3.23.99 \(\int \frac {2+x}{(-3+x) \sqrt [4]{1-x^2} (1+x^2)} \, dx\)

Optimal. Leaf size=176 \[ -\frac {\tan ^{-1}\left (\frac {\sqrt [4]{1-x^2}}{\sqrt [4]{1-x^2}-\sqrt [4]{2} x+\sqrt [4]{2}}\right )}{2 \sqrt [4]{2}}+\frac {\tan ^{-1}\left (\frac {\sqrt [4]{1-x^2}}{\sqrt [4]{1-x^2}+\sqrt [4]{2} x-\sqrt [4]{2}}\right )}{2 \sqrt [4]{2}}-\frac {\tanh ^{-1}\left (\frac {\left (2 \sqrt [4]{2} x-2 \sqrt [4]{2}\right ) \sqrt [4]{1-x^2}}{\sqrt {2} x^2+2 \sqrt {1-x^2}-2 \sqrt {2} x+\sqrt {2}}\right )}{2 \sqrt [4]{2}} \]

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Rubi [C]  time = 0.68, antiderivative size = 378, normalized size of antiderivative = 2.15, number of steps used = 32, number of rules used = 18, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {6725, 746, 399, 490, 1213, 537, 444, 63, 297, 1162, 617, 204, 1165, 628, 1010, 298, 203, 206} \begin {gather*} \frac {\log \left (\sqrt {1-x^2}-2 \sqrt [4]{2} \sqrt [4]{1-x^2}+2 \sqrt {2}\right )}{8 \sqrt [4]{2}}-\frac {\log \left (\sqrt {1-x^2}+2 \sqrt [4]{2} \sqrt [4]{1-x^2}+2 \sqrt {2}\right )}{8 \sqrt [4]{2}}-\frac {\tan ^{-1}\left (\frac {\sqrt [4]{1-x^2}}{\sqrt [4]{2}}\right )}{2 \sqrt [4]{2}}-\frac {\tan ^{-1}\left (1-\frac {\sqrt [4]{1-x^2}}{\sqrt [4]{2}}\right )}{4 \sqrt [4]{2}}+\frac {\tan ^{-1}\left (\frac {\sqrt [4]{1-x^2}}{\sqrt [4]{2}}+1\right )}{4 \sqrt [4]{2}}+\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{1-x^2}}{\sqrt [4]{2}}\right )}{2 \sqrt [4]{2}}-\frac {\sqrt {x^2} \Pi \left (-\frac {1}{\sqrt {2}};\left .\sin ^{-1}\left (\sqrt [4]{1-x^2}\right )\right |-1\right )}{2 \sqrt {2} x}+\frac {3 i \sqrt {x^2} \Pi \left (-\frac {i}{2 \sqrt {2}};\left .\sin ^{-1}\left (\sqrt [4]{1-x^2}\right )\right |-1\right )}{4 \sqrt {2} x}-\frac {3 i \sqrt {x^2} \Pi \left (\frac {i}{2 \sqrt {2}};\left .\sin ^{-1}\left (\sqrt [4]{1-x^2}\right )\right |-1\right )}{4 \sqrt {2} x}+\frac {\sqrt {x^2} \Pi \left (\frac {1}{\sqrt {2}};\left .\sin ^{-1}\left (\sqrt [4]{1-x^2}\right )\right |-1\right )}{2 \sqrt {2} x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 + x)/((-3 + x)*(1 - x^2)^(1/4)*(1 + x^2)),x]

[Out]

