3.28.96 \(\int \frac {\sqrt [4]{2 b+a x^4} (-4 b+a x^8)}{x^6 (-4 b+c x^4+a x^8)} \, dx\) [2796]

Optimal. Leaf size=270 \[ \frac {\sqrt [4]{2 b+a x^4} \left (-4 b-2 a x^4-5 c x^4\right )}{20 b x^5}-\frac {c \text {RootSum}\left [2 a^2-2 a b+a c-4 a \text {$\#$1}^4-c \text {$\#$1}^4+2 \text {$\#$1}^8\& ,\frac {-2 a^2 \log (x)+2 a b \log (x)-a c \log (x)+2 a^2 \log \left (\sqrt [4]{2 b+a x^4}-x \text {$\#$1}\right )-2 a b \log \left (\sqrt [4]{2 b+a x^4}-x \text {$\#$1}\right )+a c \log \left (\sqrt [4]{2 b+a x^4}-x \text {$\#$1}\right )+2 a \log (x) \text {$\#$1}^4+c \log (x) \text {$\#$1}^4-2 a \log \left (\sqrt [4]{2 b+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4-c \log \left (\sqrt [4]{2 b+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4}{4 a \text {$\#$1}^3+c \text {$\#$1}^3-4 \text {$\#$1}^7}\& \right ]}{16 b} \]

[Out]

Unintegrable

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Rubi [B] Leaf count is larger than twice the leaf count of optimal. \(1308\) vs. \(2(270)=540\).
time = 6.05, antiderivative size = 1308, normalized size of antiderivative = 4.84, number of steps used = 43, number of rules used = 14, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6860, 270, 283, 338, 304, 209, 212, 1542, 525, 524, 1532, 508, 211, 214} \begin {gather*} \frac {a c^2 \sqrt [4]{a x^4+2 b} F_1\left (\frac {3}{4};1,-\frac {1}{4};\frac {7}{4};-\frac {2 a x^4}{c-\sqrt {c^2+16 a b}},-\frac {a x^4}{2 b}\right ) x^3}{3\ 2^{3/4} b \left (16 a b+c \left (c-\sqrt {c^2+16 a b}\right )\right ) \sqrt [4]{\frac {a x^4}{b}+2}}+\frac {a c^2 \sqrt [4]{a x^4+2 b} F_1\left (\frac {3}{4};1,-\frac {1}{4};\frac {7}{4};-\frac {2 a x^4}{c+\sqrt {c^2+16 a b}},-\frac {a x^4}{2 b}\right ) x^3}{3\ 2^{3/4} b \left (16 a b+c \left (c+\sqrt {c^2+16 a b}\right )\right ) \sqrt [4]{\frac {a x^4}{b}+2}}-\frac {\sqrt [4]{a} (2 b-c) c \left (\sqrt {c^2+16 a b}-c\right )^{3/4} \text {ArcTan}\left (\frac {\sqrt [4]{a} \sqrt [4]{4 b-c+\sqrt {c^2+16 a b}} x}{\sqrt [4]{\sqrt {c^2+16 a b}-c} \sqrt [4]{a x^4+2 b}}\right )}{8 b \sqrt {c^2+16 a b} \left (4 b-c+\sqrt {c^2+16 a b}\right )^{3/4}}-\frac {a^{5/4} c \text {ArcTan}\left (\frac {\sqrt [4]{a} \sqrt [4]{4 b-c+\sqrt {c^2+16 a b}} x}{\sqrt [4]{\sqrt {c^2+16 a b}-c} \sqrt [4]{a x^4+2 b}}\right )}{\sqrt {c^2+16 a b} \sqrt [4]{\sqrt {c^2+16 a b}-c} \left (4 b-c+\sqrt {c^2+16 a b}\right )^{3/4}}+\frac {\sqrt [4]{a} (2 b-c) c \left (c+\sqrt {c^2+16 a b}\right )^{3/4} \text {ArcTan}\left (\frac {\sqrt [4]{a} \sqrt [4]{-4 b+c+\sqrt {c^2+16 a b}} x}{\sqrt [4]{c+\sqrt {c^2+16 a b}} \sqrt [4]{a x^4+2 b}}\right )}{8 b \sqrt {c^2+16 a b} \left (-4 b+c+\sqrt {c^2+16 a b}\right )^{3/4}}-\frac {a^{5/4} c \text {ArcTan}\left (\frac {\sqrt [4]{a} \sqrt [4]{-4 b+c+\sqrt {c^2+16 a b}} x}{\sqrt [4]{c+\sqrt {c^2+16 a b}} \sqrt [4]{a x^4+2 b}}\right )}{\sqrt {c^2+16 a b} \sqrt [4]{c+\sqrt {c^2+16 a b}} \left (-4 b+c+\sqrt {c^2+16 a b}\right )^{3/4}}+\frac {\sqrt [4]{a} (2 b-c) c \left (\sqrt {c^2+16 a b}-c\right )^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [4]{4 b-c+\sqrt {c^2+16 a b}} x}{\sqrt [4]{\sqrt {c^2+16 a b}-c} \sqrt [4]{a x^4+2 b}}\right )}{8 b \sqrt {c^2+16 a b} \left (4 b-c+\sqrt {c^2+16 a b}\right )^{3/4}}+\frac {a^{5/4} c \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [4]{4 b-c+\sqrt {c^2+16 a b}} x}{\sqrt [4]{\sqrt {c^2+16 a b}-c} \sqrt [4]{a x^4+2 b}}\right )}{\sqrt {c^2+16 a b} \sqrt [4]{\sqrt {c^2+16 a b}-c} \left (4 b-c+\sqrt {c^2+16 a b}\right )^{3/4}}-\frac {\sqrt [4]{a} (2 b-c) c \left (c+\sqrt {c^2+16 a b}\right )^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [4]{-4 b+c+\sqrt {c^2+16 a b}} x}{\sqrt [4]{c+\sqrt {c^2+16 a b}} \sqrt [4]{a x^4+2 b}}\right )}{8 b \sqrt {c^2+16 a b} \left (-4 b+c+\sqrt {c^2+16 a b}\right )^{3/4}}+\frac {a^{5/4} c \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [4]{-4 b+c+\sqrt {c^2+16 a b}} x}{\sqrt [4]{c+\sqrt {c^2+16 a b}} \sqrt [4]{a x^4+2 b}}\right )}{\sqrt {c^2+16 a b} \sqrt [4]{c+\sqrt {c^2+16 a b}} \left (-4 b+c+\sqrt {c^2+16 a b}\right )^{3/4}}-\frac {c \sqrt [4]{a x^4+2 b}}{4 b x}-\frac {\left (a x^4+2 b\right )^{5/4}}{10 b x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((2*b + a*x^4)^(1/4)*(-4*b + a*x^8))/(x^6*(-4*b + c*x^4 + a*x^8)),x]

