\(\int \frac {1+x}{(-1+x) \sqrt {1-x^2+x^4}} \, dx\) [349]

Optimal result

Integrand size = 23, antiderivative size = 29 \[ \int \frac {1+x}{(-1+x) \sqrt {1-x^2+x^4}} \, dx=-2 \text {arctanh}\left (\frac {x}{1-2 x+x^2+\sqrt {1-x^2+x^4}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.59, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {1755, 12, 1261, 738, 212, 1712} \[ \int \frac {1+x}{(-1+x) \sqrt {1-x^2+x^4}} \, dx=-\text {arctanh}\left (\frac {x}{\sqrt {x^4-x^2+1}}\right )-\text {arctanh}\left (\frac {x^2+1}{2 \sqrt {x^4-x^2+1}}\right ) \]

[In]

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

[Out]

-ArcTanh[x/Sqrt[1 - x^2 + x^4]] - ArcTanh[(1 + x^2)/(2*Sqrt[1 - x^2 + 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 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 1261

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

Rule 1712

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

Rule 1755

Int[(Px_)/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> With[{A = Coeff[Px, x,
0], B = Coeff[Px, x, 1], C = Coeff[Px, x, 2], D = Coeff[Px, x, 3]}, Int[(x*(B*d - A*e + (d*D - C*e)*x^2))/((d^
2 - e^2*x^2)*Sqrt[a + b*x^2 + c*x^4]), x] + Int[(A*d + (C*d - B*e)*x^2 - D*e*x^4)/((d^2 - e^2*x^2)*Sqrt[a + b*
x^2 + c*x^4]), x]] /; FreeQ[{a, b, c, d, e}, x] && PolyQ[Px, x] && LeQ[Expon[Px, x], 3] && NeQ[c*d^4 + b*d^2*e
^2 + a*e^4, 0]

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

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {1+x}{(-1+x) \sqrt {1-x^2+x^4}} \, dx=-2 \text {arctanh}\left (\frac {x}{1-2 x+x^2+\sqrt {1-x^2+x^4}}\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.89 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93

method	result	size
default	\(-\operatorname {arctanh}\left (\frac {2 x^{2}-3 x +2}{\sqrt {x^{4}-x^{2}+1}}\right )\)	\(27\)
pseudoelliptic	\(-\operatorname {arctanh}\left (\frac {2 x^{2}-3 x +2}{\sqrt {x^{4}-x^{2}+1}}\right )\)	\(27\)
trager	\(\ln \left (-\frac {-2 x^{2}+\sqrt {x^{4}-x^{2}+1}+3 x -2}{\left (x -1\right )^{2}}\right )\)	\(31\)
elliptic	\(-\operatorname {arctanh}\left (\frac {x^{2}+1}{2 \sqrt {\left (x^{2}-1\right )^{2}+x^{2}}}\right )-\operatorname {arctanh}\left (\frac {\sqrt {x^{4}-x^{2}+1}}{x}\right )\)	\(44\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24 \[ \int \frac {1+x}{(-1+x) \sqrt {1-x^2+x^4}} \, dx=\log \left (\frac {2 \, x^{2} - 3 \, x - \sqrt {x^{4} - x^{2} + 1} + 2}{x^{2} - 2 \, x + 1}\right ) \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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