3.25.15 \(\int \frac {b-a x^4+2 x^8}{\sqrt [4]{-b+a x^4} (b+3 a x^4)} \, dx\)

Optimal. Leaf size=194 \[ \frac {x \left (a x^4-b\right )^{3/4}}{6 a^2}+\frac {\left (-6 a^2-b\right ) \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{36 a^{9/4}}+\frac {\left (6 a^2+b\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{9 \sqrt {2} a^{9/4}}+\frac {\left (-6 a^2-b\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{36 a^{9/4}}+\frac {\left (6 a^2+b\right ) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{9 \sqrt {2} a^{9/4}} \]

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Rubi [A]  time = 0.43, antiderivative size = 256, normalized size of antiderivative = 1.32, number of steps used = 15, number of rules used = 7, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.189, Rules used = {6725, 240, 212, 206, 203, 321, 377} \begin {gather*} \frac {b \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{12 a^{9/4}}+\frac {b \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{12 a^{9/4}}+\frac {x \left (a x^4-b\right )^{3/4}}{6 a^2}-\frac {\left (3 a^2+2 b\right ) \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{18 a^{9/4}}+\frac {\left (6 a^2+b\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{9 \sqrt {2} a^{9/4}}-\frac {\left (3 a^2+2 b\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{18 a^{9/4}}+\frac {\left (6 a^2+b\right ) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{9 \sqrt {2} a^{9/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b - a*x^4 + 2*x^8)/((-b + a*x^4)^(1/4)*(b + 3*a*x^4)),x]

[Out]

(x*(-b + a*x^4)^(3/4))/(6*a^2) + (b*ArcTan[(a^(1/4)*x)/(-b + a*x^4)^(1/4)])/(12*a^(9/4)) - ((3*a^2 + 2*b)*ArcT
an[(a^(1/4)*x)/(-b + a*x^4)^(1/4)])/(18*a^(9/4)) + ((6*a^2 + b)*ArcTan[(Sqrt[2]*a^(1/4)*x)/(-b + a*x^4)^(1/4)]
)/(9*Sqrt[2]*a^(9/4)) + (b*ArcTanh[(a^(1/4)*x)/(-b + a*x^4)^(1/4)])/(12*a^(9/4)) - ((3*a^2 + 2*b)*ArcTanh[(a^(
1/4)*x)/(-b + a*x^4)^(1/4)])/(18*a^(9/4)) + ((6*a^2 + b)*ArcTanh[(Sqrt[2]*a^(1/4)*x)/(-b + a*x^4)^(1/4)])/(9*S
qrt[2]*a^(9/4))

Rule 203

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

Rule 206

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

Rule 212

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] &&
!GtQ[a/b, 0]

Rule 240

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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 377

