Integrand size = 16, antiderivative size = 133 \[ \int \left (1+\frac {2 x}{1+x^2}\right )^{5/2} \, dx=-\frac {4}{3} (1-2 x) (1+x) \sqrt {1+\frac {2 x}{1+x^2}}-\frac {(1-x) (1+x)^3 \sqrt {1+\frac {2 x}{1+x^2}}}{3 \left (1+x^2\right )}-\frac {(4+3 x) \left (1+x^2\right ) \sqrt {1+\frac {2 x}{1+x^2}}}{1+x}+\frac {5 \sqrt {1+x^2} \sqrt {1+\frac {2 x}{1+x^2}} \text {arcsinh}(x)}{1+x} \]
-4/3*(1-2*x)*(1+x)*(1+2*x/(x^2+1))^(1/2)-1/3*(1-x)*(1+x)^3*(1+2*x/(x^2+1)) ^(1/2)/(x^2+1)-(4+3*x)*(x^2+1)*(1+2*x/(x^2+1))^(1/2)/(1+x)+5*arcsinh(x)*(x ^2+1)^(1/2)*(1+2*x/(x^2+1))^(1/2)/(1+x)
Time = 0.26 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.57 \[ \int \left (1+\frac {2 x}{1+x^2}\right )^{5/2} \, dx=\frac {(1+x) \left (-17-12 x-18 x^2-8 x^3+3 x^4-15 \left (1+x^2\right )^{3/2} \log \left (-x+\sqrt {1+x^2}\right )\right )}{3 \sqrt {\frac {(1+x)^2}{1+x^2}} \left (1+x^2\right )^2} \]
((1 + x)*(-17 - 12*x - 18*x^2 - 8*x^3 + 3*x^4 - 15*(1 + x^2)^(3/2)*Log[-x + Sqrt[1 + x^2]]))/(3*Sqrt[(1 + x)^2/(1 + x^2)]*(1 + x^2)^2)
Time = 0.29 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.86, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.562, Rules used = {7274, 1298, 27, 495, 27, 684, 27, 676, 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.
| \(\displaystyle \int \left (\frac {2 x}{x^2+1}+1\right )^{5/2} \, dx\) |
| \(\Big \downarrow \) 7274 |
| \(\displaystyle \frac {\sqrt {x^2+1} \sqrt {\frac {2 x}{x^2+1}+1} \int \frac {\left (x^2+2 x+1\right )^{5/2}}{\left (x^2+1\right )^{5/2}}dx}{\sqrt {x^2+2 x+1}}\) |
| \(\Big \downarrow \) 1298 |
| \(\displaystyle \frac {\sqrt {x^2+1} \sqrt {\frac {2 x}{x^2+1}+1} \int \frac {32 (x+1)^5}{\left (x^2+1\right )^{5/2}}dx}{32 (x+1)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sqrt {x^2+1} \sqrt {\frac {2 x}{x^2+1}+1} \int \frac {(x+1)^5}{\left (x^2+1\right )^{5/2}}dx}{x+1}\) |
| \(\Big \downarrow \) 495 |
| \(\displaystyle \frac {\sqrt {x^2+1} \sqrt {\frac {2 x}{x^2+1}+1} \left (\frac {1}{3} \int \frac {2 (3-x) (x+1)^3}{\left (x^2+1\right )^{3/2}}dx-\frac {(1-x) (x+1)^4}{3 \left (x^2+1\right )^{3/2}}\right )}{x+1}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sqrt {x^2+1} \sqrt {\frac {2 x}{x^2+1}+1} \left (\frac {2}{3} \int \frac {(3-x) (x+1)^3}{\left (x^2+1\right )^{3/2}}dx-\frac {(1-x) (x+1)^4}{3 \left (x^2+1\right )^{3/2}}\right )}{x+1}\) |
| \(\Big \downarrow \) 684 |
