\(\int \frac {(-1+x^2) \sqrt [4]{x^2+x^6}}{x^2 (1+x^2)} \, dx\) [1754]

Optimal result

Integrand size = 27, antiderivative size = 118 \[ \int \frac {\left (-1+x^2\right ) \sqrt [4]{x^2+x^6}}{x^2 \left (1+x^2\right )} \, dx=\frac {2 \sqrt [4]{x^2+x^6}}{x}-\frac {\arctan \left (\frac {2^{3/4} x \sqrt [4]{x^2+x^6}}{\sqrt {2} x^2-\sqrt {x^2+x^6}}\right )}{\sqrt [4]{2}}-\frac {\text {arctanh}\left (\frac {\frac {x^2}{\sqrt [4]{2}}+\frac {\sqrt {x^2+x^6}}{2^{3/4}}}{x \sqrt [4]{x^2+x^6}}\right )}{\sqrt [4]{2}} \]

[Out]

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

Rubi [C] (verified)

Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.

Time = 0.39 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.25, number of steps used = 20, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.407, Rules used = {2081, 6857, 283, 335, 371, 1326, 1350, 331, 1351, 477, 524} \[ \int \frac {\left (-1+x^2\right ) \sqrt [4]{x^2+x^6}}{x^2 \left (1+x^2\right )} \, dx=\frac {8 \sqrt [4]{x^6+x^2} x \operatorname {AppellF1}\left (\frac {3}{8},1,\frac {3}{4},\frac {11}{8},x^4,-x^4\right )}{3 \sqrt [4]{x^4+1}}-\frac {8 \sqrt [4]{x^6+x^2} x^3 \operatorname {AppellF1}\left (\frac {7}{8},1,\frac {3}{4},\frac {15}{8},x^4,-x^4\right )}{7 \sqrt [4]{x^4+1}}-\frac {4 \sqrt [4]{x^6+x^2} x \operatorname {Hypergeometric2F1}\left (\frac {3}{8},\frac {3}{4},\frac {11}{8},-x^4\right )}{3 \sqrt [4]{x^4+1}}+\frac {2 \sqrt [4]{x^6+x^2}}{x} \]

[In]

Int[((-1 + x^2)*(x^2 + x^6)^(1/4))/(x^2*(1 + x^2)),x]

[Out]

(2*(x^2 + x^6)^(1/4))/x + (8*x*(x^2 + x^6)^(1/4)*AppellF1[3/8, 1, 3/4, 11/8, x^4, -x^4])/(3*(1 + x^4)^(1/4)) -
(8*x^3*(x^2 + x^6)^(1/4)*AppellF1[7/8, 1, 3/4, 15/8, x^4, -x^4])/(7*(1 + x^4)^(1/4)) - (4*x*(x^2 + x^6)^(1/4)
*Hypergeometric2F1[3/8, 3/4, 11/8, -x^4])/(3*(1 + x^4)^(1/4))

Rule 283

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

Rule 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 335

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 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 477

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = Deno
minator[m]}, Dist[k/e, Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/e^n))^p*(c + d*(x^(k*n)/e^n))^q, x], x, (e*
x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && FractionQ[m] && Intege
rQ[p]

Rule 524

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*
((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 1326

Int[(((f_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^4)^(p_.))/((d_.) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/(d*e), Int[(f*
x)^m*(a*e + c*d*x^2)*(a + c*x^4)^(p - 1), x], x] - Dist[(c*d^2 + a*e^2)/(d*e*f^2), Int[(f*x)^(m + 2)*((a + c*x
^4)^(p - 1)/(d + e*x^2)), x], x] /; FreeQ[{a, c, d, e, f}, x] && GtQ[p, 0] && LtQ[m, 0]

Rule 1350

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(f*x)^m*(d + e*x^2)^q*(a + c*x^4)^p, x], x] /; FreeQ[{a, c, d, e, f, m, p, q}, x] && (IGtQ[p, 0] || IGtQ[q,
0] || IntegersQ[m, q])

