3.9.62 \(\int \frac {\sqrt {1-x^4} (1+x^4)}{4-7 x^4+4 x^8} \, dx\)

Optimal. Leaf size=65 \[ \frac {1}{8} \tan ^{-1}\left (\frac {x^4+\frac {x^2}{2}-1}{x \sqrt {1-x^4}}\right )-\frac {1}{8} \tanh ^{-1}\left (\frac {x^4-\frac {x^2}{2}-1}{x \sqrt {1-x^4}}\right ) \]

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Rubi [C]  time = 0.54, antiderivative size = 155, normalized size of antiderivative = 2.38, number of steps used = 16, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {6728, 406, 221, 409, 1213, 537} \begin {gather*} -\frac {1}{8} \left (1+i \sqrt {15}\right ) F\left (\left .\sin ^{-1}(x)\right |-1\right )-\frac {1}{8} \left (1-i \sqrt {15}\right ) F\left (\left .\sin ^{-1}(x)\right |-1\right )+\frac {1}{8} \Pi \left (-\frac {2}{\sqrt {\frac {1}{2} \left (7-i \sqrt {15}\right )}};\left .\sin ^{-1}(x)\right |-1\right )+\frac {1}{8} \Pi \left (\frac {2}{\sqrt {\frac {1}{2} \left (7-i \sqrt {15}\right )}};\left .\sin ^{-1}(x)\right |-1\right )+\frac {1}{8} \Pi \left (-\frac {2}{\sqrt {\frac {1}{2} \left (7+i \sqrt {15}\right )}};\left .\sin ^{-1}(x)\right |-1\right )+\frac {1}{8} \Pi \left (\frac {2}{\sqrt {\frac {1}{2} \left (7+i \sqrt {15}\right )}};\left .\sin ^{-1}(x)\right |-1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sqrt[1 - x^4]*(1 + x^4))/(4 - 7*x^4 + 4*x^8),x]

[Out]

-1/8*((1 - I*Sqrt[15])*EllipticF[ArcSin[x], -1]) - ((1 + I*Sqrt[15])*EllipticF[ArcSin[x], -1])/8 + EllipticPi[
-2/Sqrt[(7 - I*Sqrt[15])/2], ArcSin[x], -1]/8 + EllipticPi[2/Sqrt[(7 - I*Sqrt[15])/2], ArcSin[x], -1]/8 + Elli
pticPi[-2/Sqrt[(7 + I*Sqrt[15])/2], ArcSin[x], -1]/8 + EllipticPi[2/Sqrt[(7 + I*Sqrt[15])/2], ArcSin[x], -1]/8

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 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 537

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[(1*Ellipt
icPi[(b*c)/(a*d), ArcSin[Rt[-(d/c), 2]*x], (c*f)/(d*e)])/(a*Sqrt[c]*Sqrt[e]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b,
c, d, e, f}, x] &&  !GtQ[d/c, 0] && GtQ[c, 0] && GtQ[e, 0] &&  !( !GtQ[f/e, 0] && SimplerSqrtQ[-(f/e), -(d/c)
])

