Integrand size = 76, antiderivative size = 22 \[ \int \frac {20 x+\left (50+60 x+16 x^2+\left (25+40 x+16 x^2\right ) \log (8)\right ) \log \left (\frac {-15-12 x}{10+4 x+(5+4 x) \log (8)}\right )}{50+60 x+16 x^2+\left (25+40 x+16 x^2\right ) \log (8)} \, dx=x \log \left (\frac {3}{-1-\frac {5}{5+4 x}-\log (8)}\right ) \] Output:
ln(3/(-1-5/(5+4*x)-3*ln(2)))*x
Leaf count is larger than twice the leaf count of optimal. \(60\) vs. \(2(22)=44\).
Time = 0.03 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.73 \[ \int \frac {20 x+\left (50+60 x+16 x^2+\left (25+40 x+16 x^2\right ) \log (8)\right ) \log \left (\frac {-15-12 x}{10+4 x+(5+4 x) \log (8)}\right )}{50+60 x+16 x^2+\left (25+40 x+16 x^2\right ) \log (8)} \, dx=\frac {1}{4} \left (-5 \log (5+4 x)+(5+4 x) \log \left (-\frac {3 (5+4 x)}{4 x (1+\log (8))+5 (2+\log (8))}\right )+5 \log (4 x (1+\log (8))+5 (2+\log (8)))\right ) \] Input:
Integrate[(20*x + (50 + 60*x + 16*x^2 + (25 + 40*x + 16*x^2)*Log[8])*Log[( -15 - 12*x)/(10 + 4*x + (5 + 4*x)*Log[8])])/(50 + 60*x + 16*x^2 + (25 + 40 *x + 16*x^2)*Log[8]),x]
Output:
(-5*Log[5 + 4*x] + (5 + 4*x)*Log[(-3*(5 + 4*x))/(4*x*(1 + Log[8]) + 5*(2 + Log[8]))] + 5*Log[4*x*(1 + Log[8]) + 5*(2 + Log[8])])/4
Leaf count is larger than twice the leaf count of optimal. \(158\) vs. \(2(22)=44\).
Time = 0.75 (sec) , antiderivative size = 158, normalized size of antiderivative = 7.18, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.039, Rules used = {7292, 7279, 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.
| \(\displaystyle \int \frac {\left (16 x^2+\left (16 x^2+40 x+25\right ) \log (8)+60 x+50\right ) \log \left (\frac {-12 x-15}{4 x+(4 x+5) \log (8)+10}\right )+20 x}{16 x^2+\left (16 x^2+40 x+25\right ) \log (8)+60 x+50} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {\left (16 x^2+\left (16 x^2+40 x+25\right ) \log (8)+60 x+50\right ) \log \left (\frac {-12 x-15}{4 x+(4 x+5) \log (8)+10}\right )+20 x}{16 x^2 (1+\log (8))+20 x (3+\log (64))+25 (2+\log (8))}dx\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle \int \left (\frac {20 x}{16 x^2 (1+\log (8))+20 x (3+\log (64))+25 (2+\log (8))}+\frac {(4 x+5) (4 x (1+\log (8))+5 (2+\log (8))) \log \left (-\frac {3 (4 x+5)}{4 x (1+\log (8))+5 (2+\log (8))}\right )}{16 x^2 (1+\log (8))+20 x (3+\log (64))+25 (2+\log (8))}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {5 (3+\log (64)) \text {arctanh}\left (\frac {8 x (1+\log (8))+5 (3+\log (64))}{5 \sqrt {1-4 \log ^2(8)+\log ^2(64)}}\right )}{4 (1+\log (8)) \sqrt {1-4 \log ^2(8)+\log ^2(64)}}+\frac {5 \log \left (16 x^2 (1+\log (8))+20 x (3+\log (64))+25 (2+\log (8))\right )}{8 (1+\log (8))}+\frac {1}{4} (4 x+5) \log \left (-\frac {3 (4 x+5)}{4 x (1+\log (8))+5 (2+\log (8))}\right )-\frac {5 \log (4 x (1+\log (8))+5 (2+\log (8)))}{4 (1+\log (8))}\) |
