (28 x^3-16 x^4+\left (28 x^2-16 x^3\right ) \log \left (2 x^2\right )\right ) \log \left (x+\log \left (2 x^2\right )\right )+\left (4 x^3+4 x^2 \log \left (2 x^2\right )\right ) \log ^2\left (x+\log \left (2 x^2\right )\right )} \, dx=\frac {1}{x\,\left (2\,\ln \left (x+\ln \left (2\,x^2\right )\right )-4\,x+7\right )} \] Input:
int(-(9*x - log(2*x^2)*(8*x - 7) + log(x + log(2*x^2))*(2*x + 2*log(2*x^2) ) - 8*x^2 + 4)/(log(2*x^2)*(49*x^2 - 56*x^3 + 16*x^4) + log(x + log(2*x^2) )^2*(4*x^3 + 4*x^2*log(2*x^2)) + 49*x^3 - 56*x^4 + 16*x^5 + log(x + log(2* x^2))*(log(2*x^2)*(28*x^2 - 16*x^3) + 28*x^3 - 16*x^4)),x)
Output:
1/(x*(2*log(x + log(2*x^2)) - 4*x + 7))
Time = 0.21 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {-4-9 x+8 x^2+(-7+8 x) \log \left (2 x^2\right )+\left (-2 x-2 \log \left (2 x^2\right )\right ) \log \left (x+\log \left (2 x^2\right )\right )}{49 x^3-56 x^4+16 x^5+\left (49 x^2-56 x^3+16 x^4\right ) \log \left (2 x^2\right )+\left (28 x^3-16 x^4+\left (28 x^2-16 x^3\right ) \log \left (2 x^2\right )\right ) \log \left (x+\log \left (2 x^2\right )\right )+\left (4 x^3+4 x^2 \log \left (2 x^2\right )\right ) \log ^2\left (x+\log \left (2 x^2\right )\right )} \, dx=\frac {1}{x \left (2 \,\mathrm {log}\left (\mathrm {log}\left (2 x^{2}\right )+x \right )-4 x +7\right )} \] Input:
int(((-2*log(2*x^2)-2*x)*log(log(2*x^2)+x)+(8*x-7)*log(2*x^2)+8*x^2-9*x-4) /((4*x^2*log(2*x^2)+4*x^3)*log(log(2*x^2)+x)^2+((-16*x^3+28*x^2)*log(2*x^2 )-16*x^4+28*x^3)*log(log(2*x^2)+x)+(16*x^4-56*x^3+49*x^2)*log(2*x^2)+16*x^ 5-56*x^4+49*x^3),x)
Output:
1/(x*(2*log(log(2*x**2) + x) - 4*x + 7))