[Out]

-1/4*(c*(2*b + a*x^4)^(1/4))/(b*x) - (2*b + a*x^4)^(5/4)/(10*b*x^5) + (a*c^2*x^3*(2*b + a*x^4)^(1/4)*AppellF1[
3/4, 1, -1/4, 7/4, (-2*a*x^4)/(c - Sqrt[16*a*b + c^2]), -1/2*(a*x^4)/b])/(3*2^(3/4)*b*(16*a*b + c*(c - Sqrt[16
*a*b + c^2]))*(2 + (a*x^4)/b)^(1/4)) + (a*c^2*x^3*(2*b + a*x^4)^(1/4)*AppellF1[3/4, 1, -1/4, 7/4, (-2*a*x^4)/(
c + Sqrt[16*a*b + c^2]), -1/2*(a*x^4)/b])/(3*2^(3/4)*b*(16*a*b + c*(c + Sqrt[16*a*b + c^2]))*(2 + (a*x^4)/b)^(
1/4)) - (a^(5/4)*c*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)*(2
*b + a*x^4)^(1/4))])/(Sqrt[16*a*b + c^2]*(-c + Sqrt[16*a*b + c^2])^(1/4)*(4*b - c + Sqrt[16*a*b + c^2])^(3/4))
- (a^(1/4)*(2*b - c)*c*(-c + Sqrt[16*a*b + c^2])^(3/4)*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)*(2*b + a*x^4)^(1/4))])/(8*b*Sqrt[16*a*b + c^2]*(4*b - c + Sqrt[16*a*b + c^2
])^(3/4)) - (a^(5/4)*c*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
)*(2*b + a*x^4)^(1/4))])/(Sqrt[16*a*b + c^2]*(c + Sqrt[16*a*b + c^2])^(1/4)*(-4*b + c + Sqrt[16*a*b + c^2])^(3
/4)) + (a^(1/4)*(2*b - c)*c*(c + Sqrt[16*a*b + c^2])^(3/4)*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)*(2*b + a*x^4)^(1/4))])/(8*b*Sqrt[16*a*b + c^2]*(-4*b + c + Sqrt[16*a*b +
c^2])^(3/4)) + (a^(5/4)*c*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)*(2*b + a*x^4)^(1/4))])/(Sqrt[16*a*b + c^2]*(-c + Sqrt[16*a*b + c^2])^(1/4)*(4*b - c + Sqrt[16*a*b + c^2
])^(3/4)) + (a^(1/4)*(2*b - c)*c*(-c + Sqrt[16*a*b + c^2])^(3/4)*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)*(2*b + a*x^4)^(1/4))])/(8*b*Sqrt[16*a*b + c^2]*(4*b - c + Sqrt[16
*a*b + c^2])^(3/4)) + (a^(5/4)*c*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)*(2*b + a*x^4)^(1/4))])/(Sqrt[16*a*b + c^2]*(c + Sqrt[16*a*b + c^2])^(1/4)*(-4*b + c + Sqrt[16*a*b
+ c^2])^(3/4)) - (a^(1/4)*(2*b - c)*c*(c + Sqrt[16*a*b + c^2])^(3/4)*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)*(2*b + a*x^4)^(1/4))])/(8*b*Sqrt[16*a*b + c^2]*(-4*b + c + S
qrt[16*a*b + c^2])^(3/4))

