\(\int \frac {b+a x^4}{(-b+x^2+a x^4) \sqrt [4]{-b x^2+a x^6}} \, dx\) [1721]

Optimal result

Integrand size = 38, antiderivative size = 116 \[ \int \frac {b+a x^4}{\left (-b+x^2+a x^4\right ) \sqrt [4]{-b x^2+a x^6}} \, dx=-\frac {\arctan \left (\frac {\sqrt {2} x \sqrt [4]{-b x^2+a x^6}}{-x^2+\sqrt {-b x^2+a x^6}}\right )}{\sqrt {2}}-\frac {\text {arctanh}\left (\frac {\frac {x^2}{\sqrt {2}}+\frac {\sqrt {-b x^2+a x^6}}{\sqrt {2}}}{x \sqrt [4]{-b x^2+a x^6}}\right )}{\sqrt {2}} \]

[Out]

-1/2*arctan(2^(1/2)*x*(a*x^6-b*x^2)^(1/4)/(-x^2+(a*x^6-b*x^2)^(1/2)))*2^(1/2)-1/2*arctanh((1/2*2^(1/2)*x^2+1/2
*(a*x^6-b*x^2)^(1/2)*2^(1/2))/x/(a*x^6-b*x^2)^(1/4))*2^(1/2)

Rubi [C] (verified)

Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.

Time = 1.39 (sec) , antiderivative size = 447, normalized size of antiderivative = 3.85, number of steps used = 20, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {2081, 6847, 6860, 252, 251, 1452, 441, 440, 525, 524} \[ \int \frac {b+a x^4}{\left (-b+x^2+a x^4\right ) \sqrt [4]{-b x^2+a x^6}} \, dx=-\frac {2 x \sqrt [4]{1-\frac {a x^4}{b}} \operatorname {AppellF1}\left (\frac {1}{8},1,\frac {1}{4},\frac {9}{8},\frac {2 a^2 x^4}{2 a b-\sqrt {4 a b+1}+1},\frac {a x^4}{b}\right )}{\sqrt [4]{a x^6-b x^2}}-\frac {2 x \sqrt [4]{1-\frac {a x^4}{b}} \operatorname {AppellF1}\left (\frac {1}{8},1,\frac {1}{4},\frac {9}{8},\frac {2 a^2 x^4}{2 a b+\sqrt {4 a b+1}+1},\frac {a x^4}{b}\right )}{\sqrt [4]{a x^6-b x^2}}+\frac {2 a x^3 \left (1-\sqrt {4 a b+1}\right ) \sqrt [4]{1-\frac {a x^4}{b}} \operatorname {AppellF1}\left (\frac {5}{8},1,\frac {1}{4},\frac {13}{8},\frac {2 a^2 x^4}{2 a b-\sqrt {4 a b+1}+1},\frac {a x^4}{b}\right )}{5 \left (2 a b-\sqrt {4 a b+1}+1\right ) \sqrt [4]{a x^6-b x^2}}+\frac {2 a x^3 \left (\sqrt {4 a b+1}+1\right ) \sqrt [4]{1-\frac {a x^4}{b}} \operatorname {AppellF1}\left (\frac {5}{8},1,\frac {1}{4},\frac {13}{8},\frac {2 a^2 x^4}{2 a b+\sqrt {4 a b+1}+1},\frac {a x^4}{b}\right )}{5 \left (2 a b+\sqrt {4 a b+1}+1\right ) \sqrt [4]{a x^6-b x^2}}+\frac {2 x \sqrt [4]{1-\frac {a x^4}{b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{4},\frac {9}{8},\frac {a x^4}{b}\right )}{\sqrt [4]{a x^6-b x^2}} \]

[In]

Int[(b + a*x^4)/((-b + x^2 + a*x^4)*(-(b*x^2) + a*x^6)^(1/4)),x]

[Out]

