3.3.66 \(\int \frac {-1+2 x}{\sqrt {4+x^2-2 x^3+x^4}} \, dx\) [266]

3.3.66.1 Optimal result

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

output

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

3.3.66.2 Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {-1+2 x}{\sqrt {4+x^2-2 x^3+x^4}} \, dx=\log \left (-x+x^2+\sqrt {4+x^2-2 x^3+x^4}\right ) \]

input

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

output

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

3.3.66.3 Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.64, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {2459, 27, 27, 1432, 1090, 222}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 2459

\(\displaystyle \int \frac {2 \left (x-\frac {1}{2}\right )}{\sqrt {\left (x-\frac {1}{2}\right )^4-\frac {1}{2} \left (x-\frac {1}{2}\right )^2+\frac {65}{16}}}d\left (x-\frac {1}{2}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle 2 \int \frac {4 \left (x-\frac {1}{2}\right )}{\sqrt {16 \left (x-\frac {1}{2}\right )^4-8 \left (x-\frac {1}{2}\right )^2+65}}d\left (x-\frac {1}{2}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle 8 \int \frac {x-\frac {1}{2}}{\sqrt {16 \left (x-\frac {1}{2}\right )^4-8 \left (x-\frac {1}{2}\right )^2+65}}d\left (x-\frac {1}{2}\right )\)

\(\Big \downarrow \) 1432

\(\displaystyle 4 \int \frac {1}{\sqrt {16 \left (x-\frac {1}{2}\right )^4-8 \left (x-\frac {1}{2}\right )^2+65}}d\left (x-\frac {1}{2}\right )^2\)

\(\Big \downarrow \) 1090

\(\displaystyle \frac {1}{64} \int \frac {1}{\sqrt {\frac {\left (x-\frac {1}{2}\right )^4}{4096}+1}}d\left (32 \left (x-\frac {1}{2}\right )^2-8\right )\)

\(\Big \downarrow \) 222

\(\displaystyle \text {arcsinh}\left (\frac {1}{64} \left (32 \left (x-\frac {1}{2}\right )^2-8\right )\right )\)

input

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

output

ArcSinh[(-8 + 32*(-1/2 + x)^2)/64]
 

3.3.66.3.1 Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 222

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 1090

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/(2*c*(-4* 
(c/(b^2 - 4*a*c)))^p)   Subst[Int[Simp[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 1432

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

rule 2459

Int[(Pn_)^(p_.)*(Qx_), x_Symbol] :> With[{S = Coeff[Pn, x, Expon[Pn, x] - 1 
]/(Expon[Pn, x]*Coeff[Pn, x, Expon[Pn, x]])}, Subst[Int[ExpandToSum[Pn /. x 
 -> x - S, x]^p*ExpandToSum[Qx /. x -> x - S, x], x], x, x + S] /; Binomial 
Q[Pn /. x -> x - S, x] || (IntegerQ[Expon[Pn, x]/2] && TrinomialQ[Pn /. x - 
> x - S, x])] /; FreeQ[p, x] && PolyQ[Pn, x] && GtQ[Expon[Pn, x], 2] && NeQ 
[Coeff[Pn, x, Expon[Pn, x] - 1], 0] && PolyQ[Qx, x] &&  !(MonomialQ[Qx, x] 
&& IGtQ[p, 0])
 

3.3.66.4 Maple [A] (verified)

Time = 1.43 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.32

method	result	size
default	\(\operatorname {arcsinh}\left (\frac {x \left (x -1\right )}{2}\right )\)	\(8\)
pseudoelliptic	\(\operatorname {arcsinh}\left (\frac {x \left (x -1\right )}{2}\right )\)	\(8\)
trager	\(\ln \left (-x +x^{2}+\sqrt {x^{4}-2 x^{3}+x^{2}+4}\right )\)	\(24\)
elliptic	\(\text {Expression too large to display}\)	\(888\)

input

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

output

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

3.3.66.5 Fricas [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {-1+2 x}{\sqrt {4+x^2-2 x^3+x^4}} \, dx=\log \left (x^{2} - x + \sqrt {x^{4} - 2 \, x^{3} + x^{2} + 4}\right ) \]

input

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

output

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

3.3.66.6 Sympy [F]

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

input

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

output

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

3.3.66.7 Maxima [F]

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

input

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

output

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

3.3.66.8 Giac [A] (verification not implemented)

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

input

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

output

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

3.3.66.9 Mupad [F(-1)]

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

input

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

output

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