3.10.19 \(\int \frac {x}{(a+b x^8)^2 \sqrt {c+d x^8}} \, dx\) [919]

Optimal. Leaf size=999 \[ \frac {b x^2 \sqrt {c+d x^8}}{8 a (b c-a d) \left (a+b x^8\right )}+\frac {\sqrt [4]{b} (3 b c-5 a d) \tan ^{-1}\left (\frac {\sqrt {b c-a d} x^2}{\sqrt [4]{-a} \sqrt [4]{b} \sqrt {c+d x^8}}\right )}{32 (-a)^{7/4} (b c-a d)^{3/2}}-\frac {\sqrt [4]{b} (3 b c-5 a d) \tan ^{-1}\left (\frac {\sqrt {-b c+a d} x^2}{\sqrt [4]{-a} \sqrt [4]{b} \sqrt {c+d x^8}}\right )}{32 (-a)^{7/4} (-b c+a d)^{3/2}}+\frac {d^{3/4} \left (\sqrt {c}+\sqrt {d} x^4\right ) \sqrt {\frac {c+d x^8}{\left (\sqrt {c}+\sqrt {d} x^4\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right )|\frac {1}{2}\right )}{16 a \sqrt [4]{c} (b c-a d) \sqrt {c+d x^8}}+\frac {\left (\frac {\sqrt {b} \sqrt {c}}{\sqrt {-a}}+\sqrt {d}\right ) \sqrt [4]{d} (3 b c-5 a d) \left (\sqrt {c}+\sqrt {d} x^4\right ) \sqrt {\frac {c+d x^8}{\left (\sqrt {c}+\sqrt {d} x^4\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right )|\frac {1}{2}\right )}{32 a \sqrt [4]{c} (b c-a d) (b c+a d) \sqrt {c+d x^8}}+\frac {\left (\sqrt {b} \sqrt {c}-\sqrt {-a} \sqrt {d}\right ) \sqrt [4]{d} (3 b c-5 a d) \left (\sqrt {c}+\sqrt {d} x^4\right ) \sqrt {\frac {c+d x^8}{\left (\sqrt {c}+\sqrt {d} x^4\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right )|\frac {1}{2}\right )}{32 (-a)^{3/2} \sqrt [4]{c} (b c-a d) (b c+a d) \sqrt {c+d x^8}}+\frac {\left (\sqrt {b} \sqrt {c}+\sqrt {-a} \sqrt {d}\right )^2 (3 b c-5 a d) \left (\sqrt {c}+\sqrt {d} x^4\right ) \sqrt {\frac {c+d x^8}{\left (\sqrt {c}+\sqrt {d} x^4\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^2}{\sqrt [4]{c}}\right )|\frac {1}{2}\right )}{64 a^2 \sqrt [4]{c} \sqrt [4]{d} (b c-a d) (b c+a d) \sqrt {c+d x^8}}+\frac {\left (\sqrt {b} \sqrt {c}-\sqrt {-a} \sqrt {d}\right )^2 (3 b c-5 a d) \left (\sqrt {c}+\sqrt {d} x^4\right ) \sqrt {\frac {c+d x^8}{\left (\sqrt {c}+\sqrt {d} x^4\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^2}{\sqrt [4]{c}}\right )|\frac {1}{2}\right )}{64 a^2 \sqrt [4]{c} \sqrt [4]{d} (b c-a d) (b c+a d) \sqrt {c+d x^8}} \]

[Out]

