3.450 \(\int \frac {(-\sqrt {1-x}-\sqrt {1+x}) (\sqrt {1-x}+\sqrt {1+x})}{x^3} \, dx\)

Optimal. Leaf size=33 \[ \frac {\sqrt {1-x^2}}{x^2}+\frac {1}{x^2}-\tanh ^{-1}\left (\sqrt {1-x^2}\right ) \]

[Out]

1/x^2-arctanh((-x^2+1)^(1/2))+(-x^2+1)^(1/2)/x^2

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Rubi [A]  time = 0.22, antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {6688, 6742, 266, 47, 63, 206} \[ \frac {\sqrt {1-x^2}}{x^2}+\frac {1}{x^2}-\tanh ^{-1}\left (\sqrt {1-x^2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 46, normalized size = 1.39 \[ \frac {1}{x^2 \sqrt {1-x^2}}-\frac {1}{\sqrt {1-x^2}}+\frac {1}{x^2}-\tanh ^{-1}\left (\sqrt {1-x^2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.45, size = 43, normalized size = 1.30 \[ \frac {x^{2} \log \left (\frac {\sqrt {x + 1} \sqrt {-x + 1} - 1}{x}\right ) + \sqrt {x + 1} \sqrt {-x + 1} + 1}{x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Warning, choosing root of [1,0,-4,0,%%%{
4,[2]%%%}] at parameters values [-93.616423693]Warning, choosing root of [1,0,-4,0,%%%{4,[2]%%%}] at parameter
s values [-17.8804557086](4*(2*sqrt(x+1)/(-2*sqrt(-x+1)+2*sqrt(2))-1/2*(-2*sqrt(-x+1)+2*sqrt(2))/sqrt(x+1))^3+
16*(2*sqrt(x+1)/(-2*sqrt(-x+1)+2*sqrt(2))-1/2*(-2*sqrt(-x+1)+2*sqrt(2))/sqrt(x+1)))/((2*sqrt(x+1)/(-2*sqrt(-x+
1)+2*sqrt(2))-1/2*(-2*sqrt(-x+1)+2*sqrt(2))/sqrt(x+1))^2-4)^2-ln(abs(2*sqrt(x+1)/(-2*sqrt(-x+1)+2*sqrt(2))+2-1
/2*(-2*sqrt(-x+1)+2*sqrt(2))/sqrt(x+1)))+ln(abs(2*sqrt(x+1)/(-2*sqrt(-x+1)+2*sqrt(2))-2-1/2*(-2*sqrt(-x+1)+2*s
qrt(2))/sqrt(x+1)))+1/x^2

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maple [A]  time = 0.02, size = 57, normalized size = 1.73 \[ \frac {1}{x^{2}}-\frac {\sqrt {x +1}\, \sqrt {-x +1}\, \left (x^{2} \arctanh \left (\frac {1}{\sqrt {-x^{2}+1}}\right )-\sqrt {-x^{2}+1}\right )}{\sqrt {-x^{2}+1}\, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/x^2-(x+1)^(1/2)*(-x+1)^(1/2)*(arctanh(1/(-x^2+1)^(1/2))*x^2-(-x^2+1)^(1/2))/x^2/(-x^2+1)^(1/2)

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maxima [A]  time = 2.02, size = 51, normalized size = 1.55 \[ \sqrt {-x^{2} + 1} + \frac {{\left (-x^{2} + 1\right )}^{\frac {3}{2}}}{x^{2}} + \frac {1}{x^{2}} - \log \left (\frac {2 \, \sqrt {-x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.78, size = 186, normalized size = 5.64 \[ \ln \left (\frac {{\left (\sqrt {1-x}-1\right )}^2}{{\left (\sqrt {x+1}-1\right )}^2}-1\right )-\ln \left (\frac {\sqrt {1-x}-1}{\sqrt {x+1}-1}\right )-\frac {{\left (\sqrt {1-x}-1\right )}^2}{16\,{\left (\sqrt {x+1}-1\right )}^2}+\frac {\frac {{\left (\sqrt {1-x}-1\right )}^2}{8\,{\left (\sqrt {x+1}-1\right )}^2}+\frac {15\,{\left (\sqrt {1-x}-1\right )}^4}{16\,{\left (\sqrt {x+1}-1\right )}^4}-\frac {1}{16}}{\frac {{\left (\sqrt {1-x}-1\right )}^2}{{\left (\sqrt {x+1}-1\right )}^2}-\frac {2\,{\left (\sqrt {1-x}-1\right )}^4}{{\left (\sqrt {x+1}-1\right )}^4}+\frac {{\left (\sqrt {1-x}-1\right )}^6}{{\left (\sqrt {x+1}-1\right )}^6}}+\frac {1}{x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(((1 - x)^(1/2) - 1)^2/((x + 1)^(1/2) - 1)^2 - 1) - log(((1 - x)^(1/2) - 1)/((x + 1)^(1/2) - 1)) - ((1 - x)
^(1/2) - 1)^2/(16*((x + 1)^(1/2) - 1)^2) + (((1 - x)^(1/2) - 1)^2/(8*((x + 1)^(1/2) - 1)^2) + (15*((1 - x)^(1/
2) - 1)^4)/(16*((x + 1)^(1/2) - 1)^4) - 1/16)/(((1 - x)^(1/2) - 1)^2/((x + 1)^(1/2) - 1)^2 - (2*((1 - x)^(1/2)
- 1)^4)/((x + 1)^(1/2) - 1)^4 + ((1 - x)^(1/2) - 1)^6/((x + 1)^(1/2) - 1)^6) + 1/x^2

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {2}{x^{3}}\, dx - \int \frac {2 \sqrt {1 - x} \sqrt {x + 1}}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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