Rule 1213

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

Rule 6728

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 {1-x^4} \left (1+x^4\right )}{4-7 x^4+4 x^8} \, dx &=\int \left (\frac {\left (1-i \sqrt {15}\right ) \sqrt {1-x^4}}{-7-i \sqrt {15}+8 x^4}+\frac {\left (1+i \sqrt {15}\right ) \sqrt {1-x^4}}{-7+i \sqrt {15}+8 x^4}\right ) \, dx\\ &=\left (1-i \sqrt {15}\right ) \int \frac {\sqrt {1-x^4}}{-7-i \sqrt {15}+8 x^4} \, dx+\left (1+i \sqrt {15}\right ) \int \frac {\sqrt {1-x^4}}{-7+i \sqrt {15}+8 x^4} \, dx\\ &=\frac {1}{8} \left (-1-i \sqrt {15}\right ) \int \frac {1}{\sqrt {1-x^4}} \, dx+\frac {1}{4} \left (-7+i \sqrt {15}\right ) \int \frac {1}{\sqrt {1-x^4} \left (-7+i \sqrt {15}+8 x^4\right )} \, dx+\frac {1}{8} \left (-1+i \sqrt {15}\right ) \int \frac {1}{\sqrt {1-x^4}} \, dx-\frac {1}{8} \left (i+\sqrt {15}\right )^2 \int \frac {1}{\sqrt {1-x^4} \left (-7-i \sqrt {15}+8 x^4\right )} \, dx\\ &=-\frac {1}{8} \left (1-i \sqrt {15}\right ) F\left (\left .\sin ^{-1}(x)\right |-1\right )-\frac {1}{8} \left (1+i \sqrt {15}\right ) F\left (\left .\sin ^{-1}(x)\right |-1\right )+\frac {1}{8} \int \frac {1}{\left (1-\frac {2 x^2}{\sqrt {\frac {1}{2} \left (7-i \sqrt {15}\right )}}\right ) \sqrt {1-x^4}} \, dx+\frac {1}{8} \int \frac {1}{\left (1+\frac {2 x^2}{\sqrt {\frac {1}{2} \left (7-i \sqrt {15}\right )}}\right ) \sqrt {1-x^4}} \, dx+\frac {1}{8} \int \frac {1}{\left (1-\frac {2 x^2}{\sqrt {\frac {1}{2} \left (7+i \sqrt {15}\right )}}\right ) \sqrt {1-x^4}} \, dx+\frac {1}{8} \int \frac {1}{\left (1+\frac {2 x^2}{\sqrt {\frac {1}{2} \left (7+i \sqrt {15}\right )}}\right ) \sqrt {1-x^4}} \, dx\\ &=-\frac {1}{8} \left (1-i \sqrt {15}\right ) F\left (\left .\sin ^{-1}(x)\right |-1\right )-\frac {1}{8} \left (1+i \sqrt {15}\right ) F\left (\left .\sin ^{-1}(x)\right |-1\right )+\frac {1}{8} \int \frac {1}{\sqrt {1-x^2} \sqrt {1+x^2} \left (1-\frac {2 x^2}{\sqrt {\frac {1}{2} \left (7-i \sqrt {15}\right )}}\right )} \, dx+\frac {1}{8} \int \frac {1}{\sqrt {1-x^2} \sqrt {1+x^2} \left (1+\frac {2 x^2}{\sqrt {\frac {1}{2} \left (7-i \sqrt {15}\right )}}\right )} \, dx+\frac {1}{8} \int \frac {1}{\sqrt {1-x^2} \sqrt {1+x^2} \left (1-\frac {2 x^2}{\sqrt {\frac {1}{2} \left (7+i \sqrt {15}\right )}}\right )} \, dx+\frac {1}{8} \int \frac {1}{\sqrt {1-x^2} \sqrt {1+x^2} \left (1+\frac {2 x^2}{\sqrt {\frac {1}{2} \left (7+i \sqrt {15}\right )}}\right )} \, dx\\ &=-\frac {1}{8} \left (1-i \sqrt {15}\right ) F\left (\left .\sin ^{-1}(x)\right |-1\right )-\frac {1}{8} \left (1+i \sqrt {15}\right ) F\left (\left .\sin ^{-1}(x)\right |-1\right )+\frac {1}{8} \Pi \left (-\frac {2}{\sqrt {\frac {1}{2} \left (7-i \sqrt {15}\right )}};\left .\sin ^{-1}(x)\right |-1\right )+\frac {1}{8} \Pi \left (\frac {2}{\sqrt {\frac {1}{2} \left (7-i \sqrt {15}\right )}};\left .\sin ^{-1}(x)\right |-1\right )+\frac {1}{8} \Pi \left (-\frac {2}{\sqrt {\frac {1}{2} \left (7+i \sqrt {15}\right )}};\left .\sin ^{-1}(x)\right |-1\right )+\frac {1}{8} \Pi \left (\frac {2}{\sqrt {\frac {1}{2} \left (7+i \sqrt {15}\right )}};\left .\sin ^{-1}(x)\right |-1\right )\\ \end {align*}

