3.815 \(\int \frac {x^8}{(a+b x^4) \sqrt {c+d x^4}} \, dx\)

Optimal. Leaf size=872 \[ \frac {\left (\frac {\sqrt {b} \sqrt {c}}{\sqrt {-a}}+\sqrt {d}\right ) \sqrt [4]{d} \left (\sqrt {d} x^2+\sqrt {c}\right ) \sqrt {\frac {d x^4+c}{\left (\sqrt {d} x^2+\sqrt {c}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} x}{\sqrt [4]{c}}\right )|\frac {1}{2}\right ) a^2}{4 b^2 \sqrt [4]{c} (b c+a d) \sqrt {d x^4+c}}+\frac {\left (\sqrt {d} a+\sqrt {-a} \sqrt {b} \sqrt {c}\right ) \sqrt [4]{d} \left (\sqrt {d} x^2+\sqrt {c}\right ) \sqrt {\frac {d x^4+c}{\left (\sqrt {d} x^2+\sqrt {c}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} x}{\sqrt [4]{c}}\right )|\frac {1}{2}\right ) a}{4 b^2 \sqrt [4]{c} (b c+a d) \sqrt {d x^4+c}}+\frac {\left (\sqrt {b} \sqrt {c}+\sqrt {-a} \sqrt {d}\right )^2 \left (\sqrt {d} x^2+\sqrt {c}\right ) \sqrt {\frac {d x^4+c}{\left (\sqrt {d} x^2+\sqrt {c}\right )^2}} \Pi \left (-\frac {\left (\sqrt {b} \sqrt {c}-\sqrt {-a} \sqrt {d}\right )^2}{4 \sqrt {-a} \sqrt {b} \sqrt {c} \sqrt {d}};2 \tan ^{-1}\left (\frac {\sqrt [4]{d} x}{\sqrt [4]{c}}\right )|\frac {1}{2}\right ) a}{8 b^2 \sqrt [4]{c} \sqrt [4]{d} (b c+a d) \sqrt {d x^4+c}}+\frac {\left (\sqrt {b} \sqrt {c}-\sqrt {-a} \sqrt {d}\right )^2 \left (\sqrt {d} x^2+\sqrt {c}\right ) \sqrt {\frac {d x^4+c}{\left (\sqrt {d} x^2+\sqrt {c}\right )^2}} \Pi \left (\frac {\left (\sqrt {b} \sqrt {c}+\sqrt {-a} \sqrt {d}\right )^2}{4 \sqrt {-a} \sqrt {b} \sqrt {c} \sqrt {d}};2 \tan ^{-1}\left (\frac {\sqrt [4]{d} x}{\sqrt [4]{c}}\right )|\frac {1}{2}\right ) a}{8 b^2 \sqrt [4]{c} \sqrt [4]{d} (b c+a d) \sqrt {d x^4+c}}-\frac {(-a)^{5/4} \tan ^{-1}\left (\frac {\sqrt {b c-a d} x}{\sqrt [4]{-a} \sqrt [4]{b} \sqrt {d x^4+c}}\right )}{4 b^{7/4} \sqrt {b c-a d}}-\frac {(-a)^{5/4} \tan ^{-1}\left (\frac {\sqrt {a d-b c} x}{\sqrt [4]{-a} \sqrt [4]{b} \sqrt {d x^4+c}}\right )}{4 b^{7/4} \sqrt {a d-b c}}-\frac {(b c+3 a d) \left (\sqrt {d} x^2+\sqrt {c}\right ) \sqrt {\frac {d x^4+c}{\left (\sqrt {d} x^2+\sqrt {c}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} x}{\sqrt [4]{c}}\right )|\frac {1}{2}\right )}{6 b^2 \sqrt [4]{c} d^{5/4} \sqrt {d x^4+c}}+\frac {x \sqrt {d x^4+c}}{3 b d} \]

[Out]

