\(\int \frac {e^{\frac {1}{2} (-10+\log (15-12 x \log (2+\log ^2(4+x^2))))} (8 x^2 \log (4+x^2)+(16+4 x^2+(8+2 x^2) \log ^2(4+x^2)) \log (2+\log ^2(4+x^2)))}{-40-10 x^2+(-20-5 x^2) \log ^2(4+x^2)+(32 x+8 x^3+(16 x+4 x^3) \log ^2(4+x^2)) \log (2+\log ^2(4+x^2))} \, dx\) [878]

Optimal result

Integrand size = 137, antiderivative size = 29 \[ \int \frac {e^{\frac {1}{2} \left (-10+\log \left (15-12 x \log \left (2+\log ^2\left (4+x^2\right )\right )\right )\right )} \left (8 x^2 \log \left (4+x^2\right )+\left (16+4 x^2+\left (8+2 x^2\right ) \log ^2\left (4+x^2\right )\right ) \log \left (2+\log ^2\left (4+x^2\right )\right )\right )}{-40-10 x^2+\left (-20-5 x^2\right ) \log ^2\left (4+x^2\right )+\left (32 x+8 x^3+\left (16 x+4 x^3\right ) \log ^2\left (4+x^2\right )\right ) \log \left (2+\log ^2\left (4+x^2\right )\right )} \, dx=\frac {\sqrt {3} \sqrt {5-4 x \log \left (2+\log ^2\left (4+x^2\right )\right )}}{e^5} \] Output:

exp(1/2*ln(-12*x*ln(ln(x^2+4)^2+2)+15)-5)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {1}{2} \left (-10+\log \left (15-12 x \log \left (2+\log ^2\left (4+x^2\right )\right )\right )\right )} \left (8 x^2 \log \left (4+x^2\right )+\left (16+4 x^2+\left (8+2 x^2\right ) \log ^2\left (4+x^2\right )\right ) \log \left (2+\log ^2\left (4+x^2\right )\right )\right )}{-40-10 x^2+\left (-20-5 x^2\right ) \log ^2\left (4+x^2\right )+\left (32 x+8 x^3+\left (16 x+4 x^3\right ) \log ^2\left (4+x^2\right )\right ) \log \left (2+\log ^2\left (4+x^2\right )\right )} \, dx=\frac {\sqrt {3} \sqrt {5-4 x \log \left (2+\log ^2\left (4+x^2\right )\right )}}{e^5} \] Input:

Integrate[(E^((-10 + Log[15 - 12*x*Log[2 + Log[4 + x^2]^2]])/2)*(8*x^2*Log 
[4 + x^2] + (16 + 4*x^2 + (8 + 2*x^2)*Log[4 + x^2]^2)*Log[2 + Log[4 + x^2] 
^2]))/(-40 - 10*x^2 + (-20 - 5*x^2)*Log[4 + x^2]^2 + (32*x + 8*x^3 + (16*x 
 + 4*x^3)*Log[4 + x^2]^2)*Log[2 + Log[4 + x^2]^2]),x]
 

Output:

(Sqrt[3]*Sqrt[5 - 4*x*Log[2 + Log[4 + x^2]^2]])/E^5
 

Rubi [A] (verified)

Time = 1.14 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {2704, 27, 27, 7292, 7237}

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.

\(\displaystyle \int \frac {\left (\left (4 x^2+\left (2 x^2+8\right ) \log ^2\left (x^2+4\right )+16\right ) \log \left (\log ^2\left (x^2+4\right )+2\right )+8 x^2 \log \left (x^2+4\right )\right ) \exp \left (\frac {1}{2} \left (\log \left (15-12 x \log \left (\log ^2\left (x^2+4\right )+2\right )\right )-10\right )\right )}{-10 x^2+\left (-5 x^2-20\right ) \log ^2\left (x^2+4\right )+\left (8 x^3+\left (4 x^3+16 x\right ) \log ^2\left (x^2+4\right )+32 x\right ) \log \left (\log ^2\left (x^2+4\right )+2\right )-40} \, dx\)

