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

Optimal. Leaf size=482 \[ \frac {3 \left (1+\sqrt [4]{-1}\right ) \tan ^{-1}\left (\frac {(-1)^{7/8} \sqrt {2+\sqrt {2}} x \sqrt [8]{a^2+b} \sqrt [4]{a x^4+b}}{(-1)^{3/4} x^2 \sqrt [4]{a^2+b}+\sqrt {a x^4+b}}\right )}{8 \sqrt [8]{a^2+b}}-\frac {3 i \left (\sqrt {2 \left (3-2 \sqrt {2}\right )}-i \sqrt {2}\right ) \tan ^{-1}\left (\frac {(-1)^{7/8} \left (\sqrt {2}-2\right ) x \sqrt [8]{a^2+b} \sqrt [4]{a x^4+b}}{(-1)^{3/4} \sqrt {2-\sqrt {2}} x^2 \sqrt [4]{a^2+b}+\sqrt {2-\sqrt {2}} \sqrt {a x^4+b}}\right )}{16 \sqrt [8]{a^2+b}}+\frac {3 \left (\sqrt {2}+i \sqrt {2 \left (3-2 \sqrt {2}\right )}\right ) \tanh ^{-1}\left (\frac {(-1)^{7/8} x^2 \sqrt [4]{a^2+b}-\sqrt [8]{-1} \sqrt {a x^4+b}}{\sqrt {2-\sqrt {2}} x \sqrt [8]{a^2+b} \sqrt [4]{a x^4+b}}\right )}{16 \sqrt [8]{a^2+b}}+\frac {3 \left (1+\sqrt [4]{-1}\right ) \tanh ^{-1}\left (\frac {(-1)^{7/8} x^2 \sqrt [4]{a^2+b}-\sqrt [8]{-1} \sqrt {a x^4+b}}{\sqrt {2+\sqrt {2}} x \sqrt [8]{a^2+b} \sqrt [4]{a x^4+b}}\right )}{8 \sqrt [8]{a^2+b}}+\frac {\tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{\sqrt [4]{a}}+\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{\sqrt [4]{a}} \]

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Rubi [A]  time = 1.05, antiderivative size = 451, normalized size of antiderivative = 0.94, number of steps used = 25, number of rules used = 10, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6728, 240, 212, 206, 203, 1428, 408, 377, 208, 205} \begin {gather*} \frac {3 \left (-a \sqrt {a^2+b}+a^2+b\right )^{3/4} \tan ^{-1}\left (\frac {x \sqrt [4]{-a \sqrt {a^2+b}+a^2+b}}{\sqrt [4]{a-\sqrt {a^2+b}} \sqrt [4]{a x^4+b}}\right )}{4 \sqrt {a^2+b} \left (a-\sqrt {a^2+b}\right )^{3/4}}-\frac {3 \left (a \sqrt {a^2+b}+a^2+b\right )^{3/4} \tan ^{-1}\left (\frac {x \sqrt [4]{a \sqrt {a^2+b}+a^2+b}}{\sqrt [4]{\sqrt {a^2+b}+a} \sqrt [4]{a x^4+b}}\right )}{4 \sqrt {a^2+b} \left (\sqrt {a^2+b}+a\right )^{3/4}}+\frac {3 \left (-a \sqrt {a^2+b}+a^2+b\right )^{3/4} \tanh ^{-1}\left (\frac {x \sqrt [4]{-a \sqrt {a^2+b}+a^2+b}}{\sqrt [4]{a-\sqrt {a^2+b}} \sqrt [4]{a x^4+b}}\right )}{4 \sqrt {a^2+b} \left (a-\sqrt {a^2+b}\right )^{3/4}}-\frac {3 \left (a \sqrt {a^2+b}+a^2+b\right )^{3/4} \tanh ^{-1}\left (\frac {x \sqrt [4]{a \sqrt {a^2+b}+a^2+b}}{\sqrt [4]{\sqrt {a^2+b}+a} \sqrt [4]{a x^4+b}}\right )}{4 \sqrt {a^2+b} \left (\sqrt {a^2+b}+a\right )^{3/4}}+\frac {\tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{\sqrt [4]{a}}+\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{\sqrt [4]{a}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 205

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

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 208

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

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 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 408

