Integrand size = 30, antiderivative size = 22 \[ \int \frac {25 x^4 \log ^4\left (\frac {4}{3}\right )}{-32 e^{50}+5 x^5 \log ^4\left (\frac {4}{3}\right )} \, dx=\log \left (5 \left (-2 e^{50}+\frac {5}{16} x^5 \log ^4\left (\frac {4}{3}\right )\right )\right ) \] Output:
ln(25/16*x^5*ln(3/4)^4-10*exp(25)^2)
Time = 0.00 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {25 x^4 \log ^4\left (\frac {4}{3}\right )}{-32 e^{50}+5 x^5 \log ^4\left (\frac {4}{3}\right )} \, dx=\log \left (-32 e^{50}+5 x^5 \log ^4\left (\frac {4}{3}\right )\right ) \] Input:
Integrate[(25*x^4*Log[4/3]^4)/(-32*E^50 + 5*x^5*Log[4/3]^4),x]
Output:
Log[-32*E^50 + 5*x^5*Log[4/3]^4]
Time = 0.16 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {27, 25, 792}
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.
\(\displaystyle \int \frac {25 x^4 \log ^4\left (\frac {4}{3}\right )}{5 x^5 \log ^4\left (\frac {4}{3}\right )-32 e^{50}} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 25 \log ^4\left (\frac {4}{3}\right ) \int -\frac {x^4}{32 e^{50}-5 x^5 \log ^4\left (\frac {4}{3}\right )}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -25 \log ^4\left (\frac {4}{3}\right ) \int \frac {x^4}{32 e^{50}-5 x^5 \log ^4\left (\frac {4}{3}\right )}dx\) |
\(\Big \downarrow \) 792 |
\(\displaystyle \log \left (32 e^{50}-5 x^5 \log ^4\left (\frac {4}{3}\right )\right )\) |
Input:
Int[(25*x^4*Log[4/3]^4)/(-32*E^50 + 5*x^5*Log[4/3]^4),x]
Output:
Log[32*E^50 - 5*x^5*Log[4/3]^4]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveConten t[a + b*x^n, x]]/(b*n), x] /; FreeQ[{a, b, m, n}, x] && EqQ[m, n - 1]
Time = 0.26 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82
method | result | size |
derivativedivides | \(\ln \left (5 x^{5} \ln \left (\frac {3}{4}\right )^{4}-32 \,{\mathrm e}^{50}\right )\) | \(18\) |
default | \(\ln \left (5 x^{5} \ln \left (\frac {3}{4}\right )^{4}-32 \,{\mathrm e}^{50}\right )\) | \(18\) |
norman | \(\ln \left (5 x^{5} \ln \left (\frac {3}{4}\right )^{4}-32 \,{\mathrm e}^{50}\right )\) | \(18\) |
parallelrisch | \(\ln \left (\frac {5 x^{5} \ln \left (\frac {3}{4}\right )^{4}-32 \,{\mathrm e}^{50}}{5 \ln \left (\frac {3}{4}\right )^{4}}\right )\) | \(24\) |
meijerg | \(\frac {\ln \left (\frac {3}{4}\right )^{4} \ln \left (1-\frac {5 x^{5} \left (-\ln \left (3\right )+2 \ln \left (2\right )\right )^{4} {\mathrm e}^{-50}}{32}\right )}{\left (-\ln \left (3\right )+2 \ln \left (2\right )\right )^{4}}\) | \(38\) |
