3.27.18 \(\int \frac {\sqrt {-b^4+a^4 x^4} (b^4+a^4 x^4)}{b^8-c x^4+a^8 x^8} \, dx\) [2618]

Optimal. Leaf size=229 \[ -\frac {\text {ArcTan}\left (\frac {\frac {b^4}{\sqrt {2} \sqrt [4]{2 a^4 b^4-c}}+\frac {\sqrt [4]{2 a^4 b^4-c} x^2}{\sqrt {2}}-\frac {a^4 x^4}{\sqrt {2} \sqrt [4]{2 a^4 b^4-c}}}{x \sqrt {-b^4+a^4 x^4}}\right )}{2 \sqrt {2} \sqrt [4]{2 a^4 b^4-c}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{2 a^4 b^4-c} x \sqrt {-b^4+a^4 x^4}}{-b^4+\sqrt {2 a^4 b^4-c} x^2+a^4 x^4}\right )}{2 \sqrt {2} \sqrt [4]{2 a^4 b^4-c}} \]

[Out]

-1/4*arctan((1/2*b^4*2^(1/2)/(2*a^4*b^4-c)^(1/4)+1/2*(2*a^4*b^4-c)^(1/4)*x^2*2^(1/2)-1/2*a^4*x^4*2^(1/2)/(2*a^
4*b^4-c)^(1/4))/x/(a^4*x^4-b^4)^(1/2))*2^(1/2)/(2*a^4*b^4-c)^(1/4)-1/4*arctanh(2^(1/2)*(2*a^4*b^4-c)^(1/4)*x*(
a^4*x^4-b^4)^(1/2)/(-b^4+(2*a^4*b^4-c)^(1/2)*x^2+a^4*x^4))*2^(1/2)/(2*a^4*b^4-c)^(1/4)

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Rubi [C] Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
time = 0.96, antiderivative size = 508, normalized size of antiderivative = 2.22, number of steps used = 18, number of rules used = 7, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.146, Rules used = {6860, 415, 230, 227, 418, 1233, 1232} \begin {gather*} \frac {b \left (1-\frac {2 a^4 b^4+c}{\sqrt {c^2-4 a^8 b^8}}\right ) \sqrt {1-\frac {a^4 x^4}{b^4}} F\left (\left .\text {ArcSin}\left (\frac {a x}{b}\right )\right |-1\right )}{2 a \sqrt {a^4 x^4-b^4}}+\frac {b \left (\frac {2 a^4 b^4+c}{\sqrt {c^2-4 a^8 b^8}}+1\right ) \sqrt {1-\frac {a^4 x^4}{b^4}} F\left (\left .\text {ArcSin}\left (\frac {a x}{b}\right )\right |-1\right )}{2 a \sqrt {a^4 x^4-b^4}}-\frac {b \sqrt {1-\frac {a^4 x^4}{b^4}} \Pi \left (-\frac {\sqrt {2} a^2 b^2}{\sqrt {c-\sqrt {c^2-4 a^8 b^8}}};\left .