1/32*b^(1/4)*(-5*a*d+3*b*c)*arctan(x^2*(-a*d+b*c)^(1/2)/(-a)^(1/4)/b^(1/4)/(d*x^8+c)^(1/2))/(-a)^(7/4)/(-a*d+b
*c)^(3/2)-1/32*b^(1/4)*(-5*a*d+3*b*c)*arctan(x^2*(a*d-b*c)^(1/2)/(-a)^(1/4)/b^(1/4)/(d*x^8+c)^(1/2))/(-a)^(7/4
)/(a*d-b*c)^(3/2)+1/8*b*x^2*(d*x^8+c)^(1/2)/a/(-a*d+b*c)/(b*x^8+a)+1/16*d^(3/4)*(cos(2*arctan(d^(1/4)*x^2/c^(1
/4)))^2)^(1/2)/cos(2*arctan(d^(1/4)*x^2/c^(1/4)))*EllipticF(sin(2*arctan(d^(1/4)*x^2/c^(1/4))),1/2*2^(1/2))*(c
^(1/2)+x^4*d^(1/2))*((d*x^8+c)/(c^(1/2)+x^4*d^(1/2))^2)^(1/2)/a/c^(1/4)/(-a*d+b*c)/(d*x^8+c)^(1/2)+1/32*d^(1/4
)*(-5*a*d+3*b*c)*(cos(2*arctan(d^(1/4)*x^2/c^(1/4)))^2)^(1/2)/cos(2*arctan(d^(1/4)*x^2/c^(1/4)))*EllipticF(sin
(2*arctan(d^(1/4)*x^2/c^(1/4))),1/2*2^(1/2))*(b^(1/2)*c^(1/2)/(-a)^(1/2)+d^(1/2))*(c^(1/2)+x^4*d^(1/2))*((d*x^
8+c)/(c^(1/2)+x^4*d^(1/2))^2)^(1/2)/a/c^(1/4)/(-a*d+b*c)/(a*d+b*c)/(d*x^8+c)^(1/2)+1/32*d^(1/4)*(-5*a*d+3*b*c)
*(cos(2*arctan(d^(1/4)*x^2/c^(1/4)))^2)^(1/2)/cos(2*arctan(d^(1/4)*x^2/c^(1/4)))*EllipticF(sin(2*arctan(d^(1/4
)*x^2/c^(1/4))),1/2*2^(1/2))*(c^(1/2)+x^4*d^(1/2))*(b^(1/2)*c^(1/2)-(-a)^(1/2)*d^(1/2))*((d*x^8+c)/(c^(1/2)+x^
4*d^(1/2))^2)^(1/2)/(-a)^(3/2)/c^(1/4)/(-a*d+b*c)/(a*d+b*c)/(d*x^8+c)^(1/2)+1/64*(-5*a*d+3*b*c)*(cos(2*arctan(
d^(1/4)*x^2/c^(1/4)))^2)^(1/2)/cos(2*arctan(d^(1/4)*x^2/c^(1/4)))*EllipticPi(sin(2*arctan(d^(1/4)*x^2/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^4*d^(1
/2))*(b^(1/2)*c^(1/2)-(-a)^(1/2)*d^(1/2))^2*((d*x^8+c)/(c^(1/2)+x^4*d^(1/2))^2)^(1/2)/a^2/c^(1/4)/d^(1/4)/(-a^
2*d^2+b^2*c^2)/(d*x^8+c)^(1/2)+1/64*(-5*a*d+3*b*c)*(cos(2*arctan(d^(1/4)*x^2/c^(1/4)))^2)^(1/2)/cos(2*arctan(d
^(1/4)*x^2/c^(1/4)))*EllipticPi(sin(2*arctan(d^(1/4)*x^2/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^4*d^(1/2))*(b^(1/2)*c^(1/2)+(-a)^(1/2)*d^(1/2))^2*
((d*x^8+c)/(c^(1/2)+x^4*d^(1/2))^2)^(1/2)/a^2/c^(1/4)/d^(1/4)/(-a^2*d^2+b^2*c^2)/(d*x^8+c)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 1.02, antiderivative size = 999, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {476, 425, 537, 226, 418, 1231, 1721} \begin {gather*} \frac {(3 b c-5 a d) \left (\sqrt {d} x^4+\sqrt {c}\right ) \sqrt {\frac {d x^8+c}{\left (\sqrt {d} x^4+\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 \text {ArcTan}\left (\frac {\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right )|\frac {1}{2}\right ) \left (\sqrt {b} \sqrt {c}-\sqrt {-a} \sqrt {d}\right )^2}{64 a^2 \sqrt [4]{c} \sqrt [4]{d} (b c-a d) (b c+a d) \sqrt {d x^8+c}}+\frac {\sqrt [4]{d} (3 b c-5 a d) \left (\sqrt {d} x^4+\sqrt {c}\right ) \sqrt {\frac {d x^8+c}{\left (\sqrt {d} x^4+\sqrt {c}\right )^2}} F\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right )|\frac {1}{2}\right ) \left (\sqrt {b} \sqrt {c}-\sqrt {-a} \sqrt {d}\right )}{32 (-a)^{3/2} \sqrt [4]{c} (b c-a d) (b c+a d) \sqrt {d x^8+c}}+\frac {\sqrt [4]{b} (3 b c-5 a d) \text {ArcTan}\left (\frac {\sqrt {b c-a d} x^2}{\sqrt [4]{-a} \sqrt [4]{b} \sqrt {d x^8+c}}\right )}{32 (-a)^{7/4} (b c-a d)^{3/2}}-\frac {\sqrt [4]{b} (3 b c-5 a d) \text {ArcTan}\left (\frac {\sqrt {a d-b c} x^2}{\sqrt [4]{-a} \sqrt [4]{b} \sqrt {d x^8+c}}\right )}{32 (-a)^{7/4} (a d-b c)^{3/2}}+\frac {d^{3/4} \left (\sqrt {d} x^4+\sqrt {c}\right ) \sqrt {\frac {d x^8+c}{\left (\sqrt {d} x^4+\sqrt {c}\right )^2}} F\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right )|\frac {1}{2}\right )}{16 a \sqrt [4]{c} (b c-a d) \sqrt {d x^8+c}}+\frac {\left (\frac {\sqrt {b} \sqrt {c}}{\sqrt {-a}}+\sqrt {d}\right ) \sqrt [4]{d} (3 b c-5 a d) \left (\sqrt {d} x^4+\sqrt {c}\right ) \sqrt {\frac {d x^8+c}{\left (\sqrt {d} x^4+\sqrt {c}\right )^2}} F\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right )|\frac {1}{2}\right )}{32 a \sqrt [4]{c} (b c-a d) (b c+a d) \sqrt {d x^8+c}}+\frac {\left (\sqrt {b} \sqrt {c}+\sqrt {-a} \sqrt {d}\right )^2 (3 b c-5 a d) \left (\sqrt {d} x^4+\sqrt {c}\right ) \sqrt {\frac {d x^8+c}{\left (\sqrt {d} x^4+\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 \text {ArcTan}\left (\frac {\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right )|\frac {1}{2}\right )}{64 a^2 \sqrt [4]{c} \sqrt [4]{d} (b c-a d) (b c+a d) \sqrt {d x^8+c}}+\frac {b x^2 \sqrt {d x^8+c}}{8 a (b c-a d) \left (b x^8+a\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/((a + b*x^8)^2*Sqrt[c + d*x^8]),x]

