\(\int \frac {-2 b-a x^4+2 x^8}{x^4 \sqrt [4]{-b+a x^4} (-b+2 a x^4)} \, dx\) [2669]

Optimal result

Integrand size = 44, antiderivative size = 239 \[ \int \frac {-2 b-a x^4+2 x^8}{x^4 \sqrt [4]{-b+a x^4} \left (-b+2 a x^4\right )} \, dx=\frac {2 \left (-b+a x^4\right )^{3/4}}{3 b x^3}+\frac {\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{2 a^{5/4}}+\frac {\left (5 a^2-b\right ) \arctan \left (\frac {\sqrt {2} \sqrt [4]{a} x \sqrt [4]{-b+a x^4}}{-\sqrt {a} x^2+\sqrt {-b+a x^4}}\right )}{2 \sqrt {2} a^{5/4} b}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{2 a^{5/4}}+\frac {\left (5 a^2-b\right ) \text {arctanh}\left (\frac {\sqrt {a} x^2+\sqrt {-b+a x^4}}{\sqrt {2} \sqrt [4]{a} x \sqrt [4]{-b+a x^4}}\right )}{2 \sqrt {2} a^{5/4} b} \]

[Out]

2/3*(a*x^4-b)^(3/4)/b/x^3+1/2*arctan(a^(1/4)*x/(a*x^4-b)^(1/4))/a^(5/4)+1/4*(5*a^2-b)*arctan(2^(1/2)*a^(1/4)*x
*(a*x^4-b)^(1/4)/(-a^(1/2)*x^2+(a*x^4-b)^(1/2)))*2^(1/2)/a^(5/4)/b+1/2*arctanh(a^(1/4)*x/(a*x^4-b)^(1/4))/a^(5
/4)+1/4*(5*a^2-b)*arctanh(1/2*(a^(1/2)*x^2+(a*x^4-b)^(1/2))*2^(1/2)/a^(1/4)/x/(a*x^4-b)^(1/4))*2^(1/2)/a^(5/4)
/b

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 346, normalized size of antiderivative = 1.45, number of steps used = 17, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.295, Rules used = {6857, 246, 218, 212, 209, 270, 385, 217, 1179, 642, 1176, 631, 210} \[ \int \frac {-2 b-a x^4+2 x^8}{x^4 \sqrt [4]{-b+a x^4} \left (-b+2 a x^4\right )} \, dx=\frac {\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{2 a^{5/4}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{2 a^{5/4}}-\frac {\left (5 a^2-b\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{2 \sqrt {2} a^{5/4} b}+\frac {\left (5 a^2-b\right ) \arctan \left (\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}+1\right )}{2 \sqrt {2} a^{5/4} b}-\frac {\left (5 a^2-b\right ) \log \left (-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}+\frac {\sqrt {a} x^2}{\sqrt {a x^4-b}}+1\right )}{4 \sqrt {2} a^{5/4} b}+\frac {\left (5 a^2-b\right ) \log \left (\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}+\frac {\sqrt {a} x^2}{\sqrt {a x^4-b}}+1\right )}{4 \sqrt {2} a^{5/4} b}+\frac {2 \left (a x^4-b\right )^{3/4}}{3 b x^3} \]

[In]

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

[Out]

(2*(-b + a*x^4)^(3/4))/(3*b*x^3) + ArcTan[(a^(1/4)*x)/(-b + a*x^4)^(1/4)]/(2*a^(5/4)) - ((5*a^2 - b)*ArcTan[1
- (Sqrt[2]*a^(1/4)*x)/(-b + a*x^4)^(1/4)])/(2*Sqrt[2]*a^(5/4)*b) + ((5*a^2 - b)*ArcTan[1 + (Sqrt[2]*a^(1/4)*x)
/(-b + a*x^4)^(1/4)])/(2*Sqrt[2]*a^(5/4)*b) + ArcTanh[(a^(1/4)*x)/(-b + a*x^4)^(1/4)]/(2*a^(5/4)) - ((5*a^2 -
b)*Log[1 + (Sqrt[a]*x^2)/Sqrt[-b + a*x^4] - (Sqrt[2]*a^(1/4)*x)/(-b + a*x^4)^(1/4)])/(4*Sqrt[2]*a^(5/4)*b) + (
(5*a^2 - b)*Log[1 + (Sqrt[a]*x^2)/Sqrt[-b + a*x^4] + (Sqrt[2]*a^(1/4)*x)/(-b + a*x^4)^(1/4)])/(4*Sqrt[2]*a^(5/
4)*b)

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 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[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 217

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

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 270

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] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

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 631

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 642

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 1176

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 1179

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

Mathematica [A] (verified)

