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

Optimal. Leaf size=29 \[ -2 \tanh ^{-1}\left (\frac {x}{1-2 x+x^2+\sqrt {1-x^2+x^4}}\right ) \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 46, normalized size of antiderivative = 1.59, number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {1755, 12, 1261, 738, 212, 1712} \begin {gather*} -\tanh ^{-1}\left (\frac {x}{\sqrt {x^4-x^2+1}}\right )-\tanh ^{-1}\left (\frac {x^2+1}{2 \sqrt {x^4-x^2+1}}\right ) \end {gather*}

Antiderivative was successfully verified.

[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*} \int \frac {1+x}{(-1+x) \sqrt {1-x^2+x^4}} \, dx &=\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 )\\ &=-\tanh ^{-1}\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 )\\ &=-\tanh ^{-1}\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 )\\ &=-\tanh ^{-1}\left (\frac {x}{\sqrt {1-x^2+x^4}}\right )-\tanh ^{-1}\left (\frac {1+x^2}{2 \sqrt {1-x^2+x^4}}\right )\\ \end {align*}

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Mathematica [A]
time = 0.24, size = 29, normalized size = 1.00 \begin {gather*} -2 \tanh ^{-1}\left (\frac {x}{1-2 x+x^2+\sqrt {1-x^2+x^4}}\right ) \end {gather*}

Antiderivative was successfully verified.

[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])]

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 1.20, size = 210, normalized size = 7.24

method	result	size
trager	\(\ln \left (-\frac {-2 x^{2}+\sqrt {x^{4}-x^{2}+1}+3 x -2}{\left (-1+x \right )^{2}}\right )\)	\(31\)
elliptic	\(-\arctanh \left (\frac {x^{2}+1}{2 \sqrt {\left (x^{2}-1\right )^{2}+x^{2}}}\right )-\arctanh \left (\frac {\sqrt {x^{4}-x^{2}+1}}{x}\right )\)	\(44\)
default	\(\frac {2 \sqrt {1-\left (\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) x^{2}}\, \sqrt {1-\left (\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) x^{2}}\, \EllipticF \left (\frac {x \sqrt {2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2-2 i \sqrt {3}}}{2}\right )}{\sqrt {2+2 i \sqrt {3}}\, \sqrt {x^{4}-x^{2}+1}}-\arctanh \left (\frac {x^{2}+1}{2 \sqrt {x^{4}-x^{2}+1}}\right )-\frac {2 \sqrt {1-\left (\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) x^{2}}\, \sqrt {1-\left (\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) x^{2}}\, \EllipticPi \left (\sqrt {\frac {1}{2}+\frac {i \sqrt {3}}{2}}\, x , \frac {1}{\frac {1}{2}+\frac {i \sqrt {3}}{2}}, \frac {\sqrt {\frac {1}{2}-\frac {i \sqrt {3}}{2}}}{\sqrt {\frac {1}{2}+\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {\frac {1}{2}+\frac {i \sqrt {3}}{2}}\, \sqrt {x^{4}-x^{2}+1}}\)	\(210\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/(2+2*I*3^(1/2))^(1/2)*(1-(1/2+1/2*I*3^(1/2))*x^2)^(1/2)*(1-(1/2-1/2*I*3^(1/2))*x^2)^(1/2)/(x^4-x^2+1)^(1/2)*
EllipticF(1/2*x*(2+2*I*3^(1/2))^(1/2),1/2*(-2-2*I*3^(1/2))^(1/2))-arctanh(1/2*(x^2+1)/(x^4-x^2+1)^(1/2))-2/(1/
2+1/2*I*3^(1/2))^(1/2)*(1-(1/2+1/2*I*3^(1/2))*x^2)^(1/2)*(1-(1/2-1/2*I*3^(1/2))*x^2)^(1/2)/(x^4-x^2+1)^(1/2)*E
llipticPi((1/2+1/2*I*3^(1/2))^(1/2)*x,1/(1/2+1/2*I*3^(1/2)),(1/2-1/2*I*3^(1/2))^(1/2)/(1/2+1/2*I*3^(1/2))^(1/2
))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]
time = 0.40, size = 36, normalized size = 1.24 \begin {gather*} \log \left (\frac {2 \, x^{2} - 3 \, x - \sqrt {x^{4} - x^{2} + 1} + 2}{x^{2} - 2 \, x + 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x + 1}{\left (x - 1\right ) \sqrt {x^{4} - x^{2} + 1}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {x+1}{\left (x-1\right )\,\sqrt {x^4-x^2+1}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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