Integrand size = 15, antiderivative size = 90 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^2 x^{11/2}} \, dx=-\frac {2}{5 b^2 x^{5/2}}+\frac {4 a}{3 b^3 x^{3/2}}-\frac {6 a^2}{b^4 \sqrt {x}}-\frac {a^3}{b^4 \left (a+\frac {b}{x}\right ) \sqrt {x}}+\frac {7 a^{5/2} \arctan \left (\frac {\sqrt {b}}{\sqrt {a} \sqrt {x}}\right )}{b^{9/2}} \] Output:
-2/5/b^2/x^(5/2)+4/3*a/b^3/x^(3/2)-6*a^2/b^4/x^(1/2)-a^3/b^4/(a+b/x)/x^(1/ 2)+7*a^(5/2)*arctan(1/a^(1/2)/x^(1/2)*b^(1/2))/b^(9/2)
Time = 0.09 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.88 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^2 x^{11/2}} \, dx=\frac {-6 b^3+14 a b^2 x-70 a^2 b x^2-105 a^3 x^3}{15 b^4 x^{5/2} (b+a x)}-\frac {7 a^{5/2} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{b^{9/2}} \] Input:
Integrate[1/((a + b/x)^2*x^(11/2)),x]
Output:
(-6*b^3 + 14*a*b^2*x - 70*a^2*b*x^2 - 105*a^3*x^3)/(15*b^4*x^(5/2)*(b + a* x)) - (7*a^(5/2)*ArcTan[(Sqrt[a]*Sqrt[x])/Sqrt[b]])/b^(9/2)
Time = 0.32 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.13, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {795, 52, 61, 61, 61, 73, 218}
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.
\(\displaystyle \int \frac {1}{x^{11/2} \left (a+\frac {b}{x}\right )^2} \, dx\) |
\(\Big \downarrow \) 795 |
\(\displaystyle \int \frac {1}{x^{7/2} (a x+b)^2}dx\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {7 \int \frac {1}{x^{7/2} (b+a x)}dx}{2 b}+\frac {1}{b x^{5/2} (a x+b)}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {7 \left (-\frac {a \int \frac {1}{x^{5/2} (b+a x)}dx}{b}-\frac {2}{5 b x^{5/2}}\right )}{2 b}+\frac {1}{b x^{5/2} (a x+b)}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {7 \left (-\frac {a \left (-\frac {a \int \frac {1}{x^{3/2} (b+a x)}dx}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{5 b x^{5/2}}\right )}{2 b}+\frac {1}{b x^{5/2} (a x+b)}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {7 \left (-\frac {a \left (-\frac {a \left (-\frac {a \int \frac {1}{\sqrt {x} (b+a x)}dx}{b}-\frac {2}{b \sqrt {x}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{5 b x^{5/2}}\right )}{2 b}+\frac {1}{b x^{5/2} (a x+b)}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {7 \left (-\frac {a \left (-\frac {a \left (-\frac {2 a \int \frac {1}{b+a x}d\sqrt {x}}{b}-\frac {2}{b \sqrt {x}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{5 b x^{5/2}}\right )}{2 b}+\frac {1}{b x^{5/2} (a x+b)}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {7 \left (-\frac {a \left (-\frac {a \left (-\frac {2 \sqrt {a} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{b^{3/2}}-\frac {2}{b \sqrt {x}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{5 b x^{5/2}}\right )}{2 b}+\frac {1}{b x^{5/2} (a x+b)}\) |
Input:
