\(\int \frac {b+2 a x^4}{(-b+a x^4) \sqrt [4]{b x^2+a x^4}} \, dx\) [1766]

Optimal result

Integrand size = 35, antiderivative size = 119 \[ \int \frac {b+2 a x^4}{\left (-b+a x^4\right ) \sqrt [4]{b x^2+a x^4}} \, dx=\frac {2 \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^2+a x^4}}\right )}{\sqrt [4]{a}}+\frac {2 \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^2+a x^4}}\right )}{\sqrt [4]{a}}+\frac {3}{4} \text {RootSum}\left [a^2-a b-2 a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-\log (x)+\log \left (\sqrt [4]{b x^2+a x^4}-x \text {$\#$1}\right )}{\text {$\#$1}}\&\right ] \]

[Out]

Unintegrable

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(503\) vs. \(2(119)=238\).

Time = 0.65 (sec) , antiderivative size = 503, normalized size of antiderivative = 4.23, number of steps used = 17, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.257, Rules used = {2081, 6847, 6857, 246, 218, 212, 209, 1443, 385} \[ \int \frac {b+2 a x^4}{\left (-b+a x^4\right ) \sqrt [4]{b x^2+a x^4}} \, dx=\frac {2 \sqrt {x} \sqrt [4]{a x^2+b} \arctan \left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{\sqrt [4]{a} \sqrt [4]{a x^4+b x^2}}-\frac {3 \sqrt {x} \sqrt [4]{a x^2+b} \arctan \left (\frac {\sqrt [8]{a} \sqrt {x} \sqrt [4]{\sqrt {a}-\sqrt {b}}}{\sqrt [4]{a x^2+b}}\right )}{2 \sqrt [8]{a} \sqrt [4]{\sqrt {a}-\sqrt {b}} \sqrt [4]{a x^4+b x^2}}-\frac {3 \sqrt {x} \sqrt [4]{a x^2+b} \arctan \left (\frac {\sqrt [8]{a} \sqrt {x} \sqrt [4]{\sqrt {a}+\sqrt {b}}}{\sqrt [4]{a x^2+b}}\right )}{2 \sqrt [8]{a} \sqrt [4]{\sqrt {a}+\sqrt {b}} \sqrt [4]{a x^4+b x^2}}+\frac {2 \sqrt {x} \sqrt [4]{a x^2+b} \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{\sqrt [4]{a} \sqrt [4]{a x^4+b x^2}}-\frac {3 \sqrt {x} \sqrt [4]{a x^2+b} \text {arctanh}\left (\frac {\sqrt [8]{a} \sqrt {x} \sqrt [4]{\sqrt {a}-\sqrt {b}}}{\sqrt [4]{a x^2+b}}\right )}{2 \sqrt [8]{a} \sqrt [4]{\sqrt {a}-\sqrt {b}} \sqrt [4]{a x^4+b x^2}}-\frac {3 \sqrt {x} \sqrt [4]{a x^2+b} \text {arctanh}\left (\frac {\sqrt [8]{a} \sqrt {x} \sqrt [4]{\sqrt {a}+\sqrt {b}}}{\sqrt [4]{a x^2+b}}\right )}{2 \sqrt [8]{a} \sqrt [4]{\sqrt {a}+\sqrt {b}} \sqrt [4]{a x^4+b x^2}} \]

[In]

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

[Out]

(2*Sqrt[x]*(b + a*x^2)^(1/4)*ArcTan[(a^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)])/(a^(1/4)*(b*x^2 + a*x^4)^(1/4)) - (3
*Sqrt[x]*(b + a*x^2)^(1/4)*ArcTan[(a^(1/8)*(Sqrt[a] - Sqrt[b])^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)])/(2*a^(1/8)*(
Sqrt[a] - Sqrt[b])^(1/4)*(b*x^2 + a*x^4)^(1/4)) - (3*Sqrt[x]*(b + a*x^2)^(1/4)*ArcTan[(a^(1/8)*(Sqrt[a] + Sqrt
[b])^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)])/(2*a^(1/8)*(Sqrt[a] + Sqrt[b])^(1/4)*(b*x^2 + a*x^4)^(1/4)) + (2*Sqrt[
x]*(b + a*x^2)^(1/4)*ArcTanh[(a^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)])/(a^(1/4)*(b*x^2 + a*x^4)^(1/4)) - (3*Sqrt[x
]*(b + a*x^2)^(1/4)*ArcTanh[(a^(1/8)*(Sqrt[a] - Sqrt[b])^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)])/(2*a^(1/8)*(Sqrt[a
] - Sqrt[b])^(1/4)*(b*x^2 + a*x^4)^(1/4)) - (3*Sqrt[x]*(b + a*x^2)^(1/4)*ArcTanh[(a^(1/8)*(Sqrt[a] + Sqrt[b])^
(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)])/(2*a^(1/8)*(Sqrt[a] + Sqrt[b])^(1/4)*(b*x^2 + a*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 1443