| \(\displaystyle \frac {\sqrt {x^2+1} \sqrt {\frac {2 x}{x^2+1}+1} \left (\frac {2}{3} \left (\int \frac {3 (1-3 x) (x+1)}{\sqrt {x^2+1}}dx-\frac {2 (1-2 x) (x+1)^2}{\sqrt {x^2+1}}\right )-\frac {(1-x) (x+1)^4}{3 \left (x^2+1\right )^{3/2}}\right )}{x+1}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sqrt {x^2+1} \sqrt {\frac {2 x}{x^2+1}+1} \left (\frac {2}{3} \left (3 \int \frac {(1-3 x) (x+1)}{\sqrt {x^2+1}}dx-\frac {2 (1-2 x) (x+1)^2}{\sqrt {x^2+1}}\right )-\frac {(1-x) (x+1)^4}{3 \left (x^2+1\right )^{3/2}}\right )}{x+1}\) |
| \(\Big \downarrow \) 676 |
| \(\displaystyle \frac {\sqrt {x^2+1} \sqrt {\frac {2 x}{x^2+1}+1} \left (\frac {2}{3} \left (3 \left (\frac {5}{2} \int \frac {1}{\sqrt {x^2+1}}dx-\frac {3}{2} \sqrt {x^2+1} x-2 \sqrt {x^2+1}\right )-\frac {2 (1-2 x) (x+1)^2}{\sqrt {x^2+1}}\right )-\frac {(1-x) (x+1)^4}{3 \left (x^2+1\right )^{3/2}}\right )}{x+1}\) |
| \(\Big \downarrow \) 222 |
| \(\displaystyle \frac {\sqrt {x^2+1} \sqrt {\frac {2 x}{x^2+1}+1} \left (\frac {2}{3} \left (3 \left (\frac {5 \text {arcsinh}(x)}{2}-\frac {3}{2} \sqrt {x^2+1} x-2 \sqrt {x^2+1}\right )-\frac {2 (1-2 x) (x+1)^2}{\sqrt {x^2+1}}\right )-\frac {(1-x) (x+1)^4}{3 \left (x^2+1\right )^{3/2}}\right )}{x+1}\) |
(Sqrt[1 + x^2]*Sqrt[1 + (2*x)/(1 + x^2)]*(-1/3*((1 - x)*(1 + x)^4)/(1 + x^ 2)^(3/2) + (2*((-2*(1 - 2*x)*(1 + x)^2)/Sqrt[1 + x^2] + 3*(-2*Sqrt[1 + x^2 ] - (3*x*Sqrt[1 + x^2])/2 + (5*ArcSinh[x])/2)))/3))/(1 + x)
3.9.95.3.1 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[((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[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x _Symbol] :> Simp[(e*f + d*g)*((a + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + (Sim p[e*g*x*((a + c*x^2)^(p + 1)/(c*(2*p + 3))), x] - Simp[(a*e*g - c*d*f*(2*p + 3))/(c*(2*p + 3)) Int[(a + c*x^2)^p, x], x]) /; FreeQ[{a, c, d, e, f, g , p}, x] && !LeQ[p, -1]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p _.), x_Symbol] :> Simp[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1)*((a*(e*f + d*g ) - (c*d*f - a*e*g)*x)/(2*a*c*(p + 1))), x] - Simp[1/(2*a*c*(p + 1)) Int[ (d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^ 2*f*(2*p + 3) + e*(a*e*g*m - c*d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a , c, d, e, f, g}, x] && LtQ[p, -1] && GtQ[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) || !ILtQ[m + 2*p + 3, 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 = 0.33 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.47
| method | result | size |
| default | \(\frac {\left (\frac {x^{2}+2 x +1}{x^{2}+1}\right )^{\frac {5}{2}} \left (x^{2}+1\right ) \left (15 \,\operatorname {arcsinh}\left (x \right ) \left (x^{2}+1\right )^{\frac {3}{2}}+3 x^{4}-8 x^{3}-18 x^{2}-12 x -17\right )}{3 \left (x +1\right )^{5}}\) | \(62\) |