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Mathematica [C]  time = 0.63, size = 107, normalized size = 1.65 \begin {gather*} \frac {1}{8} \left (-2 F\left (\left .\sin ^{-1}(x)\right |-1\right )+\Pi \left (\frac {1}{\sqrt {\frac {7}{8}-\frac {i \sqrt {15}}{8}}};\left .\sin ^{-1}(x)\right |-1\right )+\Pi \left (\frac {1}{\sqrt {\frac {7}{8}+\frac {i \sqrt {15}}{8}}};\left .\sin ^{-1}(x)\right |-1\right )+\Pi \left (-\frac {2}{\sqrt {\frac {1}{2} \left (7-i \sqrt {15}\right )}};\left .\sin ^{-1}(x)\right |-1\right )+\Pi \left (-\frac {2}{\sqrt {\frac {1}{2} \left (7+i \sqrt {15}\right )}};\left .\sin ^{-1}(x)\right |-1\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[1 - x^4]*(1 + x^4))/(4 - 7*x^4 + 4*x^8),x]

[Out]

(-2*EllipticF[ArcSin[x], -1] + EllipticPi[1/Sqrt[7/8 - (I/8)*Sqrt[15]], ArcSin[x], -1] + EllipticPi[1/Sqrt[7/8
+ (I/8)*Sqrt[15]], ArcSin[x], -1] + EllipticPi[-2/Sqrt[(7 - I*Sqrt[15])/2], ArcSin[x], -1] + EllipticPi[-2/Sq
rt[(7 + I*Sqrt[15])/2], ArcSin[x], -1])/8

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IntegrateAlgebraic [C]  time = 0.27, size = 57, normalized size = 0.88 \begin {gather*} \left (\frac {1}{8}-\frac {i}{8}\right ) \tan ^{-1}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) x}{\sqrt {1-x^4}}\right )-\left (\frac {1}{8}+\frac {i}{8}\right ) \tan ^{-1}\left (\frac {(1+i) \sqrt {1-x^4}}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(Sqrt[1 - x^4]*(1 + x^4))/(4 - 7*x^4 + 4*x^8),x]

[Out]

(1/8 - I/8)*ArcTan[((1/2 + I/2)*x)/Sqrt[1 - x^4]] - (1/8 + I/8)*ArcTan[((1 + I)*Sqrt[1 - x^4])/x]

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fricas [B]  time = 0.59, size = 130, normalized size = 2.00 \begin {gather*} -\frac {1}{8} \, \arctan \left (-\frac {4 \, x^{8} - 7 \, x^{4} - 4 \, {\left (2 \, x^{5} + x^{3} - 2 \, x\right )} \sqrt {-x^{4} + 1} + 4}{4 \, x^{8} + 8 \, x^{6} - 7 \, x^{4} - 8 \, x^{2} + 4}\right ) + \frac {1}{16} \, \log \left (\frac {4 \, x^{8} - 8 \, x^{6} - 7 \, x^{4} + 8 \, x^{2} - 4 \, {\left (2 \, x^{5} - x^{3} - 2 \, x\right )} \sqrt {-x^{4} + 1} + 4}{4 \, x^{8} - 7 \, x^{4} + 4}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^4+1)^(1/2)*(x^4+1)/(4*x^8-7*x^4+4),x, algorithm="fricas")

[Out]

-1/8*arctan(-(4*x^8 - 7*x^4 - 4*(2*x^5 + x^3 - 2*x)*sqrt(-x^4 + 1) + 4)/(4*x^8 + 8*x^6 - 7*x^4 - 8*x^2 + 4)) +
1/16*log((4*x^8 - 8*x^6 - 7*x^4 + 8*x^2 - 4*(2*x^5 - x^3 - 2*x)*sqrt(-x^4 + 1) + 4)/(4*x^8 - 7*x^4 + 4))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{4} + 1\right )} \sqrt {-x^{4} + 1}}{4 \, x^{8} - 7 \, x^{4} + 4}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^4+1)^(1/2)*(x^4+1)/(4*x^8-7*x^4+4),x, algorithm="giac")

[Out]

integrate((x^4 + 1)*sqrt(-x^4 + 1)/(4*x^8 - 7*x^4 + 4), x)

