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

Optimal. Leaf size=703 \[ -\frac {(b c-a d) \tan ^{-1}\left (\frac {x \sqrt {\frac {\sqrt {-a} \left (\frac {b c}{a}-d\right )}{\sqrt {b}}}}{\sqrt {c+d x^4}}\right )}{4 a^2 \sqrt {-\frac {b c-a d}{\sqrt {-a} \sqrt {b}}}}-\frac {(b c-a d) \tan ^{-1}\left (\frac {x \sqrt {\frac {b c-a d}{\sqrt {-a} \sqrt {b}}}}{\sqrt {c+d x^4}}\right )}{4 a^2 \sqrt {\frac {b c-a d}{\sqrt {-a} \sqrt {b}}}}-\frac {\left (\sqrt {c}+\sqrt {d} x^2\right ) \sqrt {\frac {c+d x^4}{\left (\sqrt {c}+\sqrt {d} x^2\right )^2}} \left (\sqrt {-a} \sqrt {d}+\sqrt {b} \sqrt {c}\right ) (b c-a d) \Pi \left (-\frac {\left (\sqrt {b} \sqrt {c}-\sqrt {-a} \sqrt {d}\right )^2}{4 \sqrt {-a} \sqrt {b} \sqrt {c} \sqrt {d}};2 \tan ^{-1}\left (\frac {\sqrt [4]{d} x}{\sqrt [4]{c}}\right )|\frac {1}{2}\right )}{8 a^2 \sqrt [4]{c} \sqrt [4]{d} \sqrt {c+d x^4} \left (\sqrt {b} \sqrt {c}-\sqrt {-a} \sqrt {d}\right )}-\frac {\left (\sqrt {c}+\sqrt {d} x^2\right ) \sqrt {\frac {c+d x^4}{\left (\sqrt {c}+\sqrt {d} x^2\right )^2}} \left (\sqrt {b} \sqrt {c}-\sqrt {-a} \sqrt {d}\right ) (b c-a d) \Pi \left (\frac {\left (\sqrt {b} \sqrt {c}+\sqrt {-a} \sqrt {d}\right )^2}{4 \sqrt {-a} \sqrt {b} \sqrt {c} \sqrt {d}};2 \tan ^{-1}\left (\frac {\sqrt [4]{d} x}{\sqrt [4]{c}}\right )|\frac {1}{2}\right )}{8 a^2 \sqrt [4]{c} \sqrt [4]{d} \sqrt {c+d x^4} \left (\sqrt {-a} \sqrt {d}+\sqrt {b} \sqrt {c}\right )}-\frac {d^{3/4} \left (\sqrt {c}+\sqrt {d} x^2\right ) \sqrt {\frac {c+d x^4}{\left (\sqrt {c}+\sqrt {d} x^2\right )^2}} (2 b c-a d) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} x}{\sqrt [4]{c}}\right )|\frac {1}{2}\right )}{3 a \sqrt [4]{c} \sqrt {c+d x^4} (a d+b c)}-\frac {\sqrt {c+d x^4}}{3 a x^3} \]

[Out]

-1/3*(d*x^4+c)^(1/2)/a/x^3-1/4*(-a*d+b*c)*arctan(x*((b*c/a-d)*(-a)^(1/2)/b^(1/2))^(1/2)/(d*x^4+c)^(1/2))/a^2/(
(a*d-b*c)/(-a)^(1/2)/b^(1/2))^(1/2)-1/4*(-a*d+b*c)*arctan(x*((-a*d+b*c)/(-a)^(1/2)/b^(1/2))^(1/2)/(d*x^4+c)^(1
/2))/a^2/((-a*d+b*c)/(-a)^(1/2)/b^(1/2))^(1/2)-1/3*d^(3/4)*(-a*d+2*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)/a/c^(1/4)/(a*d+b*c)/(d*x^4+c)^(1/2)-1/8*(-a*d+b*c)*(cos(2*arct
an(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))*((d*x^4+c)/(c^(1/2)+x^2*d^(1/2))^2)^(1/2)/a^2/c^(1/4)/d^(1/4)/(b^(1/2)*
c^(1/2)+(-a)^(1/2)*d^(1/2))/(d*x^4+c)^(1/2)-1/8*(-a*d+b*c)*(cos(2*arctan(d^(1/4)*x/c^(1/4)))^2)^(1/2)/cos(2*ar
ctan(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))*
((d*x^4+c)/(c^(1/2)+x^2*d^(1/2))^2)^(1/2)/a^2/c^(1/4)/d^(1/4)/(b^(1/2)*c^(1/2)-(-a)^(1/2)*d^(1/2))/(d*x^4+c)^(
1/2)

