Integrand size = 79, antiderivative size = 29 \[ \int \frac {37375-24500 x+4802 x^2+\left (74875-49500 x+9800 x^2\right ) \log (4)+\left (37500-25000 x+5000 x^2\right ) \log ^2(4)}{15625-12250 x+2401 x^2+\left (31250-24750 x+4900 x^2\right ) \log (4)+\left (15625-12500 x+2500 x^2\right ) \log ^2(4)} \, dx=2 x-\frac {x}{x+\frac {5}{-2+\frac {4}{25 (4+4 \log (4))}}} \] Output:
2*x-x/(x+5/(4/25/(8*ln(2)+4)-2))
Leaf count is larger than twice the leaf count of optimal. \(60\) vs. \(2(29)=58\).
Time = 0.02 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.07 \[ \int \frac {37375-24500 x+4802 x^2+\left (74875-49500 x+9800 x^2\right ) \log (4)+\left (37500-25000 x+5000 x^2\right ) \log ^2(4)}{15625-12250 x+2401 x^2+\left (31250-24750 x+4900 x^2\right ) \log (4)+\left (15625-12500 x+2500 x^2\right ) \log ^2(4)} \, dx=\frac {-125 \left (49+99 \log (4)+50 \log ^2(4)\right )+2 (-125 (1+\log (4))+x (49+50 \log (4)))^2}{(49+50 \log (4)) (-125 (1+\log (4))+x (49+50 \log (4)))} \] Input:
Integrate[(37375 - 24500*x + 4802*x^2 + (74875 - 49500*x + 9800*x^2)*Log[4 ] + (37500 - 25000*x + 5000*x^2)*Log[4]^2)/(15625 - 12250*x + 2401*x^2 + ( 31250 - 24750*x + 4900*x^2)*Log[4] + (15625 - 12500*x + 2500*x^2)*Log[4]^2 ),x]
Output:
(-125*(49 + 99*Log[4] + 50*Log[4]^2) + 2*(-125*(1 + Log[4]) + x*(49 + 50*L og[4]))^2)/((49 + 50*Log[4])*(-125*(1 + Log[4]) + x*(49 + 50*Log[4])))
Time = 0.33 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.076, Rules used = {2083, 1294, 25, 27, 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.
| \(\displaystyle \int \frac {4802 x^2+\left (5000 x^2-25000 x+37500\right ) \log ^2(4)+\left (9800 x^2-49500 x+74875\right ) \log (4)-24500 x+37375}{2401 x^2+\left (2500 x^2-12500 x+15625\right ) \log ^2(4)+\left (4900 x^2-24750 x+31250\right ) \log (4)-12250 x+15625} \, dx\) |
| \(\Big \downarrow \) 2083 |
| \(\displaystyle \int \frac {2 x^2 (49+50 \log (4))^2-500 x (1+\log (4)) (49+50 \log (4))+125 (1+\log (4)) (299+300 \log (4))}{x^2 (49+50 \log (4))^2-250 x (1+\log (4)) (49+50 \log (4))+15625 (1+\log (4))^2}dx\) |
| \(\Big \downarrow \) 1294 |
| \(\displaystyle (49+50 \log (4))^2 \int -\frac {-2 (49+50 \log (4))^2 x^2+500 (1+\log (4)) (49+50 \log (4)) x-125 (1+\log (4)) (299+300 \log (4))}{(49+50 \log (4))^2 (125 (1+\log (4))-x (49+50 \log (4)))^2}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -(49+50 \log (4))^2 \int \frac {-2 (49+50 \log (4))^2 x^2+500 (1+\log (4)) (49+50 \log (4)) x-125 (1+\log (4)) (299+300 \log (4))}{(49+50 \log (4))^2 (125 (1+\log (4))-x (49+50 \log (4)))^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\int \frac {-2 (49+50 \log (4))^2 x^2+500 (1+\log (4)) (49+50 \log (4)) x-125 (1+\log (4)) (299+300 \log (4))}{(125 (1+\log (4))-x (49+50 \log (4)))^2}dx\) |
| \(\Big \downarrow \) 1107 |
| \(\displaystyle -\int \left (\frac {125 (-49-50 \log (4)) (1+\log (4))}{(125 (1+\log (4))-x (49+50 \log (4)))^2}-2\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 2 x+\frac {125 (1+\log (4))}{125 (1+\log (4))-x (49+50 \log (4))}\) |
Input:
Int[(37375 - 24500*x + 4802*x^2 + (74875 - 49500*x + 9800*x^2)*Log[4] + (3 7500 - 25000*x + 5000*x^2)*Log[4]^2)/(15625 - 12250*x + 2401*x^2 + (31250 - 24750*x + 4900*x^2)*Log[4] + (15625 - 12500*x + 2500*x^2)*Log[4]^2),x]
Output:
2*x + (125*(1 + Log[4]))/(125*(1 + Log[4]) - x*(49 + 50*Log[4]))
