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

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

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Rubi [B]  time = 1.12, antiderivative size = 262, normalized size of antiderivative = 9.03, number of steps used = 42, number of rules used = 16, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.615, Rules used = {6742, 749, 748, 746, 399, 490, 1213, 537, 444, 63, 297, 1162, 617, 204, 1165, 628} \begin {gather*} -\frac {\sqrt [4]{-x^2+2 x+2} \log \left (\sqrt {-x^2+2 x+2}+\sqrt {2} \sqrt [4]{-x^2+2 x+2}+1\right )}{2 \sqrt {2} \sqrt [4]{x^2-2 x-2}}+\frac {\sqrt [4]{-x^2+2 x+2} \log \left (3 \sqrt {-x^2+2 x+2}-3 \sqrt {2} \sqrt [4]{-x^2+2 x+2}+3\right )}{2 \sqrt {2} \sqrt [4]{x^2-2 x-2}}-\frac {\sqrt [4]{-x^2+2 x+2} \tan ^{-1}\left (1-\sqrt {2} \sqrt [4]{3-(1-x)^2}\right )}{\sqrt {2} \sqrt [4]{x^2-2 x-2}}+\frac {\sqrt [4]{-x^2+2 x+2} \tan ^{-1}\left (\sqrt {2} \sqrt [4]{3-(1-x)^2}+1\right )}{\sqrt {2} \sqrt [4]{x^2-2 x-2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(((2 + 2*x - x^2)^(1/4)*ArcTan[1 - Sqrt[2]*(3 - (1 - x)^2)^(1/4)])/(Sqrt[2]*(-2 - 2*x + x^2)^(1/4))) + ((2 +
2*x - x^2)^(1/4)*ArcTan[1 + Sqrt[2]*(3 - (1 - x)^2)^(1/4)])/(Sqrt[2]*(-2 - 2*x + x^2)^(1/4)) - ((2 + 2*x - x^2
)^(1/4)*Log[1 + Sqrt[2]*(2 + 2*x - x^2)^(1/4) + Sqrt[2 + 2*x - x^2]])/(2*Sqrt[2]*(-2 - 2*x + x^2)^(1/4)) + ((2
+ 2*x - x^2)^(1/4)*Log[3 - 3*Sqrt[2]*(2 + 2*x - x^2)^(1/4) + 3*Sqrt[2 + 2*x - x^2]])/(2*Sqrt[2]*(-2 - 2*x + x
^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 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 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 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 748

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)/((d_.) + (e_.)*(x_)), x_Symbol] :> Dist[1/((-4*c)/(b^2 - 4*a*c))^
p, Subst[Int[Simp[1 - x^2/(b^2 - 4*a*c), x]^p/Simp[2*c*d - b*e + e*x, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b
, c, d, e, p}, x] && GtQ[4*a - b^2/c, 0] && IntegerQ[4*p]

Rule 749

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)/((d_.) + (e_.)*(x_)), x_Symbol] :> Dist[(a + b*x + c*x^2)^p/(-((c
*(a + b*x + c*x^2))/(b^2 - 4*a*c)))^p, Int[(-((a*c)/(b^2 - 4*a*c)) - (b*c*x)/(b^2 - 4*a*c) - (c^2*x^2)/(b^2 -
4*a*c))^p/(d + e*x), x], x] /; FreeQ[{a, b, c, d, e, p}, x] &&  !GtQ[4*a - b^2/c, 0] && IntegerQ[4*p]