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 270

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

Rule 283

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

Rule 304

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}
, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !
GtQ[a/b, 0]

Rule 338

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 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)] && IntegersQ[m, p + (m + 1)/n]

Rule 508

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> With[{k = Denominato
r[p]}, Dist[k*(a^(p + (m + 1)/n)/n), Subst[Int[x^(k*((m + 1)/n) - 1)*((c - (b*c - a*d)*x^k)^q/(1 - b*x^k)^(p +
q + (m + 1)/n + 1)), x], x, x^(n/k)/(a + b*x^n)^(1/k)], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && Ration
alQ[m, p] && IntegersQ[p + (m + 1)/n, q] && LtQ[-1, p, 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 1532

Int[(((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_)^(n_))^(q_))/((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_)), x_Symbol
] :> Dist[e*(f^n/c), Int[(f*x)^(m - n)*(d + e*x^n)^(q - 1), x], x] - Dist[f^n/c, Int[(f*x)^(m - n)*(d + e*x^n)
^(q - 1)*(Simp[a*e - (c*d - b*e)*x^n, x]/(a + b*x^n + c*x^(2*n))), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && E
qQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] &&  !IntegerQ[q] && GtQ[q, 0] && GtQ[m, n - 1] && LeQ[m, 2*n
- 1]