-1/4*(-a)^(5/4)*arctan(x*(-a*d+b*c)^(1/2)/(-a)^(1/4)/b^(1/4)/(d*x^4+c)^(1/2))/b^(7/4)/(-a*d+b*c)^(1/2)-1/4*(-a
)^(5/4)*arctan(x*(a*d-b*c)^(1/2)/(-a)^(1/4)/b^(1/4)/(d*x^4+c)^(1/2))/b^(7/4)/(a*d-b*c)^(1/2)+1/3*x*(d*x^4+c)^(
1/2)/b/d-1/6*(3*a*d+b*c)*(cos(2*arctan(d^(1/4)*x/c^(1/4)))^2)^(1/2)/cos(2*arctan(d^(1/4)*x/c^(1/4)))*EllipticF
(sin(2*arctan(d^(1/4)*x/c^(1/4))),1/2*2^(1/2))*(c^(1/2)+x^2*d^(1/2))*((d*x^4+c)/(c^(1/2)+x^2*d^(1/2))^2)^(1/2)
/b^2/c^(1/4)/d^(5/4)/(d*x^4+c)^(1/2)+1/4*a^2*d^(1/4)*(cos(2*arctan(d^(1/4)*x/c^(1/4)))^2)^(1/2)/cos(2*arctan(d
^(1/4)*x/c^(1/4)))*EllipticF(sin(2*arctan(d^(1/4)*x/c^(1/4))),1/2*2^(1/2))*(b^(1/2)*c^(1/2)/(-a)^(1/2)+d^(1/2)
)*(c^(1/2)+x^2*d^(1/2))*((d*x^4+c)/(c^(1/2)+x^2*d^(1/2))^2)^(1/2)/b^2/c^(1/4)/(a*d+b*c)/(d*x^4+c)^(1/2)+1/4*a*
d^(1/4)*(cos(2*arctan(d^(1/4)*x/c^(1/4)))^2)^(1/2)/cos(2*arctan(d^(1/4)*x/c^(1/4)))*EllipticF(sin(2*arctan(d^(
1/4)*x/c^(1/4))),1/2*2^(1/2))*((-a)^(1/2)*b^(1/2)*c^(1/2)+a*d^(1/2))*(c^(1/2)+x^2*d^(1/2))*((d*x^4+c)/(c^(1/2)
+x^2*d^(1/2))^2)^(1/2)/b^2/c^(1/4)/(a*d+b*c)/(d*x^4+c)^(1/2)+1/8*a*(cos(2*arctan(d^(1/4)*x/c^(1/4)))^2)^(1/2)/
cos(2*arctan(d^(1/4)*x/c^(1/4)))*EllipticPi(sin(2*arctan(d^(1/4)*x/c^(1/4))),1/4*(b^(1/2)*c^(1/2)+(-a)^(1/2)*d
^(1/2))^2/(-a)^(1/2)/b^(1/2)/c^(1/2)/d^(1/2),1/2*2^(1/2))*(c^(1/2)+x^2*d^(1/2))*(b^(1/2)*c^(1/2)-(-a)^(1/2)*d^
(1/2))^2*((d*x^4+c)/(c^(1/2)+x^2*d^(1/2))^2)^(1/2)/b^2/c^(1/4)/d^(1/4)/(a*d+b*c)/(d*x^4+c)^(1/2)+1/8*a*(cos(2*
arctan(d^(1/4)*x/c^(1/4)))^2)^(1/2)/cos(2*arctan(d^(1/4)*x/c^(1/4)))*EllipticPi(sin(2*arctan(d^(1/4)*x/c^(1/4)
)),-1/4*(b^(1/2)*c^(1/2)-(-a)^(1/2)*d^(1/2))^2/(-a)^(1/2)/b^(1/2)/c^(1/2)/d^(1/2),1/2*2^(1/2))*(c^(1/2)+x^2*d^
(1/2))*(b^(1/2)*c^(1/2)+(-a)^(1/2)*d^(1/2))^2*((d*x^4+c)/(c^(1/2)+x^2*d^(1/2))^2)^(1/2)/b^2/c^(1/4)/d^(1/4)/(a
*d+b*c)/(d*x^4+c)^(1/2)

