\(\int \frac {4 x^2+7 x^3+x^4-x \log (\log (2))+\log ^2(\log (2))}{4 x^3+4 x^4+x^5+(4 x^2+2 x^3) \log (\log (2))+x \log ^2(\log (2))} \, dx\) [1419]

Optimal result

Integrand size = 64, antiderivative size = 24 \[ \int \frac {4 x^2+7 x^3+x^4-x \log (\log (2))+\log ^2(\log (2))}{4 x^3+4 x^4+x^5+\left (4 x^2+2 x^3\right ) \log (\log (2))+x \log ^2(\log (2))} \, dx=\log (x)-\frac {5+x}{3+x+\frac {-x+\log (\log (2))}{x}} \]

[Out]

ln(x)-(5+x)/(x+(ln(ln(2))-x)/x+3)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {2099, 632, 212, 652} \[ \int \frac {4 x^2+7 x^3+x^4-x \log (\log (2))+\log ^2(\log (2))}{4 x^3+4 x^4+x^5+\left (4 x^2+2 x^3\right ) \log (\log (2))+x \log ^2(\log (2))} \, dx=\log (x)-\frac {3 x-\log (\log (2))}{x^2+2 x+\log (\log (2))} \]

[In]

Int[(4*x^2 + 7*x^3 + x^4 - x*Log[Log[2]] + Log[Log[2]]^2)/(4*x^3 + 4*x^4 + x^5 + (4*x^2 + 2*x^3)*Log[Log[2]] +
x*Log[Log[2]]^2),x]

[Out]

Log[x] - (3*x - Log[Log[2]])/(2*x + x^2 + Log[Log[2]])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 652

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b*d - 2*a*e + (2*c*d -
b*e)*x)/((p + 1)*(b^2 - 4*a*c)))*(a + b*x + c*x^2)^(p + 1), x] - Dist[(2*p + 3)*((2*c*d - b*e)/((p + 1)*(b^2 -
4*a*c))), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^
2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2]

Rule 2099

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Q^q, x], x] /;  !SumQ[N
onfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x] && PolyQ[Q, x] && IntegerQ[p] && NeQ[P, x]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{x}+\frac {3}{2 x+x^2+\log (\log (2))}+\frac {2 (-4 \log (\log (2))-x (3+\log (\log (2))))}{\left (2 x+x^2+\log (\log (2))\right )^2}\right ) \, dx \\ & = \log (x)+2 \int \frac {-4 \log (\log (2))-x (3+\log (\log (2)))}{\left (2 x+x^2+\log (\log (2))\right )^2} \, dx+3 \int \frac {1}{2 x+x^2+\log (\log (2))} \, dx \\ & = \log (x)-\frac {3 x-\log (\log (2))}{2 x+x^2+\log (\log (2))}-3 \int \frac {1}{2 x+x^2+\log (\log (2))} \, dx-6 \text {Subst}\left (\int \frac {1}{-x^2+4 (1-\log (\log (2)))} \, dx,x,2+2 x\right ) \\ & = \log (x)-\frac {3 \text {arctanh}\left (\frac {1+x}{\sqrt {1-\log (\log (2))}}\right )}{\sqrt {1-\log (\log (2))}}-\frac {3 x-\log (\log (2))}{2 x+x^2+\log (\log (2))}+6 \text {Subst}\left (\int \frac {1}{-x^2+4 (1-\log (\log (2)))} \, dx,x,2+2 x\right ) \\ & = \log (x)-\frac {3 x-\log (\log (2))}{2 x+x^2+\log (\log (2))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {4 x^2+7 x^3+x^4-x \log (\log (2))+\log ^2(\log (2))}{4 x^3+4 x^4+x^5+\left (4 x^2+2 x^3\right ) \log (\log (2))+x \log ^2(\log (2))} \, dx=\log (x)+\frac {-3 x+\log (\log (2))}{2 x+x^2+\log (\log (2))} \]

[In]

Integrate[(4*x^2 + 7*x^3 + x^4 - x*Log[Log[2]] + Log[Log[2]]^2)/(4*x^3 + 4*x^4 + x^5 + (4*x^2 + 2*x^3)*Log[Log
[2]] + x*Log[Log[2]]^2),x]

[Out]

Log[x] + (-3*x + Log[Log[2]])/(2*x + x^2 + Log[Log[2]])

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00

method	result	size
norman	\(\frac {-3 x +\ln \left (\ln \left (2\right )\right )}{x^{2}+\ln \left (\ln \left (2\right )\right )+2 x}+\ln \left (x \right )\)	\(24\)
risch	\(\frac {-3 x +\ln \left (\ln \left (2\right )\right )}{x^{2}+\ln \left (\ln \left (2\right )\right )+2 x}+\ln \left (x \right )\)	\(24\)
default	\(-\frac {3 x -\ln \left (\ln \left (2\right )\right )}{x^{2}+\ln \left (\ln \left (2\right )\right )+2 x}+\ln \left (x \right )\)	\(27\)
parallelrisch	\(\frac {x^{2} \ln \left (x \right )+\ln \left (x \right ) \ln \left (\ln \left (2\right )\right )+2 x \ln \left (x \right )+\ln \left (\ln \left (2\right )\right )-3 x}{x^{2}+\ln \left (\ln \left (2\right )\right )+2 x}\)	\(38\)