\(\Big \downarrow \) 2704

\(\displaystyle \int \frac {\sqrt {15-12 x \log \left (\log ^2\left (x^2+4\right )+2\right )} \left (\left (4 x^2+\left (2 x^2+8\right ) \log ^2\left (x^2+4\right )+16\right ) \log \left (\log ^2\left (x^2+4\right )+2\right )+8 x^2 \log \left (x^2+4\right )\right )}{e^5 \left (-10 x^2+\left (-5 x^2-20\right ) \log ^2\left (x^2+4\right )+\left (8 x^3+\left (4 x^3+16 x\right ) \log ^2\left (x^2+4\right )+32 x\right ) \log \left (\log ^2\left (x^2+4\right )+2\right )-40\right )}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int -\frac {2 \sqrt {3} \sqrt {5-4 x \log \left (\log ^2\left (x^2+4\right )+2\right )} \left (4 \log \left (x^2+4\right ) x^2+\left (2 x^2+\left (x^2+4\right ) \log ^2\left (x^2+4\right )+8\right ) \log \left (\log ^2\left (x^2+4\right )+2\right )\right )}{10 x^2+5 \left (x^2+4\right ) \log ^2\left (x^2+4\right )-4 \left (2 x^3+8 x+\left (x^3+4 x\right ) \log ^2\left (x^2+4\right )\right ) \log \left (\log ^2\left (x^2+4\right )+2\right )+40}dx}{e^5}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {2 \sqrt {3} \int \frac {\sqrt {5-4 x \log \left (\log ^2\left (x^2+4\right )+2\right )} \left (4 \log \left (x^2+4\right ) x^2+\left (2 x^2+\left (x^2+4\right ) \log ^2\left (x^2+4\right )+8\right ) \log \left (\log ^2\left (x^2+4\right )+2\right )\right )}{10 x^2+5 \left (x^2+4\right ) \log ^2\left (x^2+4\right )-4 \left (2 x^3+8 x+\left (x^3+4 x\right ) \log ^2\left (x^2+4\right )\right ) \log \left (\log ^2\left (x^2+4\right )+2\right )+40}dx}{e^5}\)

\(\Big \downarrow \) 7292

\(\displaystyle -\frac {2 \sqrt {3} \int \frac {4 \log \left (x^2+4\right ) x^2+\left (2 x^2+\left (x^2+4\right ) \log ^2\left (x^2+4\right )+8\right ) \log \left (\log ^2\left (x^2+4\right )+2\right )}{\left (x^2+4\right ) \left (\log ^2\left (x^2+4\right )+2\right ) \sqrt {5-4 x \log \left (\log ^2\left (x^2+4\right )+2\right )}}dx}{e^5}\)

\(\Big \downarrow \) 7237

\(\displaystyle \frac {\sqrt {3} \sqrt {5-4 x \log \left (\log ^2\left (x^2+4\right )+2\right )}}{e^5}\)

Input:

Int[(E^((-10 + Log[15 - 12*x*Log[2 + Log[4 + x^2]^2]])/2)*(8*x^2*Log[4 + x 
^2] + (16 + 4*x^2 + (8 + 2*x^2)*Log[4 + x^2]^2)*Log[2 + Log[4 + x^2]^2]))/ 
(-40 - 10*x^2 + (-20 - 5*x^2)*Log[4 + x^2]^2 + (32*x + 8*x^3 + (16*x + 4*x 
^3)*Log[4 + x^2]^2)*Log[2 + Log[4 + x^2]^2]),x]
 

Output:

(Sqrt[3]*Sqrt[5 - 4*x*Log[2 + Log[4 + x^2]^2]])/E^5
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2704

