Integrand size = 58, antiderivative size = 25 \[ \int \frac {\left (-2 x^2+e^2 (12+8 x)+\left (-10 x^2+e^2 (60+40 x)\right ) \log (5)\right ) \log (\log (25))}{\left (4 e^4-4 e^2 x+x^2\right ) \log (5)} \, dx=\frac {x (3+x) \left (5+\frac {1}{\log (5)}\right ) \log (\log (25))}{e^2-\frac {x}{2}} \] Output:
ln(2*ln(5))*(5+1/ln(5))/(exp(2)-1/2*x)*x*(3+x)
Time = 0.01 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.60 \[ \int \frac {\left (-2 x^2+e^2 (12+8 x)+\left (-10 x^2+e^2 (60+40 x)\right ) \log (5)\right ) \log (\log (25))}{\left (4 e^4-4 e^2 x+x^2\right ) \log (5)} \, dx=\frac {2 \left (-x-\frac {2 e^2 \left (3+2 e^2\right )}{-2 e^2+x}\right ) (1+5 \log (5)) \log (\log (25))}{\log (5)} \] Input:
Integrate[((-2*x^2 + E^2*(12 + 8*x) + (-10*x^2 + E^2*(60 + 40*x))*Log[5])* Log[Log[25]])/((4*E^4 - 4*E^2*x + x^2)*Log[5]),x]
Output:
(2*(-x - (2*E^2*(3 + 2*E^2))/(-2*E^2 + x))*(1 + 5*Log[5])*Log[Log[25]])/Lo g[5]
Time = 0.32 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.92, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {27, 27, 2083, 1294, 25, 1107, 2009}
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.
\(\displaystyle \int \frac {\log (\log (25)) \left (-2 x^2+\left (e^2 (40 x+60)-10 x^2\right ) \log (5)+e^2 (8 x+12)\right )}{\left (x^2-4 e^2 x+4 e^4\right ) \log (5)} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\log (\log (25)) \int -\frac {2 \left (x^2-2 e^2 (2 x+3)+5 \left (x^2-2 e^2 (2 x+3)\right ) \log (5)\right )}{x^2-4 e^2 x+4 e^4}dx}{\log (5)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {2 \log (\log (25)) \int \frac {x^2-2 e^2 (2 x+3)+5 \left (x^2-2 e^2 (2 x+3)\right ) \log (5)}{x^2-4 e^2 x+4 e^4}dx}{\log (5)}\) |
\(\Big \downarrow \) 2083 |
\(\displaystyle -\frac {2 \log (\log (25)) \int \frac {(1+5 \log (5)) x^2-4 e^2 (1+5 \log (5)) x-6 e^2 (1+5 \log (5))}{x^2-4 e^2 x+4 e^4}dx}{\log (5)}\) |
\(\Big \downarrow \) 1294 |
\(\displaystyle -\frac {2 \log (\log (25)) \int -\frac {-\left ((1+5 \log (5)) x^2\right )+4 e^2 (1+5 \log (5)) x+6 e^2 (1+5 \log (5))}{\left (2 e^2-x\right )^2}dx}{\log (5)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {2 \log (\log (25)) \int \frac {-\left ((1+5 \log (5)) x^2\right )+4 e^2 (1+5 \log (5)) x+6 e^2 (1+5 \log (5))}{\left (2 e^2-x\right )^2}dx}{\log (5)}\) |
\(\Big \downarrow \) 1107 |
\(\displaystyle \frac {2 \log (\log (25)) \int \left (-5 \log (5)-1+\frac {2 e^2 \left (3+2 e^2\right ) (1+5 \log (5))}{\left (2 e^2-x\right )^2}\right )dx}{\log (5)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 \log (\log (25)) \left (\frac {2 e^2 \left (3+2 e^2\right ) (1+5 \log (5))}{2 e^2-x}-x (1+5 \log (5))\right )}{\log (5)}\) |
Input:
Int[((-2*x^2 + E^2*(12 + 8*x) + (-10*x^2 + E^2*(60 + 40*x))*Log[5])*Log[Lo g[25]])/((4*E^4 - 4*E^2*x + x^2)*Log[5]),x]
Output:
(2*((2*E^2*(3 + 2*E^2)*(1 + 5*Log[5]))/(2*E^2 - x) - x*(1 + 5*Log[5]))*Log [Log[25]])/Log[5]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_ Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(a + b*x + c*x^2)^p, x], x] /; F reeQ[{a, b, c, d, e, m}, x] && EqQ[2*c*d - b*e, 0] && IGtQ[p, 0] && !(EqQ[ m, 3] && NeQ[p, 1])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.)*((d_.) + (e_.)*(x_) + (f_.)*(x _)^2)^(q_.), x_Symbol] :> Simp[1/c^p Int[(b/2 + c*x)^(2*p)*(d + e*x + f*x ^2)^q, x], x] /; FreeQ[{a, b, c, d, e, f, q}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]
