Integrand size = 15, antiderivative size = 70 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{3/2}} \, dx=-\frac {x^{3/2}}{2 a (b+a x)^2}-\frac {3 \sqrt {x}}{4 a^2 (b+a x)}+\frac {3 \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{4 a^{5/2} \sqrt {b}} \]
-1/2*x^(3/2)/a/(a*x+b)^2+3/4*arctan(a^(1/2)*x^(1/2)/b^(1/2))/a^(5/2)/b^(1/ 2)-3/4*x^(1/2)/a^2/(a*x+b)
Time = 0.16 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.84 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{3/2}} \, dx=-\frac {\sqrt {x} (3 b+5 a x)}{4 a^2 (b+a x)^2}+\frac {3 \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{4 a^{5/2} \sqrt {b}} \]
-1/4*(Sqrt[x]*(3*b + 5*a*x))/(a^2*(b + a*x)^2) + (3*ArcTan[(Sqrt[a]*Sqrt[x ])/Sqrt[b]])/(4*a^(5/2)*Sqrt[b])
Time = 0.18 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {795, 51, 51, 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.
\(\displaystyle \int \frac {1}{x^{3/2} \left (a+\frac {b}{x}\right )^3} \, dx\) |
\(\Big \downarrow \) 795 |
\(\displaystyle \int \frac {x^{3/2}}{(a x+b)^3}dx\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {3 \int \frac {\sqrt {x}}{(b+a x)^2}dx}{4 a}-\frac {x^{3/2}}{2 a (a x+b)^2}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {3 \left (\frac {\int \frac {1}{\sqrt {x} (b+a x)}dx}{2 a}-\frac {\sqrt {x}}{a (a x+b)}\right )}{4 a}-\frac {x^{3/2}}{2 a (a x+b)^2}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {3 \left (\frac {\int \frac {1}{b+a x}d\sqrt {x}}{a}-\frac {\sqrt {x}}{a (a x+b)}\right )}{4 a}-\frac {x^{3/2}}{2 a (a x+b)^2}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {3 \left (\frac {\arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{a^{3/2} \sqrt {b}}-\frac {\sqrt {x}}{a (a x+b)}\right )}{4 a}-\frac {x^{3/2}}{2 a (a x+b)^2}\) |
-1/2*x^(3/2)/(a*(b + a*x)^2) + (3*(-(Sqrt[x]/(a*(b + a*x))) + ArcTan[(Sqrt [a]*Sqrt[x])/Sqrt[b]]/(a^(3/2)*Sqrt[b])))/(4*a)
3.17.85.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x ] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
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 = 0.04 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.71
method | result | size |
derivativedivides | \(\frac {-\frac {5 x^{\frac {3}{2}}}{4 a}-\frac {3 b \sqrt {x}}{4 a^{2}}}{\left (a x +b \right )^{2}}+\frac {3 \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{4 a^{2} \sqrt {a b}}\) | \(50\) |
default | \(\frac {-\frac {5 x^{\frac {3}{2}}}{4 a}-\frac {3 b \sqrt {x}}{4 a^{2}}}{\left (a x +b \right )^{2}}+\frac {3 \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{4 a^{2} \sqrt {a b}}\) | \(50\) |
2*(-5/8/a*x^(3/2)-3/8*b*x^(1/2)/a^2)/(a*x+b)^2+3/4/a^2/(a*b)^(1/2)*arctan( a*x^(1/2)/(a*b)^(1/2))
Time = 0.26 (sec) , antiderivative size = 185, normalized size of antiderivative = 2.64 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{3/2}} \, dx=\left [-\frac {3 \, {\left (a^{2} x^{2} + 2 \, a b x + b^{2}\right )} \sqrt {-a b} \log \left (\frac {a x - b - 2 \, \sqrt {-a b} \sqrt {x}}{a x + b}\right ) + 2 \, {\left (5 \, a^{2} b x + 3 \, a b^{2}\right )} \sqrt {x}}{8 \, {\left (a^{5} b x^{2} + 2 \, a^{4} b^{2} x + a^{3} b^{3}\right )}}, -\frac {3 \, {\left (a^{2} x^{2} + 2 \, a b x + b^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b}}{a \sqrt {x}}\right ) + {\left (5 \, a^{2} b x + 3 \, a b^{2}\right )} \sqrt {x}}{4 \, {\left (a^{5} b x^{2} + 2 \, a^{4} b^{2} x + a^{3} b^{3}\right )}}\right ] \]
[-1/8*(3*(a^2*x^2 + 2*a*b*x + b^2)*sqrt(-a*b)*log((a*x - b - 2*sqrt(-a*b)* sqrt(x))/(a*x + b)) + 2*(5*a^2*b*x + 3*a*b^2)*sqrt(x))/(a^5*b*x^2 + 2*a^4* b^2*x + a^3*b^3), -1/4*(3*(a^2*x^2 + 2*a*b*x + b^2)*sqrt(a*b)*arctan(sqrt( a*b)/(a*sqrt(x))) + (5*a^2*b*x + 3*a*b^2)*sqrt(x))/(a^5*b*x^2 + 2*a^4*b^2* x + a^3*b^3)]
Leaf count of result is larger than twice the leaf count of optimal. 605 vs. \(2 (61) = 122\).
