3.28.25 \(\int \frac {-20 x-94 x^2-110 x^3+(-3 x^2-3 \log (2)) \log (x^2+\log (2))+(-20 x^2-100 x^3-125 x^4+(-20-100 x-125 x^2) \log (2)) \log ^2(x^2+\log (2))}{(20 x^2+100 x^3+125 x^4+(20+100 x+125 x^2) \log (2)) \log ^2(x^2+\log (2))} \, dx\) [2725]

3.28.25.1 Optimal result

Integrand size = 112, antiderivative size = 34 \[ \int \frac {-20 x-94 x^2-110 x^3+\left (-3 x^2-3 \log (2)\right ) \log \left (x^2+\log (2)\right )+\left (-20 x^2-100 x^3-125 x^4+\left (-20-100 x-125 x^2\right ) \log (2)\right ) \log ^2\left (x^2+\log (2)\right )}{\left (20 x^2+100 x^3+125 x^4+\left (20+100 x+125 x^2\right ) \log (2)\right ) \log ^2\left (x^2+\log (2)\right )} \, dx=-x+\frac {1+\frac {3}{2+\frac {x}{1+2 x}}}{5 \log \left (x^2+\log (2)\right )} \]

output

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

3.28.25.2 Mathematica [A] (verified)

Time = 0.60 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int \frac {-20 x-94 x^2-110 x^3+\left (-3 x^2-3 \log (2)\right ) \log \left (x^2+\log (2)\right )+\left (-20 x^2-100 x^3-125 x^4+\left (-20-100 x-125 x^2\right ) \log (2)\right ) \log ^2\left (x^2+\log (2)\right )}{\left (20 x^2+100 x^3+125 x^4+\left (20+100 x+125 x^2\right ) \log (2)\right ) \log ^2\left (x^2+\log (2)\right )} \, dx=-x-\frac {-5-11 x}{5 (2+5 x) \log \left (x^2+\log (2)\right )} \]

input

Integrate[(-20*x - 94*x^2 - 110*x^3 + (-3*x^2 - 3*Log[2])*Log[x^2 + Log[2] 
] + (-20*x^2 - 100*x^3 - 125*x^4 + (-20 - 100*x - 125*x^2)*Log[2])*Log[x^2 
 + Log[2]]^2)/((20*x^2 + 100*x^3 + 125*x^4 + (20 + 100*x + 125*x^2)*Log[2] 
)*Log[x^2 + Log[2]]^2),x]
 

output

-x - (-5 - 11*x)/(5*(2 + 5*x)*Log[x^2 + Log[2]])
 

3.28.25.3 Rubi [F]

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.

\(\displaystyle \int \frac {-110 x^3-94 x^2+\left (-3 x^2-3 \log (2)\right ) \log \left (x^2+\log (2)\right )+\left (-125 x^4-100 x^3-20 x^2+\left (-125 x^2-100 x-20\right ) \log (2)\right ) \log ^2\left (x^2+\log (2)\right )-20 x}{\left (125 x^4+100 x^3+20 x^2+\left (125 x^2+100 x+20\right ) \log (2)\right ) \log ^2\left (x^2+\log (2)\right )} \, dx\)