-1/2*ArcTan[(1 - x^2)^(1/4)/2^(1/4)]/2^(1/4) - ArcTan[1 - (1 - x^2)^(1/4)/2^(1/4)]/(4*2^(1/4)) + ArcTan[1 + (1
- x^2)^(1/4)/2^(1/4)]/(4*2^(1/4)) + ArcTanh[(1 - x^2)^(1/4)/2^(1/4)]/(2*2^(1/4)) - (Sqrt[x^2]*EllipticPi[-(1/
Sqrt[2]), ArcSin[(1 - x^2)^(1/4)], -1])/(2*Sqrt[2]*x) + (((3*I)/4)*Sqrt[x^2]*EllipticPi[(-1/2*I)/Sqrt[2], ArcS
in[(1 - x^2)^(1/4)], -1])/(Sqrt[2]*x) - (((3*I)/4)*Sqrt[x^2]*EllipticPi[(I/2)/Sqrt[2], ArcSin[(1 - x^2)^(1/4)]
, -1])/(Sqrt[2]*x) + (Sqrt[x^2]*EllipticPi[1/Sqrt[2], ArcSin[(1 - x^2)^(1/4)], -1])/(2*Sqrt[2]*x) + Log[2*Sqrt
[2] - 2*2^(1/4)*(1 - x^2)^(1/4) + Sqrt[1 - x^2]]/(8*2^(1/4)) - Log[2*Sqrt[2] + 2*2^(1/4)*(1 - x^2)^(1/4) + Sqr
t[1 - x^2]]/(8*2^(1/4))

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 297

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

Rule 399

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

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

Rule 490

Int[(x_)^2/(((a_) + (b_.)*(x_)^4)*Sqrt[(c_) + (d_.)*(x_)^4]), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]],
s = Denominator[Rt[-(a/b), 2]]}, Dist[s/(2*b), Int[1/((r + s*x^2)*Sqrt[c + d*x^4]), x], x] - Dist[s/(2*b), In
t[1/((r - s*x^2)*Sqrt[c + d*x^4]), x], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*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 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 746

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

Rule 1010

Int[((g_) + (h_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_)*((d_) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Dist[g, Int[(a + c
*x^2)^p*(d + f*x^2)^q, x], x] + Dist[h, Int[x*(a + c*x^2)^p*(d + f*x^2)^q, x], x] /; FreeQ[{a, c, d, f, g, h,
p, q}, x]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 1213

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[Sqrt[-c],
Int[1/((d + e*x^2)*Sqrt[q + c*x^2]*Sqrt[q - c*x^2]), x], x]] /; FreeQ[{a, c, d, e}, x] && GtQ[a, 0] && LtQ[c,
0]

