\(\int \frac {20 x+(50+60 x+16 x^2+(25+40 x+16 x^2) \log (8)) \log (\frac {-15-12 x}{10+4 x+(5+4 x) \log (8)})}{50+60 x+16 x^2+(25+40 x+16 x^2) \log (8)} \, dx\) [9611]

Optimal result

Integrand size = 76, antiderivative size = 22 \[ \int \frac {20 x+\left (50+60 x+16 x^2+\left (25+40 x+16 x^2\right ) \log (8)\right ) \log \left (\frac {-15-12 x}{10+4 x+(5+4 x) \log (8)}\right )}{50+60 x+16 x^2+\left (25+40 x+16 x^2\right ) \log (8)} \, dx=x \log \left (\frac {3}{-1-\frac {5}{5+4 x}-\log (8)}\right ) \]

[Out]

ln(3/(-1-5/(5+4*x)-3*ln(2)))*x

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(98\) vs. \(2(22)=44\).

Time = 0.10 (sec) , antiderivative size = 98, normalized size of antiderivative = 4.45, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {6820, 78, 2536, 31} \[ \int \frac {20 x+\left (50+60 x+16 x^2+\left (25+40 x+16 x^2\right ) \log (8)\right ) \log \left (\frac {-15-12 x}{10+4 x+(5+4 x) \log (8)}\right )}{50+60 x+16 x^2+\left (25+40 x+16 x^2\right ) \log (8)} \, dx=-\frac {5}{4} \log (4 x+5)+\frac {1}{4} (4 x+5) \log \left (-\frac {3 (4 x+5)}{4 x (1+\log (8))+5 (2+\log (8))}\right )+\frac {5 (2+\log (8)) \log (4 x (1+\log (8))+5 (2+\log (8)))}{4 (1+\log (8))}-\frac {5 \log (4 x (1+\log (8))+5 (2+\log (8)))}{4 (1+\log (8))} \]

[In]

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 + 16*x^2)*Log[8]),x]

[Out]

(-5*Log[5 + 4*x])/4 + ((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*(2 + Log[8])*Log[4*x*(1 + Log[8]) + 5*(2 + Log[8])])/(4*(1 +
Log[8]))

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 2536

Int[((A_.) + Log[(e_.)*((a_.) + (b_.)*(x_))^(n_.)*((c_.) + (d_.)*(x_))^(mn_)]*(B_.))^(p_.), x_Symbol] :> Simp[
(a + b*x)*((A + B*Log[e*((a + b*x)^n/(c + d*x)^n)])^p/b), x] - Dist[B*n*p*((b*c - a*d)/b), Int[(A + B*Log[e*((
a + b*x)^n/(c + d*x)^n)])^(p - 1)/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, A, B, n}, x] && EqQ[n + mn, 0] &&
NeQ[b*c - a*d, 0] && IGtQ[p, 0]

Rule 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {20 x}{(5+4 x) (4 x (1+\log (8))+5 (2+\log (8)))}+\log \left (-\frac {3 (5+4 x)}{4 x (1+\log (8))+5 (2+\log (8))}\right )\right ) \, dx \\ & = 20 \int \frac {x}{(5+4 x) (4 x (1+\log (8))+5 (2+\log (8)))} \, dx+\int \log \left (-\frac {3 (5+4 x)}{4 x (1+\log (8))+5 (2+\log (8))}\right ) \, dx \\ & = \frac {1}{4} (5+4 x) \log \left (-\frac {3 (5+4 x)}{4 x (1+\log (8))+5 (2+\log (8))}\right )-5 \int \frac {1}{4 x (1+\log (8))+5 (2+\log (8))} \, dx+20 \int \left (-\frac {1}{4 (5+4 x)}+\frac {2+\log (8)}{4 (4 x (1+\log (8))+5 (2+\log (8)))}\right ) \, dx \\ & = -\frac {5}{4} \log (5+4 x)+\frac {1}{4} (5+4 x) \log \left (-\frac {3 (5+4 x)}{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))}+\frac {5 (2+\log (8)) \log (4 x (1+\log (8))+5 (2+\log (8)))}{4 (1+\log (8))} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(60\) vs. \(2(22)=44\).