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Rubi [A]  time = 1.02, antiderivative size = 872, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {479, 523, 220, 409, 1217, 1707} \[ \frac {\left (\frac {\sqrt {b} \sqrt {c}}{\sqrt {-a}}+\sqrt {d}\right ) \sqrt [4]{d} \left (\sqrt {d} x^2+\sqrt {c}\right ) \sqrt {\frac {d x^4+c}{\left (\sqrt {d} x^2+\sqrt {c}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} x}{\sqrt [4]{c}}\right )|\frac {1}{2}\right ) a^2}{4 b^2 \sqrt [4]{c} (b c+a d) \sqrt {d x^4+c}}+\frac {\left (\sqrt {d} a+\sqrt {-a} \sqrt {b} \sqrt {c}\right ) \sqrt [4]{d} \left (\sqrt {d} x^2+\sqrt {c}\right ) \sqrt {\frac {d x^4+c}{\left (\sqrt {d} x^2+\sqrt {c}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} x}{\sqrt [4]{c}}\right )|\frac {1}{2}\right ) a}{4 b^2 \sqrt [4]{c} (b c+a d) \sqrt {d x^4+c}}+\frac {\left (\sqrt {b} \sqrt {c}+\sqrt {-a} \sqrt {d}\right )^2 \left (\sqrt {d} x^2+\sqrt {c}\right ) \sqrt {\frac {d x^4+c}{\left (\sqrt {d} x^2+\sqrt {c}\right )^2}} \Pi \left (-\frac {\left (\sqrt {b} \sqrt {c}-\sqrt {-a} \sqrt {d}\right )^2}{4 \sqrt {-a} \sqrt {b} \sqrt {c} \sqrt {d}};2 \tan ^{-1}\left (\frac {\sqrt [4]{d} x}{\sqrt [4]{c}}\right )|\frac {1}{2}\right ) a}{8 b^2 \sqrt [4]{c} \sqrt [4]{d} (b c+a d) \sqrt {d x^4+c}}+\frac {\left (\sqrt {b} \sqrt {c}-\sqrt {-a} \sqrt {d}\right )^2 \left (\sqrt {d} x^2+\sqrt {c}\right ) \sqrt {\frac {d x^4+c}{\left (\sqrt {d} x^2+\sqrt {c}\right )^2}} \Pi \left (\frac {\left (\sqrt {b} \sqrt {c}+\sqrt {-a} \sqrt {d}\right )^2}{4 \sqrt {-a} \sqrt {b} \sqrt {c} \sqrt {d}};2 \tan ^{-1}\left (\frac {\sqrt [4]{d} x}{\sqrt [4]{c}}\right )|\frac {1}{2}\right ) a}{8 b^2 \sqrt [4]{c} \sqrt [4]{d} (b c+a d) \sqrt {d x^4+c}}-\frac {(-a)^{5/4} \tan ^{-1}\left (\frac {\sqrt {b c-a d} x}{\sqrt [4]{-a} \sqrt [4]{b} \sqrt {d x^4+c}}\right )}{4 b^{7/4} \sqrt {b c-a d}}-\frac {(-a)^{5/4} \tan ^{-1}\left (\frac {\sqrt {a d-b c} x}{\sqrt [4]{-a} \sqrt [4]{b} \sqrt {d x^4+c}}\right )}{4 b^{7/4} \sqrt {a d-b c}}-\frac {(b c+3 a d) \left (\sqrt {d} x^2+\sqrt {c}\right ) \sqrt {\frac {d x^4+c}{\left (\sqrt {d} x^2+\sqrt {c}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} x}{\sqrt [4]{c}}\right )|\frac {1}{2}\right )}{6 b^2 \sqrt [4]{c} d^{5/4} \sqrt {d x^4+c}}+\frac {x \sqrt {d x^4+c}}{3 b d} \]

Antiderivative was successfully verified.