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 {2+x}{(-3+x) \sqrt [4]{1-x^2} \left (1+x^2\right )} \, dx &=\int \left (\frac {1}{2 (-3+x) \sqrt [4]{1-x^2}}+\frac {-1-x}{2 \sqrt [4]{1-x^2} \left (1+x^2\right )}\right ) \, dx\\ &=\frac {1}{2} \int \frac {1}{(-3+x) \sqrt [4]{1-x^2}} \, dx+\frac {1}{2} \int \frac {-1-x}{\sqrt [4]{1-x^2} \left (1+x^2\right )} \, dx\\ &=-\left (\frac {1}{2} \int \frac {x}{\sqrt [4]{1-x^2} \left (9-x^2\right )} \, dx\right )-\frac {1}{2} \int \frac {1}{\sqrt [4]{1-x^2} \left (1+x^2\right )} \, dx-\frac {1}{2} \int \frac {x}{\sqrt [4]{1-x^2} \left (1+x^2\right )} \, dx-\frac {3}{2} \int \frac {1}{\sqrt [4]{1-x^2} \left (9-x^2\right )} \, dx\\ &=-\left (\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{1-x} (9-x)} \, dx,x,x^2\right )\right )-\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{1-x} (1+x)} \, dx,x,x^2\right )-\frac {\sqrt {x^2} \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1-x^4} \left (-2+x^4\right )} \, dx,x,\sqrt [4]{1-x^2}\right )}{x}-\frac {\left (3 \sqrt {x^2}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\left (-8-x^4\right ) \sqrt {1-x^4}} \, dx,x,\sqrt [4]{1-x^2}\right )}{x}\\ &=\frac {\sqrt {x^2} \operatorname {Subst}\left (\int \frac {1}{\left (\sqrt {2}-x^2\right ) \sqrt {1-x^4}} \, dx,x,\sqrt [4]{1-x^2}\right )}{2 x}-\frac {\sqrt {x^2} \operatorname {Subst}\left (\int \frac {1}{\left (\sqrt {2}+x^2\right ) \sqrt {1-x^4}} \, dx,x,\sqrt [4]{1-x^2}\right )}{2 x}-\frac {\left (3 \sqrt {x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (2 i \sqrt {2}-x^2\right ) \sqrt {1-x^4}} \, dx,x,\sqrt [4]{1-x^2}\right )}{2 x}+\frac {\left (3 \sqrt {x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (2 i \sqrt {2}+x^2\right ) \sqrt {1-x^4}} \, dx,x,\sqrt [4]{1-x^2}\right )}{2 x}+\operatorname {Subst}\left (\int \frac {x^2}{2-x^4} \, dx,x,\sqrt [4]{1-x^2}\right )+\operatorname {Subst}\left (\int \frac {x^2}{8+x^4} \, dx,x,\sqrt [4]{1-x^2}\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{\sqrt {2}-x^2} \, dx,x,\sqrt [4]{1-x^2}\right )-\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{\sqrt {2}+x^2} \, dx,x,\sqrt [4]{1-x^2}\right )-\frac {1}{2} \operatorname {Subst}\left (\int \frac {2 \sqrt {2}-x^2}{8+x^4} \, dx,x,\sqrt [4]{1-x^2}\right )+\frac {1}{2} \operatorname {Subst}\left (\int \frac {2 \sqrt {2}+x^2}{8+x^4} \, dx,x,\sqrt [4]{1-x^2}\right )+\frac {\sqrt {x^2} \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \left (\sqrt {2}-x^2\right ) \sqrt {1+x^2}} \, dx,x,\sqrt [4]{1-x^2}\right )}{2 x}-\frac {\sqrt {x^2} \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {1+x^2} \left (\sqrt {2}+x^2\right )} \, dx,x,\sqrt [4]{1-x^2}\right )}{2 x}-\frac {\left (3 \sqrt {x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \left (2 i \sqrt {2}-x^2\right ) \sqrt {1+x^2}} \, dx,x,\sqrt [4]{1-x^2}\right )}{2 x}+\frac {\left (3 \sqrt {x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {1+x^2} \left (2 i \sqrt {2}+x^2\right )} \, dx,x,\sqrt [4]{1-x^2}\right )}{2 x}\\ &=-\frac {\tan ^{-1}\left (\frac {\sqrt [4]{1-x^2}}{\sqrt [4]{2}}\right )}{2 \sqrt [4]{2}}+\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{1-x^2}}{\sqrt [4]{2}}\right )}{2 \sqrt [4]{2}}-\frac {\sqrt {x^2} \Pi \left (-\frac {1}{\sqrt {2}};\left .