Int[1/((a + b/x)^2*x^(11/2)),x]
Output:
1/(b*x^(5/2)*(b + a*x)) + (7*(-2/(5*b*x^(5/2)) - (a*(-2/(3*b*x^(3/2)) - (a *(-2/(b*Sqrt[x]) - (2*Sqrt[a]*ArcTan[(Sqrt[a]*Sqrt[x])/Sqrt[b]])/b^(3/2))) /b))/b))/(2*b)
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d , m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)* (b + a/x^n)^p, x] /; FreeQ[{a, b, m, n}, x] && IntegerQ[p] && NegQ[n]
Time = 0.20 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.77
method | result | size |
derivativedivides | \(-\frac {2 a^{3} \left (\frac {\sqrt {x}}{2 a x +2 b}+\frac {7 \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{b^{4}}-\frac {2}{5 b^{2} x^{\frac {5}{2}}}-\frac {6 a^{2}}{b^{4} \sqrt {x}}+\frac {4 a}{3 b^{3} x^{\frac {3}{2}}}\) | \(69\) |
default | \(-\frac {2 a^{3} \left (\frac {\sqrt {x}}{2 a x +2 b}+\frac {7 \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{b^{4}}-\frac {2}{5 b^{2} x^{\frac {5}{2}}}-\frac {6 a^{2}}{b^{4} \sqrt {x}}+\frac {4 a}{3 b^{3} x^{\frac {3}{2}}}\) | \(69\) |
risch | \(-\frac {2 \left (45 a^{2} x^{2}-10 a b x +3 b^{2}\right )}{15 b^{4} x^{\frac {5}{2}}}-\frac {a^{3} \sqrt {x}}{b^{4} \left (a x +b \right )}-\frac {7 a^{3} \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{b^{4} \sqrt {a b}}\) | \(71\) |
Input:
int(1/(a+b/x)^2/x^(11/2),x,method=_RETURNVERBOSE)
Output:
-2*a^3/b^4*(1/2*x^(1/2)/(a*x+b)+7/2/(a*b)^(1/2)*arctan(a*x^(1/2)/(a*b)^(1/ 2)))-2/5/b^2/x^(5/2)-6*a^2/b^4/x^(1/2)+4/3*a/b^3/x^(3/2)
Time = 0.09 (sec) , antiderivative size = 205, normalized size of antiderivative = 2.28 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^2 x^{11/2}} \, dx=\left [\frac {105 \, {\left (a^{3} x^{4} + a^{2} b x^{3}\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {a x - 2 \, b \sqrt {x} \sqrt {-\frac {a}{b}} - b}{a x + b}\right ) - 2 \, {\left (105 \, a^{3} x^{3} + 70 \, a^{2} b x^{2} - 14 \, a b^{2} x + 6 \, b^{3}\right )} \sqrt {x}}{30 \, {\left (a b^{4} x^{4} + b^{5} x^{3}\right )}}, -\frac {105 \, {\left (a^{3} x^{4} + a^{2} b x^{3}\right )} \sqrt {\frac {a}{b}} \arctan \left (\sqrt {x} \sqrt {\frac {a}{b}}\right ) + {\left (105 \, a^{3} x^{3} + 70 \, a^{2} b x^{2} - 14 \, a b^{2} x + 6 \, b^{3}\right )} \sqrt {x}}{15 \, {\left (a b^{4} x^{4} + b^{5} x^{3}\right )}}\right ] \] Input:
integrate(1/(a+b/x)^2/x^(11/2),x, algorithm="fricas")
Output:
[1/30*(105*(a^3*x^4 + a^2*b*x^3)*sqrt(-a/b)*log((a*x - 2*b*sqrt(x)*sqrt(-a /b) - b)/(a*x + b)) - 2*(105*a^3*x^3 + 70*a^2*b*x^2 - 14*a*b^2*x + 6*b^3)* sqrt(x))/(a*b^4*x^4 + b^5*x^3), -1/15*(105*(a^3*x^4 + a^2*b*x^3)*sqrt(a/b) *arctan(sqrt(x)*sqrt(a/b)) + (105*a^3*x^3 + 70*a^2*b*x^2 - 14*a*b^2*x + 6* b^3)*sqrt(x))/(a*b^4*x^4 + b^5*x^3)]
Timed out. \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^2 x^{11/2}} \, dx=\text {Timed out} \] Input:
integrate(1/(a+b/x)**2/x**(11/2),x)
Output:
Timed out
Time = 0.10 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.86 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^2 x^{11/2}} \, dx=-\frac {a^{3}}{{\left (a b^{4} + \frac {b^{5}}{x}\right )} \sqrt {x}} + \frac {7 \, a^{3} \arctan \left (\frac {b}{\sqrt {a b} \sqrt {x}}\right )}{\sqrt {a b} b^{4}} - \frac {2 \, {\left (\frac {45 \, a^{2}}{\sqrt {x}} - \frac {10 \, a b}{x^{\frac {3}{2}}} + \frac {3 \, b^{2}}{x^{\frac {5}{2}}}\right )}}{15 \, b^{4}} \] Input:
integrate(1/(a+b/x)^2/x^(11/2),x, algorithm="maxima")
Output:
-a^3/((a*b^4 + b^5/x)*sqrt(x)) + 7*a^3*arctan(b/(sqrt(a*b)*sqrt(x)))/(sqrt (a*b)*b^4) - 2/15*(45*a^2/sqrt(x) - 10*a*b/x^(3/2) + 3*b^2/x^(5/2))/b^4
Time = 0.12 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.78 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^2 x^{11/2}} \, dx=-\frac {7 \, a^{3} \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{4}} - \frac {a^{3} \sqrt {x}}{{\left (a x + b\right )} b^{4}} - \frac {2 \, {\left (45 \, a^{2} x^{2} - 10 \, a b x + 3 \, b^{2}\right )}}{15 \, b^{4} x^{\frac {5}{2}}} \] Input:
integrate(1/(a+b/x)^2/x^(11/2),x, algorithm="giac")
Output:
-7*a^3*arctan(a*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^4) - a^3*sqrt(x)/((a*x + b )*b^4) - 2/15*(45*a^2*x^2 - 10*a*b*x + 3*b^2)/(b^4*x^(5/2))
Time = 0.32 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.78 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^2 x^{11/2}} \, dx=-\frac {\frac {2}{5\,b}+\frac {14\,a^2\,x^2}{3\,b^3}+\frac {7\,a^3\,x^3}{b^4}-\frac {14\,a\,x}{15\,b^2}}{a\,x^{7/2}+b\,x^{5/2}}-\frac {7\,a^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {x}}{\sqrt {b}}\right )}{b^{9/2}} \] Input:
int(1/(x^(11/2)*(a + b/x)^2),x)
Output:
- (2/(5*b) + (14*a^2*x^2)/(3*b^3) + (7*a^3*x^3)/b^4 - (14*a*x)/(15*b^2))/( a*x^(7/2) + b*x^(5/2)) - (7*a^(5/2)*atan((a^(1/2)*x^(1/2))/b^(1/2)))/b^(9/ 2)
Time = 0.24 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.19 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^2 x^{11/2}} \, dx=\frac {-105 \sqrt {x}\, \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, a}{\sqrt {b}\, \sqrt {a}}\right ) a^{3} x^{3}-105 \sqrt {x}\, \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, a}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b \,x^{2}-105 a^{3} b \,x^{3}-70 a^{2} b^{2} x^{2}+14 a \,b^{3} x -6 b^{4}}{15 \sqrt {x}\, b^{5} x^{2} \left (a x +b \right )} \] Input:
int(1/(a+b/x)^2/x^(11/2),x)
Output:
( - 105*sqrt(x)*sqrt(b)*sqrt(a)*atan((sqrt(x)*a)/(sqrt(b)*sqrt(a)))*a**3*x **3 - 105*sqrt(x)*sqrt(b)*sqrt(a)*atan((sqrt(x)*a)/(sqrt(b)*sqrt(a)))*a**2 *b*x**2 - 105*a**3*b*x**3 - 70*a**2*b**2*x**2 + 14*a*b**3*x - 6*b**4)/(15* sqrt(x)*b**5*x**2*(a*x + b))