Integrand size = 15, antiderivative size = 73 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{5/2}} \, dx=-\frac {\sqrt {x}}{2 a (b+a x)^2}+\frac {\sqrt {x}}{4 a b (b+a x)}+\frac {\arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{4 a^{3/2} b^{3/2}} \]
1/4*arctan(a^(1/2)*x^(1/2)/b^(1/2))/a^(3/2)/b^(3/2)-1/2*x^(1/2)/a/(a*x+b)^ 2+1/4*x^(1/2)/a/b/(a*x+b)
Time = 0.15 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.84 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{5/2}} \, dx=\frac {\sqrt {x} (-b+a x)}{4 a b (b+a x)^2}+\frac {\arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{4 a^{3/2} b^{3/2}} \]
(Sqrt[x]*(-b + a*x))/(4*a*b*(b + a*x)^2) + ArcTan[(Sqrt[a]*Sqrt[x])/Sqrt[b ]]/(4*a^(3/2)*b^(3/2))
Time = 0.18 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {795, 51, 52, 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.
\(\displaystyle \int \frac {1}{x^{5/2} \left (a+\frac {b}{x}\right )^3} \, dx\) |
\(\Big \downarrow \) 795 |
\(\displaystyle \int \frac {\sqrt {x}}{(a x+b)^3}dx\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {\int \frac {1}{\sqrt {x} (b+a x)^2}dx}{4 a}-\frac {\sqrt {x}}{2 a (a x+b)^2}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {\frac {\int \frac {1}{\sqrt {x} (b+a x)}dx}{2 b}+\frac {\sqrt {x}}{b (a x+b)}}{4 a}-\frac {\sqrt {x}}{2 a (a x+b)^2}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {\frac {\int \frac {1}{b+a x}d\sqrt {x}}{b}+\frac {\sqrt {x}}{b (a x+b)}}{4 a}-\frac {\sqrt {x}}{2 a (a x+b)^2}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\frac {\arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{\sqrt {a} b^{3/2}}+\frac {\sqrt {x}}{b (a x+b)}}{4 a}-\frac {\sqrt {x}}{2 a (a x+b)^2}\) |
-1/2*Sqrt[x]/(a*(b + a*x)^2) + (Sqrt[x]/(b*(b + a*x)) + ArcTan[(Sqrt[a]*Sq rt[x])/Sqrt[b]]/(Sqrt[a]*b^(3/2)))/(4*a)
3.17.86.3.1 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] :> 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] :> 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 = 52, normalized size of antiderivative = 0.71
method | result | size |
derivativedivides | \(\frac {\frac {x^{\frac {3}{2}}}{4 b}-\frac {\sqrt {x}}{4 a}}{\left (a x +b \right )^{2}}+\frac {\arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{4 b a \sqrt {a b}}\) | \(52\) |
default | \(\frac {\frac {x^{\frac {3}{2}}}{4 b}-\frac {\sqrt {x}}{4 a}}{\left (a x +b \right )^{2}}+\frac {\arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{4 b a \sqrt {a b}}\) | \(52\) |
Time = 0.26 (sec) , antiderivative size = 186, normalized size of antiderivative = 2.55 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{5/2}} \, dx=\left [-\frac {{\left (a^{2} x^{2} + 2 \, a b x + b^{2}\right )} \sqrt {-a b} \log \left (\frac {a x - b - 2 \, \sqrt {-a b} \sqrt {x}}{a x + b}\right ) - 2 \, {\left (a^{2} b x - a b^{2}\right )} \sqrt {x}}{8 \, {\left (a^{4} b^{2} x^{2} + 2 \, a^{3} b^{3} x + a^{2} b^{4}\right )}}, -\frac {{\left (a^{2} x^{2} + 2 \, a b x + b^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b}}{a \sqrt {x}}\right ) - {\left (a^{2} b x - a b^{2}\right )} \sqrt {x}}{4 \, {\left (a^{4} b^{2} x^{2} + 2 \, a^{3} b^{3} x + a^{2} b^{4}\right )}}\right ] \]
[-1/8*((a^2*x^2 + 2*a*b*x + b^2)*sqrt(-a*b)*log((a*x - b - 2*sqrt(-a*b)*sq rt(x))/(a*x + b)) - 2*(a^2*b*x - a*b^2)*sqrt(x))/(a^4*b^2*x^2 + 2*a^3*b^3* x + a^2*b^4), -1/4*((a^2*x^2 + 2*a*b*x + b^2)*sqrt(a*b)*arctan(sqrt(a*b)/( a*sqrt(x))) - (a^2*b*x - a*b^2)*sqrt(x))/(a^4*b^2*x^2 + 2*a^3*b^3*x + a^2* b^4)]
Leaf count of result is larger than twice the leaf count of optimal. 627 vs. \(2 (58) = 116\).
