\(\int \frac {-1+2 x}{\sqrt {13+x^2-2 x^3+x^4}} \, dx\) [267]

Optimal result

Integrand size = 23, antiderivative size = 25 \[ \int \frac {-1+2 x}{\sqrt {13+x^2-2 x^3+x^4}} \, dx=\log \left (-x+x^2+\sqrt {13+x^2-2 x^3+x^4}\right ) \]

[Out]

ln(-x+x^2+(x^4-2*x^3+x^2+13)^(1/2))

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.60, 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, 633, 221} \[ \int \frac {-1+2 x}{\sqrt {13+x^2-2 x^3+x^4}} \, dx=-\text {arcsinh}\left (\frac {(1-x) x}{\sqrt {13}}\right ) \]

[In]

Int[(-1 + 2*x)/Sqrt[13 + x^2 - 2*x^3 + x^4],x]

[Out]

-ArcSinh[((1 - x)*x)/Sqrt[13]]

Rule 12

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

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt[a])]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 633

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*(-4*(c/(b^2 - 4*a*c)))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/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 {209-8 x^2+16 x^4}} \, dx,x,-\frac {1}{2}+x\right ) \\ & = 8 \text {Subst}\left (\int \frac {x}{\sqrt {209-8 x^2+16 x^4}} \, dx,x,-\frac {1}{2}+x\right ) \\ & = 4 \text {Subst}\left (\int \frac {1}{\sqrt {209-8 x+16 x^2}} \, dx,x,\left (-\frac {1}{2}+x\right )^2\right ) \\ & = \frac {\text {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{13312}}} \, dx,x,32 (-1+x) x\right )}{32 \sqrt {13}} \\ & = -\text {arcsinh}\left (\frac {(1-x) x}{\sqrt {13}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {-1+2 x}{\sqrt {13+x^2-2 x^3+x^4}} \, dx=\log \left (-x+x^2+\sqrt {13+x^2-2 x^3+x^4}\right ) \]

[In]

Integrate[(-1 + 2*x)/Sqrt[13 + x^2 - 2*x^3 + x^4],x]

[Out]

Log[-x + x^2 + Sqrt[13 + x^2 - 2*x^3 + x^4]]

Maple [A] (verified)

Time = 1.66 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.44

method	result	size
default	\(\operatorname {arcsinh}\left (\frac {\sqrt {13}\, x \left (x -1\right )}{13}\right )\)	\(11\)
pseudoelliptic	\(\operatorname {arcsinh}\left (\frac {\sqrt {13}\, x \left (x -1\right )}{13}\right )\)	\(11\)
trager	\(-\ln \left (-x^{2}+\sqrt {x^{4}-2 x^{3}+x^{2}+13}+x \right )\)	\(26\)
elliptic	\(\text {Expression too large to display}\)	\(1358\)

[In]

int((-1+2*x)/(x^4-2*x^3+x^2+13)^(1/2),x,method=_RETURNVERBOSE)

[Out]

arcsinh(1/13*13^(1/2)*x*(x-1))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {-1+2 x}{\sqrt {13+x^2-2 x^3+x^4}} \, dx=\log \left (x^{2} - x + \sqrt {x^{4} - 2 \, x^{3} + x^{2} + 13}\right ) \]

[In]

integrate((-1+2*x)/(x^4-2*x^3+x^2+13)^(1/2),x, algorithm="fricas")

[Out]

log(x^2 - x + sqrt(x^4 - 2*x^3 + x^2 + 13))

Sympy [F]

\[ \int \frac {-1+2 x}{\sqrt {13+x^2-2 x^3+x^4}} \, dx=\int \frac {2 x - 1}{\sqrt {x^{4} - 2 x^{3} + x^{2} + 13}}\, dx \]

[In]

integrate((-1+2*x)/(x**4-2*x**3+x**2+13)**(1/2),x)

[Out]

Integral((2*x - 1)/sqrt(x**4 - 2*x**3 + x**2 + 13), x)

Maxima [F]

\[ \int \frac {-1+2 x}{\sqrt {13+x^2-2 x^3+x^4}} \, dx=\int { \frac {2 \, x - 1}{\sqrt {x^{4} - 2 \, x^{3} + x^{2} + 13}} \,d x } \]

[In]

integrate((-1+2*x)/(x^4-2*x^3+x^2+13)^(1/2),x, algorithm="maxima")

[Out]

integrate((2*x - 1)/sqrt(x^4 - 2*x^3 + x^2 + 13), x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.84 \[ \int \frac {-1+2 x}{\sqrt {13+x^2-2 x^3+x^4}} \, dx=\frac {1}{2} \, \sqrt {{\left (x^{2} - x\right )}^{2} + 13} {\left (x^{2} - x\right )} - \frac {13}{2} \, \log \left (-x^{2} + x + \sqrt {{\left (x^{2} - x\right )}^{2} + 13}\right ) \]

[In]

integrate((-1+2*x)/(x^4-2*x^3+x^2+13)^(1/2),x, algorithm="giac")

[Out]

1/2*sqrt((x^2 - x)^2 + 13)*(x^2 - x) - 13/2*log(-x^2 + x + sqrt((x^2 - x)^2 + 13))

Mupad [F(-1)]

Timed out. \[ \int \frac {-1+2 x}{\sqrt {13+x^2-2 x^3+x^4}} \, dx=\int \frac {2\,x-1}{\sqrt {x^4-2\,x^3+x^2+13}} \,d x \]

[In]

int((2*x - 1)/(x^2 - 2*x^3 + x^4 + 13)^(1/2),x)

[Out]

int((2*x - 1)/(x^2 - 2*x^3 + x^4 + 13)^(1/2), x)