Int[((d_) + (e_.)*(x_)^(n_))^(q_)/((a_) + (c_.)*(x_)^(n2_)), x_Symbol] :> With[{r = Rt[(-a)*c, 2]}, Dist[-c/(2
*r), Int[(d + e*x^n)^q/(r - c*x^n), x], x] - Dist[c/(2*r), Int[(d + e*x^n)^q/(r + c*x^n), x], x]] /; FreeQ[{a,
c, d, e, n, q}, x] && EqQ[n2, 2*n] && NeQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[q]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.30 \[ \int \frac {b+2 a x^4}{\left (-b+a x^4\right ) \sqrt [4]{b x^2+a x^4}} \, dx=\frac {\sqrt {x} \sqrt [4]{b+a x^2} \left (8 \left (\arctan \left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{b+a x^2}}\right )+\text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{b+a x^2}}\right )\right )+3 \sqrt [4]{a} \text {RootSum}\left [a^2-a b-2 a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-\log \left (\sqrt {x}\right )+\log \left (\sqrt [4]{b+a x^2}-\sqrt {x} \text {$\#$1}\right )}{\text {$\#$1}}\&\right ]\right )}{4 \sqrt [4]{a} \sqrt [4]{x^2 \left (b+a x^2\right )}} \]

[In]

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

[Out]

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

Maple [N/A] (verified)

Time = 0.37 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.08

method	result	size
pseudoelliptic	\(\frac {3 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}-2 a \,\textit {\_Z}^{4}+a^{2}-a b \right )}{\sum }\frac {\ln \left (\frac {-\textit {\_R} x +\left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}{x}\right )}{\textit {\_R}}\right ) a^{\frac {1}{4}}-8 \arctan \left (\frac {\left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )+4 \ln \left (\frac {a^{\frac {1}{4}} x +\left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}{-a^{\frac {1}{4}} x +\left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}\right )}{4 a^{\frac {1}{4}}}\)	\(128\)

[In]

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

[Out]

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

Fricas [F(-1)]

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

[In]

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

[Out]

Timed out

Sympy [N/A]

Not integrable

Time = 5.28 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.24 \[ \int \frac {b+2 a x^4}{\left (-b+a x^4\right ) \sqrt [4]{b x^2+a x^4}} \, dx=\int \frac {2 a x^{4} + b}{\sqrt [4]{x^{2} \left (a x^{2} + b\right )} \left (a x^{4} - b\right )}\, dx \]

[In]

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

[Out]

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

Maxima [N/A]

Not integrable

Time = 0.24 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.29 \[ \int \frac {b+2 a x^4}{\left (-b+a x^4\right ) \sqrt [4]{b x^2+a x^4}} \, dx=\int { \frac {2 \, a x^{4} + b}{{\left (a x^{4} + b x^{2}\right )}^{\frac {1}{4}} {\left (a x^{4} - b\right )}} \,d x } \]

[In]

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

[Out]

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

Giac [N/A]

Not integrable

Time = 0.87 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.29 \[ \int \frac {b+2 a x^4}{\left (-b+a x^4\right ) \sqrt [4]{b x^2+a x^4}} \, dx=\int { \frac {2 \, a x^{4} + b}{{\left (a x^{4} + b x^{2}\right )}^{\frac {1}{4}} {\left (a x^{4} - b\right )}} \,d x } \]

[In]

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

[Out]

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

Mupad [N/A]

Not integrable

Time = 0.00 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.29 \[ \int \frac {b+2 a x^4}{\left (-b+a x^4\right ) \sqrt [4]{b x^2+a x^4}} \, dx=\int -\frac {2\,a\,x^4+b}{\left (b-a\,x^4\right )\,{\left (a\,x^4+b\,x^2\right )}^{1/4}} \,d x \]

[In]

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

[Out]

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