3.835 \(\int \frac {x^4}{(a+b x^4)^2 \sqrt {c+d x^4}} \, dx\)

Optimal. Leaf size=908 \[ \frac {\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 ) \left (\sqrt {b} \sqrt {c}-\sqrt {-a} \sqrt {d}\right )^2}{32 a b \sqrt [4]{c} \sqrt [4]{d} (b c-a d) \sqrt {d x^4+c}}-\frac {(b c+a d) \tan ^{-1}\left (\frac {\sqrt {b c-a d} x}{\sqrt [4]{-a} \sqrt [4]{b} \sqrt {d x^4+c}}\right )}{16 (-a)^{3/4} b^{3/4} (b c-a d)^{3/2}}+\frac {(b c+a d) \tan ^{-1}\left (\frac {\sqrt {a d-b c} x}{\sqrt [4]{-a} \sqrt [4]{b} \sqrt {d x^4+c}}\right )}{16 (-a)^{3/4} b^{3/4} (a d-b c)^{3/2}}-\frac {d^{3/4} \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 )}{8 b \sqrt [4]{c} (b c-a d) \sqrt {d x^4+c}}+\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 )}{16 b \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 )}{16 a b \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 )}{32 a b \sqrt [4]{c} \sqrt [4]{d} (b c-a d) \sqrt {d x^4+c}}-\frac {x \sqrt {d x^4+c}}{4 (b c-a d) \left (b x^4+a\right )} \]

[Out]

-1/16*(a*d+b*c)*arctan(x*(-a*d+b*c)^(1/2)/(-a)^(1/4)/b^(1/4)/(d*x^4+c)^(1/2))/(-a)^(3/4)/b^(3/4)/(-a*d+b*c)^(3
/2)+1/16*(a*d+b*c)*arctan(x*(a*d-b*c)^(1/2)/(-a)^(1/4)/b^(1/4)/(d*x^4+c)^(1/2))/(-a)^(3/4)/b^(3/4)/(a*d-b*c)^(
3/2)-1/4*x*(d*x^4+c)^(1/2)/(-a*d+b*c)/(b*x^4+a)-1/8*d^(3/4)*(cos(2*arctan(d^(1/4)*x/c^(1/4)))^2)^(1/2)/cos(2*a
rctan(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/c^(1/4)/(-a*d+b*c)/(d*x^4+c)^(1/2)+1/16*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/c^(1/4)/(
-a*d+b*c)/(d*x^4+c)^(1/2)+1/16*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)/a/b/c^(1/4)/(-a*d+b*c)/(d*x^4+c)^(1/2)+1/32*(cos(2*arcta
n(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)/a/b/c^(1/4)/d^(1/4)/(-a*d+b*
c)/(d*x^4+c)^(1/2)+1/32*(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)/a/b/c^(1/4)/d^(1/4)/(-a*d+b*c)/(d*x^4+c)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.91, antiderivative size = 908, 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 = {471, 523, 220, 409, 1217, 1707} \[ \frac {\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 ) \left (\sqrt {b} \sqrt {c}-\sqrt {-a} \sqrt {d}\right )^2}{32 a b \sqrt [4]{c} \sqrt [4]{d} (b c-a d) \sqrt {d x^4+c}}-\frac {(b c+a d) \tan ^{-1}\left (\frac {\sqrt {b c-a d} x}{\sqrt [4]{-a} \sqrt [4]{b} \sqrt {d x^4+c}}\right )}{16 (-a)^{3/4} b^{3/4} (b c-a d)^{3/2}}+\frac {(b c+a d) \tan ^{-1}\left (\frac {\sqrt {a d-b c} x}{\sqrt [4]{-a} \sqrt [4]{b} \sqrt {d x^4+c}}\right )}{16 (-a)^{3/4} b^{3/4} (a d-b c)^{3/2}}-\frac {d^{3/4} \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 )}{8 b \sqrt [4]{c} (b c-a d) \sqrt {d x^4+c}}+\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 )}{16 b \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 )}{16 a b \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 )}{32 a b \sqrt [4]{c} \sqrt [4]{d} (b c-a d) \sqrt {d x^4+c}}-\frac {x \sqrt {d x^4+c}}{4 (b c-a d) \left (b x^4+a\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^4/((a + b*x^4)^2*Sqrt[c + d*x^4]),x]