Rule 1351

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

Rule 2081

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 6857

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*} \text {integral}& = \frac {\sqrt [4]{x^2+x^6} \int \frac {\left (-1+x^2\right ) \sqrt [4]{1+x^4}}{x^{3/2} \left (1+x^2\right )} \, dx}{\sqrt {x} \sqrt [4]{1+x^4}} \\ & = \frac {\sqrt [4]{x^2+x^6} \int \left (\frac {\sqrt [4]{1+x^4}}{x^{3/2}}-\frac {2 \sqrt [4]{1+x^4}}{x^{3/2} \left (1+x^2\right )}\right ) \, dx}{\sqrt {x} \sqrt [4]{1+x^4}} \\ & = \frac {\sqrt [4]{x^2+x^6} \int \frac {\sqrt [4]{1+x^4}}{x^{3/2}} \, dx}{\sqrt {x} \sqrt [4]{1+x^4}}-\frac {\left (2 \sqrt [4]{x^2+x^6}\right ) \int \frac {\sqrt [4]{1+x^4}}{x^{3/2} \left (1+x^2\right )} \, dx}{\sqrt {x} \sqrt [4]{1+x^4}} \\ & = -\frac {2 \sqrt [4]{x^2+x^6}}{x}+\frac {\left (2 \sqrt [4]{x^2+x^6}\right ) \int \frac {x^{5/2}}{\left (1+x^4\right )^{3/4}} \, dx}{\sqrt {x} \sqrt [4]{1+x^4}}-\frac {\left (2 \sqrt [4]{x^2+x^6}\right ) \int \frac {1+x^2}{x^{3/2} \left (1+x^4\right )^{3/4}} \, dx}{\sqrt {x} \sqrt [4]{1+x^4}}+\frac {\left (4 \sqrt [4]{x^2+x^6}\right ) \int \frac {\sqrt {x}}{\left (1+x^2\right ) \left (1+x^4\right )^{3/4}} \, dx}{\sqrt {x} \sqrt [4]{1+x^4}} \\ & = -\frac {2 \sqrt [4]{x^2+x^6}}{x}-\frac {\left (2 \sqrt [4]{x^2+x^6}\right ) \int \left (\frac {1}{x^{3/2} \left (1+x^4\right )^{3/4}}+\frac {\sqrt {x}}{\left (1+x^4\right )^{3/4}}\right ) \, dx}{\sqrt {x} \sqrt [4]{1+x^4}}+\frac {\left (4 \sqrt [4]{x^2+x^6}\right ) \int \left (\frac {\sqrt {x}}{\left (1-x^4\right ) \left (1+x^4\right )^{3/4}}+\frac {x^{5/2}}{\left (-1+x^4\right ) \left (1+x^4\right )^{3/4}}\right ) \, dx}{\sqrt {x} \sqrt [4]{1+x^4}}+\frac {\left (4 \sqrt [4]{x^2+x^6}\right ) \text {Subst}\left (\int \frac {x^6}{\left (1+x^8\right )^{3/4}} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt [4]{1+x^4}} \\ & = -\frac {2 \sqrt [4]{x^2+x^6}}{x}+\frac {4 x^3 \sqrt [4]{x^2+x^6} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {7}{8},\frac {15}{8},-x^4\right )}{7 \sqrt [4]{1+x^4}}-\frac {\left (2 \sqrt [4]{x^2+x^6}\right ) \int \frac {1}{x^{3/2} \left (1+x^4\right )^{3/4}} \, dx}{\sqrt {x} \sqrt [4]{1+x^4}}-\frac {\left (2 \sqrt [4]{x^2+x^6}\right ) \int \frac {\sqrt {x}}{\left (1+x^4\right )^{3/4}} \, dx}{\sqrt {x} \sqrt [4]{1+x^4}}+\frac {\left (4 \sqrt [4]{x^2+x^6}\right ) \int \frac {\sqrt {x}}{\left (1-x^4\right ) \left (1+x^4\right )^{3/4}} \, dx}{\sqrt {x} \sqrt [4]{1+x^4}}+\frac {\left (4 \sqrt [4]{x^2+x^6}\right ) \int \frac {x^{5/2}}{\left (-1+x^4\right ) \left (1+x^4\right )^{3/4}} \, dx}{\sqrt {x} \sqrt [4]{1+x^4}} \\ & = \frac {2 \sqrt [4]{x^2+x^6}}{x}+\frac {4 x^3 \sqrt [4]{x^2+x^6} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {7}{8},\frac {15}{8},-x^4\right )}{7 \sqrt [4]{1+x^4}}-\frac {\left (2 \sqrt [4]{x^2+x^6}\right ) \int \frac {x^{5/2}}{\left (1+x^4\right )^{3/4}} \, dx}{\sqrt {x} \sqrt [4]{1+x^4}}-\frac {\left (4 \sqrt [4]{x^2+x^6}\right ) \text {Subst}\left (\int \frac {x^2}{\left (1+x^8\right )^{3/4}} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt [4]{1+x^4}}+\frac {\left (8 \sqrt [4]{x^2+x^6}\right ) \text {Subst}\left (\int \frac {x^2}{\left (1-x^8\right ) \left (1+x^8\right )^{3/4}} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt [4]{1+x^4}}+\frac {\left (8 \sqrt [4]{x^2+x^6}\right ) \text {Subst}\left (\int \frac {x^6}{\left (-1+x^8\right ) \left (1+x^8\right )^{3/4}} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt [4]{1+x^4}} \\ & = \frac {2 \sqrt [4]{x^2+x^6}}{x}+\frac {8 x \sqrt [4]{x^2+x^6} \operatorname {AppellF1}\left (\frac {3}{8},1,\frac {3}{4},\frac {11}{8},x^4,-x^4\right )}{3 \sqrt [4]{1+x^4}}-\frac {8 x^3 \sqrt [4]{x^2+x^6} \operatorname {AppellF1}\left (\frac {7}{8},1,\frac {3}{4},\frac {15}{8},x^4,-x^4\right )}{7 \sqrt [4]{1+x^4}}-\frac {4 x \sqrt [4]{x^2+x^6} \operatorname {Hypergeometric2F1}\left (\frac {3}{8},\frac {3}{4},\frac {11}{8},-x^4\right )}{3 \sqrt [4]{1+x^4}}+\frac {4 x^3 \sqrt [4]{x^2+x^6} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {7}{8},\frac {15}{8},-x^4\right )}{7 \sqrt [4]{1+x^4}}-\frac {\left (4 \sqrt [4]{x^2+x^6}\right ) \text {Subst}\left (\int \frac {x^6}{\left (1+x^8\right )^{3/4}} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt [4]{1+x^4}} \\ & = \frac {2 \sqrt [4]{x^2+x^6}}{x}+\frac {8 x \sqrt [4]{x^2+x^6} \operatorname {AppellF1}\left (\frac {3}{8},1,\frac {3}{4},\frac {11}{8},x^4,-x^4\right )}{3 \sqrt [4]{1+x^4}}-\frac {8 x^3 \sqrt [4]{x^2+x^6} \operatorname {AppellF1}\left (\frac {7}{8},1,\frac {3}{4},\frac {15}{8},x^4,-x^4\right )}{7 \sqrt [4]{1+x^4}}-\frac {4 x \sqrt [4]{x^2+x^6} \operatorname {Hypergeometric2F1}\left (\frac {3}{8},\frac {3}{4},\frac {11}{8},-x^4\right )}{3 \sqrt [4]{1+x^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.18 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.24 \[ \int \frac {\left (-1+x^2\right ) \sqrt [4]{x^2+x^6}}{x^2 \left (1+x^2\right )} \, dx=-\frac {\sqrt [4]{x^2+x^6} \left (-4 \sqrt [4]{1+x^4}+2^{3/4} \sqrt {x} \arctan \left (\frac {2^{3/4} \sqrt {x} \sqrt [4]{1+x^4}}{\sqrt {2} x-\sqrt {1+x^4}}\right )+2^{3/4} \sqrt {x} \text {arctanh}\left (\frac {2 \sqrt [4]{2} \sqrt {x} \sqrt [4]{1+x^4}}{2 x+\sqrt {2} \sqrt {1+x^4}}\right )\right )}{2 x \sqrt [4]{1+x^4}} \]