(-2*x*(1 - (a*x^4)/b)^(1/4)*AppellF1[1/8, 1, 1/4, 9/8, (2*a^2*x^4)/(1 + 2*a*b - Sqrt[1 + 4*a*b]), (a*x^4)/b])/
(-(b*x^2) + a*x^6)^(1/4) - (2*x*(1 - (a*x^4)/b)^(1/4)*AppellF1[1/8, 1, 1/4, 9/8, (2*a^2*x^4)/(1 + 2*a*b + Sqrt
[1 + 4*a*b]), (a*x^4)/b])/(-(b*x^2) + a*x^6)^(1/4) + (2*a*(1 - Sqrt[1 + 4*a*b])*x^3*(1 - (a*x^4)/b)^(1/4)*Appe
llF1[5/8, 1, 1/4, 13/8, (2*a^2*x^4)/(1 + 2*a*b - Sqrt[1 + 4*a*b]), (a*x^4)/b])/(5*(1 + 2*a*b - Sqrt[1 + 4*a*b]
)*(-(b*x^2) + a*x^6)^(1/4)) + (2*a*(1 + Sqrt[1 + 4*a*b])*x^3*(1 - (a*x^4)/b)^(1/4)*AppellF1[5/8, 1, 1/4, 13/8,
(2*a^2*x^4)/(1 + 2*a*b + Sqrt[1 + 4*a*b]), (a*x^4)/b])/(5*(1 + 2*a*b + Sqrt[1 + 4*a*b])*(-(b*x^2) + a*x^6)^(1
/4)) + (2*x*(1 - (a*x^4)/b)^(1/4)*Hypergeometric2F1[1/8, 1/4, 9/8, (a*x^4)/b])/(-(b*x^2) + a*x^6)^(1/4)

Rule 251

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-p, 1/n, 1/n + 1, (-b)*(x^n/a)],
x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p
] || GtQ[a, 0])

Rule 252

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^Fra
cPart[p]), Int[(1 + b*(x^n/a))^p, x], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILt
Q[Simplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 440

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
-q, 1 + 1/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 441

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^F
racPart[p]/(1 + b*(x^n/a))^FracPart[p]), Int[(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n,
p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 524

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*
((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 525

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPar
t[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p]), Int[(e*x)^m*(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x
] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] &&  !(IntegerQ[
p] || GtQ[a, 0])

