3.2.29 \(\int \frac {1+x+x^2}{(-1+x)^2 \sqrt {-1+x^4}} \, dx\)

Optimal. Leaf size=18 \[ -\frac {\sqrt {x^4-1}}{2 (x-1)^2} \]

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Rubi [B]  time = 0.41, antiderivative size = 68, normalized size of antiderivative = 3.78, number of steps used = 32, number of rules used = 19, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.905, Rules used = {6742, 222, 2153, 1152, 414, 527, 524, 427, 424, 253, 1248, 659, 651, 1256, 471, 21, 1725, 423, 426} \begin {gather*} \frac {x \left (x^2+1\right )}{\left (1-x^2\right ) \sqrt {x^4-1}}+\frac {\sqrt {x^4-1}}{2 \left (1-x^2\right )}-\frac {\sqrt {x^4-1}}{\left (1-x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

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 253

Int[((a1_.) + (b1_.)*(x_)^(n_))^(p_)*((a2_.) + (b2_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[((a1 + b1*x^n)^FracPa
rt[p]*(a2 + b2*x^n)^FracPart[p])/(a1*a2 + b1*b2*x^(2*n))^FracPart[p], Int[(a1*a2 + b1*b2*x^(2*n))^p, x], x] /;
FreeQ[{a1, b1, a2, b2, n, p}, x] && EqQ[a2*b1 + a1*b2, 0] &&  !IntegerQ[p]

Rule 414

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1)*
(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomial
Q[a, b, c, d, n, p, q, x]

Rule 423

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

Rule 424

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[ArcSin[Rt[-(d/c)
, 2]*x], (b*c)/(a*d)])/(Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[
a, 0]

Rule 426

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

Rule 427

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

Rule 471

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

Rule 524

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

Rule 527

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> -Simp[
((b*e - a*f)*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d
)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)
*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

Rule 651

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^m*(a + c*x^2)^(p + 1))
/(2*c*d*(p + 1)), x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + 2*p
+ 2, 0]

Rule 659

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(e*(d + e*x)^m*(a + c*x^2)^(p + 1)
)/(2*c*d*(m + p + 1)), x] + Dist[Simplify[m + 2*p + 2]/(2*d*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p,
x], x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[Simplify[m + 2*p + 2
], 0]

Rule 1152

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

Rule 1248

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[(d + e*x)^q
*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x]

Rule 1256

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a + c*x^4)^Fr
acPart[p]/((d + e*x^2)^FracPart[p]*(a/d + (c*x^2)/e)^FracPart[p]), Int[(f*x)^m*(d + e*x^2)^(q + p)*(a/d + (c*x
^2)/e)^p, x], x] /; FreeQ[{a, c, d, e, f, m, p, q}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p]

