Integrand size = 114, antiderivative size = 21 \[ \int \frac {-100-40 x-20 \log (5)}{625+1000 x+600 x^2+160 x^3+16 x^4+\left (500+600 x+240 x^2+32 x^3\right ) \log (5)+\left (150+120 x+24 x^2\right ) \log ^2(5)+(20+8 x) \log ^3(5)+\log ^4(5)+\left (-50-40 x-8 x^2+(-20-8 x) \log (5)-2 \log ^2(5)\right ) \log ^2(\log (4))+\log ^4(\log (4))} \, dx=\frac {5}{(5+2 x+\log (5))^2-\log ^2(\log (4))} \] Output:
5/((5+2*x+ln(5))^2-ln(2*ln(2))^2)
Time = 0.02 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.62 \[ \int \frac {-100-40 x-20 \log (5)}{625+1000 x+600 x^2+160 x^3+16 x^4+\left (500+600 x+240 x^2+32 x^3\right ) \log (5)+\left (150+120 x+24 x^2\right ) \log ^2(5)+(20+8 x) \log ^3(5)+\log ^4(5)+\left (-50-40 x-8 x^2+(-20-8 x) \log (5)-2 \log ^2(5)\right ) \log ^2(\log (4))+\log ^4(\log (4))} \, dx=\frac {5}{25+20 x+4 x^2+10 \log (5)+4 x \log (5)+\log ^2(5)-\log ^2(\log (4))} \] Input:
Integrate[(-100 - 40*x - 20*Log[5])/(625 + 1000*x + 600*x^2 + 160*x^3 + 16 *x^4 + (500 + 600*x + 240*x^2 + 32*x^3)*Log[5] + (150 + 120*x + 24*x^2)*Lo g[5]^2 + (20 + 8*x)*Log[5]^3 + Log[5]^4 + (-50 - 40*x - 8*x^2 + (-20 - 8*x )*Log[5] - 2*Log[5]^2)*Log[Log[4]]^2 + Log[Log[4]]^4),x]
Output:
5/(25 + 20*x + 4*x^2 + 10*Log[5] + 4*x*Log[5] + Log[5]^2 - Log[Log[4]]^2)
Time = 0.30 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.33, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.044, Rules used = {2459, 27, 1380, 27, 241}
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.
\(\displaystyle \int \frac {-40 x-100-20 \log (5)}{16 x^4+160 x^3+600 x^2+\log ^2(\log (4)) \left (-8 x^2-40 x+(-8 x-20) \log (5)-50-2 \log ^2(5)\right )+\left (24 x^2+120 x+150\right ) \log ^2(5)+\left (32 x^3+240 x^2+600 x+500\right ) \log (5)+1000 x+(8 x+20) \log ^3(5)+625+\log ^4(\log (4))+\log ^4(5)} \, dx\) |
\(\Big \downarrow \) 2459 |
\(\displaystyle \int -\frac {40 \left (x+\frac {1}{64} (160+32 \log (5))\right )}{-8 \log ^2(\log (4)) \left (x+\frac {1}{64} (160+32 \log (5))\right )^2+16 \left (x+\frac {1}{64} (160+32 \log (5))\right )^4+\log ^4(\log (4))}d\left (x+\frac {1}{64} (160+32 \log (5))\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -40 \int \frac {x+\frac {1}{64} (160+32 \log (5))}{16 \left (x+\frac {1}{64} (160+32 \log (5))\right )^4-8 \log ^2(\log (4)) \left (x+\frac {1}{64} (160+32 \log (5))\right )^2+\log ^4(\log (4))}d\left (x+\frac {1}{64} (160+32 \log (5))\right )\) |
\(\Big \downarrow \) 1380 |
\(\displaystyle -640 \int \frac {x+\frac {1}{64} (160+32 \log (5))}{16 \left (4 \left (x+\frac {1}{64} (160+32 \log (5))\right )^2-\log ^2(\log (4))\right )^2}d\left (x+\frac {1}{64} (160+32 \log (5))\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -40 \int \frac {x+\frac {1}{64} (160+32 \log (5))}{\left (4 \left (x+\frac {1}{64} (160+32 \log (5))\right )^2-\log ^2(\log (4))\right )^2}d\left (x+\frac {1}{64} (160+32 \log (5))\right )\) |
\(\Big \downarrow \) 241 |
\(\displaystyle \frac {5}{4 \left (x+\frac {1}{64} (160+32 \log (5))\right )^2-\log ^2(\log (4))}\) |
Input:
Int[(-100 - 40*x - 20*Log[5])/(625 + 1000*x + 600*x^2 + 160*x^3 + 16*x^4 + (500 + 600*x + 240*x^2 + 32*x^3)*Log[5] + (150 + 120*x + 24*x^2)*Log[5]^2 + (20 + 8*x)*Log[5]^3 + Log[5]^4 + (-50 - 40*x - 8*x^2 + (-20 - 8*x)*Log[ 5] - 2*Log[5]^2)*Log[Log[4]]^2 + Log[Log[4]]^4),x]
Output:
