\(\int \frac {-1+2 x}{\sqrt {-3+x^2-2 x^3+x^4}} \, dx\) [265]

Optimal result

Integrand size = 23, antiderivative size = 25 \[ \int \frac {-1+2 x}{\sqrt {-3+x^2-2 x^3+x^4}} \, dx=\log \left (-x+x^2+\sqrt {-3+x^2-2 x^3+x^4}\right ) \]

[Out]

ln(-x+x^2+(x^4-2*x^3+x^2-3)^(1/2))

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {1694, 12, 1121, 635, 212} \[ \int \frac {-1+2 x}{\sqrt {-3+x^2-2 x^3+x^4}} \, dx=-\text {arctanh}\left (\frac {(1-x) x}{\sqrt {x^4-2 x^3+x^2-3}}\right ) \]

[In]

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

[Out]

-ArcTanh[((1 - x)*x)/Sqrt[-3 + x^2 - 2*x^3 + x^4]]

Rule 12

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

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 635

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

Rule 1121

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

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}& = \text {Subst}\left (\int \frac {8 x}{\sqrt {-47-8 x^2+16 x^4}} \, dx,x,-\frac {1}{2}+x\right ) \\ & = 8 \text {Subst}\left (\int \frac {x}{\sqrt {-47-8 x^2+16 x^4}} \, dx,x,-\frac {1}{2}+x\right ) \\ & = 4 \text {Subst}\left (\int \frac {1}{\sqrt {-47-8 x+16 x^2}} \, dx,x,\left (-\frac {1}{2}+x\right )^2\right ) \\ & = 8 \text {Subst}\left (\int \frac {1}{64-x^2} \, dx,x,\frac {8 (-1+x) x}{\sqrt {-3+x^2-2 x^3+x^4}}\right ) \\ & = -\text {arctanh}\left (\frac {(1-x) x}{\sqrt {-3+x^2-2 x^3+x^4}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {-1+2 x}{\sqrt {-3+x^2-2 x^3+x^4}} \, dx=\log \left (-x+x^2+\sqrt {-3+x^2-2 x^3+x^4}\right ) \]

[In]

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

[Out]

Log[-x + x^2 + Sqrt[-3 + x^2 - 2*x^3 + x^4]]

Maple [A] (verified)

Time = 3.11 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96

method	result	size
default	\(\ln \left (-x +x^{2}+\sqrt {x^{4}-2 x^{3}+x^{2}-3}\right )\)	\(24\)
trager	\(\ln \left (-x +x^{2}+\sqrt {x^{4}-2 x^{3}+x^{2}-3}\right )\)	\(24\)
pseudoelliptic	\(\ln \left (-x +x^{2}+\sqrt {x^{4}-2 x^{3}+x^{2}-3}\right )\)	\(24\)
elliptic	\(\text {Expression too large to display}\)	\(1358\)

[In]

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

[Out]

ln(-x+x^2+(x^4-2*x^3+x^2-3)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {-1+2 x}{\sqrt {-3+x^2-2 x^3+x^4}} \, dx=\log \left (x^{2} - x + \sqrt {x^{4} - 2 \, x^{3} + x^{2} - 3}\right ) \]

[In]

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

[Out]

log(x^2 - x + sqrt(x^4 - 2*x^3 + x^2 - 3))

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 47 vs. \(2 (23) = 46\).

Time = 0.28 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.88 \[ \int \frac {-1+2 x}{\sqrt {-3+x^2-2 x^3+x^4}} \, dx=\frac {1}{2} \, \sqrt {{\left (x^{2} - x\right )}^{2} - 3} {\left (x^{2} - x\right )} + \frac {3}{2} \, \log \left ({\left | -x^{2} + x + \sqrt {{\left (x^{2} - x\right )}^{2} - 3} \right |}\right ) \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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