\(\int \frac {b-2 c x^4+2 a x^8}{\sqrt [4]{-b+a x^4} (-2 b-c x^4+2 a x^8)} \, dx\) [2520]

Optimal result

Integrand size = 46, antiderivative size = 210 \[ \int \frac {b-2 c x^4+2 a x^8}{\sqrt [4]{-b+a x^4} \left (-2 b-c x^4+2 a x^8\right )} \, dx=\frac {\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{2 \sqrt [4]{a}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{2 \sqrt [4]{a}}+\frac {1}{4} \text {RootSum}\left [2 a^2-2 a b+a c-4 a \text {$\#$1}^4-c \text {$\#$1}^4+2 \text {$\#$1}^8\&,\frac {-3 a \log (x)+c \log (x)+3 a \log \left (\sqrt [4]{-b+a x^4}-x \text {$\#$1}\right )-c \log \left (\sqrt [4]{-b+a x^4}-x \text {$\#$1}\right )+3 \log (x) \text {$\#$1}^4-3 \log \left (\sqrt [4]{-b+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4}{4 a \text {$\#$1}+c \text {$\#$1}-4 \text {$\#$1}^5}\&\right ] \]

[Out]

Unintegrable

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(611\) vs. \(2(210)=420\).

Time = 1.68 (sec) , antiderivative size = 611, normalized size of antiderivative = 2.91, number of steps used = 16, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {6860, 246, 218, 212, 209, 385, 214, 211} \[ \int \frac {b-2 c x^4+2 a x^8}{\sqrt [4]{-b+a x^4} \left (-2 b-c x^4+2 a x^8\right )} \, dx=-\frac {\left (\frac {12 a b-c^2}{\sqrt {16 a b+c^2}}+c\right ) \arctan \left (\frac {\sqrt [4]{a} x \sqrt [4]{\sqrt {16 a b+c^2}+4 b-c}}{\sqrt [4]{\sqrt {16 a b+c^2}-c} \sqrt [4]{a x^4-b}}\right )}{2 \sqrt [4]{a} \left (\sqrt {16 a b+c^2}-c\right )^{3/4} \sqrt [4]{\sqrt {16 a b+c^2}+4 b-c}}+\frac {\left (c-\frac {12 a b-c^2}{\sqrt {16 a b+c^2}}\right ) \arctan \left (\frac {\sqrt [4]{a} x \sqrt [4]{\sqrt {16 a b+c^2}-4 b+c}}{\sqrt [4]{\sqrt {16 a b+c^2}+c} \sqrt [4]{a x^4-b}}\right )}{2 \sqrt [4]{a} \left (\sqrt {16 a b+c^2}+c\right )^{3/4} \sqrt [4]{\sqrt {16 a b+c^2}-4 b+c}}+\frac {\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{2 \sqrt [4]{a}}-\frac {\left (\frac {12 a b-c^2}{\sqrt {16 a b+c^2}}+c\right ) \text {arctanh}\left (\frac {\sqrt [4]{a} x \sqrt [4]{\sqrt {16 a b+c^2}+4 b-c}}{\sqrt [4]{\sqrt {16 a b+c^2}-c} \sqrt [4]{a x^4-b}}\right )}{2 \sqrt [4]{a} \left (\sqrt {16 a b+c^2}-c\right )^{3/4} \sqrt [4]{\sqrt {16 a b+c^2}+4 b-c}}+\frac {\left (c-\frac {12 a b-c^2}{\sqrt {16 a b+c^2}}\right ) \text {arctanh}\left (\frac {\sqrt [4]{a} x \sqrt [4]{\sqrt {16 a b+c^2}-4 b+c}}{\sqrt [4]{\sqrt {16 a b+c^2}+c} \sqrt [4]{a x^4-b}}\right )}{2 \sqrt [4]{a} \left (\sqrt {16 a b+c^2}+c\right )^{3/4} \sqrt [4]{\sqrt {16 a b+c^2}-4 b+c}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{2 \sqrt [4]{a}} \]

[In]

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

[Out]