risch | \(\frac {\ln \left (\left (5 \ln \left (3\right )^{4}-40 \ln \left (2\right ) \ln \left (3\right )^{3}+120 \ln \left (2\right )^{2} \ln \left (3\right )^{2}-160 \ln \left (3\right ) \ln \left (2\right )^{3}+80 \ln \left (2\right )^{4}\right ) x^{5}-32 \,{\mathrm e}^{50}\right ) \ln \left (3\right )^{4}}{\ln \left (3\right )^{4}-8 \ln \left (2\right ) \ln \left (3\right )^{3}+24 \ln \left (2\right )^{2} \ln \left (3\right )^{2}-32 \ln \left (3\right ) \ln \left (2\right )^{3}+16 \ln \left (2\right )^{4}}-\frac {8 \ln \left (\left (5 \ln \left (3\right )^{4}-40 \ln \left (2\right ) \ln \left (3\right )^{3}+120 \ln \left (2\right )^{2} \ln \left (3\right )^{2}-160 \ln \left (3\right ) \ln \left (2\right )^{3}+80 \ln \left (2\right )^{4}\right ) x^{5}-32 \,{\mathrm e}^{50}\right ) \ln \left (3\right )^{3} \ln \left (2\right )}{\ln \left (3\right )^{4}-8 \ln \left (2\right ) \ln \left (3\right )^{3}+24 \ln \left (2\right )^{2} \ln \left (3\right )^{2}-32 \ln \left (3\right ) \ln \left (2\right )^{3}+16 \ln \left (2\right )^{4}}+\frac {24 \ln \left (\left (5 \ln \left (3\right )^{4}-40 \ln \left (2\right ) \ln \left (3\right )^{3}+120 \ln \left (2\right )^{2} \ln \left (3\right )^{2}-160 \ln \left (3\right ) \ln \left (2\right )^{3}+80 \ln \left (2\right )^{4}\right ) x^{5}-32 \,{\mathrm e}^{50}\right ) \ln \left (3\right )^{2} \ln \left (2\right )^{2}}{\ln \left (3\right )^{4}-8 \ln \left (2\right ) \ln \left (3\right )^{3}+24 \ln \left (2\right )^{2} \ln \left (3\right )^{2}-32 \ln \left (3\right ) \ln \left (2\right )^{3}+16 \ln \left (2\right )^{4}}-\frac {32 \ln \left (\left (5 \ln \left (3\right )^{4}-40 \ln \left (2\right ) \ln \left (3\right )^{3}+120 \ln \left (2\right )^{2} \ln \left (3\right )^{2}-160 \ln \left (3\right ) \ln \left (2\right )^{3}+80 \ln \left (2\right )^{4}\right ) x^{5}-32 \,{\mathrm e}^{50}\right ) \ln \left (3\right ) \ln \left (2\right )^{3}}{\ln \left (3\right )^{4}-8 \ln \left (2\right ) \ln \left (3\right )^{3}+24 \ln \left (2\right )^{2} \ln \left (3\right )^{2}-32 \ln \left (3\right ) \ln \left (2\right )^{3}+16 \ln \left (2\right )^{4}}+\frac {16 \ln \left (\left (5 \ln \left (3\right )^{4}-40 \ln \left (2\right ) \ln \left (3\right )^{3}+120 \ln \left (2\right )^{2} \ln \left (3\right )^{2}-160 \ln \left (3\right ) \ln \left (2\right )^{3}+80 \ln \left (2\right )^{4}\right ) x^{5}-32 \,{\mathrm e}^{50}\right ) \ln \left (2\right )^{4}}{\ln \left (3\right )^{4}-8 \ln \left (2\right ) \ln \left (3\right )^{3}+24 \ln \left (2\right )^{2} \ln \left (3\right )^{2}-32 \ln \left (3\right ) \ln \left (2\right )^{3}+16 \ln \left (2\right )^{4}}\) | \(479\) |
Input:
int(25*x^4*ln(3/4)^4/(5*x^5*ln(3/4)^4-32*exp(25)^2),x,method=_RETURNVERBOS E)
Output:
ln(5*x^5*ln(3/4)^4-32*exp(25)^2)
Time = 0.06 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.68 \[ \int \frac {25 x^4 \log ^4\left (\frac {4}{3}\right )}{-32 e^{50}+5 x^5 \log ^4\left (\frac {4}{3}\right )} \, dx=\log \left (5 \, x^{5} \log \left (\frac {3}{4}\right )^{4} - 32 \, e^{50}\right ) \] Input:
integrate(25*x^4*log(3/4)^4/(5*x^5*log(3/4)^4-32*exp(25)^2),x, algorithm=" fricas")
Output:
log(5*x^5*log(3/4)^4 - 32*e^50)
Leaf count of result is larger than twice the leaf count of optimal. 112 vs. \(2 (19) = 38\).
