\(\int \frac {\sqrt [4]{x^2+x^4} (-1-x^4+x^8)}{-1+x^4} \, dx\) [1740]

Optimal result

Integrand size = 29, antiderivative size = 117 \[ \int \frac {\sqrt [4]{x^2+x^4} \left (-1-x^4+x^8\right )}{-1+x^4} \, dx=\frac {1}{192} \sqrt [4]{x^2+x^4} \left (-7 x+4 x^3+32 x^5\right )-\frac {7}{128} \arctan \left (\frac {x}{\sqrt [4]{x^2+x^4}}\right )-\frac {\arctan \left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^2+x^4}}\right )}{2^{3/4}}+\frac {7}{128} \text {arctanh}\left (\frac {x}{\sqrt [4]{x^2+x^4}}\right )+\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^2+x^4}}\right )}{2^{3/4}} \]

[Out]

1/192*(x^4+x^2)^(1/4)*(32*x^5+4*x^3-7*x)-7/128*arctan(x/(x^4+x^2)^(1/4))-1/2*arctan(2^(1/4)*x/(x^4+x^2)^(1/4))
*2^(1/4)+7/128*arctanh(x/(x^4+x^2)^(1/4))+1/2*arctanh(2^(1/4)*x/(x^4+x^2)^(1/4))*2^(1/4)

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(248\) vs. \(2(117)=234\).

Time = 0.37 (sec) , antiderivative size = 248, normalized size of antiderivative = 2.12, number of steps used = 24, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.379, Rules used = {2081, 1600, 6857, 327, 335, 338, 304, 209, 212, 477, 508} \[ \int \frac {\sqrt [4]{x^2+x^4} \left (-1-x^4+x^8\right )}{-1+x^4} \, dx=-\frac {7 \sqrt [4]{x^4+x^2} \arctan \left (\frac {\sqrt {x}}{\sqrt [4]{x^2+1}}\right )}{128 \sqrt [4]{x^2+1} \sqrt {x}}-\frac {\sqrt [4]{x^4+x^2} \arctan \left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{x^2+1}}\right )}{2^{3/4} \sqrt [4]{x^2+1} \sqrt {x}}+\frac {7 \sqrt [4]{x^4+x^2} \text {arctanh}\left (\frac {\sqrt {x}}{\sqrt [4]{x^2+1}}\right )}{128 \sqrt [4]{x^2+1} \sqrt {x}}+\frac {\sqrt [4]{x^4+x^2} \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{x^2+1}}\right )}{2^{3/4} \sqrt [4]{x^2+1} \sqrt {x}}-\frac {7}{192} \sqrt [4]{x^4+x^2} x+\frac {1}{6} \sqrt [4]{x^4+x^2} x^5+\frac {1}{48} \sqrt [4]{x^4+x^2} x^3 \]

[In]

Int[((x^2 + x^4)^(1/4)*(-1 - x^4 + x^8))/(-1 + x^4),x]

[Out]

(-7*x*(x^2 + x^4)^(1/4))/192 + (x^3*(x^2 + x^4)^(1/4))/48 + (x^5*(x^2 + x^4)^(1/4))/6 - (7*(x^2 + x^4)^(1/4)*A
rcTan[Sqrt[x]/(1 + x^2)^(1/4)])/(128*Sqrt[x]*(1 + x^2)^(1/4)) - ((x^2 + x^4)^(1/4)*ArcTan[(2^(1/4)*Sqrt[x])/(1
+ x^2)^(1/4)])/(2^(3/4)*Sqrt[x]*(1 + x^2)^(1/4)) + (7*(x^2 + x^4)^(1/4)*ArcTanh[Sqrt[x]/(1 + x^2)^(1/4)])/(12
8*Sqrt[x]*(1 + x^2)^(1/4)) + ((x^2 + x^4)^(1/4)*ArcTanh[(2^(1/4)*Sqrt[x])/(1 + x^2)^(1/4)])/(2^(3/4)*Sqrt[x]*(
1 + x^2)^(1/4))

Rule 209

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

Rule 212

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

Rule 304

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 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && 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 338

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 1)/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[
p, -2^(-1)] && IntegersQ[m, p + (m + 1)/n]

