\(\int \frac {1+2 x^4}{(-1+x^4) \sqrt [4]{x^2+x^4}} \, dx\) [1656]

Optimal result

Integrand size = 26, antiderivative size = 112 \[ \int \frac {1+2 x^4}{\left (-1+x^4\right ) \sqrt [4]{x^2+x^4}} \, dx=-\frac {3 \left (x^2+x^4\right )^{3/4}}{x \left (1+x^2\right )}+2 \arctan \left (\frac {x}{\sqrt [4]{x^2+x^4}}\right )-\frac {3 \arctan \left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^2+x^4}}\right )}{2 \sqrt [4]{2}}+2 \text {arctanh}\left (\frac {x}{\sqrt [4]{x^2+x^4}}\right )-\frac {3 \text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^2+x^4}}\right )}{2 \sqrt [4]{2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.88, number of steps used = 14, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {2081, 6847, 6857, 246, 218, 212, 209, 1418, 390, 385} \[ \int \frac {1+2 x^4}{\left (-1+x^4\right ) \sqrt [4]{x^2+x^4}} \, dx=\frac {2 \sqrt [4]{x^2+1} \sqrt {x} \arctan \left (\frac {\sqrt {x}}{\sqrt [4]{x^2+1}}\right )}{\sqrt [4]{x^4+x^2}}-\frac {3 \sqrt [4]{x^2+1} \sqrt {x} \arctan \left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{x^2+1}}\right )}{2 \sqrt [4]{2} \sqrt [4]{x^4+x^2}}+\frac {2 \sqrt [4]{x^2+1} \sqrt {x} \text {arctanh}\left (\frac {\sqrt {x}}{\sqrt [4]{x^2+1}}\right )}{\sqrt [4]{x^4+x^2}}-\frac {3 \sqrt [4]{x^2+1} \sqrt {x} \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{x^2+1}}\right )}{2 \sqrt [4]{2} \sqrt [4]{x^4+x^2}}-\frac {3 x}{\sqrt [4]{x^4+x^2}} \]

[In]

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

[Out]

(-3*x)/(x^2 + x^4)^(1/4) + (2*Sqrt[x]*(1 + x^2)^(1/4)*ArcTan[Sqrt[x]/(1 + x^2)^(1/4)])/(x^2 + x^4)^(1/4) - (3*
Sqrt[x]*(1 + x^2)^(1/4)*ArcTan[(2^(1/4)*Sqrt[x])/(1 + x^2)^(1/4)])/(2*2^(1/4)*(x^2 + x^4)^(1/4)) + (2*Sqrt[x]*
(1 + x^2)^(1/4)*ArcTanh[Sqrt[x]/(1 + x^2)^(1/4)])/(x^2 + x^4)^(1/4) - (3*Sqrt[x]*(1 + x^2)^(1/4)*ArcTanh[(2^(1
/4)*Sqrt[x])/(1 + x^2)^(1/4)])/(2*2^(1/4)*(x^2 + x^4)^(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 218

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

Rule 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + 1/n), Subst[Int[1/(1 - b*x^n)^(p + 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)] && IntegerQ[p
+ 1/n]

Rule 385

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

Rule 390

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[(b*c + n*(p + 1)*(b*c - a*d))/(a*n*(p + 1)*(b*c - a
*d)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, q}, x] && NeQ[b*c - a*d, 0] && Eq
Q[n*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  !LtQ[q, -1]) && NeQ[p, -1]

Rule 1418

Int[((d_) + (e_.)*(x_)^(n_))^(q_.)*((a_) + (c_.)*(x_)^(n2_))^(p_.), x_Symbol] :> Int[(d + e*x^n)^(p + q)*(a/d
+ (c/e)*x^n)^p, x] /; FreeQ[{a, c, d, e, n, q}, x] && EqQ[n2, 2*n] && EqQ[c*d^2 + a*e^2, 0] && IntegerQ[p]