5/(4*(x + (160 + 32*Log[5])/64)^2 - Log[Log[4]]^2)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(x_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(a + b*x^2)^(p + 1)/ (2*b*(p + 1)), x] /; FreeQ[{a, b, p}, x] && NeQ[p, -1]
Int[(u_)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> S imp[1/c^p Int[u*(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]
Int[(Pn_)^(p_.)*(Qx_), x_Symbol] :> With[{S = Coeff[Pn, x, Expon[Pn, x] - 1 ]/(Expon[Pn, x]*Coeff[Pn, x, Expon[Pn, x]])}, Subst[Int[ExpandToSum[Pn /. x -> x - S, x]^p*ExpandToSum[Qx /. x -> x - S, x], x], x, x + S] /; Binomial Q[Pn /. x -> x - S, x] || (IntegerQ[Expon[Pn, x]/2] && TrinomialQ[Pn /. x - > x - S, x])] /; FreeQ[p, x] && PolyQ[Pn, x] && GtQ[Expon[Pn, x], 2] && NeQ [Coeff[Pn, x, Expon[Pn, x] - 1], 0] && PolyQ[Qx, x] && !(MonomialQ[Qx, x] && IGtQ[p, 0])
Time = 0.63 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.76
method | result | size |
gosper | \(-\frac {5}{\ln \left (2 \ln \left (2\right )\right )^{2}-\ln \left (5\right )^{2}-4 x \ln \left (5\right )-4 x^{2}-10 \ln \left (5\right )-20 x -25}\) | \(37\) |
norman | \(-\frac {5}{\ln \left (2 \ln \left (2\right )\right )^{2}-\ln \left (5\right )^{2}-4 x \ln \left (5\right )-4 x^{2}-10 \ln \left (5\right )-20 x -25}\) | \(37\) |
parallelrisch | \(-\frac {5}{\ln \left (2 \ln \left (2\right )\right )^{2}-\ln \left (5\right )^{2}-4 x \ln \left (5\right )-4 x^{2}-10 \ln \left (5\right )-20 x -25}\) | \(37\) |
risch | \(-\frac {5}{\ln \left (\ln \left (2\right )\right )^{2}+2 \ln \left (2\right ) \ln \left (\ln \left (2\right )\right )+\ln \left (2\right )^{2}-\ln \left (5\right )^{2}-4 x \ln \left (5\right )-4 x^{2}-10 \ln \left (5\right )-20 x -25}\) | \(46\) |
default | \(\frac {5}{2 \ln \left (2 \ln \left (2\right )\right ) \left (2 x +5-\ln \left (2 \ln \left (2\right )\right )+\ln \left (5\right )\right )}-\frac {5}{2 \ln \left (2 \ln \left (2\right )\right ) \left (2 x +5+\ln \left (2 \ln \left (2\right )\right )+\ln \left (5\right )\right )}\) | \(50\) |
Input:
int((-20*ln(5)-40*x-100)/(ln(2*ln(2))^4+(-2*ln(5)^2+(-8*x-20)*ln(5)-8*x^2- 40*x-50)*ln(2*ln(2))^2+ln(5)^4+(8*x+20)*ln(5)^3+(24*x^2+120*x+150)*ln(5)^2 +(32*x^3+240*x^2+600*x+500)*ln(5)+16*x^4+160*x^3+600*x^2+1000*x+625),x,met hod=_RETURNVERBOSE)
Output:
-5/(ln(2*ln(2))^2-ln(5)^2-4*x*ln(5)-4*x^2-10*ln(5)-20*x-25)
Time = 0.09 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.71 \[ \int \frac {-100-40 x-20 \log (5)}{625+1000 x+600 x^2+160 x^3+16 x^4+\left (500+600 x+240 x^2+32 x^3\right ) \log (5)+\left (150+120 x+24 x^2\right ) \log ^2(5)+(20+8 x) \log ^3(5)+\log ^4(5)+\left (-50-40 x-8 x^2+(-20-8 x) \log (5)-2 \log ^2(5)\right ) \log ^2(\log (4))+\log ^4(\log (4))} \, dx=\frac {5}{4 \, x^{2} + 2 \, {\left (2 \, x + 5\right )} \log \left (5\right ) + \log \left (5\right )^{2} - \log \left (2 \, \log \left (2\right )\right )^{2} + 20 \, x + 25} \] Input:
integrate((-20*log(5)-40*x-100)/(log(2*log(2))^4+(-2*log(5)^2+(-8*x-20)*lo g(5)-8*x^2-40*x-50)*log(2*log(2))^2+log(5)^4+(8*x+20)*log(5)^3+(24*x^2+120 *x+150)*log(5)^2+(32*x^3+240*x^2+600*x+500)*log(5)+16*x^4+160*x^3+600*x^2+ 1000*x+625),x, algorithm="fricas")
Output:
5/(4*x^2 + 2*(2*x + 5)*log(5) + log(5)^2 - log(2*log(2))^2 + 20*x + 25)
Leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (19) = 38\).
