Integrand size = 19, antiderivative size = 93 \[ \int \frac {\sqrt {1-\frac {1}{1+2 x}}}{x^5} \, dx=-\frac {2 \sqrt {2} \sqrt {1+2 x}}{7 x^{7/2}}+\frac {24 \sqrt {2} \sqrt {1+2 x}}{35 x^{5/2}}-\frac {64 \sqrt {2} \sqrt {1+2 x}}{35 x^{3/2}}+\frac {256 \sqrt {2} \sqrt {1+2 x}}{35 \sqrt {x}} \] Output:
-2/7*2^(1/2)*(1+2*x)^(1/2)/x^(7/2)+24/35*2^(1/2)*(1+2*x)^(1/2)/x^(5/2)-64/ 35*2^(1/2)*(1+2*x)^(1/2)/x^(3/2)+256/35*2^(1/2)*(1+2*x)^(1/2)/x^(1/2)
Time = 0.09 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.38 \[ \int \frac {\sqrt {1-\frac {1}{1+2 x}}}{x^5} \, dx=\frac {2 \left (-5+12 x-32 x^2+128 x^3\right )}{35 x^3 \sqrt {\frac {x}{2+4 x}}} \] Input:
Integrate[Sqrt[1 - (1 + 2*x)^(-1)]/x^5,x]
Output:
(2*(-5 + 12*x - 32*x^2 + 128*x^3))/(35*x^3*Sqrt[x/(2 + 4*x)])
Time = 0.39 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.67, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {896, 25, 941, 281, 798, 53, 2009}
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.
\(\displaystyle \int \frac {\sqrt {1-\frac {1}{2 x+1}}}{x^5} \, dx\) |
\(\Big \downarrow \) 896 |
\(\displaystyle 16 \int \frac {\sqrt {1-\frac {1}{2 x+1}}}{32 x^5}d(2 x+1)\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -16 \int -\frac {\sqrt {1-\frac {1}{2 x+1}}}{32 x^5}d(2 x+1)\) |
\(\Big \downarrow \) 941 |
\(\displaystyle -16 \int \frac {\sqrt {1-\frac {1}{2 x+1}}}{(2 x+1)^5 \left (\frac {1}{2 x+1}-1\right )^5}d(2 x+1)\) |
\(\Big \downarrow \) 281 |
\(\displaystyle 16 \int \frac {1}{(2 x+1)^5 \left (1-\frac {1}{2 x+1}\right )^{9/2}}d(2 x+1)\) |
\(\Big \downarrow \) 798 |
\(\displaystyle -16 \int \frac {1}{16 \sqrt {2} (-x)^{9/2} (2 x+1)^3}d\frac {1}{2 x+1}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle -16 \int \left (-\frac {1}{2 \sqrt {2} (-x)^{3/2}}+\frac {3}{4 \sqrt {2} (-x)^{5/2}}-\frac {3}{8 \sqrt {2} (-x)^{7/2}}+\frac {1}{16 \sqrt {2} (-x)^{9/2}}\right )d\frac {1}{2 x+1}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -16 \left (-\frac {\sqrt {2}}{\sqrt {-x}}+\frac {1}{\sqrt {2} (-x)^{3/2}}-\frac {3}{10 \sqrt {2} (-x)^{5/2}}+\frac {1}{28 \sqrt {2} (-x)^{7/2}}\right )\) |
Input:
Int[Sqrt[1 - (1 + 2*x)^(-1)]/x^5,x]
Output:
-16*(1/(28*Sqrt[2]*(-x)^(7/2)) - 3/(10*Sqrt[2]*(-x)^(5/2)) + 1/(Sqrt[2]*(- x)^(3/2)) - Sqrt[2]/Sqrt[-x])
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Int[(u_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_ Symbol] :> Simp[(b/d)^p Int[u*(c + d*x^n)^(p + q), x], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && EqQ[b*c - a*d, 0] && IntegerQ[p] && !(IntegerQ[q] & & SimplerQ[a + b*x^n, c + d*x^n])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/n Subst [Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
