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

Optimal. Leaf size=85 \[ \frac {1}{4} \left (\sqrt {2}-2\right ) \tan ^{-1}\left (\frac {\sqrt {3-2 \sqrt {2}} \sqrt {x^3-x}}{x+1}\right )+\frac {1}{4} \left (2+\sqrt {2}\right ) \tan ^{-1}\left (\frac {\sqrt {3+2 \sqrt {2}} \sqrt {x^3-x}}{x+1}\right ) \]

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Rubi [C]  time = 0.56, antiderivative size = 137, normalized size of antiderivative = 1.61, number of steps used = 9, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2056, 6728, 933, 168, 537} \begin {gather*} -\frac {\sqrt {x} \sqrt {1-x^2} \Pi \left (\frac {1}{2+\sqrt {2}};\sin ^{-1}\left (\sqrt {1-x}\right )|\frac {1}{2}\right )}{\sqrt {2} \left (2+\sqrt {2}\right ) \sqrt {x^3-x}}-\frac {\sqrt {x} \sqrt {1-x^2} \Pi \left (\frac {1}{2} \left (2+\sqrt {2}\right );\sin ^{-1}\left (\sqrt {1-x}\right )|\frac {1}{2}\right )}{\sqrt {2} \left (2-\sqrt {2}\right ) \sqrt {x^3-x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[x]*Sqrt[1 - x^2]*EllipticPi[(2 + Sqrt[2])^(-1), ArcSin[Sqrt[1 - x]], 1/2])/(Sqrt[2]*(2 + Sqrt[2])*Sqrt
[-x + x^3])) - (Sqrt[x]*Sqrt[1 - x^2]*EllipticPi[(2 + Sqrt[2])/2, ArcSin[Sqrt[1 - x]], 1/2])/(Sqrt[2]*(2 - Sqr
t[2])*Sqrt[-x + x^3])

Rule 168

Int[1/(((a_.) + (b_.)*(x_))*Sqrt[(c_.) + (d_.)*(x_)]*Sqrt[(e_.) + (f_.)*(x_)]*Sqrt[(g_.) + (h_.)*(x_)]), x_Sym
bol] :> Dist[-2, Subst[Int[1/(Simp[b*c - a*d - b*x^2, x]*Sqrt[Simp[(d*e - c*f)/d + (f*x^2)/d, x]]*Sqrt[Simp[(d
*g - c*h)/d + (h*x^2)/d, x]]), x], x, Sqrt[c + d*x]], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x] && GtQ[(d*e - c
*f)/d, 0]

Rule 537

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[(1*Ellipt
icPi[(b*c)/(a*d), ArcSin[Rt[-(d/c), 2]*x], (c*f)/(d*e)])/(a*Sqrt[c]*Sqrt[e]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b,
c, d, e, f}, x] &&  !GtQ[d/c, 0] && GtQ[c, 0] && GtQ[e, 0] &&  !( !GtQ[f/e, 0] && SimplerSqrtQ[-(f/e), -(d/c)
])

Rule 933

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(f_.) + (g_.)*(x_)]*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> With[{q = Rt[-(c
/a), 2]}, Dist[Sqrt[1 + (c*x^2)/a]/Sqrt[a + c*x^2], Int[1/((d + e*x)*Sqrt[f + g*x]*Sqrt[1 - q*x]*Sqrt[1 + q*x]
), x], x]] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[c*d^2 + a*e^2, 0] &&  !GtQ[a, 0]

Rule 2056

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart
[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] &&  !IntegerQ[p
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

