Integrand size = 16, antiderivative size = 90 \[ \int \left (1+\frac {2 x}{1+x^2}\right )^{3/2} \, dx=-\left ((1-x) (1+x) \sqrt {1+\frac {2 x}{1+x^2}}\right )-\frac {x \left (1+x^2\right ) \sqrt {1+\frac {2 x}{1+x^2}}}{1+x}+\frac {3 \sqrt {1+x^2} \sqrt {1+\frac {2 x}{1+x^2}} \text {arcsinh}(x)}{1+x} \]
-(1-x)*(1+x)*(1+2*x/(x^2+1))^(1/2)-x*(x^2+1)*(1+2*x/(x^2+1))^(1/2)/(1+x)+3 *arcsinh(x)*(x^2+1)^(1/2)*(1+2*x/(x^2+1))^(1/2)/(1+x)
Time = 0.14 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.62 \[ \int \left (1+\frac {2 x}{1+x^2}\right )^{3/2} \, dx=\frac {\sqrt {\frac {(1+x)^2}{1+x^2}} \left (-1-2 x+x^2-3 \sqrt {1+x^2} \log \left (-x+\sqrt {1+x^2}\right )\right )}{1+x} \]
(Sqrt[(1 + x)^2/(1 + x^2)]*(-1 - 2*x + x^2 - 3*Sqrt[1 + x^2]*Log[-x + Sqrt [1 + x^2]]))/(1 + x)
Time = 0.26 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.84, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {7274, 1298, 27, 495, 27, 643, 299, 222}
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.
\(\displaystyle \int \left (\frac {2 x}{x^2+1}+1\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 7274 |
\(\displaystyle \frac {\sqrt {x^2+1} \sqrt {\frac {2 x}{x^2+1}+1} \int \frac {\left (x^2+2 x+1\right )^{3/2}}{\left (x^2+1\right )^{3/2}}dx}{\sqrt {x^2+2 x+1}}\) |
\(\Big \downarrow \) 1298 |
\(\displaystyle \frac {\sqrt {x^2+1} \sqrt {\frac {2 x}{x^2+1}+1} \int \frac {8 (x+1)^3}{\left (x^2+1\right )^{3/2}}dx}{8 (x+1)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {x^2+1} \sqrt {\frac {2 x}{x^2+1}+1} \int \frac {(x+1)^3}{\left (x^2+1\right )^{3/2}}dx}{x+1}\) |
\(\Big \downarrow \) 495 |
\(\displaystyle \frac {\sqrt {x^2+1} \sqrt {\frac {2 x}{x^2+1}+1} \left (\int \frac {2 (1-x) (x+1)}{\sqrt {x^2+1}}dx-\frac {(1-x) (x+1)^2}{\sqrt {x^2+1}}\right )}{x+1}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {x^2+1} \sqrt {\frac {2 x}{x^2+1}+1} \left (2 \int \frac {(1-x) (x+1)}{\sqrt {x^2+1}}dx-\frac {(1-x) (x+1)^2}{\sqrt {x^2+1}}\right )}{x+1}\) |
\(\Big \downarrow \) 643 |
\(\displaystyle \frac {\sqrt {x^2+1} \sqrt {\frac {2 x}{x^2+1}+1} \left (2 \int \frac {1-x^2}{\sqrt {x^2+1}}dx-\frac {(1-x) (x+1)^2}{\sqrt {x^2+1}}\right )}{x+1}\) |
\(\Big \downarrow \) 299 |
\(\displaystyle \frac {\sqrt {x^2+1} \sqrt {\frac {2 x}{x^2+1}+1} \left (2 \left (\frac {3}{2} \int \frac {1}{\sqrt {x^2+1}}dx-\frac {1}{2} x \sqrt {x^2+1}\right )-\frac {(1-x) (x+1)^2}{\sqrt {x^2+1}}\right )}{x+1}\) |
\(\Big \downarrow \) 222 |
\(\displaystyle \frac {\sqrt {x^2+1} \sqrt {\frac {2 x}{x^2+1}+1} \left (2 \left (\frac {3 \text {arcsinh}(x)}{2}-\frac {1}{2} x \sqrt {x^2+1}\right )-\frac {(1-x) (x+1)^2}{\sqrt {x^2+1}}\right )}{x+1}\) |
(Sqrt[1 + x^2]*Sqrt[1 + (2*x)/(1 + x^2)]*(-(((1 - x)*(1 + x)^2)/Sqrt[1 + x ^2]) + 2*(-1/2*(x*Sqrt[1 + x^2]) + (3*ArcSinh[x])/2)))/(1 + x)