Input:
Int[(20*x + (50 + 60*x + 16*x^2 + (25 + 40*x + 16*x^2)*Log[8])*Log[(-15 - 12*x)/(10 + 4*x + (5 + 4*x)*Log[8])])/(50 + 60*x + 16*x^2 + (25 + 40*x + 1 6*x^2)*Log[8]),x]
Output:
(5*ArcTanh[(8*x*(1 + Log[8]) + 5*(3 + Log[64]))/(5*Sqrt[1 - 4*Log[8]^2 + L og[64]^2])]*(3 + Log[64]))/(4*(1 + Log[8])*Sqrt[1 - 4*Log[8]^2 + Log[64]^2 ]) + ((5 + 4*x)*Log[(-3*(5 + 4*x))/(4*x*(1 + Log[8]) + 5*(2 + Log[8]))])/4 - (5*Log[4*x*(1 + Log[8]) + 5*(2 + Log[8])])/(4*(1 + Log[8])) + (5*Log[16 *x^2*(1 + Log[8]) + 25*(2 + Log[8]) + 20*x*(3 + Log[64])])/(8*(1 + Log[8]) )
Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ {v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
Time = 1.33 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18
| method | result | size |
| norman | \(x \ln \left (\frac {-12 x -15}{3 \left (5+4 x \right ) \ln \left (2\right )+4 x +10}\right )\) | \(26\) |
| risch | \(x \ln \left (\frac {-12 x -15}{3 \left (5+4 x \right ) \ln \left (2\right )+4 x +10}\right )\) | \(26\) |
| parallelrisch | \(-\frac {-5625 \ln \left (2\right )^{2} \ln \left (-\frac {3 \left (5+4 x \right )}{12 x \ln \left (2\right )+15 \ln \left (2\right )+4 x +10}\right ) x -7500 \ln \left (-\frac {3 \left (5+4 x \right )}{12 x \ln \left (2\right )+15 \ln \left (2\right )+4 x +10}\right ) \ln \left (2\right ) x -2500 \ln \left (-\frac {3 \left (5+4 x \right )}{12 x \ln \left (2\right )+15 \ln \left (2\right )+4 x +10}\right ) x}{625 \left (3 \ln \left (2\right )+2\right )^{2}}\) | \(99\) |
| parts | \(\frac {20 \left (\frac {3 \ln \left (2\right )}{4}+\frac {1}{2}\right ) \ln \left (12 x \ln \left (2\right )+15 \ln \left (2\right )+4 x +10\right )}{12 \ln \left (2\right )+4}-\frac {5 \ln \left (5+4 x \right )}{4}-\frac {15 \left (9 \ln \left (2\right )^{2}+6 \ln \left (2\right )+1\right ) \left (-\frac {\ln \left (3+\left (3 \ln \left (2\right )+1\right ) \left (-\frac {3}{3 \ln \left (2\right )+1}+\frac {15}{\left (3 \ln \left (2\right )+1\right ) \left (12 x \ln \left (2\right )+15 \ln \left (2\right )+4 x +10\right )}\right )\right )}{3 \left (3 \ln \left (2\right )+1\right )}+\frac {\ln \left (-\frac {3}{3 \ln \left (2\right )+1}+\frac {15}{\left (3 \ln \left (2\right )+1\right ) \left (12 x \ln \left (2\right )+15 \ln \left (2\right )+4 x +10\right )}\right ) \left (-\frac {3}{3 \ln \left (2\right )+1}+\frac {15}{\left (3 \ln \left (2\right )+1\right ) \left (12 x \ln \left (2\right )+15 \ln \left (2\right )+4 x +10\right )}\right )}{9 \left (-\frac {3}{3 \ln \left (2\right )+1}+\frac {15}{\left (3 \ln \left (2\right )+1\right ) \left (12 x \ln \left (2\right )+15 \ln \left (2\right )+4 x +10\right )}\right ) \ln \left (2\right )+\frac {45}{\left (3 \ln \left (2\right )+1\right ) \left (12 x \ln \left (2\right )+15 \ln \left (2\right )+4 x +10\right )}-\frac {9}{3 \ln \left (2\right )+1}+9}\right )}{4 \left (3 \ln \left (2\right )+1\right )^{2}}\) | \(279\) |