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 6725

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*} \int \frac {b-a x^4+2 x^8}{\sqrt [4]{-b+a x^4} \left (b+3 a x^4\right )} \, dx &=\int \left (\frac {-3-\frac {2 b}{a^2}}{9 \sqrt [4]{-b+a x^4}}+\frac {2 x^4}{3 a \sqrt [4]{-b+a x^4}}+\frac {2 \left (6 a^2 b+b^2\right )}{9 a^2 \sqrt [4]{-b+a x^4} \left (b+3 a x^4\right )}\right ) \, dx\\ &=\frac {2 \int \frac {x^4}{\sqrt [4]{-b+a x^4}} \, dx}{3 a}+\frac {\left (2 b \left (6 a^2+b\right )\right ) \int \frac {1}{\sqrt [4]{-b+a x^4} \left (b+3 a x^4\right )} \, dx}{9 a^2}+\frac {1}{9} \left (-3-\frac {2 b}{a^2}\right ) \int \frac {1}{\sqrt [4]{-b+a x^4}} \, dx\\ &=\frac {x \left (-b+a x^4\right )^{3/4}}{6 a^2}+\frac {b \int \frac {1}{\sqrt [4]{-b+a x^4}} \, dx}{6 a^2}+\frac {\left (2 b \left (6 a^2+b\right )\right ) \operatorname {Subst}\left (\int \frac {1}{b-4 a b x^4} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right )}{9 a^2}+\frac {1}{9} \left (-3-\frac {2 b}{a^2}\right ) \operatorname {Subst}\left (\int \frac {1}{1-a x^4} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right )\\ &=\frac {x \left (-b+a x^4\right )^{3/4}}{6 a^2}+\frac {b \operatorname {Subst}\left (\int \frac {1}{1-a x^4} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right )}{6 a^2}+\frac {\left (6 a^2+b\right ) \operatorname {Subst}\left (\int \frac {1}{1-2 \sqrt {a} x^2} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right )}{9 a^2}+\frac {\left (6 a^2+b\right ) \operatorname {Subst}\left (\int \frac {1}{1+2 \sqrt {a} x^2} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right )}{9 a^2}+\frac {1}{18} \left (-3-\frac {2 b}{a^2}\right ) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {a} x^2} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right )+\frac {1}{18} \left (-3-\frac {2 b}{a^2}\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {a} x^2} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right )\\ &=\frac {x \left (-b+a x^4\right )^{3/4}}{6 a^2}-\frac {\left (3 a^2+2 b\right ) \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{18 a^{9/4}}+\frac {\left (6 a^2+b\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{9 \sqrt {2} a^{9/4}}-\frac {\left (3 a^2+2 b\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{18 a^{9/4}}+\frac {\left (6 a^2+b\right ) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{9 \sqrt {2} a^{9/4}}+\frac {b \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {a} x^2} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right )}{12 a^2}+\frac {b \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {a} x^2} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right )}{12 a^2}\\ &=\frac {x \left (-b+a x^4\right )^{3/4}}{6 a^2}+\frac {b \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{12 a^{9/4}}-\frac {\left (3 a^2+2 b\right ) \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{18 a^{9/4}}+\frac {\left (6 a^2+b\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{9 \sqrt {2} a^{9/4}}+\frac {b \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{12 a^{9/4}}-\frac {\left (3 a^2+2 b\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{18 a^{9/4}}+\frac {\left (6 a^2+b\right ) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{9 \sqrt {2} a^{9/4}}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 221, normalized size = 1.14 \begin {gather*} \frac {-6 a^2 \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )+12 \sqrt {2} a^2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )-\left (6 a^2+b\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )+2 \sqrt {2} \left (6 a^2+b\right ) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )+6 \sqrt [4]{a} x \left (a x^4-b\right )^{3/4}-b \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )+2 \sqrt {2} b \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{36 a^{9/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b - a*x^4 + 2*x^8)/((-b + a*x^4)^(1/4)*(b + 3*a*x^4)),x]

[Out]

(6*a^(1/4)*x*(-b + a*x^4)^(3/4) - 6*a^2*ArcTan[(a^(1/4)*x)/(-b + a*x^4)^(1/4)] - b*ArcTan[(a^(1/4)*x)/(-b + a*
x^4)^(1/4)] + 12*Sqrt[2]*a^2*ArcTan[(Sqrt[2]*a^(1/4)*x)/(-b + a*x^4)^(1/4)] + 2*Sqrt[2]*b*ArcTan[(Sqrt[2]*a^(1
/4)*x)/(-b + a*x^4)^(1/4)] - (6*a^2 + b)*ArcTanh[(a^(1/4)*x)/(-b + a*x^4)^(1/4)] + 2*Sqrt[2]*(6*a^2 + b)*ArcTa
nh[(Sqrt[2]*a^(1/4)*x)/(-b + a*x^4)^(1/4)])/(36*a^(9/4))

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IntegrateAlgebraic [A]  time = 1.07, size = 194, normalized size = 1.00 \begin {gather*} \frac {x \left (-b+a x^4\right )^{3/4}}{6 a^2}+\frac {\left (-6 a^2-b\right ) \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{36 a^{9/4}}+\frac {\left (6 a^2+b\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{9 \sqrt {2} a^{9/4}}+\frac {\left (-6 a^2-b\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{36 a^{9/4}}+\frac {\left (6 a^2+b\right ) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{9 \sqrt {2} a^{9/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(b - a*x^4 + 2*x^8)/((-b + a*x^4)^(1/4)*(b + 3*a*x^4)),x]