Int[((a_) + (b_.)*(v_)^(n_))^(p_.)*(x_)^(m_.), x_Symbol] :> With[{c = Coeff icient[v, x, 0], d = Coefficient[v, x, 1]}, Simp[1/d^(m + 1) Subst[Int[Si mplifyIntegrand[(x - c)^m*(a + b*x^n)^p, x], x], x, v], x] /; NeQ[c, 0]] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && IntegerQ[m]
Int[((c_) + (d_.)*(x_)^(mn_.))^(q_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Sym bol] :> Int[(a + b*x^n)^p*((d + c*x^n)^q/x^(n*q)), x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[mn, -n] && IntegerQ[q] && (PosQ[n] || !IntegerQ[p])
Time = 0.06 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.42
method | result | size |
orering | \(\frac {2 \left (128 x^{3}-32 x^{2}+12 x -5\right ) \left (1+2 x \right ) \sqrt {1-\frac {1}{1+2 x}}}{35 x^{4}}\) | \(39\) |
gosper | \(\frac {2 \left (1+2 x \right ) \left (128 x^{3}-32 x^{2}+12 x -5\right ) \sqrt {2}\, \sqrt {\frac {x}{1+2 x}}}{35 x^{4}}\) | \(40\) |
trager | \(\frac {2 \left (1+2 x \right ) \left (128 x^{3}-32 x^{2}+12 x -5\right ) \sqrt {2}\, \sqrt {\frac {x}{1+2 x}}}{35 x^{4}}\) | \(40\) |
risch | \(\frac {2 \sqrt {2}\, \sqrt {\frac {x}{1+2 x}}\, \left (256 x^{4}+64 x^{3}-8 x^{2}+2 x -5\right )}{35 x^{4}}\) | \(40\) |
default | \(\frac {2 \sqrt {2}\, \sqrt {\frac {x}{1+2 x}}\, \left (1+2 x \right ) \sqrt {2 x^{2}+x}\, \left (128 x^{3}-32 x^{2}+12 x -5\right )}{35 x^{4} \sqrt {x \left (1+2 x \right )}}\) | \(58\) |
Input:
int((1-1/(1+2*x))^(1/2)/x^5,x,method=_RETURNVERBOSE)
Output:
2/35*(128*x^3-32*x^2+12*x-5)*(1+2*x)/x^4*(1-1/(1+2*x))^(1/2)
Time = 0.09 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.42 \[ \int \frac {\sqrt {1-\frac {1}{1+2 x}}}{x^5} \, dx=\frac {2 \, \sqrt {2} {\left (256 \, x^{4} + 64 \, x^{3} - 8 \, x^{2} + 2 \, x - 5\right )} \sqrt {\frac {x}{2 \, x + 1}}}{35 \, x^{4}} \] Input:
integrate((1-1/(1+2*x))^(1/2)/x^5,x, algorithm="fricas")
Output:
2/35*sqrt(2)*(256*x^4 + 64*x^3 - 8*x^2 + 2*x - 5)*sqrt(x/(2*x + 1))/x^4
\[ \int \frac {\sqrt {1-\frac {1}{1+2 x}}}{x^5} \, dx=\sqrt {2} \int \frac {\sqrt {\frac {x}{2 x + 1}}}{x^{5}}\, dx \] Input:
integrate((1-1/(1+2*x))**(1/2)/x**5,x)
Output:
sqrt(2)*Integral(sqrt(x/(2*x + 1))/x**5, x)
\[ \int \frac {\sqrt {1-\frac {1}{1+2 x}}}{x^5} \, dx=\int { \frac {\sqrt {-\frac {1}{2 \, x + 1} + 1}}{x^{5}} \,d x } \] Input:
integrate((1-1/(1+2*x))^(1/2)/x^5,x, algorithm="maxima")
Output:
integrate(sqrt(-1/(2*x + 1) + 1)/x^5, x)
Leaf count of result is larger than twice the leaf count of optimal. 135 vs. \(2 (61) = 122\).