[In]

Int[x^8/((a + b*x^4)*Sqrt[c + d*x^4]),x]

[Out]

(x*Sqrt[c + d*x^4])/(3*b*d) - ((-a)^(5/4)*ArcTan[(Sqrt[b*c - a*d]*x)/((-a)^(1/4)*b^(1/4)*Sqrt[c + d*x^4])])/(4
*b^(7/4)*Sqrt[b*c - a*d]) - ((-a)^(5/4)*ArcTan[(Sqrt[-(b*c) + a*d]*x)/((-a)^(1/4)*b^(1/4)*Sqrt[c + d*x^4])])/(
4*b^(7/4)*Sqrt[-(b*c) + a*d]) + (a^2*((Sqrt[b]*Sqrt[c])/Sqrt[-a] + Sqrt[d])*d^(1/4)*(Sqrt[c] + Sqrt[d]*x^2)*Sq
rt[(c + d*x^4)/(Sqrt[c] + Sqrt[d]*x^2)^2]*EllipticF[2*ArcTan[(d^(1/4)*x)/c^(1/4)], 1/2])/(4*b^2*c^(1/4)*(b*c +
a*d)*Sqrt[c + d*x^4]) + (a*(Sqrt[-a]*Sqrt[b]*Sqrt[c] + a*Sqrt[d])*d^(1/4)*(Sqrt[c] + Sqrt[d]*x^2)*Sqrt[(c + d
*x^4)/(Sqrt[c] + Sqrt[d]*x^2)^2]*EllipticF[2*ArcTan[(d^(1/4)*x)/c^(1/4)], 1/2])/(4*b^2*c^(1/4)*(b*c + a*d)*Sqr
t[c + d*x^4]) - ((b*c + 3*a*d)*(Sqrt[c] + Sqrt[d]*x^2)*Sqrt[(c + d*x^4)/(Sqrt[c] + Sqrt[d]*x^2)^2]*EllipticF[2
*ArcTan[(d^(1/4)*x)/c^(1/4)], 1/2])/(6*b^2*c^(1/4)*d^(5/4)*Sqrt[c + d*x^4]) + (a*(Sqrt[b]*Sqrt[c] + Sqrt[-a]*S
qrt[d])^2*(Sqrt[c] + Sqrt[d]*x^2)*Sqrt[(c + d*x^4)/(Sqrt[c] + Sqrt[d]*x^2)^2]*EllipticPi[-(Sqrt[b]*Sqrt[c] - S
qrt[-a]*Sqrt[d])^2/(4*Sqrt[-a]*Sqrt[b]*Sqrt[c]*Sqrt[d]), 2*ArcTan[(d^(1/4)*x)/c^(1/4)], 1/2])/(8*b^2*c^(1/4)*d
^(1/4)*(b*c + a*d)*Sqrt[c + d*x^4]) + (a*(Sqrt[b]*Sqrt[c] - Sqrt[-a]*Sqrt[d])^2*(Sqrt[c] + Sqrt[d]*x^2)*Sqrt[(
c + d*x^4)/(Sqrt[c] + Sqrt[d]*x^2)^2]*EllipticPi[(Sqrt[b]*Sqrt[c] + Sqrt[-a]*Sqrt[d])^2/(4*Sqrt[-a]*Sqrt[b]*Sq
rt[c]*Sqrt[d]), 2*ArcTan[(d^(1/4)*x)/c^(1/4)], 1/2])/(8*b^2*c^(1/4)*d^(1/4)*(b*c + a*d)*Sqrt[c + d*x^4])

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 409

Int[1/(Sqrt[(a_) + (b_.)*(x_)^4]*((c_) + (d_.)*(x_)^4)), x_Symbol] :> Dist[1/(2*c), Int[1/(Sqrt[a + b*x^4]*(1
- Rt[-(d/c), 2]*x^2)), x], x] + Dist[1/(2*c), Int[1/(Sqrt[a + b*x^4]*(1 + Rt[-(d/c), 2]*x^2)), x], x] /; FreeQ
[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 479