3.9.96.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x *((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 *p + 3)) Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && NeQ[2*p + 3, 0]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (a*d - b*c*x)*(c + d*x)^(n - 1)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[1/(2*a*b*(p + 1)) Int[(c + d*x)^(n - 2)*(a + b*x^2)^(p + 1)*Simp[a* d^2*(n - 1) - b*c^2*(2*p + 3) - b*c*d*(n + 2*p + 2)*x, x], x], x] /; FreeQ[ {a, b, c, d}, x] && LtQ[p, -1] && GtQ[n, 1] && IntQuadraticQ[a, 0, b, c, d, n, p, x]
Int[((c_) + (d_.)*(x_))^(m_.)*((e_) + (f_.)*(x_))^(n_.)*((a_.) + (b_.)*(x_) ^2)^(p_.), x_Symbol] :> Int[(c*e + d*f*x^2)^m*(a + b*x^2)^p, x] /; FreeQ[{a , b, c, d, e, f, m, n, p}, x] && EqQ[m, n] && EqQ[d*e + c*f, 0] && (Integer Q[m] || (GtQ[c, 0] && GtQ[e, 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 = 0.14 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.54
method | result | size |
default | \(\frac {\left (\frac {x^{2}+2 x +1}{x^{2}+1}\right )^{\frac {3}{2}} \left (x^{2}+1\right ) \left (3 \,\operatorname {arcsinh}\left (x \right ) \sqrt {x^{2}+1}+x^{2}-2 x -1\right )}{\left (x +1\right )^{3}}\) | \(49\) |
risch | \(\frac {\left (x^{2}-2 x -1\right ) \sqrt {\frac {\left (x +1\right )^{2}}{x^{2}+1}}}{x +1}+\frac {3 \,\operatorname {arcsinh}\left (x \right ) \sqrt {x^{2}+1}\, \sqrt {\frac {\left (x +1\right )^{2}}{x^{2}+1}}}{x +1}\) | \(62\) |
trager | \(\frac {\left (x^{2}-2 x -1\right ) \sqrt {-\frac {-x^{2}-2 x -1}{x^{2}+1}}}{x +1}+3 \ln \left (\frac {\sqrt {-\frac {-x^{2}-2 x -1}{x^{2}+1}}\, x^{2}+x^{2}+\sqrt {-\frac {-x^{2}-2 x -1}{x^{2}+1}}+x}{x +1}\right )\) | \(97\) |
Time = 0.26 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.92 \[ \int \left (1+\frac {2 x}{1+x^2}\right )^{3/2} \, dx=-\frac {3 \, {\left (x + 1\right )} \log \left (-\frac {x^{2} - {\left (x^{2} + 1\right )} \sqrt {\frac {x^{2} + 2 \, x + 1}{x^{2} + 1}} + x}{x + 1}\right ) - {\left (x^{2} - 2 \, x - 1\right )} \sqrt {\frac {x^{2} + 2 \, x + 1}{x^{2} + 1}} + 2 \, x + 2}{x + 1} \]
-(3*(x + 1)*log(-(x^2 - (x^2 + 1)*sqrt((x^2 + 2*x + 1)/(x^2 + 1)) + x)/(x + 1)) - (x^2 - 2*x - 1)*sqrt((x^2 + 2*x + 1)/(x^2 + 1)) + 2*x + 2)/(x + 1)
\[ \int \left (1+\frac {2 x}{1+x^2}\right )^{3/2} \, dx=\int \left (\frac {2 x}{x^{2} + 1} + 1\right )^{\frac {3}{2}}\, dx \]
\[ \int \left (1+\frac {2 x}{1+x^2}\right )^{3/2} \, dx=\int { {\left (\frac {2 \, x}{x^{2} + 1} + 1\right )}^{\frac {3}{2}} \,d x } \]
Time = 0.32 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.74 \[ \int \left (1+\frac {2 x}{1+x^2}\right )^{3/2} \, dx=-{\left (\sqrt {2} - 3 \, \log \left (\sqrt {2} + 1\right )\right )} \mathrm {sgn}\left (x + 1\right ) - 3 \, \log \left (-x + \sqrt {x^{2} + 1}\right ) \mathrm {sgn}\left (x + 1\right ) + \frac {{\left (x \mathrm {sgn}\left (x + 1\right ) - 2 \, \mathrm {sgn}\left (x + 1\right )\right )} x - \mathrm {sgn}\left (x + 1\right )}{\sqrt {x^{2} + 1}} \]
-(sqrt(2) - 3*log(sqrt(2) + 1))*sgn(x + 1) - 3*log(-x + sqrt(x^2 + 1))*sgn (x + 1) + ((x*sgn(x + 1) - 2*sgn(x + 1))*x - sgn(x + 1))/sqrt(x^2 + 1)
Timed out. \[ \int \left (1+\frac {2 x}{1+x^2}\right )^{3/2} \, dx=\int {\left (\frac {2\,x}{x^2+1}+1\right )}^{3/2} \,d x \]