\sin ^{-1}\left (\sqrt [4]{1-x^2}\right )\right |-1\right )}{2 \sqrt {2} x}+\frac {3 i \sqrt {x^2} \Pi \left (-\frac {i}{2 \sqrt {2}};\left .\sin ^{-1}\left (\sqrt [4]{1-x^2}\right )\right |-1\right )}{4 \sqrt {2} x}-\frac {3 i \sqrt {x^2} \Pi \left (\frac {i}{2 \sqrt {2}};\left .\sin ^{-1}\left (\sqrt [4]{1-x^2}\right )\right |-1\right )}{4 \sqrt {2} x}+\frac {\sqrt {x^2} \Pi \left (\frac {1}{\sqrt {2}};\left .\sin ^{-1}\left (\sqrt [4]{1-x^2}\right )\right |-1\right )}{2 \sqrt {2} x}+\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{2 \sqrt {2}-2 \sqrt [4]{2} x+x^2} \, dx,x,\sqrt [4]{1-x^2}\right )+\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{2 \sqrt {2}+2 \sqrt [4]{2} x+x^2} \, dx,x,\sqrt [4]{1-x^2}\right )+\frac {\operatorname {Subst}\left (\int \frac {2 \sqrt [4]{2}+2 x}{-2 \sqrt {2}-2 \sqrt [4]{2} x-x^2} \, dx,x,\sqrt [4]{1-x^2}\right )}{8 \sqrt [4]{2}}+\frac {\operatorname {Subst}\left (\int \frac {2 \sqrt [4]{2}-2 x}{-2 \sqrt {2}+2 \sqrt [4]{2} x-x^2} \, dx,x,\sqrt [4]{1-x^2}\right )}{8 \sqrt [4]{2}}\\ &=-\frac {\tan ^{-1}\left (\frac {\sqrt [4]{1-x^2}}{\sqrt [4]{2}}\right )}{2 \sqrt [4]{2}}+\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{1-x^2}}{\sqrt [4]{2}}\right )}{2 \sqrt [4]{2}}-\frac {\sqrt {x^2} \Pi \left (-\frac {1}{\sqrt {2}};\left .\sin ^{-1}\left (\sqrt [4]{1-x^2}\right )\right |-1\right )}{2 \sqrt {2} x}+\frac {3 i \sqrt {x^2} \Pi \left (-\frac {i}{2 \sqrt {2}};\left .\sin ^{-1}\left (\sqrt [4]{1-x^2}\right )\right |-1\right )}{4 \sqrt {2} x}-\frac {3 i \sqrt {x^2} \Pi \left (\frac {i}{2 \sqrt {2}};\left .\sin ^{-1}\left (\sqrt [4]{1-x^2}\right )\right |-1\right )}{4 \sqrt {2} x}+\frac {\sqrt {x^2} \Pi \left (\frac {1}{\sqrt {2}};\left .\sin ^{-1}\left (\sqrt [4]{1-x^2}\right )\right |-1\right )}{2 \sqrt {2} x}+\frac {\log \left (2 \sqrt {2}-2 \sqrt [4]{2} \sqrt [4]{1-x^2}+\sqrt {1-x^2}\right )}{8 \sqrt [4]{2}}-\frac {\log \left (2 \sqrt {2}+2 \sqrt [4]{2} \sqrt [4]{1-x^2}+\sqrt {1-x^2}\right )}{8 \sqrt [4]{2}}+\frac {\operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt [4]{1-x^2}}{\sqrt [4]{2}}\right )}{4 \sqrt [4]{2}}-\frac {\operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt [4]{1-x^2}}{\sqrt [4]{2}}\right )}{4 \sqrt [4]{2}}\\ &=-\frac {\tan ^{-1}\left (\frac {\sqrt [4]{1-x^2}}{\sqrt [4]{2}}\right )}{2 \sqrt [4]{2}}-\frac {\tan ^{-1}\left (1-\frac {\sqrt [4]{1-x^2}}{\sqrt [4]{2}}\right )}{4 \sqrt [4]{2}}+\frac {\tan ^{-1}\left (1+\frac {\sqrt [4]{1-x^2}}{\sqrt [4]{2}}\right )}{4 \sqrt [4]{2}}+\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{1-x^2}}{\sqrt [4]{2}}\right )}{2 \sqrt [4]{2}}-\frac {\sqrt {x^2} \Pi \left (-\frac {1}{\sqrt {2}};\left .\sin ^{-1}\left (\sqrt [4]{1-x^2}\right )\right |-1\right )}{2 \sqrt {2} x}+\frac {3 i \sqrt {x^2} \Pi \left (-\frac {i}{2 \sqrt {2}};\left .\sin ^{-1}\left (\sqrt [4]{1-x^2}\right )\right |-1\right )}{4 \sqrt {2} x}-\frac {3 i \sqrt {x^2} \Pi \left (\frac {i}{2 \sqrt {2}};\left .\sin ^{-1}\left (\sqrt [4]{1-x^2}\right )\right |-1\right )}{4 \sqrt {2} x}+\frac {\sqrt {x^2} \Pi \left (\frac {1}{\sqrt {2}};\left .\sin ^{-1}\left (\sqrt [4]{1-x^2}\right )\right |-1\right )}{2 \sqrt {2} x}+\frac {\log \left (2 \sqrt {2}-2 \sqrt [4]{2} \sqrt [4]{1-x^2}+\sqrt {1-x^2}\right )}{8 \sqrt [4]{2}}-\frac {\log \left (2 \sqrt {2}+2 \sqrt [4]{2} \sqrt [4]{1-x^2}+\sqrt {1-x^2}\right )}{8 \sqrt [4]{2}}\\ \end {align*}