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(e^(2*n
- 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*d*(m + n*(p + q) + 1)), x] - Dist[e^(2*n)
/(b*d*(m + n*(p + q) + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1) + (a*d*(m +
n*(q - 1) + 1) + b*c*(m + n*(p - 1) + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d
, 0] && IGtQ[n, 0] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 523

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]

Rule 1217

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[c/a, 2]}, Dist[(c*d + a*e*q
)/(c*d^2 - a*e^2), Int[1/Sqrt[a + c*x^4], x], x] - Dist[(a*e*(e + d*q))/(c*d^2 - a*e^2), Int[(1 + q*x^2)/((d +
e*x^2)*Sqrt[a + c*x^4]), x], x]] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0]
&& PosQ[c/a]

Rule 1707

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[B/A, 2]
}, -Simp[((B*d - A*e)*ArcTan[(Rt[(c*d)/e + (a*e)/d, 2]*x)/Sqrt[a + c*x^4]])/(2*d*e*Rt[(c*d)/e + (a*e)/d, 2]),
x] + Simp[((B*d + A*e)*(A + B*x^2)*Sqrt[(A^2*(a + c*x^4))/(a*(A + B*x^2)^2)]*EllipticPi[Cancel[-((B*d - A*e)^2
/(4*d*e*A*B))], 2*ArcTan[q*x], 1/2])/(4*d*e*A*q*Sqrt[a + c*x^4]), x]] /; FreeQ[{a, c, d, e, A, B}, x] && NeQ[c
*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[c/a] && EqQ[c*A^2 - a*B^2, 0]