Time = 17.03 (sec) , antiderivative size = 605, normalized size of antiderivative = 8.64 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{3/2}} \, dx=\begin {cases} \tilde {\infty } x^{\frac {5}{2}} & \text {for}\: a = 0 \wedge b = 0 \\\frac {2 x^{\frac {5}{2}}}{5 b^{3}} & \text {for}\: a = 0 \\- \frac {2}{a^{3} \sqrt {x}} & \text {for}\: b = 0 \\- \frac {10 a^{2} x^{\frac {3}{2}} \sqrt {- \frac {b}{a}}}{8 a^{5} x^{2} \sqrt {- \frac {b}{a}} + 16 a^{4} b x \sqrt {- \frac {b}{a}} + 8 a^{3} b^{2} \sqrt {- \frac {b}{a}}} + \frac {3 a^{2} x^{2} \log {\left (\sqrt {x} - \sqrt {- \frac {b}{a}} \right )}}{8 a^{5} x^{2} \sqrt {- \frac {b}{a}} + 16 a^{4} b x \sqrt {- \frac {b}{a}} + 8 a^{3} b^{2} \sqrt {- \frac {b}{a}}} - \frac {3 a^{2} x^{2} \log {\left (\sqrt {x} + \sqrt {- \frac {b}{a}} \right )}}{8 a^{5} x^{2} \sqrt {- \frac {b}{a}} + 16 a^{4} b x \sqrt {- \frac {b}{a}} + 8 a^{3} b^{2} \sqrt {- \frac {b}{a}}} - \frac {6 a b \sqrt {x} \sqrt {- \frac {b}{a}}}{8 a^{5} x^{2} \sqrt {- \frac {b}{a}} + 16 a^{4} b x \sqrt {- \frac {b}{a}} + 8 a^{3} b^{2} \sqrt {- \frac {b}{a}}} + \frac {6 a b x \log {\left (\sqrt {x} - \sqrt {- \frac {b}{a}} \right )}}{8 a^{5} x^{2} \sqrt {- \frac {b}{a}} + 16 a^{4} b x \sqrt {- \frac {b}{a}} + 8 a^{3} b^{2} \sqrt {- \frac {b}{a}}} - \frac {6 a b x \log {\left (\sqrt {x} + \sqrt {- \frac {b}{a}} \right )}}{8 a^{5} x^{2} \sqrt {- \frac {b}{a}} + 16 a^{4} b x \sqrt {- \frac {b}{a}} + 8 a^{3} b^{2} \sqrt {- \frac {b}{a}}} + \frac {3 b^{2} \log {\left (\sqrt {x} - \sqrt {- \frac {b}{a}} \right )}}{8 a^{5} x^{2} \sqrt {- \frac {b}{a}} + 16 a^{4} b x \sqrt {- \frac {b}{a}} + 8 a^{3} b^{2} \sqrt {- \frac {b}{a}}} - \frac {3 b^{2} \log {\left (\sqrt {x} + \sqrt {- \frac {b}{a}} \right )}}{8 a^{5} x^{2} \sqrt {- \frac {b}{a}} + 16 a^{4} b x \sqrt {- \frac {b}{a}} + 8 a^{3} b^{2} \sqrt {- \frac {b}{a}}} & \text {otherwise} \end {cases} \]