Time = 0.05 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.73 \[ \int \frac {20 x+\left (50+60 x+16 x^2+\left (25+40 x+16 x^2\right ) \log (8)\right ) \log \left (\frac {-15-12 x}{10+4 x+(5+4 x) \log (8)}\right )}{50+60 x+16 x^2+\left (25+40 x+16 x^2\right ) \log (8)} \, dx=\frac {1}{4} \left (-5 \log (5+4 x)+(5+4 x) \log \left (-\frac {3 (5+4 x)}{4 x (1+\log (8))+5 (2+\log (8))}\right )+5 \log (4 x (1+\log (8))+5 (2+\log (8)))\right ) \]

[In]

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)*Lo
g[8])])/(50 + 60*x + 16*x^2 + (25 + 40*x + 16*x^2)*Log[8]),x]

[Out]

(-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

Maple [A] (verified)

Time = 2.08 (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	\(-\frac {-5625 \ln \left (2\right )^{2} x \ln \left (-\frac {3 \left (5+4 x \right )}{12 x \ln \left (2\right )+15 \ln \left (2\right )+4 x +10}\right )-7500 \ln \left (2\right ) \ln \left (-\frac {3 \left (5+4 x \right )}{12 x \ln \left (2\right )+15 \ln \left (2\right )+4 x +10}\right ) x -2500 \ln \left (-\frac {3 \left (5+4 x \right )}{12 x \ln \left (2\right )+15 \ln \left (2\right )+4 x +10}\right ) x}{625 \left (3 \ln \left (2\right )+2\right )^{2}}\)	\(99\)
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	\(-\frac {15 \left (\left (9 \ln \left (2\right )^{2}+6 \ln \left (2\right )+1\right ) \left (-\frac {\ln \left (3+\left (3 \ln \left (2\right )+1\right ) \left (-\frac {3}{3 \ln \left (2\right )+1}+\frac {15}{\left (3 \ln \left (2\right )+1\right ) \left (12 x \ln \left (2\right )+15 \ln \left (2\right )+4 x +10\right )}\right )\right )}{3 \left (3 \ln \left (2\right )+1\right )}+\frac {\ln \left (-\frac {3}{3 \ln \left (2\right )+1}+\frac {15}{\left (3 \ln \left (2\right )+1\right ) \left (12 x \ln \left (2\right )+15 \ln \left (2\right )+4 x +10\right )}\right ) \left (-\frac {3}{3 \ln \left (2\right )+1}+\frac {15}{\left (3 \ln \left (2\right )+1\right ) \left (12 x \ln \left (2\right )+15 \ln \left (2\right )+4 x +10\right )}\right )}{9 \left (-\frac {3}{3 \ln \left (2\right )+1}+\frac {15}{\left (3 \ln \left (2\right )+1\right ) \left (12 x \ln \left (2\right )+15 \ln \left (2\right )+4 x +10\right )}\right ) \ln \left (2\right )+\frac {45}{\left (3 \ln \left (2\right )+1\right ) \left (12 x \ln \left (2\right )+15 \ln \left (2\right )+4 x +10\right )}-\frac {9}{3 \ln \left (2\right )+1}+9}\right )+\frac {\left (9 \ln \left (2\right )^{2}+6 \ln \left (2\right )+1\right ) \left (\frac {\ln \left (3 \left (-\frac {3}{3 \ln \left (2\right )+1}+\frac {15}{\left (3 \ln \left (2\right )+1\right ) \left (12 x \ln \left (2\right )+15 \ln \left (2\right )+4 x +10\right )}\right ) \ln \left (2\right )+\frac {15}{\left (3 \ln \left (2\right )+1\right ) \left (12 x \ln \left (2\right )+15 \ln \left (2\right )+4 x +10\right )}-\frac {3}{3 \ln \left (2\right )+1}+3\right )}{3 \ln \left (2\right )+1}+\ln \left (-\frac {3}{3 \ln \left (2\right )+1}+\frac {15}{\left (3 \ln \left (2\right )+1\right ) \left (12 x \ln \left (2\right )+15 \ln \left (2\right )+4 x +10\right )}\right )\right )}{3}\right )}{4 \left (3 \ln \left (2\right )+1\right )^{2}}\)	\(383\)