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 508

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> With[{k = Denominato
r[p]}, Dist[k*(a^(p + (m + 1)/n)/n), Subst[Int[x^(k*((m + 1)/n) - 1)*((c - (b*c - a*d)*x^k)^q/(1 - b*x^k)^(p +
q + (m + 1)/n + 1)), x], x, x^(n/k)/(a + b*x^n)^(1/k)], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && Ration
alQ[m, p] && IntegersQ[p + (m + 1)/n, q] && LtQ[-1, p, 0]

Rule 1600

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*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^4} \int \frac {\sqrt {x} \sqrt [4]{1+x^2} \left (-1-x^4+x^8\right )}{-1+x^4} \, dx}{\sqrt {x} \sqrt [4]{1+x^2}} \\ & = \frac {\sqrt [4]{x^2+x^4} \int \frac {\sqrt {x} \left (-1-x^4+x^8\right )}{\left (-1+x^2\right ) \left (1+x^2\right )^{3/4}} \, dx}{\sqrt {x} \sqrt [4]{1+x^2}} \\ & = \frac {\sqrt [4]{x^2+x^4} \int \left (\frac {x^{9/2}}{\left (1+x^2\right )^{3/4}}+\frac {x^{13/2}}{\left (1+x^2\right )^{3/4}}-\frac {\sqrt {x}}{\left (-1+x^2\right ) \left (1+x^2\right )^{3/4}}\right ) \, dx}{\sqrt {x} \sqrt [4]{1+x^2}} \\ & = \frac {\sqrt [4]{x^2+x^4} \int \frac {x^{9/2}}{\left (1+x^2\right )^{3/4}} \, dx}{\sqrt {x} \sqrt [4]{1+x^2}}+\frac {\sqrt [4]{x^2+x^4} \int \frac {x^{13/2}}{\left (1+x^2\right )^{3/4}} \, dx}{\sqrt {x} \sqrt [4]{1+x^2}}-\frac {\sqrt [4]{x^2+x^4} \int \frac {\sqrt {x}}{\left (-1+x^2\right ) \left (1+x^2\right )^{3/4}} \, dx}{\sqrt {x} \sqrt [4]{1+x^2}} \\ & = \frac {1}{4} x^3 \sqrt [4]{x^2+x^4}+\frac {1}{6} x^5 \sqrt [4]{x^2+x^4}-\frac {\left (7 \sqrt [4]{x^2+x^4}\right ) \int \frac {x^{5/2}}{\left (1+x^2\right )^{3/4}} \, dx}{8 \sqrt {x} \sqrt [4]{1+x^2}}-\frac {\left (11 \sqrt [4]{x^2+x^4}\right ) \int \frac {x^{9/2}}{\left (1+x^2\right )^{3/4}} \, dx}{12 \sqrt {x} \sqrt [4]{1+x^2}}-\frac {\left (2 \sqrt [4]{x^2+x^4}\right ) \text {Subst}\left (\int \frac {x^2}{\left (-1+x^4\right ) \left (1+x^4\right )^{3/4}} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt [4]{1+x^2}} \\ & = -\frac {7}{16} x \sqrt [4]{x^2+x^4}+\frac {1}{48} x^3 \sqrt [4]{x^2+x^4}+\frac {1}{6} x^5 \sqrt [4]{x^2+x^4}+\frac {\left (21 \sqrt [4]{x^2+x^4}\right ) \int \frac {\sqrt {x}}{\left (1+x^2\right )^{3/4}} \, dx}{32 \sqrt {x} \sqrt [4]{1+x^2}}+\frac {\left (77 \sqrt [4]{x^2+x^4}\right ) \int \frac {x^{5/2}}{\left (1+x^2\right )^{3/4}} \, dx}{96 \sqrt {x} \sqrt [4]{1+x^2}}-\frac {\left (2 \sqrt [4]{x^2+x^4}\right ) \text {Subst}\left (\int \frac {x^2}{-1+2 x^4} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{\sqrt {x} \sqrt [4]{1+x^2}} \\ & = -\frac {7}{192} x \sqrt [4]{x^2+x^4}+\frac {1}{48} x^3 \sqrt [4]{x^2+x^4}+\frac {1}{6} x^5 \sqrt [4]{x^2+x^4}-\frac {\left (77 \sqrt [4]{x^2+x^4}\right ) \int \frac {\sqrt {x}}{\left (1+x^2\right )^{3/4}} \, dx}{128 \sqrt {x} \sqrt [4]{1+x^2}}+\frac {\left (21 \sqrt [4]{x^2+x^4}\right ) \text {Subst}\left (\int \frac {x^2}{\left (1+x^4\right )^{3/4}} \, dx,x,\sqrt {x}\right )}{16 \sqrt {x} \sqrt [4]{1+x^2}}+\frac {\sqrt [4]{x^2+x^4} \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{\sqrt {2} \sqrt {x} \sqrt [4]{1+x^2}}-\frac {\sqrt [4]{x^2+x^4} \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{\sqrt {2} \sqrt {x} \sqrt [4]{1+x^2}} \\ & = -\frac {7}{192} x \sqrt [4]{x^2+x^4}+\frac {1}{48} x^3 \sqrt [4]{x^2+x^4}+\frac {1}{6} x^5 \sqrt [4]{x^2+x^4}-\frac {\sqrt [4]{x^2+x^4} \arctan \left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{2^{3/4} \sqrt {x} \sqrt [4]{1+x^2}}+\frac {\sqrt [4]{x^2+x^4} \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{2^{3/4} \sqrt {x} \sqrt [4]{1+x^2}}-\frac {\left (77 \sqrt [4]{x^2+x^4}\right ) \text {Subst}\left (\int \frac {x^2}{\left (1+x^4\right )^{3/4}} \, dx,x,\sqrt {x}\right )}{64 \sqrt {x} \sqrt [4]{1+x^2}}+\frac {\left (21 \sqrt [4]{x^2+x^4}\right ) \text {Subst}\left (\int \frac {x^2}{1-x^4} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{16 \sqrt {x} \sqrt [4]{1+x^2}} \\ & = -\frac {7}{192} x \sqrt [4]{x^2+x^4}+\frac {1}{48} x^3 \sqrt [4]{x^2+x^4}+\frac {1}{6} x^5 \sqrt [4]{x^2+x^4}-\frac {\sqrt [4]{x^2+x^4} \arctan \left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{2^{3/4} \sqrt {x} \sqrt [4]{1+x^2}}+\frac {\sqrt [4]{x^2+x^4} \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{2^{3/4} \sqrt {x} \sqrt [4]{1+x^2}}+\frac {\left (21 \sqrt [4]{x^2+x^4}\right ) \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{32 \sqrt {x} \sqrt [4]{1+x^2}}-\frac {\left (21 \sqrt [4]{x^2+x^4}\right ) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{32 \sqrt {x} \sqrt [4]{1+x^2}}-\frac {\left (77 \sqrt [4]{x^2+x^4}\right ) \text {Subst}\left (\int \frac {x^2}{1-x^4} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{64 \sqrt {x} \sqrt [4]{1+x^2}} \\ & = -\frac {7}{192} x \sqrt [4]{x^2+x^4}+\frac {1}{48} x^3 \sqrt [4]{x^2+x^4}+\frac {1}{6} x^5 \sqrt [4]{x^2+x^4}-\frac {21 \sqrt [4]{x^2+x^4} \arctan \left (\frac {\sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{32 \sqrt {x} \sqrt [4]{1+x^2}}-\frac {\sqrt [4]{x^2+x^4} \arctan \left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{2^{3/4} \sqrt {x} \sqrt [4]{1+x^2}}+\frac {21 \sqrt [4]{x^2+x^4} \text {arctanh}\left (\frac {\sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{32 \sqrt {x} \sqrt [4]{1+x^2}}+\frac {\sqrt [4]{x^2+x^4} \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{2^{3/4} \sqrt {x} \sqrt [4]{1+x^2}}-\frac {\left (77 \sqrt [4]{x^2+x^4}\right ) \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{128 \sqrt {x} \sqrt [4]{1+x^2}}+\frac {\left (77 \sqrt [4]{x^2+x^4}\right ) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{128 \sqrt {x} \sqrt [4]{1+x^2}} \\ & = -\frac {7}{192} x \sqrt [4]{x^2+x^4}+\frac {1}{48} x^3 \sqrt [4]{x^2+x^4}+\frac {1}{6} x^5 \sqrt [4]{x^2+x^4}-\frac {7 \sqrt [4]{x^2+x^4} \arctan \left (\frac {\sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{128 \sqrt {x} \sqrt [4]{1+x^2}}-\frac {\sqrt [4]{x^2+x^4} \arctan \left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{2^{3/4} \sqrt {x} \sqrt [4]{1+x^2}}+\frac {7 \sqrt [4]{x^2+x^4} \text {arctanh}\left (\frac {\sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{128 \sqrt {x} \sqrt [4]{1+x^2}}+\frac {\sqrt [4]{x^2+x^4} \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{2^{3/4} \sqrt {x} \sqrt [4]{1+x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.45 \[ \int \frac {\sqrt [4]{x^2+x^4} \left (-1-x^4+x^8\right )}{-1+x^4} \, dx=\frac {\sqrt [4]{x^2+x^4} \left (-14 x^{3/2} \sqrt [4]{1+x^2}+8 x^{7/2} \sqrt [4]{1+x^2}+64 x^{11/2} \sqrt [4]{1+x^2}-21 \arctan \left (\frac {\sqrt {x}}{\sqrt [4]{1+x^2}}\right )-192 \sqrt [4]{2} \arctan \left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{1+x^2}}\right )+21 \text {arctanh}\left (\frac {\sqrt {x}}{\sqrt [4]{1+x^2}}\right )+192 \sqrt [4]{2} \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{1+x^2}}\right )\right )}{384 \sqrt {x} \sqrt [4]{1+x^2}} \]