[Out]

(b*x^2*Sqrt[c + d*x^8])/(8*a*(b*c - a*d)*(a + b*x^8)) + (b^(1/4)*(3*b*c - 5*a*d)*ArcTan[(Sqrt[b*c - a*d]*x^2)/
((-a)^(1/4)*b^(1/4)*Sqrt[c + d*x^8])])/(32*(-a)^(7/4)*(b*c - a*d)^(3/2)) - (b^(1/4)*(3*b*c - 5*a*d)*ArcTan[(Sq
rt[-(b*c) + a*d]*x^2)/((-a)^(1/4)*b^(1/4)*Sqrt[c + d*x^8])])/(32*(-a)^(7/4)*(-(b*c) + a*d)^(3/2)) + (d^(3/4)*(
Sqrt[c] + Sqrt[d]*x^4)*Sqrt[(c + d*x^8)/(Sqrt[c] + Sqrt[d]*x^4)^2]*EllipticF[2*ArcTan[(d^(1/4)*x^2)/c^(1/4)],
1/2])/(16*a*c^(1/4)*(b*c - a*d)*Sqrt[c + d*x^8]) + (((Sqrt[b]*Sqrt[c])/Sqrt[-a] + Sqrt[d])*d^(1/4)*(3*b*c - 5*
a*d)*(Sqrt[c] + Sqrt[d]*x^4)*Sqrt[(c + d*x^8)/(Sqrt[c] + Sqrt[d]*x^4)^2]*EllipticF[2*ArcTan[(d^(1/4)*x^2)/c^(1
/4)], 1/2])/(32*a*c^(1/4)*(b*c - a*d)*(b*c + a*d)*Sqrt[c + d*x^8]) + ((Sqrt[b]*Sqrt[c] - Sqrt[-a]*Sqrt[d])*d^(
1/4)*(3*b*c - 5*a*d)*(Sqrt[c] + Sqrt[d]*x^4)*Sqrt[(c + d*x^8)/(Sqrt[c] + Sqrt[d]*x^4)^2]*EllipticF[2*ArcTan[(d
^(1/4)*x^2)/c^(1/4)], 1/2])/(32*(-a)^(3/2)*c^(1/4)*(b*c - a*d)*(b*c + a*d)*Sqrt[c + d*x^8]) + ((Sqrt[b]*Sqrt[c
] + Sqrt[-a]*Sqrt[d])^2*(3*b*c - 5*a*d)*(Sqrt[c] + Sqrt[d]*x^4)*Sqrt[(c + d*x^8)/(Sqrt[c] + Sqrt[d]*x^4)^2]*El
lipticPi[-1/4*(Sqrt[b]*Sqrt[c] - Sqrt[-a]*Sqrt[d])^2/(Sqrt[-a]*Sqrt[b]*Sqrt[c]*Sqrt[d]), 2*ArcTan[(d^(1/4)*x^2
)/c^(1/4)], 1/2])/(64*a^2*c^(1/4)*d^(1/4)*(b*c - a*d)*(b*c + a*d)*Sqrt[c + d*x^8]) + ((Sqrt[b]*Sqrt[c] - Sqrt[
-a]*Sqrt[d])^2*(3*b*c - 5*a*d)*(Sqrt[c] + Sqrt[d]*x^4)*Sqrt[(c + d*x^8)/(Sqrt[c] + Sqrt[d]*x^4)^2]*EllipticPi[
(Sqrt[b]*Sqrt[c] + Sqrt[-a]*Sqrt[d])^2/(4*Sqrt[-a]*Sqrt[b]*Sqrt[c]*Sqrt[d]), 2*ArcTan[(d^(1/4)*x^2)/c^(1/4)],
1/2])/(64*a^2*c^(1/4)*d^(1/4)*(b*c - a*d)*(b*c + a*d)*Sqrt[c + d*x^8])

Rule 226

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)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 418

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 425

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-b)*x*(a + b*x^n)^(p + 1)*
((c + d*x^n)^(q + 1)/(a*n*(p + 1)*(b*c - a*d))), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1
)*(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c,
d, n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomi
alQ[a, b, c, d, n, p, q, x]

