\(\int \frac {-2+5 x+7 x^2+e (x+x^2)+(3 x+21 x^2+e (x+3 x^2)) \log (x)}{(-2 x+5 x^2+7 x^3+e (x^2+x^3)) \log (x)} \, dx\) [8764]

Optimal result

Integrand size = 68, antiderivative size = 20 \[ \int \frac {-2+5 x+7 x^2+e \left (x+x^2\right )+\left (3 x+21 x^2+e \left (x+3 x^2\right )\right ) \log (x)}{\left (-2 x+5 x^2+7 x^3+e \left (x^2+x^3\right )\right ) \log (x)} \, dx=\log \left ((1+x)^2\right )+\log \left (\left (7+e-\frac {2}{x}\right ) x \log (x)\right ) \]

[Out]

ln((1+x)^2)+ln(x*ln(x)*(exp(1)+7-2/x))

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {6820, 78, 2339, 29} \[ \int \frac {-2+5 x+7 x^2+e \left (x+x^2\right )+\left (3 x+21 x^2+e \left (x+3 x^2\right )\right ) \log (x)}{\left (-2 x+5 x^2+7 x^3+e \left (x^2+x^3\right )\right ) \log (x)} \, dx=2 \log (x+1)+\log (2-(7+e) x)+\log (\log (x)) \]

[In]

Int[(-2 + 5*x + 7*x^2 + E*(x + x^2) + (3*x + 21*x^2 + E*(x + 3*x^2))*Log[x])/((-2*x + 5*x^2 + 7*x^3 + E*(x^2 +
x^3))*Log[x]),x]

[Out]

2*Log[1 + x] + Log[2 - (7 + E)*x] + Log[Log[x]]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 2339

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 6820

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

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {3+e+3 (7+e) x}{(1+x) (-2+(7+e) x)}+\frac {1}{x \log (x)}\right ) \, dx \\ & = \int \frac {3+e+3 (7+e) x}{(1+x) (-2+(7+e) x)} \, dx+\int \frac {1}{x \log (x)} \, dx \\ & = \int \left (\frac {2}{1+x}+\frac {-7-e}{2-(7+e) x}\right ) \, dx+\text {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right ) \\ & = 2 \log (1+x)+\log (2-(7+e) x)+\log (\log (x)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.25 \[ \int \frac {-2+5 x+7 x^2+e \left (x+x^2\right )+\left (3 x+21 x^2+e \left (x+3 x^2\right )\right ) \log (x)}{\left (-2 x+5 x^2+7 x^3+e \left (x^2+x^3\right )\right ) \log (x)} \, dx=2 \log (1+x)+\log (9+e-7 (1+x)-e (1+x))+\log (\log (x)) \]

[In]

Integrate[(-2 + 5*x + 7*x^2 + E*(x + x^2) + (3*x + 21*x^2 + E*(x + 3*x^2))*Log[x])/((-2*x + 5*x^2 + 7*x^3 + E*
(x^2 + x^3))*Log[x]),x]

[Out]

2*Log[1 + x] + Log[9 + E - 7*(1 + x) - E*(1 + x)] + Log[Log[x]]

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05

method	result	size
default	\(\ln \left (\ln \left (x \right )\right )+\ln \left (x \,{\mathrm e}+7 x -2\right )+2 \ln \left (1+x \right )\)	\(21\)
norman	\(\ln \left (\ln \left (x \right )\right )+\ln \left (x \,{\mathrm e}+7 x -2\right )+2 \ln \left (1+x \right )\)	\(21\)
parts	\(\ln \left (\ln \left (x \right )\right )+\ln \left (x \,{\mathrm e}+7 x -2\right )+2 \ln \left (1+x \right )\)	\(21\)
risch	\(\ln \left (x \left (7+{\mathrm e}\right )-2\right )+2 \ln \left (-1-x \right )+\ln \left (\ln \left (x \right )\right )\)	\(22\)
parallelrisch	\(\ln \left (\ln \left (x \right )\right )+\ln \left (\frac {x \,{\mathrm e}+7 x -2}{7+{\mathrm e}}\right )+2 \ln \left (1+x \right )\)	\(28\)

[In]

int((((3*x^2+x)*exp(1)+21*x^2+3*x)*ln(x)+(x^2+x)*exp(1)+7*x^2+5*x-2)/((x^3+x^2)*exp(1)+7*x^3+5*x^2-2*x)/ln(x),
x,method=_RETURNVERBOSE)

