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

Optimal. Leaf size=25 \[ 3+\frac {15}{\log \left (\left (e^x-\frac {3}{4} x \left (x+\log \left (\log \left (x^2\right )\right )\right )\right )^2\right )} \]

________________________________________________________________________________________

Rubi [A]  time = 0.57, antiderivative size = 30, normalized size of antiderivative = 1.20, number of steps used = 3, number of rules used = 3, integrand size = 117, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {6688, 12, 6686} \begin {gather*} \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 )} \end {gather*}

Antiderivative was successfully verified.

[In]

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]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6686

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

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {30 \left (6-\log \left (x^2\right ) \left (4 e^x-6 x-3 \log \left (\log \left (x^2\right )\right )\right )\right )}{\log \left (x^2\right ) \left (4 e^x-3 x^2-3 x \log \left (\log \left (x^2\right )\right )\right ) \log ^2\left (\frac {1}{16} \left (-4 e^x+3 x^2+3 x \log \left (\log \left (x^2\right )\right )\right )^2\right )} \, dx\\ &=30 \int \frac {6-\log \left (x^2\right ) \left (4 e^x-6 x-3 \log \left (\log \left (x^2\right )\right )\right )}{\log \left (x^2\right ) \left (4 e^x-3 x^2-3 x \log \left (\log \left (x^2\right )\right )\right ) \log ^2\left (\frac {1}{16} \left (-4 e^x+3 x^2+3 x \log \left (\log \left (x^2\right )\right )\right )^2\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 )}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.08, size = 30, normalized size = 1.20 \begin {gather*} \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 )} \end {gather*}

Antiderivative was successfully verified.

[In]

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*Lo
g[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[Lo
g[x^2]]^2)/16]^2),x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.54, size = 52, normalized size = 2.08 \begin {gather*} \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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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")

[Out]

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))

________________________________________________________________________________________

giac [B]  time = 0.70, size = 63, normalized size = 2.52 \begin {gather*} -\frac {15}{4 \, \log \relax (2) - \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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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")

[Out]

-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)))

________________________________________________________________________________________

maple [C]  time = 0.90, size = 334, normalized size = 13.36

method	result	size
risch	\(\frac {30 i}{\pi \mathrm {csgn}\left (i \left (-x^{2}-x \ln \left (2 \ln \relax (x )-\frac {i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \left (-\mathrm {csgn}\left (i x^{2}\right )+\mathrm {csgn}\left (i x \right )\right )^{2}}{2}\right )+\frac {4 \,{\mathrm e}^{x}}{3}\right )\right )^{2} \mathrm {csgn}\left (i \left (-x^{2}-x \ln \left (2 \ln \relax (x )-\frac {i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \left (-\mathrm {csgn}\left (i x^{2}\right )+\mathrm {csgn}\left (i x \right )\right )^{2}}{2}\right )+\frac {4 \,{\mathrm e}^{x}}{3}\right )^{2}\right )+2 \pi \,\mathrm {csgn}\left (i \left (-x^{2}-x \ln \left (2 \ln \relax (x )-\frac {i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \left (-\mathrm {csgn}\left (i x^{2}\right )+\mathrm {csgn}\left (i x \right )\right )^{2}}{2}\right )+\frac {4 \,{\mathrm e}^{x}}{3}\right )\right ) \mathrm {csgn}\left (i \left (-x^{2}-x \ln \left (2 \ln \relax (x )-\frac {i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \left (-\mathrm {csgn}\left (i x^{2}\right )+\mathrm {csgn}\left (i x \right )\right )^{2}}{2}\right )+\frac {4 \,{\mathrm e}^{x}}{3}\right )^{2}\right )^{2}+\pi \mathrm {csgn}\left (i \left (-x^{2}-x \ln \left (2 \ln \relax (x )-\frac {i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \left (-\mathrm {csgn}\left (i x^{2}\right )+\mathrm {csgn}\left (i x \right )\right )^{2}}{2}\right )+\frac {4 \,{\mathrm e}^{x}}{3}\right )^{2}\right )^{3}-8 i \ln \relax (2)+4 i \ln \left (x^{2}+x \ln \left (2 \ln \relax (x )-\frac {i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \left (-\mathrm {csgn}\left (i x^{2}\right )+\mathrm {csgn}\left (i x \right )\right )^{2}}{2}\right )-\frac {4 \,{\mathrm e}^{x}}{3}\right )}\)	\(334\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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,meth
od=_RETURNVERBOSE)

[Out]

30*I/(Pi*csgn(I*(-x^2-x*ln(2*ln(x)-1/2*I*Pi*csgn(I*x^2)*(-csgn(I*x^2)+csgn(I*x))^2)+4/3*exp(x)))^2*csgn(I*(-x^
2-x*ln(2*ln(x)-1/2*I*Pi*csgn(I*x^2)*(-csgn(I*x^2)+csgn(I*x))^2)+4/3*exp(x))^2)+2*Pi*csgn(I*(-x^2-x*ln(2*ln(x)-
1/2*I*Pi*csgn(I*x^2)*(-csgn(I*x^2)+csgn(I*x))^2)+4/3*exp(x)))*csgn(I*(-x^2-x*ln(2*ln(x)-1/2*I*Pi*csgn(I*x^2)*(
-csgn(I*x^2)+csgn(I*x))^2)+4/3*exp(x))^2)^2+Pi*csgn(I*(-x^2-x*ln(2*ln(x)-1/2*I*Pi*csgn(I*x^2)*(-csgn(I*x^2)+cs
gn(I*x))^2)+4/3*exp(x))^2)^3-8*I*ln(2)+4*I*ln(x^2+x*ln(2*ln(x)-1/2*I*Pi*csgn(I*x^2)*(-csgn(I*x^2)+csgn(I*x))^2
)-4/3*exp(x)))

________________________________________________________________________________________

maxima [A]  time = 0.56, size = 33, normalized size = 1.32 \begin {gather*} -\frac {15}{2 \, {\left (2 \, \log \relax (2) - \log \left (-3 \, x^{2} - 3 \, x \log \relax (2) - 3 \, x \log \left (\log \relax (x)\right ) + 4 \, e^{x}\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 3.90, size = 52, normalized size = 2.08 \begin {gather*} \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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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)

[Out]

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)

________________________________________________________________________________________

sympy [B]  time = 12.60, size = 61, normalized size = 2.44 \begin {gather*} \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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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)

[Out]

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))

________________________________________________________________________________________