\(\int \frac {-180+(120 e^x-180 x) \log (x^2)-90 \log (x^2) \log (\log (x^2))}{((-4 e^x+3 x^2) \log (x^2)+3 x \log (x^2) \log (\log (x^2))) \log ^2(\frac {1}{16} (16 e^{2 x}-24 e^x x^2+9 x^4+(-24 e^x x+18 x^3) \log (\log (x^2))+9 x^2 \log ^2(\log (x^2))))} \, dx\) [2654]

Optimal result

Integrand size = 117, antiderivative size = 25 \[ \int \frac {-180+\left (120 e^x-180 x\right ) \log \left (x^2\right )-90 \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{\left (\left (-4 e^x+3 x^2\right ) \log \left (x^2\right )+3 x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )\right ) \log ^2\left (\frac {1}{16} \left (16 e^{2 x}-24 e^x x^2+9 x^4+\left (-24 e^x x+18 x^3\right ) \log \left (\log \left (x^2\right )\right )+9 x^2 \log ^2\left (\log \left (x^2\right )\right )\right )\right )} \, dx=3+\frac {15}{\log \left (\left (e^x-\frac {3}{4} x \left (x+\log \left (\log \left (x^2\right )\right )\right )\right )^2\right )} \] Output:

3+15/ln((exp(x)-3/4*(x+ln(ln(x^2)))*x)^2)
 

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int \frac {-180+\left (120 e^x-180 x\right ) \log \left (x^2\right )-90 \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{\left (\left (-4 e^x+3 x^2\right ) \log \left (x^2\right )+3 x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )\right ) \log ^2\left (\frac {1}{16} \left (16 e^{2 x}-24 e^x x^2+9 x^4+\left (-24 e^x x+18 x^3\right ) \log \left (\log \left (x^2\right )\right )+9 x^2 \log ^2\left (\log \left (x^2\right )\right )\right )\right )} \, dx=\frac {15}{\log \left (\frac {1}{16} \left (-4 e^x+3 x^2+3 x \log \left (\log \left (x^2\right )\right )\right )^2\right )} \] Input:

Integrate[(-180 + (120*E^x - 180*x)*Log[x^2] - 90*Log[x^2]*Log[Log[x^2]])/ 
(((-4*E^x + 3*x^2)*Log[x^2] + 3*x*Log[x^2]*Log[Log[x^2]])*Log[(16*E^(2*x) 
- 24*E^x*x^2 + 9*x^4 + (-24*E^x*x + 18*x^3)*Log[Log[x^2]] + 9*x^2*Log[Log[ 
x^2]]^2)/16]^2),x]
 

Output:

15/Log[(-4*E^x + 3*x^2 + 3*x*Log[Log[x^2]])^2/16]
 

Rubi [A] (verified)

Time = 0.92 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.017, Rules used = {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.

\(\displaystyle \int \frac {\left (120 e^x-180 x\right ) \log \left (x^2\right )-90 \log \left (\log \left (x^2\right )\right ) \log \left (x^2\right )-180}{\left (\left (3 x^2-4 e^x\right ) \log \left (x^2\right )+3 x \log \left (\log \left (x^2\right )\right ) \log \left (x^2\right )\right ) \log ^2\left (\frac {1}{16} \left (9 x^4-24 e^x x^2+9 x^2 \log ^2\left (\log \left (x^2\right )\right )+\left (18 x^3-24 e^x x\right ) \log \left (\log \left (x^2\right )\right )+16 e^{2 x}\right )\right )} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {-\left (\left (120 e^x-180 x\right ) \log \left (x^2\right )\right )+90 \log \left (\log \left (x^2\right )\right ) \log \left (x^2\right )+180}{\log \left (x^2\right ) \left (-3 x^2-3 x \log \left (\log \left (x^2\right )\right )+4 e^x\right ) \log ^2\left (\frac {1}{16} \left (-3 x^2-3 x \log \left (\log \left (x^2\right )\right )+4 e^x\right )^2\right )}dx\)