[Out]

-(x*Sqrt[c + d*x^4])/(4*(b*c - a*d)*(a + b*x^4)) - ((b*c + a*d)*ArcTan[(Sqrt[b*c - a*d]*x)/((-a)^(1/4)*b^(1/4)
*Sqrt[c + d*x^4])])/(16*(-a)^(3/4)*b^(3/4)*(b*c - a*d)^(3/2)) + ((b*c + a*d)*ArcTan[(Sqrt[-(b*c) + a*d]*x)/((-
a)^(1/4)*b^(1/4)*Sqrt[c + d*x^4])])/(16*(-a)^(3/4)*b^(3/4)*(-(b*c) + a*d)^(3/2)) + (((Sqrt[b]*Sqrt[c])/Sqrt[-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])/(16*b*c^(1/4)*(b*c - a*d)*Sqrt[c + d*x^4]) + ((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])/(16*a*b*c^(1/4)*(b*c - a*d)*Sqrt[c + d*x^4]) - (d^(3/4)*(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])/(8*b*c^(1/4)*(b*c - a*d)*Sqrt[c + d
*x^4]) + ((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]*Sqrt[c]*Sqrt[d]), 2*ArcTan[(d^(1
/4)*x)/c^(1/4)], 1/2])/(32*a*b*c^(1/4)*d^(1/4)*(b*c - a*d)*Sqrt[c + d*x^4]) + ((Sqrt[b]*Sqrt[c] - Sqrt[-a]*Sqr
t[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]*Sqrt[c]*Sqrt[d]), 2*ArcTan[(d^(1/4)*x)/c^(1/4)], 1/2])/(32*a*b*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 471