Int[(u_)^(p_.)*(v_)^(q_.), x_Symbol] :> Int[ExpandToSum[u, x]^p*ExpandToSum [v, x]^q, x] /; FreeQ[{p, q}, x] && QuadraticQ[{u, v}, x] && !QuadraticMat chQ[{u, v}, x]
Time = 0.36 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.44
method | result | size |
gosper | \(\frac {2 \left (x^{2}+6 \,{\mathrm e}^{2}\right ) \left (5 \ln \left (5\right )+1\right ) \ln \left (2 \ln \left (5\right )\right )}{\ln \left (5\right ) \left (2 \,{\mathrm e}^{2}-x \right )}\) | \(36\) |
parallelrisch | \(\frac {\ln \left (2 \ln \left (5\right )\right ) \left (10 x^{2} \ln \left (5\right )+60 \ln \left (5\right ) {\mathrm e}^{2}+2 x^{2}+12 \,{\mathrm e}^{2}\right )}{\ln \left (5\right ) \left (2 \,{\mathrm e}^{2}-x \right )}\) | \(44\) |
norman | \(\frac {\frac {2 \left (5 \ln \left (5\right ) \ln \left (\ln \left (5\right )\right )+5 \ln \left (2\right ) \ln \left (5\right )+\ln \left (\ln \left (5\right )\right )+\ln \left (2\right )\right ) x^{2}}{\ln \left (5\right )}+\frac {12 \,{\mathrm e}^{2} \left (5 \ln \left (5\right ) \ln \left (\ln \left (5\right )\right )+5 \ln \left (2\right ) \ln \left (5\right )+\ln \left (\ln \left (5\right )\right )+\ln \left (2\right )\right )}{\ln \left (5\right )}}{2 \,{\mathrm e}^{2}-x}\) | \(68\) |
meijerg | \(-\frac {8 \,{\mathrm e}^{2} \left (-\frac {5 \ln \left (5\right )}{2}-\frac {1}{2}\right ) \ln \left (2 \ln \left (5\right )\right ) \left (-\frac {x \,{\mathrm e}^{-2} \left (-\frac {3 x \,{\mathrm e}^{-2}}{2}+6\right )}{6 \left (1-\frac {x \,{\mathrm e}^{-2}}{2}\right )}-2 \ln \left (1-\frac {x \,{\mathrm e}^{-2}}{2}\right )\right )}{\ln \left (5\right )}+\frac {4 \left (10 \ln \left (5\right ) {\mathrm e}^{2}+2 \,{\mathrm e}^{2}\right ) \ln \left (2 \ln \left (5\right )\right ) \left (\frac {x \,{\mathrm e}^{-2}}{2-x \,{\mathrm e}^{-2}}+\ln \left (1-\frac {x \,{\mathrm e}^{-2}}{2}\right )\right )}{\ln \left (5\right )}+\frac {15 \,{\mathrm e}^{-2} \ln \left (2 \ln \left (5\right )\right ) x}{1-\frac {x \,{\mathrm e}^{-2}}{2}}+\frac {3 \,{\mathrm e}^{-2} \ln \left (2 \ln \left (5\right )\right ) x}{\ln \left (5\right ) \left (1-\frac {x \,{\mathrm e}^{-2}}{2}\right )}\) | \(140\) |
risch | \(-10 x \ln \left (\ln \left (5\right )\right )-10 x \ln \left (2\right )-\frac {2 x \ln \left (\ln \left (5\right )\right )}{\ln \left (5\right )}-\frac {2 x \ln \left (2\right )}{\ln \left (5\right )}+\frac {20 \ln \left (\ln \left (5\right )\right ) {\mathrm e}^{4}}{{\mathrm e}^{2}-\frac {x}{2}}+\frac {30 \,{\mathrm e}^{2} \ln \left (\ln \left (5\right )\right )}{{\mathrm e}^{2}-\frac {x}{2}}+\frac {20 \ln \left (2\right ) {\mathrm e}^{4}}{{\mathrm e}^{2}-\frac {x}{2}}+\frac {30 \,{\mathrm e}^{2} \ln \left (2\right )}{{\mathrm e}^{2}-\frac {x}{2}}+\frac {4 \ln \left (\ln \left (5\right )\right ) {\mathrm e}^{4}}{\ln \left (5\right ) \left ({\mathrm e}^{2}-\frac {x}{2}\right )}+\frac {6 \,{\mathrm e}^{2} \ln \left (\ln \left (5\right )\right )}{\ln \left (5\right ) \left ({\mathrm e}^{2}-\frac {x}{2}\right )}+\frac {4 \ln \left (2\right ) {\mathrm e}^{4}}{\ln \left (5\right ) \left ({\mathrm e}^{2}-\frac {x}{2}\right )}+\frac {6 \,{\mathrm e}^{2} \ln \left (2\right )}{\ln \left (5\right ) \left ({\mathrm e}^{2}-\frac {x}{2}\right )}\) | \(164\) |
Input:
int((((40*x+60)*exp(2)-10*x^2)*ln(5)+(8*x+12)*exp(2)-2*x^2)*ln(2*ln(5))/(4 *exp(2)^2-4*exp(2)*x+x^2)/ln(5),x,method=_RETURNVERBOSE)
Output:
2*(x^2+6*exp(2))*(5*ln(5)+1)*ln(2*ln(5))/ln(5)/(2*exp(2)-x)
Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (25) = 50\).