\text {ArcSin}\left (\frac {a x}{b}\right )\right |-1\right )}{2 a \sqrt {a^4 x^4-b^4}}-\frac {b \sqrt {1-\frac {a^4 x^4}{b^4}} \Pi \left (\frac {\sqrt {2} a^2 b^2}{\sqrt {c-\sqrt {c^2-4 a^8 b^8}}};\left .\text {ArcSin}\left (\frac {a x}{b}\right )\right |-1\right )}{2 a \sqrt {a^4 x^4-b^4}}-\frac {b \sqrt {1-\frac {a^4 x^4}{b^4}} \Pi \left (-\frac {\sqrt {2} a^2 b^2}{\sqrt {c+\sqrt {c^2-4 a^8 b^8}}};\left .\text {ArcSin}\left (\frac {a x}{b}\right )\right |-1\right )}{2 a \sqrt {a^4 x^4-b^4}}-\frac {b \sqrt {1-\frac {a^4 x^4}{b^4}} \Pi \left (\frac {\sqrt {2} a^2 b^2}{\sqrt {c+\sqrt {c^2-4 a^8 b^8}}};\left .\text {ArcSin}\left (\frac {a x}{b}\right )\right |-1\right )}{2 a \sqrt {a^4 x^4-b^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(1 - (2*a^4*b^4 + c)/Sqrt[-4*a^8*b^8 + c^2])*Sqrt[1 - (a^4*x^4)/b^4]*EllipticF[ArcSin[(a*x)/b], -1])/(2*a*S
qrt[-b^4 + a^4*x^4]) + (b*(1 + (2*a^4*b^4 + c)/Sqrt[-4*a^8*b^8 + c^2])*Sqrt[1 - (a^4*x^4)/b^4]*EllipticF[ArcSi
n[(a*x)/b], -1])/(2*a*Sqrt[-b^4 + a^4*x^4]) - (b*Sqrt[1 - (a^4*x^4)/b^4]*EllipticPi[-((Sqrt[2]*a^2*b^2)/Sqrt[c
- Sqrt[-4*a^8*b^8 + c^2]]), ArcSin[(a*x)/b], -1])/(2*a*Sqrt[-b^4 + a^4*x^4]) - (b*Sqrt[1 - (a^4*x^4)/b^4]*Ell
ipticPi[(Sqrt[2]*a^2*b^2)/Sqrt[c - Sqrt[-4*a^8*b^8 + c^2]], ArcSin[(a*x)/b], -1])/(2*a*Sqrt[-b^4 + a^4*x^4]) -
(b*Sqrt[1 - (a^4*x^4)/b^4]*EllipticPi[-((Sqrt[2]*a^2*b^2)/Sqrt[c + Sqrt[-4*a^8*b^8 + c^2]]), ArcSin[(a*x)/b],
-1])/(2*a*Sqrt[-b^4 + a^4*x^4]) - (b*Sqrt[1 - (a^4*x^4)/b^4]*EllipticPi[(Sqrt[2]*a^2*b^2)/Sqrt[c + Sqrt[-4*a^
8*b^8 + c^2]], ArcSin[(a*x)/b], -1])/(2*a*Sqrt[-b^4 + a^4*x^4])

