\(\int \frac {(-3 e^6+e^3 (-5-2 x)) \log (4)+(-3 e^6+e^3 (-5-2 x)) \log (\log (4 \log (2)))}{25 x^2+9 e^6 x^2+10 x^3+x^4+e^3 (30 x^2+6 x^3)} \, dx\) [9448]

Optimal result

Integrand size = 81, antiderivative size = 24 \[ \int \frac {\left (-3 e^6+e^3 (-5-2 x)\right ) \log (4)+\left (-3 e^6+e^3 (-5-2 x)\right ) \log (\log (4 \log (2)))}{25 x^2+9 e^6 x^2+10 x^3+x^4+e^3 \left (30 x^2+6 x^3\right )} \, dx=\frac {\log (4)+\log (\log (4 \log (2)))}{x \left (3+\frac {5+x}{e^3}\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.049, Rules used = {6, 12, 1694, 267} \[ \int \frac {\left (-3 e^6+e^3 (-5-2 x)\right ) \log (4)+\left (-3 e^6+e^3 (-5-2 x)\right ) \log (\log (4 \log (2)))}{25 x^2+9 e^6 x^2+10 x^3+x^4+e^3 \left (30 x^2+6 x^3\right )} \, dx=\frac {e^3 \log (4 \log (\log (16)))}{x \left (x+3 e^3+5\right )} \]

[In]

Int[((-3*E^6 + E^3*(-5 - 2*x))*Log[4] + (-3*E^6 + E^3*(-5 - 2*x))*Log[Log[4*Log[2]]])/(25*x^2 + 9*E^6*x^2 + 10
*x^3 + x^4 + E^3*(30*x^2 + 6*x^3)),x]

[Out]

(E^3*Log[4*Log[Log[16]]])/(x*(5 + 3*E^3 + x))

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 12

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

Rule 267

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 1694

Int[(Pq_)*(Q4_)^(p_), x_Symbol] :> With[{a = Coeff[Q4, x, 0], b = Coeff[Q4, x, 1], c = Coeff[Q4, x, 2], d = Co
eff[Q4, x, 3], e = Coeff[Q4, x, 4]}, Subst[Int[SimplifyIntegrand[(Pq /. x -> -d/(4*e) + x)*(a + d^4/(256*e^3)
- b*(d/(8*e)) + (c - 3*(d^2/(8*e)))*x^2 + e*x^4)^p, x], x], x, d/(4*e) + x] /; EqQ[d^3 - 4*c*d*e + 8*b*e^2, 0]
&& NeQ[d, 0]] /; FreeQ[p, x] && PolyQ[Pq, x] && PolyQ[Q4, x, 4] &&  !IGtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\left (-3 e^6+e^3 (-5-2 x)\right ) \log (4)+\left (-3 e^6+e^3 (-5-2 x)\right ) \log (\log (4 \log (2)))}{\left (25+9 e^6\right ) x^2+10 x^3+x^4+e^3 \left (30 x^2+6 x^3\right )} \, dx \\ & = \int \frac {\left (-3 e^6+e^3 (-5-2 x)\right ) (\log (4)+\log (\log (4 \log (2))))}{\left (25+9 e^6\right ) x^2+10 x^3+x^4+e^3 \left (30 x^2+6 x^3\right )} \, dx \\ & = \log (4 \log (\log (16))) \int \frac {-3 e^6+e^3 (-5-2 x)}{\left (25+9 e^6\right ) x^2+10 x^3+x^4+e^3 \left (30 x^2+6 x^3\right )} \, dx \\ & = \log (4 \log (\log (16))) \text {Subst}\left (\int -\frac {32 e^3 x}{\left (25+30 e^3+9 e^6-4 x^2\right )^2} \, dx,x,\frac {1}{4} \left (10+6 e^3\right )+x\right ) \\ & = -\left (\left (32 e^3 \log (4 \log (\log (16)))\right ) \text {Subst}\left (\int \frac {x}{\left (25+30 e^3+9 e^6-4 x^2\right )^2} \, dx,x,\frac {1}{4} \left (10+6 e^3\right )+x\right )\right ) \\ & = \frac {e^3 \log (4 \log (\log (16)))}{x \left (5+3 e^3+x\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {\left (-3 e^6+e^3 (-5-2 x)\right ) \log (4)+\left (-3 e^6+e^3 (-5-2 x)\right ) \log (\log (4 \log (2)))}{25 x^2+9 e^6 x^2+10 x^3+x^4+e^3 \left (30 x^2+6 x^3\right )} \, dx=\frac {e^3 \log (4 \log (\log (16)))}{x \left (5+3 e^3+x\right )} \]

