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

Optimal. Leaf size=35 \[ -\frac {2 \tanh ^{-1}\left (\frac {\frac {x}{\sqrt {3}}-\frac {1}{\sqrt {3}}}{\sqrt {x^3-x}}\right )}{\sqrt {3}} \]

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Rubi [C]  time = 0.92, antiderivative size = 229, normalized size of antiderivative = 6.54, number of steps used = 13, number of rules used = 7, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.219, Rules used = {2056, 6728, 329, 222, 933, 168, 537} \begin {gather*} \frac {\sqrt {2} \sqrt {x-1} \sqrt {x} \sqrt {x+1} F\left (\sin ^{-1}\left (\frac {\sqrt {2} \sqrt {x}}{\sqrt {x-1}}\right )|\frac {1}{2}\right )}{3 \sqrt {x^3-x}}+\frac {2 \left (1+2 i \sqrt {2}\right ) \sqrt {x} \sqrt {1-x^2} \Pi \left (\frac {3}{4-i \sqrt {2}};\sin ^{-1}\left (\sqrt {1-x}\right )|\frac {1}{2}\right )}{3 \left (\sqrt {2}+4 i\right ) \sqrt {x^3-x}}-\frac {2 \left (1-2 i \sqrt {2}\right ) \sqrt {x} \sqrt {1-x^2} \Pi \left (\frac {3}{4+i \sqrt {2}};\sin ^{-1}\left (\sqrt {1-x}\right )|\frac {1}{2}\right )}{3 \left (-\sqrt {2}+4 i\right ) \sqrt {x^3-x}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[2]*Sqrt[-1 + x]*Sqrt[x]*Sqrt[1 + x]*EllipticF[ArcSin[(Sqrt[2]*Sqrt[x])/Sqrt[-1 + x]], 1/2])/(3*Sqrt[-x +
x^3]) + (2*(1 + (2*I)*Sqrt[2])*Sqrt[x]*Sqrt[1 - x^2]*EllipticPi[3/(4 - I*Sqrt[2]), ArcSin[Sqrt[1 - x]], 1/2])
/(3*(4*I + Sqrt[2])*Sqrt[-x + x^3]) - (2*(1 - (2*I)*Sqrt[2])*Sqrt[x]*Sqrt[1 - x^2]*EllipticPi[3/(4 + I*Sqrt[2]
), ArcSin[Sqrt[1 - x]], 1/2])/(3*(4*I - Sqrt[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 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[-(a*b), 2]}, Simp[(Sqrt[-a + q*x^2]*Sqrt[(a + q*x^2
)/q]*EllipticF[ArcSin[x/Sqrt[(a + q*x^2)/(2*q)]], 1/2])/(Sqrt[2]*Sqrt[-a]*Sqrt[a + b*x^4]), x] /; IntegerQ[q]]
/; FreeQ[{a, b}, x] && LtQ[a, 0] && GtQ[b, 0]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

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 {-1-2 x+x^2}{\left (1+2 x+3 x^2\right ) \sqrt {-x+x^3}} \, dx &=\frac {\left (\sqrt {x} \sqrt {-1+x^2}\right ) \int \frac {-1-2 x+x^2}{\sqrt {x} \sqrt {-1+x^2} \left (1+2 x+3 x^2\right )} \, dx}{\sqrt {-x+x^3}}\\ &=\frac {\left (\sqrt {x} \sqrt {-1+x^2}\right ) \int \left (\frac {1}{3 \sqrt {x} \sqrt {-1+x^2}}-\frac {4 (1+2 x)}{3 \sqrt {x} \sqrt {-1+x^2} \left (1+2 x+3 x^2\right )}\right ) \, dx}{\sqrt {-x+x^3}}\\ &=\frac {\left (\sqrt {x} \sqrt {-1+x^2}\right ) \int \frac {1}{\sqrt {x} \sqrt {-1+x^2}} \, dx}{3 \sqrt {-x+x^3}}-\frac {\left (4 \sqrt {x} \sqrt {-1+x^2}\right ) \int \frac {1+2 x}{\sqrt {x} \sqrt {-1+x^2} \left (1+2 x+3 x^2\right )} \, dx}{3 \sqrt {-x+x^3}}\\ &=\frac {\left (2 \sqrt {x} \sqrt {-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {-1+x^4}} \, dx,x,\sqrt {x}\right )}{3 \sqrt {-x+x^3}}-\frac {\left (4 \sqrt {x} \sqrt {-1+x^2}\right ) \int \left (\frac {2-\frac {i}{\sqrt {2}}}{\sqrt {x} \left (2-2 i \sqrt {2}+6 x\right ) \sqrt {-1+x^2}}+\frac {2+\frac {i}{\sqrt {2}}}{\sqrt {x} \left (2+2 i \sqrt {2}+6 x\right ) \sqrt {-1+x^2}}\right ) \, dx}{3 \sqrt {-x+x^3}}\\ &=\frac {\sqrt {2} \sqrt {-1+x} \sqrt {x} \sqrt {1+x} F\left (\sin ^{-1}\left (\frac {\sqrt {2} \sqrt {x}}{\sqrt {-1+x}}\right )|\frac {1}{2}\right )}{3 \sqrt {-x+x^3}}-\frac {\left (2 \left (4-i \sqrt {2}\right ) \sqrt {x} \sqrt {-1+x^2}\right ) \int \frac {1}{\sqrt {x} \left (2-2 i \sqrt {2}+6 x\right ) \sqrt {-1+x^2}} \, dx}{3 \sqrt {-x+x^3}}-\frac {\left (2 \left (4+i \sqrt {2}\right ) \sqrt {x} \sqrt {-1+x^2}\right ) \int \frac {1}{\sqrt {x} \left (2+2 i \sqrt {2}+6 x\right ) \sqrt {-1+x^2}} \, dx}{3 \sqrt {-x+x^3}}\\ &=\frac {\sqrt {2} \sqrt {-1+x} \sqrt {x} \sqrt {1+x} F\left (\sin ^{-1}\left (\frac {\sqrt {2} \sqrt {x}}{\sqrt {-1+x}}\right )|\frac {1}{2}\right )}{3 \sqrt {-x+x^3}}-\frac {\left (2 \left (4-i \sqrt {2}\right ) \sqrt {x} \sqrt {1-x^2}\right ) \int \frac {1}{\sqrt {1-x} \sqrt {x} \sqrt {1+x} \left (2-2 i \sqrt {2}+6 x\right )} \, dx}{3 \sqrt {-x+x^3}}-\frac {\left (2 \left (4+i \sqrt {2}\right ) \sqrt {x} \sqrt {1-x^2}\right ) \int \frac {1}{\sqrt {1-x} \sqrt {x} \sqrt {1+x} \left (2+2 i \sqrt {2}+6 x\right )} \, dx}{3 \sqrt {-x+x^3}}\\ &=\frac {\sqrt {2} \sqrt {-1+x} \sqrt {x} \sqrt {1+x} F\left (\sin ^{-1}\left (\frac {\sqrt {2} \sqrt {x}}{\sqrt {-1+x}}\right )|\frac {1}{2}\right )}{3 \sqrt {-x+x^3}}+\frac {\left (4 \left (4-i \sqrt {2}\right ) \sqrt {x} \sqrt {1-x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (2 \left (4-i \sqrt {2}\right )-6 x^2\right ) \sqrt {1-x^2} \sqrt {2-x^2}} \, dx,x,\sqrt {1-x}\right )}{3 \sqrt {-x+x^3}}+\frac {\left (4 \left (4+i \sqrt {2}\right ) \sqrt {x} \sqrt {1-x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (2 \left (4+i \sqrt {2}\right )-6 x^2\right ) \sqrt {1-x^2} \sqrt {2-x^2}} \, dx,x,\sqrt {1-x}\right )}{3 \sqrt {-x+x^3}}\\ &=\frac {\sqrt {2} \sqrt {-1+x} \sqrt {x} \sqrt {1+x} F\left (\sin ^{-1}\left (\frac {\sqrt {2} \sqrt {x}}{\sqrt {-1+x}}\right )|\frac {1}{2}\right )}{3 \sqrt {-x+x^3}}+\frac {\sqrt {2} \sqrt {x} \sqrt {1-x^2} \Pi \left (\frac {3}{4-i \sqrt {2}};\sin ^{-1}\left (\sqrt {1-x}\right )|\frac {1}{2}\right )}{3 \sqrt {-x+x^3}}+\frac {\sqrt {2} \sqrt {x} \sqrt {1-x^2} \Pi \left (\frac {3}{4+i \sqrt {2}};\sin ^{-1}\left (\sqrt {1-x}\right )|\frac {1}{2}\right )}{3 \sqrt {-x+x^3}}\\ \end {align*}