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 6847

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

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 {\left (\sqrt {x} \sqrt [4]{1+x^2}\right ) \int \frac {1+2 x^4}{\sqrt {x} \sqrt [4]{1+x^2} \left (-1+x^4\right )} \, dx}{\sqrt [4]{x^2+x^4}} \\ & = \frac {\left (2 \sqrt {x} \sqrt [4]{1+x^2}\right ) \text {Subst}\left (\int \frac {1+2 x^8}{\sqrt [4]{1+x^4} \left (-1+x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2+x^4}} \\ & = \frac {\left (2 \sqrt {x} \sqrt [4]{1+x^2}\right ) \text {Subst}\left (\int \left (\frac {2}{\sqrt [4]{1+x^4}}+\frac {3}{\sqrt [4]{1+x^4} \left (-1+x^8\right )}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2+x^4}} \\ & = \frac {\left (4 \sqrt {x} \sqrt [4]{1+x^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [4]{1+x^4}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2+x^4}}+\frac {\left (6 \sqrt {x} \sqrt [4]{1+x^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [4]{1+x^4} \left (-1+x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2+x^4}} \\ & = \frac {\left (4 \sqrt {x} \sqrt [4]{1+x^2}\right ) \text {Subst}\left (\int \frac {1}{1-x^4} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{\sqrt [4]{x^2+x^4}}+\frac {\left (6 \sqrt {x} \sqrt [4]{1+x^2}\right ) \text {Subst}\left (\int \frac {1}{\left (-1+x^4\right ) \left (1+x^4\right )^{5/4}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2+x^4}} \\ & = -\frac {3 x}{\sqrt [4]{x^2+x^4}}+\frac {\left (2 \sqrt {x} \sqrt [4]{1+x^2}\right ) \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{\sqrt [4]{x^2+x^4}}+\frac {\left (2 \sqrt {x} \sqrt [4]{1+x^2}\right ) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{\sqrt [4]{x^2+x^4}}+\frac {\left (3 \sqrt {x} \sqrt [4]{1+x^2}\right ) \text {Subst}\left (\int \frac {1}{\left (-1+x^4\right ) \sqrt [4]{1+x^4}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2+x^4}} \\ & = -\frac {3 x}{\sqrt [4]{x^2+x^4}}+\frac {2 \sqrt {x} \sqrt [4]{1+x^2} \arctan \left (\frac {\sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{\sqrt [4]{x^2+x^4}}+\frac {2 \sqrt {x} \sqrt [4]{1+x^2} \text {arctanh}\left (\frac {\sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{\sqrt [4]{x^2+x^4}}+\frac {\left (3 \sqrt {x} \sqrt [4]{1+x^2}\right ) \text {Subst}\left (\int \frac {1}{-1+2 x^4} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{\sqrt [4]{x^2+x^4}} \\ & = -\frac {3 x}{\sqrt [4]{x^2+x^4}}+\frac {2 \sqrt {x} \sqrt [4]{1+x^2} \arctan \left (\frac {\sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{\sqrt [4]{x^2+x^4}}+\frac {2 \sqrt {x} \sqrt [4]{1+x^2} \text {arctanh}\left (\frac {\sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{\sqrt [4]{x^2+x^4}}-\frac {\left (3 \sqrt {x} \sqrt [4]{1+x^2}\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{2 \sqrt [4]{x^2+x^4}}-\frac {\left (3 \sqrt {x} \sqrt [4]{1+x^2}\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{2 \sqrt [4]{x^2+x^4}} \\ & = -\frac {3 x}{\sqrt [4]{x^2+x^4}}+\frac {2 \sqrt {x} \sqrt [4]{1+x^2} \arctan \left (\frac {\sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{\sqrt [4]{x^2+x^4}}-\frac {3 \sqrt {x} \sqrt [4]{1+x^2} \arctan \left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{2 \sqrt [4]{2} \sqrt [4]{x^2+x^4}}+\frac {2 \sqrt {x} \sqrt [4]{1+x^2} \text {arctanh}\left (\frac {\sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{\sqrt [4]{x^2+x^4}}-\frac {3 \sqrt {x} \sqrt [4]{1+x^2} \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{2 \sqrt [4]{2} \sqrt [4]{x^2+x^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.39 \[ \int \frac {1+2 x^4}{\left (-1+x^4\right ) \sqrt [4]{x^2+x^4}} \, dx=\frac {\sqrt {x} \left (-12 \sqrt {x}+8 \sqrt [4]{1+x^2} \arctan \left (\frac {\sqrt {x}}{\sqrt [4]{1+x^2}}\right )-3\ 2^{3/4} \sqrt [4]{1+x^2} \arctan \left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{1+x^2}}\right )+8 \sqrt [4]{1+x^2} \text {arctanh}\left (\frac {\sqrt {x}}{\sqrt [4]{1+x^2}}\right )-3\ 2^{3/4} \sqrt [4]{1+x^2} \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{1+x^2}}\right )\right )}{4 \sqrt [4]{x^2+x^4}} \]

