3.255 \(\int \frac {\sqrt {1+x^2}}{2+x^2} \, dx\)

Optimal. Leaf size=27 \[ \sinh ^{-1}(x)-\frac {\tanh ^{-1}\left (\frac {x}{\sqrt {2} \sqrt {x^2+1}}\right )}{\sqrt {2}} \]

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Rubi [A]  time = 0.01, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {402, 215, 377, 206} \[ \sinh ^{-1}(x)-\frac {\tanh ^{-1}\left (\frac {x}{\sqrt {2} \sqrt {x^2+1}}\right )}{\sqrt {2}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 + x^2]/(2 + x^2),x]

[Out]

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

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 215

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

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 402

Int[((a_) + (b_.)*(x_)^2)^(p_.)/((c_) + (d_.)*(x_)^2), x_Symbol] :> Dist[b/d, Int[(a + b*x^2)^(p - 1), x], x]
- Dist[(b*c - a*d)/d, Int[(a + b*x^2)^(p - 1)/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d,
0] && GtQ[p, 0] && (EqQ[p, 1/2] || EqQ[Denominator[p], 4])

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 27, normalized size = 1.00 \[ \sinh ^{-1}(x)-\frac {\tanh ^{-1}\left (\frac {x}{\sqrt {2} \sqrt {x^2+1}}\right )}{\sqrt {2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[1 + x^2]/(2 + x^2),x]

[Out]

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

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IntegrateAlgebraic [B]  time = 0.09, size = 57, normalized size = 2.11 \[ -\log \left (\sqrt {x^2+1}-x\right )-\frac {\tanh ^{-1}\left (\frac {x^2}{\sqrt {2}}-\frac {\sqrt {x^2+1} x}{\sqrt {2}}+\sqrt {2}\right )}{\sqrt {2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.57, size = 67, normalized size = 2.48 \[ \frac {1}{4} \, \sqrt {2} \log \left (\frac {9 \, x^{2} - 2 \, \sqrt {2} {\left (3 \, x^{2} + 2\right )} - 2 \, \sqrt {x^{2} + 1} {\left (3 \, \sqrt {2} x - 4 \, x\right )} + 6}{x^{2} + 2}\right ) - \log \left (-x + \sqrt {x^{2} + 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(2)*log((9*x^2 - 2*sqrt(2)*(3*x^2 + 2) - 2*sqrt(x^2 + 1)*(3*sqrt(2)*x - 4*x) + 6)/(x^2 + 2)) - log(-x
+ sqrt(x^2 + 1))

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giac [B]  time = 0.63, size = 64, normalized size = 2.37 \[ \frac {1}{4} \, \sqrt {2} \log \left (\frac {{\left (x - \sqrt {x^{2} + 1}\right )}^{2} - 2 \, \sqrt {2} + 3}{{\left (x - \sqrt {x^{2} + 1}\right )}^{2} + 2 \, \sqrt {2} + 3}\right ) - \log \left (-x + \sqrt {x^{2} + 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(2)*log(((x - sqrt(x^2 + 1))^2 - 2*sqrt(2) + 3)/((x - sqrt(x^2 + 1))^2 + 2*sqrt(2) + 3)) - log(-x + sq
rt(x^2 + 1))

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maple [A]  time = 0.28, size = 23, normalized size = 0.85

method	result	size
default	\(\arcsinh \relax (x )-\frac {\arctanh \left (\frac {x \sqrt {2}}{2 \sqrt {x^{2}+1}}\right ) \sqrt {2}}{2}\)	\(23\)
trager	\(-\ln \left (x -\sqrt {x^{2}+1}\right )+\frac {\RootOf \left (\textit {\_Z}^{2}-2\right ) \ln \left (-\frac {3 \RootOf \left (\textit {\_Z}^{2}-2\right ) x^{2}-4 x \sqrt {x^{2}+1}+2 \RootOf \left (\textit {\_Z}^{2}-2\right )}{x^{2}+2}\right )}{4}\)	\(64\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\sqrt {x^{2} + 1} x}{x^{2} + 2} + \int \frac {\sqrt {x^{2} + 1} x^{4}}{x^{6} + 5 \, x^{4} + 8 \, x^{2} + 4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(x^2 + 1)*x/(x^2 + 2) + integrate(sqrt(x^2 + 1)*x^4/(x^6 + 5*x^4 + 8*x^2 + 4), x)

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mupad [B]  time = 0.17, size = 77, normalized size = 2.85 \[ \mathrm {asinh}\relax (x)+\frac {\sqrt {2}\,\left (\ln \left (x-\sqrt {2}\,1{}\mathrm {i}\right )-\ln \left (1+\sqrt {2}\,x\,1{}\mathrm {i}+\sqrt {x^2+1}\,1{}\mathrm {i}\right )\right )}{4}-\frac {\sqrt {2}\,\left (\ln \left (x+\sqrt {2}\,1{}\mathrm {i}\right )-\ln \left (1-\sqrt {2}\,x\,1{}\mathrm {i}+\sqrt {x^2+1}\,1{}\mathrm {i}\right )\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2 + 1)^(1/2)/(x^2 + 2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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