\(\Big \downarrow \) 7237

\(\displaystyle \frac {15}{\log \left (\frac {1}{16} \left (-3 x^2-3 x \log \left (\log \left (x^2\right )\right )+4 e^x\right )^2\right )}\)

Input:

Int[(-180 + (120*E^x - 180*x)*Log[x^2] - 90*Log[x^2]*Log[Log[x^2]])/(((-4* 
E^x + 3*x^2)*Log[x^2] + 3*x*Log[x^2]*Log[Log[x^2]])*Log[(16*E^(2*x) - 24*E 
^x*x^2 + 9*x^4 + (-24*E^x*x + 18*x^3)*Log[Log[x^2]] + 9*x^2*Log[Log[x^2]]^ 
2)/16]^2),x]
 

Output:

15/Log[(4*E^x - 3*x^2 - 3*x*Log[Log[x^2]])^2/16]
 

Defintions of rubi rules used

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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(52\) vs. \(2(22)=44\).

Time = 30.28 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.12

method	result	size
parallelrisch	\(\frac {15}{\ln \left (\frac {9 x^{2} {\ln \left (\ln \left (x^{2}\right )\right )}^{2}}{16}+\frac {\left (-24 \,{\mathrm e}^{x} x +18 x^{3}\right ) \ln \left (\ln \left (x^{2}\right )\right )}{16}+{\mathrm e}^{2 x}-\frac {3 \,{\mathrm e}^{x} x^{2}}{2}+\frac {9 x^{4}}{16}\right )}\)	\(53\)
risch	\(\frac {30 i}{\pi {\operatorname {csgn}\left (i \left (-x^{2}-x \ln \left (2 \ln \left (x \right )-\frac {i \pi \,\operatorname {csgn}\left (i x^{2}\right ) {\left (-\operatorname {csgn}\left (i x^{2}\right )+\operatorname {csgn}\left (i x \right )\right )}^{2}}{2}\right )+\frac {4 \,{\mathrm e}^{x}}{3}\right )\right )}^{2} \operatorname {csgn}\left (i \left (-x^{2}-x \ln \left (2 \ln \left (x \right )-\frac {i \pi \,\operatorname {csgn}\left (i x^{2}\right ) {\left (-\operatorname {csgn}\left (i x^{2}\right )+\operatorname {csgn}\left (i x \right )\right )}^{2}}{2}\right )+\frac {4 \,{\mathrm e}^{x}}{3}\right )^{2}\right )+2 \pi \,\operatorname {csgn}\left (i \left (-x^{2}-x \ln \left (2 \ln \left (x \right )-\frac {i \pi \,\operatorname {csgn}\left (i x^{2}\right ) {\left (-\operatorname {csgn}\left (i x^{2}\right )+\operatorname {csgn}\left (i x \right )\right )}^{2}}{2}\right )+\frac {4 \,{\mathrm e}^{x}}{3}\right )\right ) {\operatorname {csgn}\left (i \left (-x^{2}-x \ln \left (2 \ln \left (x \right )-\frac {i \pi \,\operatorname {csgn}\left (i x^{2}\right ) {\left (-\operatorname {csgn}\left (i x^{2}\right )+\operatorname {csgn}\left (i x \right )\right )}^{2}}{2}\right )+\frac {4 \,{\mathrm e}^{x}}{3}\right )^{2}\right )}^{2}+\pi {\operatorname {csgn}\left (i \left (-x^{2}-x \ln \left (2 \ln \left (x \right )-\frac {i \pi \,\operatorname {csgn}\left (i x^{2}\right ) {\left (-\operatorname {csgn}\left (i x^{2}\right )+\operatorname {csgn}\left (i x \right )\right )}^{2}}{2}\right )+\frac {4 \,{\mathrm e}^{x}}{3}\right )^{2}\right )}^{3}-8 i \ln \left (2\right )+4 i \ln \left (x^{2}+x \ln \left (2 \ln \left (x \right )-\frac {i \pi \,\operatorname {csgn}\left (i x^{2}\right ) {\left (-\operatorname {csgn}\left (i x^{2}\right )+\operatorname {csgn}\left (i x \right )\right )}^{2}}{2}\right )-\frac {4 \,{\mathrm e}^{x}}{3}\right )}\)	\(334\)

