Integrand size = 16, antiderivative size = 109 \[ \int \frac {1}{\sqrt {1+\frac {2 x}{1+x^2}}} \, dx=\frac {1+x}{\sqrt {1+\frac {2 x}{1+x^2}}}-\frac {(1+x) \text {arcsinh}(x)}{\sqrt {1+x^2} \sqrt {1+\frac {2 x}{1+x^2}}}-\frac {\sqrt {2} (1+x) \text {arctanh}\left (\frac {1-x}{\sqrt {2} \sqrt {1+x^2}}\right )}{\sqrt {1+x^2} \sqrt {1+\frac {2 x}{1+x^2}}} \] Output:
(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)-2^(1/2)*(1+x)*arctanh(1/2*(1-x)*2^(1/2)/(x^2+1)^(1/2))/(x^2+1)^(1/2 )/(1+2*x/(x^2+1))^(1/2)
Time = 0.19 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.75 \[ \int \frac {1}{\sqrt {1+\frac {2 x}{1+x^2}}} \, dx=\frac {(1+x) \left (\sqrt {1+x^2}+2 \sqrt {2} \text {arctanh}\left (\frac {1+x-\sqrt {1+x^2}}{\sqrt {2}}\right )+\log \left (-x+\sqrt {1+x^2}\right )\right )}{\sqrt {\frac {(1+x)^2}{1+x^2}} \sqrt {1+x^2}} \] Input:
Integrate[1/Sqrt[1 + (2*x)/(1 + x^2)],x]
Output:
((1 + x)*(Sqrt[1 + x^2] + 2*Sqrt[2]*ArcTanh[(1 + x - Sqrt[1 + x^2])/Sqrt[2 ]] + Log[-x + Sqrt[1 + x^2]]))/(Sqrt[(1 + x)^2/(1 + x^2)]*Sqrt[1 + x^2])
Time = 0.44 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.65, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {7274, 1298, 27, 493, 719, 222, 488, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{\sqrt {\frac {2 x}{x^2+1}+1}} \, dx\) |
\(\Big \downarrow \) 7274 |
\(\displaystyle \frac {\sqrt {x^2+2 x+1} \int \frac {\sqrt {x^2+1}}{\sqrt {x^2+2 x+1}}dx}{\sqrt {x^2+1} \sqrt {\frac {2 x}{x^2+1}+1}}\) |
\(\Big \downarrow \) 1298 |
\(\displaystyle \frac {2 (x+1) \int \frac {\sqrt {x^2+1}}{2 (x+1)}dx}{\sqrt {x^2+1} \sqrt {\frac {2 x}{x^2+1}+1}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(x+1) \int \frac {\sqrt {x^2+1}}{x+1}dx}{\sqrt {x^2+1} \sqrt {\frac {2 x}{x^2+1}+1}}\) |
\(\Big \downarrow \) 493 |
\(\displaystyle \frac {(x+1) \left (\int \frac {1-x}{(x+1) \sqrt {x^2+1}}dx+\sqrt {x^2+1}\right )}{\sqrt {x^2+1} \sqrt {\frac {2 x}{x^2+1}+1}}\) |
\(\Big \downarrow \) 719 |
\(\displaystyle \frac {(x+1) \left (-\int \frac {1}{\sqrt {x^2+1}}dx+2 \int \frac {1}{(x+1) \sqrt {x^2+1}}dx+\sqrt {x^2+1}\right )}{\sqrt {x^2+1} \sqrt {\frac {2 x}{x^2+1}+1}}\) |
\(\Big \downarrow \) 222 |
\(\displaystyle \frac {(x+1) \left (2 \int \frac {1}{(x+1) \sqrt {x^2+1}}dx-\text {arcsinh}(x)+\sqrt {x^2+1}\right )}{\sqrt {x^2+1} \sqrt {\frac {2 x}{x^2+1}+1}}\) |
\(\Big \downarrow \) 488 |
\(\displaystyle \frac {(x+1) \left (-2 \int \frac {1}{2-\frac {(1-x)^2}{x^2+1}}d\frac {1-x}{\sqrt {x^2+1}}-\text {arcsinh}(x)+\sqrt {x^2+1}\right )}{\sqrt {x^2+1} \sqrt {\frac {2 x}{x^2+1}+1}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {(x+1) \left (-\text {arcsinh}(x)-\sqrt {2} \text {arctanh}\left (\frac {1-x}{\sqrt {2} \sqrt {x^2+1}}\right )+\sqrt {x^2+1}\right )}{\sqrt {x^2+1} \sqrt {\frac {2 x}{x^2+1}+1}}\) |