| derivativedivides | \(-\frac {15 \left (\left (9 \ln \left (2\right )^{2}+6 \ln \left (2\right )+1\right ) \left (-\frac {\ln \left (3+\left (3 \ln \left (2\right )+1\right ) \left (-\frac {3}{3 \ln \left (2\right )+1}+\frac {15}{\left (3 \ln \left (2\right )+1\right ) \left (12 x \ln \left (2\right )+15 \ln \left (2\right )+4 x +10\right )}\right )\right )}{3 \left (3 \ln \left (2\right )+1\right )}+\frac {\ln \left (-\frac {3}{3 \ln \left (2\right )+1}+\frac {15}{\left (3 \ln \left (2\right )+1\right ) \left (12 x \ln \left (2\right )+15 \ln \left (2\right )+4 x +10\right )}\right ) \left (-\frac {3}{3 \ln \left (2\right )+1}+\frac {15}{\left (3 \ln \left (2\right )+1\right ) \left (12 x \ln \left (2\right )+15 \ln \left (2\right )+4 x +10\right )}\right )}{9 \left (-\frac {3}{3 \ln \left (2\right )+1}+\frac {15}{\left (3 \ln \left (2\right )+1\right ) \left (12 x \ln \left (2\right )+15 \ln \left (2\right )+4 x +10\right )}\right ) \ln \left (2\right )+\frac {45}{\left (3 \ln \left (2\right )+1\right ) \left (12 x \ln \left (2\right )+15 \ln \left (2\right )+4 x +10\right )}-\frac {9}{3 \ln \left (2\right )+1}+9}\right )+\frac {\left (9 \ln \left (2\right )^{2}+6 \ln \left (2\right )+1\right ) \left (\ln \left (-\frac {3}{3 \ln \left (2\right )+1}+\frac {15}{\left (3 \ln \left (2\right )+1\right ) \left (12 x \ln \left (2\right )+15 \ln \left (2\right )+4 x +10\right )}\right )+\frac {\ln \left (3 \left (-\frac {3}{3 \ln \left (2\right )+1}+\frac {15}{\left (3 \ln \left (2\right )+1\right ) \left (12 x \ln \left (2\right )+15 \ln \left (2\right )+4 x +10\right )}\right ) \ln \left (2\right )+\frac {15}{\left (3 \ln \left (2\right )+1\right ) \left (12 x \ln \left (2\right )+15 \ln \left (2\right )+4 x +10\right )}-\frac {3}{3 \ln \left (2\right )+1}+3\right )}{3 \ln \left (2\right )+1}\right )}{3}\right )}{4 \left (3 \ln \left (2\right )+1\right )^{2}}\) | \(383\) |
| default | \(-\frac {15 \left (\left (9 \ln \left (2\right )^{2}+6 \ln \left (2\right )+1\right ) \left (-\frac {\ln \left (3+\left (3 \ln \left (2\right )+1\right ) \left (-\frac {3}{3 \ln \left (2\right )+1}+\frac {15}{\left (3 \ln \left (2\right )+1\right ) \left (12 x \ln \left (2\right )+15 \ln \left (2\right )+4 x +10\right )}\right )\right )}{3 \left (3 \ln \left (2\right )+1\right )}+\frac {\ln \left (-\frac {3}{3 \ln \left (2\right )+1}+\frac {15}{\left (3 \ln \left (2\right )+1\right ) \left (12 x \ln \left (2\right )+15 \ln \left (2\right )+4 x +10\right )}\right ) \left (-\frac {3}{3 \ln \left (2\right )+1}+\frac {15}{\left (3 \ln \left (2\right )+1\right ) \left (12 x \ln \left (2\right )+15 \ln \left (2\right )+4 x +10\right )}\right )}{9 \left (-\frac {3}{3 \ln \left (2\right )+1}+\frac {15}{\left (3 \ln \left (2\right )+1\right ) \left (12 x \ln \left (2\right )+15 \ln \left (2\right )+4 x +10\right )}\right ) \ln \left (2\right )+\frac {45}{\left (3 \ln \left (2\right )+1\right ) \left (12 x \ln \left (2\right )+15 \ln \left (2\right )+4 x +10\right )}-\frac {9}{3 \ln \left (2\right )+1}+9}\right )+\frac {\left (9 \ln \left (2\right )^{2}+6 \ln \left (2\right )+1\right ) \left (\ln \left (-\frac {3}{3 \ln \left (2\right )+1}+\frac {15}{\left (3 \ln \left (2\right )+1\right ) \left (12 x \ln \left (2\right )+15 \ln \left (2\right )+4 x +10\right )}\right )+\frac {\ln \left (3 \left (-\frac {3}{3 \ln \left (2\right )+1}+\frac {15}{\left (3 \ln \left (2\right )+1\right ) \left (12 x \ln \left (2\right )+15 \ln \left (2\right )+4 x +10\right )}\right ) \ln \left (2\right )+\frac {15}{\left (3 \ln \left (2\right )+1\right ) \left (12 x \ln \left (2\right )+15 \ln \left (2\right )+4 x +10\right )}-\frac {3}{3 \ln \left (2\right )+1}+3\right )}{3 \ln \left (2\right )+1}\right )}{3}\right )}{4 \left (3 \ln \left (2\right )+1\right )^{2}}\) | \(383\) |
Input:
int(((3*(16*x^2+40*x+25)*ln(2)+16*x^2+60*x+50)*ln((-12*x-15)/(3*(5+4*x)*ln (2)+4*x+10))+20*x)/(3*(16*x^2+40*x+25)*ln(2)+16*x^2+60*x+50),x,method=_RET URNVERBOSE)
Output:
x*ln((-12*x-15)/(3*(5+4*x)*ln(2)+4*x+10))
Time = 0.08 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {20 x+\left (50+60 x+16 x^2+\left (25+40 x+16 x^2\right ) \log (8)\right ) \log \left (\frac {-15-12 x}{10+4 x+(5+4 x) \log (8)}\right )}{50+60 x+16 x^2+\left (25+40 x+16 x^2\right ) \log (8)} \, dx=x \log \left (-\frac {3 \, {\left (4 \, x + 5\right )}}{3 \, {\left (4 \, x + 5\right )} \log \left (2\right ) + 4 \, x + 10}\right ) \] Input:
integrate(((3*(16*x^2+40*x+25)*log(2)+16*x^2+60*x+50)*log((-12*x-15)/(3*(5 +4*x)*log(2)+4*x+10))+20*x)/(3*(16*x^2+40*x+25)*log(2)+16*x^2+60*x+50),x, algorithm="fricas")
Output:
x*log(-3*(4*x + 5)/(3*(4*x + 5)*log(2) + 4*x + 10))
Time = 0.10 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {20 x+\left (50+60 x+16 x^2+\left (25+40 x+16 x^2\right ) \log (8)\right ) \log \left (\frac {-15-12 x}{10+4 x+(5+4 x) \log (8)}\right )}{50+60 x+16 x^2+\left (25+40 x+16 x^2\right ) \log (8)} \, dx=x \log {\left (\frac {- 12 x - 15}{4 x + \left (12 x + 15\right ) \log {\left (2 \right )} + 10} \right )} \] Input:
integrate(((3*(16*x**2+40*x+25)*ln(2)+16*x**2+60*x+50)*ln((-12*x-15)/(3*(5 +4*x)*ln(2)+4*x+10))+20*x)/(3*(16*x**2+40*x+25)*ln(2)+16*x**2+60*x+50),x)
Output:
x*log((-12*x - 15)/(4*x + (12*x + 15)*log(2) + 10))
Result contains complex when optimal does not.
