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

Optimal. Leaf size=65 \[ -\frac {3 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {x^3+x}}{x^2+1}\right )}{2 \sqrt {2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {x^3+x}}{x^2+1}\right )}{2 \sqrt {2}} \]

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Rubi [C]  time = 0.79, antiderivative size = 172, normalized size of antiderivative = 2.65, number of steps used = 15, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {2056, 6725, 329, 220, 932, 168, 538, 537} \begin {gather*} \frac {\sqrt {x} (x+1) \sqrt {\frac {x^2+1}{(x+1)^2}} F\left (2 \tan ^{-1}\left (\sqrt {x}\right )|\frac {1}{2}\right )}{\sqrt {x^3+x}}+\frac {\left (\frac {3}{2}-\frac {3 i}{2}\right ) \sqrt {i x} \sqrt {x^2+1} \Pi \left (\frac {1}{2}-\frac {i}{2};\sin ^{-1}\left (\sqrt {1-i x}\right )|\frac {1}{2}\right )}{\sqrt {2} \sqrt {x^3+x}}-\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {i x} \sqrt {x^2+1} \Pi \left (\frac {1}{2}+\frac {i}{2};\sin ^{-1}\left (\sqrt {1-i x}\right )|\frac {1}{2}\right )}{\sqrt {2} \sqrt {x^3+x}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[x]*(1 + x)*Sqrt[(1 + x^2)/(1 + x)^2]*EllipticF[2*ArcTan[Sqrt[x]], 1/2])/Sqrt[x + x^3] + ((3/2 - (3*I)/2)
*Sqrt[I*x]*Sqrt[1 + x^2]*EllipticPi[1/2 - I/2, ArcSin[Sqrt[1 - I*x]], 1/2])/(Sqrt[2]*Sqrt[x + x^3]) - ((1/2 +
I/2)*Sqrt[I*x]*Sqrt[1 + x^2]*EllipticPi[1/2 + I/2, ArcSin[Sqrt[1 - I*x]], 1/2])/(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 220

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

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 538

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

Rule 932

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(f_.) + (g_.)*(x_)]*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> With[{q = Rt[-(c
/a), 2]}, Dist[1/Sqrt[a], 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 6725

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

Rubi steps

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

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Mathematica [C]  time = 0.55, size = 177, normalized size = 2.72 \begin {gather*} \frac {2 \left (5 \sqrt {\frac {1}{x^2}+1} F_1\left (\frac {3}{4};\frac {1}{2},1;\frac {7}{4};-\frac {1}{x^2},\frac {1}{x^2}\right )-\frac {3 \sqrt {\frac {1}{x^2}+1} F_1\left (\frac {5}{4};\frac {1}{2},1;\frac {9}{4};-\frac {1}{x^2},\frac {1}{x^2}\right )}{x}-\frac {75 x^5 F_1\left (\frac {1}{4};\frac {1}{2},1;\frac {5}{4};-\frac {1}{x^2},\frac {1}{x^2}\right )}{\left (x^2-1\right ) \left (5 x^2 F_1\left (\frac {1}{4};\frac {1}{2},1;\frac {5}{4};-\frac {1}{x^2},\frac {1}{x^2}\right )+4 F_1\left (\frac {5}{4};\frac {1}{2},2;\frac {9}{4};-\frac {1}{x^2},\frac {1}{x^2}\right )-2 F_1\left (\frac {5}{4};\frac {3}{2},1;\frac {9}{4};-\frac {1}{x^2},\frac {1}{x^2}\right )\right )}\right )}{15 \sqrt {x^3+x}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*(5*Sqrt[1 + x^(-2)]*AppellF1[3/4, 1/2, 1, 7/4, -x^(-2), x^(-2)] - (3*Sqrt[1 + x^(-2)]*AppellF1[5/4, 1/2, 1,
9/4, -x^(-2), x^(-2)])/x - (75*x^5*AppellF1[1/4, 1/2, 1, 5/4, -x^(-2), x^(-2)])/((-1 + x^2)*(5*x^2*AppellF1[1
/4, 1/2, 1, 5/4, -x^(-2), x^(-2)] + 4*AppellF1[5/4, 1/2, 2, 9/4, -x^(-2), x^(-2)] - 2*AppellF1[5/4, 3/2, 1, 9/
4, -x^(-2), x^(-2)]))))/(15*Sqrt[x + x^3])

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IntegrateAlgebraic [A]  time = 0.27, size = 65, normalized size = 1.00 \begin {gather*} -\frac {3 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {x+x^3}}{1+x^2}\right )}{2 \sqrt {2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {x+x^3}}{1+x^2}\right )}{2 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.51, size = 92, normalized size = 1.42 \begin {gather*} \frac {3}{8} \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (x^{2} - 2 \, x + 1\right )}}{4 \, \sqrt {x^{3} + x}}\right ) + \frac {1}{16} \, \sqrt {2} \log \left (\frac {x^{4} + 12 \, x^{3} - 4 \, \sqrt {2} \sqrt {x^{3} + x} {\left (x^{2} + 2 \, x + 1\right )} + 6 \, x^{2} + 12 \, x + 1}{x^{4} - 4 \, x^{3} + 6 \, x^{2} - 4 \, x + 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/8*sqrt(2)*arctan(1/4*sqrt(2)*(x^2 - 2*x + 1)/sqrt(x^3 + x)) + 1/16*sqrt(2)*log((x^4 + 12*x^3 - 4*sqrt(2)*sqr
t(x^3 + x)*(x^2 + 2*x + 1) + 6*x^2 + 12*x + 1)/(x^4 - 4*x^3 + 6*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} - x + 1}{\sqrt {x^{3} + x} {\left (x^{2} - 1\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.56, size = 102, normalized size = 1.57

