3.6.53 \(\int \frac {-1+x^6}{\sqrt {1+x^4} (1+x^6)} \, dx\)

Optimal. Leaf size=43 \[ -\frac {\tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {x^4+1}}\right )}{3 \sqrt {2}}-\frac {2}{3} \tanh ^{-1}\left (\frac {x}{\sqrt {x^4+1}}\right ) \]

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Rubi [C]  time = 0.68, antiderivative size = 216, normalized size of antiderivative = 5.02, number of steps used = 17, number of rules used = 9, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {6725, 220, 2073, 1211, 1699, 203, 6728, 1217, 1707} \begin {gather*} -\frac {\tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {x^4+1}}\right )}{3 \sqrt {2}}-\frac {2}{3} \tanh ^{-1}\left (\frac {x}{\sqrt {x^4+1}}\right )-\frac {\left (\sqrt {3}+i\right ) \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{3 \left (\sqrt {3}+3 i\right ) \sqrt {x^4+1}}-\frac {\left (-\sqrt {3}+i\right ) \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{3 \left (-\sqrt {3}+3 i\right ) \sqrt {x^4+1}}+\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{3 \sqrt {x^4+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 + x^6)/(Sqrt[1 + x^4]*(1 + x^6)),x]

[Out]

-1/3*ArcTan[(Sqrt[2]*x)/Sqrt[1 + x^4]]/Sqrt[2] - (2*ArcTanh[x/Sqrt[1 + x^4]])/3 + ((1 + x^2)*Sqrt[(1 + x^4)/(1
+ x^2)^2]*EllipticF[2*ArcTan[x], 1/2])/(3*Sqrt[1 + x^4]) - ((I - Sqrt[3])*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^
2]*EllipticF[2*ArcTan[x], 1/2])/(3*(3*I - Sqrt[3])*Sqrt[1 + x^4]) - ((I + Sqrt[3])*(1 + x^2)*Sqrt[(1 + x^4)/(1
+ x^2)^2]*EllipticF[2*ArcTan[x], 1/2])/(3*(3*I + Sqrt[3])*Sqrt[1 + x^4])

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

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 1211

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> Dist[1/(2*d), Int[1/Sqrt[a + c*x^4], x],
x] + Dist[1/(2*d), Int[(d - e*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] && EqQ[c*d^2 - a*e^2, 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 1699

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> Dist[A, Subst[Int[1/
(d + 2*a*e*x^2), x], x, x/Sqrt[a + c*x^4]], x] /; FreeQ[{a, c, d, e, A, B}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ
[c*d^2 - a*e^2, 0] && EqQ[B*d + A*e, 0]

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]

Rule 2073

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P /. x -> Sqrt[x]]}, Int[ExpandIntegrand[(PP /. x ->
x^2)^p*Q^q, x], x] /;  !SumQ[NonfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x^2] && PolyQ[Q, x] && ILtQ[p,
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]

