3.9.98 \(\int \frac {1}{\sqrt {1+\frac {2 x}{1+x^2}}} \, dx\) [898]

Optimal. Leaf size=109 \[ \frac {1+x}{\sqrt {1+\frac {2 x}{1+x^2}}}-\frac {(1+x) \sinh ^{-1}(x)}{\sqrt {1+x^2} \sqrt {1+\frac {2 x}{1+x^2}}}-\frac {\sqrt {2} (1+x) \tanh ^{-1}\left (\frac {1-x}{\sqrt {2} \sqrt {1+x^2}}\right )}{\sqrt {1+x^2} \sqrt {1+\frac {2 x}{1+x^2}}} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {6855, 984, 749, 858, 221, 739, 212} \begin {gather*} \frac {x+1}{\sqrt {\frac {2 x}{x^2+1}+1}}-\frac {(x+1) \sinh ^{-1}(x)}{\sqrt {x^2+1} \sqrt {\frac {2 x}{x^2+1}+1}}-\frac {\sqrt {2} (x+1) \tanh ^{-1}\left (\frac {1-x}{\sqrt {2} \sqrt {x^2+1}}\right )}{\sqrt {x^2+1} \sqrt {\frac {2 x}{x^2+1}+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[1 + (2*x)/(1 + x^2)],x]

[Out]

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

Rule 212

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

Rule 221

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 739

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 749

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*((a + c*x^2)^p/(e
*(m + 2*p + 1))), x] + Dist[2*(p/(e*(m + 2*p + 1))), Int[(d + e*x)^m*Simp[a*e - c*d*x, x]*(a + c*x^2)^(p - 1),
x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && NeQ[m + 2*p + 1, 0] && ( !Ration
alQ[m] || LtQ[m, 1]) &&  !ILtQ[m + 2*p, 0] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 858

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rule 984

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_) + (f_.)*(x_)^2)^(q_.), x_Symbol] :> Dist[(a + b*x + c*x^2)^F
racPart[p]/((4*c)^IntPart[p]*(b + 2*c*x)^(2*FracPart[p])), Int[(b + 2*c*x)^(2*p)*(d + f*x^2)^q, x], x] /; Free
Q[{a, b, c, d, f, p, q}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p]

Rule 6855

Int[(u_.)*((a_.) + (b_.)*(v_)^(n_)*(x_)^(m_.))^(p_), x_Symbol] :> Dist[(a + b*x^m*v^n)^FracPart[p]/(v^(n*FracP
art[p])*(b*x^m + a/v^n)^FracPart[p]), Int[u*v^(n*p)*(b*x^m + a/v^n)^p, x], x] /; FreeQ[{a, b, m, p}, x] &&  !I
ntegerQ[p] && ILtQ[n, 0] && BinomialQ[v, x]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {1+\frac {2 x}{1+x^2}}} \, dx &=\frac {\sqrt {1+2 x+x^2} \int \frac {\sqrt {1+x^2}}{\sqrt {1+2 x+x^2}} \, dx}{\sqrt {1+x^2} \sqrt {1+\frac {2 x}{1+x^2}}}\\ &=\frac {(2+2 x) \int \frac {\sqrt {1+x^2}}{2+2 x} \, dx}{\sqrt {1+x^2} \sqrt {1+\frac {2 x}{1+x^2}}}\\ &=\frac {1+x}{\sqrt {1+\frac {2 x}{1+x^2}}}+\frac {(2+2 x) \int \frac {2-2 x}{(2+2 x) \sqrt {1+x^2}} \, dx}{2 \sqrt {1+x^2} \sqrt {1+\frac {2 x}{1+x^2}}}\\ &=\frac {1+x}{\sqrt {1+\frac {2 x}{1+x^2}}}-\frac {(2+2 x) \int \frac {1}{\sqrt {1+x^2}} \, dx}{2 \sqrt {1+x^2} \sqrt {1+\frac {2 x}{1+x^2}}}+\frac {(2 (2+2 x)) \int \frac {1}{(2+2 x) \sqrt {1+x^2}} \, dx}{\sqrt {1+x^2} \sqrt {1+\frac {2 x}{1+x^2}}}\\ &=\frac {1+x}{\sqrt {1+\frac {2 x}{1+x^2}}}-\frac {(1+x) \sinh ^{-1}(x)}{\sqrt {1+x^2} \sqrt {1+\frac {2 x}{1+x^2}}}-\frac {(2 (2+2 x)) \text {Subst}\left (\int \frac {1}{8-x^2} \, dx,x,\frac {2-2 x}{\sqrt {1+x^2}}\right )}{\sqrt {1+x^2} \sqrt {1+\frac {2 x}{1+x^2}}}\\ &=\frac {1+x}{\sqrt {1+\frac {2 x}{1+x^2}}}-\frac {(1+x) \sinh ^{-1}(x)}{\sqrt {1+x^2} \sqrt {1+\frac {2 x}{1+x^2}}}-\frac {\sqrt {2} (1+x) \tanh ^{-1}\left (\frac {1-x}{\sqrt {2} \sqrt {1+x^2}}\right )}{\sqrt {1+x^2} \sqrt {1+\frac {2 x}{1+x^2}}}\\ \end {align*}