Time = 0.14 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.45 \[ \int \frac {\sqrt {1-\frac {1}{1+2 x}}}{x^5} \, dx=-\frac {1}{35} \, \sqrt {2} {\left (256 \, \sqrt {2} \mathrm {sgn}\left (2 \, x + 1\right ) - \frac {\sqrt {2} {\left (280 \, {\left (\sqrt {2} x - \sqrt {2 \, x^{2} + x}\right )}^{3} \mathrm {sgn}\left (2 \, x + 1\right ) + 168 \, \sqrt {2} {\left (\sqrt {2} x - \sqrt {2 \, x^{2} + x}\right )}^{2} \mathrm {sgn}\left (2 \, x + 1\right ) + 70 \, {\left (\sqrt {2} x - \sqrt {2 \, x^{2} + x}\right )} \mathrm {sgn}\left (2 \, x + 1\right ) + 5 \, \sqrt {2} \mathrm {sgn}\left (2 \, x + 1\right )\right )}}{{\left (\sqrt {2} x - \sqrt {2 \, x^{2} + x}\right )}^{7}}\right )} \] Input:
integrate((1-1/(1+2*x))^(1/2)/x^5,x, algorithm="giac")
Output:
-1/35*sqrt(2)*(256*sqrt(2)*sgn(2*x + 1) - sqrt(2)*(280*(sqrt(2)*x - sqrt(2 *x^2 + x))^3*sgn(2*x + 1) + 168*sqrt(2)*(sqrt(2)*x - sqrt(2*x^2 + x))^2*sg n(2*x + 1) + 70*(sqrt(2)*x - sqrt(2*x^2 + x))*sgn(2*x + 1) + 5*sqrt(2)*sgn (2*x + 1))/(sqrt(2)*x - sqrt(2*x^2 + x))^7)
Time = 8.71 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.91 \[ \int \frac {\sqrt {1-\frac {1}{1+2 x}}}{x^5} \, dx=\frac {512\,\left (x+\frac {1}{2}\right )\,\sqrt {1-\frac {1}{2\,x+1}}}{35\,x}-\frac {128\,\left (x+\frac {1}{2}\right )\,\sqrt {1-\frac {1}{2\,x+1}}}{35\,x^2}+\frac {48\,\left (x+\frac {1}{2}\right )\,\sqrt {1-\frac {1}{2\,x+1}}}{35\,x^3}-\frac {4\,\left (x+\frac {1}{2}\right )\,\sqrt {1-\frac {1}{2\,x+1}}}{7\,x^4} \] Input:
int((1 - 1/(2*x + 1))^(1/2)/x^5,x)
Output:
(512*(x + 1/2)*(1 - 1/(2*x + 1))^(1/2))/(35*x) - (128*(x + 1/2)*(1 - 1/(2* x + 1))^(1/2))/(35*x^2) + (48*(x + 1/2)*(1 - 1/(2*x + 1))^(1/2))/(35*x^3) - (4*(x + 1/2)*(1 - 1/(2*x + 1))^(1/2))/(7*x^4)
Time = 0.25 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.71 \[ \int \frac {\sqrt {1-\frac {1}{1+2 x}}}{x^5} \, dx=\frac {\frac {256 \sqrt {x}\, \sqrt {2 x +1}\, \sqrt {2}\, x^{3}}{35}-\frac {64 \sqrt {x}\, \sqrt {2 x +1}\, \sqrt {2}\, x^{2}}{35}+\frac {24 \sqrt {x}\, \sqrt {2 x +1}\, \sqrt {2}\, x}{35}-\frac {2 \sqrt {x}\, \sqrt {2 x +1}\, \sqrt {2}}{7}-\frac {512 x^{4}}{35}}{x^{4}} \] Input:
int((1-1/(1+2*x))^(1/2)/x^5,x)
Output:
(2*(128*sqrt(x)*sqrt(2*x + 1)*sqrt(2)*x**3 - 32*sqrt(x)*sqrt(2*x + 1)*sqrt (2)*x**2 + 12*sqrt(x)*sqrt(2*x + 1)*sqrt(2)*x - 5*sqrt(x)*sqrt(2*x + 1)*sq rt(2) - 256*x**4))/(35*x**4)