Rule 1725

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

Rule 2153

Int[((c_) + (d_.)*(x_)^(n_.))^(q_)*((a_) + (b_.)*(x_)^(nn_.))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^
nn)^p, (c/(c^2 - d^2*x^(2*n)) - (d*x^n)/(c^2 - d^2*x^(2*n)))^(-q), x], x] /; FreeQ[{a, b, c, d, n, nn, p}, x]
&&  !IntegerQ[p] && ILtQ[q, 0] && IGtQ[Log[2, nn/n], 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+x^2}{(-1+x)^2 \sqrt {-1+x^4}} \, dx &=\int \left (\frac {1}{\sqrt {-1+x^4}}+\frac {3}{(-1+x)^2 \sqrt {-1+x^4}}+\frac {3}{(-1+x) \sqrt {-1+x^4}}\right ) \, dx\\ &=3 \int \frac {1}{(-1+x)^2 \sqrt {-1+x^4}} \, dx+3 \int \frac {1}{(-1+x) \sqrt {-1+x^4}} \, dx+\int \frac {1}{\sqrt {-1+x^4}} \, dx\\ &=\frac {\sqrt {-1+x^2} \sqrt {1+x^2} F\left (\sin ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {-1+x^2}}\right )|\frac {1}{2}\right )}{\sqrt {2} \sqrt {-1+x^4}}-3 \int \frac {1}{\left (1-x^2\right ) \sqrt {-1+x^4}} \, dx-3 \int \frac {x}{\left (1-x^2\right ) \sqrt {-1+x^4}} \, dx+3 \int \left (\frac {1}{\left (-1+x^2\right )^2 \sqrt {-1+x^4}}+\frac {2 x}{\left (-1+x^2\right )^2 \sqrt {-1+x^4}}+\frac {x^2}{\left (-1+x^2\right )^2 \sqrt {-1+x^4}}\right ) \, dx\\ &=\frac {\sqrt {-1+x^2} \sqrt {1+x^2} F\left (\sin ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {-1+x^2}}\right )|\frac {1}{2}\right )}{\sqrt {2} \sqrt {-1+x^4}}-\frac {3}{2} \operatorname {Subst}\left (\int \frac {1}{(1-x) \sqrt {-1+x^2}} \, dx,x,x^2\right )+3 \int \frac {1}{\left (-1+x^2\right )^2 \sqrt {-1+x^4}} \, dx+3 \int \frac {x^2}{\left (-1+x^2\right )^2 \sqrt {-1+x^4}} \, dx+6 \int \frac {x}{\left (-1+x^2\right )^2 \sqrt {-1+x^4}} \, dx-\frac {\left (3 \sqrt {-1-x^2} \sqrt {1-x^2}\right ) \int \frac {1}{\sqrt {-1-x^2} \left (1-x^2\right )^{3/2}} \, dx}{\sqrt {-1+x^4}}\\ &=-\frac {3 x \left (1+x^2\right )}{2 \sqrt {-1+x^4}}+\frac {3 \sqrt {-1+x^4}}{2 \left (1-x^2\right )}+\frac {\sqrt {-1+x^2} \sqrt {1+x^2} F\left (\sin ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {-1+x^2}}\right )|\frac {1}{2}\right )}{\sqrt {2} \sqrt {-1+x^4}}+3 \operatorname {Subst}\left (\int \frac {1}{(-1+x)^2 \sqrt {-1+x^2}} \, dx,x,x^2\right )+\frac {\left (3 \sqrt {-1-x^2} \sqrt {1-x^2}\right ) \int \frac {-1+x^2}{\sqrt {-1-x^2} \sqrt {1-x^2}} \, dx}{2 \sqrt {-1+x^4}}+\frac {\left (3 \sqrt {-1+x^2} \sqrt {1+x^2}\right ) \int \frac {1}{\left (-1+x^2\right )^{5/2} \sqrt {1+x^2}} \, dx}{\sqrt {-1+x^4}}+\frac {\left (3 \sqrt {-1+x^2} \sqrt {1+x^2}\right ) \int \frac {x^2}{\left (-1+x^2\right )^{5/2} \sqrt {1+x^2}} \, dx}{\sqrt {-1+x^4}}\\ &=-\frac {3 x \left (1+x^2\right )}{2 \sqrt {-1+x^4}}+\frac {x \left (1+x^2\right )}{\left (1-x^2\right ) \sqrt {-1+x^4}}-\frac {\sqrt {-1+x^4}}{\left (1-x^2\right )^2}+\frac {3 \sqrt {-1+x^4}}{2 \left (1-x^2\right )}+\frac {\sqrt {-1+x^2} \sqrt {1+x^2} F\left (\sin ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {-1+x^2}}\right )|\frac {1}{2}\right )}{\sqrt {2} \sqrt {-1+x^4}}-\frac {\left (3 \sqrt {-1-x^2} \sqrt {1-x^2}\right ) \int \frac {\sqrt {1-x^2}}{\sqrt {-1-x^2}} \, dx}{2 \sqrt {-1+x^4}}+\frac {\left (\sqrt {-1+x^2} \sqrt {1+x^2}\right ) \int \frac {-5-x^2}{\left (-1+x^2\right )^{3/2} \sqrt {1+x^2}} \, dx}{2 \sqrt {-1+x^4}}+\frac {\left (\sqrt {-1+x^2} \sqrt {1+x^2}\right ) \int \frac {1-x^2}{\left (-1+x^2\right )^{3/2} \sqrt {1+x^2}} \, dx}{2 \sqrt {-1+x^4}}-\operatorname {Subst}\left (\int \frac {1}{(-1+x) \sqrt {-1+x^2}} \, dx,x,x^2\right )\\ &=\frac {x \left (1+x^2\right )}{\left (1-x^2\right ) \sqrt {-1+x^4}}-\frac {\sqrt {-1+x^4}}{\left (1-x^2\right )^2}+\frac {\sqrt {-1+x^4}}{2 \left (1-x^2\right )}+\frac {\sqrt {-1+x^2} \sqrt {1+x^2} F\left (\sin ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {-1+x^2}}\right )|\frac {1}{2}\right )}{\sqrt {2} \sqrt {-1+x^4}}-\frac {\left (3 \sqrt {-1-x^2} \sqrt {1-x^2}\right ) \int \frac {\sqrt {-1-x^2}}{\sqrt {1-x^2}} \, dx}{2 \sqrt {-1+x^4}}-\frac {\left (3 \sqrt {-1-x^2} \sqrt {1-x^2}\right ) \int \frac {1}{\sqrt {-1-x^2} \sqrt {1-x^2}} \, dx}{\sqrt {-1+x^4}}+\frac {\left (\sqrt {-1+x^2} \sqrt {1+x^2}\right ) \int \frac {4-6 x^2}{\sqrt {-1+x^2} \sqrt {1+x^2}} \, dx}{4 \sqrt {-1+x^4}}-\frac {\left (\sqrt {-1+x^2} \sqrt {1+x^2}\right ) \int \frac {1}{\sqrt {-1+x^2} \sqrt {1+x^2}} \, dx}{2 \sqrt {-1+x^4}}\\ &=\frac {x \left (1+x^2\right )}{\left (1-x^2\right ) \sqrt {-1+x^4}}-\frac {\sqrt {-1+x^4}}{\left (1-x^2\right )^2}+\frac {\sqrt {-1+x^4}}{2 \left (1-x^2\right )}+\frac {\sqrt {-1+x^2} \sqrt {1+x^2} F\left (\sin ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {-1+x^2}}\right )|\frac {1}{2}\right )}{\sqrt {2} \sqrt {-1+x^4}}-\frac {1}{2} \int \frac {1}{\sqrt {-1+x^4}} \, dx-3 \int \frac {1}{\sqrt {-1+x^4}} \, dx-\frac {\left (3 \left (-1-x^2\right ) \sqrt {1-x^2}\right ) \int \frac {\sqrt {1+x^2}}{\sqrt {1-x^2}} \, dx}{2 \sqrt {1+x^2} \sqrt {-1+x^4}}-\frac {\left (3 \sqrt {-1+x^2} \sqrt {1+x^2}\right ) \int \frac {\sqrt {1+x^2}}{\sqrt {-1+x^2}} \, dx}{2 \sqrt {-1+x^4}}+\frac {\left (5 \sqrt {-1+x^2} \sqrt {1+x^2}\right ) \int \frac {1}{\sqrt {-1+x^2} \sqrt {1+x^2}} \, dx}{2 \sqrt {-1+x^4}}\\ &=\frac {x \left (1+x^2\right )}{\left (1-x^2\right ) \sqrt {-1+x^4}}-\frac {\sqrt {-1+x^4}}{\left (1-x^2\right )^2}+\frac {\sqrt {-1+x^4}}{2 \left (1-x^2\right )}+\frac {3 \sqrt {1-x^2} \sqrt {1+x^2} E\left (\left .\sin ^{-1}(x)\right |-1\right )}{2 \sqrt {-1+x^4}}-\frac {5 \sqrt {-1+x^2} \sqrt {1+x^2} F\left (\sin ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {-1+x^2}}\right )|\frac {1}{2}\right )}{2 \sqrt {2} \sqrt {-1+x^4}}+\frac {5}{2} \int \frac {1}{\sqrt {-1+x^4}} \, dx-\frac {\left (3 \sqrt {1-x^2} \sqrt {1+x^2}\right ) \int \frac {\sqrt {1+x^2}}{\sqrt {1-x^2}} \, dx}{2 \sqrt {-1+x^4}}\\ &=\frac {x \left (1+x^2\right )}{\left (1-x^2\right ) \sqrt {-1+x^4}}-\frac {\sqrt {-1+x^4}}{\left (1-x^2\right )^2}+\frac {\sqrt {-1+x^4}}{2 \left (1-x^2\right )}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 33, normalized size = 1.83 \begin {gather*} \frac {-x^3-x^2-x-1}{2 (x-1) \sqrt {x^4-1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 3.33, size = 18, normalized size = 1.00 \begin {gather*} -\frac {\sqrt {-1+x^4}}{2 (-1+x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*Sqrt[-1 + x^4]/(-1 + x)^2