default	\(-\frac {15 \left (\left (9 \ln \left (2\right )^{2}+6 \ln \left (2\right )+1\right ) \left (-\frac {\ln \left (3+\left (3 \ln \left (2\right )+1\right ) \left (-\frac {3}{3 \ln \left (2\right )+1}+\frac {15}{\left (3 \ln \left (2\right )+1\right ) \left (12 x \ln \left (2\right )+15 \ln \left (2\right )+4 x +10\right )}\right )\right )}{3 \left (3 \ln \left (2\right )+1\right )}+\frac {\ln \left (-\frac {3}{3 \ln \left (2\right )+1}+\frac {15}{\left (3 \ln \left (2\right )+1\right ) \left (12 x \ln \left (2\right )+15 \ln \left (2\right )+4 x +10\right )}\right ) \left (-\frac {3}{3 \ln \left (2\right )+1}+\frac {15}{\left (3 \ln \left (2\right )+1\right ) \left (12 x \ln \left (2\right )+15 \ln \left (2\right )+4 x +10\right )}\right )}{9 \left (-\frac {3}{3 \ln \left (2\right )+1}+\frac {15}{\left (3 \ln \left (2\right )+1\right ) \left (12 x \ln \left (2\right )+15 \ln \left (2\right )+4 x +10\right )}\right ) \ln \left (2\right )+\frac {45}{\left (3 \ln \left (2\right )+1\right ) \left (12 x \ln \left (2\right )+15 \ln \left (2\right )+4 x +10\right )}-\frac {9}{3 \ln \left (2\right )+1}+9}\right )+\frac {\left (9 \ln \left (2\right )^{2}+6 \ln \left (2\right )+1\right ) \left (\frac {\ln \left (3 \left (-\frac {3}{3 \ln \left (2\right )+1}+\frac {15}{\left (3 \ln \left (2\right )+1\right ) \left (12 x \ln \left (2\right )+15 \ln \left (2\right )+4 x +10\right )}\right ) \ln \left (2\right )+\frac {15}{\left (3 \ln \left (2\right )+1\right ) \left (12 x \ln \left (2\right )+15 \ln \left (2\right )+4 x +10\right )}-\frac {3}{3 \ln \left (2\right )+1}+3\right )}{3 \ln \left (2\right )+1}+\ln \left (-\frac {3}{3 \ln \left (2\right )+1}+\frac {15}{\left (3 \ln \left (2\right )+1\right ) \left (12 x \ln \left (2\right )+15 \ln \left (2\right )+4 x +10\right )}\right )\right )}{3}\right )}{4 \left (3 \ln \left (2\right )+1\right )^{2}}\)	\(383\)

[In]

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+2
5)*ln(2)+16*x^2+60*x+50),x,method=_RETURNVERBOSE)

[Out]

x*ln((-12*x-15)/(3*(5+4*x)*ln(2)+4*x+10))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {20 x+\left (50+60 x+16 x^2+\left (25+40 x+16 x^2\right ) \log (8)\right ) \log \left (\frac {-15-12 x}{10+4 x+(5+4 x) \log (8)}\right )}{50+60 x+16 x^2+\left (25+40 x+16 x^2\right ) \log (8)} \, dx=x \log \left (-\frac {3 \, {\left (4 \, x + 5\right )}}{3 \, {\left (4 \, x + 5\right )} \log \left (2\right ) + 4 \, x + 10}\right ) \]

[In]

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")

[Out]

x*log(-3*(4*x + 5)/(3*(4*x + 5)*log(2) + 4*x + 10))