| risch | \(\frac {\left (3 x^{4}-8 x^{3}-18 x^{2}-12 x -17\right ) \sqrt {\frac {\left (x +1\right )^{2}}{x^{2}+1}}}{3 \left (x^{2}+1\right ) \left (x +1\right )}+\frac {5 \,\operatorname {arcsinh}\left (x \right ) \sqrt {x^{2}+1}\, \sqrt {\frac {\left (x +1\right )^{2}}{x^{2}+1}}}{x +1}\) | \(82\) |
| trager | \(\frac {\left (3 x^{4}-8 x^{3}-18 x^{2}-12 x -17\right ) \sqrt {-\frac {-x^{2}-2 x -1}{x^{2}+1}}}{3 \left (x^{2}+1\right ) \left (x +1\right )}+5 \ln \left (\frac {\sqrt {-\frac {-x^{2}-2 x -1}{x^{2}+1}}\, x^{2}+x^{2}+\sqrt {-\frac {-x^{2}-2 x -1}{x^{2}+1}}+x}{x +1}\right )\) | \(117\) |
1/3*((x^2+2*x+1)/(x^2+1))^(5/2)/(x+1)^5*(x^2+1)*(15*arcsinh(x)*(x^2+1)^(3/ 2)+3*x^4-8*x^3-18*x^2-12*x-17)
Time = 0.26 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.88 \[ \int \left (1+\frac {2 x}{1+x^2}\right )^{5/2} \, dx=-\frac {8 \, x^{3} + 8 \, x^{2} + 15 \, {\left (x^{3} + x^{2} + x + 1\right )} \log \left (-\frac {x^{2} - {\left (x^{2} + 1\right )} \sqrt {\frac {x^{2} + 2 \, x + 1}{x^{2} + 1}} + x}{x + 1}\right ) - {\left (3 \, x^{4} - 8 \, x^{3} - 18 \, x^{2} - 12 \, x - 17\right )} \sqrt {\frac {x^{2} + 2 \, x + 1}{x^{2} + 1}} + 8 \, x + 8}{3 \, {\left (x^{3} + x^{2} + x + 1\right )}} \]
-1/3*(8*x^3 + 8*x^2 + 15*(x^3 + x^2 + x + 1)*log(-(x^2 - (x^2 + 1)*sqrt((x ^2 + 2*x + 1)/(x^2 + 1)) + x)/(x + 1)) - (3*x^4 - 8*x^3 - 18*x^2 - 12*x - 17)*sqrt((x^2 + 2*x + 1)/(x^2 + 1)) + 8*x + 8)/(x^3 + x^2 + x + 1)
\[ \int \left (1+\frac {2 x}{1+x^2}\right )^{5/2} \, dx=\int \left (\frac {2 x}{x^{2} + 1} + 1\right )^{\frac {5}{2}}\, dx \]
\[ \int \left (1+\frac {2 x}{1+x^2}\right )^{5/2} \, dx=\int { {\left (\frac {2 \, x}{x^{2} + 1} + 1\right )}^{\frac {5}{2}} \,d x } \]
Time = 0.32 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.65 \[ \int \left (1+\frac {2 x}{1+x^2}\right )^{5/2} \, dx={\left (\sqrt {2} + 5 \, \log \left (\sqrt {2} + 1\right )\right )} \mathrm {sgn}\left (x + 1\right ) - 5 \, \log \left (-x + \sqrt {x^{2} + 1}\right ) \mathrm {sgn}\left (x + 1\right ) + \frac {{\left ({\left ({\left (3 \, x \mathrm {sgn}\left (x + 1\right ) - 8 \, \mathrm {sgn}\left (x + 1\right )\right )} x - 18 \, \mathrm {sgn}\left (x + 1\right )\right )} x - 12 \, \mathrm {sgn}\left (x + 1\right )\right )} x - 17 \, \mathrm {sgn}\left (x + 1\right )}{3 \, {\left (x^{2} + 1\right )}^{\frac {3}{2}}} \]
(sqrt(2) + 5*log(sqrt(2) + 1))*sgn(x + 1) - 5*log(-x + sqrt(x^2 + 1))*sgn( x + 1) + 1/3*((((3*x*sgn(x + 1) - 8*sgn(x + 1))*x - 18*sgn(x + 1))*x - 12* sgn(x + 1))*x - 17*sgn(x + 1))/(x^2 + 1)^(3/2)
Timed out. \[ \int \left (1+\frac {2 x}{1+x^2}\right )^{5/2} \, dx=\int {\left (\frac {2\,x}{x^2+1}+1\right )}^{5/2} \,d x \]