[Out]

ln(ln(x))+ln(x*exp(1)+7*x-2)+2*ln(1+x)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {-2+5 x+7 x^2+e \left (x+x^2\right )+\left (3 x+21 x^2+e \left (x+3 x^2\right )\right ) \log (x)}{\left (-2 x+5 x^2+7 x^3+e \left (x^2+x^3\right )\right ) \log (x)} \, dx=\log \left (x e + 7 \, x - 2\right ) + 2 \, \log \left (x + 1\right ) + \log \left (\log \left (x\right )\right ) \]

[In]

integrate((((3*x^2+x)*exp(1)+21*x^2+3*x)*log(x)+(x^2+x)*exp(1)+7*x^2+5*x-2)/((x^3+x^2)*exp(1)+7*x^3+5*x^2-2*x)
/log(x),x, algorithm="fricas")

[Out]

log(x*e + 7*x - 2) + 2*log(x + 1) + log(log(x))

Sympy [A] (verification not implemented)

Time = 0.46 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {-2+5 x+7 x^2+e \left (x+x^2\right )+\left (3 x+21 x^2+e \left (x+3 x^2\right )\right ) \log (x)}{\left (-2 x+5 x^2+7 x^3+e \left (x^2+x^3\right )\right ) \log (x)} \, dx=2 \log {\left (x + 1 \right )} + \log {\left (x - \frac {2}{e + 7} \right )} + \log {\left (\log {\left (x \right )} \right )} \]

[In]

integrate((((3*x**2+x)*exp(1)+21*x**2+3*x)*ln(x)+(x**2+x)*exp(1)+7*x**2+5*x-2)/((x**3+x**2)*exp(1)+7*x**3+5*x*
*2-2*x)/ln(x),x)

[Out]

2*log(x + 1) + log(x - 2/(E + 7)) + log(log(x))

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {-2+5 x+7 x^2+e \left (x+x^2\right )+\left (3 x+21 x^2+e \left (x+3 x^2\right )\right ) \log (x)}{\left (-2 x+5 x^2+7 x^3+e \left (x^2+x^3\right )\right ) \log (x)} \, dx=\log \left (x {\left (e + 7\right )} - 2\right ) + 2 \, \log \left (x + 1\right ) + \log \left (\log \left (x\right )\right ) \]

[In]

integrate((((3*x^2+x)*exp(1)+21*x^2+3*x)*log(x)+(x^2+x)*exp(1)+7*x^2+5*x-2)/((x^3+x^2)*exp(1)+7*x^3+5*x^2-2*x)
/log(x),x, algorithm="maxima")

[Out]

log(x*(e + 7) - 2) + 2*log(x + 1) + log(log(x))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {-2+5 x+7 x^2+e \left (x+x^2\right )+\left (3 x+21 x^2+e \left (x+3 x^2\right )\right ) \log (x)}{\left (-2 x+5 x^2+7 x^3+e \left (x^2+x^3\right )\right ) \log (x)} \, dx=\log \left (x e + 7 \, x - 2\right ) + 2 \, \log \left (x + 1\right ) + \log \left (\log \left (x\right )\right ) \]

[In]

integrate((((3*x^2+x)*exp(1)+21*x^2+3*x)*log(x)+(x^2+x)*exp(1)+7*x^2+5*x-2)/((x^3+x^2)*exp(1)+7*x^3+5*x^2-2*x)
/log(x),x, algorithm="giac")

[Out]

log(x*e + 7*x - 2) + 2*log(x + 1) + log(log(x))

Mupad [B] (verification not implemented)

Time = 15.05 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {-2+5 x+7 x^2+e \left (x+x^2\right )+\left (3 x+21 x^2+e \left (x+3 x^2\right )\right ) \log (x)}{\left (-2 x+5 x^2+7 x^3+e \left (x^2+x^3\right )\right ) \log (x)} \, dx=2\,\ln \left (x+1\right )+\ln \left (\ln \left (x\right )\right )+\ln \left (7\,x+x\,\mathrm {e}-2\right ) \]

[In]

int((5*x + exp(1)*(x + x^2) + log(x)*(3*x + exp(1)*(x + 3*x^2) + 21*x^2) + 7*x^2 - 2)/(log(x)*(exp(1)*(x^2 + x
^3) - 2*x + 5*x^2 + 7*x^3)),x)

[Out]

2*log(x + 1) + log(log(x)) + log(7*x + x*exp(1) - 2)