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {20 x+\left (50+60 x+16 x^2+\left (25+40 x+16 x^2\right ) \log (8)\right ) \log \left (\frac {-15-12 x}{10+4 x+(5+4 x) \log (8)}\right )}{50+60 x+16 x^2+\left (25+40 x+16 x^2\right ) \log (8)} \, dx=x \log {\left (\frac {- 12 x - 15}{4 x + \left (12 x + 15\right ) \log {\left (2 \right )} + 10} \right )} \]

[In]

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)

[Out]

x*log((-12*x - 15)/(4*x + (12*x + 15)*log(2) + 10))

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.37 (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=-\frac {15}{4} \, {\left (\log \left (4 \, x {\left (3 \, \log \left (2\right ) + 1\right )} + 15 \, \log \left (2\right ) + 10\right ) - \log \left (4 \, x + 5\right )\right )} \log \left (2\right ) \log \left (-\frac {12 \, x}{12 \, x \log \left (2\right ) + 4 \, x + 15 \, \log \left (2\right ) + 10} - \frac {15}{12 \, x \log \left (2\right ) + 4 \, x + 15 \, \log \left (2\right ) + 10}\right ) - \frac {15}{8} \, {\left (\log \left (4 \, x {\left (3 \, \log \left (2\right ) + 1\right )} + 15 \, \log \left (2\right ) + 10\right )^{2} - 2 \, \log \left (4 \, x {\left (3 \, \log \left (2\right ) + 1\right )} + 15 \, \log \left (2\right ) + 10\right ) \log \left (4 \, x + 5\right ) + \log \left (4 \, x + 5\right )^{2}\right )} \log \left (2\right ) - \frac {5}{4} \, \log \left (4 \, x {\left (3 \, \log \left (2\right ) + 1\right )} + 15 \, \log \left (2\right ) + 10\right )^{2} + \frac {5}{2} \, \log \left (4 \, x {\left (3 \, \log \left (2\right ) + 1\right )} + 15 \, \log \left (2\right ) + 10\right ) \log \left (4 \, x + 5\right ) - \frac {5}{4} \, \log \left (4 \, x + 5\right )^{2} - \frac {5}{2} \, {\left (\log \left (4 \, x {\left (3 \, \log \left (2\right ) + 1\right )} + 15 \, \log \left (2\right ) + 10\right ) - \log \left (4 \, x + 5\right )\right )} \log \left (-\frac {12 \, x}{12 \, x \log \left (2\right ) + 4 \, x + 15 \, \log \left (2\right ) + 10} - \frac {15}{12 \, x \log \left (2\right ) + 4 \, x + 15 \, \log \left (2\right ) + 10}\right ) + \frac {5 \, {\left (3 \, \log \left (2\right ) + 2\right )} \log \left (4 \, x {\left (3 \, \log \left (2\right ) + 1\right )} + 15 \, \log \left (2\right ) + 10\right )}{4 \, {\left (3 \, \log \left (2\right ) + 1\right )}} - \frac {5 \, {\left (9 \, \log \left (2\right )^{2} + 9 \, \log \left (2\right ) + 2\right )} \log \left (4 \, x {\left (3 \, \log \left (2\right ) + 1\right )} + 15 \, \log \left (2\right ) + 10\right )^{2} + 5 \, {\left (9 \, \log \left (2\right )^{2} + 9 \, \log \left (2\right ) + 2\right )} \log \left (4 \, x + 5\right )^{2} - 8 \, {\left (i \, \pi {\left (3 \, \log \left (2\right ) + 1\right )} + 3 \, \log \left (3\right ) \log \left (2\right ) + \log \left (3\right )\right )} x - 2 \, {\left (45 \, \log \left (3\right ) \log \left (2\right )^{2} + 5 i \, \pi {\left (9 \, \log \left (2\right )^{2} + 9 \, \log \left (2\right ) + 2\right )} - 4 \, x {\left (3 \, \log \left (2\right ) + 1\right )} + 15 \, {\left (3 \, \log \left (3\right ) - 1\right )} \log \left (2\right ) + 5 \, {\left (9 \, \log \left (2\right )^{2} + 9 \, \log \left (2\right ) + 2\right )} \log \left (4 \, x + 5\right ) + 10 \, \log \left (3\right ) - 10\right )} \log \left (4 \, x {\left (3 \, \log \left (2\right ) + 1\right )} + 15 \, \log \left (2\right ) + 10\right ) + 2 \, {\left (45 \, \log \left (3\right ) \log \left (2\right )^{2} + 5 i \, \pi {\left (9 \, \log \left (2\right )^{2} + 9 \, \log \left (2\right ) + 2\right )} - 4 \, x {\left (3 \, \log \left (2\right ) + 1\right )} + 15 \, {\left (3 \, \log \left (3\right ) - 1\right )} \log \left (2\right ) + 10 \, \log \left (3\right ) - 5\right )} \log \left (4 \, x + 5\right )}{8 \, {\left (3 \, \log \left (2\right ) + 1\right )}} - \frac {5}{4} \, \log \left (4 \, x + 5\right ) \]

