\(\int \frac {1-x}{\sqrt {3+2 x-5 x^2-4 x^3+x^4+2 x^5+x^6}} \, dx\) [953]

Optimal result

Integrand size = 36, antiderivative size = 72 \[ \int \frac {1-x}{\sqrt {3+2 x-5 x^2-4 x^3+x^4+2 x^5+x^6}} \, dx=-\sqrt {2} \text {arctanh}\left (\frac {(1+x) \left (-\sqrt {2}+\sqrt {2} x\right )}{-1-x+x^2+x^3-\sqrt {3+2 x-5 x^2-4 x^3+x^4+2 x^5+x^6}}\right ) \]

[Out]

-2^(1/2)*arctanh((1+x)*(-2^(1/2)+x*2^(1/2))/(-1-x+x^2+x^3-(x^6+2*x^5+x^4-4*x^3-5*x^2+2*x+3)^(1/2)))

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.93, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {6820, 6851, 1047, 738, 212, 702, 213} \[ \int \frac {1-x}{\sqrt {3+2 x-5 x^2-4 x^3+x^4+2 x^5+x^6}} \, dx=-\frac {\left (1-x^2\right ) \sqrt {x^2+2 x+3} \text {arctanh}\left (\frac {\sqrt {x^2+2 x+3}}{\sqrt {2}}\right )}{\sqrt {2} \sqrt {\left (1-x^2\right )^2 \left (x^2+2 x+3\right )}} \]

[In]

Int[(1 - x)/Sqrt[3 + 2*x - 5*x^2 - 4*x^3 + x^4 + 2*x^5 + x^6],x]

[Out]

-(((1 - x^2)*Sqrt[3 + 2*x + x^2]*ArcTanh[Sqrt[3 + 2*x + x^2]/Sqrt[2]])/(Sqrt[2]*Sqrt[(1 - x^2)^2*(3 + 2*x + x^
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 213

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

Rule 702

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

Rule 738

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

Rule 1047

Int[((g_.) + (h_.)*(x_))/(((a_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Symbol] :> With[{q
= Rt[(-a)*c, 2]}, Dist[h/2 + c*(g/(2*q)), Int[1/((-q + c*x)*Sqrt[d + e*x + f*x^2]), x], x] + Dist[h/2 - c*(g/
(2*q)), Int[1/((q + c*x)*Sqrt[d + e*x + f*x^2]), x], x]] /; FreeQ[{a, c, d, e, f, g, h}, x] && NeQ[e^2 - 4*d*f
, 0] && PosQ[(-a)*c]

Rule 6820

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

Rule 6851

Int[(u_.)*((a_.)*(v_)^(m_.)*(w_)^(n_.))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a*v^m*w^n)^FracPart[p]/(v^(m*Fr
acPart[p])*w^(n*FracPart[p]))), Int[u*v^(m*p)*w^(n*p), x], x] /; FreeQ[{a, m, n, p}, x] &&  !IntegerQ[p] &&  !
FreeQ[v, x] &&  !FreeQ[w, x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {1-x}{\sqrt {\left (-1+x^2\right )^2 \left (3+2 x+x^2\right )}} \, dx \\ & = \frac {\left (\left (-1+x^2\right ) \sqrt {3+2 x+x^2}\right ) \int \frac {1-x}{\left (-1+x^2\right ) \sqrt {3+2 x+x^2}} \, dx}{\sqrt {\left (-1+x^2\right )^2 \left (3+2 x+x^2\right )}} \\ & = -\frac {\left (\left (-1+x^2\right ) \sqrt {3+2 x+x^2}\right ) \int \frac {1}{(1+x) \sqrt {3+2 x+x^2}} \, dx}{\sqrt {\left (-1+x^2\right )^2 \left (3+2 x+x^2\right )}} \\ & = -\frac {\left (4 \left (-1+x^2\right ) \sqrt {3+2 x+x^2}\right ) \text {Subst}\left (\int \frac {1}{-8+4 x^2} \, dx,x,\sqrt {3+2 x+x^2}\right )}{\sqrt {\left (-1+x^2\right )^2 \left (3+2 x+x^2\right )}} \\ & = -\frac {\left (1-x^2\right ) \sqrt {3+2 x+x^2} \text {arctanh}\left (\frac {\sqrt {3+2 x+x^2}}{\sqrt {2}}\right )}{\sqrt {2} \sqrt {\left (1-x^2\right )^2 \left (3+2 x+x^2\right )}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.94 \[ \int \frac {1-x}{\sqrt {3+2 x-5 x^2-4 x^3+x^4+2 x^5+x^6}} \, dx=-\frac {\sqrt {2} \left (-1+x^2\right ) \sqrt {3+2 x+x^2} \text {arctanh}\left (\frac {1+x-\sqrt {3+2 x+x^2}}{\sqrt {2}}\right )}{\sqrt {\left (-1+x^2\right )^2 \left (3+2 x+x^2\right )}} \]

[In]

Integrate[(1 - x)/Sqrt[3 + 2*x - 5*x^2 - 4*x^3 + x^4 + 2*x^5 + x^6],x]

[Out]