ArcTan[(a^(1/4)*x)/(-b + a*x^4)^(1/4)]/(2*a^(1/4)) - ((c + (12*a*b - c^2)/Sqrt[16*a*b + c^2])*ArcTan[(a^(1/4)*
(4*b - c + Sqrt[16*a*b + c^2])^(1/4)*x)/((-c + Sqrt[16*a*b + c^2])^(1/4)*(-b + a*x^4)^(1/4))])/(2*a^(1/4)*(-c
+ Sqrt[16*a*b + c^2])^(3/4)*(4*b - c + Sqrt[16*a*b + c^2])^(1/4)) + ((c - (12*a*b - c^2)/Sqrt[16*a*b + c^2])*A
rcTan[(a^(1/4)*(-4*b + c + Sqrt[16*a*b + c^2])^(1/4)*x)/((c + Sqrt[16*a*b + c^2])^(1/4)*(-b + a*x^4)^(1/4))])/
(2*a^(1/4)*(c + Sqrt[16*a*b + c^2])^(3/4)*(-4*b + c + Sqrt[16*a*b + c^2])^(1/4)) + ArcTanh[(a^(1/4)*x)/(-b + a
*x^4)^(1/4)]/(2*a^(1/4)) - ((c + (12*a*b - c^2)/Sqrt[16*a*b + c^2])*ArcTanh[(a^(1/4)*(4*b - c + Sqrt[16*a*b +
c^2])^(1/4)*x)/((-c + Sqrt[16*a*b + c^2])^(1/4)*(-b + a*x^4)^(1/4))])/(2*a^(1/4)*(-c + Sqrt[16*a*b + c^2])^(3/
4)*(4*b - c + Sqrt[16*a*b + c^2])^(1/4)) + ((c - (12*a*b - c^2)/Sqrt[16*a*b + c^2])*ArcTanh[(a^(1/4)*(-4*b + c
+ Sqrt[16*a*b + c^2])^(1/4)*x)/((c + Sqrt[16*a*b + c^2])^(1/4)*(-b + a*x^4)^(1/4))])/(2*a^(1/4)*(c + Sqrt[16*
a*b + c^2])^(3/4)*(-4*b + c + Sqrt[16*a*b + c^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 211

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

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 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/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 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 6860