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Rubi [A]  time = 1.19, antiderivative size = 893, normalized size of antiderivative = 1.27, 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 = {475, 523, 220, 409, 1217, 1707} \[ -\frac {(b c-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}} \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}{8 a^2 \sqrt [4]{c} \sqrt [4]{d} (b c+a d) \sqrt {d x^4+c}}-\frac {\sqrt [4]{d} (b c-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 ) \left (\sqrt {b} \sqrt {c}-\sqrt {-a} \sqrt {d}\right )}{4 (-a)^{3/2} \sqrt [4]{c} (b c+a d) \sqrt {d x^4+c}}-\frac {\sqrt [4]{b} \sqrt {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 )}{4 (-a)^{7/4}}+\frac {\sqrt [4]{b} \sqrt {a d-b c} \tan ^{-1}\left (\frac {\sqrt {a d-b c} x}{\sqrt [4]{-a} \sqrt [4]{b} \sqrt {d x^4+c}}\right )}{4 (-a)^{7/4}}-\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 )}{6 a \sqrt [4]{c} \sqrt {d x^4+c}}-\frac {\left (\frac {\sqrt {b} \sqrt {c}}{\sqrt {-a}}+\sqrt {d}\right ) \sqrt [4]{d} (b c-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 )}{4 a \sqrt [4]{c} (b c+a d) \sqrt {d x^4+c}}-\frac {\left (\sqrt {b} \sqrt {c}+\sqrt {-a} \sqrt {d}\right )^2 (b c-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}} \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 a^2 \sqrt [4]{c} \sqrt [4]{d} (b c+a d) \sqrt {d x^4+c}}-\frac {\sqrt {d x^4+c}}{3 a x^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-Sqrt[c + d*x^4]/(3*a*x^3) - (b^(1/4)*Sqrt[b*c - a*d]*ArcTan[(Sqrt[b*c - a*d]*x)/((-a)^(1/4)*b^(1/4)*Sqrt[c +
d*x^4])])/(4*(-a)^(7/4)) + (b^(1/4)*Sqrt[-(b*c) + a*d]*ArcTan[(Sqrt[-(b*c) + a*d]*x)/((-a)^(1/4)*b^(1/4)*Sqrt[
c + d*x^4])])/(4*(-a)^(7/4)) - (d^(3/4)*(Sqrt[c] + Sqrt[d]*x^2)*Sqrt[(c + d*x^4)/(Sqrt[c] + Sqrt[d]*x^2)^2]*El
lipticF[2*ArcTan[(d^(1/4)*x)/c^(1/4)], 1/2])/(6*a*c^(1/4)*Sqrt[c + d*x^4]) - (((Sqrt[b]*Sqrt[c])/Sqrt[-a] + Sq
rt[d])*d^(1/4)*(b*c - a*d)*(Sqrt[c] + Sqrt[d]*x^2)*Sqrt[(c + d*x^4)/(Sqrt[c] + Sqrt[d]*x^2)^2]*EllipticF[2*Arc
Tan[(d^(1/4)*x)/c^(1/4)], 1/2])/(4*a*c^(1/4)*(b*c + a*d)*Sqrt[c + d*x^4]) - ((Sqrt[b]*Sqrt[c] - Sqrt[-a]*Sqrt[
d])*d^(1/4)*(b*c - 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])/(4*(-a)^(3/2)*c^(1/4)*(b*c + a*d)*Sqrt[c + d*x^4]) - ((Sqrt[b]*Sqrt[c] + Sqrt[-a]
*Sqrt[d])^2*(b*c - a*d)*(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])/(8
*a^2*c^(1/4)*d^(1/4)*(b*c + a*d)*Sqrt[c + d*x^4]) - ((Sqrt[b]*Sqrt[c] - Sqrt[-a]*Sqrt[d])^2*(b*c - a*d)*(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])/(8*a^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 475

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[((e*x)^(m
+ 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q)/(a*e*(m + 1)), x] - Dist[1/(a*e^n*(m + 1)), Int[(e*x)^(m + n)*(a + b*
x^n)^p*(c + d*x^n)^(q - 1)*Simp[c*b*(m + 1) + n*(b*c*(p + 1) + a*d*q) + d*(b*(m + 1) + b*n*(p + q + 1))*x^n, x
], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[0, q, 1] && LtQ[m, -1] &&