Int[(u_.)*(F_)^((a_.)*(Log[z_]*(b_.) + (v_.))), x_Symbol] :> Int[u*F^(a*v)* 
z^(a*b*Log[F]), x] /; FreeQ[{F, a, b}, x]
 

rule 7237

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Si 
mp[q*(y^(m + 1)/(m + 1)), x] /;  !FalseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]
 

rule 7292

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

Maple [F]

\[\int \frac {\left (\left (\left (2 x^{2}+8\right ) \ln \left (x^{2}+4\right )^{2}+4 x^{2}+16\right ) \ln \left (\ln \left (x^{2}+4\right )^{2}+2\right )+8 x^{2} \ln \left (x^{2}+4\right )\right ) {\mathrm e}^{\frac {\ln \left (-12 x \ln \left (\ln \left (x^{2}+4\right )^{2}+2\right )+15\right )}{2}-5}}{\left (\left (4 x^{3}+16 x \right ) \ln \left (x^{2}+4\right )^{2}+8 x^{3}+32 x \right ) \ln \left (\ln \left (x^{2}+4\right )^{2}+2\right )+\left (-5 x^{2}-20\right ) \ln \left (x^{2}+4\right )^{2}-10 x^{2}-40}d x\]

Input:

int((((2*x^2+8)*ln(x^2+4)^2+4*x^2+16)*ln(ln(x^2+4)^2+2)+8*x^2*ln(x^2+4))*e 
xp(1/2*ln(-12*x*ln(ln(x^2+4)^2+2)+15)-5)/(((4*x^3+16*x)*ln(x^2+4)^2+8*x^3+ 
32*x)*ln(ln(x^2+4)^2+2)+(-5*x^2-20)*ln(x^2+4)^2-10*x^2-40),x)
 

Output:

int((((2*x^2+8)*ln(x^2+4)^2+4*x^2+16)*ln(ln(x^2+4)^2+2)+8*x^2*ln(x^2+4))*e 
xp(1/2*ln(-12*x*ln(ln(x^2+4)^2+2)+15)-5)/(((4*x^3+16*x)*ln(x^2+4)^2+8*x^3+ 
32*x)*ln(ln(x^2+4)^2+2)+(-5*x^2-20)*ln(x^2+4)^2-10*x^2-40),x)
 

Fricas [F(-2)]

Exception generated. \[ \int \frac {e^{\frac {1}{2} \left (-10+\log \left (15-12 x \log \left (2+\log ^2\left (4+x^2\right )\right )\right )\right )} \left (8 x^2 \log \left (4+x^2\right )+\left (16+4 x^2+\left (8+2 x^2\right ) \log ^2\left (4+x^2\right )\right ) \log \left (2+\log ^2\left (4+x^2\right )\right )\right )}{-40-10 x^2+\left (-20-5 x^2\right ) \log ^2\left (4+x^2\right )+\left (32 x+8 x^3+\left (16 x+4 x^3\right ) \log ^2\left (4+x^2\right )\right ) \log \left (2+\log ^2\left (4+x^2\right )\right )} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((((2*x^2+8)*log(x^2+4)^2+4*x^2+16)*log(log(x^2+4)^2+2)+8*x^2*log 
(x^2+4))*exp(1/2*log(-12*x*log(log(x^2+4)^2+2)+15)-5)/(((4*x^3+16*x)*log(x 
^2+4)^2+8*x^3+32*x)*log(log(x^2+4)^2+2)+(-5*x^2-20)*log(x^2+4)^2-10*x^2-40 
),x, algorithm="fricas")
 

Output:

Exception raised: TypeError >>  Error detected within library code:   inte 
grate: implementation incomplete (constant residues)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{\frac {1}{2} \left (-10+\log \left (15-12 x \log \left (2+\log ^2\left (4+x^2\right )\right )\right )\right )} \left (8 x^2 \log \left (4+x^2\right )+\left (16+4 x^2+\left (8+2 x^2\right ) \log ^2\left (4+x^2\right )\right ) \log \left (2+\log ^2\left (4+x^2\right )\right )\right )}{-40-10 x^2+\left (-20-5 x^2\right ) \log ^2\left (4+x^2\right )+\left (32 x+8 x^3+\left (16 x+4 x^3\right ) \log ^2\left (4+x^2\right )\right ) \log \left (2+\log ^2\left (4+x^2\right )\right )} \, dx=\text {Timed out} \] Input:

integrate((((2*x**2+8)*ln(x**2+4)**2+4*x**2+16)*ln(ln(x**2+4)**2+2)+8*x**2 
*ln(x**2+4))*exp(1/2*ln(-12*x*ln(ln(x**2+4)**2+2)+15)-5)/(((4*x**3+16*x)*l 
n(x**2+4)**2+8*x**3+32*x)*ln(ln(x**2+4)**2+2)+(-5*x**2-20)*ln(x**2+4)**2-1 
0*x**2-40),x)
 

Output:

Timed out
 

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.38 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {e^{\frac {1}{2} \left (-10+\log \left (15-12 x \log \left (2+\log ^2\left (4+x^2\right )\right )\right )\right )} \left (8 x^2 \log \left (4+x^2\right )+\left (16+4 x^2+\left (8+2 x^2\right ) \log ^2\left (4+x^2\right )\right ) \log \left (2+\log ^2\left (4+x^2\right )\right )\right )}{-40-10 x^2+\left (-20-5 x^2\right ) \log ^2\left (4+x^2\right )+\left (32 x+8 x^3+\left (16 x+4 x^3\right ) \log ^2\left (4+x^2\right )\right ) \log \left (2+\log ^2\left (4+x^2\right )\right )} \, dx=i \, \sqrt {3} \sqrt {4 \, x \log \left (\log \left (x^{2} + 4\right )^{2} + 2\right ) - 5} e^{\left (-5\right )} \] Input:

integrate((((2*x^2+8)*log(x^2+4)^2+4*x^2+16)*log(log(x^2+4)^2+2)+8*x^2*log 
(x^2+4))*exp(1/2*log(-12*x*log(log(x^2+4)^2+2)+15)-5)/(((4*x^3+16*x)*log(x 
^2+4)^2+8*x^3+32*x)*log(log(x^2+4)^2+2)+(-5*x^2-20)*log(x^2+4)^2-10*x^2-40 
),x, algorithm="maxima")
 

Output:

I*sqrt(3)*sqrt(4*x*log(log(x^2 + 4)^2 + 2) - 5)*e^(-5)
 

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {e^{\frac {1}{2} \left (-10+\log \left (15-12 x \log \left (2+\log ^2\left (4+x^2\right )\right )\right )\right )} \left (8 x^2 \log \left (4+x^2\right )+\left (16+4 x^2+\left (8+2 x^2\right ) \log ^2\left (4+x^2\right )\right ) \log \left (2+\log ^2\left (4+x^2\right )\right )\right )}{-40-10 x^2+\left (-20-5 x^2\right ) \log ^2\left (4+x^2\right )+\left (32 x+8 x^3+\left (16 x+4 x^3\right ) \log ^2\left (4+x^2\right )\right ) \log \left (2+\log ^2\left (4+x^2\right )\right )} \, dx=e^{\left (\frac {1}{2} \, \log \left (-12 \, x \log \left (\log \left (x^{2} + 4\right )^{2} + 2\right ) + 15\right ) - 5\right )} \] Input:

integrate((((2*x^2+8)*log(x^2+4)^2+4*x^2+16)*log(log(x^2+4)^2+2)+8*x^2*log 
(x^2+4))*exp(1/2*log(-12*x*log(log(x^2+4)^2+2)+15)-5)/(((4*x^3+16*x)*log(x 
^2+4)^2+8*x^3+32*x)*log(log(x^2+4)^2+2)+(-5*x^2-20)*log(x^2+4)^2-10*x^2-40 
),x, algorithm="giac")
 