Input:

int((-90*ln(x^2)*ln(ln(x^2))+(120*exp(x)-180*x)*ln(x^2)-180)/(3*x*ln(x^2)* 
ln(ln(x^2))+(-4*exp(x)+3*x^2)*ln(x^2))/ln(9/16*x^2*ln(ln(x^2))^2+1/16*(-24 
*exp(x)*x+18*x^3)*ln(ln(x^2))+exp(x)^2-3/2*exp(x)*x^2+9/16*x^4)^2,x,method 
=_RETURNVERBOSE)
 

Output:

15/ln(9/16*x^2*ln(ln(x^2))^2+1/16*(-24*exp(x)*x+18*x^3)*ln(ln(x^2))+exp(x) 
^2-3/2*exp(x)*x^2+9/16*x^4)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.08 \[ \int \frac {-180+\left (120 e^x-180 x\right ) \log \left (x^2\right )-90 \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{\left (\left (-4 e^x+3 x^2\right ) \log \left (x^2\right )+3 x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )\right ) \log ^2\left (\frac {1}{16} \left (16 e^{2 x}-24 e^x x^2+9 x^4+\left (-24 e^x x+18 x^3\right ) \log \left (\log \left (x^2\right )\right )+9 x^2 \log ^2\left (\log \left (x^2\right )\right )\right )\right )} \, dx=\frac {15}{\log \left (\frac {9}{16} \, x^{4} + \frac {9}{16} \, x^{2} \log \left (\log \left (x^{2}\right )\right )^{2} - \frac {3}{2} \, x^{2} e^{x} + \frac {3}{8} \, {\left (3 \, x^{3} - 4 \, x e^{x}\right )} \log \left (\log \left (x^{2}\right )\right ) + e^{\left (2 \, x\right )}\right )} \] Input:

integrate((-90*log(x^2)*log(log(x^2))+(120*exp(x)-180*x)*log(x^2)-180)/(3* 
x*log(x^2)*log(log(x^2))+(-4*exp(x)+3*x^2)*log(x^2))/log(9/16*x^2*log(log( 
x^2))^2+1/16*(-24*exp(x)*x+18*x^3)*log(log(x^2))+exp(x)^2-3/2*exp(x)*x^2+9 
/16*x^4)^2,x, algorithm="fricas")
 

Output:

15/log(9/16*x^4 + 9/16*x^2*log(log(x^2))^2 - 3/2*x^2*e^x + 3/8*(3*x^3 - 4* 
x*e^x)*log(log(x^2)) + e^(2*x))
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 61 vs. \(2 (26) = 52\).

Time = 6.05 (sec) , antiderivative size = 61, normalized size of antiderivative = 2.44 \[ \int \frac {-180+\left (120 e^x-180 x\right ) \log \left (x^2\right )-90 \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{\left (\left (-4 e^x+3 x^2\right ) \log \left (x^2\right )+3 x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )\right ) \log ^2\left (\frac {1}{16} \left (16 e^{2 x}-24 e^x x^2+9 x^4+\left (-24 e^x x+18 x^3\right ) \log \left (\log \left (x^2\right )\right )+9 x^2 \log ^2\left (\log \left (x^2\right )\right )\right )\right )} \, dx=\frac {15}{\log {\left (\frac {9 x^{4}}{16} - \frac {3 x^{2} e^{x}}{2} + \frac {9 x^{2} \log {\left (\log {\left (x^{2} \right )} \right )}^{2}}{16} + \left (\frac {9 x^{3}}{8} - \frac {3 x e^{x}}{2}\right ) \log {\left (\log {\left (x^{2} \right )} \right )} + e^{2 x} \right )}} \] Input:

integrate((-90*ln(x**2)*ln(ln(x**2))+(120*exp(x)-180*x)*ln(x**2)-180)/(3*x 
*ln(x**2)*ln(ln(x**2))+(-4*exp(x)+3*x**2)*ln(x**2))/ln(9/16*x**2*ln(ln(x** 
2))**2+1/16*(-24*exp(x)*x+18*x**3)*ln(ln(x**2))+exp(x)**2-3/2*exp(x)*x**2+ 
9/16*x**4)**2,x)
 

Output:

15/log(9*x**4/16 - 3*x**2*exp(x)/2 + 9*x**2*log(log(x**2))**2/16 + (9*x**3 
/8 - 3*x*exp(x)/2)*log(log(x**2)) + exp(2*x))
 

Maxima [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32 \[ \int \frac {-180+\left (120 e^x-180 x\right ) \log \left (x^2\right )-90 \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{\left (\left (-4 e^x+3 x^2\right ) \log \left (x^2\right )+3 x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )\right ) \log ^2\left (\frac {1}{16} \left (16 e^{2 x}-24 e^x x^2+9 x^4+\left (-24 e^x x+18 x^3\right ) \log \left (\log \left (x^2\right )\right )+9 x^2 \log ^2\left (\log \left (x^2\right )\right )\right )\right )} \, dx=-\frac {15}{2 \, {\left (2 \, \log \left (2\right ) - \log \left (-3 \, x^{2} - 3 \, x \log \left (2\right ) - 3 \, x \log \left (\log \left (x\right )\right ) + 4 \, e^{x}\right )\right )}} \] Input:

integrate((-90*log(x^2)*log(log(x^2))+(120*exp(x)-180*x)*log(x^2)-180)/(3* 
x*log(x^2)*log(log(x^2))+(-4*exp(x)+3*x^2)*log(x^2))/log(9/16*x^2*log(log( 
x^2))^2+1/16*(-24*exp(x)*x+18*x^3)*log(log(x^2))+exp(x)^2-3/2*exp(x)*x^2+9 
/16*x^4)^2,x, algorithm="maxima")
 

Output:

-15/2/(2*log(2) - log(-3*x^2 - 3*x*log(2) - 3*x*log(log(x)) + 4*e^x))
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (26) = 52\).

Time = 3.27 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.52 \[ \int \frac {-180+\left (120 e^x-180 x\right ) \log \left (x^2\right )-90 \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{\left (\left (-4 e^x+3 x^2\right ) \log \left (x^2\right )+3 x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )\right ) \log ^2\left (\frac {1}{16} \left (16 e^{2 x}-24 e^x x^2+9 x^4+\left (-24 e^x x+18 x^3\right ) \log \left (\log \left (x^2\right )\right )+9 x^2 \log ^2\left (\log \left (x^2\right )\right )\right )\right )} \, dx=-\frac {15}{4 \, \log \left (2\right ) - \log \left (9 \, x^{4} + 18 \, x^{3} \log \left (\log \left (x^{2}\right )\right ) + 9 \, x^{2} \log \left (\log \left (x^{2}\right )\right )^{2} - 24 \, x^{2} e^{x} - 24 \, x e^{x} \log \left (\log \left (x^{2}\right )\right ) + 16 \, e^{\left (2 \, x\right )}\right )} \] Input:

integrate((-90*log(x^2)*log(log(x^2))+(120*exp(x)-180*x)*log(x^2)-180)/(3* 
x*log(x^2)*log(log(x^2))+(-4*exp(x)+3*x^2)*log(x^2))/log(9/16*x^2*log(log( 
x^2))^2+1/16*(-24*exp(x)*x+18*x^3)*log(log(x^2))+exp(x)^2-3/2*exp(x)*x^2+9 
/16*x^4)^2,x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