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Mathematica [A]
time = 0.11, size = 82, normalized size = 0.75 \begin {gather*} \frac {(1+x) \left (\sqrt {1+x^2}-\tanh ^{-1}\left (\frac {x}{\sqrt {1+x^2}}\right )+2 \sqrt {2} \tanh ^{-1}\left (\frac {1+x-\sqrt {1+x^2}}{\sqrt {2}}\right )\right )}{\sqrt {\frac {(1+x)^2}{1+x^2}} \sqrt {1+x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.16, size = 79, normalized size = 0.72

method	result	size
risch	\(\frac {1+x}{\sqrt {\frac {\left (1+x \right )^{2}}{x^{2}+1}}}+\frac {\left (-\arcsinh \left (x \right )-\sqrt {2}\, \arctanh \left (\frac {\left (2-2 x \right ) \sqrt {2}}{4 \sqrt {\left (1+x \right )^{2}-2 x}}\right )\right ) \left (1+x \right )}{\sqrt {\frac {\left (1+x \right )^{2}}{x^{2}+1}}\, \sqrt {x^{2}+1}}\)	\(79\)
trager	\(\frac {\sqrt {-\frac {-x^{2}-2 x -1}{x^{2}+1}}\, \left (x^{2}+1\right )}{1+x}+\RootOf \left (\textit {\_Z}^{2}-2\right ) \ln \left (\frac {\RootOf \left (\textit {\_Z}^{2}-2\right ) x^{2}+2 \sqrt {-\frac {-x^{2}-2 x -1}{x^{2}+1}}\, x^{2}-\RootOf \left (\textit {\_Z}^{2}-2\right )+2 \sqrt {-\frac {-x^{2}-2 x -1}{x^{2}+1}}}{\left (1+x \right )^{2}}\right )+\ln \left (-\frac {\sqrt {-\frac {-x^{2}-2 x -1}{x^{2}+1}}\, x^{2}-x^{2}+\sqrt {-\frac {-x^{2}-2 x -1}{x^{2}+1}}-x}{1+x}\right )\)	\(179\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.35, size = 142, normalized size = 1.30 \begin {gather*} \frac {\sqrt {2} {\left (x + 1\right )} \log \left (-\frac {x^{2} + \sqrt {2} {\left (x^{2} - 1\right )} + {\left (2 \, x^{2} + \sqrt {2} {\left (x^{2} + 1\right )} + 2\right )} \sqrt {\frac {x^{2} + 2 \, x + 1}{x^{2} + 1}} - 1}{x^{2} + 2 \, x + 1}\right ) + {\left (x + 1\right )} \log \left (-\frac {x^{2} - {\left (x^{2} + 1\right )} \sqrt {\frac {x^{2} + 2 \, x + 1}{x^{2} + 1}} + x}{x + 1}\right ) + {\left (x^{2} + 1\right )} \sqrt {\frac {x^{2} + 2 \, x + 1}{x^{2} + 1}}}{x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {\frac {2 x}{x^{2} + 1} + 1}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 5.72, size = 88, normalized size = 0.81 \begin {gather*} \frac {\sqrt {2} \log \left (\frac {{\left | -2 \, x - 2 \, \sqrt {2} + 2 \, \sqrt {x^{2} + 1} - 2 \right |}}{{\left | -2 \, x + 2 \, \sqrt {2} + 2 \, \sqrt {x^{2} + 1} - 2 \right |}}\right )}{\mathrm {sgn}\left (x + 1\right )} + \frac {\log \left (-x + \sqrt {x^{2} + 1}\right )}{\mathrm {sgn}\left (x + 1\right )} + \frac {\sqrt {x^{2} + 1}}{\mathrm {sgn}\left (x + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{\sqrt {\frac {2\,x}{x^2+1}+1}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((2*x)/(x^2 + 1) + 1)^(1/2),x)

[Out]

int(1/((2*x)/(x^2 + 1) + 1)^(1/2), x)

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