[In]

int((ln(ln(2))^2-x*ln(ln(2))+x^4+7*x^3+4*x^2)/(x*ln(ln(2))^2+(2*x^3+4*x^2)*ln(ln(2))+x^5+4*x^4+4*x^3),x,method
=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46 \[ \int \frac {4 x^2+7 x^3+x^4-x \log (\log (2))+\log ^2(\log (2))}{4 x^3+4 x^4+x^5+\left (4 x^2+2 x^3\right ) \log (\log (2))+x \log ^2(\log (2))} \, dx=\frac {{\left (x^{2} + 2 \, x\right )} \log \left (x\right ) + {\left (\log \left (x\right ) + 1\right )} \log \left (\log \left (2\right )\right ) - 3 \, x}{x^{2} + 2 \, x + \log \left (\log \left (2\right )\right )} \]

[In]

integrate((log(log(2))^2-x*log(log(2))+x^4+7*x^3+4*x^2)/(x*log(log(2))^2+(2*x^3+4*x^2)*log(log(2))+x^5+4*x^4+4
*x^3),x, algorithm="fricas")

[Out]

((x^2 + 2*x)*log(x) + (log(x) + 1)*log(log(2)) - 3*x)/(x^2 + 2*x + log(log(2)))

Sympy [A] (verification not implemented)

Time = 0.43 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {4 x^2+7 x^3+x^4-x \log (\log (2))+\log ^2(\log (2))}{4 x^3+4 x^4+x^5+\left (4 x^2+2 x^3\right ) \log (\log (2))+x \log ^2(\log (2))} \, dx=\frac {- 3 x + \log {\left (\log {\left (2 \right )} \right )}}{x^{2} + 2 x + \log {\left (\log {\left (2 \right )} \right )}} + \log {\left (x \right )} \]

[In]

integrate((ln(ln(2))**2-x*ln(ln(2))+x**4+7*x**3+4*x**2)/(x*ln(ln(2))**2+(2*x**3+4*x**2)*ln(ln(2))+x**5+4*x**4+
4*x**3),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {4 x^2+7 x^3+x^4-x \log (\log (2))+\log ^2(\log (2))}{4 x^3+4 x^4+x^5+\left (4 x^2+2 x^3\right ) \log (\log (2))+x \log ^2(\log (2))} \, dx=-\frac {3 \, x - \log \left (\log \left (2\right )\right )}{x^{2} + 2 \, x + \log \left (\log \left (2\right )\right )} + \log \left (x\right ) \]

[In]

integrate((log(log(2))^2-x*log(log(2))+x^4+7*x^3+4*x^2)/(x*log(log(2))^2+(2*x^3+4*x^2)*log(log(2))+x^5+4*x^4+4
*x^3),x, algorithm="maxima")

[Out]

-(3*x - log(log(2)))/(x^2 + 2*x + log(log(2))) + log(x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {4 x^2+7 x^3+x^4-x \log (\log (2))+\log ^2(\log (2))}{4 x^3+4 x^4+x^5+\left (4 x^2+2 x^3\right ) \log (\log (2))+x \log ^2(\log (2))} \, dx=-\frac {3 \, x - \log \left (\log \left (2\right )\right )}{x^{2} + 2 \, x + \log \left (\log \left (2\right )\right )} + \log \left ({\left | x \right |}\right ) \]

[In]

integrate((log(log(2))^2-x*log(log(2))+x^4+7*x^3+4*x^2)/(x*log(log(2))^2+(2*x^3+4*x^2)*log(log(2))+x^5+4*x^4+4
*x^3),x, algorithm="giac")

[Out]

-(3*x - log(log(2)))/(x^2 + 2*x + log(log(2))) + log(abs(x))

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {4 x^2+7 x^3+x^4-x \log (\log (2))+\log ^2(\log (2))}{4 x^3+4 x^4+x^5+\left (4 x^2+2 x^3\right ) \log (\log (2))+x \log ^2(\log (2))} \, dx=\ln \left (x\right )-\frac {3\,x-\ln \left (\ln \left (2\right )\right )}{x^2+2\,x+\ln \left (\ln \left (2\right )\right )} \]

[In]

int((log(log(2))^2 - x*log(log(2)) + 4*x^2 + 7*x^3 + x^4)/(log(log(2))*(4*x^2 + 2*x^3) + x*log(log(2))^2 + 4*x
^3 + 4*x^4 + x^5),x)

[Out]

log(x) - (3*x - log(log(2)))/(2*x + log(log(2)) + x^2)