3.181 \(\int \frac {1}{\sqrt {a+b x^4} (c+d x^4)} \, dx\)

Optimal. Leaf size=742 \[ -\frac {\sqrt [4]{d} \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt [4]{-c} \sqrt [4]{d} \sqrt {a+b x^4}}\right )}{4 (-c)^{3/4} \sqrt {b c-a d}}-\frac {\sqrt [4]{d} \tan ^{-1}\left (\frac {x \sqrt {a d-b c}}{\sqrt [4]{-c} \sqrt [4]{d} \sqrt {a+b x^4}}\right )}{4 (-c)^{3/4} \sqrt {a d-b c}}+\frac {\sqrt [4]{b} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (\frac {\sqrt {a} \sqrt {d}}{\sqrt {-c}}+\sqrt {b}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{4 \sqrt [4]{a} \sqrt {a+b x^4} (a d+b c)}+\frac {\sqrt [4]{b} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (\sqrt {a} \sqrt {-c} \sqrt {d}+\sqrt {b} c\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{4 \sqrt [4]{a} c \sqrt {a+b x^4} (a d+b c)}+\frac {\left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (\sqrt {b} \sqrt {-c}-\sqrt {a} \sqrt {d}\right )^2 \Pi \left (\frac {\left (\sqrt {b} \sqrt {-c}+\sqrt {a} \sqrt {d}\right )^2}{4 \sqrt {a} \sqrt {b} \sqrt {-c} \sqrt {d}};2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{8 \sqrt [4]{a} \sqrt [4]{b} c \sqrt {a+b x^4} (a d+b c)}+\frac {\left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {-c}\right )^2 \Pi \left (-\frac {\left (\sqrt {b} \sqrt {-c}-\sqrt {a} \sqrt {d}\right )^2}{4 \sqrt {a} \sqrt {b} \sqrt {-c} \sqrt {d}};2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{8 \sqrt [4]{a} \sqrt [4]{b} c \sqrt {a+b x^4} (a d+b c)} \]

[Out]

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

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Rubi [A]  time = 0.63, antiderivative size = 742, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {409, 1217, 220, 1707} \[ -\frac {\sqrt [4]{d} \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt [4]{-c} \sqrt [4]{d} \sqrt {a+b x^4}}\right )}{4 (-c)^{3/4} \sqrt {b c-a d}}-\frac {\sqrt [4]{d} \tan ^{-1}\left (\frac {x \sqrt {a d-b c}}{\sqrt [4]{-c} \sqrt [4]{d} \sqrt {a+b x^4}}\right )}{4 (-c)^{3/4} \sqrt {a d-b c}}+\frac {\sqrt [4]{b} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (\frac {\sqrt {a} \sqrt {d}}{\sqrt {-c}}+\sqrt {b}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{4 \sqrt [4]{a} \sqrt {a+b x^4} (a d+b c)}+\frac {\sqrt [4]{b} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (\sqrt {a} \sqrt {-c} \sqrt {d}+\sqrt {b} c\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{4 \sqrt [4]{a} c \sqrt {a+b x^4} (a d+b c)}+\frac {\left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (\sqrt {b} \sqrt {-c}-\sqrt {a} \sqrt {d}\right )^2 \Pi \left (\frac {\left (\sqrt {b} \sqrt {-c}+\sqrt {a} \sqrt {d}\right )^2}{4 \sqrt {a} \sqrt {b} \sqrt {-c} \sqrt {d}};2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{8 \sqrt [4]{a} \sqrt [4]{b} c \sqrt {a+b x^4} (a d+b c)}+\frac {\left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {-c}\right )^2 \Pi \left (-\frac {\left (\sqrt {b} \sqrt {-c}-\sqrt {a} \sqrt {d}\right )^2}{4 \sqrt {a} \sqrt {b} \sqrt {-c} \sqrt {d}};2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{8 \sqrt [4]{a} \sqrt [4]{b} c \sqrt {a+b x^4} (a d+b c)} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[a + b*x^4]*(c + d*x^4)),x]

[Out]

