Integrand size = 90, antiderivative size = 29 \[ \int \frac {20 \log ^2(2)}{640 x+448 x^2+\left (480 x+336 x^2\right ) \log \left (\frac {-10-7 x}{2 x}\right )+\left (120 x+84 x^2\right ) \log ^2\left (\frac {-10-7 x}{2 x}\right )+\left (10 x+7 x^2\right ) \log ^3\left (\frac {-10-7 x}{2 x}\right )} \, dx=\frac {x+\frac {x \log ^2(2)}{\left (4+\log \left (\frac {3}{2}-\frac {5 (1+x)}{x}\right )\right )^2}}{x} \]
Time = 0.02 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.66 \[ \int \frac {20 \log ^2(2)}{640 x+448 x^2+\left (480 x+336 x^2\right ) \log \left (\frac {-10-7 x}{2 x}\right )+\left (120 x+84 x^2\right ) \log ^2\left (\frac {-10-7 x}{2 x}\right )+\left (10 x+7 x^2\right ) \log ^3\left (\frac {-10-7 x}{2 x}\right )} \, dx=\frac {\log ^2(2)}{\left (4+\log \left (-\frac {7}{2}-\frac {5}{x}\right )\right )^2} \]
Integrate[(20*Log[2]^2)/(640*x + 448*x^2 + (480*x + 336*x^2)*Log[(-10 - 7* x)/(2*x)] + (120*x + 84*x^2)*Log[(-10 - 7*x)/(2*x)]^2 + (10*x + 7*x^2)*Log [(-10 - 7*x)/(2*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.
| \(\displaystyle \int \frac {20 \log ^2(2)}{448 x^2+\left (7 x^2+10 x\right ) \log ^3\left (\frac {-7 x-10}{2 x}\right )+\left (84 x^2+120 x\right ) \log ^2\left (\frac {-7 x-10}{2 x}\right )+\left (336 x^2+480 x\right ) \log \left (\frac {-7 x-10}{2 x}\right )+640 x} \, dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 20 \log ^2(2) \int \frac {1}{\left (7 x^2+10 x\right ) \log ^3\left (-\frac {7 x+10}{2 x}\right )+12 \left (7 x^2+10 x\right ) \log ^2\left (-\frac {7 x+10}{2 x}\right )+48 \left (7 x^2+10 x\right ) \log \left (-\frac {7 x+10}{2 x}\right )+448 x^2+640 x}dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle 20 \log ^2(2) \int \frac {1}{x (7 x+10) \left (\log \left (-\frac {7}{2}-\frac {5}{x}\right )+4\right )^3}dx\) |
| \(\Big \downarrow \) 2916 |
| \(\displaystyle 20 \log ^2(2) \int \left (\frac {1}{10 x \left (\log \left (-\frac {7}{2}-\frac {5}{x}\right )+4\right )^3}-\frac {7}{10 (7 x+10) \left (\log \left (-\frac {7}{2}-\frac {5}{x}\right )+4\right )^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 20 \log ^2(2) \left (\frac {1}{10} \int \frac {1}{x \left (\log \left (-\frac {7}{2}-\frac {5}{x}\right )+4\right )^3}dx-\frac {7}{10} \int \frac {1}{(7 x+10) \left (\log \left (-\frac {7}{2}-\frac {5}{x}\right )+4\right )^3}dx\right )\) |
Int[(20*Log[2]^2)/(640*x + 448*x^2 + (480*x + 336*x^2)*Log[(-10 - 7*x)/(2* x)] + (120*x + 84*x^2)*Log[(-10 - 7*x)/(2*x)]^2 + (10*x + 7*x^2)*Log[(-10 - 7*x)/(2*x)]^3),x]
3.30.2.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m _.)*((f_.) + (g_.)*(x_))^(r_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log [c*(d + e*x^n)^p])^q, x^m*(f + g*x)^r, x], x] /; FreeQ[{a, b, c, d, e, f, g , n, p, q}, x] && IntegerQ[m] && IntegerQ[r]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.43 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.62
| method | result | size |
| derivativedivides | \(\frac {\ln \left (2\right )^{2}}{\left (\ln \left (-\frac {7}{2}-\frac {5}{x}\right )+4\right )^{2}}\) | \(18\) |
| default | \(\frac {\ln \left (2\right )^{2}}{\left (\ln \left (-\frac {7}{2}-\frac {5}{x}\right )+4\right )^{2}}\) | \(18\) |
| norman | \(\frac {\ln \left (2\right )^{2}}{\left (\ln \left (\frac {-7 x -10}{2 x}\right )+4\right )^{2}}\) | \(21\) |
| risch | \(\frac {\ln \left (2\right )^{2}}{\left (\ln \left (\frac {-7 x -10}{2 x}\right )+4\right )^{2}}\) | \(21\) |
| parallelrisch | \(\frac {\ln \left (2\right )^{2}}{\ln \left (-\frac {7 x +10}{2 x}\right )^{2}+8 \ln \left (-\frac {7 x +10}{2 x}\right )+16}\) | \(36\) |