Rule 1542

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

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*} \int \frac {\sqrt [4]{2 b+a x^4} \left (-4 b+a x^8\right )}{x^6 \left (-4 b+c x^4+a x^8\right )} \, dx &=\int \left (\frac {\sqrt [4]{2 b+a x^4}}{x^6}+\frac {c \sqrt [4]{2 b+a x^4}}{4 b x^2}+\frac {c x^2 \sqrt [4]{2 b+a x^4} \left (c+a x^4\right )}{4 b \left (4 b-c x^4-a x^8\right )}\right ) \, dx\\ &=\frac {c \int \frac {\sqrt [4]{2 b+a x^4}}{x^2} \, dx}{4 b}+\frac {c \int \frac {x^2 \sqrt [4]{2 b+a x^4} \left (c+a x^4\right )}{4 b-c x^4-a x^8} \, dx}{4 b}+\int \frac {\sqrt [4]{2 b+a x^4}}{x^6} \, dx\\ &=-\frac {c \sqrt [4]{2 b+a x^4}}{4 b x}-\frac {\left (2 b+a x^4\right )^{5/4}}{10 b x^5}+\frac {c \int \left (-\frac {c x^2 \sqrt [4]{2 b+a x^4}}{-4 b+c x^4+a x^8}-\frac {a x^6 \sqrt [4]{2 b+a x^4}}{-4 b+c x^4+a x^8}\right ) \, dx}{4 b}+\frac {(a c) \int \frac {x^2}{\left (2 b+a x^4\right )^{3/4}} \, dx}{4 b}\\ &=-\frac {c \sqrt [4]{2 b+a x^4}}{4 b x}-\frac {\left (2 b+a x^4\right )^{5/4}}{10 b x^5}-\frac {(a c) \int \frac {x^6 \sqrt [4]{2 b+a x^4}}{-4 b+c x^4+a x^8} \, dx}{4 b}+\frac {(a c) \text {Subst}\left (\int \frac {x^2}{1-a x^4} \, dx,x,\frac {x}{\sqrt [4]{2 b+a x^4}}\right )}{4 b}-\frac {c^2 \int \frac {x^2 \sqrt [4]{2 b+a x^4}}{-4 b+c x^4+a x^8} \, dx}{4 b}\\ &=-\frac {c \sqrt [4]{2 b+a x^4}}{4 b x}-\frac {\left (2 b+a x^4\right )^{5/4}}{10 b x^5}+\frac {c \int \frac {x^2 \left (-4 a b-a (2 b-c) x^4\right )}{\left (2 b+a x^4\right )^{3/4} \left (-4 b+c x^4+a x^8\right )} \, dx}{4 b}+\frac {\left (\sqrt {a} c\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt {a} x^2} \, dx,x,\frac {x}{\sqrt [4]{2 b+a x^4}}\right )}{8 b}-\frac {\left (\sqrt {a} c\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt {a} x^2} \, dx,x,\frac {x}{\sqrt [4]{2 b+a x^4}}\right )}{8 b}-\frac {(a c) \int \frac {x^2}{\left (2 b+a x^4\right )^{3/4}} \, dx}{4 b}-\frac {c^2 \int \left (\frac {2 a x^2 \sqrt [4]{2 b+a x^4}}{\sqrt {16 a b+c^2} \left (c-\sqrt {16 a b+c^2}+2 a x^4\right )}-\frac {2 a x^2 \sqrt [4]{2 b+a x^4}}{\sqrt {16 a b+c^2} \left (c+\sqrt {16 a b+c^2}+2 a x^4\right )}\right ) \, dx}{4 b}\\ &=-\frac {c \sqrt [4]{2 b+a x^4}}{4 b x}-\frac {\left (2 b+a x^4\right )^{5/4}}{10 b x^5}-\frac {\sqrt [4]{a} c \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{2 b+a x^4}}\right )}{8 b}+\frac {\sqrt [4]{a} c \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{2 b+a x^4}}\right )}{8 b}+\frac {c \int \left (-\frac {4 a b x^2}{\left (2 b+a x^4\right )^{3/4} \left (-4 b+c x^4+a x^8\right )}-\frac {a (2 b-c) x^6}{\left (2 b+a x^4\right )^{3/4} \left (-4 b+c x^4+a x^8\right )}\right ) \, dx}{4 b}-\frac {(a c) \text {Subst}\left (\int \frac {x^2}{1-a x^4} \, dx,x,\frac {x}{\sqrt [4]{2 b+a x^4}}\right )}{4 b}-\frac {\left (a c^2\right ) \int \frac {x^2 \sqrt [4]{2 b+a x^4}}{c-\sqrt {16 a b+c^2}+2 a x^4} \, dx}{2 b \sqrt {16 a b+c^2}}+\frac {\left (a c^2\right ) \int \frac {x^2 \sqrt [4]{2 b+a x^4}}{c+\sqrt {16 a b+c^2}+2 a x^4} \, dx}{2 b \sqrt {16 a b+c^2}}\\ &=-\frac {c \sqrt [4]{2 b+a x^4}}{4 b x}-\frac {\left (2 b+a x^4\right )^{5/4}}{10 b x^5}-\frac {\sqrt [4]{a} c \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{2 b+a x^4}}\right )}{8 b}+\frac {\sqrt [4]{a} c \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{2 b+a x^4}}\right )}{8 b}-(a c) \int \frac {x^2}{\left (2 b+a x^4\right )^{3/4} \left (-4 b+c x^4+a x^8\right )} \, dx-\frac {\left (\sqrt {a} c\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt {a} x^2} \, dx,x,\frac {x}{\sqrt [4]{2 b+a x^4}}\right )}{8 b}+\frac {\left (\sqrt {a} c\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt {a} x^2} \, dx,x,\frac {x}{\sqrt [4]{2 b+a x^4}}\right )}{8 b}-\frac {(a (2 b-c) c) \int \frac {x^6}{\left (2 b+a x^4\right )^{3/4} \left (-4 b+c x^4+a x^8\right )} \, dx}{4 b}-\frac {\left (a c^2 \sqrt [4]{2 b+a x^4}\right ) \int \frac {x^2 \sqrt [4]{1+\frac {a x^4}{2 b}}}{c-\sqrt {16 a b+c^2}+2 a x^4} \, dx}{2 b \sqrt {16 a b+c^2} \sqrt [4]{1+\frac {a x^4}{2 b}}}+\frac {\left (a c^2 \sqrt [4]{2 b+a x^4}\right ) \int \frac {x^2 \sqrt [4]{1+\frac {a x^4}{2 b}}}{c+\sqrt {16 a b+c^2}+2 a x^4} \, dx}{2 b \sqrt {16 a b+c^2} \sqrt [4]{1+\frac {a x^4}{2 b}}}\\ &=-\frac {c \sqrt [4]{2 b+a x^4}}{4 b x}-\frac {\left (2 b+a x^4\right )^{5/4}}{10 b x^5}+\frac {a c^2 x^3 \sqrt [4]{2 b+a x^4} F_1\left (\frac {3}{4};1,-\frac {1}{4};\frac {7}{4};-\frac {2 a x^4}{c-\sqrt {16 a b+c^2}},-\frac {a x^4}{2 b}\right )}{3\ 2^{3/4} b \left (16 a b+c \left (c-\sqrt {16 a b+c^2}\right )\right ) \sqrt [4]{2+\frac {a x^4}{b}}}+\frac {a c^2 x^3 \sqrt [4]{2 b+a x^4} F_1\left (\frac {3}{4};1,-\frac {1}{4};\frac {7}{4};-\frac {2 a x^4}{c+\sqrt {16 a b+c^2}},-\frac {a x^4}{2 b}\right )}{3\ 2^{3/4} b \left (16 a b+c \left (c+\sqrt {16 a b+c^2}\right )\right ) \sqrt [4]{2+\frac {a x^4}{b}}}-(a c) \int \left (\frac {2 a x^2}{\sqrt {16 a b+c^2} \left (2 b+a x^4\right )^{3/4} \left (c-\sqrt {16 a b+c^2}+2 a x^4\right )}-\frac {2 a x^2}{\sqrt {16 a b+c^2} \left (2 b+a x^4\right )^{3/4} \left (c+\sqrt {16 a b+c^2}+2 a x^4\right )}\right ) \, dx-\frac {(a (2 b-c) c) \int \left (\frac {\left (-c+\sqrt {16 a b+c^2}\right ) x^2}{\sqrt {16 a b+c^2} \left (2 b+a x^4\right )^{3/4} \left (c-\sqrt {16 a b+c^2}+2 a x^4\right )}+\frac {\left (c+\sqrt {16 a b+c^2}\right ) x^2}{\sqrt {16 a b+c^2} \left (2 b+a x^4\right )^{3/4} \left (c+\sqrt {16 a b+c^2}+2 a x^4\right )}\right ) \, dx}{4 b}\\ &=-\frac {c \sqrt [4]{2 b+a x^4}}{4 b x}-\frac {\left (2 b+a x^4\right )^{5/4}}{10 b x^5}+\frac {a c^2 x^3 \sqrt [4]{2 b+a x^4} F_1\left (\frac {3}{4};1,-\frac {1}{4};\frac {7}{4};-\frac {2 a x^4}{c-\sqrt {16 a b+c^2}},-\frac {a x^4}{2 b}\right )}{3\ 2^{3/4} b \left (16 a b+c \left (c-\sqrt {16 a b+c^2}\right )\right ) \sqrt [4]{2+\frac {a x^4}{b}}}+\frac {a c^2 x^3 \sqrt [4]{2 b+a x^4} F_1\left (\frac {3}{4};1,-\frac {1}{4};\frac {7}{4};-\frac {2 a x^4}{c+\sqrt {16 a b+c^2}},-\frac {a x^4}{2 b}\right )}{3\ 2^{3/4} b \left (16 a b+c \left (c+\sqrt {16 a b+c^2}\right )\right ) \sqrt [4]{2+\frac {a x^4}{b}}}-\frac {\left (2 a^2 c\right ) \int \frac {x^2}{\left (2 b+a x^4\right )^{3/4} \left (c-\sqrt {16 a b+c^2}+2 a x^4\right )} \, dx}{\sqrt {16 a b+c^2}}+\frac {\left (2 a^2 c\right ) \int \frac {x^2}{\left (2 b+a x^4\right )^{3/4} \left (c+\sqrt {16 a b+c^2}+2 a x^4\right )} \, dx}{\sqrt {16 a b+c^2}}-\frac {\left (a (2 b-c) c \left (1-\frac {c}{\sqrt {16 a b+c^2}}\right )\right ) \int \frac {x^2}{\left (2 b+a x^4\right )^{3/4} \left (c-\sqrt {16 a b+c^2}+2 a x^4\right )} \, dx}{4 b}-\frac {\left (a (2 b-c) c \left (1+\frac {c}{\sqrt {16 a b+c^2}}\right )\right ) \int \frac {x^2}{\left (2 b+a x^4\right )^{3/4} \left (c+\sqrt {16 a b+c^2}+2 