-(d^(1/4)*ArcTan[(Sqrt[b*c - a*d]*x)/((-c)^(1/4)*d^(1/4)*Sqrt[a + b*x^4])])/(4*(-c)^(3/4)*Sqrt[b*c - a*d]) - (
d^(1/4)*ArcTan[(Sqrt[-(b*c) + a*d]*x)/((-c)^(1/4)*d^(1/4)*Sqrt[a + b*x^4])])/(4*(-c)^(3/4)*Sqrt[-(b*c) + a*d])
+ (b^(1/4)*(Sqrt[b] + (Sqrt[a]*Sqrt[d])/Sqrt[-c])*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]
*x^2)^2]*EllipticF[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(4*a^(1/4)*(b*c + a*d)*Sqrt[a + b*x^4]) + (b^(1/4)*(Sq
rt[b]*c + Sqrt[a]*Sqrt[-c]*Sqrt[d])*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*Ellipt
icF[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(4*a^(1/4)*c*(b*c + a*d)*Sqrt[a + b*x^4]) + ((Sqrt[b]*Sqrt[-c] + Sqrt
[a]*Sqrt[d])^2*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*EllipticPi[-(Sqrt[b]*Sqrt[-
c] - Sqrt[a]*Sqrt[d])^2/(4*Sqrt[a]*Sqrt[b]*Sqrt[-c]*Sqrt[d]), 2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(8*a^(1/4)*
b^(1/4)*c*(b*c + a*d)*Sqrt[a + b*x^4]) + ((Sqrt[b]*Sqrt[-c] - Sqrt[a]*Sqrt[d])^2*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[
(a + b*x^4)/(Sqrt[a] + Sqrt[b]*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[(b^(1/4)*x)/a^(1/4)], 1/2])/(8*a^(1/4)*b^(1/4)*c*(b*c + a*d)*Sqrt[a + b*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 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 {1}{\sqrt {a+b x^4} \left (c+d x^4\right )} \, dx &=\frac {\int \frac {1}{\left (1-\frac {\sqrt {d} x^2}{\sqrt {-c}}\right ) \sqrt {a+b x^4}} \, dx}{2 c}+\frac {\int \frac {1}{\left (1+\frac {\sqrt {d} x^2}{\sqrt {-c}}\right ) \sqrt {a+b x^4}} \, dx}{2 c}\\ &=\frac {\left (\sqrt {b} \left (\sqrt {b}+\frac {\sqrt {a} \sqrt {d}}{\sqrt {-c}}\right )\right ) \int \frac {1}{\sqrt {a+b x^4}} \, dx}{2 (b c+a d)}+\frac {\left (\sqrt {b} \left (\sqrt {b} c+\sqrt {a} \sqrt {-c} \sqrt {d}\right )\right ) \int \frac {1}{\sqrt {a+b x^4}} \, dx}{2 c (b c+a d)}-\frac {\left (\sqrt {a} \left (\sqrt {b} \sqrt {-c}-\sqrt {a} \sqrt {d}\right ) \sqrt {d}\right ) \int \frac {1+\frac {\sqrt {b} x^2}{\sqrt {a}}}{\left (1-\frac {\sqrt {d} x^2}{\sqrt {-c}}\right ) \sqrt {a+b x^4}} \, dx}{2 c (b c+a d)}+\frac {\left (\sqrt {a} \left (\sqrt {b} \sqrt {-c}+\sqrt {a} \sqrt {d}\right ) \sqrt {d}\right ) \int \frac {1+\frac {\sqrt {b} x^2}{\sqrt {a}}}{\left (1+\frac {\sqrt {d} x^2}{\sqrt {-c}}\right ) \sqrt {a+b x^4}} \, dx}{2 c (b c+a d)}\\ &=-\frac {\sqrt [4]{d} \tan ^{-1}\left (\frac {\sqrt {b c-a d} x}{\sqrt [4]{-c} \sqrt [4]{d} \sqrt {a+b x^4}}\right )}{4 (-c)^{3/4} \sqrt {b c-a d}}-\frac {\sqrt [4]{d} \tan ^{-1}\left (\frac {\sqrt {-b c+a d} x}{\sqrt [4]{-c} \sqrt [4]{d} \sqrt {a+b x^4}}\right )}{4 (-c)^{3/4} \sqrt {-b c+a d}}+\frac {\sqrt [4]{b} \left (\sqrt {b}+\frac {\sqrt {a} \sqrt {d}}{\sqrt {-c}}\right ) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{4 \sqrt [4]{a} (b c+a d) \sqrt {a+b x^4}}+\frac {\sqrt [4]{b} \left (\sqrt {b} c+\sqrt {a} \sqrt {-c} \sqrt {d}\right ) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{4 \sqrt [4]{a} c (b c+a d) \sqrt {a+b x^4}}+\frac {\left (\sqrt {b} \sqrt {-c}+\sqrt {a} \sqrt {d}\right )^2 \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} 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]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{8 \sqrt [4]{a} \sqrt [4]{b} c (b c+a d) \sqrt {a+b x^4}}+\frac {\left (\sqrt {b} \sqrt {-c}-\sqrt {a} \sqrt {d}\right )^2 \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} 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]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{8 \sqrt [4]{a} \sqrt [4]{b} c (b c+a d) \sqrt {a+b x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 161, normalized size = 0.22 \[ -\frac {5 a c x F_1\left (\frac {1}{4};\frac {1}{2},1;\frac {5}{4};-\frac {b x^4}{a},-\frac {d x^4}{c}\right )}{\sqrt {a+b x^4} \left (c+d x^4\right ) \left (2 x^4 \left (2 a d F_1\left (\frac {5}{4};\frac {1}{2},2;\frac {9}{4};-\frac {b x^4}{a},-\frac {d x^4}{c}\right )+b c F_1\left (\frac {5}{4};\frac {3}{2},1;\frac {9}{4};-\frac {b x^4}{a},-\frac {d x^4}{c}\right )\right )-5 a c F_1\left (\frac {1}{4};\frac {1}{2},1;\frac {5}{4};-\frac {b x^4}{a},-\frac {d x^4}{c}\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(Sqrt[a + b*x^4]*(c + d*x^4)),x]

[Out]

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

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

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {b x^{4} + a} {\left (d x^{4} + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.33, size = 191, normalized size = 0.26 \[ \frac {\frac {2 \sqrt {-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \RootOf \left (d \,\textit {\_Z}^{4}+c \right )^{3} d \EllipticPi \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , \frac {i \RootOf \left (d \,\textit {\_Z}^{4}+c \right )^{2} \sqrt {a}\, d}{\sqrt {b}\, c}, \frac {\sqrt {-\frac {i \sqrt {b}}{\sqrt {a}}}}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}}\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, c}-\frac {\arctanh \left (\frac {2 \RootOf \left (d \,\textit {\_Z}^{4}+c \right )^{2} b \,x^{2}+2 a}{2 \sqrt {\frac {a d -b c}{d}}\, \sqrt {b \,x^{4}+a}}\right )}{\sqrt {\frac {a d -b c}{d}}}}{8 d \RootOf \left (d \,\textit {\_Z}^{4}+c \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {b x^{4} + a} {\left (d x^{4} + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {a + b x^{4}} \left (c + d x^{4}\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(sqrt(a + b*x**4)*(c + d*x**4)), x)

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