int(20*ln(2)^2/((7*x^2+10*x)*ln(1/2*(-7*x-10)/x)^3+(84*x^2+120*x)*ln(1/2*( -7*x-10)/x)^2+(336*x^2+480*x)*ln(1/2*(-7*x-10)/x)+448*x^2+640*x),x,method= _RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.21 \[ \int \frac {20 \log ^2(2)}{640 x+448 x^2+\left (480 x+336 x^2\right ) \log \left (\frac {-10-7 x}{2 x}\right )+\left (120 x+84 x^2\right ) \log ^2\left (\frac {-10-7 x}{2 x}\right )+\left (10 x+7 x^2\right ) \log ^3\left (\frac {-10-7 x}{2 x}\right )} \, dx=\frac {\log \left (2\right )^{2}}{\log \left (-\frac {7 \, x + 10}{2 \, x}\right )^{2} + 8 \, \log \left (-\frac {7 \, x + 10}{2 \, x}\right ) + 16} \]
integrate(20*log(2)^2/((7*x^2+10*x)*log(1/2*(-7*x-10)/x)^3+(84*x^2+120*x)* log(1/2*(-7*x-10)/x)^2+(336*x^2+480*x)*log(1/2*(-7*x-10)/x)+448*x^2+640*x) ,x, algorithm=\
Time = 0.07 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10 \[ \int \frac {20 \log ^2(2)}{640 x+448 x^2+\left (480 x+336 x^2\right ) \log \left (\frac {-10-7 x}{2 x}\right )+\left (120 x+84 x^2\right ) \log ^2\left (\frac {-10-7 x}{2 x}\right )+\left (10 x+7 x^2\right ) \log ^3\left (\frac {-10-7 x}{2 x}\right )} \, dx=\frac {\log {\left (2 \right )}^{2}}{\log {\left (\frac {- \frac {7 x}{2} - 5}{x} \right )}^{2} + 8 \log {\left (\frac {- \frac {7 x}{2} - 5}{x} \right )} + 16} \]
integrate(20*ln(2)**2/((7*x**2+10*x)*ln(1/2*(-7*x-10)/x)**3+(84*x**2+120*x )*ln(1/2*(-7*x-10)/x)**2+(336*x**2+480*x)*ln(1/2*(-7*x-10)/x)+448*x**2+640 *x),x)
Time = 0.27 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.76 \[ \int \frac {20 \log ^2(2)}{640 x+448 x^2+\left (480 x+336 x^2\right ) \log \left (\frac {-10-7 x}{2 x}\right )+\left (120 x+84 x^2\right ) \log ^2\left (\frac {-10-7 x}{2 x}\right )+\left (10 x+7 x^2\right ) \log ^3\left (\frac {-10-7 x}{2 x}\right )} \, dx=\frac {\log \left (2\right )^{2}}{\log \left (2\right )^{2} + 2 \, {\left (\log \left (2\right ) - 4\right )} \log \left (x\right ) + \log \left (x\right )^{2} - 2 \, {\left (\log \left (2\right ) + \log \left (x\right ) - 4\right )} \log \left (-7 \, x - 10\right ) + \log \left (-7 \, x - 10\right )^{2} - 8 \, \log \left (2\right ) + 16} \]
integrate(20*log(2)^2/((7*x^2+10*x)*log(1/2*(-7*x-10)/x)^3+(84*x^2+120*x)* log(1/2*(-7*x-10)/x)^2+(336*x^2+480*x)*log(1/2*(-7*x-10)/x)+448*x^2+640*x) ,x, algorithm=\
log(2)^2/(log(2)^2 + 2*(log(2) - 4)*log(x) + log(x)^2 - 2*(log(2) + log(x) - 4)*log(-7*x - 10) + log(-7*x - 10)^2 - 8*log(2) + 16)
Time = 0.28 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.69 \[ \int \frac {20 \log ^2(2)}{640 x+448 x^2+\left (480 x+336 x^2\right ) \log \left (\frac {-10-7 x}{2 x}\right )+\left (120 x+84 x^2\right ) \log ^2\left (\frac {-10-7 x}{2 x}\right )+\left (10 x+7 x^2\right ) \log ^3\left (\frac {-10-7 x}{2 x}\right )} \, dx=\frac {\log \left (2\right )^{2}}{{\left (\log \left (-\frac {7 \, x + 10}{2 \, x}\right ) + 4\right )}^{2}} \]
integrate(20*log(2)^2/((7*x^2+10*x)*log(1/2*(-7*x-10)/x)^3+(84*x^2+120*x)* log(1/2*(-7*x-10)/x)^2+(336*x^2+480*x)*log(1/2*(-7*x-10)/x)+448*x^2+640*x) ,x, algorithm=\
Time = 13.68 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.69 \[ \int \frac {20 \log ^2(2)}{640 x+448 x^2+\left (480 x+336 x^2\right ) \log \left (\frac {-10-7 x}{2 x}\right )+\left (120 x+84 x^2\right ) \log ^2\left (\frac {-10-7 x}{2 x}\right )+\left (10 x+7 x^2\right ) \log ^3\left (\frac {-10-7 x}{2 x}\right )} \, dx=\frac {{\ln \left (2\right )}^2}{{\left (\ln \left (-\frac {7\,x+10}{2\,x}\right )+4\right )}^2} \]