Time = 0.10 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.12 \[ \int \frac {\left (-2 x^2+e^2 (12+8 x)+\left (-10 x^2+e^2 (60+40 x)\right ) \log (5)\right ) \log (\log (25))}{\left (4 e^4-4 e^2 x+x^2\right ) \log (5)} \, dx=-\frac {2 \, {\left (x^{2} - 2 \, {\left (x - 3\right )} e^{2} + 5 \, {\left (x^{2} - 2 \, {\left (x - 3\right )} e^{2} + 4 \, e^{4}\right )} \log \left (5\right ) + 4 \, e^{4}\right )} \log \left (2 \, \log \left (5\right )\right )}{{\left (x - 2 \, e^{2}\right )} \log \left (5\right )} \] Input:
integrate((((40*x+60)*exp(2)-10*x^2)*log(5)+(8*x+12)*exp(2)-2*x^2)*log(2*l og(5))/(4*exp(2)^2-4*exp(2)*x+x^2)/log(5),x, algorithm="fricas")
Output:
-2*(x^2 - 2*(x - 3)*e^2 + 5*(x^2 - 2*(x - 3)*e^2 + 4*e^4)*log(5) + 4*e^4)* log(2*log(5))/((x - 2*e^2)*log(5))
Leaf count of result is larger than twice the leaf count of optimal. 134 vs. \(2 (24) = 48\).
Time = 0.32 (sec) , antiderivative size = 134, normalized size of antiderivative = 5.36 \[ \int \frac {\left (-2 x^2+e^2 (12+8 x)+\left (-10 x^2+e^2 (60+40 x)\right ) \log (5)\right ) \log (\log (25))}{\left (4 e^4-4 e^2 x+x^2\right ) \log (5)} \, dx=- x \left (\frac {2 \log {\left (\log {\left (5 \right )} \right )}}{\log {\left (5 \right )}} + \frac {2 \log {\left (2 \right )}}{\log {\left (5 \right )}} + 10 \log {\left (\log {\left (5 \right )} \right )} + 10 \log {\left (2 \right )}\right ) - \frac {12 e^{2} \log {\left (\log {\left (5 \right )} \right )} + 12 e^{2} \log {\left (2 \right )} + 8 e^{4} \log {\left (\log {\left (5 \right )} \right )} + 8 e^{4} \log {\left (2 \right )} + 60 e^{2} \log {\left (5 \right )} \log {\left (\log {\left (5 \right )} \right )} + 60 e^{2} \log {\left (2 \right )} \log {\left (5 \right )} + 40 e^{4} \log {\left (5 \right )} \log {\left (\log {\left (5 \right )} \right )} + 40 e^{4} \log {\left (2 \right )} \log {\left (5 \right )}}{x \log {\left (5 \right )} - 2 e^{2} \log {\left (5 \right )}} \] Input:
integrate((((40*x+60)*exp(2)-10*x**2)*ln(5)+(8*x+12)*exp(2)-2*x**2)*ln(2*l n(5))/(4*exp(2)**2-4*exp(2)*x+x**2)/ln(5),x)
Output:
-x*(2*log(log(5))/log(5) + 2*log(2)/log(5) + 10*log(log(5)) + 10*log(2)) - (12*exp(2)*log(log(5)) + 12*exp(2)*log(2) + 8*exp(4)*log(log(5)) + 8*exp( 4)*log(2) + 60*exp(2)*log(5)*log(log(5)) + 60*exp(2)*log(2)*log(5) + 40*ex p(4)*log(5)*log(log(5)) + 40*exp(4)*log(2)*log(5))/(x*log(5) - 2*exp(2)*lo g(5))
Leaf count of result is larger than twice the leaf count of optimal. 52 vs. \(2 (25) = 50\).