[In]

Integrate[((-3*E^6 + E^3*(-5 - 2*x))*Log[4] + (-3*E^6 + E^3*(-5 - 2*x))*Log[Log[4*Log[2]]])/(25*x^2 + 9*E^6*x^
2 + 10*x^3 + x^4 + E^3*(30*x^2 + 6*x^3)),x]

[Out]

(E^3*Log[4*Log[Log[16]]])/(x*(5 + 3*E^3 + x))

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12

method	result	size
gosper	\(\frac {{\mathrm e}^{3} \left (\ln \left (\ln \left (4 \ln \left (2\right )\right )\right )+2 \ln \left (2\right )\right )}{x \left (5+3 \,{\mathrm e}^{3}+x \right )}\)	\(27\)
parallelrisch	\(\frac {2 \,{\mathrm e}^{3} \ln \left (2\right )+{\mathrm e}^{3} \ln \left (\ln \left (4 \ln \left (2\right )\right )\right )}{x \left (5+3 \,{\mathrm e}^{3}+x \right )}\)	\(30\)
norman	\(\frac {2 \,{\mathrm e}^{3} \ln \left (2\right )+{\mathrm e}^{3} \ln \left (2 \ln \left (2\right )+\ln \left (\ln \left (2\right )\right )\right )}{x \left (5+3 \,{\mathrm e}^{3}+x \right )}\)	\(33\)
risch	\(\frac {2 \,{\mathrm e}^{3} \ln \left (2\right )+{\mathrm e}^{3} \ln \left (2 \ln \left (2\right )+\ln \left (\ln \left (2\right )\right )\right )}{x \left (5+3 \,{\mathrm e}^{3}+x \right )}\)	\(35\)

[In]

int(((-3*exp(3)^2+(-2*x-5)*exp(3))*ln(ln(4*ln(2)))+2*(-3*exp(3)^2+(-2*x-5)*exp(3))*ln(2))/(9*x^2*exp(3)^2+(6*x
^3+30*x^2)*exp(3)+x^4+10*x^3+25*x^2),x,method=_RETURNVERBOSE)

[Out]

exp(3)*(ln(ln(4*ln(2)))+2*ln(2))/x/(5+3*exp(3)+x)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {\left (-3 e^6+e^3 (-5-2 x)\right ) \log (4)+\left (-3 e^6+e^3 (-5-2 x)\right ) \log (\log (4 \log (2)))}{25 x^2+9 e^6 x^2+10 x^3+x^4+e^3 \left (30 x^2+6 x^3\right )} \, dx=\frac {2 \, e^{3} \log \left (2\right ) + e^{3} \log \left (\log \left (4 \, \log \left (2\right )\right )\right )}{x^{2} + 3 \, x e^{3} + 5 \, x} \]

[In]

integrate(((-3*exp(3)^2+(-2*x-5)*exp(3))*log(log(4*log(2)))+2*(-3*exp(3)^2+(-2*x-5)*exp(3))*log(2))/(9*x^2*exp
(3)^2+(6*x^3+30*x^2)*exp(3)+x^4+10*x^3+25*x^2),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.62 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54 \[ \int \frac {\left (-3 e^6+e^3 (-5-2 x)\right ) \log (4)+\left (-3 e^6+e^3 (-5-2 x)\right ) \log (\log (4 \log (2)))}{25 x^2+9 e^6 x^2+10 x^3+x^4+e^3 \left (30 x^2+6 x^3\right )} \, dx=- \frac {- 2 e^{3} \log {\left (2 \right )} - e^{3} \log {\left (\log {\left (\log {\left (2 \right )} \right )} + 2 \log {\left (2 \right )} \right )}}{x^{2} + x \left (5 + 3 e^{3}\right )} \]

