3.9.52 \(\int \frac {1+x^2}{(-1+x^2) (1+2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=65 \[ -\frac {x}{3 \sqrt {2 x^2+1}}-\frac {2 \tanh ^{-1}\left (-\sqrt {\frac {2}{3}} x^2+\frac {\sqrt {2 x^2+1} x}{\sqrt {3}}+\sqrt {\frac {2}{3}}\right )}{3 \sqrt {3}} \]

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Rubi [A]  time = 0.02, antiderivative size = 45, normalized size of antiderivative = 0.69, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {527, 12, 377, 207} \begin {gather*} -\frac {x}{3 \sqrt {2 x^2+1}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {3} x}{\sqrt {2 x^2+1}}\right )}{3 \sqrt {3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + x^2)/((-1 + x^2)*(1 + 2*x^2)^(3/2)),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[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 527

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> -Simp[
((b*e - a*f)*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d
)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)
*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

Rubi steps

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

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Mathematica [C]  time = 1.31, size = 159, normalized size = 2.45 \begin {gather*} \frac {x \left (\frac {72 \left (2 x^2+1\right ) x^2 \, _2F_1\left (2,2;\frac {7}{2};\frac {3 x^2}{x^2-1}\right )}{\left (x^2-1\right )^2}+\frac {10 \left (4 x^2+3\right ) \left (\sqrt {\frac {6 x^2+3}{1-x^2}} x^2+\sqrt {\frac {x^2}{x^2-1}} \left (2 x^2+1\right ) \sin ^{-1}\left (\sqrt {3} \sqrt {\frac {x^2}{x^2-1}}\right )\right )}{\sqrt {\frac {2 x^2+1}{3-3 x^2}} x^4}+45\right )}{45 \sqrt {2 x^2+1}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(1 + x^2)/((-1 + x^2)*(1 + 2*x^2)^(3/2)),x]

[Out]

(x*(45 + (10*(3 + 4*x^2)*(x^2*Sqrt[(3 + 6*x^2)/(1 - x^2)] + Sqrt[x^2/(-1 + x^2)]*(1 + 2*x^2)*ArcSin[Sqrt[3]*Sq
rt[x^2/(-1 + x^2)]]))/(x^4*Sqrt[(1 + 2*x^2)/(3 - 3*x^2)]) + (72*x^2*(1 + 2*x^2)*Hypergeometric2F1[2, 2, 7/2, (
3*x^2)/(-1 + x^2)])/(-1 + x^2)^2))/(45*Sqrt[1 + 2*x^2])

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IntegrateAlgebraic [A]  time = 0.17, size = 65, normalized size = 1.00 \begin {gather*} -\frac {x}{3 \sqrt {1+2 x^2}}-\frac {2 \tanh ^{-1}\left (\sqrt {\frac {2}{3}}-\sqrt {\frac {2}{3}} x^2+\frac {x \sqrt {1+2 x^2}}{\sqrt {3}}\right )}{3 \sqrt {3}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(1 + x^2)/((-1 + x^2)*(1 + 2*x^2)^(3/2)),x]

[Out]

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

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fricas [A]  time = 0.44, size = 66, normalized size = 1.02 \begin {gather*} \frac {\sqrt {3} {\left (2 \, x^{2} + 1\right )} \log \left (\frac {2 \, \sqrt {3} \sqrt {2 \, x^{2} + 1} x - 5 \, x^{2} - 1}{x^{2} - 1}\right ) - 3 \, \sqrt {2 \, x^{2} + 1} x}{9 \, {\left (2 \, x^{2} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+1)/(x^2-1)/(2*x^2+1)^(3/2),x, algorithm="fricas")

[Out]

1/9*(sqrt(3)*(2*x^2 + 1)*log((2*sqrt(3)*sqrt(2*x^2 + 1)*x - 5*x^2 - 1)/(x^2 - 1)) - 3*sqrt(2*x^2 + 1)*x)/(2*x^
2 + 1)

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giac [A]  time = 0.35, size = 83, normalized size = 1.28 \begin {gather*} -\frac {1}{18} \, \sqrt {6} \sqrt {2} \log \left (\frac {{\left | 2 \, {\left (\sqrt {2} x - \sqrt {2 \, x^{2} + 1}\right )}^{2} - 4 \, \sqrt {6} - 10 \right |}}{{\left | 2 \, {\left (\sqrt {2} x - \sqrt {2 \, x^{2} + 1}\right )}^{2} + 4 \, \sqrt {6} - 10 \right |}}\right ) - \frac {x}{3 \, \sqrt {2 \, x^{2} + 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+1)/(x^2-1)/(2*x^2+1)^(3/2),x, algorithm="giac")

[Out]

-1/18*sqrt(6)*sqrt(2)*log(abs(2*(sqrt(2)*x - sqrt(2*x^2 + 1))^2 - 4*sqrt(6) - 10)/abs(2*(sqrt(2)*x - sqrt(2*x^
2 + 1))^2 + 4*sqrt(6) - 10)) - 1/3*x/sqrt(2*x^2 + 1)