Int[((a_) + (b_.)*(x_)^4)^(p_)/((c_) + (d_.)*(x_)^4), x_Symbol] :> Dist[b/d, Int[(a + b*x^4)^(p - 1), x], x] -
Dist[(b*c - a*d)/d, Int[(a + b*x^4)^(p - 1)/(c + d*x^4), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0
] && (EqQ[p, 3/4] || EqQ[p, 5/4])

Rule 1428

Int[((d_) + (e_.)*(x_)^(n_))^(q_)/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_)), x_Symbol] :> With[{r = Rt[b^2 -
4*a*c, 2]}, Dist[(2*c)/r, Int[(d + e*x^n)^q/(b - r + 2*c*x^n), x], x] - Dist[(2*c)/r, Int[(d + e*x^n)^q/(b +
r + 2*c*x^n), x], x]] /; FreeQ[{a, b, c, d, e, n, q}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 0] &&  !IntegerQ[q]

Rule 6728

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

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Mathematica [F]  time = 0.29, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {b-a x^4+2 x^8}{\sqrt [4]{b+a x^4} \left (-b-2 a x^4+x^8\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 133.15, size = 536, normalized size = 1.11 \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{\sqrt [4]{a}}+\frac {3 i \left (i \sqrt {2}+\sqrt {2 \left (3+2 \sqrt {2}\right )}\right ) \tan ^{-1}\left (\frac {\left ((-1+i)-(1+i) (-1)^{3/4}\right ) \sqrt [4]{a^2+b} x^2+(1+i) \sqrt {b+a x^4}+(1+i) (-1)^{3/4} \sqrt {b+a x^4}}{2 \sqrt [8]{a^2+b} x \sqrt [4]{b+a x^4}}\right )}{16 \sqrt [8]{a^2+b}}+\frac {3 \left (\sqrt {2}+i \sqrt {2 \left (3-2 \sqrt {2}\right )}\right ) \tan ^{-1}\left (\frac {2 \sqrt [8]{a^2+b} x \sqrt [4]{b+a x^4}}{\left ((1-i)+\sqrt {2}\right ) \sqrt [4]{a^2+b} x^2-(1+i) \sqrt {b+a x^4}-\sqrt {2} \sqrt {b+a x^4}}\right )}{16 \sqrt [8]{a^2+b}}+\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{\sqrt [4]{a}}+\frac {3 \left (i+(-1)^{3/4}\right ) \tanh ^{-1}\left (\frac {\left ((-2+2 i)-(2+2 i) (-1)^{3/4}\right ) \sqrt [4]{a^2+b} x^2-(2+2 i) \sqrt {b+a x^4}-(2+2 i) (-1)^{3/4} \sqrt {b+a x^4}}{4 \sqrt [8]{a^2+b} x \sqrt [4]{b+a x^4}}\right )}{8 \sqrt [8]{a^2+b}}-\frac {3 i \left (-i \sqrt {2}+\sqrt {2 \left (3-2 \sqrt {2}\right )}\right ) \tanh ^{-1}\left (\frac {\left ((1-i)+\sqrt {2}\right ) \sqrt [4]{a^2+b} x^2+(1+i) \sqrt {b+a x^4}+\sqrt {2} \sqrt {b+a x^4}}{2 \sqrt [8]{a^2+b} x \sqrt [4]{b+a x^4}}\right )}{16 \sqrt [8]{a^2+b}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTan[(a^(1/4)*x)/(b + a*x^4)^(1/4)]/a^(1/4) + (((3*I)/16)*(I*Sqrt[2] + Sqrt[2*(3 + 2*Sqrt[2])])*ArcTan[(((-1
+ I) - (1 + I)*(-1)^(3/4))*(a^2 + b)^(1/4)*x^2 + (1 + I)*Sqrt[b + a*x^4] + (1 + I)*(-1)^(3/4)*Sqrt[b + a*x^4]
)/(2*(a^2 + b)^(1/8)*x*(b + a*x^4)^(1/4))])/(a^2 + b)^(1/8) + (3*(Sqrt[2] + I*Sqrt[2*(3 - 2*Sqrt[2])])*ArcTan[
(2*(a^2 + b)^(1/8)*x*(b + a*x^4)^(1/4))/(((1 - I) + Sqrt[2])*(a^2 + b)^(1/4)*x^2 - (1 + I)*Sqrt[b + a*x^4] - S
qrt[2]*Sqrt[b + a*x^4])])/(16*(a^2 + b)^(1/8)) + ArcTanh[(a^(1/4)*x)/(b + a*x^4)^(1/4)]/a^(1/4) + (3*(I + (-1)
^(3/4))*ArcTanh[(((-2 + 2*I) - (2 + 2*I)*(-1)^(3/4))*(a^2 + b)^(1/4)*x^2 - (2 + 2*I)*Sqrt[b + a*x^4] - (2 + 2*
I)*(-1)^(3/4)*Sqrt[b + a*x^4])/(4*(a^2 + b)^(1/8)*x*(b + a*x^4)^(1/4))])/(8*(a^2 + b)^(1/8)) - (((3*I)/16)*((-
I)*Sqrt[2] + Sqrt[2*(3 - 2*Sqrt[2])])*ArcTanh[(((1 - I) + Sqrt[2])*(a^2 + b)^(1/4)*x^2 + (1 + I)*Sqrt[b + a*x^
4] + Sqrt[2]*Sqrt[b + a*x^4])/(2*(a^2 + b)^(1/8)*x*(b + a*x^4)^(1/4))])/(a^2 + b)^(1/8)