[In]

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

[Out]

(Sqrt[x]*(-12*Sqrt[x] + 8*(1 + x^2)^(1/4)*ArcTan[Sqrt[x]/(1 + x^2)^(1/4)] - 3*2^(3/4)*(1 + x^2)^(1/4)*ArcTan[(
2^(1/4)*Sqrt[x])/(1 + x^2)^(1/4)] + 8*(1 + x^2)^(1/4)*ArcTanh[Sqrt[x]/(1 + x^2)^(1/4)] - 3*2^(3/4)*(1 + x^2)^(
1/4)*ArcTanh[(2^(1/4)*Sqrt[x])/(1 + x^2)^(1/4)]))/(4*(x^2 + x^4)^(1/4))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(205\) vs. \(2(90)=180\).

Time = 2.36 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.84

method	result	size
pseudoelliptic	\(\frac {6 \arctan \left (\frac {\left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}} 2^{\frac {3}{4}}}{2 x}\right ) 2^{\frac {3}{4}} \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}-3 \ln \left (\frac {-2^{\frac {1}{4}} x -\left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}{2^{\frac {1}{4}} x -\left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}\right ) 2^{\frac {3}{4}} \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}+8 \ln \left (\frac {x +\left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}{x}\right ) \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}-8 \ln \left (\frac {\left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}-x}{x}\right ) \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}-16 \arctan \left (\frac {\left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}{x}\right ) \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}-24 x}{8 \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}\)	\(206\)

[In]

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

[Out]

1/8*(6*arctan(1/2*(x^2*(x^2+1))^(1/4)/x*2^(3/4))*2^(3/4)*(x^2*(x^2+1))^(1/4)-3*ln((-2^(1/4)*x-(x^2*(x^2+1))^(1
/4))/(2^(1/4)*x-(x^2*(x^2+1))^(1/4)))*2^(3/4)*(x^2*(x^2+1))^(1/4)+8*ln((x+(x^2*(x^2+1))^(1/4))/x)*(x^2*(x^2+1)
)^(1/4)-8*ln(((x^2*(x^2+1))^(1/4)-x)/x)*(x^2*(x^2+1))^(1/4)-16*arctan((x^2*(x^2+1))^(1/4)/x)*(x^2*(x^2+1))^(1/
4)-24*x)/(x^2*(x^2+1))^(1/4)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 10.41 (sec) , antiderivative size = 434, normalized size of antiderivative = 3.88 \[ \int \frac {1+2 x^4}{\left (-1+x^4\right ) \sqrt [4]{x^2+x^4}} \, dx=-\frac {3 \cdot 2^{\frac {3}{4}} {\left (x^{3} + x\right )} \log \left (\frac {4 \, \sqrt {2} {\left (x^{4} + x^{2}\right )}^{\frac {1}{4}} x^{2} + 2^{\frac {3}{4}} {\left (3 \, x^{3} + x\right )} + 4 \cdot 2^{\frac {1}{4}} \sqrt {x^{4} + x^{2}} x + 4 \, {\left (x^{4} + x^{2}\right )}^{\frac {3}{4}}}{x^{3} - x}\right ) - 3 \cdot 2^{\frac {3}{4}} {\left (x^{3} + x\right )} \log \left (\frac {4 \, \sqrt {2} {\left (x^{4} + x^{2}\right )}^{\frac {1}{4}} x^{2} - 2^{\frac {3}{4}} {\left (3 \, x^{3} + x\right )} - 4 \cdot 2^{\frac {1}{4}} \sqrt {x^{4} + x^{2}} x + 4 \, {\left (x^{4} + x^{2}\right )}^{\frac {3}{4}}}{x^{3} - x}\right ) + 3 \cdot 2^{\frac {3}{4}} {\left (i \, x^{3} + i \, x\right )} \log \left (-\frac {4 \, \sqrt {2} {\left (x^{4} + x^{2}\right )}^{\frac {1}{4}} x^{2} - 2^{\frac {3}{4}} {\left (3 i \, x^{3} + i \, x\right )} + 4 i \cdot 2^{\frac {1}{4}} \sqrt {x^{4} + x^{2}} x - 4 \, {\left (x^{4} + x^{2}\right )}^{\frac {3}{4}}}{x^{3} - x}\right ) + 3 \cdot 2^{\frac {3}{4}} {\left (-i \, x^{3} - i \, x\right )} \log \left (-\frac {4 \, \sqrt {2} {\left (x^{4} + x^{2}\right )}^{\frac {1}{4}} x^{2} - 2^{\frac {3}{4}} {\left (-3 i \, x^{3} - i \, x\right )} - 4 i \cdot 2^{\frac {1}{4}} \sqrt {x^{4} + x^{2}} x - 4 \, {\left (x^{4} + x^{2}\right )}^{\frac {3}{4}}}{x^{3} - x}\right ) - 16 \, {\left (x^{3} + x\right )} \arctan \left (\frac {2 \, {\left ({\left (x^{4} + x^{2}\right )}^{\frac {1}{4}} x^{2} + {\left (x^{4} + x^{2}\right )}^{\frac {3}{4}}\right )}}{x}\right ) - 16 \, {\left (x^{3} + x\right )} \log \left (\frac {2 \, x^{3} + 2 \, {\left (x^{4} + x^{2}\right )}^{\frac {1}{4}} x^{2} + 2 \, \sqrt {x^{4} + x^{2}} x + x + 2 \, {\left (x^{4} + x^{2}\right )}^{\frac {3}{4}}}{x}\right ) + 48 \, {\left (x^{4} + x^{2}\right )}^{\frac {3}{4}}}{16 \, {\left (x^{3} + x\right )}} \]