[In]

Integrate[((x^2 + x^4)^(1/4)*(-1 - x^4 + x^8))/(-1 + x^4),x]

[Out]

((x^2 + x^4)^(1/4)*(-14*x^(3/2)*(1 + x^2)^(1/4) + 8*x^(7/2)*(1 + x^2)^(1/4) + 64*x^(11/2)*(1 + x^2)^(1/4) - 21
*ArcTan[Sqrt[x]/(1 + x^2)^(1/4)] - 192*2^(1/4)*ArcTan[(2^(1/4)*Sqrt[x])/(1 + x^2)^(1/4)] + 21*ArcTanh[Sqrt[x]/
(1 + x^2)^(1/4)] + 192*2^(1/4)*ArcTanh[(2^(1/4)*Sqrt[x])/(1 + x^2)^(1/4)]))/(384*Sqrt[x]*(1 + x^2)^(1/4))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(234\) vs. \(2(94)=188\).

Time = 3.23 (sec) , antiderivative size = 235, normalized size of antiderivative = 2.01

method	result	size
pseudoelliptic	\(\frac {x^{6} \left (128 x^{5} \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}+16 \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}} x^{3}+192 \,2^{\frac {1}{4}} \ln \left (\frac {-2^{\frac {1}{4}} x -\left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}{2^{\frac {1}{4}} x -\left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}\right )+384 \,2^{\frac {1}{4}} \arctan \left (\frac {\left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}} 2^{\frac {3}{4}}}{2 x}\right )-28 \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}} x +21 \ln \left (\frac {x +\left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}{x}\right )-21 \ln \left (\frac {\left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}-x}{x}\right )+42 \arctan \left (\frac {\left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}{x}\right )\right )}{768 {\left (x +\left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}\right )}^{3} {\left (\left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}-x \right )}^{3} \left (x^{2}+\sqrt {x^{2} \left (x^{2}+1\right )}\right )^{3}}\)	\(235\)

[In]

int((x^4+x^2)^(1/4)*(x^8-x^4-1)/(x^4-1),x,method=_RETURNVERBOSE)

[Out]

1/768*x^6*(128*x^5*(x^2*(x^2+1))^(1/4)+16*(x^2*(x^2+1))^(1/4)*x^3+192*2^(1/4)*ln((-2^(1/4)*x-(x^2*(x^2+1))^(1/
4))/(2^(1/4)*x-(x^2*(x^2+1))^(1/4)))+384*2^(1/4)*arctan(1/2*(x^2*(x^2+1))^(1/4)/x*2^(3/4))-28*(x^2*(x^2+1))^(1
/4)*x+21*ln((x+(x^2*(x^2+1))^(1/4))/x)-21*ln(((x^2*(x^2+1))^(1/4)-x)/x)+42*arctan((x^2*(x^2+1))^(1/4)/x))/(x+(
x^2*(x^2+1))^(1/4))^3/((x^2*(x^2+1))^(1/4)-x)^3/(x^2+(x^2*(x^2+1))^(1/2))^3