a x^4\right )} \, dx}{4 b}\\ &=-\frac {c \sqrt [4]{2 b+a x^4}}{4 b x}-\frac {\left (2 b+a x^4\right )^{5/4}}{10 b x^5}+\frac {a c^2 x^3 \sqrt [4]{2 b+a x^4} F_1\left (\frac {3}{4};1,-\frac {1}{4};\frac {7}{4};-\frac {2 a x^4}{c-\sqrt {16 a b+c^2}},-\frac {a x^4}{2 b}\right )}{3\ 2^{3/4} b \left (16 a b+c \left (c-\sqrt {16 a b+c^2}\right )\right ) \sqrt [4]{2+\frac {a x^4}{b}}}+\frac {a c^2 x^3 \sqrt [4]{2 b+a x^4} F_1\left (\frac {3}{4};1,-\frac {1}{4};\frac {7}{4};-\frac {2 a x^4}{c+\sqrt {16 a b+c^2}},-\frac {a x^4}{2 b}\right )}{3\ 2^{3/4} b \left (16 a b+c \left (c+\sqrt {16 a b+c^2}\right )\right ) \sqrt [4]{2+\frac {a x^4}{b}}}-\frac {\left (2 a^2 c\right ) \text {Subst}\left (\int \frac {x^2}{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]{2 b+a x^4}}\right )}{\sqrt {16 a b+c^2}}+\frac {\left (2 a^2 c\right ) \text {Subst}\left (\int \frac {x^2}{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]{2 b+a x^4}}\right )}{\sqrt {16 a b+c^2}}-\frac {\left (a (2 b-c) c \left (1-\frac {c}{\sqrt {16 a b+c^2}}\right )\right ) \text {Subst}\left (\int \frac {x^2}{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]{2 b+a x^4}}\right )}{4 b}-\frac {\left (a (2 b-c) c \left (1+\frac {c}{\sqrt {16 a b+c^2}}\right )\right ) \text {Subst}\left (\int \frac {x^2}{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]{2 b+a x^4}}\right )}{4 b}\\ &=-\frac {c \sqrt [4]{2 b+a x^4}}{4 b x}-\frac {\left (2 b+a x^4\right )^{5/4}}{10 b x^5}+\frac {a c^2 x^3 \sqrt [4]{2 b+a x^4} F_1\left (\frac {3}{4};1,-\frac {1}{4};\frac {7}{4};-\frac {2 a x^4}{c-\sqrt {16 a b+c^2}},-\frac {a x^4}{2 b}\right )}{3\ 2^{3/4} b \left (16 a b+c \left (c-\sqrt {16 a b+c^2}\right )\right ) \sqrt [4]{2+\frac {a x^4}{b}}}+\frac {a c^2 x^3 \sqrt [4]{2 b+a x^4} F_1\left (\frac {3}{4};1,-\frac {1}{4};\frac {7}{4};-\frac {2 a x^4}{c+\sqrt {16 a b+c^2}},-\frac {a x^4}{2 b}\right )}{3\ 2^{3/4} b \left (16 a b+c \left (c+\sqrt {16 a b+c^2}\right )\right ) \sqrt [4]{2+\frac {a x^4}{b}}}+\frac {\left (a^{3/2} c\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]{2 b+a x^4}}\right )}{\sqrt {16 a b+c^2} \sqrt {4 b-c+\sqrt {16 a b+c^2}}}-\frac {\left (a^{3/2} c\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]{2 b+a x^4}}\right )}{\sqrt {16 a b+c^2} \sqrt {4 b-c+\sqrt {16 a b+c^2}}}+\frac {\left (\sqrt {a} (2 b-c) c \left (1-\frac {c}{\sqrt {16 a b+c^2}}\right )\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]{2 b+a x^4}}\right )}{8 b \sqrt {4 b-c+\sqrt {16 a b+c^2}}}-\frac {\left (\sqrt {a} (2 b-c) c \left (1-\frac {c}{\sqrt {16 a b+c^2}}\right )\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]{2 b+a x^4}}\right )}{8 b \sqrt {4 b-c+\sqrt {16 a b+c^2}}}+\frac {\left (a^{3/2} c\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]{2 b+a x^4}}\right )}{\sqrt {16 a b+c^2} \sqrt {-4 b+c+\sqrt {16 a b+c^2}}}-\frac {\left (a^{3/2} c\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]{2 b+a x^4}}\right )}{\sqrt {16 a b+c^2} \sqrt {-4 b+c+\sqrt {16 a b+c^2}}}-\frac {\left (\sqrt {a} (2 b-c) c \left (1+\frac {c}{\sqrt {16 a b+c^2}}\right )\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]{2 b+a x^4}}\right )}{8 b \sqrt {-4 b+c+\sqrt {16 a b+c^2}}}+\frac {\left (\sqrt {a} (2 b-c) c \left (1+\frac {c}{\sqrt {16 a b+c^2}}\right )\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]{2 b+a x^4}}\right )}{8 b \sqrt {-4 b+c+\sqrt {16 a b+c^2}}}\\ &=-\frac {c \sqrt [4]{2 b+a x^4}}{4 b x}-\frac {\left (2 b+a x^4\right )^{5/4}}{10 b x^5}+\frac {a c^2 x^3 \sqrt [4]{2 b+a x^4} F_1\left (\frac {3}{4};1,-\frac {1}{4};\frac {7}{4};-\frac {2 a x^4}{c-\sqrt {16 a b+c^2}},-\frac {a x^4}{2 b}\right )}{3\ 2^{3/4} b \left (16 a b+c \left (c-\sqrt {16 a b+c^2}\right )\right ) \sqrt [4]{2+\frac {a x^4}{b}}}+\frac {a c^2 x^3 \sqrt [4]{2 b+a x^4} F_1\left (\frac {3}{4};1,-\frac {1}{4};\frac {7}{4};-\frac {2 a x^4}{c+\sqrt {16 a b+c^2}},-\frac {a x^4}{2 b}\right )}{3\ 2^{3/4} b \left (16 a b+c \left (c+\sqrt {16 a b+c^2}\right )\right ) \sqrt [4]{2+\frac {a x^4}{b}}}-\frac {a^{5/4} c \tan ^{-1}\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]{2 b+a x^4}}\right )}{\sqrt {16 a b+c^2} \sqrt [4]{-c+\sqrt {16 a b+c^2}} \left (4 b-c+\sqrt {16 a b+c^2}\right )^{3/4}}-\frac {\sqrt [4]{a} (2 b-c) c \left (-c+\sqrt {16 a b+c^2}\right )^{3/4} \tan ^{-1}\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]{2 b+a x^4}}\right )}{8 b \sqrt {16 a b+c^2} \left (4 b-c+\sqrt {16 a b+c^2}\right )^{3/4}}-\frac {a^{5/4} c \tan ^{-1}\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]{2 b+a x^4}}\right )}{\sqrt {16 a b+c^2} \sqrt [4]{c+\sqrt {16 a b+c^2}} \left (-4 b+c+\sqrt {16 a b+c^2}\right )^{3/4}}+\frac {\sqrt [4]{a} (2 b-c) c \left (c+\sqrt {16 a b+c^2}\right )^{3/4} \tan ^{-1}\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]{2 b+a x^4}}\right )}{8 b \sqrt {16 a b+c^2} \left (-4 b+c+\sqrt {16 a b+c^2}\right )^{3/4}}+\frac {a^{5/4} c \tanh ^{-1}\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]{2 b+a x^4}}\right )}{\sqrt {16 a b+c^2} \sqrt [4]{-c+\sqrt {16 a b+c^2}} \left (4 b-c+\sqrt {16 a b+c^2}\right )^{3/4}}+\frac {\sqrt [4]{a} (2 b-c) c \left (-c+\sqrt {16 a b+c^2}\right )^{3/4} \tanh ^{-1}\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]{2 b+a x^4}}\right )}{8 b \sqrt {16 a b+c^2} \left (4 b-c+\sqrt {16 a b+c^2}\right )^{3/4}}+\frac {a^{5/4} c \tanh ^{-1}\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]{2 b+a x^4}}\right )}{\sqrt {16 a b+c^2} \sqrt [4]{c+\sqrt {16 a b+c^2}} \left (-4 b+c+\sqrt {16 a b+c^2}\right )^{3/4}}-\frac {\sqrt [4]{a} (2 b-c) c \left (c+\sqrt {16 a b+c^2}\right )^{3/4} \tanh ^{-1}\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]{2 b+a x^4}}\right )}{8 b \sqrt {16 a b+c^2} \left (-4 b+c+\sqrt {16 a b+c^2}\right )^{3/4}}\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 271, normalized size = 1.00 \begin {gather*} \frac {\sqrt [4]{2 b+a x^4} \left (-4 b-2 a x^4-5 c x^4\right )}{20 b x^5}-\frac {c \text {RootSum}\left [2 a^2-2 a b+a c-4 a \text {$\#$1}^4-c \text {$\#$1}^4+2 \text {$\#$1}^8\&,\frac {2 a^2 \log (x)-2 a b \log (x)+a c \log (x)-2 a^2 \log \left (\sqrt [4]{2 b+a x^4}-x \text {$\#$1}\right )+2 a b \log \left (\sqrt [4]{2 b+a x^4}-x \text {$\#$1}\right )-a c \log \left (\sqrt [4]{2 b+a x^4}-x \text {$\#$1}\right )-2 a \log (x) \text {$\#$1}^4-c \log (x) \text {$\#$1}^4+2 a \log \left (\sqrt [4]{2 b+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4+c \log \left (\sqrt [4]{2 b+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4}{-4 a \text {$\#$1}^3-c \text {$\#$1}^3+4 \text {$\#$1}^7}\&\right ]}{16 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((2*b + a*x^4)^(1/4)*(-4*b + a*x^8))/(x^6*(-4*b + c*x^4 + a*x^8)),x]

