3.21.83 \(\int \frac {-b^8+a^8 x^8}{\sqrt {-b^4+a^4 x^4} (b^8+a^8 x^8)} \, dx\)

Optimal. Leaf size=150 \[ -\frac {\tan ^{-1}\left (\frac {-\frac {a^3 x^4}{2^{3/4} b}+\frac {b^3}{2^{3/4} a}+\frac {a b x^2}{\sqrt [4]{2}}}{x \sqrt {a^4 x^4-b^4}}\right )}{2\ 2^{3/4} a b}-\frac {\tanh ^{-1}\left (\frac {2^{3/4} a b x \sqrt {a^4 x^4-b^4}}{a^4 x^4+\sqrt {2} a^2 b^2 x^2-b^4}\right )}{2\ 2^{3/4} a b} \]

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Rubi [C]  time = 0.65, antiderivative size = 400, normalized size of antiderivative = 2.67, number of steps used = 19, number of rules used = 8, integrand size = 44, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {1586, 6725, 406, 224, 221, 409, 1219, 1218} \begin {gather*} -\frac {b \sqrt {1-\frac {a^4 x^4}{b^4}} \Pi \left (\frac {a^6}{\left (-a^8\right )^{3/4}};\left .\sin ^{-1}\left (\frac {a x}{b}\right )\right |-1\right )}{2 a \sqrt {a^4 x^4-b^4}}+\frac {\left (a^4-\sqrt {-a^8}\right ) b \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^5 \sqrt {a^4 x^4-b^4}}+\frac {\left (\sqrt {-a^8}+a^4\right ) b \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^5 \sqrt {a^4 x^4-b^4}}-\frac {b \sqrt {1-\frac {a^4 x^4}{b^4}} \Pi \left (\frac {\sqrt [4]{-a^8}}{a^2};\left .\sin ^{-1}\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 {-\sqrt {-a^8}}}{a^2};\left .\sin ^{-1}\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 {-\sqrt {-a^8}}}{a^2};\left .\sin ^{-1}\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[(-b^8 + a^8*x^8)/(Sqrt[-b^4 + a^4*x^4]*(b^8 + a^8*x^8)),x]

[Out]

((a^4 - Sqrt[-a^8])*b*Sqrt[1 - (a^4*x^4)/b^4]*EllipticF[ArcSin[(a*x)/b], -1])/(2*a^5*Sqrt[-b^4 + a^4*x^4]) + (
(a^4 + Sqrt[-a^8])*b*Sqrt[1 - (a^4*x^4)/b^4]*EllipticF[ArcSin[(a*x)/b], -1])/(2*a^5*Sqrt[-b^4 + a^4*x^4]) - (b
*Sqrt[1 - (a^4*x^4)/b^4]*EllipticPi[a^6/(-a^8)^(3/4), ArcSin[(a*x)/b], -1])/(2*a*Sqrt[-b^4 + a^4*x^4]) - (b*Sq
rt[1 - (a^4*x^4)/b^4]*EllipticPi[(-a^8)^(1/4)/a^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[-Sqrt[-a^8]]/a^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[-Sqrt[-a^8]]/a^2, ArcSin[(a*x)/b], -1])/(2*a*Sqrt[-b^4 + a^4*x^4])

Rule 221

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 224

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 406

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 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 1218

