3.242 \(\int \frac {1}{\sqrt {2+x^2} (-1+x^4)} \, dx\)

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

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Rubi [A]  time = 0.02, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {1175, 377, 206, 203} \[ -\frac {1}{2} \tan ^{-1}\left (\frac {x}{\sqrt {x^2+2}}\right )-\frac {\tanh ^{-1}\left (\frac {\sqrt {3} x}{\sqrt {x^2+2}}\right )}{2 \sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[2 + x^2]*(-1 + x^4)),x]

[Out]

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

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 206

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

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 1175

Int[((d_) + (e_.)*(x_)^2)^(q_)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{r = Rt[-(a*c), 2]}, -Dist[c/(2*r), In
t[(d + e*x^2)^q/(r - c*x^2), x], x] - Dist[c/(2*r), Int[(d + e*x^2)^q/(r + c*x^2), x], x]] /; FreeQ[{a, c, d,
e, q}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[q]

Rubi steps

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

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Mathematica [C]  time = 0.05, size = 96, normalized size = 2.23 \[ \frac {1}{12} \left (-3 \tan ^{-1}\left (\frac {-x+2 i}{\sqrt {x^2+2}}\right )+3 \tan ^{-1}\left (\frac {x+2 i}{\sqrt {x^2+2}}\right )+\sqrt {3} \tanh ^{-1}\left (\frac {2-x}{\sqrt {3} \sqrt {x^2+2}}\right )-\sqrt {3} \tanh ^{-1}\left (\frac {x+2}{\sqrt {3} \sqrt {x^2+2}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[2 + x^2]*(-1 + x^4)),x]

[Out]

(-3*ArcTan[(2*I - x)/Sqrt[2 + x^2]] + 3*ArcTan[(2*I + x)/Sqrt[2 + x^2]] + Sqrt[3]*ArcTanh[(2 - x)/(Sqrt[3]*Sqr
t[2 + x^2])] - Sqrt[3]*ArcTanh[(2 + x)/(Sqrt[3]*Sqrt[2 + x^2])])/12

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IntegrateAlgebraic [A]  time = 0.09, size = 65, normalized size = 1.51 \[ \frac {1}{2} \tan ^{-1}\left (x^2-\sqrt {x^2+2} x+1\right )-\frac {\tanh ^{-1}\left (-\frac {x^2}{\sqrt {3}}+\frac {\sqrt {x^2+2} x}{\sqrt {3}}+\frac {1}{\sqrt {3}}\right )}{2 \sqrt {3}} \]

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(Sqrt[2 + x^2]*(-1 + x^4)),x]

[Out]

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

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fricas [B]  time = 0.99, size = 72, normalized size = 1.67 \[ \frac {1}{12} \, \sqrt {3} \log \left (\frac {4 \, x^{2} - \sqrt {3} {\left (2 \, x^{2} + 1\right )} - \sqrt {x^{2} + 2} {\left (2 \, \sqrt {3} x - 3 \, x\right )} + 2}{x^{2} - 1}\right ) - \frac {1}{2} \, \arctan \left (-x^{2} + \sqrt {x^{2} + 2} x - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*sqrt(3)*log((4*x^2 - sqrt(3)*(2*x^2 + 1) - sqrt(x^2 + 2)*(2*sqrt(3)*x - 3*x) + 2)/(x^2 - 1)) - 1/2*arctan
(-x^2 + sqrt(x^2 + 2)*x - 1)

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giac [B]  time = 0.65, size = 74, normalized size = 1.72 \[ -\frac {1}{12} \, \sqrt {3} \log \left (\frac {{\left | 2 \, {\left (x - \sqrt {x^{2} + 2}\right )}^{2} - 4 \, \sqrt {3} - 8 \right |}}{{\left | 2 \, {\left (x - \sqrt {x^{2} + 2}\right )}^{2} + 4 \, \sqrt {3} - 8 \right |}}\right ) + \frac {1}{2} \, \arctan \left (\frac {1}{2} \, {\left (x - \sqrt {x^{2} + 2}\right )}^{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*sqrt(3)*log(abs(2*(x - sqrt(x^2 + 2))^2 - 4*sqrt(3) - 8)/abs(2*(x - sqrt(x^2 + 2))^2 + 4*sqrt(3) - 8)) +
1/2*arctan(1/2*(x - sqrt(x^2 + 2))^2)

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maple [B]  time = 0.28, size = 70, normalized size = 1.63

method	result	size
default	\(-\frac {\arctan \left (\frac {x}{\sqrt {x^{2}+2}}\right )}{2}-\frac {\sqrt {3}\, \arctanh \left (\frac {\left (4+2 x \right ) \sqrt {3}}{6 \sqrt {\left (-1+x \right )^{2}+1+2 x}}\right )}{12}+\frac {\sqrt {3}\, \arctanh \left (\frac {\left (4-2 x \right ) \sqrt {3}}{6 \sqrt {\left (1+x \right )^{2}+1-2 x}}\right )}{12}\)	\(70\)
trager	\(\frac {\RootOf \left (\textit {\_Z}^{2}-3\right ) \ln \left (-\frac {-2 \RootOf \left (\textit {\_Z}^{2}-3\right ) x^{2}+3 \sqrt {x^{2}+2}\, x -\RootOf \left (\textit {\_Z}^{2}-3\right )}{\left (1+x \right ) \left (-1+x \right )}\right )}{12}-\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {\sqrt {x^{2}+2}\, x +\RootOf \left (\textit {\_Z}^{2}+1\right )}{x^{2}+1}\right )}{4}\)	\(86\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*arctan(x/(x^2+2)^(1/2))-1/12*3^(1/2)*arctanh(1/6*(4+2*x)*3^(1/2)/((-1+x)^2+1+2*x)^(1/2))+1/12*3^(1/2)*arc
tanh(1/6*(4-2*x)*3^(1/2)/((1+x)^2+1-2*x)^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (x^{4} - 1\right )} \sqrt {x^{2} + 2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.11, size = 107, normalized size = 2.49 \[ \frac {\sqrt {3}\,\left (\ln \left (x-1\right )-\ln \left (x+\sqrt {3}\,\sqrt {x^2+2}+2\right )\right )}{12}-\frac {\sqrt {3}\,\left (\ln \left (x+1\right )-\ln \left (\sqrt {3}\,\sqrt {x^2+2}-x+2\right )\right )}{12}+\frac {\ln \left (\sqrt {x^2+2}+2-x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4}-\frac {\ln \left (\sqrt {x^2+2}+2+x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4}+\frac {\ln \left (x-\mathrm {i}\right )\,1{}\mathrm {i}}{4}-\frac {\ln \left (x+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((x^2 + 2)^(1/2)*(x^4 - 1)),x)

[Out]

(log((x^2 + 2)^(1/2) - x*1i + 2)*1i)/4 - (log(x*1i + (x^2 + 2)^(1/2) + 2)*1i)/4 + (log(x - 1i)*1i)/4 - (log(x
+ 1i)*1i)/4 + (3^(1/2)*(log(x - 1) - log(x + 3^(1/2)*(x^2 + 2)^(1/2) + 2)))/12 - (3^(1/2)*(log(x + 1) - log(3^
(1/2)*(x^2 + 2)^(1/2) - x + 2)))/12

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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