[In]

Integrate[((-1 + x^2)*(x^2 + x^6)^(1/4))/(x^2*(1 + x^2)),x]

[Out]

-1/2*((x^2 + x^6)^(1/4)*(-4*(1 + x^4)^(1/4) + 2^(3/4)*Sqrt[x]*ArcTan[(2^(3/4)*Sqrt[x]*(1 + x^4)^(1/4))/(Sqrt[2
]*x - Sqrt[1 + x^4])] + 2^(3/4)*Sqrt[x]*ArcTanh[(2*2^(1/4)*Sqrt[x]*(1 + x^4)^(1/4))/(2*x + Sqrt[2]*Sqrt[1 + x^
4])]))/(x*(1 + x^4)^(1/4))

Maple [A] (verified)

Time = 6.56 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.35

method	result	size
pseudoelliptic	\(\frac {-\ln \left (\frac {2^{\frac {3}{4}} \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}} x +\sqrt {2}\, x^{2}+\sqrt {x^{2} \left (x^{4}+1\right )}}{-2^{\frac {3}{4}} \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}} x +\sqrt {2}\, x^{2}+\sqrt {x^{2} \left (x^{4}+1\right )}}\right ) 2^{\frac {3}{4}} x -2 \arctan \left (\frac {2^{\frac {1}{4}} \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}+x}{x}\right ) 2^{\frac {3}{4}} x -2 \arctan \left (\frac {2^{\frac {1}{4}} \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}-x}{x}\right ) 2^{\frac {3}{4}} x +8 \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}}{4 x}\)	\(159\)

[In]

int((x^2-1)*(x^6+x^2)^(1/4)/x^2/(x^2+1),x,method=_RETURNVERBOSE)

[Out]

1/4*(-ln((2^(3/4)*(x^2*(x^4+1))^(1/4)*x+2^(1/2)*x^2+(x^2*(x^4+1))^(1/2))/(-2^(3/4)*(x^2*(x^4+1))^(1/4)*x+2^(1/
2)*x^2+(x^2*(x^4+1))^(1/2)))*2^(3/4)*x-2*arctan((2^(1/4)*(x^2*(x^4+1))^(1/4)+x)/x)*2^(3/4)*x-2*arctan((2^(1/4)
*(x^2*(x^4+1))^(1/4)-x)/x)*2^(3/4)*x+8*(x^2*(x^4+1))^(1/4))/x

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 4.67 (sec) , antiderivative size = 332, normalized size of antiderivative = 2.81 \[ \int \frac {\left (-1+x^2\right ) \sqrt [4]{x^2+x^6}}{x^2 \left (1+x^2\right )} \, dx=\frac {\left (-2\right )^{\frac {1}{4}} x \log \left (\frac {4 \, \left (-2\right )^{\frac {1}{4}} {\left (x^{6} + x^{2}\right )}^{\frac {1}{4}} x^{2} - 2 \, \left (-2\right )^{\frac {3}{4}} {\left (x^{6} + x^{2}\right )}^{\frac {3}{4}} + \sqrt {-2} {\left (x^{5} - 2 \, x^{3} + x\right )} - 4 \, \sqrt {x^{6} + x^{2}} x}{x^{5} + 2 \, x^{3} + x}\right ) - \left (-2\right )^{\frac {1}{4}} x \log \left (-\frac {4 \, \left (-2\right )^{\frac {1}{4}} {\left (x^{6} + x^{2}\right )}^{\frac {1}{4}} x^{2} - 2 \, \left (-2\right )^{\frac {3}{4}} {\left (x^{6} + x^{2}\right )}^{\frac {3}{4}} - \sqrt {-2} {\left (x^{5} - 2 \, x^{3} + x\right )} + 4 \, \sqrt {x^{6} + x^{2}} x}{x^{5} + 2 \, x^{3} + x}\right ) + i \, \left (-2\right )^{\frac {1}{4}} x \log \left (\frac {4 i \, \left (-2\right )^{\frac {1}{4}} {\left (x^{6} + x^{2}\right )}^{\frac {1}{4}} x^{2} + 2 i \, \left (-2\right )^{\frac {3}{4}} {\left (x^{6} + x^{2}\right )}^{\frac {3}{4}} - \sqrt {-2} {\left (x^{5} - 2 \, x^{3} + x\right )} - 4 \, \sqrt {x^{6} + x^{2}} x}{x^{5} + 2 \, x^{3} + x}\right ) - i \, \left (-2\right )^{\frac {1}{4}} x \log \left (\frac {-4 i \, \left (-2\right )^{\frac {1}{4}} {\left (x^{6} + x^{2}\right )}^{\frac {1}{4}} x^{2} - 2 i \, \left (-2\right )^{\frac {3}{4}} {\left (x^{6} + x^{2}\right )}^{\frac {3}{4}} - \sqrt {-2} {\left (x^{5} - 2 \, x^{3} + x\right )} - 4 \, \sqrt {x^{6} + x^{2}} x}{x^{5} + 2 \, x^{3} + x}\right ) + 8 \, {\left (x^{6} + x^{2}\right )}^{\frac {1}{4}}}{4 \, x} \]