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

Rule 1219

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 1586

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

Rule 6725

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

Rubi steps

\begin {align*} \int \frac {-b^8+a^8 x^8}{\sqrt {-b^4+a^4 x^4} \left (b^8+a^8 x^8\right )} \, dx &=\int \frac {\sqrt {-b^4+a^4 x^4} \left (b^4+a^4 x^4\right )}{b^8+a^8 x^8} \, dx\\ &=\int \left (-\frac {\sqrt {-a^8} \left (a^4 b^4+\sqrt {-a^8} b^4\right ) \sqrt {-b^4+a^4 x^4}}{2 a^8 b^4 \left (b^4-\sqrt {-a^8} x^4\right )}+\frac {\sqrt {-a^8} \left (a^4 b^4-\sqrt {-a^8} b^4\right ) \sqrt {-b^4+a^4 x^4}}{2 a^8 b^4 \left (b^4+\sqrt {-a^8} x^4\right )}\right ) \, dx\\ &=\frac {\left (a^4+\sqrt {-a^8}\right ) \int \frac {\sqrt {-b^4+a^4 x^4}}{b^4+\sqrt {-a^8} x^4} \, dx}{2 a^4}-\frac {\left (\sqrt {-a^8} \left (a^4 b^4+\sqrt {-a^8} b^4\right )\right ) \int \frac {\sqrt {-b^4+a^4 x^4}}{b^4-\sqrt {-a^8} x^4} \, dx}{2 a^8 b^4}\\ &=\frac {1}{2} \left (1+\frac {a^4}{\sqrt {-a^8}}\right ) \int \frac {1}{\sqrt {-b^4+a^4 x^4}} \, dx+\frac {\left (a^4+\sqrt {-a^8}\right ) \int \frac {1}{\sqrt {-b^4+a^4 x^4}} \, dx}{2 a^4}-b^4 \int \frac {1}{\sqrt {-b^4+a^4 x^4} \left (b^4-\sqrt {-a^8} x^4\right )} \, dx-b^4 \int \frac {1}{\sqrt {-b^4+a^4 x^4} \left (b^4+\sqrt {-a^8} x^4\right )} \, dx\\ &=-\left (\frac {1}{2} \int \frac {1}{\left (1-\frac {\sqrt [4]{-a^8} x^2}{b^2}\right ) \sqrt {-b^4+a^4 x^4}} \, dx\right )-\frac {1}{2} \int \frac {1}{\left (1+\frac {\sqrt [4]{-a^8} x^2}{b^2}\right ) \sqrt {-b^4+a^4 x^4}} \, dx-\frac {1}{2} \int \frac {1}{\left (1-\frac {\sqrt {-\sqrt {-a^8}} x^2}{b^2}\right ) \sqrt {-b^4+a^4 x^4}} \, dx-\frac {1}{2} \int \frac {1}{\left (1+\frac {\sqrt {-\sqrt {-a^8}} x^2}{b^2}\right ) \sqrt {-b^4+a^4 x^4}} \, dx+\frac {\left (\left (1+\frac {a^4}{\sqrt {-a^8}}\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 (a^4+\sqrt {-a^8}\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 a^4 \sqrt {-b^4+a^4 x^4}}\\ &=\frac {\left (1+\frac {a^4}{\sqrt {-a^8}}\right ) b \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 {\left (a^4+\sqrt {-a^8}\right ) b \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^5 \sqrt {-b^4+a^4 x^4}}-\frac {\sqrt {1-\frac {a^4 x^4}{b^4}} \int \frac {1}{\left (1-\frac {\sqrt [4]{-a^8} x^2}{b^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 [4]{-a^8} x^2}{b^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 {-\sqrt {-a^8}} x^2}{b^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 {-\sqrt {-a^8}} x^2}{b^2}\right ) \sqrt {1-\frac {a^4 x^4}{b^4}}} \, dx}{2 \sqrt {-b^4+a^4 x^4}}\\ &=\frac {\left (1+\frac {a^4}{\sqrt {-a^8}}\right ) b \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 {\left (a^4+\sqrt {-a^8}\right ) b \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^5 \sqrt {-b^4+a^4 x^4}}-\frac {b \sqrt {1-\frac {a^4 x^4}{b^4}} \Pi \left (\frac {a^6}{\left (-a^8\right )^{3/4}};\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 [4]{-a^8}}{a^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 {-\sqrt {-a^8}}}{a^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 {-\sqrt {-a^8}}}{a^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 [C]  time = 0.60, size = 192, normalized size = 1.28 \begin {gather*} -\frac {i \sqrt {1-\frac {a^4 x^4}{b^4}} \left (2 F\left (\left .i \sinh ^{-1}\left (\sqrt {-\frac {a^2}{b^2}} x\right )\right |-1\right )-\Pi \left (-\sqrt [4]{-1};\left .i \sinh ^{-1}\left (\sqrt {-\frac {a^2}{b^2}} x\right )\right |-1\right )-\Pi \left (\sqrt [4]{-1};\left .i \sinh ^{-1}\left (\sqrt {-\frac {a^2}{b^2}} x\right )\right |-1\right )-\Pi \left (-(-1)^{3/4};\left .i \sinh ^{-1}\left (\sqrt {-\frac {a^2}{b^2}} x\right )\right |-1\right )-\Pi \left ((-1)^{3/4};\left .i \sinh ^{-1}\left (\sqrt {-\frac {a^2}{b^2}} x\right )\right |-1\right )\right )}{2 \sqrt {-\frac {a^2}{b^2}} \sqrt {a^4 x^4-b^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/2*I)*Sqrt[1 - (a^4*x^4)/b^4]*(2*EllipticF[I*ArcSinh[Sqrt[-(a^2/b^2)]*x], -1] - EllipticPi[-(-1)^(1/4), I*
ArcSinh[Sqrt[-(a^2/b^2)]*x], -1] - EllipticPi[(-1)^(1/4), I*ArcSinh[Sqrt[-(a^2/b^2)]*x], -1] - EllipticPi[-(-1
)^(3/4), I*ArcSinh[Sqrt[-(a^2/b^2)]*x], -1] - EllipticPi[(-1)^(3/4), I*ArcSinh[Sqrt[-(a^2/b^2)]*x], -1]))/(Sqr
t[-(a^2/b^2)]*Sqrt[-b^4 + a^4*x^4])