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 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {align*} \int \frac {-1+x}{(-3+x) (1+x) \sqrt [4]{-2-2 x+x^2}} \, dx &=\int \left (\frac {1}{2 (-3+x) \sqrt [4]{-2-2 x+x^2}}+\frac {1}{2 (1+x) \sqrt [4]{-2-2 x+x^2}}\right ) \, dx\\ &=\frac {1}{2} \int \frac {1}{(-3+x) \sqrt [4]{-2-2 x+x^2}} \, dx+\frac {1}{2} \int \frac {1}{(1+x) \sqrt [4]{-2-2 x+x^2}} \, dx\\ &=\frac {\sqrt [4]{2+2 x-x^2} \int \frac {1}{(-3+x) \sqrt [4]{\frac {1}{6}+\frac {x}{6}-\frac {x^2}{12}}} \, dx}{2 \sqrt {2} \sqrt [4]{3} \sqrt [4]{-2-2 x+x^2}}+\frac {\sqrt [4]{2+2 x-x^2} \int \frac {1}{(1+x) \sqrt [4]{\frac {1}{6}+\frac {x}{6}-\frac {x^2}{12}}} \, dx}{2 \sqrt {2} \sqrt [4]{3} \sqrt [4]{-2-2 x+x^2}}\\ &=\frac {\sqrt [4]{2+2 x-x^2} \operatorname {Subst}\left (\int \frac {1}{\left (-\frac {1}{3}+x\right ) \sqrt [4]{1-12 x^2}} \, dx,x,\frac {1}{6}-\frac {x}{6}\right )}{2 \sqrt [4]{3} \sqrt [4]{-2-2 x+x^2}}+\frac {\sqrt [4]{2+2 x-x^2} \operatorname {Subst}\left (\int \frac {1}{\left (\frac {1}{3}+x\right ) \sqrt [4]{1-12 x^2}} \, dx,x,\frac {1}{6}-\frac {x}{6}\right )}{2 \sqrt [4]{3} \sqrt [4]{-2-2 x+x^2}}\\ &=-2 \frac {\sqrt [4]{2+2 x-x^2} \operatorname {Subst}\left (\int \frac {x}{\sqrt [4]{1-12 x^2} \left (\frac {1}{9}-x^2\right )} \, dx,x,\frac {1}{6}-\frac {x}{6}\right )}{2 \sqrt [4]{3} \sqrt [4]{-2-2 x+x^2}}\\ &=-2 \frac {\sqrt [4]{2+2 x-x^2} \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{1-12 x} \left (\frac {1}{9}-x\right )} \, dx,x,\left (\frac {1}{6}-\frac {x}{6}\right )^2\right )}{4 \sqrt [4]{3} \sqrt [4]{-2-2 x+x^2}}\\ &=2 \frac {\sqrt [4]{2+2 x-x^2} \operatorname {Subst}\left (\int \frac {x^2}{\frac {1}{36}+\frac {x^4}{12}} \, dx,x,\frac {\sqrt [4]{3-(-1+x)^2}}{\sqrt [4]{3}}\right )}{12 \sqrt [4]{3} \sqrt [4]{-2-2 x+x^2}}\\ &=2 \left (-\frac {\sqrt [4]{2+2 x-x^2} \operatorname {Subst}\left (\int \frac {1-\sqrt {3} x^2}{\frac {1}{36}+\frac {x^4}{12}} \, dx,x,\frac {\sqrt [4]{3-(-1+x)^2}}{\sqrt [4]{3}}\right )}{24\ 3^{3/4} \sqrt [4]{-2-2 x+x^2}}+\frac {\sqrt [4]{2+2 x-x^2} \operatorname {Subst}\left (\int \frac {1+\sqrt {3} x^2}{\frac {1}{36}+\frac {x^4}{12}} \, dx,x,\frac {\sqrt [4]{3-(-1+x)^2}}{\sqrt [4]{3}}\right )}{24\ 3^{3/4} \sqrt [4]{-2-2 x+x^2}}\right )\\ &=2 \left (\frac {\sqrt [4]{2+2 x-x^2} \operatorname {Subst}\left (\int \frac {\frac {\sqrt {2}}{\sqrt [4]{3}}+2 x}{-\frac {1}{\sqrt {3}}-\frac {\sqrt {2} x}{\sqrt [4]{3}}-x^2} \, dx,x,\frac {\sqrt [4]{3-(-1+x)^2}}{\sqrt [4]{3}}\right )}{4 \sqrt {2} \sqrt [4]{-2-2 x+x^2}}+\frac {\sqrt [4]{2+2 x-x^2} \operatorname {Subst}\left (\int \frac {\frac {\sqrt {2}}{\sqrt [4]{3}}-2 x}{-\frac {1}{\sqrt {3}}+\frac {\sqrt {2} x}{\sqrt [4]{3}}-x^2} \, dx,x,\frac {\sqrt [4]{3-(-1+x)^2}}{\sqrt [4]{3}}\right )}{4 \sqrt {2} \sqrt [4]{-2-2 x+x^2}}+\frac {\sqrt [4]{2+2 x-x^2} \operatorname {Subst}\left (\int \frac {1}{\frac {1}{\sqrt {3}}-\frac {\sqrt {2} x}{\sqrt [4]{3}}+x^2} \, dx,x,\frac {\sqrt [4]{3-(-1+x)^2}}{\sqrt [4]{3}}\right )}{4 \sqrt [4]{3} \sqrt [4]{-2-2 x+x^2}}+\frac {\sqrt [4]{2+2 x-x^2} \operatorname {Subst}\left (\int \frac {1}{\frac {1}{\sqrt {3}}+\frac {\sqrt {2} x}{\sqrt [4]{3}}+x^2} \, dx,x,\frac {\sqrt [4]{3-(-1+x)^2}}{\sqrt [4]{3}}\right )}{4 \sqrt [4]{3} \sqrt [4]{-2-2 x+x^2}}\right )\\ &=2 \left (\frac {\sqrt [4]{2+2 x-x^2} \log \left (\sqrt {3}-\sqrt {6} \sqrt [4]{3-(1-x)^2}+\sqrt {3} \sqrt {3-(1-x)^2}\right )}{4 \sqrt {2} \sqrt [4]{-2-2 x+x^2}}-\frac {\sqrt [4]{2+2 x-x^2} \log \left (\sqrt {3}+\sqrt {6} \sqrt [4]{3-(1-x)^2}+\sqrt {3} \sqrt {3-(1-x)^2}\right )}{4 \sqrt {2} \sqrt [4]{-2-2 x+x^2}}+\frac {\sqrt [4]{2+2 x-x^2} \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt [4]{3-(-1+x)^2}\right )}{2 \sqrt {2} \sqrt [4]{-2-2 x+x^2}}-\frac {\sqrt [4]{2+2 x-x^2} \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt [4]{3-(-1+x)^2}\right )}{2 \sqrt {2} \sqrt [4]{-2-2 x+x^2}}\right )\\ &=2 \left (-\frac {\sqrt [4]{2+2 x-x^2} \tan ^{-1}\left (1-\sqrt {2} \sqrt [4]{3-(1-x)^2}\right )}{2 \sqrt {2} \sqrt [4]{-2-2 x+x^2}}+\frac {\sqrt [4]{2+2 x-x^2} \tan ^{-1}\left (1+\sqrt {2} \sqrt [4]{3-(1-x)^2}\right )}{2 \sqrt {2} \sqrt [4]{-2-2 x+x^2}}+\frac {\sqrt [4]{2+2 x-x^2} \log \left (\sqrt {3}-\sqrt {6} \sqrt [4]{3-(1-x)^2}+\sqrt {3} \sqrt {3-(1-x)^2}\right )}{4 \sqrt {2} \sqrt [4]{-2-2 x+x^2}}-\frac {\sqrt [4]{2+2 x-x^2} \log \left (\sqrt {3}+\sqrt {6} \sqrt [4]{3-(1-x)^2}+\sqrt {3} \sqrt {3-(1-x)^2}\right )}{4 \sqrt {2} \sqrt [4]{-2-2 x+x^2}}\right )\\ \end {align*}

