Integrand size = 90, antiderivative size = 27 \[ \int \frac {-160 x+68 x^2-8 x^3-4 x \log (7)+\left (-72 x+16 x^2\right ) \log (7) \log \left (\frac {4}{x}\right )-8 x \log ^2(7) \log ^2\left (\frac {4}{x}\right )}{16-8 x+x^2+(8-2 x) \log (7) \log \left (\frac {4}{x}\right )+\log ^2(7) \log ^2\left (\frac {4}{x}\right )} \, dx=5-x^2 \left (4+\frac {4}{4-x+\log (7) \log \left (\frac {4}{x}\right )}\right ) \]
Time = 0.32 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {-160 x+68 x^2-8 x^3-4 x \log (7)+\left (-72 x+16 x^2\right ) \log (7) \log \left (\frac {4}{x}\right )-8 x \log ^2(7) \log ^2\left (\frac {4}{x}\right )}{16-8 x+x^2+(8-2 x) \log (7) \log \left (\frac {4}{x}\right )+\log ^2(7) \log ^2\left (\frac {4}{x}\right )} \, dx=-4 \left (x^2+\frac {x^2}{4-x+\log (7) \log \left (\frac {4}{x}\right )}\right ) \]
Integrate[(-160*x + 68*x^2 - 8*x^3 - 4*x*Log[7] + (-72*x + 16*x^2)*Log[7]* Log[4/x] - 8*x*Log[7]^2*Log[4/x]^2)/(16 - 8*x + x^2 + (8 - 2*x)*Log[7]*Log [4/x] + Log[7]^2*Log[4/x]^2),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.
\(\displaystyle \int \frac {-8 x^3+68 x^2+\left (16 x^2-72 x\right ) \log (7) \log \left (\frac {4}{x}\right )-160 x-8 x \log ^2(7) \log ^2\left (\frac {4}{x}\right )-4 x \log (7)}{x^2-8 x+\log ^2(7) \log ^2\left (\frac {4}{x}\right )+(8-2 x) \log (7) \log \left (\frac {4}{x}\right )+16} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {-8 x^3+68 x^2+\left (16 x^2-72 x\right ) \log (7) \log \left (\frac {4}{x}\right )-8 x \log ^2(7) \log ^2\left (\frac {4}{x}\right )+x (-160-4 \log (7))}{x^2-8 x+\log ^2(7) \log ^2\left (\frac {4}{x}\right )+(8-2 x) \log (7) \log \left (\frac {4}{x}\right )+16}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {4 x \left (-2 x^2+17 x-2 \log ^2(7) \log ^2\left (\frac {4}{x}\right )+4 x \log (7) \log \left (\frac {4}{x}\right )-18 \log (7) \log \left (\frac {4}{x}\right )-40 \left (1+\frac {\log (7)}{40}\right )\right )}{\left (-x+\log (7) \log \left (\frac {4}{x}\right )+4\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 4 \int -\frac {x \left (2 x^2-4 \log (7) \log \left (\frac {4}{x}\right ) x-17 x+2 \log ^2(7) \log ^2\left (\frac {4}{x}\right )+18 \log (7) \log \left (\frac {4}{x}\right )+\log (7)+40\right )}{\left (-x+\log (7) \log \left (\frac {4}{x}\right )+4\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -4 \int \frac {x \left (2 x^2-4 \log (7) \log \left (\frac {4}{x}\right ) x-17 x+2 \log ^2(7) \log ^2\left (\frac {4}{x}\right )+18 \log (7) \log \left (\frac {4}{x}\right )+\log (7)+40\right )}{\left (-x+\log (7) \log \left (\frac {4}{x}\right )+4\right )^2}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle -4 \int \frac {x \left (2 x^2-4 \log (7) \log \left (\frac {4}{x}\right ) x-17 x+2 \log ^2(7) \log ^2\left (\frac {4}{x}\right )+18 \log (7) \log \left (\frac {4}{x}\right )+40 \left (1+\frac {\log (7)}{40}\right )\right )}{\left (-x+\log (7) \log \left (\frac {4}{x}\right )+4\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -4 \int \left (-\frac {2 x}{x-\log (7) \log \left (\frac {4}{x}\right )-4}+\frac {(x+\log (7)) x}{\left (x-\log (7) \log \left (\frac {4}{x}\right )-4\right )^2}+2 x\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -4 \left (\int \frac {x^2}{\left (x-\log (7) \log \left (\frac {4}{x}\right )-4\right )^2}dx+\log (7) \int \frac {x}{\left (x-\log (7) \log \left (\frac {4}{x}\right )-4\right )^2}dx-2 \int \frac {x}{x-\log (7) \log \left (\frac {4}{x}\right )-4}dx+x^2\right )\) |
Int[(-160*x + 68*x^2 - 8*x^3 - 4*x*Log[7] + (-72*x + 16*x^2)*Log[7]*Log[4/ x] - 8*x*Log[7]^2*Log[4/x]^2)/(16 - 8*x + x^2 + (8 - 2*x)*Log[7]*Log[4/x] + Log[7]^2*Log[4/x]^2),x]
3.1.54.3.1 Defintions of rubi rules used
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 1.65 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04
method | result | size |
risch | \(-4 x^{2}-\frac {4 x^{2}}{4+\ln \left (7\right ) \ln \left (\frac {4}{x}\right )-x}\) | \(28\) |
norman | \(\frac {-20 x^{2}+4 x^{3}-4 \ln \left (7\right ) \ln \left (\frac {4}{x}\right ) x^{2}}{4+\ln \left (7\right ) \ln \left (\frac {4}{x}\right )-x}\) | \(42\) |
parallelrisch | \(\frac {-20 x^{2}+4 x^{3}-4 \ln \left (7\right ) \ln \left (\frac {4}{x}\right ) x^{2}}{4+\ln \left (7\right ) \ln \left (\frac {4}{x}\right )-x}\) | \(42\) |
derivativedivides | \(\frac {4 \left (4-\frac {4 \ln \left (7\right ) \ln \left (\frac {4}{x}\right )}{x}-\frac {20}{x}\right ) x^{2}}{\frac {4 \ln \left (7\right ) \ln \left (\frac {4}{x}\right )}{x}+\frac {16}{x}-4}\) | \(48\) |
default | \(\frac {4 \left (4-\frac {4 \ln \left (7\right ) \ln \left (\frac {4}{x}\right )}{x}-\frac {20}{x}\right ) x^{2}}{\frac {4 \ln \left (7\right ) \ln \left (\frac {4}{x}\right )}{x}+\frac {16}{x}-4}\) | \(48\) |
int((-8*x*ln(7)^2*ln(4/x)^2+(16*x^2-72*x)*ln(7)*ln(4/x)-4*x*ln(7)-8*x^3+68 *x^2-160*x)/(ln(7)^2*ln(4/x)^2+(-2*x+8)*ln(7)*ln(4/x)+x^2-8*x+16),x,method =_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.52 \[ \int \frac {-160 x+68 x^2-8 x^3-4 x \log (7)+\left (-72 x+16 x^2\right ) \log (7) \log \left (\frac {4}{x}\right )-8 x \log ^2(7) \log ^2\left (\frac {4}{x}\right )}{16-8 x+x^2+(8-2 x) \log (7) \log \left (\frac {4}{x}\right )+\log ^2(7) \log ^2\left (\frac {4}{x}\right )} \, dx=-\frac {4 \, {\left (x^{2} \log \left (7\right ) \log \left (\frac {4}{x}\right ) - x^{3} + 5 \, x^{2}\right )}}{\log \left (7\right ) \log \left (\frac {4}{x}\right ) - x + 4} \]
integrate((-8*x*log(7)^2*log(4/x)^2+(16*x^2-72*x)*log(7)*log(4/x)-4*x*log( 7)-8*x^3+68*x^2-160*x)/(log(7)^2*log(4/x)^2+(-2*x+8)*log(7)*log(4/x)+x^2-8 *x+16),x, algorithm=\
Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {-160 x+68 x^2-8 x^3-4 x \log (7)+\left (-72 x+16 x^2\right ) \log (7) \log \left (\frac {4}{x}\right )-8 x \log ^2(7) \log ^2\left (\frac {4}{x}\right )}{16-8 x+x^2+(8-2 x) \log (7) \log \left (\frac {4}{x}\right )+\log ^2(7) \log ^2\left (\frac {4}{x}\right )} \, dx=- 4 x^{2} - \frac {4 x^{2}}{- x + \log {\left (7 \right )} \log {\left (\frac {4}{x} \right )} + 4} \]
integrate((-8*x*ln(7)**2*ln(4/x)**2+(16*x**2-72*x)*ln(7)*ln(4/x)-4*x*ln(7) -8*x**3+68*x**2-160*x)/(ln(7)**2*ln(4/x)**2+(-2*x+8)*ln(7)*ln(4/x)+x**2-8* x+16),x)
Time = 0.30 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.70 \[ \int \frac {-160 x+68 x^2-8 x^3-4 x \log (7)+\left (-72 x+16 x^2\right ) \log (7) \log \left (\frac {4}{x}\right )-8 x \log ^2(7) \log ^2\left (\frac {4}{x}\right )}{16-8 x+x^2+(8-2 x) \log (7) \log \left (\frac {4}{x}\right )+\log ^2(7) \log ^2\left (\frac {4}{x}\right )} \, dx=\frac {4 \, {\left (x^{2} \log \left (7\right ) \log \left (x\right ) - {\left (2 \, \log \left (7\right ) \log \left (2\right ) + 5\right )} x^{2} + x^{3}\right )}}{2 \, \log \left (7\right ) \log \left (2\right ) - \log \left (7\right ) \log \left (x\right ) - x + 4} \]
integrate((-8*x*log(7)^2*log(4/x)^2+(16*x^2-72*x)*log(7)*log(4/x)-4*x*log( 7)-8*x^3+68*x^2-160*x)/(log(7)^2*log(4/x)^2+(-2*x+8)*log(7)*log(4/x)+x^2-8 *x+16),x, algorithm=\
4*(x^2*log(7)*log(x) - (2*log(7)*log(2) + 5)*x^2 + x^3)/(2*log(7)*log(2) - log(7)*log(x) - x + 4)
Time = 0.29 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \frac {-160 x+68 x^2-8 x^3-4 x \log (7)+\left (-72 x+16 x^2\right ) \log (7) \log \left (\frac {4}{x}\right )-8 x \log ^2(7) \log ^2\left (\frac {4}{x}\right )}{16-8 x+x^2+(8-2 x) \log (7) \log \left (\frac {4}{x}\right )+\log ^2(7) \log ^2\left (\frac {4}{x}\right )} \, dx=-4 \, x^{2} - \frac {4}{\frac {\log \left (7\right ) \log \left (\frac {4}{x}\right )}{x^{2}} - \frac {1}{x} + \frac {4}{x^{2}}} \]
integrate((-8*x*log(7)^2*log(4/x)^2+(16*x^2-72*x)*log(7)*log(4/x)-4*x*log( 7)-8*x^3+68*x^2-160*x)/(log(7)^2*log(4/x)^2+(-2*x+8)*log(7)*log(4/x)+x^2-8 *x+16),x, algorithm=\
Time = 11.78 (sec) , antiderivative size = 85, normalized size of antiderivative = 3.15 \[ \int \frac {-160 x+68 x^2-8 x^3-4 x \log (7)+\left (-72 x+16 x^2\right ) \log (7) \log \left (\frac {4}{x}\right )-8 x \log ^2(7) \log ^2\left (\frac {4}{x}\right )}{16-8 x+x^2+(8-2 x) \log (7) \log \left (\frac {4}{x}\right )+\log ^2(7) \log ^2\left (\frac {4}{x}\right )} \, dx=8\,x-\frac {\frac {8\,x^2\,\ln \left (\frac {4}{x}\right )}{x+\ln \left (7\right )}+\frac {4\,x\,\left (8\,x+x\,\ln \left (7\right )-x^2\right )}{\ln \left (7\right )\,\left (x+\ln \left (7\right )\right )}}{\ln \left (\frac {4}{x}\right )-\frac {x-4}{\ln \left (7\right )}}+\frac {8\,{\ln \left (7\right )}^2}{x+\ln \left (7\right )}-4\,x^2 \]
int(-(160*x + 4*x*log(7) - 68*x^2 + 8*x^3 + 8*x*log(7)^2*log(4/x)^2 + log( 7)*log(4/x)*(72*x - 16*x^2))/(log(7)^2*log(4/x)^2 - 8*x + x^2 - log(7)*log (4/x)*(2*x - 8) + 16),x)