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Mathematica [F]  time = 0.35, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2+x}{(-3+x) \sqrt [4]{1-x^2} \left (1+x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(2 + x)/((-3 + x)*(1 - x^2)^(1/4)*(1 + x^2)),x]

[Out]

Integrate[(2 + x)/((-3 + x)*(1 - x^2)^(1/4)*(1 + x^2)), x]

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IntegrateAlgebraic [A]  time = 0.33, size = 176, normalized size = 1.00 \begin {gather*} -\frac {\tan ^{-1}\left (\frac {\sqrt [4]{1-x^2}}{\sqrt [4]{2}-\sqrt [4]{2} x+\sqrt [4]{1-x^2}}\right )}{2 \sqrt [4]{2}}+\frac {\tan ^{-1}\left (\frac {\sqrt [4]{1-x^2}}{-\sqrt [4]{2}+\sqrt [4]{2} x+\sqrt [4]{1-x^2}}\right )}{2 \sqrt [4]{2}}-\frac {\tanh ^{-1}\left (\frac {\left (-2 \sqrt [4]{2}+2 \sqrt [4]{2} x\right ) \sqrt [4]{1-x^2}}{\sqrt {2}-2 \sqrt {2} x+\sqrt {2} x^2+2 \sqrt {1-x^2}}\right )}{2 \sqrt [4]{2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(2 + x)/((-3 + x)*(1 - x^2)^(1/4)*(1 + x^2)),x]

[Out]

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

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fricas [B]  time = 8.81, size = 986, normalized size = 5.60

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+x)/(-3+x)/(-x^2+1)^(1/4)/(x^2+1),x, algorithm="fricas")

[Out]