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(e^(n -
1)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(n*(b*c - a*d)*(p + 1)), x] - Dist[e^n/(n*(b*c -
a*d)*(p + 1)), Int[(e*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(m - n + 1) + d*(m + n*(p + q + 1)
+ 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GeQ[n
, m - n + 1] && GtQ[m - n + 1, 0] && 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^4}{\left (a+b x^4\right )^2 \sqrt {c+d x^4}} \, dx &=-\frac {x \sqrt {c+d x^4}}{4 (b c-a d) \left (a+b x^4\right )}+\frac {\int \frac {c-d x^4}{\left (a+b x^4\right ) \sqrt {c+d x^4}} \, dx}{4 (b c-a d)}\\ &=-\frac {x \sqrt {c+d x^4}}{4 (b c-a d) \left (a+b x^4\right )}-\frac {d \int \frac {1}{\sqrt {c+d x^4}} \, dx}{4 b (b c-a d)}+\frac {(b c+a d) \int \frac {1}{\left (a+b x^4\right ) \sqrt {c+d x^4}} \, dx}{4 b (b c-a d)}\\ &=-\frac {x \sqrt {c+d x^4}}{4 (b c-a d) \left (a+b x^4\right )}-\frac {d^{3/4} \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 )}{8 b \sqrt [4]{c} (b c-a d) \sqrt {c+d x^4}}+\frac {(b c+a d) \int \frac {1}{\left (1-\frac {\sqrt {b} x^2}{\sqrt {-a}}\right ) \sqrt {c+d x^4}} \, dx}{8 a b (b c-a d)}+\frac {(b c+a d) \int \frac {1}{\left (1+\frac {\sqrt {b} x^2}{\sqrt {-a}}\right ) \sqrt {c+d x^4}} \, dx}{8 a b (b c-a d)}\\ &=-\frac {x \sqrt {c+d x^4}}{4 (b c-a d) \left (a+b x^4\right )}-\frac {d^{3/4} \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 )}{8 b \sqrt [4]{c} (b c-a d) \sqrt {c+d x^4}}+\frac {\left (\sqrt {c} \left (\sqrt {c}-\frac {\sqrt {-a} \sqrt {d}}{\sqrt {b}}\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}{8 a (b c-a d)}+\frac {\left (\sqrt {c} \left (\sqrt {c}+\frac {\sqrt {-a} \sqrt {d}}{\sqrt {b}}\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}{8 a (b c-a d)}+\frac {\left (\left (\frac {\sqrt {b} \sqrt {c}}{\sqrt {-a}}+\sqrt {d}\right ) \sqrt {d}\right ) \int \frac {1}{\sqrt {c+d x^4}} \, dx}{8 b (b c-a d)}+\frac {\left (\left (\sqrt {-a} \sqrt {b} \sqrt {c}+a \sqrt {d}\right ) \sqrt {d}\right ) \int \frac {1}{\sqrt {c+d x^4}} \, dx}{8 a b (b c-a d)}\\ &=-\frac {x \sqrt {c+d x^4}}{4 (b c-a d) \left (a+b x^4\right )}-\frac {(b c+a d) \tan ^{-1}\left (\frac {\sqrt {b c-a d} x}{\sqrt [4]{-a} \sqrt [4]{b} \sqrt {c+d x^4}}\right )}{16 (-a)^{3/4} b^{3/4} (b c-a d)^{3/2}}+\frac {(b c+a d) \tan ^{-1}\left (\frac {\sqrt {-b c+a d} x}{\sqrt [4]{-a} \sqrt [4]{b} \sqrt {c+d x^4}}\right )}{16 (-a)^{3/4} b^{3/4} (-b c+a d)^{3/2}}+\frac {\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 )}{16 b \sqrt [4]{c} (b c-a d) \sqrt {c+d x^4}}+\frac {\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 )}{16 a b \sqrt [4]{c} (b c-a d) \sqrt {c+d x^4}}-\frac {d^{3/4} \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 )}{8 b \sqrt [4]{c} (b c-a d) \sqrt {c+d x^4}}+\frac {\left (\sqrt {c}+\frac {\sqrt {-a} \sqrt {d}}{\sqrt {b}}\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 )}{32 a \sqrt [4]{c} \sqrt [4]{d} (b c-a d) \sqrt {c+d x^4}}+\frac {\left (\sqrt {c}-\frac {\sqrt {-a} \sqrt {d}}{\sqrt {b}}\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 )}{32 a \sqrt [4]{c} \sqrt [4]{d} (b c-a d) \sqrt {c+d x^4}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.19, size = 238, normalized size = 0.26 \[ \frac {x \left (\frac {5 \left (\frac {5 a c^2 F_1\left (\frac {1}{4};\frac {1}{2},1;\frac {5}{4};-\frac {d x^4}{c},-\frac {b x^4}{a}\right )}{2 x^4 \left (2 b c F_1\left (\frac {5}{4};\frac {1}{2},2;\frac {9}{4};-\frac {d x^4}{c},-\frac {b x^4}{a}\right )+a d F_1\left (\frac {5}{4};\frac {3}{2},1;\frac {9}{4};-\frac {d x^4}{c},-\frac {b x^4}{a}\right )\right )-5 a c F_1\left (\frac {1}{4};\frac {1}{2},1;\frac {5}{4};-\frac {d x^4}{c},-\frac {b x^4}{a}\right )}+c+d x^4\right )}{a+b x^4}+\frac {d x^4 \sqrt {\frac {d x^4}{c}+1} F_1\left (\frac {5}{4};\frac {1}{2},1;\frac {9}{4};-\frac {d x^4}{c},-\frac {b x^4}{a}\right )}{a}\right )}{20 \sqrt {c+d x^4} (a d-b c)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^4/((a + b*x^4)^2*Sqrt[c + d*x^4]),x]

[Out]

(x*((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 + (5*(c + d*x^4 + (5*
a*c^2*AppellF1[1/4, 1/2, 1, 5/4, -((d*x^4)/c), -((b*x^4)/a)])/(-5*a*c*AppellF1[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)]))))/(a + b*x^4)))/(20*(-(b*c) + a*d)*Sqrt[c + d*x^4])

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{{\left (b x^{4} + a\right )}^{2} \sqrt {d x^{4} + c}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^4/((b*x^4 + a)^2*sqrt(d*x^4 + c)), x)