Rule 6728

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {1+x}{\left (-1+2 x+x^2\right ) \sqrt {-x+x^3}} \, dx &=\frac {\left (\sqrt {x} \sqrt {-1+x^2}\right ) \int \frac {1+x}{\sqrt {x} \sqrt {-1+x^2} \left (-1+2 x+x^2\right )} \, dx}{\sqrt {-x+x^3}}\\ &=\frac {\left (\sqrt {x} \sqrt {-1+x^2}\right ) \int \left (\frac {1}{\sqrt {x} \left (2-2 \sqrt {2}+2 x\right ) \sqrt {-1+x^2}}+\frac {1}{\sqrt {x} \left (2+2 \sqrt {2}+2 x\right ) \sqrt {-1+x^2}}\right ) \, dx}{\sqrt {-x+x^3}}\\ &=\frac {\left (\sqrt {x} \sqrt {-1+x^2}\right ) \int \frac {1}{\sqrt {x} \left (2-2 \sqrt {2}+2 x\right ) \sqrt {-1+x^2}} \, dx}{\sqrt {-x+x^3}}+\frac {\left (\sqrt {x} \sqrt {-1+x^2}\right ) \int \frac {1}{\sqrt {x} \left (2+2 \sqrt {2}+2 x\right ) \sqrt {-1+x^2}} \, dx}{\sqrt {-x+x^3}}\\ &=\frac {\left (\sqrt {x} \sqrt {1-x^2}\right ) \int \frac {1}{\sqrt {1-x} \sqrt {x} \sqrt {1+x} \left (2-2 \sqrt {2}+2 x\right )} \, dx}{\sqrt {-x+x^3}}+\frac {\left (\sqrt {x} \sqrt {1-x^2}\right ) \int \frac {1}{\sqrt {1-x} \sqrt {x} \sqrt {1+x} \left (2+2 \sqrt {2}+2 x\right )} \, dx}{\sqrt {-x+x^3}}\\ &=-\frac {\left (2 \sqrt {x} \sqrt {1-x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (2 \left (2-\sqrt {2}\right )-2 x^2\right ) \sqrt {1-x^2} \sqrt {2-x^2}} \, dx,x,\sqrt {1-x}\right )}{\sqrt {-x+x^3}}-\frac {\left (2 \sqrt {x} \sqrt {1-x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (2 \left (2+\sqrt {2}\right )-2 x^2\right ) \sqrt {1-x^2} \sqrt {2-x^2}} \, dx,x,\sqrt {1-x}\right )}{\sqrt {-x+x^3}}\\ &=-\frac {\sqrt {x} \sqrt {1-x^2} \Pi \left (\frac {1}{2+\sqrt {2}};\sin ^{-1}\left (\sqrt {1-x}\right )|\frac {1}{2}\right )}{\sqrt {2} \left (2+\sqrt {2}\right ) \sqrt {-x+x^3}}-\frac {\sqrt {x} \sqrt {1-x^2} \Pi \left (\frac {1}{2} \left (2+\sqrt {2}\right );\sin ^{-1}\left (\sqrt {1-x}\right )|\frac {1}{2}\right )}{\sqrt {2} \left (2-\sqrt {2}\right ) \sqrt {-x+x^3}}\\ \end {align*}

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Mathematica [C]  time = 0.78, size = 89, normalized size = 1.05 \begin {gather*} -\frac {\sqrt {1-\frac {1}{x^2}} x^{3/2} \left (-2 F\left (\left .\sin ^{-1}\left (\frac {1}{\sqrt {x}}\right )\right |-1\right )-\left (\sqrt {2}-1\right ) \Pi \left (-1-\sqrt {2};\left .\sin ^{-1}\left (\frac {1}{\sqrt {x}}\right )\right |-1\right )+\left (1+\sqrt {2}\right ) \Pi \left (-1+\sqrt {2};\left .\sin ^{-1}\left (\frac {1}{\sqrt {x}}\right )\right |-1\right )\right )}{\sqrt {x \left (x^2-1\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[1 - x^(-2)]*x^(3/2)*(-2*EllipticF[ArcSin[1/Sqrt[x]], -1] - (-1 + Sqrt[2])*EllipticPi[-1 - Sqrt[2], Arc
Sin[1/Sqrt[x]], -1] + (1 + Sqrt[2])*EllipticPi[-1 + Sqrt[2], ArcSin[1/Sqrt[x]], -1]))/Sqrt[x*(-1 + x^2)])

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IntegrateAlgebraic [A]  time = 0.61, size = 73, normalized size = 0.86 \begin {gather*} \frac {1}{4} \left (-2+\sqrt {2}\right ) \tan ^{-1}\left (\frac {\left (-1+\sqrt {2}\right ) \sqrt {-x+x^3}}{1+x}\right )+\frac {1}{4} \left (2+\sqrt {2}\right ) \tan ^{-1}\left (\frac {\left (1+\sqrt {2}\right ) \sqrt {-x+x^3}}{1+x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.51, size = 101, normalized size = 1.19 \begin {gather*} \frac {1}{4} \, \sqrt {2} \sqrt {2 \, \sqrt {2} + 3} \arctan \left (\frac {\sqrt {x^{3} - x} \sqrt {2 \, \sqrt {2} + 3} {\left (2 \, \sqrt {2} - 3\right )}}{x^{2} - x}\right ) + \frac {1}{4} \, \sqrt {2} \sqrt {-2 \, \sqrt {2} + 3} \arctan \left (\frac {\sqrt {x^{3} - x} {\left (2 \, \sqrt {2} + 3\right )} \sqrt {-2 \, \sqrt {2} + 3}}{x^{2} - x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x + 1}{\sqrt {x^{3} - x} {\left (x^{2} + 2 \, x - 1\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 1.79, size = 96, normalized size = 1.13