-1/16*8^(3/4)*sqrt(2)*arctan(-1/2*(2*x^6 - 12*x^5 + 22*x^4 - 24*x^3 + 8^(3/4)*sqrt(2)*(x^5 - 5*x^4 + 16*x^3 -
16*x^2 - x + 5)*(-x^2 + 1)^(1/4) + 4*8^(1/4)*sqrt(2)*(3*x^3 - 9*x^2 + 11*x - 1)*(-x^2 + 1)^(3/4) + 8*sqrt(2)*(
x^4 - 4*x^3 + 4*x^2 - 4*x + 3)*sqrt(-x^2 + 1) + 38*x^2 - (8^(3/4)*sqrt(2)*(3*x^4 - 12*x^3 + 20*x^2 - 12*x + 1)
*sqrt(-x^2 + 1) + 32*sqrt(2)*(x^3 - 3*x^2 + 3*x - 1)*(-x^2 + 1)^(3/4) + 8^(1/4)*sqrt(2)*(x^6 - 6*x^5 - x^4 + 1
2*x^3 + 11*x^2 - 46*x + 13) + 8*(x^5 - 5*x^4 + 8*x^3 - 8*x^2 + 7*x - 3)*(-x^2 + 1)^(1/4))*sqrt(-(2*8^(1/4)*sqr
t(2)*(x^2 - 2*x + 1)*(-x^2 + 1)^(1/4) + 8^(3/4)*sqrt(2)*(-x^2 + 1)^(3/4) - sqrt(2)*(x^3 - 3*x^2 + x - 3) - 8*s
qrt(-x^2 + 1)*(x - 1))/(x^3 - 3*x^2 + x - 3)) - 12*x + 18)/(x^6 - 6*x^5 + 43*x^4 - 76*x^3 + 19*x^2 + 58*x - 23
)) + 1/16*8^(3/4)*sqrt(2)*arctan(-1/2*(2*x^6 - 12*x^5 + 22*x^4 - 24*x^3 - 8^(3/4)*sqrt(2)*(x^5 - 5*x^4 + 16*x^
3 - 16*x^2 - x + 5)*(-x^2 + 1)^(1/4) - 4*8^(1/4)*sqrt(2)*(3*x^3 - 9*x^2 + 11*x - 1)*(-x^2 + 1)^(3/4) + 8*sqrt(
2)*(x^4 - 4*x^3 + 4*x^2 - 4*x + 3)*sqrt(-x^2 + 1) + 38*x^2 + (8^(3/4)*sqrt(2)*(3*x^4 - 12*x^3 + 20*x^2 - 12*x
+ 1)*sqrt(-x^2 + 1) - 32*sqrt(2)*(x^3 - 3*x^2 + 3*x - 1)*(-x^2 + 1)^(3/4) + 8^(1/4)*sqrt(2)*(x^6 - 6*x^5 - x^4
+ 12*x^3 + 11*x^2 - 46*x + 13) - 8*(x^5 - 5*x^4 + 8*x^3 - 8*x^2 + 7*x - 3)*(-x^2 + 1)^(1/4))*sqrt((2*8^(1/4)*
sqrt(2)*(x^2 - 2*x + 1)*(-x^2 + 1)^(1/4) + 8^(3/4)*sqrt(2)*(-x^2 + 1)^(3/4) + sqrt(2)*(x^3 - 3*x^2 + x - 3) +
8*sqrt(-x^2 + 1)*(x - 1))/(x^3 - 3*x^2 + x - 3)) - 12*x + 18)/(x^6 - 6*x^5 + 43*x^4 - 76*x^3 + 19*x^2 + 58*x -
23)) - 1/64*8^(3/4)*sqrt(2)*log(64*(2*8^(1/4)*sqrt(2)*(x^2 - 2*x + 1)*(-x^2 + 1)^(1/4) + 8^(3/4)*sqrt(2)*(-x^
2 + 1)^(3/4) + sqrt(2)*(x^3 - 3*x^2 + x - 3) + 8*sqrt(-x^2 + 1)*(x - 1))/(x^3 - 3*x^2 + x - 3)) + 1/64*8^(3/4)
*sqrt(2)*log(-64*(2*8^(1/4)*sqrt(2)*(x^2 - 2*x + 1)*(-x^2 + 1)^(1/4) + 8^(3/4)*sqrt(2)*(-x^2 + 1)^(3/4) - sqrt
(2)*(x^3 - 3*x^2 + x - 3) - 8*sqrt(-x^2 + 1)*(x - 1))/(x^3 - 3*x^2 + x - 3))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x + 2}{{\left (x^{2} + 1\right )} {\left (-x^{2} + 1\right )}^{\frac {1}{4}} {\left (x - 3\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+x)/(-3+x)/(-x^2+1)^(1/4)/(x^2+1),x, algorithm="giac")

[Out]

