Integrand size = 74, antiderivative size = 28 \[ \int \frac {-225-300 \log (2)+(-60 x+75 \log (2)) \log \left (49 x^4\right )+\left (-4 x^2+20 x \log (2)\right ) \log ^2\left (49 x^4\right )}{225 x^2+60 x^3 \log \left (49 x^4\right )+4 x^4 \log ^2\left (49 x^4\right )} \, dx=\frac {1-\frac {\log (2)}{\frac {2 x}{5}+\frac {3}{\log \left (49 x^4\right )}}}{x} \] Output:
(1-ln(2)/(2/5*x+3/ln(49*x^4)))/x
Time = 0.18 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21 \[ \int \frac {-225-300 \log (2)+(-60 x+75 \log (2)) \log \left (49 x^4\right )+\left (-4 x^2+20 x \log (2)\right ) \log ^2\left (49 x^4\right )}{225 x^2+60 x^3 \log \left (49 x^4\right )+4 x^4 \log ^2\left (49 x^4\right )} \, dx=\frac {15+(2 x-5 \log (2)) \log \left (49 x^4\right )}{x \left (15+2 x \log \left (49 x^4\right )\right )} \] Input:
Integrate[(-225 - 300*Log[2] + (-60*x + 75*Log[2])*Log[49*x^4] + (-4*x^2 + 20*x*Log[2])*Log[49*x^4]^2)/(225*x^2 + 60*x^3*Log[49*x^4] + 4*x^4*Log[49* x^4]^2),x]
Output:
(15 + (2*x - 5*Log[2])*Log[49*x^4])/(x*(15 + 2*x*Log[49*x^4]))
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.
\(\displaystyle \int \frac {(75 \log (2)-60 x) \log \left (49 x^4\right )+\left (20 x \log (2)-4 x^2\right ) \log ^2\left (49 x^4\right )-225-300 \log (2)}{4 x^4 \log ^2\left (49 x^4\right )+225 x^2+60 x^3 \log \left (49 x^4\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {(75 \log (2)-60 x) \log \left (49 x^4\right )+\left (20 x \log (2)-4 x^2\right ) \log ^2\left (49 x^4\right )-225 \left (1+\frac {4 \log (2)}{3}\right )}{x^2 \left (2 x \log \left (49 x^4\right )+15\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {\log (32)-x}{x^3}-\frac {75 (8 x-15) \log (2)}{2 x^3 \left (2 x \log \left (49 x^4\right )+15\right )^2}-\frac {225 \log (2)}{2 x^3 \left (2 x \log \left (49 x^4\right )+15\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1125}{2} \log (2) \int \frac {1}{x^3 \left (2 x \log \left (49 x^4\right )+15\right )^2}dx-\frac {225}{2} \log (2) \int \frac {1}{x^3 \left (2 x \log \left (49 x^4\right )+15\right )}dx-300 \log (2) \int \frac {1}{x^2 \left (2 x \log \left (49 x^4\right )+15\right )^2}dx-\frac {(x-\log (32))^2}{2 x^2 \log (32)}\) |
Input:
Int[(-225 - 300*Log[2] + (-60*x + 75*Log[2])*Log[49*x^4] + (-4*x^2 + 20*x* Log[2])*Log[49*x^4]^2)/(225*x^2 + 60*x^3*Log[49*x^4] + 4*x^4*Log[49*x^4]^2 ),x]
Output:
$Aborted
Time = 0.70 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.25
method | result | size |
risch | \(-\frac {5 \ln \left (2\right )-2 x}{2 x^{2}}+\frac {75 \ln \left (2\right )}{2 x^{2} \left (2 \ln \left (49 x^{4}\right ) x +15\right )}\) | \(35\) |