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IntegrateAlgebraic [A]  time = 0.72, size = 156, normalized size = 1.04 \begin {gather*} -\frac {\tan ^{-1}\left (\frac {\frac {b^3}{2^{3/4} a}+\frac {a b x^2}{\sqrt [4]{2}}-\frac {a^3 x^4}{2^{3/4} b}}{x \sqrt {-b^4+a^4 x^4}}\right )}{2\ 2^{3/4} a b}+\frac {\tanh ^{-1}\left (\frac {\frac {b^3}{2^{3/4} a}-\frac {a b x^2}{\sqrt [4]{2}}-\frac {a^3 x^4}{2^{3/4} b}}{x \sqrt {-b^4+a^4 x^4}}\right )}{2\ 2^{3/4} a b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*ArcTan[(b^3/(2^(3/4)*a) + (a*b*x^2)/2^(1/4) - (a^3*x^4)/(2^(3/4)*b))/(x*Sqrt[-b^4 + a^4*x^4])]/(2^(3/4)*a
*b) + ArcTanh[(b^3/(2^(3/4)*a) - (a*b*x^2)/2^(1/4) - (a^3*x^4)/(2^(3/4)*b))/(x*Sqrt[-b^4 + a^4*x^4])]/(2*2^(3/
4)*a*b)

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fricas [F(-1)]  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^8*x^8-b^8)/(a^4*x^4-b^4)^(1/2)/(a^8*x^8+b^8),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a^{8} x^{8} - b^{8}}{{\left (a^{8} x^{8} + b^{8}\right )} \sqrt {a^{4} x^{4} - b^{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.14, size = 281, normalized size = 1.87