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 = 0.40 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.41
| method | result | size |
| risch | \(2 x -\frac {5 \ln \left (2\right )}{2 \left (x \ln \left (2\right )-\frac {5 \ln \left (2\right )}{2}+\frac {49 x}{100}-\frac {5}{4}\right )}-\frac {5}{4 \left (x \ln \left (2\right )-\frac {5 \ln \left (2\right )}{2}+\frac {49 x}{100}-\frac {5}{4}\right )}\) | \(41\) |
| default | \(2 x -\frac {25000 \ln \left (2\right )^{2}+24750 \ln \left (2\right )+6125}{\left (100 \ln \left (2\right )+49\right ) \left (100 x \ln \left (2\right )-250 \ln \left (2\right )+49 x -125\right )}\) | \(43\) |
| norman | \(\frac {\left (200 \ln \left (2\right )+98\right ) x^{2}-\frac {125 \left (1200 \ln \left (2\right )^{2}+1198 \ln \left (2\right )+299\right )}{100 \ln \left (2\right )+49}}{100 x \ln \left (2\right )-250 \ln \left (2\right )+49 x -125}\) | \(51\) |
| gosper | \(\frac {20000 x^{2} \ln \left (2\right )^{2}+19600 x^{2} \ln \left (2\right )-150000 \ln \left (2\right )^{2}+4802 x^{2}-149750 \ln \left (2\right )-37375}{\left (100 x \ln \left (2\right )-250 \ln \left (2\right )+49 x -125\right ) \left (100 \ln \left (2\right )+49\right )}\) | \(59\) |
| parallelrisch | \(\frac {20000 x^{2} \ln \left (2\right )^{2}+19600 x^{2} \ln \left (2\right )-150000 \ln \left (2\right )^{2}+4802 x^{2}-149750 \ln \left (2\right )-37375}{\left (100 x \ln \left (2\right )-250 \ln \left (2\right )+49 x -125\right ) \left (100 \ln \left (2\right )+49\right )}\) | \(59\) |
| meijerg | \(\frac {299 \left (100 \ln \left (2\right )+49\right )^{2} x}{125 \left (10000 \ln \left (2\right )^{2}+9800 \ln \left (2\right )+2401\right ) \left (1+2 \ln \left (2\right )\right )^{2} \left (1-\frac {x \left (100 \ln \left (2\right )+49\right )}{125 \left (1+2 \ln \left (2\right )\right )}\right )}-\frac {1953125 \left (\frac {32 \ln \left (2\right )^{2}}{25}+\frac {784 \ln \left (2\right )}{625}+\frac {4802}{15625}\right ) \left (1+2 \ln \left (2\right )\right ) \left (-\frac {x \left (100 \ln \left (2\right )+49\right ) \left (-\frac {3 x \left (100 \ln \left (2\right )+49\right )}{125 \left (1+2 \ln \left (2\right )\right )}+6\right )}{375 \left (1+2 \ln \left (2\right )\right ) \left (1-\frac {x \left (100 \ln \left (2\right )+49\right )}{125 \left (1+2 \ln \left (2\right )\right )}\right )}-2 \ln \left (1-\frac {x \left (100 \ln \left (2\right )+49\right )}{125 \left (1+2 \ln \left (2\right )\right )}\right )\right )}{\left (10000 \ln \left (2\right )^{2}+9800 \ln \left (2\right )+2401\right ) \left (100 \ln \left (2\right )+49\right )}+\frac {15625 \left (-\frac {32 \ln \left (2\right )^{2}}{5}-\frac {792 \ln \left (2\right )}{125}-\frac {196}{125}\right ) \left (\frac {x \left (100 \ln \left (2\right )+49\right )}{125 \left (1+2 \ln \left (2\right )\right ) \left (1-\frac {x \left (100 \ln \left (2\right )+49\right )}{125 \left (1+2 \ln \left (2\right )\right )}\right )}+\ln \left (1-\frac {x \left (100 \ln \left (2\right )+49\right )}{125 \left (1+2 \ln \left (2\right )\right )}\right )\right )}{10000 \ln \left (2\right )^{2}+9800 \ln \left (2\right )+2401}+\frac {48 \ln \left (2\right )^{2} \left (100 \ln \left (2\right )+49\right )^{2} x}{5 \left (10000 \ln \left (2\right )^{2}+9800 \ln \left (2\right )+2401\right ) \left (1+2 \ln \left (2\right )\right )^{2} \left (1-\frac {x \left (100 \ln \left (2\right )+49\right )}{125 \left (1+2 \ln \left (2\right )\right )}\right )}+\frac {1198 \ln \left (2\right ) \left (100 \ln \left (2\right )+49\right )^{2} x}{125 \left (10000 \ln \left (2\right )^{2}+9800 \ln \left (2\right )+2401\right ) \left (1+2 \ln \left (2\right )\right )^{2} \left (1-\frac {x \left (100 \ln \left (2\right )+49\right )}{125 \left (1+2 \ln \left (2\right )\right )}\right )}\) | \(379\) |
Input:
int((4*(5000*x^2-25000*x+37500)*ln(2)^2+2*(9800*x^2-49500*x+74875)*ln(2)+4 802*x^2-24500*x+37375)/(4*(2500*x^2-12500*x+15625)*ln(2)^2+2*(4900*x^2-247 50*x+31250)*ln(2)+2401*x^2-12250*x+15625),x,method=_RETURNVERBOSE)
Output:
2*x-5/2/(x*ln(2)-5/2*ln(2)+49/100*x-5/4)*ln(2)-5/4/(x*ln(2)-5/2*ln(2)+49/1 00*x-5/4)
Time = 0.07 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.41 \[ \int \frac {37375-24500 x+4802 x^2+\left (74875-49500 x+9800 x^2\right ) \log (4)+\left (37500-25000 x+5000 x^2\right ) \log ^2(4)}{15625-12250 x+2401 x^2+\left (31250-24750 x+4900 x^2\right ) \log (4)+\left (15625-12500 x+2500 x^2\right ) \log ^2(4)} \, dx=\frac {98 \, x^{2} + 50 \, {\left (4 \, x^{2} - 10 \, x - 5\right )} \log \left (2\right ) - 250 \, x - 125}{50 \, {\left (2 \, x - 5\right )} \log \left (2\right ) + 49 \, x - 125} \] Input:
integrate((4*(5000*x^2-25000*x+37500)*log(2)^2+2*(9800*x^2-49500*x+74875)* log(2)+4802*x^2-24500*x+37375)/(4*(2500*x^2-12500*x+15625)*log(2)^2+2*(490 0*x^2-24750*x+31250)*log(2)+2401*x^2-12250*x+15625),x, algorithm="fricas")
Output:
(98*x^2 + 50*(4*x^2 - 10*x - 5)*log(2) - 250*x - 125)/(50*(2*x - 5)*log(2) + 49*x - 125)
Time = 0.49 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {37375-24500 x+4802 x^2+\left (74875-49500 x+9800 x^2\right ) \log (4)+\left (37500-25000 x+5000 x^2\right ) \log ^2(4)}{15625-12250 x+2401 x^2+\left (31250-24750 x+4900 x^2\right ) \log (4)+\left (15625-12500 x+2500 x^2\right ) \log ^2(4)} \, dx=2 x + \frac {- 250 \log {\left (2 \right )} - 125}{x \left (49 + 100 \log {\left (2 \right )}\right ) - 250 \log {\left (2 \right )} - 125} \] Input:
integrate((4*(5000*x**2-25000*x+37500)*ln(2)**2+2*(9800*x**2-49500*x+74875 )*ln(2)+4802*x**2-24500*x+37375)/(4*(2500*x**2-12500*x+15625)*ln(2)**2+2*( 4900*x**2-24750*x+31250)*ln(2)+2401*x**2-12250*x+15625),x)
Output:
2*x + (-250*log(2) - 125)/(x*(49 + 100*log(2)) - 250*log(2) - 125)
Time = 0.03 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {37375-24500 x+4802 x^2+\left (74875-49500 x+9800 x^2\right ) \log (4)+\left (37500-25000 x+5000 x^2\right ) \log ^2(4)}{15625-12250 x+2401 x^2+\left (31250-24750 x+4900 x^2\right ) \log (4)+\left (15625-12500 x+2500 x^2\right ) \log ^2(4)} \, dx=2 \, x - \frac {125 \, {\left (2 \, \log \left (2\right ) + 1\right )}}{x {\left (100 \, \log \left (2\right ) + 49\right )} - 250 \, \log \left (2\right ) - 125} \] Input:
integrate((4*(5000*x^2-25000*x+37500)*log(2)^2+2*(9800*x^2-49500*x+74875)* log(2)+4802*x^2-24500*x+37375)/(4*(2500*x^2-12500*x+15625)*log(2)^2+2*(490 0*x^2-24750*x+31250)*log(2)+2401*x^2-12250*x+15625),x, algorithm="maxima")
Output:
2*x - 125*(2*log(2) + 1)/(x*(100*log(2) + 49) - 250*log(2) - 125)
Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (25) = 50\).