\(\Big \downarrow \) 2463

\(\displaystyle \int \left (\frac {5 \left (-110 x^3-94 x^2+\left (-3 x^2-3 \log (2)\right ) \log \left (x^2+\log (2)\right )+\left (-125 x^4-100 x^3-20 x^2+\left (-125 x^2-100 x-20\right ) \log (2)\right ) \log ^2\left (x^2+\log (2)\right )-20 x\right )}{(5 x+2)^2 (4+25 \log (2)) \log ^2\left (x^2+\log (2)\right )}+\frac {20 \left (-110 x^3-94 x^2+\left (-3 x^2-3 \log (2)\right ) \log \left (x^2+\log (2)\right )+\left (-125 x^4-100 x^3-20 x^2+\left (-125 x^2-100 x-20\right ) \log (2)\right ) \log ^2\left (x^2+\log (2)\right )-20 x\right )}{(5 x+2) (4+25 \log (2))^2 \log ^2\left (x^2+\log (2)\right )}+\frac {(-20 x+4-25 \log (2)) \left (-110 x^3-94 x^2+\left (-3 x^2-3 \log (2)\right ) \log \left (x^2+\log (2)\right )+\left (-125 x^4-100 x^3-20 x^2+\left (-125 x^2-100 x-20\right ) \log (2)\right ) \log ^2\left (x^2+\log (2)\right )-20 x\right )}{5 (4+25 \log (2))^2 \left (x^2+\log (2)\right ) \log ^2\left (x^2+\log (2)\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {12 \int \frac {1}{(5 x+2) \log ^2\left (x^2+\log (2)\right )}dx}{5 (4+25 \log (2))}-\frac {2 \log (8) \int \frac {1}{\left (x^2+\log (2)\right ) \log ^2\left (x^2+\log (2)\right )}dx}{5 (4+25 \log (2))}-\frac {3 \int \frac {1}{\log \left (x^2+\log (2)\right )}dx}{5 (4+25 \log (2))}+\frac {24 \int \frac {1}{\log \left (x^2+\log (2)\right )}dx}{5 (4+25 \log (2))^2}-\frac {3}{5} \int \frac {1}{(5 x+2)^2 \log \left (x^2+\log (2)\right )}dx+\frac {3 \int \frac {20 x+25 \log (2)-4}{\log \left (x^2+\log (2)\right )}dx}{5 (4+25 \log (2))^2}+\frac {(144+275 \log (2)) \operatorname {LogIntegral}\left (x^2+\log (2)\right )}{(4+25 \log (2))^2}-\frac {11 \operatorname {LogIntegral}\left (x^2+\log (2)\right )}{4+25 \log (2)}-\frac {106 \operatorname {LogIntegral}\left (x^2+\log (2)\right )}{(4+25 \log (2))^2}-\frac {25 x^3}{3 (4+25 \log (2))}-\frac {200 x^3}{3 (4+25 \log (2))^2}+\frac {2+\log (2048)}{(4+25 \log (2)) \log \left (x^2+\log (2)\right )}-\frac {(144+275 \log (2)) \left (x^2+\log (2)\right )}{(4+25 \log (2))^2 \log \left (x^2+\log (2)\right )}+\frac {11 \left (x^2+\log (2)\right )}{(4+25 \log (2)) \log \left (x^2+\log (2)\right )}+\frac {100 \left (x^2+\log (2)\right )}{(4+25 \log (2))^2 \log \left (x^2+\log (2)\right )}-\frac {125 \left (x^2+\log (2)\right )^2}{(4+25 \log (2))^2}+\frac {(5 x+2)^4}{5 (4+25 \log (2))^2}-\frac {(5 x+2)^3 (12-25 \log (2))}{15 (4+25 \log (2))^2}-\frac {25 x \log (2)}{4+25 \log (2)}-\frac {200 x \log (2)}{(4+25 \log (2))^2}\)

input

Int[(-20*x - 94*x^2 - 110*x^3 + (-3*x^2 - 3*Log[2])*Log[x^2 + Log[2]] + (- 
20*x^2 - 100*x^3 - 125*x^4 + (-20 - 100*x - 125*x^2)*Log[2])*Log[x^2 + Log 
[2]]^2)/((20*x^2 + 100*x^3 + 125*x^4 + (20 + 100*x + 125*x^2)*Log[2])*Log[ 
x^2 + Log[2]]^2),x]
 

output

$Aborted
 

3.28.25.3.1 Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2463

Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px]}, Int[ExpandIntegr 
and[u, Qx^p, x], x] /;  !SumQ[NonfreeFactors[Qx, x]]] /; PolyQ[Px, x] && Gt 
Q[Expon[Px, x], 2] &&  !BinomialQ[Px, x] &&  !TrinomialQ[Px, x] && ILtQ[p, 
0]
 