-((Sqrt[2]*(-1 + x^2)*Sqrt[3 + 2*x + x^2]*ArcTanh[(1 + x - Sqrt[3 + 2*x + x^2])/Sqrt[2]])/Sqrt[(-1 + x^2)^2*(3
+ 2*x + x^2)])

Maple [A] (verified)

Time = 1.91 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.89

method	result	size
default	\(\frac {\left (x^{2}-1\right ) \sqrt {x^{2}+2 x +3}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {2}}{\sqrt {x^{2}+2 x +3}}\right )}{2 \sqrt {x^{6}+2 x^{5}+x^{4}-4 x^{3}-5 x^{2}+2 x +3}}\)	\(64\)
trager	\(\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) x^{2}-\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right )+\sqrt {x^{6}+2 x^{5}+x^{4}-4 x^{3}-5 x^{2}+2 x +3}}{\left (1+x \right )^{2} \left (x -1\right )}\right )}{2}\)	\(69\)

[In]

int((1-x)/(x^6+2*x^5+x^4-4*x^3-5*x^2+2*x+3)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/2/(x^6+2*x^5+x^4-4*x^3-5*x^2+2*x+3)^(1/2)*(x^2-1)*(x^2+2*x+3)^(1/2)*2^(1/2)*arctanh(2^(1/2)/(x^2+2*x+3)^(1/2
))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.81 \[ \int \frac {1-x}{\sqrt {3+2 x-5 x^2-4 x^3+x^4+2 x^5+x^6}} \, dx=\frac {1}{2} \, \sqrt {2} \log \left (\frac {\sqrt {2} {\left (x^{2} - 1\right )} + \sqrt {x^{6} + 2 \, x^{5} + x^{4} - 4 \, x^{3} - 5 \, x^{2} + 2 \, x + 3}}{x^{3} + x^{2} - x - 1}\right ) \]

[In]

integrate((1-x)/(x^6+2*x^5+x^4-4*x^3-5*x^2+2*x+3)^(1/2),x, algorithm="fricas")

[Out]

1/2*sqrt(2)*log((sqrt(2)*(x^2 - 1) + sqrt(x^6 + 2*x^5 + x^4 - 4*x^3 - 5*x^2 + 2*x + 3))/(x^3 + x^2 - x - 1))

Sympy [F]

\[ \int \frac {1-x}{\sqrt {3+2 x-5 x^2-4 x^3+x^4+2 x^5+x^6}} \, dx=- \int \frac {x}{\sqrt {x^{6} + 2 x^{5} + x^{4} - 4 x^{3} - 5 x^{2} + 2 x + 3}}\, dx - \int \left (- \frac {1}{\sqrt {x^{6} + 2 x^{5} + x^{4} - 4 x^{3} - 5 x^{2} + 2 x + 3}}\right )\, dx \]

[In]

integrate((1-x)/(x**6+2*x**5+x**4-4*x**3-5*x**2+2*x+3)**(1/2),x)

[Out]

-Integral(x/sqrt(x**6 + 2*x**5 + x**4 - 4*x**3 - 5*x**2 + 2*x + 3), x) - Integral(-1/sqrt(x**6 + 2*x**5 + x**4
- 4*x**3 - 5*x**2 + 2*x + 3), x)

Maxima [F]

\[ \int \frac {1-x}{\sqrt {3+2 x-5 x^2-4 x^3+x^4+2 x^5+x^6}} \, dx=\int { -\frac {x - 1}{\sqrt {x^{6} + 2 \, x^{5} + x^{4} - 4 \, x^{3} - 5 \, x^{2} + 2 \, x + 3}} \,d x } \]

[In]

integrate((1-x)/(x^6+2*x^5+x^4-4*x^3-5*x^2+2*x+3)^(1/2),x, algorithm="maxima")

[Out]

-integrate((x - 1)/sqrt(x^6 + 2*x^5 + x^4 - 4*x^3 - 5*x^2 + 2*x + 3), x)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.85 \[ \int \frac {1-x}{\sqrt {3+2 x-5 x^2-4 x^3+x^4+2 x^5+x^6}} \, dx=-\frac {\sqrt {2} \log \left (-\frac {{\left | -2 \, x - 2 \, \sqrt {2} + 2 \, \sqrt {x^{2} + 2 \, x + 3} - 2 \right |}}{2 \, {\left (x - \sqrt {2} - \sqrt {x^{2} + 2 \, x + 3} + 1\right )}}\right )}{2 \, \mathrm {sgn}\left (x^{2} - 1\right )} \]

[In]

integrate((1-x)/(x^6+2*x^5+x^4-4*x^3-5*x^2+2*x+3)^(1/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1-x}{\sqrt {3+2 x-5 x^2-4 x^3+x^4+2 x^5+x^6}} \, dx=\int -\frac {x-1}{\sqrt {x^6+2\,x^5+x^4-4\,x^3-5\,x^2+2\,x+3}} \,d x \]

[In]

int(-(x - 1)/(2*x - 5*x^2 - 4*x^3 + x^4 + 2*x^5 + x^6 + 3)^(1/2),x)

[Out]

int(-(x - 1)/(2*x - 5*x^2 - 4*x^3 + x^4 + 2*x^5 + x^6 + 3)^(1/2), x)