Rule 476

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

Rule 537

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 1231

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 1721

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))]/(4*d*e*A*q*Sqrt[a + c*x^4]))*El
lipticPi[Cancel[-(B*d - A*e)^2/(4*d*e*A*B)], 2*ArcTan[q*x], 1/2], 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}{\left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {1}{\left (a+b x^4\right )^2 \sqrt {c+d x^4}} \, dx,x,x^2\right )\\ &=\frac {b x^2 \sqrt {c+d x^8}}{8 a (b c-a d) \left (a+b x^8\right )}-\frac {\text {Subst}\left (\int \frac {-3 b c+4 a d-b d x^4}{\left (a+b x^4\right ) \sqrt {c+d x^4}} \, dx,x,x^2\right )}{8 a (b c-a d)}\\ &=\frac {b x^2 \sqrt {c+d x^8}}{8 a (b c-a d) \left (a+b x^8\right )}+\frac {d \text {Subst}\left (\int \frac {1}{\sqrt {c+d x^4}} \, dx,x,x^2\right )}{8 a (b c-a d)}+\frac {(3 b c-5 a d) \text {Subst}\left (\int \frac {1}{\left (a+b x^4\right ) \sqrt {c+d x^4}} \, dx,x,x^2\right )}{8 a (b c-a d)}\\ &=\frac {b x^2 \sqrt {c+d x^8}}{8 a (b c-a d) \left (a+b x^8\right )}+\frac {d^{3/4} \left (\sqrt {c}+\sqrt {d} x^4\right ) \sqrt {\frac {c+d x^8}{\left (\sqrt {c}+\sqrt {d} x^4\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right )|\frac {1}{2}\right )}{16 a \sqrt [4]{c} (b c-a d) \sqrt {c+d x^8}}+\frac {(3 b c-5 a d) \text {Subst}\left (\int \frac {1}{\left (1-\frac {\sqrt {b} x^2}{\sqrt {-a}}\right ) \sqrt {c+d x^4}} \, dx,x,x^2\right )}{16 a^2 (b c-a d)}+\frac {(3 b c-5 a d) \text {Subst}\left (\int \frac {1}{\left (1+\frac {\sqrt {b} x^2}{\sqrt {-a}}\right ) \sqrt {c+d x^4}} \, dx,x,x^2\right )}{16 a^2 (b c-a d)}\\ &=\frac {b x^2 \sqrt {c+d x^8}}{8 a (b c-a d) \left (a+b x^8\right )}+\frac {d^{3/4} \left (\sqrt {c}+\sqrt {d} x^4\right ) \sqrt {\frac {c+d x^8}{\left (\sqrt {c}+\sqrt {d} x^4\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right )|\frac {1}{2}\right )}{16 a \sqrt [4]{c} (b c-a d) \sqrt {c+d x^8}}+\frac {\left (\sqrt {b} \sqrt {c} \left (\sqrt {b} \sqrt {c}-\sqrt {-a} \sqrt {d}\right ) (3 b c-5 a d)\right ) \text {Subst}\left (\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,x,x^2\right )}{16 a^2 (b c-a d) (b c+a d)}+\frac {\left (\sqrt {b} \sqrt {c} \left (\sqrt {b} \sqrt {c}+\sqrt {-a} \sqrt {d}\right ) (3 b c-5 a d)\right ) \text {Subst}\left (\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,x,x^2\right )}{16 a^2 (b c-a d) (b c+a d)}+\frac {\left (\left (\frac {\sqrt {b} \sqrt {c}}{\sqrt {-a}}+\sqrt {d}\right ) \sqrt {d} (3 b c-5 a d)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {c+d x^4}} \, dx,x,x^2\right )}{16 a (b c-a d) (b c+a d)}+\frac {\left (\left (\sqrt {b} \sqrt {c}-\sqrt {-a} \sqrt {d}\right ) \sqrt {d} (3 b c-5 a d)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {c+d x^4}} \, dx,x,x^2\right )}{16 (-a)^{3/2} (b c-a d) (b c+a d)}\\ &=\frac {b x^2 \sqrt {c+d x^8}}{8 a (b c-a d) \left (a+b x^8\right )}+\frac {\sqrt [4]{b} (3 b c-5 a d) \tan ^{-1}\left (\frac {\sqrt {b c-a d} x^2}{\sqrt [4]{-a} \sqrt [4]{b} \sqrt {c+d x^8}}\right )}{32 (-a)^{7/4} (b c-a d)^{3/2}}-\frac {\sqrt [4]{b} (3 b c-5 a d) \tan ^{-1}\left (\frac {\sqrt {-b c+a d} x^2}{\sqrt [4]{-a} \sqrt [4]{b} \sqrt {c+d x^8}}\right )}{32 (-a)^{7/4} (-b c+a d)^{3/2}}+\frac {d^{3/4} \left (\sqrt {c}+\sqrt {d} x^4\right ) \sqrt {\frac {c+d x^8}{\left (\sqrt {c}+\sqrt {d} x^4\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right )|\frac {1}{2}\right )}{16 a \sqrt [4]{c} (b c-a d) \sqrt {c+d x^8}}+\frac {\left (\frac {\sqrt {b} \sqrt {c}}{\sqrt {-a}}+\sqrt {d}\right ) \sqrt [4]{d} (3 b c-5 a d) \left (\sqrt {c}+\sqrt {d} x^4\right ) \sqrt {\frac {c+d x^8}{\left (\sqrt {c}+\sqrt {d} x^4\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right )|\frac {1}{2}\right )}{32 a \sqrt [4]{c} (b c-a d) (b c+a d) \sqrt {c+d x^8}}+\frac {\left (\sqrt {b} \sqrt {c}-\sqrt {-a} \sqrt {d}\right ) \sqrt [4]{d} (3 b c-5 a d) \left (\sqrt {c}+\sqrt {d} x^4\right ) \sqrt {\frac {c+d x^8}{\left (\sqrt {c}+\sqrt {d} x^4\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} x^2}{\sqrt [4]{c}}\right )|\frac {1}{2}\right )}{32 (-a)^{3/2} \sqrt [4]{c} (b c-a d) (b c+a d) \sqrt {c+d x^8}}+\frac {\left (\sqrt {b} \sqrt {c}+\sqrt {-a} \sqrt {d}\right )^2 (3 b c-5 a d) \left (\sqrt {c}+\sqrt {d} x^4\right ) \sqrt {\frac {c+d x^8}{\left (\sqrt {c}+\sqrt {d} x^4\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^2}{\sqrt [4]{c}}\right )|\frac {1}{2}\right )}{64 a^2 \sqrt [4]{c} \sqrt [4]{d} (b c-a d) (b c+a d) \sqrt {c+d x^8}}+\frac {\left (\sqrt {b} \sqrt {c}-\sqrt {-a} \sqrt {d}\right )^2 (3 b c-5 a d) \left (\sqrt {c}+\sqrt {d} x^4\right ) \sqrt {\frac {c+d x^8}{\left (\sqrt {c}+\sqrt {d} x^4\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^2}{\sqrt [4]{c}}\right )|\frac {1}{2}\right )}{64 a^2 \sqrt [4]{c} \sqrt [4]{d} (b c-a d) (b c+a d) \sqrt {c+d x^8}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
time = 10.15, size = 169, normalized size = 0.17 \begin {gather*} \frac {x^2 \left (5 a b \left (c+d x^8\right )+5 (3 b c-4 a d) \left (a+b x^8\right ) \sqrt {1+\frac {d x^8}{c}} F_1\left (\frac {1}{4};\frac {1}{2},1;\frac {5}{4};-\frac {d x^8}{c},-\frac {b x^8}{a}\right )+b d x^8 \left (a+b x^8\right ) \sqrt {1+\frac {d x^8}{c}} F_1\left (\frac {5}{4};\frac {1}{2},1;\frac {9}{4};-\frac {d x^8}{c},-\frac {b x^8}{a}\right )\right )}{40 a^2 (b c-a d) \left (a+b x^8\right ) \sqrt {c+d x^8}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x/((a + b*x^8)^2*Sqrt[c + d*x^8]),x]

[Out]

(x^2*(5*a*b*(c + d*x^8) + 5*(3*b*c - 4*a*d)*(a + b*x^8)*Sqrt[1 + (d*x^8)/c]*AppellF1[1/4, 1/2, 1, 5/4, -((d*x^
8)/c), -((b*x^8)/a)] + b*d*x^8*(a + b*x^8)*Sqrt[1 + (d*x^8)/c]*AppellF1[5/4, 1/2, 1, 9/4, -((d*x^8)/c), -((b*x
^8)/a)]))/(40*a^2*(b*c - a*d)*(a + b*x^8)*Sqrt[c + d*x^8])

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Maple [F]
time = 0.07, size = 0, normalized size = 0.00 \[\int \frac {x}{\left (b \,x^{8}+a \right )^{2} \sqrt {d \,x^{8}+c}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x/((a + b*x**8)**2*sqrt(c + d*x**8)), x)

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

[Out]

integrate(x/((b*x^8 + a)^2*sqrt(d*x^8 + c)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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