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

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maple [C]  time = 4.88, size = 403, normalized size = 2.29

method	result	size
trager	\(\frac {\RootOf \left (\textit {\_Z}^{4}+2\right ) \ln \left (\frac {2 \sqrt {-x^{2}+1}\, \RootOf \left (\textit {\_Z}^{4}+2\right )^{3} x -2 \sqrt {-x^{2}+1}\, \RootOf \left (\textit {\_Z}^{4}+2\right )^{3}-2 \left (-x^{2}+1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}+2\right )^{2} x^{2}+4 \left (-x^{2}+1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}+2\right )^{2} x +\RootOf \left (\textit {\_Z}^{4}+2\right ) x^{3}+4 \left (-x^{2}+1\right )^{\frac {3}{4}}-2 \left (-x^{2}+1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}+2\right )^{2}-3 \RootOf \left (\textit {\_Z}^{4}+2\right ) x^{2}+5 \RootOf \left (\textit {\_Z}^{4}+2\right ) x +\RootOf \left (\textit {\_Z}^{4}+2\right )}{\left (-3+x \right ) \left (x^{2}+1\right )}\right )}{4}+\frac {\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}+2\right )^{2}\right ) \ln \left (\frac {-2 \sqrt {-x^{2}+1}\, \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}+2\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{4}+2\right )^{2} x +2 \sqrt {-x^{2}+1}\, \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}+2\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{4}+2\right )^{2}+2 \left (-x^{2}+1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}+2\right )^{2} x^{2}-4 \left (-x^{2}+1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}+2\right )^{2} x +\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}+2\right )^{2}\right ) x^{3}+4 \left (-x^{2}+1\right )^{\frac {3}{4}}+2 \left (-x^{2}+1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}+2\right )^{2}-3 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}+2\right )^{2}\right ) x^{2}+5 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}+2\right )^{2}\right ) x +\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}+2\right )^{2}\right )}{\left (-3+x \right ) \left (x^{2}+1\right )}\right )}{4}\)	\(403\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2+x)/(-3+x)/(-x^2+1)^(1/4)/(x^2+1),x,method=_RETURNVERBOSE)

[Out]

1/4*RootOf(_Z^4+2)*ln((2*(-x^2+1)^(1/2)*RootOf(_Z^4+2)^3*x-2*(-x^2+1)^(1/2)*RootOf(_Z^4+2)^3-2*(-x^2+1)^(1/4)*
RootOf(_Z^4+2)^2*x^2+4*(-x^2+1)^(1/4)*RootOf(_Z^4+2)^2*x+RootOf(_Z^4+2)*x^3+4*(-x^2+1)^(3/4)-2*(-x^2+1)^(1/4)*
RootOf(_Z^4+2)^2-3*RootOf(_Z^4+2)*x^2+5*RootOf(_Z^4+2)*x+RootOf(_Z^4+2))/(-3+x)/(x^2+1))+1/4*RootOf(_Z^2+RootO
f(_Z^4+2)^2)*ln((-2*(-x^2+1)^(1/2)*RootOf(_Z^2+RootOf(_Z^4+2)^2)*RootOf(_Z^4+2)^2*x+2*(-x^2+1)^(1/2)*RootOf(_Z
^2+RootOf(_Z^4+2)^2)*RootOf(_Z^4+2)^2+2*(-x^2+1)^(1/4)*RootOf(_Z^4+2)^2*x^2-4*(-x^2+1)^(1/4)*RootOf(_Z^4+2)^2*
x+RootOf(_Z^2+RootOf(_Z^4+2)^2)*x^3+4*(-x^2+1)^(3/4)+2*(-x^2+1)^(1/4)*RootOf(_Z^4+2)^2-3*RootOf(_Z^2+RootOf(_Z
^4+2)^2)*x^2+5*RootOf(_Z^2+RootOf(_Z^4+2)^2)*x+RootOf(_Z^2+RootOf(_Z^4+2)^2))/(-3+x)/(x^2+1))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x + 2}{{\left (x^{2} + 1\right )} {\left (-x^{2} + 1\right )}^{\frac {1}{4}} {\left (x - 3\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+x)/(-3+x)/(-x^2+1)^(1/4)/(x^2+1),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x+2}{{\left (1-x^2\right )}^{1/4}\,\left (x^2+1\right )\,\left (x-3\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x + 2}{\sqrt [4]{- \left (x - 1\right ) \left (x + 1\right )} \left (x - 3\right ) \left (x^{2} + 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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