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maple [B]  time = 1.12, size = 116, normalized size = 1.78

method	result	size
default	\(\frac {\left (-\frac {\sqrt {2}\, \arctan \left (\frac {2 \sqrt {-x^{4}+1}}{x}+1\right )}{8}-\frac {\sqrt {2}\, \arctan \left (\frac {2 \sqrt {-x^{4}+1}}{x}-1\right )}{8}-\frac {\sqrt {2}\, \ln \left (\frac {\frac {-x^{4}+1}{2 x^{2}}-\frac {\sqrt {-x^{4}+1}}{2 x}+\frac {1}{4}}{\frac {-x^{4}+1}{2 x^{2}}+\frac {\sqrt {-x^{4}+1}}{2 x}+\frac {1}{4}}\right )}{16}\right ) \sqrt {2}}{2}\)	\(116\)
elliptic	\(\frac {\left (-\frac {\sqrt {2}\, \arctan \left (\frac {2 \sqrt {-x^{4}+1}}{x}+1\right )}{8}-\frac {\sqrt {2}\, \arctan \left (\frac {2 \sqrt {-x^{4}+1}}{x}-1\right )}{8}-\frac {\sqrt {2}\, \ln \left (\frac {\frac {-x^{4}+1}{2 x^{2}}-\frac {\sqrt {-x^{4}+1}}{2 x}+\frac {1}{4}}{\frac {-x^{4}+1}{2 x^{2}}+\frac {\sqrt {-x^{4}+1}}{2 x}+\frac {1}{4}}\right )}{16}\right ) \sqrt {2}}{2}\)	\(116\)
trager	\(\RootOf \left (128 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right ) \ln \left (-\frac {-8 \RootOf \left (128 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right ) x^{4}-64 x^{2} \RootOf \left (128 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right )^{2}+4 x^{2} \RootOf \left (128 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right )+\sqrt {-x^{4}+1}\, x +8 \RootOf \left (128 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right )}{-2 x^{4}+16 x^{2} \RootOf \left (128 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right )-x^{2}+2}\right )+\frac {\ln \left (-\frac {16 \RootOf \left (128 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right ) x^{4}-128 x^{2} \RootOf \left (128 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right )^{2}-2 x^{4}+24 x^{2} \RootOf \left (128 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right )+2 \sqrt {-x^{4}+1}\, x -x^{2}-16 \RootOf \left (128 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right )+2}{2 x^{4}+16 x^{2} \RootOf \left (128 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right )-x^{2}-2}\right )}{8}-\ln \left (-\frac {16 \RootOf \left (128 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right ) x^{4}-128 x^{2} \RootOf \left (128 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right )^{2}-2 x^{4}+24 x^{2} \RootOf \left (128 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right )+2 \sqrt {-x^{4}+1}\, x -x^{2}-16 \RootOf \left (128 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right )+2}{2 x^{4}+16 x^{2} \RootOf \left (128 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right )-x^{2}-2}\right ) \RootOf \left (128 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right )\)	\(377\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x^4+1)^(1/2)*(x^4+1)/(4*x^8-7*x^4+4),x,method=_RETURNVERBOSE)

[Out]

1/2*(-1/8*2^(1/2)*arctan(2*(-x^4+1)^(1/2)/x+1)-1/8*2^(1/2)*arctan(2*(-x^4+1)^(1/2)/x-1)-1/16*2^(1/2)*ln((1/2*(
-x^4+1)/x^2-1/2*(-x^4+1)^(1/2)/x+1/4)/(1/2*(-x^4+1)/x^2+1/2*(-x^4+1)^(1/2)/x+1/4)))*2^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{4} + 1\right )} \sqrt {-x^{4} + 1}}{4 \, x^{8} - 7 \, x^{4} + 4}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^4+1)^(1/2)*(x^4+1)/(4*x^8-7*x^4+4),x, algorithm="maxima")

[Out]

integrate((x^4 + 1)*sqrt(-x^4 + 1)/(4*x^8 - 7*x^4 + 4), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {\sqrt {1-x^4}\,\left (x^4+1\right )}{4\,x^8-7\,x^4+4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((1 - x^4)^(1/2)*(x^4 + 1))/(4*x^8 - 7*x^4 + 4),x)

[Out]

int(((1 - x^4)^(1/2)*(x^4 + 1))/(4*x^8 - 7*x^4 + 4), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {- \left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right )} \left (x^{4} + 1\right )}{4 x^{8} - 7 x^{4} + 4}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x**4+1)**(1/2)*(x**4+1)/(4*x**8-7*x**4+4),x)

[Out]

Integral(sqrt(-(x - 1)*(x + 1)*(x**2 + 1))*(x**4 + 1)/(4*x**8 - 7*x**4 + 4), x)

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