3.28.25.4 Maple [A] (verified)

Time = 7.71 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.82

method	result	size
risch	\(-x +\frac {11 x +5}{5 \left (2+5 x \right ) \ln \left (\ln \left (2\right )+x^{2}\right )}\)	\(28\)
norman	\(\frac {1+\frac {4 \ln \left (\ln \left (2\right )+x^{2}\right )}{5}+\frac {11 x}{5}-5 \ln \left (\ln \left (2\right )+x^{2}\right ) x^{2}}{\left (2+5 x \right ) \ln \left (\ln \left (2\right )+x^{2}\right )}\)	\(44\)
parallelrisch	\(\frac {-125 \ln \left (\ln \left (2\right )+x^{2}\right ) x^{2}+25+55 x +20 \ln \left (\ln \left (2\right )+x^{2}\right )}{25 \ln \left (\ln \left (2\right )+x^{2}\right ) \left (2+5 x \right )}\)	\(45\)

input

int((((-125*x^2-100*x-20)*ln(2)-125*x^4-100*x^3-20*x^2)*ln(ln(2)+x^2)^2+(- 
3*ln(2)-3*x^2)*ln(ln(2)+x^2)-110*x^3-94*x^2-20*x)/((125*x^2+100*x+20)*ln(2 
)+125*x^4+100*x^3+20*x^2)/ln(ln(2)+x^2)^2,x,method=_RETURNVERBOSE)
 

output

-x+1/5*(11*x+5)/(2+5*x)/ln(ln(2)+x^2)
 

3.28.25.5 Fricas [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.21 \[ \int \frac {-20 x-94 x^2-110 x^3+\left (-3 x^2-3 \log (2)\right ) \log \left (x^2+\log (2)\right )+\left (-20 x^2-100 x^3-125 x^4+\left (-20-100 x-125 x^2\right ) \log (2)\right ) \log ^2\left (x^2+\log (2)\right )}{\left (20 x^2+100 x^3+125 x^4+\left (20+100 x+125 x^2\right ) \log (2)\right ) \log ^2\left (x^2+\log (2)\right )} \, dx=-\frac {5 \, {\left (5 \, x^{2} + 2 \, x\right )} \log \left (x^{2} + \log \left (2\right )\right ) - 11 \, x - 5}{5 \, {\left (5 \, x + 2\right )} \log \left (x^{2} + \log \left (2\right )\right )} \]

input

integrate((((-125*x^2-100*x-20)*log(2)-125*x^4-100*x^3-20*x^2)*log(log(2)+ 
x^2)^2+(-3*log(2)-3*x^2)*log(log(2)+x^2)-110*x^3-94*x^2-20*x)/((125*x^2+10 
0*x+20)*log(2)+125*x^4+100*x^3+20*x^2)/log(log(2)+x^2)^2,x, algorithm=\
 

output

-1/5*(5*(5*x^2 + 2*x)*log(x^2 + log(2)) - 11*x - 5)/((5*x + 2)*log(x^2 + l 
og(2)))
 

3.28.25.6 Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.56 \[ \int \frac {-20 x-94 x^2-110 x^3+\left (-3 x^2-3 \log (2)\right ) \log \left (x^2+\log (2)\right )+\left (-20 x^2-100 x^3-125 x^4+\left (-20-100 x-125 x^2\right ) \log (2)\right ) \log ^2\left (x^2+\log (2)\right )}{\left (20 x^2+100 x^3+125 x^4+\left (20+100 x+125 x^2\right ) \log (2)\right ) \log ^2\left (x^2+\log (2)\right )} \, dx=- x + \frac {11 x + 5}{\left (25 x + 10\right ) \log {\left (x^{2} + \log {\left (2 \right )} \right )}} \]

input

integrate((((-125*x**2-100*x-20)*ln(2)-125*x**4-100*x**3-20*x**2)*ln(ln(2) 
+x**2)**2+(-3*ln(2)-3*x**2)*ln(ln(2)+x**2)-110*x**3-94*x**2-20*x)/((125*x* 
*2+100*x+20)*ln(2)+125*x**4+100*x**3+20*x**2)/ln(ln(2)+x**2)**2,x)
 

output

-x + (11*x + 5)/((25*x + 10)*log(x**2 + log(2)))
 