Rule 1452

Int[((d_) + (e_.)*(x_)^(n_))^(q_)*((a_) + (c_.)*(x_)^(n2_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + c*x^(2
*n))^p, (d/(d^2 - e^2*x^(2*n)) - e*(x^n/(d^2 - e^2*x^(2*n))))^(-q), x], x] /; FreeQ[{a, c, d, e, n, p}, x] &&
EqQ[n2, 2*n] && NeQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[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 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 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}& = \frac {\left (\sqrt {x} \sqrt [4]{-b+a x^4}\right ) \int \frac {b+a x^4}{\sqrt {x} \sqrt [4]{-b+a x^4} \left (-b+x^2+a x^4\right )} \, dx}{\sqrt [4]{-b x^2+a x^6}} \\ & = \frac {\left (2 \sqrt {x} \sqrt [4]{-b+a x^4}\right ) \text {Subst}\left (\int \frac {b+a x^8}{\sqrt [4]{-b+a x^8} \left (-b+x^4+a x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{-b x^2+a x^6}} \\ & = \frac {\left (2 \sqrt {x} \sqrt [4]{-b+a x^4}\right ) \text {Subst}\left (\int \left (\frac {1}{\sqrt [4]{-b+a x^8}}+\frac {2 b-x^4}{\sqrt [4]{-b+a x^8} \left (-b+x^4+a x^8\right )}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt [4]{-b x^2+a x^6}} \\ & = \frac {\left (2 \sqrt {x} \sqrt [4]{-b+a x^4}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [4]{-b+a x^8}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{-b x^2+a x^6}}+\frac {\left (2 \sqrt {x} \sqrt [4]{-b+a x^4}\right ) \text {Subst}\left (\int \frac {2 b-x^4}{\sqrt [4]{-b+a x^8} \left (-b+x^4+a x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{-b x^2+a x^6}} \\ & = \frac {\left (2 \sqrt {x} \sqrt [4]{-b+a x^4}\right ) \text {Subst}\left (\int \left (\frac {-1+\sqrt {1+4 a b}}{\left (1-\sqrt {1+4 a b}+2 a x^4\right ) \sqrt [4]{-b+a x^8}}+\frac {-1-\sqrt {1+4 a b}}{\left (1+\sqrt {1+4 a b}+2 a x^4\right ) \sqrt [4]{-b+a x^8}}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt [4]{-b x^2+a x^6}}+\frac {\left (2 \sqrt {x} \sqrt [4]{1-\frac {a x^4}{b}}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [4]{1-\frac {a x^8}{b}}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{-b x^2+a x^6}} \\ & = \frac {2 x \sqrt [4]{1-\frac {a x^4}{b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{4},\frac {9}{8},\frac {a x^4}{b}\right )}{\sqrt [4]{-b x^2+a x^6}}+\frac {\left (2 \left (-1-\sqrt {1+4 a b}\right ) \sqrt {x} \sqrt [4]{-b+a x^4}\right ) \text {Subst}\left (\int \frac {1}{\left (1+\sqrt {1+4 a b}+2 a x^4\right ) \sqrt [4]{-b+a x^8}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{-b x^2+a x^6}}+\frac {\left (2 \left (-1+\sqrt {1+4 a b}\right ) \sqrt {x} \sqrt [4]{-b+a x^4}\right ) \text {Subst}\left (\int \frac {1}{\left (1-\sqrt {1+4 a b}+2 a x^4\right ) \sqrt [4]{-b+a x^8}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{-b x^2+a x^6}} \\ & = \frac {2 x \sqrt [4]{1-\frac {a x^4}{b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{4},\frac {9}{8},\frac {a x^4}{b}\right )}{\sqrt [4]{-b x^2+a x^6}}+\frac {\left (2 \left (-1-\sqrt {1+4 a b}\right ) \sqrt {x} \sqrt [4]{-b+a x^4}\right ) \text {Subst}\left (\int \left (\frac {1+\sqrt {1+4 a b}}{2 \sqrt [4]{-b+a x^8} \left (1+2 a b+\sqrt {1+4 a b}-2 a^2 x^8\right )}+\frac {a x^4}{\sqrt [4]{-b+a x^8} \left (-1-2 a b-\sqrt {1+4 a b}+2 a^2 x^8\right )}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt [4]{-b x^2+a x^6}}+\frac {\left (2 \left (-1+\sqrt {1+4 a b}\right ) \sqrt {x} \sqrt [4]{-b+a x^4}\right ) \text {Subst}\left (\int \left (\frac {-1+\sqrt {1+4 a b}}{2 \sqrt [4]{-b+a x^8} \left (-1-2 a b+\sqrt {1+4 a b}+2 a^2 x^8\right )}+\frac {a x^4}{\sqrt [4]{-b+a x^8} \left (-1-2 a b+\sqrt {1+4 a b}+2 a^2 x^8\right )}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt [4]{-b x^2+a x^6}} \\ & = \frac {2 x \sqrt [4]{1-\frac {a