method	result	size
default	\(\frac {\sqrt {\frac {a^{2} x^{2}}{b^{2}}+1}\, \sqrt {1-\frac {a^{2} x^{2}}{b^{2}}}\, \EllipticF \left (x \sqrt {-\frac {a^{2}}{b^{2}}}, i\right )}{\sqrt {-\frac {a^{2}}{b^{2}}}\, \sqrt {a^{4} x^{4}-b^{4}}}-\frac {b^{8} \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (\textit {\_Z}^{8} a^{8}+b^{8}\right )}{\sum }\frac {-\frac {\arctanh \left (\frac {\underline {\hspace {1.25 ex}}\alpha ^{2} \left (a^{4} \underline {\hspace {1.25 ex}}\alpha ^{6}+b^{4} x^{2}\right ) a^{4}}{b^{4} \sqrt {a^{4} \underline {\hspace {1.25 ex}}\alpha ^{4}-b^{4}}\, \sqrt {a^{4} x^{4}-b^{4}}}\right )}{\sqrt {a^{4} \underline {\hspace {1.25 ex}}\alpha ^{4}-b^{4}}}+\frac {2 \underline {\hspace {1.25 ex}}\alpha ^{7} a^{8} \sqrt {\frac {a^{2} x^{2}}{b^{2}}+1}\, \sqrt {1-\frac {a^{2} x^{2}}{b^{2}}}\, \EllipticPi \left (x \sqrt {-\frac {a^{2}}{b^{2}}}, \frac {\underline {\hspace {1.25 ex}}\alpha ^{6} a^{6}}{b^{6}}, \frac {\sqrt {\frac {a^{2}}{b^{2}}}}{\sqrt {-\frac {a^{2}}{b^{2}}}}\right )}{\sqrt {-\frac {a^{2}}{b^{2}}}\, b^{8} \sqrt {a^{4} x^{4}-b^{4}}}}{\underline {\hspace {1.25 ex}}\alpha ^{7}}\right )}{8 a^{8}}\)	\(281\)
elliptic	\(\frac {\left (\frac {\ln \left (\frac {\frac {a^{4} x^{4}-b^{4}}{2 x^{2}}-\frac {\sqrt {\sqrt {2}\, \sqrt {a^{4} b^{4}}}\, \sqrt {a^{4} x^{4}-b^{4}}\, \sqrt {2}}{2 x}+\frac {\sqrt {2}\, \sqrt {a^{4} b^{4}}}{2}}{\frac {a^{4} x^{4}-b^{4}}{2 x^{2}}+\frac {\sqrt {\sqrt {2}\, \sqrt {a^{4} b^{4}}}\, \sqrt {a^{4} x^{4}-b^{4}}\, \sqrt {2}}{2 x}+\frac {\sqrt {2}\, \sqrt {a^{4} b^{4}}}{2}}\right )}{4 \sqrt {\sqrt {2}\, \sqrt {a^{4} b^{4}}}}+\frac {\arctan \left (\frac {\sqrt {a^{4} x^{4}-b^{4}}\, \sqrt {2}}{\sqrt {\sqrt {2}\, \sqrt {a^{4} b^{4}}}\, x}+1\right )}{2 \sqrt {\sqrt {2}\, \sqrt {a^{4} b^{4}}}}+\frac {\arctan \left (\frac {\sqrt {a^{4} x^{4}-b^{4}}\, \sqrt {2}}{\sqrt {\sqrt {2}\, \sqrt {a^{4} b^{4}}}\, x}-1\right )}{2 \sqrt {\sqrt {2}\, \sqrt {a^{4} b^{4}}}}\right ) \sqrt {2}}{2}\)	\(284\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(-a^2/b^2)^(1/2)*(a^2*x^2/b^2+1)^(1/2)*(1-a^2*x^2/b^2)^(1/2)/(a^4*x^4-b^4)^(1/2)*EllipticF(x*(-a^2/b^2)^(1/2
),I)-1/8*b^8/a^8*sum(1/_alpha^7*(-1/(_alpha^4*a^4-b^4)^(1/2)*arctanh(_alpha^2/b^4*(_alpha^6*a^4+b^4*x^2)*a^4/(
_alpha^4*a^4-b^4)^(1/2)/(a^4*x^4-b^4)^(1/2))+2/(-a^2/b^2)^(1/2)*_alpha^7*a^8/b^8*(a^2*x^2/b^2+1)^(1/2)*(1-a^2*
x^2/b^2)^(1/2)/(a^4*x^4-b^4)^(1/2)*EllipticPi(x*(-a^2/b^2)^(1/2),_alpha^6*a^6/b^6,(a^2/b^2)^(1/2)/(-a^2/b^2)^(
1/2))),_alpha=RootOf(_Z^8*a^8+b^8))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a^{8} x^{8} - b^{8}}{{\left (a^{8} x^{8} + b^{8}\right )} \sqrt {a^{4} x^{4} - b^{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int -\frac {b^8-a^8\,x^8}{\sqrt {a^4\,x^4-b^4}\,\left (a^8\,x^8+b^8\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a x - b\right ) \left (a x + b\right ) \left (a^{2} x^{2} + b^{2}\right ) \left (a^{4} x^{4} + b^{4}\right )}{\sqrt {\left (a x - b\right ) \left (a x + b\right ) \left (a^{2} x^{2} + b^{2}\right )} \left (a^{8} x^{8} + b^{8}\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a**8*x**8-b**8)/(a**4*x**4-b**4)**(1/2)/(a**8*x**8+b**8),x)

[Out]

Integral((a*x - b)*(a*x + b)*(a**2*x**2 + b**2)*(a**4*x**4 + b**4)/(sqrt((a*x - b)*(a*x + b)*(a**2*x**2 + b**2
))*(a**8*x**8 + b**8)), x)

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