[Out]

(x*(-b + a*x^4)^(3/4))/(6*a^2) + ((-6*a^2 - b)*ArcTan[(a^(1/4)*x)/(-b + a*x^4)^(1/4)])/(36*a^(9/4)) + ((6*a^2
+ b)*ArcTan[(Sqrt[2]*a^(1/4)*x)/(-b + a*x^4)^(1/4)])/(9*Sqrt[2]*a^(9/4)) + ((-6*a^2 - b)*ArcTanh[(a^(1/4)*x)/(
-b + a*x^4)^(1/4)])/(36*a^(9/4)) + ((6*a^2 + b)*ArcTanh[(Sqrt[2]*a^(1/4)*x)/(-b + a*x^4)^(1/4)])/(9*Sqrt[2]*a^
(9/4))

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fricas [B]  time = 0.61, size = 1231, normalized size = 6.35

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^8-a*x^4+b)/(a*x^4-b)^(1/4)/(3*a*x^4+b),x, algorithm="fricas")

[Out]

1/72*(16*(1/4)^(1/4)*a^2*((1296*a^8 + 864*a^6*b + 216*a^4*b^2 + 24*a^2*b^3 + b^4)/a^9)^(1/4)*arctan(((1/4)^(1/
4)*a^2*x*sqrt((2*(1296*a^13 + 864*a^11*b + 216*a^9*b^2 + 24*a^7*b^3 + a^5*b^4)*x^2*sqrt((1296*a^8 + 864*a^6*b
+ 216*a^4*b^2 + 24*a^2*b^3 + b^4)/a^9) + (46656*a^12 + 46656*a^10*b + 19440*a^8*b^2 + 4320*a^6*b^3 + 540*a^4*b
^4 + 36*a^2*b^5 + b^6)*sqrt(a*x^4 - b))/x^2)*((1296*a^8 + 864*a^6*b + 216*a^4*b^2 + 24*a^2*b^3 + b^4)/a^9)^(1/
4) - (1/4)^(1/4)*(216*a^8 + 108*a^6*b + 18*a^4*b^2 + a^2*b^3)*(a*x^4 - b)^(1/4)*((1296*a^8 + 864*a^6*b + 216*a
^4*b^2 + 24*a^2*b^3 + b^4)/a^9)^(1/4))/((1296*a^8 + 864*a^6*b + 216*a^4*b^2 + 24*a^2*b^3 + b^4)*x)) + 4*(1/4)^
(1/4)*a^2*((1296*a^8 + 864*a^6*b + 216*a^4*b^2 + 24*a^2*b^3 + b^4)/a^9)^(1/4)*log((4*(1/4)^(3/4)*a^7*x*((1296*
a^8 + 864*a^6*b + 216*a^4*b^2 + 24*a^2*b^3 + b^4)/a^9)^(3/4) + (216*a^6 + 108*a^4*b + 18*a^2*b^2 + b^3)*(a*x^4
- b)^(1/4))/x) - 4*(1/4)^(1/4)*a^2*((1296*a^8 + 864*a^6*b + 216*a^4*b^2 + 24*a^2*b^3 + b^4)/a^9)^(1/4)*log(-(
4*(1/4)^(3/4)*a^7*x*((1296*a^8 + 864*a^6*b + 216*a^4*b^2 + 24*a^2*b^3 + b^4)/a^9)^(3/4) - (216*a^6 + 108*a^4*b
+ 18*a^2*b^2 + b^3)*(a*x^4 - b)^(1/4))/x) - 4*a^2*((1296*a^8 + 864*a^6*b + 216*a^4*b^2 + 24*a^2*b^3 + b^4)/a^
9)^(1/4)*arctan((a^2*x*sqrt(((1296*a^13 + 864*a^11*b + 216*a^9*b^2 + 24*a^7*b^3 + a^5*b^4)*x^2*sqrt((1296*a^8
+ 864*a^6*b + 216*a^4*b^2 + 24*a^2*b^3 + b^4)/a^9) + (46656*a^12 + 46656*a^10*b + 19440*a^8*b^2 + 4320*a^6*b^3
+ 540*a^4*b^4 + 36*a^2*b^5 + b^6)*sqrt(a*x^4 - b))/x^2)*((1296*a^8 + 864*a^6*b + 216*a^4*b^2 + 24*a^2*b^3 + b
^4)/a^9)^(1/4) - (216*a^8 + 108*a^6*b + 18*a^4*b^2 + a^2*b^3)*(a*x^4 - b)^(1/4)*((1296*a^8 + 864*a^6*b + 216*a
^4*b^2 + 24*a^2*b^3 + b^4)/a^9)^(1/4))/((1296*a^8 + 864*a^6*b + 216*a^4*b^2 + 24*a^2*b^3 + b^4)*x)) - a^2*((12
96*a^8 + 864*a^6*b + 216*a^4*b^2 + 24*a^2*b^3 + b^4)/a^9)^(1/4)*log((a^7*x*((1296*a^8 + 864*a^6*b + 216*a^4*b^
2 + 24*a^2*b^3 + b^4)/a^9)^(3/4) + (216*a^6 + 108*a^4*b + 18*a^2*b^2 + b^3)*(a*x^4 - b)^(1/4))/x) + a^2*((1296
*a^8 + 864*a^6*b + 216*a^4*b^2 + 24*a^2*b^3 + b^4)/a^9)^(1/4)*log(-(a^7*x*((1296*a^8 + 864*a^6*b + 216*a^4*b^2
+ 24*a^2*b^3 + b^4)/a^9)^(3/4) - (216*a^6 + 108*a^4*b + 18*a^2*b^2 + b^3)*(a*x^4 - b)^(1/4))/x) + 12*(a*x^4 -
b)^(3/4)*x)/a^2