x^4}{b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{4},\frac {9}{8},\frac {a x^4}{b}\right )}{\sqrt [4]{-b x^2+a x^6}}+\frac {\left (2 a \left (-1-\sqrt {1+4 a b}\right ) \sqrt {x} \sqrt [4]{-b+a x^4}\right ) \text {Subst}\left (\int \frac {x^4}{\sqrt [4]{-b+a x^8} \left (-1-2 a b-\sqrt {1+4 a b}+2 a^2 x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{-b x^2+a x^6}}+\frac {\left (2 a \left (-1+\sqrt {1+4 a b}\right ) \sqrt {x} \sqrt [4]{-b+a x^4}\right ) \text {Subst}\left (\int \frac {x^4}{\sqrt [4]{-b+a x^8} \left (-1-2 a b+\sqrt {1+4 a b}+2 a^2 x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{-b x^2+a x^6}}+\frac {\left (\left (-1+\sqrt {1+4 a b}\right )^2 \sqrt {x} \sqrt [4]{-b+a x^4}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [4]{-b+a x^8} \left (-1-2 a b+\sqrt {1+4 a b}+2 a^2 x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{-b x^2+a x^6}}+\frac {\left (\left (-1-\sqrt {1+4 a b}\right ) \left (1+\sqrt {1+4 a b}\right ) \sqrt {x} \sqrt [4]{-b+a x^4}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [4]{-b+a x^8} \left (1+2 a b+\sqrt {1+4 a b}-2 a^2 x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{-b x^2+a x^6}} \\ & = \frac {2 x \sqrt [4]{1-\frac {a x^4}{b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{4},\frac {9}{8},\frac {a x^4}{b}\right )}{\sqrt [4]{-b x^2+a x^6}}+\frac {\left (2 a \left (-1-\sqrt {1+4 a b}\right ) \sqrt {x} \sqrt [4]{1-\frac {a x^4}{b}}\right ) \text {Subst}\left (\int \frac {x^4}{\left (-1-2 a b-\sqrt {1+4 a b}+2 a^2 x^8\right ) \sqrt [4]{1-\frac {a x^8}{b}}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{-b x^2+a x^6}}+\frac {\left (2 a \left (-1+\sqrt {1+4 a b}\right ) \sqrt {x} \sqrt [4]{1-\frac {a x^4}{b}}\right ) \text {Subst}\left (\int \frac {x^4}{\left (-1-2 a b+\sqrt {1+4 a b}+2 a^2 x^8\right ) \sqrt [4]{1-\frac {a x^8}{b}}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{-b x^2+a x^6}}+\frac {\left (\left (-1+\sqrt {1+4 a b}\right )^2 \sqrt {x} \sqrt [4]{1-\frac {a x^4}{b}}\right ) \text {Subst}\left (\int \frac {1}{\left (-1-2 a b+\sqrt {1+4 a b}+2 a^2 x^8\right ) \sqrt [4]{1-\frac {a x^8}{b}}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{-b x^2+a x^6}}+\frac {\left (\left (-1-\sqrt {1+4 a b}\right ) \left (1+\sqrt {1+4 a b}\right ) \sqrt {x} \sqrt [4]{1-\frac {a x^4}{b}}\right ) \text {Subst}\left (\int \frac {1}{\left (1+2 a b+\sqrt {1+4 a b}-2 a^2 x^8\right ) \sqrt [4]{1-\frac {a x^8}{b}}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{-b x^2+a x^6}} \\ & = -\frac {2 x \sqrt [4]{1-\frac {a x^4}{b}} \operatorname {AppellF1}\left (\frac {1}{8},1,\frac {1}{4},\frac {9}{8},\frac {2 a^2 x^4}{1+2 a b-\sqrt {1+4 a b}},\frac {a x^4}{b}\right )}{\sqrt [4]{-b x^2+a x^6}}-\frac {2 x \sqrt [4]{1-\frac {a x^4}{b}} \operatorname {AppellF1}\left (\frac {1}{8},1,\frac {1}{4},\frac {9}{8},\frac {2 a^2 x^4}{1+2 a b+\sqrt {1+4 a b}},\frac {a x^4}{b}\right )}{\sqrt [4]{-b x^2+a x^6}}+\frac {2 a \left (1-\sqrt {1+4 a b}\right ) x^3 \sqrt [4]{1-\frac {a x^4}{b}} \operatorname {AppellF1}\left (\frac {5}{8},1,\frac {1}{4},\frac {13}{8},\frac {2 a^2 x^4}{1+2 a b-\sqrt {1+4 a b}},\frac {a x^4}{b}\right )}{5 \left (1+2 a b-\sqrt {1+4 a b}\right ) \sqrt [4]{-b x^2+a x^6}}+\frac {2 a \left (1+\sqrt {1+4 a b}\right ) x^3 \sqrt [4]{1-\frac {a x^4}{b}} \operatorname {AppellF1}\left (\frac {5}{8},1,\frac {1}{4},\frac {13}{8},\frac {2 a^2 x^4}{1+2 a b+\sqrt {1+4 a b}},\frac {a x^4}{b}\right )}{5 \left (1+2 a b+\sqrt {1+4 a b}\right ) \sqrt [4]{-b x^2+a x^6}}+\frac {2 x \sqrt [4]{1-\frac {a x^4}{b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{4},\frac {9}{8},\frac {a x^4}{b}\right )}{\sqrt [4]{-b x^2+a x^6}} \\ \end{align*}

