Integrand size = 41, antiderivative size = 33 \[ \int \frac {4 e^5 x+24 e^5 \log (3)-x \log \left (\frac {1}{25} \left (\log ^2(4)-2 \log (4) \log (5)+\log ^2(5)\right )\right )}{x^3} \, dx=\frac {-e^5 \left (4+\frac {12 \log (3)}{x}\right )+\log \left (\frac {1}{25} (-\log (4)+\log (5))^2\right )}{x} \] Output:
(ln(1/5*(ln(5)-2*ln(2))*(1/5*ln(5)-2/5*ln(2)))-(4+12*ln(3)/x)*exp(5))/x
Time = 0.01 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88 \[ \int \frac {4 e^5 x+24 e^5 \log (3)-x \log \left (\frac {1}{25} \left (\log ^2(4)-2 \log (4) \log (5)+\log ^2(5)\right )\right )}{x^3} \, dx=-\frac {2 e^5 (2 x+\log (729))+x \left (\log (25)-2 \log \left (\log \left (\frac {5}{4}\right )\right )\right )}{x^2} \] Input:
Integrate[(4*E^5*x + 24*E^5*Log[3] - x*Log[(Log[4]^2 - 2*Log[4]*Log[5] + L og[5]^2)/25])/x^3,x]
Output:
-((2*E^5*(2*x + Log[729]) + x*(Log[25] - 2*Log[Log[5/4]]))/x^2)
Time = 0.16 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.24, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.049, Rules used = {6, 48}
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.
\(\displaystyle \int \frac {4 e^5 x+x \left (-\log \left (\frac {1}{25} \left (\log ^2(4)+\log ^2(5)-2 \log (4) \log (5)\right )\right )\right )+24 e^5 \log (3)}{x^3} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {x \left (4 e^5-\log \left (\frac {1}{25} \left (\log ^2(4)+\log ^2(5)-2 \log (4) \log (5)\right )\right )\right )+24 e^5 \log (3)}{x^3}dx\) |
\(\Big \downarrow \) 48 |
\(\displaystyle -\frac {\left (x \left (4 e^5+\log (25)-2 \log \left (\log \left (\frac {5}{4}\right )\right )\right )+24 e^5 \log (3)\right )^2}{48 e^5 x^2 \log (3)}\) |
Input:
Int[(4*E^5*x + 24*E^5*Log[3] - x*Log[(Log[4]^2 - 2*Log[4]*Log[5] + Log[5]^ 2)/25])/x^3,x]
Output:
-1/48*(24*E^5*Log[3] + x*(4*E^5 + Log[25] - 2*Log[Log[5/4]]))^2/(E^5*x^2*L og[3])
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_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp [(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{ a, b, c, d, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1]
Time = 0.11 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00
method | result | size |
risch | \(\frac {\left (-2 \ln \left (5\right )+2 \ln \left (\ln \left (5\right )-2 \ln \left (2\right )\right )-4 \,{\mathrm e}^{5}\right ) x -12 \,{\mathrm e}^{5} \ln \left (3\right )}{x^{2}}\) | \(33\) |
gosper | \(\frac {x \ln \left (\frac {\ln \left (5\right )^{2}}{25}-\frac {4 \ln \left (2\right ) \ln \left (5\right )}{25}+\frac {4 \ln \left (2\right )^{2}}{25}\right )-12 \,{\mathrm e}^{5} \ln \left (3\right )-4 x \,{\mathrm e}^{5}}{x^{2}}\) | \(39\) |
parallelrisch | \(\frac {x \ln \left (\frac {\ln \left (5\right )^{2}}{25}-\frac {4 \ln \left (2\right ) \ln \left (5\right )}{25}+\frac {4 \ln \left (2\right )^{2}}{25}\right )-12 \,{\mathrm e}^{5} \ln \left (3\right )-4 x \,{\mathrm e}^{5}}{x^{2}}\) | \(39\) |
norman | \(\frac {\left (-2 \ln \left (5\right )+\ln \left (\ln \left (5\right )^{2}-4 \ln \left (2\right ) \ln \left (5\right )+4 \ln \left (2\right )^{2}\right )-4 \,{\mathrm e}^{5}\right ) x -12 \,{\mathrm e}^{5} \ln \left (3\right )}{x^{2}}\) | \(41\) |
default | \(-\frac {-\ln \left (\frac {\ln \left (5\right )^{2}}{25}-\frac {4 \ln \left (2\right ) \ln \left (5\right )}{25}+\frac {4 \ln \left (2\right )^{2}}{25}\right )+4 \,{\mathrm e}^{5}}{x}-\frac {12 \,{\mathrm e}^{5} \ln \left (3\right )}{x^{2}}\) | \(43\) |
Input:
int((-x*ln(1/25*ln(5)^2-4/25*ln(2)*ln(5)+4/25*ln(2)^2)+24*exp(5)*ln(3)+4*x *exp(5))/x^3,x,method=_RETURNVERBOSE)
Output:
((-2*ln(5)+2*ln(ln(5)-2*ln(2))-4*exp(5))*x-12*exp(5)*ln(3))/x^2
Time = 0.09 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.21 \[ \int \frac {4 e^5 x+24 e^5 \log (3)-x \log \left (\frac {1}{25} \left (\log ^2(4)-2 \log (4) \log (5)+\log ^2(5)\right )\right )}{x^3} \, dx=-\frac {4 \, x e^{5} + 12 \, e^{5} \log \left (3\right ) - x \log \left (\frac {1}{25} \, \log \left (5\right )^{2} - \frac {4}{25} \, \log \left (5\right ) \log \left (2\right ) + \frac {4}{25} \, \log \left (2\right )^{2}\right )}{x^{2}} \] Input:
integrate((-x*log(1/25*log(5)^2-4/25*log(2)*log(5)+4/25*log(2)^2)+24*exp(5 )*log(3)+4*x*exp(5))/x^3,x, algorithm="fricas")
Output:
-(4*x*e^5 + 12*e^5*log(3) - x*log(1/25*log(5)^2 - 4/25*log(5)*log(2) + 4/2 5*log(2)^2))/x^2
Time = 0.12 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.33 \[ \int \frac {4 e^5 x+24 e^5 \log (3)-x \log \left (\frac {1}{25} \left (\log ^2(4)-2 \log (4) \log (5)+\log ^2(5)\right )\right )}{x^3} \, dx=\frac {x \left (- 4 e^{5} - 2 \log {\left (5 \right )} + \log {\left (- 4 \log {\left (2 \right )} \log {\left (5 \right )} + 4 \log {\left (2 \right )}^{2} + \log {\left (5 \right )}^{2} \right )}\right ) - 12 e^{5} \log {\left (3 \right )}}{x^{2}} \] Input:
integrate((-x*ln(1/25*ln(5)**2-4/25*ln(2)*ln(5)+4/25*ln(2)**2)+24*exp(5)*l n(3)+4*x*exp(5))/x**3,x)
Output:
(x*(-4*exp(5) - 2*log(5) + log(-4*log(2)*log(5) + 4*log(2)**2 + log(5)**2) ) - 12*exp(5)*log(3))/x**2
Time = 0.04 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.24 \[ \int \frac {4 e^5 x+24 e^5 \log (3)-x \log \left (\frac {1}{25} \left (\log ^2(4)-2 \log (4) \log (5)+\log ^2(5)\right )\right )}{x^3} \, dx=-\frac {x {\left (4 \, e^{5} - \log \left (\frac {1}{25} \, \log \left (5\right )^{2} - \frac {4}{25} \, \log \left (5\right ) \log \left (2\right ) + \frac {4}{25} \, \log \left (2\right )^{2}\right )\right )} + 12 \, e^{5} \log \left (3\right )}{x^{2}} \] Input:
integrate((-x*log(1/25*log(5)^2-4/25*log(2)*log(5)+4/25*log(2)^2)+24*exp(5 )*log(3)+4*x*exp(5))/x^3,x, algorithm="maxima")
Output:
-(x*(4*e^5 - log(1/25*log(5)^2 - 4/25*log(5)*log(2) + 4/25*log(2)^2)) + 12 *e^5*log(3))/x^2
Time = 0.11 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.21 \[ \int \frac {4 e^5 x+24 e^5 \log (3)-x \log \left (\frac {1}{25} \left (\log ^2(4)-2 \log (4) \log (5)+\log ^2(5)\right )\right )}{x^3} \, dx=-\frac {4 \, x e^{5} + 12 \, e^{5} \log \left (3\right ) - x \log \left (\frac {1}{25} \, \log \left (5\right )^{2} - \frac {4}{25} \, \log \left (5\right ) \log \left (2\right ) + \frac {4}{25} \, \log \left (2\right )^{2}\right )}{x^{2}} \] Input:
integrate((-x*log(1/25*log(5)^2-4/25*log(2)*log(5)+4/25*log(2)^2)+24*exp(5 )*log(3)+4*x*exp(5))/x^3,x, algorithm="giac")
Output:
-(4*x*e^5 + 12*e^5*log(3) - x*log(1/25*log(5)^2 - 4/25*log(5)*log(2) + 4/2 5*log(2)^2))/x^2
Time = 0.10 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.18 \[ \int \frac {4 e^5 x+24 e^5 \log (3)-x \log \left (\frac {1}{25} \left (\log ^2(4)-2 \log (4) \log (5)+\log ^2(5)\right )\right )}{x^3} \, dx=\frac {\ln \left (\frac {4\,{\ln \left (2\right )}^2}{25}-\frac {4\,\ln \left (2\right )\,\ln \left (5\right )}{25}+\frac {{\ln \left (5\right )}^2}{25}\right )-4\,{\mathrm {e}}^5}{x}-\frac {12\,{\mathrm {e}}^5\,\ln \left (3\right )}{x^2} \] Input:
int((24*exp(5)*log(3) - x*log((4*log(2)^2)/25 - (4*log(2)*log(5))/25 + log (5)^2/25) + 4*x*exp(5))/x^3,x)
Output:
(log((4*log(2)^2)/25 - (4*log(2)*log(5))/25 + log(5)^2/25) - 4*exp(5))/x - (12*exp(5)*log(3))/x^2
Time = 0.16 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.21 \[ \int \frac {4 e^5 x+24 e^5 \log (3)-x \log \left (\frac {1}{25} \left (\log ^2(4)-2 \log (4) \log (5)+\log ^2(5)\right )\right )}{x^3} \, dx=\frac {\mathrm {log}\left (\frac {\mathrm {log}\left (5\right )^{2}}{25}-\frac {4 \,\mathrm {log}\left (5\right ) \mathrm {log}\left (2\right )}{25}+\frac {4 \mathrm {log}\left (2\right )^{2}}{25}\right ) x -12 \,\mathrm {log}\left (3\right ) e^{5}-4 e^{5} x}{x^{2}} \] Input:
int((-x*log(1/25*log(5)^2-4/25*log(2)*log(5)+4/25*log(2)^2)+24*exp(5)*log( 3)+4*x*exp(5))/x^3,x)
Output:
(log((log(5)**2 - 4*log(5)*log(2) + 4*log(2)**2)/25)*x - 12*log(3)*e**5 - 4*e**5*x)/x**2