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{\sqrt [4]{-b+a x^4}}+\frac {3 b-c x^4}{\sqrt [4]{-b+a x^4} \left (-2 b-c x^4+2 a x^8\right )}\right ) \, dx \\ & = \int \frac {1}{\sqrt [4]{-b+a x^4}} \, dx+\int \frac {3 b-c x^4}{\sqrt [4]{-b+a x^4} \left (-2 b-c x^4+2 a x^8\right )} \, dx \\ & = \int \left (\frac {-c+\frac {12 a b-c^2}{\sqrt {16 a b+c^2}}}{\sqrt [4]{-b+a x^4} \left (-c-\sqrt {16 a b+c^2}+4 a x^4\right )}+\frac {-c-\frac {12 a b-c^2}{\sqrt {16 a b+c^2}}}{\sqrt [4]{-b+a x^4} \left (-c+\sqrt {16 a b+c^2}+4 a x^4\right )}\right ) \, dx+\text {Subst}\left (\int \frac {1}{1-a x^4} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {1}{1-\sqrt {a} x^2} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right )+\frac {1}{2} \text {Subst}\left (\int \frac {1}{1+\sqrt {a} x^2} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right )+\left (-c-\frac {12 a b-c^2}{\sqrt {16 a b+c^2}}\right ) \int \frac {1}{\sqrt [4]{-b+a x^4} \left (-c+\sqrt {16 a b+c^2}+4 a x^4\right )} \, dx+\left (-c+\frac {12 a b-c^2}{\sqrt {16 a b+c^2}}\right ) \int \frac {1}{\sqrt [4]{-b+a x^4} \left (-c-\sqrt {16 a b+c^2}+4 a x^4\right )} \, dx \\ & = \frac {\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{2 \sqrt [4]{a}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{2 \sqrt [4]{a}}+\left (-c-\frac {12 a b-c^2}{\sqrt {16 a b+c^2}}\right ) \text {Subst}\left (\int \frac {1}{-c+\sqrt {16 a b+c^2}-\left (4 a b+a \left (-c+\sqrt {16 a b+c^2}\right )\right ) x^4} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right )+\left (-c+\frac {12 a b-c^2}{\sqrt {16 a b+c^2}}\right ) \text {Subst}\left (\int \frac {1}{-c-\sqrt {16 a b+c^2}-\left (4 a b+a \left (-c-\sqrt {16 a b+c^2}\right )\right ) x^4} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right ) \\ & = \frac {\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{2 \sqrt [4]{a}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{2 \sqrt [4]{a}}-\frac {\left (c+\frac {12 a b-c^2}{\sqrt {16 a b+c^2}}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {-c+\sqrt {16 a b+c^2}}-\sqrt {a} \sqrt {4 b-c+\sqrt {16 a b+c^2}} x^2} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right )}{2 \sqrt {-c+\sqrt {16 a b+c^2}}}-\frac {\left (c+\frac {12 a b-c^2}{\sqrt {16 a b+c^2}}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {-c+\sqrt {16 a b+c^2}}+\sqrt {a} \sqrt {4 b-c+\sqrt {16 a b+c^2}} x^2} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right )}{2 \sqrt {-c+\sqrt {16 a b+c^2}}}+\frac {\left (c-\frac {12 a b-c^2}{\sqrt {16 a b+c^2}}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {c+\sqrt {16 a b+c^2}}-\sqrt {a} \sqrt {-4 b+c+\sqrt {16 a b+c^2}} x^2} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right )}{2 \sqrt {c+\sqrt {16 a b+c^2}}}+\frac {\left (c-\frac {12 a b-c^2}{\sqrt {16 a b+c^2}}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {c+\sqrt {16 a b+c^2}}+\sqrt {a} \sqrt {-4 b+c+\sqrt {16 a b+c^2}} x^2} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right )}{2 \sqrt {c+\sqrt {16 a b+c^2}}} \\ & = \frac {\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{2 \sqrt [4]{a}}-\frac {\left (c+\frac {12 a b-c^2}{\sqrt {16 a b+c^2}}\right ) \arctan \left (\frac {\sqrt [4]{a} \sqrt [4]{4 b-c+\sqrt {16 a b+c^2}} x}{\sqrt [4]{-c+\sqrt {16 a b+c^2}} \sqrt [4]{-b+a x^4}}\right )}{2 \sqrt [4]{a} \left (-c+\sqrt {16 a b+c^2}\right )^{3/4} \sqrt [4]{4 b-c+\sqrt {16 a b+c^2}}}+\frac {\left (c-\frac {12 a b-c^2}{\sqrt {16 a b+c^2}}\right ) \arctan \left (\frac {\sqrt [4]{a} \sqrt [4]{-4 b+c+\sqrt {16 a b+c^2}} x}{\sqrt [4]{c+\sqrt {16 a b+c^2}} \sqrt [4]{-b+a x^4}}\right )}{2 \sqrt [4]{a} \left (c+\sqrt {16 a b+c^2}\right )^{3/4} \sqrt [4]{-4 b+c+\sqrt {16 a b+c^2}}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{2 \sqrt [4]{a}}-\frac {\left (c+\frac {12 a b-c^2}{\sqrt {16 a b+c^2}}\right ) \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt [4]{4 b-c+\sqrt {16 a b+c^2}} x}{\sqrt [4]{-c+\sqrt {16 a b+c^2}} \sqrt [4]{-b+a x^4}}\right )}{2 \sqrt [4]{a} \left (-c+\sqrt {16 a b+c^2}\right )^{3/4} \sqrt [4]{4 b-c+\sqrt {16 a b+c^2}}}+\frac {\left (c-\frac {12 a b-c^2}{\sqrt {16 a b+c^2}}\right ) \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt [4]{-4 b+c+\sqrt {16 a b+c^2}} x}{\sqrt [4]{c+\sqrt {16 a b+c^2}} \sqrt [4]{-b+a x^4}}\right )}{2 \sqrt [4]{a} \left (c+\sqrt {16 a b+c^2}\right )^{3/4} \sqrt [4]{-4 b+c+\sqrt {16 a b+c^2}}} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.85 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.96 \[ \int \frac {b-2 c x^4+2 a x^8}{\sqrt [4]{-b+a x^4} \left (-2 b-c x^4+2 a x^8\right )} \, dx=\frac {1}{4} \left (\frac {2 \left (\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )+\text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )\right )}{\sqrt [4]{a}}+\text {RootSum}\left [2 a^2-2 a b+a c-4 a \text {$\#$1}^4-c \text {$\#$1}^4+2 \text {$\#$1}^8\&,\frac {3 a \log (x)-c \log (x)-3 a \log \left (\sqrt [4]{-b+a x^4}-x \text {$\#$1}\right )+c \log \left (\sqrt [4]{-b+a x^4}-x \text {$\#$1}\right )-3 \log (x) \text {$\#$1}^4+3 \log \left (\sqrt [4]{-b+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4}{-4 a \text {$\#$1}-c \text {$\#$1}+4 \text {$\#$1}^5}\&\right ]\right ) \]