norman | \(\frac {15+2 \ln \left (49 x^{4}\right ) x -5 \ln \left (49 x^{4}\right ) \ln \left (2\right )}{x \left (2 \ln \left (49 x^{4}\right ) x +15\right )}\) | \(39\) |
default | \(\frac {1}{x}+\frac {-5 \ln \left (2\right ) \ln \left (x^{4}\right )-10 \ln \left (7\right ) \ln \left (2\right )}{x \left (4 x \ln \left (7\right )+2 x \ln \left (x^{4}\right )+15\right )}\) | \(40\) |
parallelrisch | \(-\frac {-60+20 \ln \left (49 x^{4}\right ) \ln \left (2\right )-8 \ln \left (49 x^{4}\right ) x}{4 x \left (2 \ln \left (49 x^{4}\right ) x +15\right )}\) | \(40\) |
Input:
int(((20*x*ln(2)-4*x^2)*ln(49*x^4)^2+(75*ln(2)-60*x)*ln(49*x^4)-300*ln(2)- 225)/(4*x^4*ln(49*x^4)^2+60*x^3*ln(49*x^4)+225*x^2),x,method=_RETURNVERBOS E)
Output:
-1/2*(5*ln(2)-2*x)/x^2+75/2/x^2*ln(2)/(2*ln(49*x^4)*x+15)
Time = 0.09 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.25 \[ \int \frac {-225-300 \log (2)+(-60 x+75 \log (2)) \log \left (49 x^4\right )+\left (-4 x^2+20 x \log (2)\right ) \log ^2\left (49 x^4\right )}{225 x^2+60 x^3 \log \left (49 x^4\right )+4 x^4 \log ^2\left (49 x^4\right )} \, dx=\frac {{\left (2 \, x - 5 \, \log \left (2\right )\right )} \log \left (49 \, x^{4}\right ) + 15}{2 \, x^{2} \log \left (49 \, x^{4}\right ) + 15 \, x} \] Input:
integrate(((20*x*log(2)-4*x^2)*log(49*x^4)^2+(75*log(2)-60*x)*log(49*x^4)- 300*log(2)-225)/(4*x^4*log(49*x^4)^2+60*x^3*log(49*x^4)+225*x^2),x, algori thm="fricas")
Output:
((2*x - 5*log(2))*log(49*x^4) + 15)/(2*x^2*log(49*x^4) + 15*x)
Time = 0.10 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21 \[ \int \frac {-225-300 \log (2)+(-60 x+75 \log (2)) \log \left (49 x^4\right )+\left (-4 x^2+20 x \log (2)\right ) \log ^2\left (49 x^4\right )}{225 x^2+60 x^3 \log \left (49 x^4\right )+4 x^4 \log ^2\left (49 x^4\right )} \, dx=\frac {75 \log {\left (2 \right )}}{4 x^{3} \log {\left (49 x^{4} \right )} + 30 x^{2}} - \frac {- 2 x + 5 \log {\left (2 \right )}}{2 x^{2}} \] Input:
integrate(((20*x*ln(2)-4*x**2)*ln(49*x**4)**2+(75*ln(2)-60*x)*ln(49*x**4)- 300*ln(2)-225)/(4*x**4*ln(49*x**4)**2+60*x**3*ln(49*x**4)+225*x**2),x)
Output:
75*log(2)/(4*x**3*log(49*x**4) + 30*x**2) - (-2*x + 5*log(2))/(2*x**2)
Time = 0.14 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.64 \[ \int \frac {-225-300 \log (2)+(-60 x+75 \log (2)) \log \left (49 x^4\right )+\left (-4 x^2+20 x \log (2)\right ) \log ^2\left (49 x^4\right )}{225 x^2+60 x^3 \log \left (49 x^4\right )+4 x^4 \log ^2\left (49 x^4\right )} \, dx=\frac {4 \, x \log \left (7\right ) - 10 \, \log \left (7\right ) \log \left (2\right ) + 4 \, {\left (2 \, x - 5 \, \log \left (2\right )\right )} \log \left (x\right ) + 15}{4 \, x^{2} \log \left (7\right ) + 8 \, x^{2} \log \left (x\right ) + 15 \, x} \] Input:
integrate(((20*x*log(2)-4*x^2)*log(49*x^4)^2+(75*log(2)-60*x)*log(49*x^4)- 300*log(2)-225)/(4*x^4*log(49*x^4)^2+60*x^3*log(49*x^4)+225*x^2),x, algori thm="maxima")
Output:
(4*x*log(7) - 10*log(7)*log(2) + 4*(2*x - 5*log(2))*log(x) + 15)/(4*x^2*lo g(7) + 8*x^2*log(x) + 15*x)
Time = 0.13 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.32 \[ \int \frac {-225-300 \log (2)+(-60 x+75 \log (2)) \log \left (49 x^4\right )+\left (-4 x^2+20 x \log (2)\right ) \log ^2\left (49 x^4\right )}{225 x^2+60 x^3 \log \left (49 x^4\right )+4 x^4 \log ^2\left (49 x^4\right )} \, dx=\frac {75 \, \log \left (2\right )}{2 \, {\left (2 \, x^{3} \log \left (49 \, x^{4}\right ) + 15 \, x^{2}\right )}} + \frac {2 \, x - 5 \, \log \left (2\right )}{2 \, x^{2}} \] Input:
integrate(((20*x*log(2)-4*x^2)*log(49*x^4)^2+(75*log(2)-60*x)*log(49*x^4)- 300*log(2)-225)/(4*x^4*log(49*x^4)^2+60*x^3*log(49*x^4)+225*x^2),x, algori thm="giac")
Output:
75/2*log(2)/(2*x^3*log(49*x^4) + 15*x^2) + 1/2*(2*x - 5*log(2))/x^2
Time = 8.94 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.36 \[ \int \frac {-225-300 \log (2)+(-60 x+75 \log (2)) \log \left (49 x^4\right )+\left (-4 x^2+20 x \log (2)\right ) \log ^2\left (49 x^4\right )}{225 x^2+60 x^3 \log \left (49 x^4\right )+4 x^4 \log ^2\left (49 x^4\right )} \, dx=\frac {2\,x\,\ln \left (49\,x^4\right )-\ln \left (32\right )\,\ln \left (49\,x^4\right )+15}{x\,\left (2\,x\,\ln \left (49\,x^4\right )+15\right )} \] Input:
int(-(300*log(2) + log(49*x^4)*(60*x - 75*log(2)) - log(49*x^4)^2*(20*x*lo g(2) - 4*x^2) + 225)/(4*x^4*log(49*x^4)^2 + 225*x^2 + 60*x^3*log(49*x^4)), x)
Output:
(2*x*log(49*x^4) - log(32)*log(49*x^4) + 15)/(x*(2*x*log(49*x^4) + 15))
Time = 0.22 (sec) , antiderivative size = 61, normalized size of antiderivative = 2.18 \[ \int \frac {-225-300 \log (2)+(-60 x+75 \log (2)) \log \left (49 x^4\right )+\left (-4 x^2+20 x \log (2)\right ) \log ^2\left (49 x^4\right )}{225 x^2+60 x^3 \log \left (49 x^4\right )+4 x^4 \log ^2\left (49 x^4\right )} \, dx=\frac {-4 \mathrm {log}\left (49 x^{4}\right )^{2} x^{2}+16 \,\mathrm {log}\left (49 x^{4}\right ) \mathrm {log}\left (x \right ) x^{2}-75 \,\mathrm {log}\left (49 x^{4}\right ) \mathrm {log}\left (2\right )+120 \,\mathrm {log}\left (x \right ) x +225}{15 x \left (2 \,\mathrm {log}\left (49 x^{4}\right ) x +15\right )} \] Input:
int(((20*x*log(2)-4*x^2)*log(49*x^4)^2+(75*log(2)-60*x)*log(49*x^4)-300*lo g(2)-225)/(4*x^4*log(49*x^4)^2+60*x^3*log(49*x^4)+225*x^2),x)
Output:
( - 4*log(49*x**4)**2*x**2 + 16*log(49*x**4)*log(x)*x**2 - 75*log(49*x**4) *log(2) + 120*log(x)*x + 225)/(15*x*(2*log(49*x**4)*x + 15))