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 {-1+x^6}{\sqrt {1+x^4} \left (1+x^6\right )} \, dx &=\int \left (\frac {1}{\sqrt {1+x^4}}-\frac {2}{\sqrt {1+x^4} \left (1+x^6\right )}\right ) \, dx\\ &=-\left (2 \int \frac {1}{\sqrt {1+x^4} \left (1+x^6\right )} \, dx\right )+\int \frac {1}{\sqrt {1+x^4}} \, dx\\ &=\frac {\left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{2 \sqrt {1+x^4}}-2 \int \left (\frac {1}{3 \left (1+x^2\right ) \sqrt {1+x^4}}+\frac {2-x^2}{3 \sqrt {1+x^4} \left (1-x^2+x^4\right )}\right ) \, dx\\ &=\frac {\left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{2 \sqrt {1+x^4}}-\frac {2}{3} \int \frac {1}{\left (1+x^2\right ) \sqrt {1+x^4}} \, dx-\frac {2}{3} \int \frac {2-x^2}{\sqrt {1+x^4} \left (1-x^2+x^4\right )} \, dx\\ &=\frac {\left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{2 \sqrt {1+x^4}}-\frac {1}{3} \int \frac {1}{\sqrt {1+x^4}} \, dx-\frac {1}{3} \int \frac {1-x^2}{\left (1+x^2\right ) \sqrt {1+x^4}} \, dx-\frac {2}{3} \int \left (\frac {-1-i \sqrt {3}}{\left (-1-i \sqrt {3}+2 x^2\right ) \sqrt {1+x^4}}+\frac {-1+i \sqrt {3}}{\left (-1+i \sqrt {3}+2 x^2\right ) \sqrt {1+x^4}}\right ) \, dx\\ &=\frac {\left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{3 \sqrt {1+x^4}}-\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{1+2 x^2} \, dx,x,\frac {x}{\sqrt {1+x^4}}\right )+\frac {1}{3} \left (2 \left (1-i \sqrt {3}\right )\right ) \int \frac {1}{\left (-1+i \sqrt {3}+2 x^2\right ) \sqrt {1+x^4}} \, dx+\frac {1}{3} \left (2 \left (1+i \sqrt {3}\right )\right ) \int \frac {1}{\left (-1-i \sqrt {3}+2 x^2\right ) \sqrt {1+x^4}} \, dx\\ &=-\frac {\tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{3 \sqrt {2}}+\frac {\left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{3 \sqrt {1+x^4}}-\frac {\left (2 \left (i-\sqrt {3}\right )\right ) \int \frac {1}{\sqrt {1+x^4}} \, dx}{3 \left (3 i-\sqrt {3}\right )}+\frac {\left (4 \left (i-\sqrt {3}\right )\right ) \int \frac {1+x^2}{\left (-1-i \sqrt {3}+2 x^2\right ) \sqrt {1+x^4}} \, dx}{3 \left (3 i-\sqrt {3}\right )}-\frac {\left (2 \left (i+\sqrt {3}\right )\right ) \int \frac {1}{\sqrt {1+x^4}} \, dx}{3 \left (3 i+\sqrt {3}\right )}+\frac {\left (4 \left (i+\sqrt {3}\right )\right ) \int \frac {1+x^2}{\left (-1+i \sqrt {3}+2 x^2\right ) \sqrt {1+x^4}} \, dx}{3 \left (3 i+\sqrt {3}\right )}\\ &=-\frac {\tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{3 \sqrt {2}}-\frac {2}{3} \tanh ^{-1}\left (\frac {x}{\sqrt {1+x^4}}\right )+\frac {\left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{3 \sqrt {1+x^4}}-\frac {\left (i-\sqrt {3}\right ) \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{3 \left (3 i-\sqrt {3}\right ) \sqrt {1+x^4}}-\frac {\left (i+\sqrt {3}\right ) \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{3 \left (3 i+\sqrt {3}\right ) \sqrt {1+x^4}}-\frac {\left (i+\sqrt {3}\right ) \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} \Pi \left (\frac {3}{4};2 \tan ^{-1}(x)|\frac {1}{2}\right )}{6 \left (3 i-\sqrt {3}\right ) \sqrt {1+x^4}}-\frac {\left (i-\sqrt {3}\right ) \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} \Pi \left (\frac {3}{4};2 \tan ^{-1}(x)|\frac {1}{2}\right )}{6 \left (3 i+\sqrt {3}\right ) \sqrt {1+x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.35, size = 125, normalized size = 2.91 \begin {gather*} \frac {(-1)^{5/12} \left (2 \left (\Pi \left (-i;\left .i \sinh ^{-1}\left (\frac {(1+i) x}{\sqrt {2}}\right )\right |-1\right )+\Pi \left (\frac {i}{2}-\frac {\sqrt {3}}{2};\left .i \sinh ^{-1}\left (\frac {(1+i) x}{\sqrt {2}}\right )\right |-1\right )+\Pi \left (\frac {1}{2} \left (i+\sqrt {3}\right );\left .i \sinh ^{-1}\left (\frac {(1+i) x}{\sqrt {2}}\right )\right |-1\right )\right )-3 F\left (\left .i \sinh ^{-1}\left (\frac {(1+i) x}{\sqrt {2}}\right )\right |-1\right )\right )}{\sqrt {3} \left (1+\sqrt [3]{-1}\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(-1 + x^6)/(Sqrt[1 + x^4]*(1 + x^6)),x]

[Out]

((-1)^(5/12)*(-3*EllipticF[I*ArcSinh[((1 + I)*x)/Sqrt[2]], -1] + 2*(EllipticPi[-I, I*ArcSinh[((1 + I)*x)/Sqrt[
2]], -1] + EllipticPi[I/2 - Sqrt[3]/2, I*ArcSinh[((1 + I)*x)/Sqrt[2]], -1] + EllipticPi[(I + Sqrt[3])/2, I*Arc
Sinh[((1 + I)*x)/Sqrt[2]], -1])))/(Sqrt[3]*(1 + (-1)^(1/3)))