[In]

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")

[Out]

-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*log(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*l
og(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*log(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)*lo
g(4*x + 5)^2 - 8*(I*pi*(3*log(2) + 1) + 3*log(3)*log(2) + log(3))*x - 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) + 5*(9*log(2)^2 + 9*log(2) + 2)*log(4*x + 5
) + 10*log(3) - 10)*log(4*x*(3*log(2) + 1) + 15*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)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 149 vs. \(2 (22) = 44\).

Time = 0.37 (sec) , antiderivative size = 149, normalized size of antiderivative = 6.77 \[ \int \frac {20 x+\left (50+60 x+16 x^2+\left (25+40 x+16 x^2\right ) \log (8)\right ) \log \left (\frac {-15-12 x}{10+4 x+(5+4 x) \log (8)}\right )}{50+60 x+16 x^2+\left (25+40 x+16 x^2\right ) \log (8)} \, dx=-\frac {5 \, {\left (3 \, \log \left (2\right ) + 2\right )} \log \left (-\frac {3 \, {\left (4 \, x + 5\right )}}{12 \, x \log \left (2\right ) + 4 \, x + 15 \, \log \left (2\right ) + 10}\right )}{4 \, {\left (3 \, \log \left (2\right ) + 1\right )}} - \frac {5 \, \log \left (-\frac {3 \, {\left (4 \, x + 5\right )}}{12 \, x \log \left (2\right ) + 4 \, x + 15 \, \log \left (2\right ) + 10}\right )}{4 \, {\left (\frac {9 \, {\left (4 \, x + 5\right )} \log \left (2\right )^{2}}{12 \, x \log \left (2\right ) + 4 \, x + 15 \, \log \left (2\right ) + 10} + \frac {6 \, {\left (4 \, x + 5\right )} \log \left (2\right )}{12 \, x \log \left (2\right ) + 4 \, x + 15 \, \log \left (2\right ) + 10} + \frac {4 \, x + 5}{12 \, x \log \left (2\right ) + 4 \, x + 15 \, \log \left (2\right ) + 10} - 3 \, \log \left (2\right ) - 1\right )}} \]

[In]

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")

[Out]

-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)

Mupad [B] (verification not implemented)

Time = 1.33 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {20 x+\left (50+60 x+16 x^2+\left (25+40 x+16 x^2\right ) \log (8)\right ) \log \left (\frac {-15-12 x}{10+4 x+(5+4 x) \log (8)}\right )}{50+60 x+16 x^2+\left (25+40 x+16 x^2\right ) \log (8)} \, dx=x\,\ln \left (-\frac {12\,x+15}{4\,x+3\,\ln \left (2\right )\,\left (4\,x+5\right )+10}\right ) \]

[In]

int((20*x + log(-(12*x + 15)/(4*x + 3*log(2)*(4*x + 5) + 10))*(60*x + 3*log(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)

[Out]

x*log(-(12*x + 15)/(4*x + 3*log(2)*(4*x + 5) + 10))