Rubi steps

\begin {align*} \int \frac {x^8}{\left (a+b x^4\right ) \sqrt {c+d x^4}} \, dx &=\frac {x \sqrt {c+d x^4}}{3 b d}-\frac {\int \frac {a c+(b c+3 a d) x^4}{\left (a+b x^4\right ) \sqrt {c+d x^4}} \, dx}{3 b d}\\ &=\frac {x \sqrt {c+d x^4}}{3 b d}+\frac {a^2 \int \frac {1}{\left (a+b x^4\right ) \sqrt {c+d x^4}} \, dx}{b^2}-\frac {(b c+3 a d) \int \frac {1}{\sqrt {c+d x^4}} \, dx}{3 b^2 d}\\ &=\frac {x \sqrt {c+d x^4}}{3 b d}-\frac {(b c+3 a d) \left (\sqrt {c}+\sqrt {d} x^2\right ) \sqrt {\frac {c+d x^4}{\left (\sqrt {c}+\sqrt {d} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} x}{\sqrt [4]{c}}\right )|\frac {1}{2}\right )}{6 b^2 \sqrt [4]{c} d^{5/4} \sqrt {c+d x^4}}+\frac {a \int \frac {1}{\left (1-\frac {\sqrt {b} x^2}{\sqrt {-a}}\right ) \sqrt {c+d x^4}} \, dx}{2 b^2}+\frac {a \int \frac {1}{\left (1+\frac {\sqrt {b} x^2}{\sqrt {-a}}\right ) \sqrt {c+d x^4}} \, dx}{2 b^2}\\ &=\frac {x \sqrt {c+d x^4}}{3 b d}-\frac {(b c+3 a d) \left (\sqrt {c}+\sqrt {d} x^2\right ) \sqrt {\frac {c+d x^4}{\left (\sqrt {c}+\sqrt {d} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} x}{\sqrt [4]{c}}\right )|\frac {1}{2}\right )}{6 b^2 \sqrt [4]{c} d^{5/4} \sqrt {c+d x^4}}+\frac {\left (a \sqrt {c} \left (\sqrt {b} \sqrt {c}-\sqrt {-a} \sqrt {d}\right )\right ) \int \frac {1+\frac {\sqrt {d} x^2}{\sqrt {c}}}{\left (1-\frac {\sqrt {b} x^2}{\sqrt {-a}}\right ) \sqrt {c+d x^4}} \, dx}{2 b^{3/2} (b c+a d)}+\frac {\left (a \sqrt {c} \left (\sqrt {b} \sqrt {c}+\sqrt {-a} \sqrt {d}\right )\right ) \int \frac {1+\frac {\sqrt {d} x^2}{\sqrt {c}}}{\left (1+\frac {\sqrt {b} x^2}{\sqrt {-a}}\right ) \sqrt {c+d x^4}} \, dx}{2 b^{3/2} (b c+a d)}+\frac {\left (a^2 \left (\frac {\sqrt {b} \sqrt {c}}{\sqrt {-a}}+\sqrt {d}\right ) \sqrt {d}\right ) \int \frac {1}{\sqrt {c+d x^4}} \, dx}{2 b^2 (b c+a d)}+\frac {\left (a \left (\sqrt {-a} \sqrt {b} \sqrt {c}+a \sqrt {d}\right ) \sqrt {d}\right ) \int \frac {1}{\sqrt {c+d x^4}} \, dx}{2 b^2 (b c+a d)}\\ &=\frac {x \sqrt {c+d x^4}}{3 b d}-\frac {(-a)^{5/4} \tan ^{-1}\left (\frac {\sqrt {b c-a d} x}{\sqrt [4]{-a} \sqrt [4]{b} \sqrt {c+d x^4}}\right )}{4 b^{7/4} \sqrt {b c-a d}}-\frac {(-a)^{5/4} \tan ^{-1}\left (\frac {\sqrt {-b c+a d} x}{\sqrt [4]{-a} \sqrt [4]{b} \sqrt {c+d x^4}}\right )}{4 b^{7/4} \sqrt {-b c+a d}}+\frac {a^2 \left (\frac {\sqrt {b} \sqrt {c}}{\sqrt {-a}}+\sqrt {d}\right ) \sqrt [4]{d} \left (\sqrt {c}+\sqrt {d} x^2\right ) \sqrt {\frac {c+d x^4}{\left (\sqrt {c}+\sqrt {d} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} x}{\sqrt [4]{c}}\right )|\frac {1}{2}\right )}{4 b^2 \sqrt [4]{c} (b c+a d) \sqrt {c+d x^4}}+\frac {a \left (\sqrt {-a} \sqrt {b} \sqrt {c}+a \sqrt {d}\right ) \sqrt [4]{d} \left (\sqrt {c}+\sqrt {d} x^2\right ) \sqrt {\frac {c+d x^4}{\left (\sqrt {c}+\sqrt {d} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} x}{\sqrt [4]{c}}\right )|\frac {1}{2}\right )}{4 b^2 \sqrt [4]{c} (b c+a d) \sqrt {c+d x^4}}-\frac {(b c+3 a d) \left (\sqrt {c}+\sqrt {d} x^2\right ) \sqrt {\frac {c+d x^4}{\left (\sqrt {c}+\sqrt {d} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} x}{\sqrt [4]{c}}\right )|\frac {1}{2}\right )}{6 b^2 \sqrt [4]{c} d^{5/4} \sqrt {c+d x^4}}+\frac {a \left (\sqrt {b} \sqrt {c}+\sqrt {-a} \sqrt {d}\right )^2 \left (\sqrt {c}+\sqrt {d} x^2\right ) \sqrt {\frac {c+d x^4}{\left (\sqrt {c}+\sqrt {d} x^2\right )^2}} \Pi \left (-\frac {\left (\sqrt {b} \sqrt {c}-\sqrt {-a} \sqrt {d}\right )^2}{4 \sqrt {-a} \sqrt {b} \sqrt {c} \sqrt {d}};2 \tan ^{-1}\left (\frac {\sqrt [4]{d} x}{\sqrt [4]{c}}\right )|\frac {1}{2}\right )}{8 b^2 \sqrt [4]{c} \sqrt [4]{d} (b c+a d) \sqrt {c+d x^4}}+\frac {a \left (\sqrt {b} \sqrt {c}-\sqrt {-a} \sqrt {d}\right )^2 \left (\sqrt {c}+\sqrt {d} x^2\right ) \sqrt {\frac {c+d x^4}{\left (\sqrt {c}+\sqrt {d} x^2\right )^2}} \Pi \left (\frac {\left (\sqrt {b} \sqrt {c}+\sqrt {-a} \sqrt {d}\right )^2}{4 \sqrt {-a} \sqrt {b} \sqrt {c} \sqrt {d}};2 \tan ^{-1}\left (\frac {\sqrt [4]{d} x}{\sqrt [4]{c}}\right )|\frac {1}{2}\right )}{8 b^2 \sqrt [4]{c} \sqrt [4]{d} (b c+a d) \sqrt {c+d x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.35, size = 249, normalized size = 0.29 \[ \frac {x \left (5 \left (\frac {5 a^2 c^2 F_1\left (\frac {1}{4};\frac {1}{2},1;\frac {5}{4};-\frac {d x^4}{c},-\frac {b x^4}{a}\right )}{d \left (a+b x^4\right ) \left (2 x^4 \left (2 b c F_1\left (\frac {5}{4};\frac {1}{2},2;\frac {9}{4};-\frac {d x^4}{c},-\frac {b x^4}{a}\right )+a d F_1\left (\frac {5}{4};\frac {3}{2},1;\frac {9}{4};-\frac {d x^4}{c},-\frac {b x^4}{a}\right )\right )-5 a c F_1\left (\frac {1}{4};\frac {1}{2},1;\frac {5}{4};-\frac {d x^4}{c},-\frac {b x^4}{a}\right )\right )}+\frac {c}{d}+x^4\right )-\frac {x^4 \sqrt {\frac {d x^4}{c}+1} (3 a d+b c) F_1\left (\frac {5}{4};\frac {1}{2},1;\frac {9}{4};-\frac {d x^4}{c},-\frac {b x^4}{a}\right )}{a d}\right )}{15 b \sqrt {c+d x^4}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^8/((a + b*x^4)*Sqrt[c + d*x^4]),x]