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Mathematica [C]  time = 1.04, size = 112, normalized size = 3.20 \begin {gather*} -\frac {2 \sqrt {1-\frac {1}{x^2}} x^{3/2} \left (-3 F\left (\left .\sin ^{-1}\left (\frac {1}{\sqrt {x}}\right )\right |-1\right )+\left (2+i \sqrt {2}\right ) \Pi \left (\frac {i}{-i+\sqrt {2}};\left .\sin ^{-1}\left (\frac {1}{\sqrt {x}}\right )\right |-1\right )+\left (2-i \sqrt {2}\right ) \Pi \left (-\frac {i}{i+\sqrt {2}};\left .\sin ^{-1}\left (\frac {1}{\sqrt {x}}\right )\right |-1\right )\right )}{3 \sqrt {x \left (x^2-1\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.27, size = 35, normalized size = 1.00 \begin {gather*} -\frac {2 \tanh ^{-1}\left (\frac {-\frac {1}{\sqrt {3}}+\frac {x}{\sqrt {3}}}{\sqrt {-x+x^3}}\right )}{\sqrt {3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.48, size = 73, normalized size = 2.09 \begin {gather*} \frac {1}{6} \, \sqrt {3} \log \left (\frac {9 \, x^{4} + 36 \, x^{3} - 4 \, \sqrt {3} \sqrt {x^{3} - x} {\left (3 \, x^{2} + 4 \, x - 1\right )} + 10 \, x^{2} - 20 \, x + 1}{9 \, x^{4} + 12 \, x^{3} + 10 \, x^{2} + 4 \, x + 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*sqrt(3)*log((9*x^4 + 36*x^3 - 4*sqrt(3)*sqrt(x^3 - x)*(3*x^2 + 4*x - 1) + 10*x^2 - 20*x + 1)/(9*x^4 + 12*x
^3 + 10*x^2 + 4*x + 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.57, size = 61, normalized size = 1.74

method	result	size
trager	\(\frac {\RootOf \left (\textit {\_Z}^{2}-3\right ) \ln \left (\frac {-3 \RootOf \left (\textit {\_Z}^{2}-3\right ) x^{2}-4 \RootOf \left (\textit {\_Z}^{2}-3\right ) x +6 \sqrt {x^{3}-x}+\RootOf \left (\textit {\_Z}^{2}-3\right )}{3 x^{2}+2 x +1}\right )}{3}\)	\(61\)
elliptic	\(\frac {\sqrt {1+x}\, \sqrt {2-2 x}\, \sqrt {-x}\, \EllipticF \left (\sqrt {1+x}, \frac {\sqrt {2}}{2}\right )}{3 \sqrt {x^{3}-x}}+\frac {\left (-\frac {4}{9}+\frac {i \sqrt {2}}{9}\right ) \sqrt {1+x}\, \sqrt {2-2 x}\, \sqrt {-x}\, \left (-1+\frac {i \sqrt {2}}{2}\right ) \EllipticPi \left (\sqrt {1+x}, 1-\frac {i \sqrt {2}}{2}, \frac {\sqrt {2}}{2}\right )}{\sqrt {x^{3}-x}}+\frac {\left (-\frac {4}{9}-\frac {i \sqrt {2}}{9}\right ) \sqrt {1+x}\, \sqrt {2-2 x}\, \sqrt {-x}\, \left (-1-\frac {i \sqrt {2}}{2}\right ) \EllipticPi \left (\sqrt {1+x}, 1+\frac {i \sqrt {2}}{2}, \frac {\sqrt {2}}{2}\right )}{\sqrt {x^{3}-x}}\)	\(165\)