Time = 0.03 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.08 \[ \int \frac {\left (-2 x^2+e^2 (12+8 x)+\left (-10 x^2+e^2 (60+40 x)\right ) \log (5)\right ) \log (\log (25))}{\left (4 e^4-4 e^2 x+x^2\right ) \log (5)} \, dx=-\frac {2 \, {\left (x {\left (5 \, \log \left (5\right ) + 1\right )} + \frac {2 \, {\left (5 \, {\left (2 \, e^{4} + 3 \, e^{2}\right )} \log \left (5\right ) + 2 \, e^{4} + 3 \, e^{2}\right )}}{x - 2 \, e^{2}}\right )} \log \left (2 \, \log \left (5\right )\right )}{\log \left (5\right )} \] Input:
integrate((((40*x+60)*exp(2)-10*x^2)*log(5)+(8*x+12)*exp(2)-2*x^2)*log(2*l og(5))/(4*exp(2)^2-4*exp(2)*x+x^2)/log(5),x, algorithm="maxima")
Output:
-2*(x*(5*log(5) + 1) + 2*(5*(2*e^4 + 3*e^2)*log(5) + 2*e^4 + 3*e^2)/(x - 2 *e^2))*log(2*log(5))/log(5)
Time = 0.11 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.96 \[ \int \frac {\left (-2 x^2+e^2 (12+8 x)+\left (-10 x^2+e^2 (60+40 x)\right ) \log (5)\right ) \log (\log (25))}{\left (4 e^4-4 e^2 x+x^2\right ) \log (5)} \, dx=-\frac {2 \, {\left (5 \, x \log \left (5\right ) + x + \frac {2 \, {\left (10 \, e^{4} \log \left (5\right ) + 15 \, e^{2} \log \left (5\right ) + 2 \, e^{4} + 3 \, e^{2}\right )}}{x - 2 \, e^{2}}\right )} \log \left (2 \, \log \left (5\right )\right )}{\log \left (5\right )} \] Input:
integrate((((40*x+60)*exp(2)-10*x^2)*log(5)+(8*x+12)*exp(2)-2*x^2)*log(2*l og(5))/(4*exp(2)^2-4*exp(2)*x+x^2)/log(5),x, algorithm="giac")
Output:
-2*(5*x*log(5) + x + 2*(10*e^4*log(5) + 15*e^2*log(5) + 2*e^4 + 3*e^2)/(x - 2*e^2))*log(2*log(5))/log(5)
Time = 0.22 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.68 \[ \int \frac {\left (-2 x^2+e^2 (12+8 x)+\left (-10 x^2+e^2 (60+40 x)\right ) \log (5)\right ) \log (\log (25))}{\left (4 e^4-4 e^2 x+x^2\right ) \log (5)} \, dx=-\frac {2\,\ln \left (2\,\ln \left (5\right )\right )\,\left (5\,\ln \left (5\right )+1\right )\,\left (x^2-2\,{\mathrm {e}}^2\,x+6\,{\mathrm {e}}^2+4\,{\mathrm {e}}^4\right )}{\ln \left (5\right )\,\left (x-2\,{\mathrm {e}}^2\right )} \] Input:
int(-(log(2*log(5))*(log(5)*(10*x^2 - exp(2)*(40*x + 60)) + 2*x^2 - exp(2) *(8*x + 12)))/(log(5)*(4*exp(4) - 4*x*exp(2) + x^2)),x)
Output:
-(2*log(2*log(5))*(5*log(5) + 1)*(6*exp(2) + 4*exp(4) - 2*x*exp(2) + x^2)) /(log(5)*(x - 2*exp(2)))
Time = 0.18 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.40 \[ \int \frac {\left (-2 x^2+e^2 (12+8 x)+\left (-10 x^2+e^2 (60+40 x)\right ) \log (5)\right ) \log (\log (25))}{\left (4 e^4-4 e^2 x+x^2\right ) \log (5)} \, dx=\frac {2 \,\mathrm {log}\left (2 \,\mathrm {log}\left (5\right )\right ) x \left (5 \,\mathrm {log}\left (5\right ) x +15 \,\mathrm {log}\left (5\right )+x +3\right )}{\mathrm {log}\left (5\right ) \left (2 e^{2}-x \right )} \] Input:
int((((40*x+60)*exp(2)-10*x^2)*log(5)+(8*x+12)*exp(2)-2*x^2)*log(2*log(5)) /(4*exp(2)^2-4*exp(2)*x+x^2)/log(5),x)
Output:
(2*log(2*log(5))*x*(5*log(5)*x + 15*log(5) + x + 3))/(log(5)*(2*e**2 - x))