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, x^{8} - a x^{4} + b}{{\left (3 \, a x^{4} + b\right )} {\left (a x^{4} - b\right )}^{\frac {1}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^8-a*x^4+b)/(a*x^4-b)^(1/4)/(3*a*x^4+b),x, algorithm="giac")

[Out]

integrate((2*x^8 - a*x^4 + b)/((3*a*x^4 + b)*(a*x^4 - b)^(1/4)), x)

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {2 x^{8}-a \,x^{4}+b}{\left (a \,x^{4}-b \right )^{\frac {1}{4}} \left (3 a \,x^{4}+b \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^8-a*x^4+b)/(a*x^4-b)^(1/4)/(3*a*x^4+b),x)

[Out]

int((2*x^8-a*x^4+b)/(a*x^4-b)^(1/4)/(3*a*x^4+b),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, x^{8} - a x^{4} + b}{{\left (3 \, a x^{4} + b\right )} {\left (a x^{4} - b\right )}^{\frac {1}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^8-a*x^4+b)/(a*x^4-b)^(1/4)/(3*a*x^4+b),x, algorithm="maxima")

[Out]

integrate((2*x^8 - a*x^4 + b)/((3*a*x^4 + b)*(a*x^4 - b)^(1/4)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {2\,x^8-a\,x^4+b}{{\left (a\,x^4-b\right )}^{1/4}\,\left (3\,a\,x^4+b\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b - a*x^4 + 2*x^8)/((a*x^4 - b)^(1/4)*(b + 3*a*x^4)),x)

[Out]

int((b - a*x^4 + 2*x^8)/((a*x^4 - b)^(1/4)*(b + 3*a*x^4)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {- a x^{4} + b + 2 x^{8}}{\sqrt [4]{a x^{4} - b} \left (3 a x^{4} + b\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x**8-a*x**4+b)/(a*x**4-b)**(1/4)/(3*a*x**4+b),x)

[Out]

Integral((-a*x**4 + b + 2*x**8)/((a*x**4 - b)**(1/4)*(3*a*x**4 + b)), x)

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