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Mathematica [F]  time = 0.24, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {-1+x}{(-3+x) (1+x) \sqrt [4]{-2-2 x+x^2}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.05, size = 29, normalized size = 1.00 \begin {gather*} \tan ^{-1}\left (\sqrt [4]{-2-2 x+x^2}\right )-\tanh ^{-1}\left (\sqrt [4]{-2-2 x+x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.46, size = 42, normalized size = 1.45 \begin {gather*} \arctan \left ({\left (x^{2} - 2 \, x - 2\right )}^{\frac {1}{4}}\right ) - \frac {1}{2} \, \log \left ({\left (x^{2} - 2 \, x - 2\right )}^{\frac {1}{4}} + 1\right ) + \frac {1}{2} \, \log \left ({\left (x^{2} - 2 \, x - 2\right )}^{\frac {1}{4}} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 1.40, size = 149, normalized size = 5.14

method	result	size
trager	\(\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {2 \RootOf \left (\textit {\_Z}^{2}+1\right ) \sqrt {x^{2}-2 x -2}-\RootOf \left (\textit {\_Z}^{2}+1\right ) x^{2}+2 \left (x^{2}-2 x -2\right )^{\frac {3}{4}}+2 \RootOf \left (\textit {\_Z}^{2}+1\right ) x +\RootOf \left (\textit {\_Z}^{2}+1\right )-2 \left (x^{2}-2 x -2\right )^{\frac {1}{4}}}{\left (1+x \right ) \left (-3+x \right )}\right )}{2}-\frac {\ln \left (\frac {2 \left (x^{2}-2 x -2\right )^{\frac {3}{4}}+2 \sqrt {x^{2}-2 x -2}+x^{2}+2 \left (x^{2}-2 x -2\right )^{\frac {1}{4}}-2 x -1}{\left (1+x \right ) \left (-3+x \right )}\right )}{2}\)	\(149\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.39, size = 25, normalized size = 0.86 \begin {gather*} \mathrm {atan}\left ({\left (x^2-2\,x-2\right )}^{1/4}\right )-\mathrm {atanh}\left ({\left (x^2-2\,x-2\right )}^{1/4}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

atan((x^2 - 2*x - 2)^(1/4)) - atanh((x^2 - 2*x - 2)^(1/4))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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