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fricas [A]  time = 0.67, size = 567, normalized size = 1.18 \begin {gather*} -\frac {3 \, \sqrt {2} \arctan \left (\frac {\frac {\sqrt {2} x \sqrt {\frac {{\left (a^{2} + b\right )}^{\frac {1}{4}} x^{2} + \sqrt {2} {\left (a x^{4} + b\right )}^{\frac {1}{4}} {\left (a^{2} + b\right )}^{\frac {1}{8}} x + \sqrt {a x^{4} + b}}{x^{2}}}}{{\left (a^{2} + b\right )}^{\frac {1}{8}}} - x - \frac {\sqrt {2} {\left (a x^{4} + b\right )}^{\frac {1}{4}}}{{\left (a^{2} + b\right )}^{\frac {1}{8}}}}{x}\right )}{4 \, {\left (a^{2} + b\right )}^{\frac {1}{8}}} - \frac {3 \, \sqrt {2} \arctan \left (\frac {\frac {\sqrt {2} x \sqrt {\frac {{\left (a^{2} + b\right )}^{\frac {1}{4}} x^{2} - \sqrt {2} {\left (a x^{4} + b\right )}^{\frac {1}{4}} {\left (a^{2} + b\right )}^{\frac {1}{8}} x + \sqrt {a x^{4} + b}}{x^{2}}}}{{\left (a^{2} + b\right )}^{\frac {1}{8}}} + x - \frac {\sqrt {2} {\left (a x^{4} + b\right )}^{\frac {1}{4}}}{{\left (a^{2} + b\right )}^{\frac {1}{8}}}}{x}\right )}{4 \, {\left (a^{2} + b\right )}^{\frac {1}{8}}} - \frac {3 \, \sqrt {2} \log \left (\frac {{\left (a^{2} + b\right )}^{\frac {1}{4}} x^{2} + \sqrt {2} {\left (a x^{4} + b\right )}^{\frac {1}{4}} {\left (a^{2} + b\right )}^{\frac {1}{8}} x + \sqrt {a x^{4} + b}}{x^{2}}\right )}{16 \, {\left (a^{2} + b\right )}^{\frac {1}{8}}} + \frac {3 \, \sqrt {2} \log \left (\frac {{\left (a^{2} + b\right )}^{\frac {1}{4}} x^{2} - \sqrt {2} {\left (a x^{4} + b\right )}^{\frac {1}{4}} {\left (a^{2} + b\right )}^{\frac {1}{8}} x + \sqrt {a x^{4} + b}}{x^{2}}\right )}{16 \, {\left (a^{2} + b\right )}^{\frac {1}{8}}} - \frac {3 \, \arctan \left (\frac {\frac {x \sqrt {\frac {{\left (a^{2} + b\right )}^{\frac {1}{4}} x^{2} + \sqrt {a x^{4} + b}}{x^{2}}}}{{\left (a^{2} + b\right )}^{\frac {1}{8}}} - \frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{{\left (a^{2} + b\right )}^{\frac {1}{8}}}}{x}\right )}{2 \, {\left (a^{2} + b\right )}^{\frac {1}{8}}} - \frac {3 \, \log \left (\frac {{\left (a^{2} + b\right )}^{\frac {1}{8}} x + {\left (a x^{4} + b\right )}^{\frac {1}{4}}}{x}\right )}{8 \, {\left (a^{2} + b\right )}^{\frac {1}{8}}} + \frac {3 \, \log \left (-\frac {{\left (a^{2} + b\right )}^{\frac {1}{8}} x - {\left (a x^{4} + b\right )}^{\frac {1}{4}}}{x}\right )}{8 \, {\left (a^{2} + b\right )}^{\frac {1}{8}}} + \frac {2 \, \arctan \left (\frac {\frac {x \sqrt {\frac {\sqrt {a} x^{2} + \sqrt {a x^{4} + b}}{x^{2}}}}{a^{\frac {1}{4}}} - \frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}}}}{x}\right )}{a^{\frac {1}{4}}} + \frac {\log \left (\frac {a^{\frac {1}{4}} x + {\left (a x^{4} + b\right )}^{\frac {1}{4}}}{x}\right )}{2 \, a^{\frac {1}{4}}} - \frac {\log \left (-\frac {a^{\frac {1}{4}} x - {\left (a x^{4} + b\right )}^{\frac {1}{4}}}{x}\right )}{2 \, a^{\frac {1}{4}}} \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)/(x^8-2*a*x^4-b),x, algorithm="fricas")