Time = 1.74 (sec) , antiderivative size = 112, normalized size of antiderivative = 5.09 \[ \int \frac {25 x^4 \log ^4\left (\frac {4}{3}\right )}{-32 e^{50}+5 x^5 \log ^4\left (\frac {4}{3}\right )} \, dx=\frac {\left (- 800 \log {\left (2 \right )}^{3} \log {\left (3 \right )} - 200 \log {\left (2 \right )} \log {\left (3 \right )}^{3} + 25 \log {\left (3 \right )}^{4} + 400 \log {\left (2 \right )}^{4} + 600 \log {\left (2 \right )}^{2} \log {\left (3 \right )}^{2}\right ) \log {\left (x^{5} \left (- 160 \log {\left (2 \right )}^{3} \log {\left (3 \right )} - 40 \log {\left (2 \right )} \log {\left (3 \right )}^{3} + 5 \log {\left (3 \right )}^{4} + 80 \log {\left (2 \right )}^{4} + 120 \log {\left (2 \right )}^{2} \log {\left (3 \right )}^{2}\right ) - 32 e^{50} \right )}}{25 \left (- \log {\left (3 \right )} + 2 \log {\left (2 \right )}\right )^{4}} \] Input:
integrate(25*x**4*ln(3/4)**4/(5*x**5*ln(3/4)**4-32*exp(25)**2),x)
Output:
(-800*log(2)**3*log(3) - 200*log(2)*log(3)**3 + 25*log(3)**4 + 400*log(2)* *4 + 600*log(2)**2*log(3)**2)*log(x**5*(-160*log(2)**3*log(3) - 40*log(2)* log(3)**3 + 5*log(3)**4 + 80*log(2)**4 + 120*log(2)**2*log(3)**2) - 32*exp (50))/(25*(-log(3) + 2*log(2))**4)
Time = 0.03 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.68 \[ \int \frac {25 x^4 \log ^4\left (\frac {4}{3}\right )}{-32 e^{50}+5 x^5 \log ^4\left (\frac {4}{3}\right )} \, dx=\log \left (5 \, x^{5} \log \left (\frac {3}{4}\right )^{4} - 32 \, e^{50}\right ) \] Input:
integrate(25*x^4*log(3/4)^4/(5*x^5*log(3/4)^4-32*exp(25)^2),x, algorithm=" maxima")
Output:
log(5*x^5*log(3/4)^4 - 32*e^50)
Time = 0.12 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.73 \[ \int \frac {25 x^4 \log ^4\left (\frac {4}{3}\right )}{-32 e^{50}+5 x^5 \log ^4\left (\frac {4}{3}\right )} \, dx=\log \left ({\left | 5 \, x^{5} \log \left (\frac {3}{4}\right )^{4} - 32 \, e^{50} \right |}\right ) \] Input:
integrate(25*x^4*log(3/4)^4/(5*x^5*log(3/4)^4-32*exp(25)^2),x, algorithm=" giac")
Output:
log(abs(5*x^5*log(3/4)^4 - 32*e^50))
Time = 0.10 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.68 \[ \int \frac {25 x^4 \log ^4\left (\frac {4}{3}\right )}{-32 e^{50}+5 x^5 \log ^4\left (\frac {4}{3}\right )} \, dx=\ln \left (5\,x^5\,{\ln \left (\frac {3}{4}\right )}^4-32\,{\mathrm {e}}^{50}\right ) \] Input:
int(-(25*x^4*log(3/4)^4)/(32*exp(50) - 5*x^5*log(3/4)^4),x)
Output:
log(5*x^5*log(3/4)^4 - 32*exp(50))
Time = 0.18 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.73 \[ \int \frac {25 x^4 \log ^4\left (\frac {4}{3}\right )}{-32 e^{50}+5 x^5 \log ^4\left (\frac {4}{3}\right )} \, dx=\mathrm {log}\left (5 \mathrm {log}\left (\frac {3}{4}\right )^{4} x^{5}-32 e^{50}\right ) \] Input:
int(25*x^4*log(3/4)^4/(5*x^5*log(3/4)^4-32*exp(25)^2),x)
Output:
log(5*log(3/4)**4*x**5 - 32*e**50)