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fricas [A]  time = 0.47, size = 19, normalized size = 1.06 \begin {gather*} -\frac {\sqrt {x^{4} - 1}}{2 \, {\left (x^{2} - 2 \, x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(x^4 - 1)/(x^2 - 2*x + 1)

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giac [A]  time = 0.45, size = 27, normalized size = 1.50 \begin {gather*} -\frac {1}{2} \, \sqrt {\frac {4}{x - 1} + \frac {6}{{\left (x - 1\right )}^{2}} + \frac {4}{{\left (x - 1\right )}^{3}} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(4/(x - 1) + 6/(x - 1)^2 + 4/(x - 1)^3 + 1)

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maple [A]  time = 0.12, size = 15, normalized size = 0.83

method	result	size
default	\(-\frac {\sqrt {x^{4}-1}}{2 \left (-1+x \right )^{2}}\)	\(15\)
trager	\(-\frac {\sqrt {x^{4}-1}}{2 \left (-1+x \right )^{2}}\)	\(15\)
elliptic	\(-\frac {\sqrt {x^{4}-1}}{2 \left (-1+x \right )^{2}}\)	\(15\)
gosper	\(-\frac {\left (1+x \right ) \left (x^{2}+1\right )}{2 \left (-1+x \right ) \sqrt {x^{4}-1}}\)	\(23\)
risch	\(-\frac {x^{3}+x^{2}+x +1}{2 \left (-1+x \right ) \sqrt {x^{4}-1}}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.59, size = 28, normalized size = 1.56 \begin {gather*} -\frac {x^{3} + x^{2} + x + 1}{2 \, \sqrt {x^{2} + 1} \sqrt {x + 1} {\left (x - 1\right )}^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.14, size = 14, normalized size = 0.78 \begin {gather*} -\frac {\sqrt {x^4-1}}{2\,{\left (x-1\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(x^4 - 1)^(1/2)/(2*(x - 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 {\left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right )} \left (x - 1\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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