Time = 0.12 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.97 \[ \int \frac {37375-24500 x+4802 x^2+\left (74875-49500 x+9800 x^2\right ) \log (4)+\left (37500-25000 x+5000 x^2\right ) \log ^2(4)}{15625-12250 x+2401 x^2+\left (31250-24750 x+4900 x^2\right ) \log (4)+\left (15625-12500 x+2500 x^2\right ) \log ^2(4)} \, dx=\frac {2 \, {\left (10000 \, x \log \left (2\right )^{2} + 9800 \, x \log \left (2\right ) + 2401 \, x\right )}}{10000 \, \log \left (2\right )^{2} + 9800 \, \log \left (2\right ) + 2401} - \frac {125 \, {\left (2 \, \log \left (2\right ) + 1\right )}}{100 \, x \log \left (2\right ) + 49 \, x - 250 \, \log \left (2\right ) - 125} \] Input:
integrate((4*(5000*x^2-25000*x+37500)*log(2)^2+2*(9800*x^2-49500*x+74875)* log(2)+4802*x^2-24500*x+37375)/(4*(2500*x^2-12500*x+15625)*log(2)^2+2*(490 0*x^2-24750*x+31250)*log(2)+2401*x^2-12250*x+15625),x, algorithm="giac")
Output:
2*(10000*x*log(2)^2 + 9800*x*log(2) + 2401*x)/(10000*log(2)^2 + 9800*log(2 ) + 2401) - 125*(2*log(2) + 1)/(100*x*log(2) + 49*x - 250*log(2) - 125)
Timed out. \[ \int \frac {37375-24500 x+4802 x^2+\left (74875-49500 x+9800 x^2\right ) \log (4)+\left (37500-25000 x+5000 x^2\right ) \log ^2(4)}{15625-12250 x+2401 x^2+\left (31250-24750 x+4900 x^2\right ) \log (4)+\left (15625-12500 x+2500 x^2\right ) \log ^2(4)} \, dx=\int \frac {2\,\ln \left (2\right )\,\left (9800\,x^2-49500\,x+74875\right )-24500\,x+4\,{\ln \left (2\right )}^2\,\left (5000\,x^2-25000\,x+37500\right )+4802\,x^2+37375}{2\,\ln \left (2\right )\,\left (4900\,x^2-24750\,x+31250\right )-12250\,x+4\,{\ln \left (2\right )}^2\,\left (2500\,x^2-12500\,x+15625\right )+2401\,x^2+15625} \,d x \] Input:
int((2*log(2)*(9800*x^2 - 49500*x + 74875) - 24500*x + 4*log(2)^2*(5000*x^ 2 - 25000*x + 37500) + 4802*x^2 + 37375)/(2*log(2)*(4900*x^2 - 24750*x + 3 1250) - 12250*x + 4*log(2)^2*(2500*x^2 - 12500*x + 15625) + 2401*x^2 + 156 25),x)
Output:
int((2*log(2)*(9800*x^2 - 49500*x + 74875) - 24500*x + 4*log(2)^2*(5000*x^ 2 - 25000*x + 37500) + 4802*x^2 + 37375)/(2*log(2)*(4900*x^2 - 24750*x + 3 1250) - 12250*x + 4*log(2)^2*(2500*x^2 - 12500*x + 15625) + 2401*x^2 + 156 25), x)
Time = 0.17 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10 \[ \int \frac {37375-24500 x+4802 x^2+\left (74875-49500 x+9800 x^2\right ) \log (4)+\left (37500-25000 x+5000 x^2\right ) \log ^2(4)}{15625-12250 x+2401 x^2+\left (31250-24750 x+4900 x^2\right ) \log (4)+\left (15625-12500 x+2500 x^2\right ) \log ^2(4)} \, dx=\frac {x \left (200 \,\mathrm {log}\left (2\right ) x -600 \,\mathrm {log}\left (2\right )+98 x -299\right )}{100 \,\mathrm {log}\left (2\right ) x -250 \,\mathrm {log}\left (2\right )+49 x -125} \] Input:
int((4*(5000*x^2-25000*x+37500)*log(2)^2+2*(9800*x^2-49500*x+74875)*log(2) +4802*x^2-24500*x+37375)/(4*(2500*x^2-12500*x+15625)*log(2)^2+2*(4900*x^2- 24750*x+31250)*log(2)+2401*x^2-12250*x+15625),x)
Output:
(x*(200*log(2)*x - 600*log(2) + 98*x - 299))/(100*log(2)*x - 250*log(2) + 49*x - 125)