Rule 227

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[EllipticF[ArcSin[Rt[-b, 4]*(x/Rt[a, 4])], -1]/(Rt[a, 4]*Rt[
-b, 4]), x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]

Rule 230

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Dist[Sqrt[1 + b*(x^4/a)]/Sqrt[a + b*x^4], Int[1/Sqrt[1 + b*(x^4/
a)], x], x] /; FreeQ[{a, b}, x] && NegQ[b/a] &&  !GtQ[a, 0]

Rule 415

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

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 1232

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

Rule 1233

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> Dist[Sqrt[1 + c*(x^4/a)]/Sqrt[a + c*x^4]
, Int[1/((d + e*x^2)*Sqrt[1 + c*(x^4/a)]), x], x] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] &&  !GtQ[a, 0]

Rule 6860

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {\sqrt {-b^4+a^4 x^4} \left (b^4+a^4 x^4\right )}{b^8-c x^4+a^8 x^8} \, dx &=\int \left (\frac {\left (a^4+\frac {a^4 \left (2 a^4 b^4+c\right )}{\sqrt {-4 a^8 b^8+c^2}}\right ) \sqrt {-b^4+a^4 x^4}}{-c-\sqrt {-4 a^8 b^8+c^2}+2 a^8 x^4}+\frac {\left (a^4-\frac {a^4 \left (2 a^4 b^4+c\right )}{\sqrt {-4 a^8 b^8+c^2}}\right ) \sqrt {-b^4+a^4 x^4}}{-c+\sqrt {-4 a^8 b^8+c^2}+2 a^8 x^4}\right ) \, dx\\ &=\left (a^4 \left (1-\frac {2 a^4 b^4+c}{\sqrt {-4 a^8 b^8+c^2}}\right )\right ) \int \frac {\sqrt {-b^4+a^4 x^4}}{-c+\sqrt {-4 a^8 b^8+c^2}+2 a^8 x^4} \, dx+\left (a^4 \left (1+\frac {2 a^4 b^4+c}{\sqrt {-4 a^8 b^8+c^2}}\right )\right ) \int \frac {\sqrt {-b^4+a^4 x^4}}{-c-\sqrt {-4 a^8 b^8+c^2}+2 a^8 x^4} \, dx\\ &=\frac {1}{2} \left (1-\frac {2 a^4 b^4+c}{\sqrt {-4 a^8 b^8+c^2}}\right ) \int \frac {1}{\sqrt {-b^4+a^4 x^4}} \, dx+\frac {1}{2} \left (1+\frac {2 a^4 b^4+c}{\sqrt {-4 a^8 b^8+c^2}}\right ) \int \frac {1}{\sqrt {-b^4+a^4 x^4}} \, dx+\frac {\left (4 a^8 b^8-c \left (c-\sqrt {-4 a^8 b^8+c^2}\right )\right ) \int \frac {1}{\sqrt {-b^4+a^4 x^4} \left (-c+\sqrt {-4 a^8 b^8+c^2}+2 a^8 x^4\right )} \, dx}{\sqrt {-4 a^8 b^8+c^2}}-\frac {\left (4 a^8 b^8-c \left (c+\sqrt {-4 a^8 b^8+c^2}\right )\right ) \int \frac {1}{\sqrt {-b^4+a^4 x^4} \left (-c-\sqrt {-4 a^8 b^8+c^2}+2 a^8 x^4\right )} \, dx}{\sqrt {-4 a^8 b^8+c^2}}\\ &=-\left (\frac {1}{2} \int \frac {1}{\left (1-\frac {\sqrt {2} a^4 x^2}{\sqrt {c-\sqrt {-4 a^8 b^8+c^2}}}\right ) \sqrt {-b^4+a^4 x^4}} \, dx\right )-\frac {1}{2} \int \frac {1}{\left (1+\frac {\sqrt {2} a^4 x^2}{\sqrt {c-\sqrt {-4 a^8 b^8+c^2}}}\right ) \sqrt {-b^4+a^4 x^4}} \, dx-\frac {1}{2} \int \frac {1}{\left (1-\frac {\sqrt {2} a^4 x^2}{\sqrt {c+\sqrt {-4 a^8 b^8+c^2}}}\right ) \sqrt {-b^4+a^4 x^4}} \, dx-\frac {1}{2} \int \frac {1}{\left (1+\frac {\sqrt {2} a^4 x^2}{\sqrt {c+\sqrt {-4 a^8 b^8+c^2}}}\right ) \sqrt {-b^4+a^4 x^4}} \, dx+\frac {\left (\left (1-\frac {2 a^4 b^4+c}{\sqrt {-4 a^8 b^8+c^2}}\right ) \sqrt {1-\frac {a^4 x^4}{b^4}}\right ) \int \frac {1}{\sqrt {1-\frac {a^4 x^4}{b^4}}} \, dx}{2 \sqrt {-b^4+a^4 x^4}}+\frac {\left (\left (1+\frac {2 a^4 b^4+c}{\sqrt {-4 a^8 b^8+c^2}}\right ) \sqrt {1-\frac {a^4 x^4}{b^4}}\right ) \int \frac {1}{\sqrt {1-\frac {a^4 x^4}{b^4}}} \, dx}{2 \sqrt {-b^4+a^4 x^4}}\\ &=\frac {b \left (1-\frac {2 a^4 b^4+c}{\sqrt {-4 a^8 b^8+c^2}}\right ) \sqrt {1-\frac {a^4 x^4}{b^4}} F\left (\left .\sin ^{-1}\left (\frac {a x}{b}\right )\right |-1\right )}{2 a \sqrt {-b^4+a^4 x^4}}+\frac {b \left (1+\frac {2 a^4 b^4+c}{\sqrt {-4 a^8 b^8+c^2}}\right ) \sqrt {1-\frac {a^4 x^4}{b^4}} F\left (\left .\sin ^{-1}\left (\frac {a x}{b}\right )\right |-1\right )}{2 a \sqrt {-b^4+a^4 x^4}}-\frac {\sqrt {1-\frac {a^4 x^4}{b^4}} \int \frac {1}{\left (1-\frac {\sqrt {2} a^4 x^2}{\sqrt {c-\sqrt {-4 a^8 b^8+c^2}}}\right ) \sqrt {1-\frac {a^4 x^4}{b^4}}} \, dx}{2 \sqrt {-b^4+a^4 x^4}}-\frac {\sqrt {1-\frac {a^4 x^4}{b^4}} \int \frac {1}{\left (1+\frac {\sqrt {2} a^4 x^2}{\sqrt {c-\sqrt {-4 a^8 b^8+c^2}}}\right ) \sqrt {1-\frac {a^4 x^4}{b^4}}} \, dx}{2 \sqrt {-b^4+a^4 x^4}}-\frac {\sqrt {1-\frac {a^4 x^4}{b^4}} \int \frac {1}{\left (1-\frac {\sqrt {2} a^4 x^2}{\sqrt {c+\sqrt {-4 a^8 b^8+c^2}}}\right ) \sqrt {1-\frac {a^4 x^4}{b^4}}} \, dx}{2 \sqrt {-b^4+a^4 x^4}}-\frac {\sqrt {1-\frac {a^4 x^4}{b^4}} \int \frac {1}{\left (1+\frac {\sqrt {2} a^4 x^2}{\sqrt {c+\sqrt {-4 a^8 b^8+c^2}}}\right ) \sqrt {1-\frac {a^4 x^4}{b^4}}} \, dx}{2 \sqrt {-b^4+a^4 x^4}}\\ &=\frac {b \left (1-\frac {2 a^4 b^4+c}{\sqrt {-4 a^8 b^8+c^2}}\right ) \sqrt {1-\frac {a^4 x^4}{b^4}} F\left (\left .\sin ^{-1}\left (\frac {a x}{b}\right )\right |-1\right )}{2 a \sqrt {-b^4+a^4 x^4}}+\frac {b \left (1+\frac {2 a^4 b^4+c}{\sqrt {-4 a^8 b^8+c^2}}\right ) \sqrt {1-\frac {a^4 x^4}{b^4}} F\left (\left .\sin ^{-1}\left (\frac {a x}{b}\right )\right |-1\right )}{2 a \sqrt {-b^4+a^4 x^4}}-\frac {b \sqrt {1-\frac {a^4 x^4}{b^4}} \Pi \left (-\frac {\sqrt {2} a^2 b^2}{\sqrt {c-\sqrt {-4 a^8 b^8+c^2}}};\left .\sin ^{-1}\left (\frac {a x}{b}\right )\right |-1\right )}{2 a \sqrt {-b^4+a^4 x^4}}-\frac {b \sqrt {1-\frac {a^4 x^4}{b^4}} \Pi \left (\frac {\sqrt {2} a^2 b^2}{\sqrt {c-\sqrt {-4 a^8 b^8+c^2}}};\left .\sin ^{-1}\left (\frac {a x}{b}\right )\right |-1\right )}{2 a \sqrt {-b^4+a^4 x^4}}-\frac {b \sqrt {1-\frac {a^4 x^4}{b^4}} \Pi \left (-\frac {\sqrt {2} a^2 b^2}{\sqrt {c+\sqrt {-4 a^8 b^8+c^2}}};\left .\sin ^{-1}\left (\frac {a x}{b}\right )\right |-1\right )}{2 a \sqrt {-b^4+a^4 x^4}}-\frac {b \sqrt {1-\frac {a^4 x^4}{b^4}} \Pi \left (\frac {\sqrt {2} a^2 b^2}{\sqrt {c+\sqrt {-4 a^8 b^8+c^2}}};\left .\sin ^{-1}\left (\frac {a x}{b}\right )\right |-1\right )}{2 a \sqrt {-b^4+a^4 x^4}}\\ \end {align*}