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 4.81 (sec) , antiderivative size = 400, normalized size of antiderivative = 3.42 \[ \int \frac {\sqrt [4]{x^2+x^4} \left (-1-x^4+x^8\right )}{-1+x^4} \, dx=\frac {1}{32} \cdot 8^{\frac {3}{4}} \log \left (\frac {4 \, \sqrt {2} {\left (x^{4} + x^{2}\right )}^{\frac {1}{4}} x^{2} + 8^{\frac {3}{4}} \sqrt {x^{4} + x^{2}} x + 8^{\frac {1}{4}} {\left (3 \, x^{3} + x\right )} + 4 \, {\left (x^{4} + x^{2}\right )}^{\frac {3}{4}}}{x^{3} - x}\right ) - \frac {1}{32} i \cdot 8^{\frac {3}{4}} \log \left (-\frac {4 \, \sqrt {2} {\left (x^{4} + x^{2}\right )}^{\frac {1}{4}} x^{2} + i \cdot 8^{\frac {3}{4}} \sqrt {x^{4} + x^{2}} x - 8^{\frac {1}{4}} {\left (3 i \, x^{3} + i \, x\right )} - 4 \, {\left (x^{4} + x^{2}\right )}^{\frac {3}{4}}}{x^{3} - x}\right ) + \frac {1}{32} i \cdot 8^{\frac {3}{4}} \log \left (-\frac {4 \, \sqrt {2} {\left (x^{4} + x^{2}\right )}^{\frac {1}{4}} x^{2} - i \cdot 8^{\frac {3}{4}} \sqrt {x^{4} + x^{2}} x - 8^{\frac {1}{4}} {\left (-3 i \, x^{3} - i \, x\right )} - 4 \, {\left (x^{4} + x^{2}\right )}^{\frac {3}{4}}}{x^{3} - x}\right ) - \frac {1}{32} \cdot 8^{\frac {3}{4}} \log \left (\frac {4 \, \sqrt {2} {\left (x^{4} + x^{2}\right )}^{\frac {1}{4}} x^{2} - 8^{\frac {3}{4}} \sqrt {x^{4} + x^{2}} x - 8^{\frac {1}{4}} {\left (3 \, x^{3} + x\right )} + 4 \, {\left (x^{4} + x^{2}\right )}^{\frac {3}{4}}}{x^{3} - x}\right ) + \frac {1}{192} \, {\left (32 \, x^{5} + 4 \, x^{3} - 7 \, x\right )} {\left (x^{4} + x^{2}\right )}^{\frac {1}{4}} - \frac {7}{256} \, \arctan \left (\frac {2 \, {\left ({\left (x^{4} + x^{2}\right )}^{\frac {1}{4}} x^{2} + {\left (x^{4} + x^{2}\right )}^{\frac {3}{4}}\right )}}{x}\right ) + \frac {7}{256} \, \log \left (\frac {2 \, x^{3} + 2 \, {\left (x^{4} + x^{2}\right )}^{\frac {1}{4}} x^{2} + 2 \, \sqrt {x^{4} + x^{2}} x + x + 2 \, {\left (x^{4} + x^{2}\right )}^{\frac {3}{4}}}{x}\right ) \]

[In]

integrate((x^4+x^2)^(1/4)*(x^8-x^4-1)/(x^4-1),x, algorithm="fricas")

[Out]

1/32*8^(3/4)*log((4*sqrt(2)*(x^4 + x^2)^(1/4)*x^2 + 8^(3/4)*sqrt(x^4 + x^2)*x + 8^(1/4)*(3*x^3 + x) + 4*(x^4 +
x^2)^(3/4))/(x^3 - x)) - 1/32*I*8^(3/4)*log(-(4*sqrt(2)*(x^4 + x^2)^(1/4)*x^2 + I*8^(3/4)*sqrt(x^4 + x^2)*x -
8^(1/4)*(3*I*x^3 + I*x) - 4*(x^4 + x^2)^(3/4))/(x^3 - x)) + 1/32*I*8^(3/4)*log(-(4*sqrt(2)*(x^4 + x^2)^(1/4)*
x^2 - I*8^(3/4)*sqrt(x^4 + x^2)*x - 8^(1/4)*(-3*I*x^3 - I*x) - 4*(x^4 + x^2)^(3/4))/(x^3 - x)) - 1/32*8^(3/4)*
log((4*sqrt(2)*(x^4 + x^2)^(1/4)*x^2 - 8^(3/4)*sqrt(x^4 + x^2)*x - 8^(1/4)*(3*x^3 + x) + 4*(x^4 + x^2)^(3/4))/
(x^3 - x)) + 1/192*(32*x^5 + 4*x^3 - 7*x)*(x^4 + x^2)^(1/4) - 7/256*arctan(2*((x^4 + x^2)^(1/4)*x^2 + (x^4 + x
^2)^(3/4))/x) + 7/256*log((2*x^3 + 2*(x^4 + x^2)^(1/4)*x^2 + 2*sqrt(x^4 + x^2)*x + x + 2*(x^4 + x^2)^(3/4))/x)