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IntegrateAlgebraic [A]  time = 0.70, size = 43, normalized size = 1.00 \begin {gather*} -\frac {\tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{3 \sqrt {2}}-\frac {2}{3} \tanh ^{-1}\left (\frac {x}{\sqrt {1+x^4}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(-1 + x^6)/(Sqrt[1 + x^4]*(1 + x^6)),x]

[Out]

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

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fricas [A]  time = 0.51, size = 53, normalized size = 1.23 \begin {gather*} -\frac {1}{6} \, \sqrt {2} \arctan \left (\frac {\sqrt {2} x}{\sqrt {x^{4} + 1}}\right ) + \frac {1}{3} \, \log \left (\frac {x^{4} + x^{2} - 2 \, \sqrt {x^{4} + 1} x + 1}{x^{4} - x^{2} + 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^6-1)/(x^4+1)^(1/2)/(x^6+1),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{6} - 1}{{\left (x^{6} + 1\right )} \sqrt {x^{4} + 1}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^6-1)/(x^4+1)^(1/2)/(x^6+1),x, algorithm="giac")

[Out]

integrate((x^6 - 1)/((x^6 + 1)*sqrt(x^4 + 1)), x)

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maple [A]  time = 0.34, size = 42, normalized size = 0.98

method	result	size
elliptic	\(\frac {\left (\frac {\arctan \left (\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{2 x}\right )}{3}-\frac {2 \sqrt {2}\, \arctanh \left (\frac {\sqrt {x^{4}+1}}{x}\right )}{3}\right ) \sqrt {2}}{2}\)	\(42\)
trager	\(\frac {\RootOf \left (\textit {\_Z}^{2}+2\right ) \ln \left (\frac {\RootOf \left (\textit {\_Z}^{2}+2\right ) x +\sqrt {x^{4}+1}}{x^{2}+1}\right )}{6}+\frac {\ln \left (-\frac {-x^{4}+2 \sqrt {x^{4}+1}\, x -x^{2}-1}{x^{4}-x^{2}+1}\right )}{3}\)	\(74\)
default	\(\frac {\sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticF \left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )}{\left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}+\frac {\left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (\textit {\_Z}^{4}-\textit {\_Z}^{2}+1\right )}{\sum }\underline {\hspace {1.25 ex}}\alpha \left (-\frac {\arctanh \left (\frac {\underline {\hspace {1.25 ex}}\alpha ^{2} \left (-\underline {\hspace {1.25 ex}}\alpha ^{2}+x^{2}+1\right )}{\sqrt {\underline {\hspace {1.25 ex}}\alpha ^{2}}\, \sqrt {x^{4}+1}}\right )}{\sqrt {\underline {\hspace {1.25 ex}}\alpha ^{2}}}+\frac {2 \left (-1\right )^{\frac {3}{4}} \left (-\underline {\hspace {1.25 ex}}\alpha ^{3}+\underline {\hspace {1.25 ex}}\alpha \right ) \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticPi \left (\left (-1\right )^{\frac {1}{4}} x , i \underline {\hspace {1.25 ex}}\alpha ^{2}-i, i\right )}{\sqrt {x^{4}+1}}\right )\right )}{6}+\frac {2 \left (-1\right )^{\frac {3}{4}} \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticPi \left (\left (-1\right )^{\frac {1}{4}} x , i, -\sqrt {-i}\, \left (-1\right )^{\frac {3}{4}}\right )}{3 \sqrt {x^{4}+1}}\)	\(220\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^6-1)/(x^4+1)^(1/2)/(x^6+1),x,method=_RETURNVERBOSE)

[Out]

1/2*(1/3*arctan(1/2*2^(1/2)/x*(x^4+1)^(1/2))-2/3*2^(1/2)*arctanh((x^4+1)^(1/2)/x))*2^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{6} - 1}{{\left (x^{6} + 1\right )} \sqrt {x^{4} + 1}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^6-1)/(x^4+1)^(1/2)/(x^6+1),x, algorithm="maxima")

[Out]

integrate((x^6 - 1)/((x^6 + 1)*sqrt(x^4 + 1)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {x^6-1}{\sqrt {x^4+1}\,\left (x^6+1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^6 - 1)/((x^4 + 1)^(1/2)*(x^6 + 1)),x)

[Out]

int((x^6 - 1)/((x^4 + 1)^(1/2)*(x^6 + 1)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**6-1)/(x**4+1)**(1/2)/(x**6+1),x)

[Out]

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

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