Time = 0.19 (sec) , antiderivative size = 512, normalized size of antiderivative = 23.27 \[ \int \frac {20 x+\left (50+60 x+16 x^2+\left (25+40 x+16 x^2\right ) \log (8)\right ) \log \left (\frac {-15-12 x}{10+4 x+(5+4 x) \log (8)}\right )}{50+60 x+16 x^2+\left (25+40 x+16 x^2\right ) \log (8)} \, dx =\text {Too large to display} \] Input:
integrate(((3*(16*x^2+40*x+25)*log(2)+16*x^2+60*x+50)*log((-12*x-15)/(3*(5 +4*x)*log(2)+4*x+10))+20*x)/(3*(16*x^2+40*x+25)*log(2)+16*x^2+60*x+50),x, algorithm="maxima")
Output:
-15/4*(log(4*x*(3*log(2) + 1) + 15*log(2) + 10) - log(4*x + 5))*log(2)*log (-12*x/(12*x*log(2) + 4*x + 15*log(2) + 10) - 15/(12*x*log(2) + 4*x + 15*l og(2) + 10)) - 15/8*(log(4*x*(3*log(2) + 1) + 15*log(2) + 10)^2 - 2*log(4* x*(3*log(2) + 1) + 15*log(2) + 10)*log(4*x + 5) + log(4*x + 5)^2)*log(2) - 5/4*log(4*x*(3*log(2) + 1) + 15*log(2) + 10)^2 + 5/2*log(4*x*(3*log(2) + 1) + 15*log(2) + 10)*log(4*x + 5) - 5/4*log(4*x + 5)^2 - 5/2*(log(4*x*(3*l og(2) + 1) + 15*log(2) + 10) - log(4*x + 5))*log(-12*x/(12*x*log(2) + 4*x + 15*log(2) + 10) - 15/(12*x*log(2) + 4*x + 15*log(2) + 10)) + 5/4*(3*log( 2) + 2)*log(4*x*(3*log(2) + 1) + 15*log(2) + 10)/(3*log(2) + 1) - 1/8*(5*( 9*log(2)^2 + 9*log(2) + 2)*log(4*x*(3*log(2) + 1) + 15*log(2) + 10)^2 + 5* (9*log(2)^2 + 9*log(2) + 2)*log(4*x + 5)^2 - 8*(I*pi*(3*log(2) + 1) + 3*lo g(3)*log(2) + log(3))*x - 2*(45*log(3)*log(2)^2 + 5*I*pi*(9*log(2)^2 + 9*l og(2) + 2) - 4*x*(3*log(2) + 1) + 15*(3*log(3) - 1)*log(2) + 5*(9*log(2)^2 + 9*log(2) + 2)*log(4*x + 5) + 10*log(3) - 10)*log(4*x*(3*log(2) + 1) + 1 5*log(2) + 10) + 2*(45*log(3)*log(2)^2 + 5*I*pi*(9*log(2)^2 + 9*log(2) + 2 ) - 4*x*(3*log(2) + 1) + 15*(3*log(3) - 1)*log(2) + 10*log(3) - 5)*log(4*x + 5))/(3*log(2) + 1) - 5/4*log(4*x + 5)
Leaf count of result is larger than twice the leaf count of optimal. 149 vs. \(2 (22) = 44\).
Time = 0.21 (sec) , antiderivative size = 149, normalized size of antiderivative = 6.77 \[ \int \frac {20 x+\left (50+60 x+16 x^2+\left (25+40 x+16 x^2\right ) \log (8)\right ) \log \left (\frac {-15-12 x}{10+4 x+(5+4 x) \log (8)}\right )}{50+60 x+16 x^2+\left (25+40 x+16 x^2\right ) \log (8)} \, dx=-\frac {5 \, {\left (3 \, \log \left (2\right ) + 2\right )} \log \left (-\frac {3 \, {\left (4 \, x + 5\right )}}{12 \, x \log \left (2\right ) + 4 \, x + 15 \, \log \left (2\right ) + 10}\right )}{4 \, {\left (3 \, \log \left (2\right ) + 1\right )}} - \frac {5 \, \log \left (-\frac {3 \, {\left (4 \, x + 5\right )}}{12 \, x \log \left (2\right ) + 4 \, x + 15 \, \log \left (2\right ) + 10}\right )}{4 \, {\left (\frac {9 \, {\left (4 \, x + 5\right )} \log \left (2\right )^{2}}{12 \, x \log \left (2\right ) + 4 \, x + 15 \, \log \left (2\right ) + 10} + \frac {6 \, {\left (4 \, x + 5\right )} \log \left (2\right )}{12 \, x \log \left (2\right ) + 4 \, x + 15 \, \log \left (2\right ) + 10} + \frac {4 \, x + 5}{12 \, x \log \left (2\right ) + 4 \, x + 15 \, \log \left (2\right ) + 10} - 3 \, \log \left (2\right ) - 1\right )}} \] Input:
integrate(((3*(16*x^2+40*x+25)*log(2)+16*x^2+60*x+50)*log((-12*x-15)/(3*(5 +4*x)*log(2)+4*x+10))+20*x)/(3*(16*x^2+40*x+25)*log(2)+16*x^2+60*x+50),x, algorithm="giac")
Output:
-5/4*(3*log(2) + 2)*log(-3*(4*x + 5)/(12*x*log(2) + 4*x + 15*log(2) + 10)) /(3*log(2) + 1) - 5/4*log(-3*(4*x + 5)/(12*x*log(2) + 4*x + 15*log(2) + 10 ))/(9*(4*x + 5)*log(2)^2/(12*x*log(2) + 4*x + 15*log(2) + 10) + 6*(4*x + 5 )*log(2)/(12*x*log(2) + 4*x + 15*log(2) + 10) + (4*x + 5)/(12*x*log(2) + 4 *x + 15*log(2) + 10) - 3*log(2) - 1)
Time = 0.72 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {20 x+\left (50+60 x+16 x^2+\left (25+40 x+16 x^2\right ) \log (8)\right ) \log \left (\frac {-15-12 x}{10+4 x+(5+4 x) \log (8)}\right )}{50+60 x+16 x^2+\left (25+40 x+16 x^2\right ) \log (8)} \, dx=x\,\ln \left (-\frac {12\,x+15}{4\,x+3\,\ln \left (2\right )\,\left (4\,x+5\right )+10}\right ) \] Input:
int((20*x + log(-(12*x + 15)/(4*x + 3*log(2)*(4*x + 5) + 10))*(60*x + 3*lo g(2)*(40*x + 16*x^2 + 25) + 16*x^2 + 50))/(60*x + 3*log(2)*(40*x + 16*x^2 + 25) + 16*x^2 + 50),x)
Output:
x*log(-(12*x + 15)/(4*x + 3*log(2)*(4*x + 5) + 10))
Time = 0.17 (sec) , antiderivative size = 170, normalized size of antiderivative = 7.73 \[ \int \frac {20 x+\left (50+60 x+16 x^2+\left (25+40 x+16 x^2\right ) \log (8)\right ) \log \left (\frac {-15-12 x}{10+4 x+(5+4 x) \log (8)}\right )}{50+60 x+16 x^2+\left (25+40 x+16 x^2\right ) \log (8)} \, dx=\frac {15 \,\mathrm {log}\left (12 \,\mathrm {log}\left (2\right ) x +15 \,\mathrm {log}\left (2\right )+4 x +10\right ) \mathrm {log}\left (2\right )+10 \,\mathrm {log}\left (12 \,\mathrm {log}\left (2\right ) x +15 \,\mathrm {log}\left (2\right )+4 x +10\right )-15 \,\mathrm {log}\left (4 x +5\right ) \mathrm {log}\left (2\right )-10 \,\mathrm {log}\left (4 x +5\right )+12 \,\mathrm {log}\left (\frac {-12 x -15}{12 \,\mathrm {log}\left (2\right ) x +15 \,\mathrm {log}\left (2\right )+4 x +10}\right ) \mathrm {log}\left (2\right ) x +15 \,\mathrm {log}\left (\frac {-12 x -15}{12 \,\mathrm {log}\left (2\right ) x +15 \,\mathrm {log}\left (2\right )+4 x +10}\right ) \mathrm {log}\left (2\right )+4 \,\mathrm {log}\left (\frac {-12 x -15}{12 \,\mathrm {log}\left (2\right ) x +15 \,\mathrm {log}\left (2\right )+4 x +10}\right ) x +10 \,\mathrm {log}\left (\frac {-12 x -15}{12 \,\mathrm {log}\left (2\right ) x +15 \,\mathrm {log}\left (2\right )+4 x +10}\right )}{12 \,\mathrm {log}\left (2\right )+4} \] Input:
int(((3*(16*x^2+40*x+25)*log(2)+16*x^2+60*x+50)*log((-12*x-15)/(3*(5+4*x)* log(2)+4*x+10))+20*x)/(3*(16*x^2+40*x+25)*log(2)+16*x^2+60*x+50),x)
Output:
(15*log(12*log(2)*x + 15*log(2) + 4*x + 10)*log(2) + 10*log(12*log(2)*x + 15*log(2) + 4*x + 10) - 15*log(4*x + 5)*log(2) - 10*log(4*x + 5) + 12*log( ( - 12*x - 15)/(12*log(2)*x + 15*log(2) + 4*x + 10))*log(2)*x + 15*log(( - 12*x - 15)/(12*log(2)*x + 15*log(2) + 4*x + 10))*log(2) + 4*log(( - 12*x - 15)/(12*log(2)*x + 15*log(2) + 4*x + 10))*x + 10*log(( - 12*x - 15)/(12* log(2)*x + 15*log(2) + 4*x + 10)))/(4*(3*log(2) + 1))