default	\(\frac {\sqrt {1+x}\, \sqrt {2-2 x}\, \sqrt {-x}\, \EllipticF \left (\sqrt {1+x}, \frac {\sqrt {2}}{2}\right )}{3 \sqrt {x^{3}-x}}-\frac {4 \left (\frac {1}{3}-\frac {i \sqrt {2}}{12}\right ) \sqrt {1+x}\, \sqrt {2-2 x}\, \sqrt {-x}\, \left (-1+\frac {i \sqrt {2}}{2}\right ) \EllipticPi \left (\sqrt {1+x}, 1-\frac {i \sqrt {2}}{2}, \frac {\sqrt {2}}{2}\right )}{3 \sqrt {x^{3}-x}}-\frac {4 \left (\frac {1}{3}+\frac {i \sqrt {2}}{12}\right ) \sqrt {1+x}\, \sqrt {2-2 x}\, \sqrt {-x}\, \left (-1-\frac {i \sqrt {2}}{2}\right ) \EllipticPi \left (\sqrt {1+x}, 1+\frac {i \sqrt {2}}{2}, \frac {\sqrt {2}}{2}\right )}{3 \sqrt {x^{3}-x}}\)	\(167\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*RootOf(_Z^2-3)*ln((-3*RootOf(_Z^2-3)*x^2-4*RootOf(_Z^2-3)*x+6*(x^3-x)^(1/2)+RootOf(_Z^2-3))/(3*x^2+2*x+1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.22, size = 175, normalized size = 5.00 \begin {gather*} -\frac {2\,\sqrt {-x}\,\sqrt {1-x}\,\sqrt {x+1}\,\mathrm {F}\left (\mathrm {asin}\left (\sqrt {-x}\right )\middle |-1\right )}{3\,\sqrt {x^3-x}}-\frac {\sqrt {2}\,\sqrt {-x}\,\left (-\frac {4}{9}+\frac {\sqrt {2}\,8{}\mathrm {i}}{9}\right )\,\sqrt {1-x}\,\sqrt {x+1}\,\Pi \left (\frac {1}{\frac {1}{3}+\frac {\sqrt {2}\,1{}\mathrm {i}}{3}};\mathrm {asin}\left (\sqrt {-x}\right )\middle |-1\right )\,1{}\mathrm {i}}{2\,\sqrt {x^3-x}\,\left (\frac {1}{3}+\frac {\sqrt {2}\,1{}\mathrm {i}}{3}\right )}+\frac {\sqrt {2}\,\sqrt {-x}\,\left (\frac {4}{9}+\frac {\sqrt {2}\,8{}\mathrm {i}}{9}\right )\,\sqrt {1-x}\,\sqrt {x+1}\,\Pi \left (-\frac {1}{-\frac {1}{3}+\frac {\sqrt {2}\,1{}\mathrm {i}}{3}};\mathrm {asin}\left (\sqrt {-x}\right )\middle |-1\right )\,1{}\mathrm {i}}{2\,\sqrt {x^3-x}\,\left (-\frac {1}{3}+\frac {\sqrt {2}\,1{}\mathrm {i}}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2^(1/2)*(-x)^(1/2)*((2^(1/2)*8i)/9 + 4/9)*(1 - x)^(1/2)*(x + 1)^(1/2)*ellipticPi(-1/((2^(1/2)*1i)/3 - 1/3), a
sin((-x)^(1/2)), -1)*1i)/(2*(x^3 - x)^(1/2)*((2^(1/2)*1i)/3 - 1/3)) - (2^(1/2)*(-x)^(1/2)*((2^(1/2)*8i)/9 - 4/
9)*(1 - x)^(1/2)*(x + 1)^(1/2)*ellipticPi(1/((2^(1/2)*1i)/3 + 1/3), asin((-x)^(1/2)), -1)*1i)/(2*(x^3 - x)^(1/
2)*((2^(1/2)*1i)/3 + 1/3)) - (2*(-x)^(1/2)*(1 - x)^(1/2)*(x + 1)^(1/2)*ellipticF(asin((-x)^(1/2)), -1))/(3*(x^
3 - x)^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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