Piecewise((zoo*x**(5/2), Eq(a, 0) & Eq(b, 0)), (2*x**(5/2)/(5*b**3), Eq(a, 0)), (-2/(a**3*sqrt(x)), Eq(b, 0)), (-10*a**2*x**(3/2)*sqrt(-b/a)/(8*a**5 *x**2*sqrt(-b/a) + 16*a**4*b*x*sqrt(-b/a) + 8*a**3*b**2*sqrt(-b/a)) + 3*a* *2*x**2*log(sqrt(x) - sqrt(-b/a))/(8*a**5*x**2*sqrt(-b/a) + 16*a**4*b*x*sq rt(-b/a) + 8*a**3*b**2*sqrt(-b/a)) - 3*a**2*x**2*log(sqrt(x) + sqrt(-b/a)) /(8*a**5*x**2*sqrt(-b/a) + 16*a**4*b*x*sqrt(-b/a) + 8*a**3*b**2*sqrt(-b/a) ) - 6*a*b*sqrt(x)*sqrt(-b/a)/(8*a**5*x**2*sqrt(-b/a) + 16*a**4*b*x*sqrt(-b /a) + 8*a**3*b**2*sqrt(-b/a)) + 6*a*b*x*log(sqrt(x) - sqrt(-b/a))/(8*a**5* x**2*sqrt(-b/a) + 16*a**4*b*x*sqrt(-b/a) + 8*a**3*b**2*sqrt(-b/a)) - 6*a*b *x*log(sqrt(x) + sqrt(-b/a))/(8*a**5*x**2*sqrt(-b/a) + 16*a**4*b*x*sqrt(-b /a) + 8*a**3*b**2*sqrt(-b/a)) + 3*b**2*log(sqrt(x) - sqrt(-b/a))/(8*a**5*x **2*sqrt(-b/a) + 16*a**4*b*x*sqrt(-b/a) + 8*a**3*b**2*sqrt(-b/a)) - 3*b**2 *log(sqrt(x) + sqrt(-b/a))/(8*a**5*x**2*sqrt(-b/a) + 16*a**4*b*x*sqrt(-b/a ) + 8*a**3*b**2*sqrt(-b/a)), True))
Time = 0.29 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.89 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{3/2}} \, dx=-\frac {\frac {5 \, a}{\sqrt {x}} + \frac {3 \, b}{x^{\frac {3}{2}}}}{4 \, {\left (a^{4} + \frac {2 \, a^{3} b}{x} + \frac {a^{2} b^{2}}{x^{2}}\right )}} - \frac {3 \, \arctan \left (\frac {b}{\sqrt {a b} \sqrt {x}}\right )}{4 \, \sqrt {a b} a^{2}} \]
-1/4*(5*a/sqrt(x) + 3*b/x^(3/2))/(a^4 + 2*a^3*b/x + a^2*b^2/x^2) - 3/4*arc tan(b/(sqrt(a*b)*sqrt(x)))/(sqrt(a*b)*a^2)
Time = 0.27 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.67 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{3/2}} \, dx=\frac {3 \, \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{2}} - \frac {5 \, a x^{\frac {3}{2}} + 3 \, b \sqrt {x}}{4 \, {\left (a x + b\right )}^{2} a^{2}} \]
3/4*arctan(a*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^2) - 1/4*(5*a*x^(3/2) + 3*b*s qrt(x))/((a*x + b)^2*a^2)
Time = 5.89 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.83 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{3/2}} \, dx=\frac {3\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {x}}{\sqrt {b}}\right )}{4\,a^{5/2}\,\sqrt {b}}-\frac {\frac {5\,x^{3/2}}{4\,a}+\frac {3\,b\,\sqrt {x}}{4\,a^2}}{a^2\,x^2+2\,a\,b\,x+b^2} \]