[In]

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

[Out]

-1/16*(3*2^(3/4)*(x^3 + x)*log((4*sqrt(2)*(x^4 + x^2)^(1/4)*x^2 + 2^(3/4)*(3*x^3 + x) + 4*2^(1/4)*sqrt(x^4 + x
^2)*x + 4*(x^4 + x^2)^(3/4))/(x^3 - x)) - 3*2^(3/4)*(x^3 + x)*log((4*sqrt(2)*(x^4 + x^2)^(1/4)*x^2 - 2^(3/4)*(
3*x^3 + x) - 4*2^(1/4)*sqrt(x^4 + x^2)*x + 4*(x^4 + x^2)^(3/4))/(x^3 - x)) + 3*2^(3/4)*(I*x^3 + I*x)*log(-(4*s
qrt(2)*(x^4 + x^2)^(1/4)*x^2 - 2^(3/4)*(3*I*x^3 + I*x) + 4*I*2^(1/4)*sqrt(x^4 + x^2)*x - 4*(x^4 + x^2)^(3/4))/
(x^3 - x)) + 3*2^(3/4)*(-I*x^3 - I*x)*log(-(4*sqrt(2)*(x^4 + x^2)^(1/4)*x^2 - 2^(3/4)*(-3*I*x^3 - I*x) - 4*I*2
^(1/4)*sqrt(x^4 + x^2)*x - 4*(x^4 + x^2)^(3/4))/(x^3 - x)) - 16*(x^3 + x)*arctan(2*((x^4 + x^2)^(1/4)*x^2 + (x
^4 + x^2)^(3/4))/x) - 16*(x^3 + x)*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) + 48*(x^4 + x^2)^(3/4))/(x^3 + x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.87 \[ \int \frac {1+2 x^4}{\left (-1+x^4\right ) \sqrt [4]{x^2+x^4}} \, dx=\frac {3}{4} \cdot 2^{\frac {3}{4}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {3}{4}} {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}\right ) - \frac {3}{8} \cdot 2^{\frac {3}{4}} \log \left (2^{\frac {1}{4}} + {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}\right ) + \frac {3}{8} \cdot 2^{\frac {3}{4}} \log \left ({\left | -2^{\frac {1}{4}} + {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}} \right |}\right ) - \frac {3}{{\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}} - 2 \, \arctan \left ({\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}\right ) + \log \left ({\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}} + 1\right ) - \log \left ({\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}} - 1\right ) \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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