-15/(4*log(2) - log(9*x^4 + 18*x^3*log(log(x^2)) + 9*x^2*log(log(x^2))^2 - 
 24*x^2*e^x - 24*x*e^x*log(log(x^2)) + 16*e^(2*x)))
 

Mupad [B] (verification not implemented)

Time = 2.90 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.08 \[ \int \frac {-180+\left (120 e^x-180 x\right ) \log \left (x^2\right )-90 \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{\left (\left (-4 e^x+3 x^2\right ) \log \left (x^2\right )+3 x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )\right ) \log ^2\left (\frac {1}{16} \left (16 e^{2 x}-24 e^x x^2+9 x^4+\left (-24 e^x x+18 x^3\right ) \log \left (\log \left (x^2\right )\right )+9 x^2 \log ^2\left (\log \left (x^2\right )\right )\right )\right )} \, dx=\frac {15}{\ln \left ({\mathrm {e}}^{2\,x}-\frac {3\,x^2\,{\mathrm {e}}^x}{2}+\frac {9\,x^2\,{\ln \left (\ln \left (x^2\right )\right )}^2}{16}-\frac {\ln \left (\ln \left (x^2\right )\right )\,\left (24\,x\,{\mathrm {e}}^x-18\,x^3\right )}{16}+\frac {9\,x^4}{16}\right )} \] Input:

int((log(x^2)*(180*x - 120*exp(x)) + 90*log(x^2)*log(log(x^2)) + 180)/(log 
(exp(2*x) - (3*x^2*exp(x))/2 + (9*x^2*log(log(x^2))^2)/16 - (log(log(x^2)) 
*(24*x*exp(x) - 18*x^3))/16 + (9*x^4)/16)^2*(log(x^2)*(4*exp(x) - 3*x^2) - 
 3*x*log(x^2)*log(log(x^2)))),x)
 

Output:

15/log(exp(2*x) - (3*x^2*exp(x))/2 + (9*x^2*log(log(x^2))^2)/16 - (log(log 
(x^2))*(24*x*exp(x) - 18*x^3))/16 + (9*x^4)/16)
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.28 \[ \int \frac {-180+\left (120 e^x-180 x\right ) \log \left (x^2\right )-90 \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{\left (\left (-4 e^x+3 x^2\right ) \log \left (x^2\right )+3 x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )\right ) \log ^2\left (\frac {1}{16} \left (16 e^{2 x}-24 e^x x^2+9 x^4+\left (-24 e^x x+18 x^3\right ) \log \left (\log \left (x^2\right )\right )+9 x^2 \log ^2\left (\log \left (x^2\right )\right )\right )\right )} \, dx=\frac {15}{\mathrm {log}\left (e^{2 x}-\frac {3 e^{x} \mathrm {log}\left (\mathrm {log}\left (x^{2}\right )\right ) x}{2}-\frac {3 e^{x} x^{2}}{2}+\frac {9 {\mathrm {log}\left (\mathrm {log}\left (x^{2}\right )\right )}^{2} x^{2}}{16}+\frac {9 \,\mathrm {log}\left (\mathrm {log}\left (x^{2}\right )\right ) x^{3}}{8}+\frac {9 x^{4}}{16}\right )} \] Input:

int((-90*log(x^2)*log(log(x^2))+(120*exp(x)-180*x)*log(x^2)-180)/(3*x*log( 
x^2)*log(log(x^2))+(-4*exp(x)+3*x^2)*log(x^2))/log(9/16*x^2*log(log(x^2))^ 
2+1/16*(-24*exp(x)*x+18*x^3)*log(log(x^2))+exp(x)^2-3/2*exp(x)*x^2+9/16*x^ 
4)^2,x)
 

Output:

15/log((16*e**(2*x) - 24*e**x*log(log(x**2))*x - 24*e**x*x**2 + 9*log(log( 
x**2))**2*x**2 + 18*log(log(x**2))*x**3 + 9*x**4)/16)