[In]

integrate(((-3*exp(3)**2+(-2*x-5)*exp(3))*ln(ln(4*ln(2)))+2*(-3*exp(3)**2+(-2*x-5)*exp(3))*ln(2))/(9*x**2*exp(
3)**2+(6*x**3+30*x**2)*exp(3)+x**4+10*x**3+25*x**2),x)

[Out]

-(-2*exp(3)*log(2) - exp(3)*log(log(log(2)) + 2*log(2)))/(x**2 + x*(5 + 3*exp(3)))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {\left (-3 e^6+e^3 (-5-2 x)\right ) \log (4)+\left (-3 e^6+e^3 (-5-2 x)\right ) \log (\log (4 \log (2)))}{25 x^2+9 e^6 x^2+10 x^3+x^4+e^3 \left (30 x^2+6 x^3\right )} \, dx=\frac {2 \, e^{3} \log \left (2\right ) + e^{3} \log \left (\log \left (4 \, \log \left (2\right )\right )\right )}{x^{2} + x {\left (3 \, e^{3} + 5\right )}} \]

[In]

integrate(((-3*exp(3)^2+(-2*x-5)*exp(3))*log(log(4*log(2)))+2*(-3*exp(3)^2+(-2*x-5)*exp(3))*log(2))/(9*x^2*exp
(3)^2+(6*x^3+30*x^2)*exp(3)+x^4+10*x^3+25*x^2),x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 79 vs. \(2 (25) = 50\).

Time = 0.26 (sec) , antiderivative size = 79, normalized size of antiderivative = 3.29 \[ \int \frac {\left (-3 e^6+e^3 (-5-2 x)\right ) \log (4)+\left (-3 e^6+e^3 (-5-2 x)\right ) \log (\log (4 \log (2)))}{25 x^2+9 e^6 x^2+10 x^3+x^4+e^3 \left (30 x^2+6 x^3\right )} \, dx=\frac {4 \, e^{6} \log \left (2\right )^{2} + 4 \, e^{6} \log \left (2\right ) \log \left (\log \left (4 \, \log \left (2\right )\right )\right ) + e^{6} \log \left (\log \left (4 \, \log \left (2\right )\right )\right )^{2}}{2 \, {\left (3 \, x e^{6} + {\left (x^{2} + 5 \, x\right )} e^{3}\right )} \log \left (2\right ) + {\left (3 \, x e^{6} + {\left (x^{2} + 5 \, x\right )} e^{3}\right )} \log \left (\log \left (4 \, \log \left (2\right )\right )\right )} \]

[In]

integrate(((-3*exp(3)^2+(-2*x-5)*exp(3))*log(log(4*log(2)))+2*(-3*exp(3)^2+(-2*x-5)*exp(3))*log(2))/(9*x^2*exp
(3)^2+(6*x^3+30*x^2)*exp(3)+x^4+10*x^3+25*x^2),x, algorithm="giac")

[Out]

(4*e^6*log(2)^2 + 4*e^6*log(2)*log(log(4*log(2))) + e^6*log(log(4*log(2)))^2)/(2*(3*x*e^6 + (x^2 + 5*x)*e^3)*l
og(2) + (3*x*e^6 + (x^2 + 5*x)*e^3)*log(log(4*log(2))))

Mupad [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {\left (-3 e^6+e^3 (-5-2 x)\right ) \log (4)+\left (-3 e^6+e^3 (-5-2 x)\right ) \log (\log (4 \log (2)))}{25 x^2+9 e^6 x^2+10 x^3+x^4+e^3 \left (30 x^2+6 x^3\right )} \, dx=\frac {{\mathrm {e}}^3\,\ln \left (4\,\ln \left (\ln \left (16\right )\right )\right )}{x^2+\left (3\,{\mathrm {e}}^3+5\right )\,x} \]

[In]

int(-(log(log(4*log(2)))*(3*exp(6) + exp(3)*(2*x + 5)) + 2*log(2)*(3*exp(6) + exp(3)*(2*x + 5)))/(exp(3)*(30*x
^2 + 6*x^3) + 9*x^2*exp(6) + 25*x^2 + 10*x^3 + x^4),x)

[Out]

(exp(3)*log(4*log(log(16))))/(x^2 + x*(3*exp(3) + 5))