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maple [C]  time = 0.18, size = 67, normalized size = 1.03

method	result	size
trager	\(-\frac {x}{3 \sqrt {2 x^{2}+1}}+\frac {\RootOf \left (\textit {\_Z}^{2}-3\right ) \ln \left (-\frac {-5 \RootOf \left (\textit {\_Z}^{2}-3\right ) x^{2}+6 \sqrt {2 x^{2}+1}\, x -\RootOf \left (\textit {\_Z}^{2}-3\right )}{\left (-1+x \right ) \left (1+x \right )}\right )}{9}\)	\(67\)
risch	\(-\frac {x}{3 \sqrt {2 x^{2}+1}}-\frac {\sqrt {3}\, \arctanh \left (\frac {\left (2+4 x \right ) \sqrt {3}}{6 \sqrt {2 \left (-1+x \right )^{2}-1+4 x}}\right )}{9}+\frac {\sqrt {3}\, \arctanh \left (\frac {\left (2-4 x \right ) \sqrt {3}}{6 \sqrt {2 \left (1+x \right )^{2}-1-4 x}}\right )}{9}\)	\(74\)
default	\(\frac {x}{\sqrt {2 x^{2}+1}}-\frac {1}{3 \sqrt {2 \left (1+x \right )^{2}-1-4 x}}-\frac {2 x}{3 \sqrt {2 \left (1+x \right )^{2}-1-4 x}}+\frac {\sqrt {3}\, \arctanh \left (\frac {\left (2-4 x \right ) \sqrt {3}}{6 \sqrt {2 \left (1+x \right )^{2}-1-4 x}}\right )}{9}+\frac {1}{3 \sqrt {2 \left (-1+x \right )^{2}-1+4 x}}-\frac {2 x}{3 \sqrt {2 \left (-1+x \right )^{2}-1+4 x}}-\frac {\sqrt {3}\, \arctanh \left (\frac {\left (2+4 x \right ) \sqrt {3}}{6 \sqrt {2 \left (-1+x \right )^{2}-1+4 x}}\right )}{9}\)	\(139\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2+1)/(x^2-1)/(2*x^2+1)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/3*x/(2*x^2+1)^(1/2)+1/9*RootOf(_Z^2-3)*ln(-(-5*RootOf(_Z^2-3)*x^2+6*(2*x^2+1)^(1/2)*x-RootOf(_Z^2-3))/(-1+x
)/(1+x))

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maxima [A]  time = 0.42, size = 80, normalized size = 1.23 \begin {gather*} -\frac {1}{9} \, \sqrt {3} \operatorname {arsinh}\left (\frac {2 \, \sqrt {2} x}{{\left | 2 \, x + 2 \right |}} - \frac {\sqrt {2}}{{\left | 2 \, x + 2 \right |}}\right ) - \frac {1}{9} \, \sqrt {3} \operatorname {arsinh}\left (\frac {2 \, \sqrt {2} x}{{\left | 2 \, x - 2 \right |}} + \frac {\sqrt {2}}{{\left | 2 \, x - 2 \right |}}\right ) - \frac {x}{3 \, \sqrt {2 \, x^{2} + 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+1)/(x^2-1)/(2*x^2+1)^(3/2),x, algorithm="maxima")

[Out]

-1/9*sqrt(3)*arcsinh(2*sqrt(2)*x/abs(2*x + 2) - sqrt(2)/abs(2*x + 2)) - 1/9*sqrt(3)*arcsinh(2*sqrt(2)*x/abs(2*
x - 2) + sqrt(2)/abs(2*x - 2)) - 1/3*x/sqrt(2*x^2 + 1)

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mupad [B]  time = 0.83, size = 111, normalized size = 1.71 \begin {gather*} \frac {\sqrt {3}\,\left (\ln \left (x-1\right )-\ln \left (x+\frac {\sqrt {2}\,\sqrt {3}\,\sqrt {x^2+\frac {1}{2}}}{2}+\frac {1}{2}\right )\right )}{9}-\frac {\sqrt {3}\,\left (\ln \left (x+1\right )-\ln \left (x-\frac {\sqrt {2}\,\sqrt {3}\,\sqrt {x^2+\frac {1}{2}}}{2}-\frac {1}{2}\right )\right )}{9}-\frac {\sqrt {2}\,\sqrt {x^2+\frac {1}{2}}}{12\,\left (x-\frac {\sqrt {2}\,1{}\mathrm {i}}{2}\right )}-\frac {\sqrt {2}\,\sqrt {x^2+\frac {1}{2}}}{12\,\left (x+\frac {\sqrt {2}\,1{}\mathrm {i}}{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2 + 1)/((x^2 - 1)*(2*x^2 + 1)^(3/2)),x)

[Out]

(3^(1/2)*(log(x - 1) - log(x + (2^(1/2)*3^(1/2)*(x^2 + 1/2)^(1/2))/2 + 1/2)))/9 - (3^(1/2)*(log(x + 1) - log(x
- (2^(1/2)*3^(1/2)*(x^2 + 1/2)^(1/2))/2 - 1/2)))/9 - (2^(1/2)*(x^2 + 1/2)^(1/2))/(12*(x - (2^(1/2)*1i)/2)) -
(2^(1/2)*(x^2 + 1/2)^(1/2))/(12*(x + (2^(1/2)*1i)/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2} + 1}{\left (x - 1\right ) \left (x + 1\right ) \left (2 x^{2} + 1\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2+1)/(x**2-1)/(2*x**2+1)**(3/2),x)

[Out]

Integral((x**2 + 1)/((x - 1)*(x + 1)*(2*x**2 + 1)**(3/2)), x)

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