Time = 0.62 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.29 \[ \int \frac {-100-40 x-20 \log (5)}{625+1000 x+600 x^2+160 x^3+16 x^4+\left (500+600 x+240 x^2+32 x^3\right ) \log (5)+\left (150+120 x+24 x^2\right ) \log ^2(5)+(20+8 x) \log ^3(5)+\log ^4(5)+\left (-50-40 x-8 x^2+(-20-8 x) \log (5)-2 \log ^2(5)\right ) \log ^2(\log (4))+\log ^4(\log (4))} \, dx=\frac {5}{4 x^{2} + x \left (4 \log {\left (5 \right )} + 20\right ) - \log {\left (2 \right )}^{2} - \log {\left (\log {\left (2 \right )} \right )}^{2} - 2 \log {\left (2 \right )} \log {\left (\log {\left (2 \right )} \right )} + \log {\left (5 \right )}^{2} + 10 \log {\left (5 \right )} + 25} \] Input:
integrate((-20*ln(5)-40*x-100)/(ln(2*ln(2))**4+(-2*ln(5)**2+(-8*x-20)*ln(5 )-8*x**2-40*x-50)*ln(2*ln(2))**2+ln(5)**4+(8*x+20)*ln(5)**3+(24*x**2+120*x +150)*ln(5)**2+(32*x**3+240*x**2+600*x+500)*ln(5)+16*x**4+160*x**3+600*x** 2+1000*x+625),x)
Output:
5/(4*x**2 + x*(4*log(5) + 20) - log(2)**2 - log(log(2))**2 - 2*log(2)*log( log(2)) + log(5)**2 + 10*log(5) + 25)
Time = 0.03 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.67 \[ \int \frac {-100-40 x-20 \log (5)}{625+1000 x+600 x^2+160 x^3+16 x^4+\left (500+600 x+240 x^2+32 x^3\right ) \log (5)+\left (150+120 x+24 x^2\right ) \log ^2(5)+(20+8 x) \log ^3(5)+\log ^4(5)+\left (-50-40 x-8 x^2+(-20-8 x) \log (5)-2 \log ^2(5)\right ) \log ^2(\log (4))+\log ^4(\log (4))} \, dx=\frac {5}{4 \, x^{2} + 4 \, x {\left (\log \left (5\right ) + 5\right )} + \log \left (5\right )^{2} - \log \left (2 \, \log \left (2\right )\right )^{2} + 10 \, \log \left (5\right ) + 25} \] Input:
integrate((-20*log(5)-40*x-100)/(log(2*log(2))^4+(-2*log(5)^2+(-8*x-20)*lo g(5)-8*x^2-40*x-50)*log(2*log(2))^2+log(5)^4+(8*x+20)*log(5)^3+(24*x^2+120 *x+150)*log(5)^2+(32*x^3+240*x^2+600*x+500)*log(5)+16*x^4+160*x^3+600*x^2+ 1000*x+625),x, algorithm="maxima")
Output:
5/(4*x^2 + 4*x*(log(5) + 5) + log(5)^2 - log(2*log(2))^2 + 10*log(5) + 25)
Time = 0.11 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.71 \[ \int \frac {-100-40 x-20 \log (5)}{625+1000 x+600 x^2+160 x^3+16 x^4+\left (500+600 x+240 x^2+32 x^3\right ) \log (5)+\left (150+120 x+24 x^2\right ) \log ^2(5)+(20+8 x) \log ^3(5)+\log ^4(5)+\left (-50-40 x-8 x^2+(-20-8 x) \log (5)-2 \log ^2(5)\right ) \log ^2(\log (4))+\log ^4(\log (4))} \, dx=\frac {5}{4 \, x^{2} + 4 \, x \log \left (5\right ) + \log \left (5\right )^{2} - \log \left (2 \, \log \left (2\right )\right )^{2} + 20 \, x + 10 \, \log \left (5\right ) + 25} \] Input:
integrate((-20*log(5)-40*x-100)/(log(2*log(2))^4+(-2*log(5)^2+(-8*x-20)*lo g(5)-8*x^2-40*x-50)*log(2*log(2))^2+log(5)^4+(8*x+20)*log(5)^3+(24*x^2+120 *x+150)*log(5)^2+(32*x^3+240*x^2+600*x+500)*log(5)+16*x^4+160*x^3+600*x^2+ 1000*x+625),x, algorithm="giac")
Output:
5/(4*x^2 + 4*x*log(5) + log(5)^2 - log(2*log(2))^2 + 20*x + 10*log(5) + 25 )
Time = 0.26 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.62 \[ \int \frac {-100-40 x-20 \log (5)}{625+1000 x+600 x^2+160 x^3+16 x^4+\left (500+600 x+240 x^2+32 x^3\right ) \log (5)+\left (150+120 x+24 x^2\right ) \log ^2(5)+(20+8 x) \log ^3(5)+\log ^4(5)+\left (-50-40 x-8 x^2+(-20-8 x) \log (5)-2 \log ^2(5)\right ) \log ^2(\log (4))+\log ^4(\log (4))} \, dx=\frac {5}{4\,x^2+\left (4\,\ln \left (5\right )+20\right )\,x+10\,\ln \left (5\right )-{\ln \left (\ln \left (4\right )\right )}^2+{\ln \left (5\right )}^2+25} \] Input:
int(-(40*x + 20*log(5) + 100)/(1000*x + log(5)^3*(8*x + 20) + log(5)*(600* x + 240*x^2 + 32*x^3 + 500) - log(2*log(2))^2*(40*x + log(5)*(8*x + 20) + 2*log(5)^2 + 8*x^2 + 50) + log(2*log(2))^4 + log(5)^2*(120*x + 24*x^2 + 15 0) + log(5)^4 + 600*x^2 + 160*x^3 + 16*x^4 + 625),x)
Output:
5/(10*log(5) + x*(4*log(5) + 20) - log(log(4))^2 + log(5)^2 + 4*x^2 + 25)
Time = 0.16 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.71 \[ \int \frac {-100-40 x-20 \log (5)}{625+1000 x+600 x^2+160 x^3+16 x^4+\left (500+600 x+240 x^2+32 x^3\right ) \log (5)+\left (150+120 x+24 x^2\right ) \log ^2(5)+(20+8 x) \log ^3(5)+\log ^4(5)+\left (-50-40 x-8 x^2+(-20-8 x) \log (5)-2 \log ^2(5)\right ) \log ^2(\log (4))+\log ^4(\log (4))} \, dx=-\frac {5}{\mathrm {log}\left (2 \,\mathrm {log}\left (2\right )\right )^{2}-\mathrm {log}\left (5\right )^{2}-4 \,\mathrm {log}\left (5\right ) x -10 \,\mathrm {log}\left (5\right )-4 x^{2}-20 x -25} \] Input:
int((-20*log(5)-40*x-100)/(log(2*log(2))^4+(-2*log(5)^2+(-8*x-20)*log(5)-8 *x^2-40*x-50)*log(2*log(2))^2+log(5)^4+(8*x+20)*log(5)^3+(24*x^2+120*x+150 )*log(5)^2+(32*x^3+240*x^2+600*x+500)*log(5)+16*x^4+160*x^3+600*x^2+1000*x +625),x)
Output:
( - 5)/(log(2*log(2))**2 - log(5)**2 - 4*log(5)*x - 10*log(5) - 4*x**2 - 2 0*x - 25)