Time = 1.00 (sec) , antiderivative size = 231, normalized size of antiderivative = 0.97 \[ \int \frac {-2 b-a x^4+2 x^8}{x^4 \sqrt [4]{-b+a x^4} \left (-b+2 a x^4\right )} \, dx=\frac {8 a^{5/4} \left (-b+a x^4\right )^{3/4}+6 b x^3 \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )+3 \sqrt {2} \left (5 a^2-b\right ) x^3 \arctan \left (\frac {\sqrt {2} \sqrt [4]{a} x \sqrt [4]{-b+a x^4}}{-\sqrt {a} x^2+\sqrt {-b+a x^4}}\right )+6 b x^3 \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )+3 \sqrt {2} \left (5 a^2-b\right ) x^3 \text {arctanh}\left (\frac {\sqrt {a} x^2+\sqrt {-b+a x^4}}{\sqrt {2} \sqrt [4]{a} x \sqrt [4]{-b+a x^4}}\right )}{12 a^{5/4} b x^3} \]

[In]

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

[Out]

(8*a^(5/4)*(-b + a*x^4)^(3/4) + 6*b*x^3*ArcTan[(a^(1/4)*x)/(-b + a*x^4)^(1/4)] + 3*Sqrt[2]*(5*a^2 - b)*x^3*Arc
Tan[(Sqrt[2]*a^(1/4)*x*(-b + a*x^4)^(1/4))/(-(Sqrt[a]*x^2) + Sqrt[-b + a*x^4])] + 6*b*x^3*ArcTanh[(a^(1/4)*x)/
(-b + a*x^4)^(1/4)] + 3*Sqrt[2]*(5*a^2 - b)*x^3*ArcTanh[(Sqrt[a]*x^2 + Sqrt[-b + a*x^4])/(Sqrt[2]*a^(1/4)*x*(-
b + a*x^4)^(1/4))])/(12*a^(5/4)*b*x^3)

Maple [A] (verified)

Time = 1.30 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.18

method	result	size
pseudoelliptic	\(\frac {-\frac {5 \left (a^{2}-\frac {b}{5}\right ) x^{3} \sqrt {2}\, \ln \left (\frac {-\left (a \,x^{4}-b \right )^{\frac {1}{4}} x \,a^{\frac {1}{4}} \sqrt {2}+\sqrt {a}\, x^{2}+\sqrt {a \,x^{4}-b}}{\left (a \,x^{4}-b \right )^{\frac {1}{4}} x \,a^{\frac {1}{4}} \sqrt {2}+\sqrt {a}\, x^{2}+\sqrt {a \,x^{4}-b}}\right )}{8}+\frac {\ln \left (\frac {a^{\frac {1}{4}} x +\left (a \,x^{4}-b \right )^{\frac {1}{4}}}{-a^{\frac {1}{4}} x +\left (a \,x^{4}-b \right )^{\frac {1}{4}}}\right ) b \,x^{3}}{4}-\frac {5 \left (a^{2}-\frac {b}{5}\right ) x^{3} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (a \,x^{4}-b \right )^{\frac {1}{4}}-a^{\frac {1}{4}} x}{a^{\frac {1}{4}} x}\right )}{4}-\frac {5 \left (a^{2}-\frac {b}{5}\right ) x^{3} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (a \,x^{4}-b \right )^{\frac {1}{4}}+a^{\frac {1}{4}} x}{a^{\frac {1}{4}} x}\right )}{4}+\frac {2 \left (a \,x^{4}-b \right )^{\frac {3}{4}} a^{\frac {5}{4}}}{3}-\frac {\arctan \left (\frac {\left (a \,x^{4}-b \right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right ) b \,x^{3}}{2}}{x^{3} a^{\frac {5}{4}} b}\)	\(283\)

[In]

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

[Out]

2/3*(-15/16*(a^2-1/5*b)*x^3*2^(1/2)*ln((-(a*x^4-b)^(1/4)*x*a^(1/4)*2^(1/2)+a^(1/2)*x^2+(a*x^4-b)^(1/2))/((a*x^
4-b)^(1/4)*x*a^(1/4)*2^(1/2)+a^(1/2)*x^2+(a*x^4-b)^(1/2)))+3/8*ln((a^(1/4)*x+(a*x^4-b)^(1/4))/(-a^(1/4)*x+(a*x
^4-b)^(1/4)))*b*x^3-15/8*(a^2-1/5*b)*x^3*2^(1/2)*arctan((2^(1/2)*(a*x^4-b)^(1/4)-a^(1/4)*x)/a^(1/4)/x)-15/8*(a
^2-1/5*b)*x^3*2^(1/2)*arctan((2^(1/2)*(a*x^4-b)^(1/4)+a^(1/4)*x)/a^(1/4)/x)+(a*x^4-b)^(3/4)*a^(5/4)-3/4*arctan
(1/a^(1/4)/x*(a*x^4-b)^(1/4))*b*x^3)/a^(5/4)/x^3/b

Fricas [F(-1)]

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

[In]

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

[Out]

Timed out

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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