[In]

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

[Out]

((2*(ArcTan[(a^(1/4)*x)/(-b + a*x^4)^(1/4)] + ArcTanh[(a^(1/4)*x)/(-b + a*x^4)^(1/4)]))/a^(1/4) + RootSum[2*a^
2 - 2*a*b + a*c - 4*a*#1^4 - c*#1^4 + 2*#1^8 & , (3*a*Log[x] - c*Log[x] - 3*a*Log[(-b + a*x^4)^(1/4) - x*#1] +
c*Log[(-b + a*x^4)^(1/4) - x*#1] - 3*Log[x]*#1^4 + 3*Log[(-b + a*x^4)^(1/4) - x*#1]*#1^4)/(-4*a*#1 - c*#1 + 4
*#1^5) & ])/4

Maple [N/A] (verified)

Time = 0.38 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.75

method	result	size
pseudoelliptic	\(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (2 \textit {\_Z}^{8}+\left (-4 a -c \right ) \textit {\_Z}^{4}+2 a^{2}-2 a b +a c \right )}{\sum }\frac {\left (3 \textit {\_R}^{4}-3 a +c \right ) \ln \left (\frac {-\textit {\_R} x +\left (a \,x^{4}-b \right )^{\frac {1}{4}}}{x}\right )}{\textit {\_R} \left (4 \textit {\_R}^{4}-4 a -c \right )}\right ) a^{\frac {1}{4}}-2 \arctan \left (\frac {\left (a \,x^{4}-b \right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )+\ln \left (\frac {-x \,a^{\frac {1}{4}}-\left (a \,x^{4}-b \right )^{\frac {1}{4}}}{x \,a^{\frac {1}{4}}-\left (a \,x^{4}-b \right )^{\frac {1}{4}}}\right )}{4 a^{\frac {1}{4}}}\)	\(157\)

[In]

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

[Out]

1/4*(sum(1/_R*(3*_R^4-3*a+c)*ln((-_R*x+(a*x^4-b)^(1/4))/x)/(4*_R^4-4*a-c),_R=RootOf(2*_Z^8+(-4*a-c)*_Z^4+2*a^2
-2*a*b+a*c))*a^(1/4)-2*arctan(1/a^(1/4)/x*(a*x^4-b)^(1/4))+ln((-x*a^(1/4)-(a*x^4-b)^(1/4))/(x*a^(1/4)-(a*x^4-b
)^(1/4))))/a^(1/4)

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 3 vs. order 1.

Time = 6.47 (sec) , antiderivative size = 16424, normalized size of antiderivative = 78.21 \[ \int \frac {b-2 c x^4+2 a x^8}{\sqrt [4]{-b+a x^4} \left (-2 b-c x^4+2 a x^8\right )} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Too large to include

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [N/A]

Not integrable

Time = 0.21 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.22 \[ \int \frac {b-2 c x^4+2 a x^8}{\sqrt [4]{-b+a x^4} \left (-2 b-c x^4+2 a x^8\right )} \, dx=\int { \frac {2 \, a x^{8} - 2 \, c x^{4} + b}{{\left (2 \, a x^{8} - c x^{4} - 2 \, b\right )} {\left (a x^{4} - b\right )}^{\frac {1}{4}}} \,d x } \]

[In]

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

[Out]

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

Giac [N/A]

Not integrable

Time = 1.77 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.22 \[ \int \frac {b-2 c x^4+2 a x^8}{\sqrt [4]{-b+a x^4} \left (-2 b-c x^4+2 a x^8\right )} \, dx=\int { \frac {2 \, a x^{8} - 2 \, c x^{4} + b}{{\left (2 \, a x^{8} - c x^{4} - 2 \, b\right )} {\left (a x^{4} - b\right )}^{\frac {1}{4}}} \,d x } \]

[In]

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

[Out]

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

Mupad [N/A]

Not integrable

Time = 7.40 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.22 \[ \int \frac {b-2 c x^4+2 a x^8}{\sqrt [4]{-b+a x^4} \left (-2 b-c x^4+2 a x^8\right )} \, dx=\int -\frac {2\,a\,x^8-2\,c\,x^4+b}{{\left (a\,x^4-b\right )}^{1/4}\,\left (-2\,a\,x^8+c\,x^4+2\,b\right )} \,d x \]

[In]

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

[Out]

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