Input:
Int[1/Sqrt[1 + (2*x)/(1 + x^2)],x]
Output:
((1 + x)*(Sqrt[1 + x^2] - ArcSinh[x] - Sqrt[2]*ArcTanh[(1 - x)/(Sqrt[2]*Sq rt[1 + x^2])]))/(Sqrt[1 + x^2]*Sqrt[1 + (2*x)/(1 + x^2)])
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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] && (Gt Q[a, 0] || LtQ[b, 0])
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]
Int[1/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> -Subst[ Int[1/(b*c^2 + a*d^2 - x^2), x], x, (a*d - b*c*x)/Sqrt[a + b*x^2]] /; FreeQ [{a, b, c, d}, x]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (c + d*x)^(n + 1)*((a + b*x^2)^p/(d*(n + 2*p + 1))), x] + Simp[2*(p/(d*(n + 2*p + 1))) Int[(c + d*x)^n*(a + b*x^2)^(p - 1)*(a*d - b*c*x), x], x] /; FreeQ[{a, b, c, d, n}, x] && GtQ[p, 0] && NeQ[n + 2*p + 1, 0] && ( !Rationa lQ[n] || LtQ[n, 1]) && !ILtQ[n + 2*p, 0] && IntQuadraticQ[a, 0, b, c, d, n , p, x]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p _.), x_Symbol] :> Simp[g/e Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Simp[(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] && !IGtQ[m, 0]
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_) + (f_.)*(x_)^2)^(q_.), x_ Symbol] :> Simp[(a + b*x + c*x^2)^FracPart[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] /; FreeQ[{a , b, c, d, f, p, q}, x] && EqQ[b^2 - 4*a*c, 0] && !IntegerQ[p]
Int[(u_.)*((a_.) + (b_.)*(v_)^(n_)*(x_)^(m_.))^(p_), x_Symbol] :> Simp[(a + b*x^m*v^n)^FracPart[p]/(v^(n*FracPart[p])*(b*x^m + a/v^n)^FracPart[p]) I nt[u*v^(n*p)*(b*x^m + a/v^n)^p, x], x] /; FreeQ[{a, b, m, p}, x] && !Integ erQ[p] && ILtQ[n, 0] && BinomialQ[v, x]
Time = 0.15 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.72
method | result | size |
risch | \(\frac {1+x}{\sqrt {\frac {\left (1+x \right )^{2}}{x^{2}+1}}}+\frac {\left (-\operatorname {arcsinh}\left (x \right )-\sqrt {2}\, \operatorname {arctanh}\left (\frac {\left (2-2 x \right ) \sqrt {2}}{4 \sqrt {\left (1+x \right )^{2}-2 x}}\right )\right ) \left (1+x \right )}{\sqrt {\frac {\left (1+x \right )^{2}}{x^{2}+1}}\, \sqrt {x^{2}+1}}\) | \(79\) |
trager | \(\frac {\sqrt {-\frac {-x^{2}-2 x -1}{x^{2}+1}}\, \left (x^{2}+1\right )}{1+x}+\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) \ln \left (\frac {2 \sqrt {-\frac {-x^{2}-2 x -1}{x^{2}+1}}\, x^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) x^{2}+2 \sqrt {-\frac {-x^{2}-2 x -1}{x^{2}+1}}-\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right )}{\left (1+x \right )^{2}}\right )+\ln \left (-\frac {\sqrt {-\frac {-x^{2}-2 x -1}{x^{2}+1}}\, x^{2}-x^{2}+\sqrt {-\frac {-x^{2}-2 x -1}{x^{2}+1}}-x}{1+x}\right )\) | \(179\) |