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 {\sqrt {c+d x^4}}{x^4 \left (a+b x^4\right )} \, dx &=-\frac {\sqrt {c+d x^4}}{3 a x^3}+\frac {\int \frac {-3 b c+2 a d-b d x^4}{\left (a+b x^4\right ) \sqrt {c+d x^4}} \, dx}{3 a}\\ &=-\frac {\sqrt {c+d x^4}}{3 a x^3}-\frac {d \int \frac {1}{\sqrt {c+d x^4}} \, dx}{3 a}-\frac {(b c-a d) \int \frac {1}{\left (a+b x^4\right ) \sqrt {c+d x^4}} \, dx}{a}\\ &=-\frac {\sqrt {c+d x^4}}{3 a x^3}-\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 )}{6 a \sqrt [4]{c} \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}{2 a^2}-\frac {(b c-a d) \int \frac {1}{\left (1+\frac {\sqrt {b} x^2}{\sqrt {-a}}\right ) \sqrt {c+d x^4}} \, dx}{2 a^2}\\ &=-\frac {\sqrt {c+d x^4}}{3 a x^3}-\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 )}{6 a \sqrt [4]{c} \sqrt {c+d x^4}}-\frac {\left (\sqrt {b} \sqrt {c} \left (\sqrt {b} \sqrt {c}-\sqrt {-a} \sqrt {d}\right ) (b c-a d)\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 a^2 (b c+a d)}-\frac {\left (\sqrt {b} \sqrt {c} \left (\sqrt {b} \sqrt {c}+\sqrt {-a} \sqrt {d}\right ) (b c-a d)\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 a^2 (b c+a d)}-\frac {\left (\left (\frac {\sqrt {b} \sqrt {c}}{\sqrt {-a}}+\sqrt {d}\right ) \sqrt {d} (b c-a d)\right ) \int \frac {1}{\sqrt {c+d x^4}} \, dx}{2 a (b c+a d)}-\frac {\left (\left (\sqrt {b} \sqrt {c}-\sqrt {-a} \sqrt {d}\right ) \sqrt {d} (b c-a d)\right ) \int \frac {1}{\sqrt {c+d x^4}} \, dx}{2 (-a)^{3/2} (b c+a d)}\\ &=-\frac {\sqrt {c+d x^4}}{3 a x^3}-\frac {\sqrt [4]{b} \sqrt {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 )}{4 (-a)^{7/4}}+\frac {\sqrt [4]{b} \sqrt {-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 )}{4 (-a)^{7/4}}-\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 )}{6 a \sqrt [4]{c} \sqrt {c+d x^4}}-\frac {\left (\frac {\sqrt {b} \sqrt {c}}{\sqrt {-a}}+\sqrt {d}\right ) \sqrt [4]{d} (b c-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 )}{4 a \sqrt [4]{c} (b c+a d) \sqrt {c+d x^4}}-\frac {\left (\sqrt {b} \sqrt {c}-\sqrt {-a} \sqrt {d}\right ) \sqrt [4]{d} (b c-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 )}{4 (-a)^{3/2} \sqrt [4]{c} (b c+a d) \sqrt {c+d x^4}}-\frac {\left (\sqrt {b} \sqrt {c}+\sqrt {-a} \sqrt {d}\right )^2 (b c-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}} \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 a^2 \sqrt [4]{c} \sqrt [4]{d} (b c+a d) \sqrt {c+d x^4}}-\frac {\left (\sqrt {b} \sqrt {c}-\sqrt {-a} \sqrt {d}\right )^2 (b c-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}} \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 a^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 = 0.30, size = 333, normalized size = 0.47 \[ \frac {\frac {a \left (25 a c \left (a c-a d x^4+4 b c x^4+b d x^8\right ) F_1\left (\frac {1}{4};\frac {1}{2},1;\frac {5}{4};-\frac {d x^4}{c},-\frac {b x^4}{a}\right )-10 x^4 \left (a+b x^4\right ) \left (c+d x^4\right ) \left (2 b c F_1\left (\frac {5}{4};\frac {1}{2},2;\frac {9}{4};-\frac {d x^4}{c},-\frac {b x^4}{a}\right )+a d F_1\left (\frac {5}{4};\frac {3}{2},1;\frac {9}{4};-\frac {d x^4}{c},-\frac {b x^4}{a}\right )\right )\right )}{\left (a+b x^4\right ) \left (2 x^4 \left (2 b