method	result	size
trager	\(\frac {3 \RootOf \left (\textit {\_Z}^{2}+2\right ) \ln \left (\frac {\RootOf \left (\textit {\_Z}^{2}+2\right ) x^{2}-2 \RootOf \left (\textit {\_Z}^{2}+2\right ) x +\RootOf \left (\textit {\_Z}^{2}+2\right )-4 \sqrt {x^{3}+x}}{\left (1+x \right )^{2}}\right )}{8}-\frac {\RootOf \left (\textit {\_Z}^{2}-2\right ) \ln \left (\frac {\RootOf \left (\textit {\_Z}^{2}-2\right ) x^{2}+2 \RootOf \left (\textit {\_Z}^{2}-2\right ) x +4 \sqrt {x^{3}+x}+\RootOf \left (\textit {\_Z}^{2}-2\right )}{\left (-1+x \right )^{2}}\right )}{8}\)	\(102\)
default	\(\frac {i \sqrt {-i \left (i+x \right )}\, \sqrt {2}\, \sqrt {i \left (-i+x \right )}\, \sqrt {i x}\, \EllipticF \left (\sqrt {-i \left (i+x \right )}, \frac {\sqrt {2}}{2}\right )}{\sqrt {x^{3}+x}}+\frac {\left (\frac {3}{4}-\frac {3 i}{4}\right ) \sqrt {-i \left (i+x \right )}\, \sqrt {2}\, \sqrt {i \left (-i+x \right )}\, \sqrt {i x}\, \EllipticPi \left (\sqrt {-i \left (i+x \right )}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )}{\sqrt {x^{3}+x}}+\frac {\left (-\frac {1}{4}-\frac {i}{4}\right ) \sqrt {-i \left (i+x \right )}\, \sqrt {2}\, \sqrt {i \left (-i+x \right )}\, \sqrt {i x}\, \EllipticPi \left (\sqrt {-i \left (i+x \right )}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )}{\sqrt {x^{3}+x}}\)	\(166\)
elliptic	\(\frac {i \sqrt {-i x +1}\, \sqrt {2}\, \sqrt {i x +1}\, \sqrt {i x}\, \EllipticF \left (\sqrt {-i \left (i+x \right )}, \frac {\sqrt {2}}{2}\right )}{\sqrt {x^{3}+x}}+\frac {3 \sqrt {-i x +1}\, \sqrt {2}\, \sqrt {i x +1}\, \sqrt {i x}\, \EllipticPi \left (\sqrt {-i \left (i+x \right )}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )}{4 \sqrt {x^{3}+x}}-\frac {3 i \sqrt {-i x +1}\, \sqrt {2}\, \sqrt {i x +1}\, \sqrt {i x}\, \EllipticPi \left (\sqrt {-i \left (i+x \right )}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )}{4 \sqrt {x^{3}+x}}-\frac {\sqrt {-i x +1}\, \sqrt {2}\, \sqrt {i x +1}\, \sqrt {i x}\, \EllipticPi \left (\sqrt {-i \left (i+x \right )}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )}{4 \sqrt {x^{3}+x}}-\frac {i \sqrt {-i x +1}\, \sqrt {2}\, \sqrt {i x +1}\, \sqrt {i x}\, \EllipticPi \left (\sqrt {-i \left (i+x \right )}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )}{4 \sqrt {x^{3}+x}}\)	\(262\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/8*RootOf(_Z^2+2)*ln((RootOf(_Z^2+2)*x^2-2*RootOf(_Z^2+2)*x+RootOf(_Z^2+2)-4*(x^3+x)^(1/2))/(1+x)^2)-1/8*Root
Of(_Z^2-2)*ln((RootOf(_Z^2-2)*x^2+2*RootOf(_Z^2-2)*x+4*(x^3+x)^(1/2)+RootOf(_Z^2-2))/(-1+x)^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.72, size = 116, normalized size = 1.78 \begin {gather*} -\frac {-\sqrt {1-x\,1{}\mathrm {i}}\,\sqrt {1+x\,1{}\mathrm {i}}\,\sqrt {-x\,1{}\mathrm {i}}\,\mathrm {F}\left (\mathrm {asin}\left (\sqrt {-x\,1{}\mathrm {i}}\right )\middle |-1\right )\,2{}\mathrm {i}+\sqrt {1-x\,1{}\mathrm {i}}\,\sqrt {1+x\,1{}\mathrm {i}}\,\sqrt {-x\,1{}\mathrm {i}}\,\Pi \left (-\mathrm {i};\mathrm {asin}\left (\sqrt {-x\,1{}\mathrm {i}}\right )\middle |-1\right )\,3{}\mathrm {i}+\sqrt {1-x\,1{}\mathrm {i}}\,\sqrt {1+x\,1{}\mathrm {i}}\,\sqrt {-x\,1{}\mathrm {i}}\,\Pi \left (1{}\mathrm {i};\mathrm {asin}\left (\sqrt {-x\,1{}\mathrm {i}}\right )\middle |-1\right )\,1{}\mathrm {i}}{\sqrt {x^3+x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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