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Mathematica [A]
time = 0.72, size = 171, normalized size = 0.75 \begin {gather*} \frac {\text {ArcTan}\left (\frac {\sqrt [4]{2 a^4 b^4-c} x \sqrt {-2 b^4+2 a^4 x^4}}{b^4+\sqrt {2 a^4 b^4-c} x^2-a^4 x^4}\right )-\tanh ^{-1}\left (\frac {-b^4+\sqrt {2 a^4 b^4-c} x^2+a^4 x^4}{\sqrt [4]{2 a^4 b^4-c} x \sqrt {-2 b^4+2 a^4 x^4}}\right )}{2 \sqrt {2} \sqrt [4]{2 a^4 b^4-c}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(ArcTan[((2*a^4*b^4 - c)^(1/4)*x*Sqrt[-2*b^4 + 2*a^4*x^4])/(b^4 + Sqrt[2*a^4*b^4 - c]*x^2 - a^4*x^4)] - ArcTan
h[(-b^4 + Sqrt[2*a^4*b^4 - c]*x^2 + a^4*x^4)/((2*a^4*b^4 - c)^(1/4)*x*Sqrt[-2*b^4 + 2*a^4*x^4])])/(2*Sqrt[2]*(
2*a^4*b^4 - c)^(1/4))

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Maple [A]
time = 0.11, size = 281, normalized size = 1.23

method	result	size
default	\(\frac {\left (\frac {\ln \left (\frac {\frac {a^{4} x^{4}-b^{4}}{2 x^{2}}-\frac {\left (2 a^{4} b^{4}-c \right )^{\frac {1}{4}} \sqrt {a^{4} x^{4}-b^{4}}\, \sqrt {2}}{2 x}+\frac {\sqrt {2 a^{4} b^{4}-c}}{2}}{\frac {a^{4} x^{4}-b^{4}}{2 x^{2}}+\frac {\left (2 a^{4} b^{4}-c \right )^{\frac {1}{4}} \sqrt {a^{4} x^{4}-b^{4}}\, \sqrt {2}}{2 x}+\frac {\sqrt {2 a^{4} b^{4}-c}}{2}}\right )}{4 \left (2 a^{4} b^{4}-c \right )^{\frac {1}{4}}}+\frac {\arctan \left (\frac {\sqrt {a^{4} x^{4}-b^{4}}\, \sqrt {2}}{\left (2 a^{4} b^{4}-c \right )^{\frac {1}{4}} x}+1\right )}{2 \left (2 a^{4} b^{4}-c \right )^{\frac {1}{4}}}+\frac {\arctan \left (\frac {\sqrt {a^{4} x^{4}-b^{4}}\, \sqrt {2}}{\left (2 a^{4} b^{4}-c \right )^{\frac {1}{4}} x}-1\right )}{2 \left (2 a^{4} b^{4}-c \right )^{\frac {1}{4}}}\right ) \sqrt {2}}{2}\)	\(281\)
elliptic	\(\frac {\left (\frac {\ln \left (\frac {\frac {a^{4} x^{4}-b^{4}}{2 x^{2}}-\frac {\left (2 a^{4} b^{4}-c \right )^{\frac {1}{4}} \sqrt {a^{4} x^{4}-b^{4}}\, \sqrt {2}}{2 x}+\frac {\sqrt {2 a^{4} b^{4}-c}}{2}}{\frac {a^{4} x^{4}-b^{4}}{2 x^{2}}+\frac {\left (2 a^{4} b^{4}-c \right )^{\frac {1}{4}} \sqrt {a^{4} x^{4}-b^{4}}\, \sqrt {2}}{2 x}+\frac {\sqrt {2 a^{4} b^{4}-c}}{2}}\right )}{4 \left (2 a^{4} b^{4}-c \right )^{\frac {1}{4}}}+\frac {\arctan \left (\frac {\sqrt {a^{4} x^{4}-b^{4}}\, \sqrt {2}}{\left (2 a^{4} b^{4}-c \right )^{\frac {1}{4}} x}+1\right )}{2 \left (2 a^{4} b^{4}-c \right )^{\frac {1}{4}}}+\frac {\arctan \left (\frac {\sqrt {a^{4} x^{4}-b^{4}}\, \sqrt {2}}{\left (2 a^{4} b^{4}-c \right )^{\frac {1}{4}} x}-1\right )}{2 \left (2 a^{4} b^{4}-c \right )^{\frac {1}{4}}}\right ) \sqrt {2}}{2}\)	\(281\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^4*x^4-b^4)^(1/2)*(a^4*x^4+b^4)/(a^8*x^8+b^8-c*x^4),x,method=_RETURNVERBOSE)