c F_1\left (\frac {5}{4};\frac {1}{2},2;\frac {9}{4};-\frac {d x^4}{c},-\frac {b x^4}{a}\right )+a d F_1\left (\frac {5}{4};\frac {3}{2},1;\frac {9}{4};-\frac {d x^4}{c},-\frac {b x^4}{a}\right )\right )-5 a c F_1\left (\frac {1}{4};\frac {1}{2},1;\frac {5}{4};-\frac {d x^4}{c},-\frac {b x^4}{a}\right )\right )}-b d x^8 \sqrt {\frac {d x^4}{c}+1} F_1\left (\frac {5}{4};\frac {1}{2},1;\frac {9}{4};-\frac {d x^4}{c},-\frac {b x^4}{a}\right )}{15 a^2 x^3 \sqrt {c+d x^4}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-(b*d*x^8*Sqrt[1 + (d*x^4)/c]*AppellF1[5/4, 1/2, 1, 9/4, -((d*x^4)/c), -((b*x^4)/a)]) + (a*(25*a*c*(a*c + 4*b
*c*x^4 - a*d*x^4 + b*d*x^8)*AppellF1[1/4, 1/2, 1, 5/4, -((d*x^4)/c), -((b*x^4)/a)] - 10*x^4*(a + b*x^4)*(c + d
*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)*(-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)]))))/(15*a^2*x^3*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((d*x^4+c)^(1/2)/x^4/(b*x^4+a),x, algorithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.30, size = 370, normalized size = 0.53 \[ -\frac {\left (\frac {\sqrt {-\frac {i \sqrt {d}\, x^{2}}{\sqrt {c}}+1}\, \sqrt {\frac {i \sqrt {d}\, x^{2}}{\sqrt {c}}+1}\, d \EllipticF \left (\sqrt {\frac {i \sqrt {d}}{\sqrt {c}}}\, x , i\right )}{\sqrt {\frac {i \sqrt {d}}{\sqrt {c}}}\, \sqrt {d \,x^{4}+c}\, b}-\frac {\left (a d -b c \right ) \left (\frac {2 \sqrt {-\frac {i \sqrt {d}\, x^{2}}{\sqrt {c}}+1}\, \sqrt {\frac {i \sqrt {d}\, x^{2}}{\sqrt {c}}+1}\, \RootOf \left (b \,\textit {\_Z}^{4}+a \right )^{3} b \EllipticPi \left (\sqrt {\frac {i \sqrt {d}}{\sqrt {c}}}\, x , \frac {i \RootOf \left (b \,\textit {\_Z}^{4}+a \right )^{2} b \sqrt {c}}{a \sqrt {d}}, \frac {\sqrt {-\frac {i \sqrt {d}}{\sqrt {c}}}}{\sqrt {\frac {i \sqrt {d}}{\sqrt {c}}}}\right )}{\sqrt {\frac {i \sqrt {d}}{\sqrt {c}}}\, \sqrt {d \,x^{4}+c}\, a}-\frac {\arctanh \left (\frac {2 \RootOf \left (b \,\textit {\_Z}^{4}+a \right )^{2} d \,x^{2}+2 c}{2 \sqrt {\frac {-a d +b c}{b}}\, \sqrt {d \,x^{4}+c}}\right )}{\sqrt {\frac {-a d +b c}{b}}}\right )}{8 b^{2} \RootOf \left (b \,\textit {\_Z}^{4}+a \right )^{3}}\right ) b}{a}+\frac {\frac {2 \sqrt {-\frac {i \sqrt {d}\, x^{2}}{\sqrt {c}}+1}\, \sqrt {\frac {i \sqrt {d}\, x^{2}}{\sqrt {c}}+1}\, d \EllipticF \left (\sqrt {\frac {i \sqrt {d}}{\sqrt {c}}}\, x , i\right )}{3 \sqrt {\frac {i \sqrt {d}}{\sqrt {c}}}\, \sqrt {d \,x^{4}+c}}-\frac {\sqrt {d \,x^{4}+c}}{3 x^{3}}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a*(-1/3*(d*x^4+c)^(1/2)/x^3+2/3*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))-1/a*b*(d/b/(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/b^2*sum((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)*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 \[ \int \frac {\sqrt {d x^{4} + c}}{{\left (b x^{4} + a\right )} x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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