Input:
int(1/(1+2*x/(x^2+1))^(1/2),x,method=_RETURNVERBOSE)
Output:
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)
Time = 0.09 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.30 \[ \int \frac {1}{\sqrt {1+\frac {2 x}{1+x^2}}} \, dx=\frac {\sqrt {2} {\left (x + 1\right )} \log \left (-\frac {x^{2} + \sqrt {2} {\left (x^{2} - 1\right )} + {\left (2 \, x^{2} + \sqrt {2} {\left (x^{2} + 1\right )} + 2\right )} \sqrt {\frac {x^{2} + 2 \, x + 1}{x^{2} + 1}} - 1}{x^{2} + 2 \, x + 1}\right ) + {\left (x + 1\right )} \log \left (-\frac {x^{2} - {\left (x^{2} + 1\right )} \sqrt {\frac {x^{2} + 2 \, x + 1}{x^{2} + 1}} + x}{x + 1}\right ) + {\left (x^{2} + 1\right )} \sqrt {\frac {x^{2} + 2 \, x + 1}{x^{2} + 1}}}{x + 1} \] Input:
integrate(1/(1+2*x/(x^2+1))^(1/2),x, algorithm="fricas")
Output:
(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)
\[ \int \frac {1}{\sqrt {1+\frac {2 x}{1+x^2}}} \, dx=\int \frac {1}{\sqrt {\frac {2 x}{x^{2} + 1} + 1}}\, dx \] Input:
integrate(1/(1+2*x/(x**2+1))**(1/2),x)
Output:
Integral(1/sqrt(2*x/(x**2 + 1) + 1), x)
\[ \int \frac {1}{\sqrt {1+\frac {2 x}{1+x^2}}} \, dx=\int { \frac {1}{\sqrt {\frac {2 \, x}{x^{2} + 1} + 1}} \,d x } \] Input:
integrate(1/(1+2*x/(x^2+1))^(1/2),x, algorithm="maxima")
Output:
integrate(1/sqrt(2*x/(x^2 + 1) + 1), x)
Time = 0.20 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.81 \[ \int \frac {1}{\sqrt {1+\frac {2 x}{1+x^2}}} \, dx=\frac {\sqrt {2} \log \left (\frac {{\left | -2 \, x - 2 \, \sqrt {2} + 2 \, \sqrt {x^{2} + 1} - 2 \right |}}{{\left | -2 \, x + 2 \, \sqrt {2} + 2 \, \sqrt {x^{2} + 1} - 2 \right |}}\right )}{\mathrm {sgn}\left (x + 1\right )} + \frac {\log \left (-x + \sqrt {x^{2} + 1}\right )}{\mathrm {sgn}\left (x + 1\right )} + \frac {\sqrt {x^{2} + 1}}{\mathrm {sgn}\left (x + 1\right )} \] Input:
integrate(1/(1+2*x/(x^2+1))^(1/2),x, algorithm="giac")
Output:
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)
Timed out. \[ \int \frac {1}{\sqrt {1+\frac {2 x}{1+x^2}}} \, dx=\int \frac {1}{\sqrt {\frac {2\,x}{x^2+1}+1}} \,d x \] Input:
int(1/((2*x)/(x^2 + 1) + 1)^(1/2),x)
Output:
int(1/((2*x)/(x^2 + 1) + 1)^(1/2), x)
Time = 0.19 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.47 \[ \int \frac {1}{\sqrt {1+\frac {2 x}{1+x^2}}} \, dx=\sqrt {x^{2}+1}+\sqrt {2}\, \mathrm {log}\left (\sqrt {x^{2}+1}-\sqrt {2}+x +1\right )-\sqrt {2}\, \mathrm {log}\left (\sqrt {x^{2}+1}+\sqrt {2}+x +1\right )-\mathrm {log}\left (\sqrt {x^{2}+1}+x \right ) \] Input:
int(1/(1+2*x/(x^2+1))^(1/2),x)
Output:
sqrt(x**2 + 1) + sqrt(2)*log(sqrt(x**2 + 1) - sqrt(2) + x + 1) - sqrt(2)*l og(sqrt(x**2 + 1) + sqrt(2) + x + 1) - log(sqrt(x**2 + 1) + x)