[Out]

(x*(-(((b*c + 3*a*d)*x^4*Sqrt[1 + (d*x^4)/c]*AppellF1[5/4, 1/2, 1, 9/4, -((d*x^4)/c), -((b*x^4)/a)])/(a*d)) +
5*(c/d + x^4 + (5*a^2*c^2*AppellF1[1/4, 1/2, 1, 5/4, -((d*x^4)/c), -((b*x^4)/a)])/(d*(a + b*x^4)*(-5*a*c*Appel
lF1[1/4, 1/2, 1, 5/4, -((d*x^4)/c), -((b*x^4)/a)] + 2*x^4*(2*b*c*AppellF1[5/4, 1/2, 2, 9/4, -((d*x^4)/c), -((b
*x^4)/a)] + a*d*AppellF1[5/4, 3/2, 1, 9/4, -((d*x^4)/c), -((b*x^4)/a)]))))))/(15*b*Sqrt[c + d*x^4])

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{8}}{{\left (b x^{4} + a\right )} \sqrt {d x^{4} + c}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^8/((b*x^4 + a)*sqrt(d*x^4 + c)), x)

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maple [C]  time = 0.32, size = 363, normalized size = 0.42 \[ \frac {a^{2} \left (\frac {2 \sqrt {-\frac {i \sqrt {d}\, x^{2}}{\sqrt {c}}+1}\, \sqrt {\frac {i \sqrt {d}\, x^{2}}{\sqrt {c}}+1}\, \RootOf \left (b \,\textit {\_Z}^{4}+a \right )^{3} b \EllipticPi \left (\sqrt {\frac {i \sqrt {d}}{\sqrt {c}}}\, x , \frac {i \RootOf \left (b \,\textit {\_Z}^{4}+a \right )^{2} b \sqrt {c}}{a \sqrt {d}}, \frac {\sqrt {-\frac {i \sqrt {d}}{\sqrt {c}}}}{\sqrt {\frac {i \sqrt {d}}{\sqrt {c}}}}\right )}{\sqrt {\frac {i \sqrt {d}}{\sqrt {c}}}\, \sqrt {d \,x^{4}+c}\, a}-\frac {\arctanh \left (\frac {2 \RootOf \left (b \,\textit {\_Z}^{4}+a \right )^{2} d \,x^{2}+2 c}{2 \sqrt {\frac {-a d +b c}{b}}\, \sqrt {d \,x^{4}+c}}\right )}{\sqrt {\frac {-a d +b c}{b}}}\right )}{8 b^{3} \RootOf \left (b \,\textit {\_Z}^{4}+a \right )^{3}}+\frac {-\frac {\sqrt {-\frac {i \sqrt {d}\, x^{2}}{\sqrt {c}}+1}\, \sqrt {\frac {i \sqrt {d}\, x^{2}}{\sqrt {c}}+1}\, a \EllipticF \left (\sqrt {\frac {i \sqrt {d}}{\sqrt {c}}}\, x , i\right )}{\sqrt {\frac {i \sqrt {d}}{\sqrt {c}}}\, \sqrt {d \,x^{4}+c}}+\left (-\frac {\sqrt {-\frac {i \sqrt {d}\, x^{2}}{\sqrt {c}}+1}\, \sqrt {\frac {i \sqrt {d}\, x^{2}}{\sqrt {c}}+1}\, c \EllipticF \left (\sqrt {\frac {i \sqrt {d}}{\sqrt {c}}}\, x , i\right )}{3 \sqrt {\frac {i \sqrt {d}}{\sqrt {c}}}\, \sqrt {d \,x^{4}+c}\, d}+\frac {\sqrt {d \,x^{4}+c}\, x}{3 d}\right ) b}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b^2*(b*(1/3/d*x*(d*x^4+c)^(1/2)-1/3*c/d/(I/c^(1/2)*d^(1/2))^(1/2)*(-I/c^(1/2)*d^(1/2)*x^2+1)^(1/2)*(I/c^(1/2
)*d^(1/2)*x^2+1)^(1/2)/(d*x^4+c)^(1/2)*EllipticF((I/c^(1/2)*d^(1/2))^(1/2)*x,I))-a/(I/c^(1/2)*d^(1/2))^(1/2)*(
-I/c^(1/2)*d^(1/2)*x^2+1)^(1/2)*(I/c^(1/2)*d^(1/2)*x^2+1)^(1/2)/(d*x^4+c)^(1/2)*EllipticF((I/c^(1/2)*d^(1/2))^
(1/2)*x,I))+1/8*a^2/b^3*sum(1/_alpha^3*(-1/((-a*d+b*c)/b)^(1/2)*arctanh(1/2*(2*_alpha^2*d*x^2+2*c)/((-a*d+b*c)
/b)^(1/2)/(d*x^4+c)^(1/2))+2/(I/c^(1/2)*d^(1/2))^(1/2)*_alpha^3*b/a*(-I/c^(1/2)*d^(1/2)*x^2+1)^(1/2)*(I/c^(1/2
)*d^(1/2)*x^2+1)^(1/2)/(d*x^4+c)^(1/2)*EllipticPi((I/c^(1/2)*d^(1/2))^(1/2)*x,I*_alpha^2/a*b*c^(1/2)/d^(1/2),(
-I/c^(1/2)*d^(1/2))^(1/2)/(I/c^(1/2)*d^(1/2))^(1/2))),_alpha=RootOf(_Z^4*b+a))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{8}}{{\left (b x^{4} + a\right )} \sqrt {d x^{4} + c}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^8/((b*x^4 + a)*sqrt(d*x^4 + c)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^8}{\left (b\,x^4+a\right )\,\sqrt {d\,x^4+c}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{8}}{\left (a + b x^{4}\right ) \sqrt {c + d x^{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**8/(b*x**4+a)/(d*x**4+c)**(1/2),x)

[Out]

Integral(x**8/((a + b*x**4)*sqrt(c + d*x**4)), x)

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