Output:

e^(1/2*log(-12*x*log(log(x^2 + 4)^2 + 2) + 15) - 5)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{\frac {1}{2} \left (-10+\log \left (15-12 x \log \left (2+\log ^2\left (4+x^2\right )\right )\right )\right )} \left (8 x^2 \log \left (4+x^2\right )+\left (16+4 x^2+\left (8+2 x^2\right ) \log ^2\left (4+x^2\right )\right ) \log \left (2+\log ^2\left (4+x^2\right )\right )\right )}{-40-10 x^2+\left (-20-5 x^2\right ) \log ^2\left (4+x^2\right )+\left (32 x+8 x^3+\left (16 x+4 x^3\right ) \log ^2\left (4+x^2\right )\right ) \log \left (2+\log ^2\left (4+x^2\right )\right )} \, dx=\int -\frac {{\mathrm {e}}^{\frac {\ln \left (15-12\,x\,\ln \left ({\ln \left (x^2+4\right )}^2+2\right )\right )}{2}-5}\,\left (\ln \left ({\ln \left (x^2+4\right )}^2+2\right )\,\left (4\,x^2+{\ln \left (x^2+4\right )}^2\,\left (2\,x^2+8\right )+16\right )+8\,x^2\,\ln \left (x^2+4\right )\right )}{10\,x^2-\ln \left ({\ln \left (x^2+4\right )}^2+2\right )\,\left (32\,x+{\ln \left (x^2+4\right )}^2\,\left (4\,x^3+16\,x\right )+8\,x^3\right )+{\ln \left (x^2+4\right )}^2\,\left (5\,x^2+20\right )+40} \,d x \] Input:

int(-(exp(log(15 - 12*x*log(log(x^2 + 4)^2 + 2))/2 - 5)*(log(log(x^2 + 4)^ 
2 + 2)*(4*x^2 + log(x^2 + 4)^2*(2*x^2 + 8) + 16) + 8*x^2*log(x^2 + 4)))/(1 
0*x^2 - log(log(x^2 + 4)^2 + 2)*(32*x + log(x^2 + 4)^2*(16*x + 4*x^3) + 8* 
x^3) + log(x^2 + 4)^2*(5*x^2 + 20) + 40),x)
 

Output:

int(-(exp(log(15 - 12*x*log(log(x^2 + 4)^2 + 2))/2 - 5)*(log(log(x^2 + 4)^ 
2 + 2)*(4*x^2 + log(x^2 + 4)^2*(2*x^2 + 8) + 16) + 8*x^2*log(x^2 + 4)))/(1 
0*x^2 - log(log(x^2 + 4)^2 + 2)*(32*x + log(x^2 + 4)^2*(16*x + 4*x^3) + 8* 
x^3) + log(x^2 + 4)^2*(5*x^2 + 20) + 40), x)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79 \[ \int \frac {e^{\frac {1}{2} \left (-10+\log \left (15-12 x \log \left (2+\log ^2\left (4+x^2\right )\right )\right )\right )} \left (8 x^2 \log \left (4+x^2\right )+\left (16+4 x^2+\left (8+2 x^2\right ) \log ^2\left (4+x^2\right )\right ) \log \left (2+\log ^2\left (4+x^2\right )\right )\right )}{-40-10 x^2+\left (-20-5 x^2\right ) \log ^2\left (4+x^2\right )+\left (32 x+8 x^3+\left (16 x+4 x^3\right ) \log ^2\left (4+x^2\right )\right ) \log \left (2+\log ^2\left (4+x^2\right )\right )} \, dx=\frac {\sqrt {-4 \,\mathrm {log}\left (\mathrm {log}\left (x^{2}+4\right )^{2}+2\right ) x +5}\, \sqrt {3}}{e^{5}} \] Input:

int((((2*x^2+8)*log(x^2+4)^2+4*x^2+16)*log(log(x^2+4)^2+2)+8*x^2*log(x^2+4 
))*exp(1/2*log(-12*x*log(log(x^2+4)^2+2)+15)-5)/(((4*x^3+16*x)*log(x^2+4)^ 
2+8*x^3+32*x)*log(log(x^2+4)^2+2)+(-5*x^2-20)*log(x^2+4)^2-10*x^2-40),x)
 

Output:

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