________________________________________________________________________________________

maple [C]  time = 0.26, size = 530, normalized size = 0.58 \[ -\frac {\left (-\frac {\sqrt {d \,x^{4}+c}\, b x}{4 \left (a d -b c \right ) \left (b \,x^{4}+a \right ) a}-\frac {\sqrt {-\frac {i \sqrt {d}\, x^{2}}{\sqrt {c}}+1}\, \sqrt {\frac {i \sqrt {d}\, x^{2}}{\sqrt {c}}+1}\, d \EllipticF \left (\sqrt {\frac {i \sqrt {d}}{\sqrt {c}}}\, x , i\right )}{4 \left (a d -b c \right ) \sqrt {\frac {i \sqrt {d}}{\sqrt {c}}}\, \sqrt {d \,x^{4}+c}\, a}-\frac {\left (-5 a d +3 b c \right ) \left (\frac {2 \sqrt {-\frac {i \sqrt {d}\, x^{2}}{\sqrt {c}}+1}\, \sqrt {\frac {i \sqrt {d}\, x^{2}}{\sqrt {c}}+1}\, \RootOf \left (b \,\textit {\_Z}^{4}+a \right )^{3} b \EllipticPi \left (\sqrt {\frac {i \sqrt {d}}{\sqrt {c}}}\, x , \frac {i \RootOf \left (b \,\textit {\_Z}^{4}+a \right )^{2} b \sqrt {c}}{a \sqrt {d}}, \frac {\sqrt {-\frac {i \sqrt {d}}{\sqrt {c}}}}{\sqrt {\frac {i \sqrt {d}}{\sqrt {c}}}}\right )}{\sqrt {\frac {i \sqrt {d}}{\sqrt {c}}}\, \sqrt {d \,x^{4}+c}\, a}-\frac {\arctanh \left (\frac {2 \RootOf \left (b \,\textit {\_Z}^{4}+a \right )^{2} d \,x^{2}+2 c}{2 \sqrt {\frac {-a d +b c}{b}}\, \sqrt {d \,x^{4}+c}}\right )}{\sqrt {\frac {-a d +b c}{b}}}\right )}{32 a b \left (a d -b c \right ) \RootOf \left (b \,\textit {\_Z}^{4}+a \right )^{3}}\right ) a}{b}+\frac {\frac {2 \sqrt {-\frac {i \sqrt {d}\, x^{2}}{\sqrt {c}}+1}\, \sqrt {\frac {i \sqrt {d}\, x^{2}}{\sqrt {c}}+1}\, \RootOf \left (b \,\textit {\_Z}^{4}+a \right )^{3} b \EllipticPi \left (\sqrt {\frac {i \sqrt {d}}{\sqrt {c}}}\, x , \frac {i \RootOf \left (b \,\textit {\_Z}^{4}+a \right )^{2} b \sqrt {c}}{a \sqrt {d}}, \frac {\sqrt {-\frac {i \sqrt {d}}{\sqrt {c}}}}{\sqrt {\frac {i \sqrt {d}}{\sqrt {c}}}}\right )}{\sqrt {\frac {i \sqrt {d}}{\sqrt {c}}}\, \sqrt {d \,x^{4}+c}\, a}-\frac {\arctanh \left (\frac {2 \RootOf \left (b \,\textit {\_Z}^{4}+a \right )^{2} d \,x^{2}+2 c}{2 \sqrt {\frac {-a d +b c}{b}}\, \sqrt {d \,x^{4}+c}}\right )}{\sqrt {\frac {-a d +b c}{b}}}}{8 b^{2} \RootOf \left (b \,\textit {\_Z}^{4}+a \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/b^2*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))-a/b*(-1/4/(a*d-b*c)*(d*x^4+c)^(1/2)/(b*x^4+a)/a
*b*x-1/4*d/(a*d-b*c)/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)-1/32/b/a*sum((-5*a*d+3*b*c)/(a*d-b*c)/_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)*El
lipticPi((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)))

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{{\left (b x^{4} + a\right )}^{2} \sqrt {d x^{4} + c}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^4/((b*x^4 + a)^2*sqrt(d*x^4 + c)), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^4}{{\left (b\,x^4+a\right )}^2\,\sqrt {d\,x^4+c}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{\left (a + b x^{4}\right )^{2} \sqrt {c + d x^{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**4/((a + b*x**4)**2*sqrt(c + d*x**4)), x)

________________________________________________________________________________________