[Out]

((2*b + a*x^4)^(1/4)*(-4*b - 2*a*x^4 - 5*c*x^4))/(20*b*x^5) - (c*RootSum[2*a^2 - 2*a*b + a*c - 4*a*#1^4 - c*#1
^4 + 2*#1^8 & , (2*a^2*Log[x] - 2*a*b*Log[x] + a*c*Log[x] - 2*a^2*Log[(2*b + a*x^4)^(1/4) - x*#1] + 2*a*b*Log[
(2*b + a*x^4)^(1/4) - x*#1] - a*c*Log[(2*b + a*x^4)^(1/4) - x*#1] - 2*a*Log[x]*#1^4 - c*Log[x]*#1^4 + 2*a*Log[
(2*b + a*x^4)^(1/4) - x*#1]*#1^4 + c*Log[(2*b + a*x^4)^(1/4) - x*#1]*#1^4)/(-4*a*#1^3 - c*#1^3 + 4*#1^7) & ])/
(16*b)

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Maple [F]
time = 0.00, size = 0, normalized size = 0.00 \[\int \frac {\left (a \,x^{4}+2 b \right )^{\frac {1}{4}} \left (a \,x^{8}-4 b \right )}{x^{6} \left (a \,x^{8}+c \,x^{4}-4 b \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^4+2*b)^(1/4)*(a*x^8-4*b)/x^6/(a*x^8+c*x^4-4*b),x)

[Out]

int((a*x^4+2*b)^(1/4)*(a*x^8-4*b)/x^6/(a*x^8+c*x^4-4*b),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^4+2*b)^(1/4)*(a*x^8-4*b)/x^6/(a*x^8+c*x^4-4*b),x, algorithm="maxima")

[Out]

integrate((a*x^8 - 4*b)*(a*x^4 + 2*b)^(1/4)/((a*x^8 + c*x^4 - 4*b)*x^6), x)

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Fricas [F(-1)] Timed out
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((a*x^4+2*b)^(1/4)*(a*x^8-4*b)/x^6/(a*x^8+c*x^4-4*b),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F(-1)] Timed out
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((a*x**4+2*b)**(1/4)*(a*x**8-4*b)/x**6/(a*x**8+c*x**4-4*b),x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^4+2*b)^(1/4)*(a*x^8-4*b)/x^6/(a*x^8+c*x^4-4*b),x, algorithm="giac")

[Out]

integrate((a*x^8 - 4*b)*(a*x^4 + 2*b)^(1/4)/((a*x^8 + c*x^4 - 4*b)*x^6), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int -\frac {{\left (a\,x^4+2\,b\right )}^{1/4}\,\left (4\,b-a\,x^8\right )}{x^6\,\left (a\,x^8+c\,x^4-4\,b\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((2*b + a*x^4)^(1/4)*(4*b - a*x^8))/(x^6*(a*x^8 - 4*b + c*x^4)),x)

[Out]

int(-((2*b + a*x^4)^(1/4)*(4*b - a*x^8))/(x^6*(a*x^8 - 4*b + c*x^4)), x)

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