[Out]

-3/4*sqrt(2)*arctan((sqrt(2)*x*sqrt(((a^2 + b)^(1/4)*x^2 + sqrt(2)*(a*x^4 + b)^(1/4)*(a^2 + b)^(1/8)*x + sqrt(
a*x^4 + b))/x^2)/(a^2 + b)^(1/8) - x - sqrt(2)*(a*x^4 + b)^(1/4)/(a^2 + b)^(1/8))/x)/(a^2 + b)^(1/8) - 3/4*sqr
t(2)*arctan((sqrt(2)*x*sqrt(((a^2 + b)^(1/4)*x^2 - sqrt(2)*(a*x^4 + b)^(1/4)*(a^2 + b)^(1/8)*x + sqrt(a*x^4 +
b))/x^2)/(a^2 + b)^(1/8) + x - sqrt(2)*(a*x^4 + b)^(1/4)/(a^2 + b)^(1/8))/x)/(a^2 + b)^(1/8) - 3/16*sqrt(2)*lo
g(((a^2 + b)^(1/4)*x^2 + sqrt(2)*(a*x^4 + b)^(1/4)*(a^2 + b)^(1/8)*x + sqrt(a*x^4 + b))/x^2)/(a^2 + b)^(1/8) +
3/16*sqrt(2)*log(((a^2 + b)^(1/4)*x^2 - sqrt(2)*(a*x^4 + b)^(1/4)*(a^2 + b)^(1/8)*x + sqrt(a*x^4 + b))/x^2)/(
a^2 + b)^(1/8) - 3/2*arctan((x*sqrt(((a^2 + b)^(1/4)*x^2 + sqrt(a*x^4 + b))/x^2)/(a^2 + b)^(1/8) - (a*x^4 + b)
^(1/4)/(a^2 + b)^(1/8))/x)/(a^2 + b)^(1/8) - 3/8*log(((a^2 + b)^(1/8)*x + (a*x^4 + b)^(1/4))/x)/(a^2 + b)^(1/8
) + 3/8*log(-((a^2 + b)^(1/8)*x - (a*x^4 + b)^(1/4))/x)/(a^2 + b)^(1/8) + 2*arctan((x*sqrt((sqrt(a)*x^2 + sqrt
(a*x^4 + b))/x^2)/a^(1/4) - (a*x^4 + b)^(1/4)/a^(1/4))/x)/a^(1/4) + 1/2*log((a^(1/4)*x + (a*x^4 + b)^(1/4))/x)
/a^(1/4) - 1/2*log(-(a^(1/4)*x - (a*x^4 + b)^(1/4))/x)/a^(1/4)

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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 (x^{8} - 2 \, 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)/(x^8-2*a*x^4-b),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[\int \frac {2 x^{8}-a \,x^{4}+b}{\left (a \,x^{4}+b \right )^{\frac {1}{4}} \left (x^{8}-2 a \,x^{4}-b \right )}\, dx\]

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

[Out]

int((2*x^8-a*x^4+b)/(a*x^4+b)^(1/4)/(x^8-2*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 (x^{8} - 2 \, 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)/(x^8-2*a*x^4-b),x, algorithm="maxima")

[Out]

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

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \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)/(x**8-2*a*x**4-b),x)

[Out]

Timed out

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