Mathematica [F]

\[ \int \frac {b+a x^4}{\left (-b+x^2+a x^4\right ) \sqrt [4]{-b x^2+a x^6}} \, dx=\int \frac {b+a x^4}{\left (-b+x^2+a x^4\right ) \sqrt [4]{-b x^2+a x^6}} \, dx \]

[In]

Integrate[(b + a*x^4)/((-b + x^2 + a*x^4)*(-(b*x^2) + a*x^6)^(1/4)),x]

[Out]

Integrate[(b + a*x^4)/((-b + x^2 + a*x^4)*(-(b*x^2) + a*x^6)^(1/4)), x]

Maple [A] (verified)

Time = 1.08 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.28

method	result	size
pseudoelliptic	\(\frac {\sqrt {2}\, \left (\ln \left (\frac {-\left (x^{2} \left (a \,x^{4}-b \right )\right )^{\frac {1}{4}} \sqrt {2}\, x +x^{2}+\sqrt {x^{2} \left (a \,x^{4}-b \right )}}{\left (x^{2} \left (a \,x^{4}-b \right )\right )^{\frac {1}{4}} \sqrt {2}\, x +x^{2}+\sqrt {x^{2} \left (a \,x^{4}-b \right )}}\right )+2 \arctan \left (\frac {\left (x^{2} \left (a \,x^{4}-b \right )\right )^{\frac {1}{4}} \sqrt {2}+x}{x}\right )+2 \arctan \left (\frac {\left (x^{2} \left (a \,x^{4}-b \right )\right )^{\frac {1}{4}} \sqrt {2}-x}{x}\right )\right )}{4}\)	\(148\)

[In]

int((a*x^4+b)/(a*x^4+x^2-b)/(a*x^6-b*x^2)^(1/4),x,method=_RETURNVERBOSE)

[Out]

1/4*2^(1/2)*(ln((-(x^2*(a*x^4-b))^(1/4)*2^(1/2)*x+x^2+(x^2*(a*x^4-b))^(1/2))/((x^2*(a*x^4-b))^(1/4)*2^(1/2)*x+
x^2+(x^2*(a*x^4-b))^(1/2)))+2*arctan(((x^2*(a*x^4-b))^(1/4)*2^(1/2)+x)/x)+2*arctan(((x^2*(a*x^4-b))^(1/4)*2^(1
/2)-x)/x))

Fricas [F(-1)]

Timed out. \[ \int \frac {b+a x^4}{\left (-b+x^2+a x^4\right ) \sqrt [4]{-b x^2+a x^6}} \, dx=\text {Timed out} \]

[In]

integrate((a*x^4+b)/(a*x^4+x^2-b)/(a*x^6-b*x^2)^(1/4),x, algorithm="fricas")

[Out]

Timed out

Sympy [F]

\[ \int \frac {b+a x^4}{\left (-b+x^2+a x^4\right ) \sqrt [4]{-b x^2+a x^6}} \, dx=\int \frac {a x^{4} + b}{\sqrt [4]{x^{2} \left (a x^{4} - b\right )} \left (a x^{4} - b + x^{2}\right )}\, dx \]

[In]

integrate((a*x**4+b)/(a*x**4+x**2-b)/(a*x**6-b*x**2)**(1/4),x)

[Out]

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

Maxima [F]

\[ \int \frac {b+a x^4}{\left (-b+x^2+a x^4\right ) \sqrt [4]{-b x^2+a x^6}} \, dx=\int { \frac {a x^{4} + b}{{\left (a x^{6} - b x^{2}\right )}^{\frac {1}{4}} {\left (a x^{4} + x^{2} - b\right )}} \,d x } \]

[In]

integrate((a*x^4+b)/(a*x^4+x^2-b)/(a*x^6-b*x^2)^(1/4),x, algorithm="maxima")

[Out]

integrate((a*x^4 + b)/((a*x^6 - b*x^2)^(1/4)*(a*x^4 + x^2 - b)), x)

Giac [F]

\[ \int \frac {b+a x^4}{\left (-b+x^2+a x^4\right ) \sqrt [4]{-b x^2+a x^6}} \, dx=\int { \frac {a x^{4} + b}{{\left (a x^{6} - b x^{2}\right )}^{\frac {1}{4}} {\left (a x^{4} + x^{2} - b\right )}} \,d x } \]

[In]

integrate((a*x^4+b)/(a*x^4+x^2-b)/(a*x^6-b*x^2)^(1/4),x, algorithm="giac")

[Out]

integrate((a*x^4 + b)/((a*x^6 - b*x^2)^(1/4)*(a*x^4 + x^2 - b)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {b+a x^4}{\left (-b+x^2+a x^4\right ) \sqrt [4]{-b x^2+a x^6}} \, dx=\int \frac {a\,x^4+b}{{\left (a\,x^6-b\,x^2\right )}^{1/4}\,\left (a\,x^4+x^2-b\right )} \,d x \]

[In]

int((b + a*x^4)/((a*x^6 - b*x^2)^(1/4)*(a*x^4 - b + x^2)),x)

[Out]

int((b + a*x^4)/((a*x^6 - b*x^2)^(1/4)*(a*x^4 - b + x^2)), x)