3.28.25.7 Maxima [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.21 \[ \int \frac {-20 x-94 x^2-110 x^3+\left (-3 x^2-3 \log (2)\right ) \log \left (x^2+\log (2)\right )+\left (-20 x^2-100 x^3-125 x^4+\left (-20-100 x-125 x^2\right ) \log (2)\right ) \log ^2\left (x^2+\log (2)\right )}{\left (20 x^2+100 x^3+125 x^4+\left (20+100 x+125 x^2\right ) \log (2)\right ) \log ^2\left (x^2+\log (2)\right )} \, dx=-\frac {5 \, {\left (5 \, x^{2} + 2 \, x\right )} \log \left (x^{2} + \log \left (2\right )\right ) - 11 \, x - 5}{5 \, {\left (5 \, x + 2\right )} \log \left (x^{2} + \log \left (2\right )\right )} \]

input

integrate((((-125*x^2-100*x-20)*log(2)-125*x^4-100*x^3-20*x^2)*log(log(2)+ 
x^2)^2+(-3*log(2)-3*x^2)*log(log(2)+x^2)-110*x^3-94*x^2-20*x)/((125*x^2+10 
0*x+20)*log(2)+125*x^4+100*x^3+20*x^2)/log(log(2)+x^2)^2,x, algorithm=\
 

output

-1/5*(5*(5*x^2 + 2*x)*log(x^2 + log(2)) - 11*x - 5)/((5*x + 2)*log(x^2 + l 
og(2)))
 

3.28.25.8 Giac [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.97 \[ \int \frac {-20 x-94 x^2-110 x^3+\left (-3 x^2-3 \log (2)\right ) \log \left (x^2+\log (2)\right )+\left (-20 x^2-100 x^3-125 x^4+\left (-20-100 x-125 x^2\right ) \log (2)\right ) \log ^2\left (x^2+\log (2)\right )}{\left (20 x^2+100 x^3+125 x^4+\left (20+100 x+125 x^2\right ) \log (2)\right ) \log ^2\left (x^2+\log (2)\right )} \, dx=-x + \frac {11 \, x + 5}{5 \, {\left (5 \, x \log \left (x^{2} + \log \left (2\right )\right ) + 2 \, \log \left (x^{2} + \log \left (2\right )\right )\right )}} \]

input

integrate((((-125*x^2-100*x-20)*log(2)-125*x^4-100*x^3-20*x^2)*log(log(2)+ 
x^2)^2+(-3*log(2)-3*x^2)*log(log(2)+x^2)-110*x^3-94*x^2-20*x)/((125*x^2+10 
0*x+20)*log(2)+125*x^4+100*x^3+20*x^2)/log(log(2)+x^2)^2,x, algorithm=\
 

output

-x + 1/5*(11*x + 5)/(5*x*log(x^2 + log(2)) + 2*log(x^2 + log(2)))
 

3.28.25.9 Mupad [B] (verification not implemented)

Time = 0.66 (sec) , antiderivative size = 88, normalized size of antiderivative = 2.59 \[ \int \frac {-20 x-94 x^2-110 x^3+\left (-3 x^2-3 \log (2)\right ) \log \left (x^2+\log (2)\right )+\left (-20 x^2-100 x^3-125 x^4+\left (-20-100 x-125 x^2\right ) \log (2)\right ) \log ^2\left (x^2+\log (2)\right )}{\left (20 x^2+100 x^3+125 x^4+\left (20+100 x+125 x^2\right ) \log (2)\right ) \log ^2\left (x^2+\log (2)\right )} \, dx=\frac {\frac {11\,x+5}{5\,\left (5\,x+2\right )}+\frac {\ln \left (x^2+\ln \left (2\right )\right )\,\left (3\,x^2+\ln \left (8\right )\right )}{10\,x\,\left (25\,x^2+20\,x+4\right )}}{\ln \left (x^2+\ln \left (2\right )\right )}-x-\frac {3\,x^2+3\,\ln \left (2\right )}{250\,x^3+200\,x^2+40\,x} \]

input

int(-(20*x + log(log(2) + x^2)^2*(log(2)*(100*x + 125*x^2 + 20) + 20*x^2 + 
 100*x^3 + 125*x^4) + log(log(2) + x^2)*(3*log(2) + 3*x^2) + 94*x^2 + 110* 
x^3)/(log(log(2) + x^2)^2*(log(2)*(100*x + 125*x^2 + 20) + 20*x^2 + 100*x^ 
3 + 125*x^4)),x)
 

output

((11*x + 5)/(5*(5*x + 2)) + (log(log(2) + x^2)*(log(8) + 3*x^2))/(10*x*(20 
*x + 25*x^2 + 4)))/log(log(2) + x^2) - x - (3*log(2) + 3*x^2)/(40*x + 200* 
x^2 + 250*x^3)