method	result	size
default	\(-\frac {\sqrt {1+x}\, \sqrt {2-2 x}\, \sqrt {-x}\, \sqrt {2}\, \EllipticPi \left (\sqrt {1+x}, \frac {\sqrt {2}}{2}, \frac {\sqrt {2}}{2}\right )}{4 \sqrt {x^{3}-x}}+\frac {\sqrt {1+x}\, \sqrt {2-2 x}\, \sqrt {-x}\, \sqrt {2}\, \EllipticPi \left (\sqrt {1+x}, -\frac {\sqrt {2}}{2}, \frac {\sqrt {2}}{2}\right )}{4 \sqrt {x^{3}-x}}\)	\(96\)
elliptic	\(-\frac {\sqrt {1+x}\, \sqrt {2-2 x}\, \sqrt {-x}\, \sqrt {2}\, \EllipticPi \left (\sqrt {1+x}, \frac {\sqrt {2}}{2}, \frac {\sqrt {2}}{2}\right )}{4 \sqrt {x^{3}-x}}+\frac {\sqrt {1+x}\, \sqrt {2-2 x}\, \sqrt {-x}\, \sqrt {2}\, \EllipticPi \left (\sqrt {1+x}, -\frac {\sqrt {2}}{2}, \frac {\sqrt {2}}{2}\right )}{4 \sqrt {x^{3}-x}}\)	\(96\)
trager	\(4 \ln \left (-\frac {-27452160 \RootOf \left (64 \textit {\_Z}^{4}+48 \textit {\_Z}^{2}+1\right )^{5} x^{2}[phone] \RootOf \left (64 \textit {\_Z}^{4}+48 \textit {\_Z}^{2}+1\right )^{5} x [phone] \RootOf \left (64 \textit {\_Z}^{4}+48 \textit {\_Z}^{2}+1\right )^{5}-37781952 \RootOf \left (64 \textit {\_Z}^{4}+48 \textit {\_Z}^{2}+1\right )^{3} x^{2}[phone] \RootOf \left (64 \textit {\_Z}^{4}+48 \textit {\_Z}^{2}+1\right )^{3} x +2278872 \sqrt {x^{3}-x}\, \RootOf \left (64 \textit {\_Z}^{4}+48 \textit {\_Z}^{2}+1\right )^{2}[phone] \RootOf \left (64 \textit {\_Z}^{4}+48 \textit {\_Z}^{2}+1\right )^{3}-12987008 \RootOf \left (64 \textit {\_Z}^{4}+48 \textit {\_Z}^{2}+1\right ) x^{2}[phone] \RootOf \left (64 \textit {\_Z}^{4}+48 \textit {\_Z}^{2}+1\right ) x +1527689 \sqrt {x^{3}-x}+2435064 \RootOf \left (64 \textit {\_Z}^{4}+48 \textit {\_Z}^{2}+1\right )}{\left (8 x \RootOf \left (64 \textit {\_Z}^{4}+48 \textit {\_Z}^{2}+1\right )^{2}-16 \RootOf \left (64 \textit {\_Z}^{4}+48 \textit {\_Z}^{2}+1\right )^{2}+9 x -4\right )^{2}}\right ) \RootOf \left (64 \textit {\_Z}^{4}+48 \textit {\_Z}^{2}+1\right )^{3}+3 \ln \left (-\frac {-27452160 \RootOf \left (64 \textit {\_Z}^{4}+48 \textit {\_Z}^{2}+1\right )^{5} x^{2}[phone] \RootOf \left (64 \textit {\_Z}^{4}+48 \textit {\_Z}^{2}+1\right )^{5} x [phone] \RootOf \left (64 \textit {\_Z}^{4}+48 \textit {\_Z}^{2}+1\right )^{5}-37781952 \RootOf \left (64 \textit {\_Z}^{4}+48 \textit {\_Z}^{2}+1\right )^{3} x^{2}[phone] \RootOf \left (64 \textit {\_Z}^{4}+48 \textit {\_Z}^{2}+1\right )^{3} x +2278872 \sqrt {x^{3}-x}\, \RootOf \left (64 \textit {\_Z}^{4}+48 \textit {\_Z}^{2}+1\right )^{2}[phone] \RootOf \left (64 \textit {\_Z}^{4}+48 \textit {\_Z}^{2}+1\right )^{3}-12987008 \RootOf \left (64 \textit {\_Z}^{4}+48 \textit {\_Z}^{2}+1\right ) x^{2}[phone] \RootOf \left (64 \textit {\_Z}^{4}+48 \textit {\_Z}^{2}+1\right ) x +1527689 \sqrt {x^{3}-x}+2435064 \RootOf \left (64 \textit {\_Z}^{4}+48 \textit {\_Z}^{2}+1\right )}{\left (8 x \RootOf \left (64 \textit {\_Z}^{4}+48 \textit {\_Z}^{2}+1\right )^{2}-16 \RootOf \left (64 \textit {\_Z}^{4}+48 \textit {\_Z}^{2}+1\right )^{2}+9 x -4\right )^{2}}\right ) \RootOf \left (64 \textit {\_Z}^{4}+48 \textit {\_Z}^{2}+1\right )+\frac {\RootOf \left (64 \textit {\_Z}^{4}+48 \textit {\_Z}^{2}+1\right ) \ln \left (\frac {8868864 \RootOf \left (64 \textit {\_Z}^{4}+48 \textit {\_Z}^{2}+1\right )^{5} x^{2}-14781440 \RootOf \left (64 \textit {\_Z}^{4}+48 \textit {\_Z}^{2}+1\right )^{5} x -5912576 \RootOf \left (64 \textit {\_Z}^{4}+48 \textit {\_Z}^{2}+1\right )^{5}+3959200 \RootOf \left (64 \textit {\_Z}^{4}+48 \textit {\_Z}^{2}+1\right )^{3} x^{2}-9431776 \RootOf \left (64 \textit {\_Z}^{4}+48 \textit {\_Z}^{2}+1\right )^{3} x +2278872 \sqrt {x^{3}-x}\, \RootOf \left (64 \textit {\_Z}^{4}+48 \textit {\_Z}^{2}+1\right )^{2}-5472576 \RootOf \left (64 \textit {\_Z}^{4}+48 \textit {\_Z}^{2}+1\right )^{3}+268344 \RootOf \left (64 \textit {\_Z}^{4}+48 \textit {\_Z}^{2}+1\right ) x^{2}-1475892 \RootOf \left (64 \textit {\_Z}^{4}+48 \textit {\_Z}^{2}+1\right ) x +181465 \sqrt {x^{3}-x}-1207548 \RootOf \left (64 \textit {\_Z}^{4}+48 \textit {\_Z}^{2}+1\right )}{\left (8 x \RootOf \left (64 \textit {\_Z}^{4}+48 \textit {\_Z}^{2}+1\right )^{2}-16 \RootOf \left (64 \textit {\_Z}^{4}+48 \textit {\_Z}^{2}+1\right )^{2}-3 x -8\right )^{2}}\right )}{2}\)	\(774\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x + 1}{\sqrt {x^{3} - x} {\left (x^{2} + 2 \, x - 1\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.11, size = 102, normalized size = 1.20 \begin {gather*} \frac {\sqrt {-x}\,\sqrt {1-x}\,\sqrt {x+1}\,\Pi \left (-\frac {1}{\sqrt {2}-1};\mathrm {asin}\left (\sqrt {-x}\right )\middle |-1\right )}{\sqrt {x^3-x}\,\left (\sqrt {2}-1\right )}-\frac {\sqrt {-x}\,\sqrt {1-x}\,\sqrt {x+1}\,\Pi \left (\frac {1}{\sqrt {2}+1};\mathrm {asin}\left (\sqrt {-x}\right )\middle |-1\right )}{\sqrt {x^3-x}\,\left (\sqrt {2}+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((-x)^(1/2)*(1 - x)^(1/2)*(x + 1)^(1/2)*ellipticPi(-1/(2^(1/2) - 1), asin((-x)^(1/2)), -1))/((x^3 - x)^(1/2)*(
2^(1/2) - 1)) - ((-x)^(1/2)*(1 - x)^(1/2)*(x + 1)^(1/2)*ellipticPi(1/(2^(1/2) + 1), asin((-x)^(1/2)), -1))/((x
^3 - x)^(1/2)*(2^(1/2) + 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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