[Out]

1/2*(1/4/(2*a^4*b^4-c)^(1/4)*ln((1/2*(a^4*x^4-b^4)/x^2-1/2*(2*a^4*b^4-c)^(1/4)*(a^4*x^4-b^4)^(1/2)*2^(1/2)/x+1
/2*(2*a^4*b^4-c)^(1/2))/(1/2*(a^4*x^4-b^4)/x^2+1/2*(2*a^4*b^4-c)^(1/4)*(a^4*x^4-b^4)^(1/2)*2^(1/2)/x+1/2*(2*a^
4*b^4-c)^(1/2)))+1/2/(2*a^4*b^4-c)^(1/4)*arctan(1/(2*a^4*b^4-c)^(1/4)*(a^4*x^4-b^4)^(1/2)*2^(1/2)/x+1)+1/2/(2*
a^4*b^4-c)^(1/4)*arctan(1/(2*a^4*b^4-c)^(1/4)*(a^4*x^4-b^4)^(1/2)*2^(1/2)/x-1))*2^(1/2)

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 746 vs. \(2 (197) = 394\).
time = 11.94, size = 746, normalized size = 3.26 \begin {gather*} \frac {1}{2} \, \left (-\frac {1}{2 \, a^{4} b^{4} - c}\right )^{\frac {1}{4}} \arctan \left (\frac {2 \, \sqrt {a^{4} x^{4} - b^{4}} {\left ({\left (2 \, a^{4} b^{4} - c\right )} x^{3} \left (-\frac {1}{2 \, a^{4} b^{4} - c}\right )^{\frac {1}{4}} + {\left ({\left (2 \, a^{8} b^{4} - a^{4} c\right )} x^{5} - {\left (2 \, a^{4} b^{8} - b^{4} c\right )} x\right )} \left (-\frac {1}{2 \, a^{4} b^{4} - c}\right )^{\frac {3}{4}}\right )} - {\left ({\left (2 \, a^{4} b^{12} - b^{8} c + {\left (2 \, a^{12} b^{4} - a^{8} c\right )} x^{8} - {\left (8 \, a^{8} b^{8} - 6 \, a^{4} b^{4} c + c^{2}\right )} x^{4}\right )} \left (-\frac {1}{2 \, a^{4} b^{4} - c}\right )^{\frac {3}{4}} + 2 \, {\left ({\left (2 \, a^{8} b^{4} - a^{4} c\right )} x^{6} - {\left (2 \, a^{4} b^{8} - b^{4} c\right )} x^{2}\right )} \left (-\frac {1}{2 \, a^{4} b^{4} - c}\right )^{\frac {1}{4}}\right )} \left (-\frac {1}{2 \, a^{4} b^{4} - c}\right )^{\frac {1}{4}}}{a^{8} x^{8} + b^{8} - c x^{4}}\right ) + \frac {1}{8} \, \left (-\frac {1}{2 \, a^{4} b^{4} - c}\right )^{\frac {1}{4}} \log \left (\frac {2 \, {\left ({\left (2 \, a^{8} b^{4} - a^{4} c\right )} x^{6} - {\left (2 \, a^{4} b^{8} - b^{4} c\right )} x^{2}\right )} \left (-\frac {1}{2 \, a^{4} b^{4} - c}\right )^{\frac {3}{4}} + 2 \, {\left (a^{4} x^{5} - b^{4} x - {\left (2 \, a^{4} b^{4} - c\right )} x^{3} \sqrt {-\frac {1}{2 \, a^{4} b^{4} - c}}\right )} \sqrt {a^{4} x^{4} - b^{4}} - {\left (a^{8} x^{8} + b^{8} - {\left (4 \, a^{4} b^{4} - c\right )} x^{4}\right )} \left (-\frac {1}{2 \, a^{4} b^{4} - c}\right )^{\frac {1}{4}}}{2 \, {\left (a^{8} x^{8} + b^{8} - c x^{4}\right )}}\right ) - \frac {1}{8} \, \left (-\frac {1}{2 \, a^{4} b^{4} - c}\right )^{\frac {1}{4}} \log \left (-\frac {2 \, {\left ({\left (2 \, a^{8} b^{4} - a^{4} c\right )} x^{6} - {\left (2 \, a^{4} b^{8} - b^{4} c\right )} x^{2}\right )} \left (-\frac {1}{2 \, a^{4} b^{4} - c}\right )^{\frac {3}{4}} - 2 \, {\left (a^{4} x^{5} - b^{4} x - {\left (2 \, a^{4} b^{4} - c\right )} x^{3} \sqrt {-\frac {1}{2 \, a^{4} b^{4} - c}}\right )} \sqrt {a^{4} x^{4} - b^{4}} - {\left (a^{8} x^{8} + b^{8} - {\left (4 \, a^{4} b^{4} - c\right )} x^{4}\right )} \left (-\frac {1}{2 \, a^{4} b^{4} - c}\right )^{\frac {1}{4}}}{2 \, {\left (a^{8} x^{8} + b^{8} - c x^{4}\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(-1/(2*a^4*b^4 - c))^(1/4)*arctan((2*sqrt(a^4*x^4 - b^4)*((2*a^4*b^4 - c)*x^3*(-1/(2*a^4*b^4 - c))^(1/4) +
((2*a^8*b^4 - a^4*c)*x^5 - (2*a^4*b^8 - b^4*c)*x)*(-1/(2*a^4*b^4 - c))^(3/4)) - ((2*a^4*b^12 - b^8*c + (2*a^1
2*b^4 - a^8*c)*x^8 - (8*a^8*b^8 - 6*a^4*b^4*c + c^2)*x^4)*(-1/(2*a^4*b^4 - c))^(3/4) + 2*((2*a^8*b^4 - a^4*c)*
x^6 - (2*a^4*b^8 - b^4*c)*x^2)*(-1/(2*a^4*b^4 - c))^(1/4))*(-1/(2*a^4*b^4 - c))^(1/4))/(a^8*x^8 + b^8 - c*x^4)
) + 1/8*(-1/(2*a^4*b^4 - c))^(1/4)*log(1/2*(2*((2*a^8*b^4 - a^4*c)*x^6 - (2*a^4*b^8 - b^4*c)*x^2)*(-1/(2*a^4*b
^4 - c))^(3/4) + 2*(a^4*x^5 - b^4*x - (2*a^4*b^4 - c)*x^3*sqrt(-1/(2*a^4*b^4 - c)))*sqrt(a^4*x^4 - b^4) - (a^8
*x^8 + b^8 - (4*a^4*b^4 - c)*x^4)*(-1/(2*a^4*b^4 - c))^(1/4))/(a^8*x^8 + b^8 - c*x^4)) - 1/8*(-1/(2*a^4*b^4 -
c))^(1/4)*log(-1/2*(2*((2*a^8*b^4 - a^4*c)*x^6 - (2*a^4*b^8 - b^4*c)*x^2)*(-1/(2*a^4*b^4 - c))^(3/4) - 2*(a^4*
x^5 - b^4*x - (2*a^4*b^4 - c)*x^3*sqrt(-1/(2*a^4*b^4 - c)))*sqrt(a^4*x^4 - b^4) - (a^8*x^8 + b^8 - (4*a^4*b^4
- c)*x^4)*(-1/(2*a^4*b^4 - c))^(1/4))/(a^8*x^8 + b^8 - c*x^4))

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Sympy [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((a**4*x**4-b**4)**(1/2)*(a**4*x**4+b**4)/(a**8*x**8+b**8-c*x**4),x)

[Out]

Timed out

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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