[In]

integrate((x^2-1)*(x^6+x^2)^(1/4)/x^2/(x^2+1),x, algorithm="fricas")

[Out]

1/4*((-2)^(1/4)*x*log((4*(-2)^(1/4)*(x^6 + x^2)^(1/4)*x^2 - 2*(-2)^(3/4)*(x^6 + x^2)^(3/4) + sqrt(-2)*(x^5 - 2
*x^3 + x) - 4*sqrt(x^6 + x^2)*x)/(x^5 + 2*x^3 + x)) - (-2)^(1/4)*x*log(-(4*(-2)^(1/4)*(x^6 + x^2)^(1/4)*x^2 -
2*(-2)^(3/4)*(x^6 + x^2)^(3/4) - sqrt(-2)*(x^5 - 2*x^3 + x) + 4*sqrt(x^6 + x^2)*x)/(x^5 + 2*x^3 + x)) + I*(-2)
^(1/4)*x*log((4*I*(-2)^(1/4)*(x^6 + x^2)^(1/4)*x^2 + 2*I*(-2)^(3/4)*(x^6 + x^2)^(3/4) - sqrt(-2)*(x^5 - 2*x^3
+ x) - 4*sqrt(x^6 + x^2)*x)/(x^5 + 2*x^3 + x)) - I*(-2)^(1/4)*x*log((-4*I*(-2)^(1/4)*(x^6 + x^2)^(1/4)*x^2 - 2
*I*(-2)^(3/4)*(x^6 + x^2)^(3/4) - sqrt(-2)*(x^5 - 2*x^3 + x) - 4*sqrt(x^6 + x^2)*x)/(x^5 + 2*x^3 + x)) + 8*(x^
6 + x^2)^(1/4))/x

Sympy [F]

\[ \int \frac {\left (-1+x^2\right ) \sqrt [4]{x^2+x^6}}{x^2 \left (1+x^2\right )} \, dx=\int \frac {\sqrt [4]{x^{2} \left (x^{4} + 1\right )} \left (x - 1\right ) \left (x + 1\right )}{x^{2} \left (x^{2} + 1\right )}\, dx \]

[In]

integrate((x**2-1)*(x**6+x**2)**(1/4)/x**2/(x**2+1),x)

[Out]

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

Maxima [F]

\[ \int \frac {\left (-1+x^2\right ) \sqrt [4]{x^2+x^6}}{x^2 \left (1+x^2\right )} \, dx=\int { \frac {{\left (x^{6} + x^{2}\right )}^{\frac {1}{4}} {\left (x^{2} - 1\right )}}{{\left (x^{2} + 1\right )} x^{2}} \,d x } \]

[In]

integrate((x^2-1)*(x^6+x^2)^(1/4)/x^2/(x^2+1),x, algorithm="maxima")

[Out]

integrate((x^6 + x^2)^(1/4)*(x^2 - 1)/((x^2 + 1)*x^2), x)

Giac [F]

\[ \int \frac {\left (-1+x^2\right ) \sqrt [4]{x^2+x^6}}{x^2 \left (1+x^2\right )} \, dx=\int { \frac {{\left (x^{6} + x^{2}\right )}^{\frac {1}{4}} {\left (x^{2} - 1\right )}}{{\left (x^{2} + 1\right )} x^{2}} \,d x } \]

[In]

integrate((x^2-1)*(x^6+x^2)^(1/4)/x^2/(x^2+1),x, algorithm="giac")

[Out]

integrate((x^6 + x^2)^(1/4)*(x^2 - 1)/((x^2 + 1)*x^2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (-1+x^2\right ) \sqrt [4]{x^2+x^6}}{x^2 \left (1+x^2\right )} \, dx=\int \frac {{\left (x^6+x^2\right )}^{1/4}\,\left (x^2-1\right )}{x^2\,\left (x^2+1\right )} \,d x \]

[In]

int(((x^2 + x^6)^(1/4)*(x^2 - 1))/(x^2*(x^2 + 1)),x)

[Out]

int(((x^2 + x^6)^(1/4)*(x^2 - 1))/(x^2*(x^2 + 1)), x)