Sympy [F]

\[ \int \frac {\sqrt [4]{x^2+x^4} \left (-1-x^4+x^8\right )}{-1+x^4} \, dx=\int \frac {\sqrt [4]{x^{2} \left (x^{2} + 1\right )} \left (x^{8} - x^{4} - 1\right )}{\left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right )}\, dx \]

[In]

integrate((x**4+x**2)**(1/4)*(x**8-x**4-1)/(x**4-1),x)

[Out]

Integral((x**2*(x**2 + 1))**(1/4)*(x**8 - x**4 - 1)/((x - 1)*(x + 1)*(x**2 + 1)), x)

Maxima [F]

\[ \int \frac {\sqrt [4]{x^2+x^4} \left (-1-x^4+x^8\right )}{-1+x^4} \, dx=\int { \frac {{\left (x^{8} - x^{4} - 1\right )} {\left (x^{4} + x^{2}\right )}^{\frac {1}{4}}}{x^{4} - 1} \,d x } \]

[In]

integrate((x^4+x^2)^(1/4)*(x^8-x^4-1)/(x^4-1),x, algorithm="maxima")

[Out]

integrate((x^8 - x^4 - 1)*(x^4 + x^2)^(1/4)/(x^4 - 1), x)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.05 \[ \int \frac {\sqrt [4]{x^2+x^4} \left (-1-x^4+x^8\right )}{-1+x^4} \, dx=-\frac {1}{192} \, {\left (7 \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {9}{4}} - 18 \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {5}{4}} - 21 \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}\right )} x^{6} + \frac {1}{2} \cdot 2^{\frac {1}{4}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {3}{4}} {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}\right ) + \frac {1}{4} \cdot 2^{\frac {1}{4}} \log \left (2^{\frac {1}{4}} + {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{4} \cdot 2^{\frac {1}{4}} \log \left ({\left | -2^{\frac {1}{4}} + {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}} \right |}\right ) + \frac {7}{128} \, \arctan \left ({\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}\right ) + \frac {7}{256} \, \log \left ({\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}} + 1\right ) - \frac {7}{256} \, \log \left ({\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}} - 1\right ) \]

[In]

integrate((x^4+x^2)^(1/4)*(x^8-x^4-1)/(x^4-1),x, algorithm="giac")

[Out]

-1/192*(7*(1/x^2 + 1)^(9/4) - 18*(1/x^2 + 1)^(5/4) - 21*(1/x^2 + 1)^(1/4))*x^6 + 1/2*2^(1/4)*arctan(1/2*2^(3/4
)*(1/x^2 + 1)^(1/4)) + 1/4*2^(1/4)*log(2^(1/4) + (1/x^2 + 1)^(1/4)) - 1/4*2^(1/4)*log(abs(-2^(1/4) + (1/x^2 +
1)^(1/4))) + 7/128*arctan((1/x^2 + 1)^(1/4)) + 7/256*log((1/x^2 + 1)^(1/4) + 1) - 7/256*log((1/x^2 + 1)^(1/4)
- 1)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt [4]{x^2+x^4} \left (-1-x^4+x^8\right )}{-1+x^4} \, dx=\int -\frac {{\left (x^4+x^2\right )}^{1/4}\,\left (-x^8+x^4+1\right )}{x^4-1} \,d x \]

[In]

int(-((x^2 + x^4)^(1/4)*(x^4 - x^8 + 1))/(x^4 - 1),x)

[Out]

int(-((x^2 + x^4)^(1/4)*(x^4 - x^8 + 1))/(x^4 - 1), x)