Time = 39.82 (sec) , antiderivative size = 627, normalized size of antiderivative = 8.59 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{5/2}} \, dx=\begin {cases} \tilde {\infty } x^{\frac {3}{2}} & \text {for}\: a = 0 \wedge b = 0 \\\frac {2 x^{\frac {3}{2}}}{3 b^{3}} & \text {for}\: a = 0 \\- \frac {2}{3 a^{3} x^{\frac {3}{2}}} & \text {for}\: b = 0 \\\frac {2 a^{2} x^{\frac {3}{2}} \sqrt {- \frac {b}{a}}}{8 a^{4} b x^{2} \sqrt {- \frac {b}{a}} + 16 a^{3} b^{2} x \sqrt {- \frac {b}{a}} + 8 a^{2} b^{3} \sqrt {- \frac {b}{a}}} + \frac {a^{2} x^{2} \log {\left (\sqrt {x} - \sqrt {- \frac {b}{a}} \right )}}{8 a^{4} b x^{2} \sqrt {- \frac {b}{a}} + 16 a^{3} b^{2} x \sqrt {- \frac {b}{a}} + 8 a^{2} b^{3} \sqrt {- \frac {b}{a}}} - \frac {a^{2} x^{2} \log {\left (\sqrt {x} + \sqrt {- \frac {b}{a}} \right )}}{8 a^{4} b x^{2} \sqrt {- \frac {b}{a}} + 16 a^{3} b^{2} x \sqrt {- \frac {b}{a}} + 8 a^{2} b^{3} \sqrt {- \frac {b}{a}}} - \frac {2 a b \sqrt {x} \sqrt {- \frac {b}{a}}}{8 a^{4} b x^{2} \sqrt {- \frac {b}{a}} + 16 a^{3} b^{2} x \sqrt {- \frac {b}{a}} + 8 a^{2} b^{3} \sqrt {- \frac {b}{a}}} + \frac {2 a b x \log {\left (\sqrt {x} - \sqrt {- \frac {b}{a}} \right )}}{8 a^{4} b x^{2} \sqrt {- \frac {b}{a}} + 16 a^{3} b^{2} x \sqrt {- \frac {b}{a}} + 8 a^{2} b^{3} \sqrt {- \frac {b}{a}}} - \frac {2 a b x \log {\left (\sqrt {x} + \sqrt {- \frac {b}{a}} \right )}}{8 a^{4} b x^{2} \sqrt {- \frac {b}{a}} + 16 a^{3} b^{2} x \sqrt {- \frac {b}{a}} + 8 a^{2} b^{3} \sqrt {- \frac {b}{a}}} + \frac {b^{2} \log {\left (\sqrt {x} - \sqrt {- \frac {b}{a}} \right )}}{8 a^{4} b x^{2} \sqrt {- \frac {b}{a}} + 16 a^{3} b^{2} x \sqrt {- \frac {b}{a}} + 8 a^{2} b^{3} \sqrt {- \frac {b}{a}}} - \frac {b^{2} \log {\left (\sqrt {x} + \sqrt {- \frac {b}{a}} \right )}}{8 a^{4} b x^{2} \sqrt {- \frac {b}{a}} + 16 a^{3} b^{2} x \sqrt {- \frac {b}{a}} + 8 a^{2} b^{3} \sqrt {- \frac {b}{a}}} & \text {otherwise} \end {cases} \]
Piecewise((zoo*x**(3/2), Eq(a, 0) & Eq(b, 0)), (2*x**(3/2)/(3*b**3), Eq(a, 0)), (-2/(3*a**3*x**(3/2)), Eq(b, 0)), (2*a**2*x**(3/2)*sqrt(-b/a)/(8*a** 4*b*x**2*sqrt(-b/a) + 16*a**3*b**2*x*sqrt(-b/a) + 8*a**2*b**3*sqrt(-b/a)) + a**2*x**2*log(sqrt(x) - sqrt(-b/a))/(8*a**4*b*x**2*sqrt(-b/a) + 16*a**3* b**2*x*sqrt(-b/a) + 8*a**2*b**3*sqrt(-b/a)) - a**2*x**2*log(sqrt(x) + sqrt (-b/a))/(8*a**4*b*x**2*sqrt(-b/a) + 16*a**3*b**2*x*sqrt(-b/a) + 8*a**2*b** 3*sqrt(-b/a)) - 2*a*b*sqrt(x)*sqrt(-b/a)/(8*a**4*b*x**2*sqrt(-b/a) + 16*a* *3*b**2*x*sqrt(-b/a) + 8*a**2*b**3*sqrt(-b/a)) + 2*a*b*x*log(sqrt(x) - sqr t(-b/a))/(8*a**4*b*x**2*sqrt(-b/a) + 16*a**3*b**2*x*sqrt(-b/a) + 8*a**2*b* *3*sqrt(-b/a)) - 2*a*b*x*log(sqrt(x) + sqrt(-b/a))/(8*a**4*b*x**2*sqrt(-b/ a) + 16*a**3*b**2*x*sqrt(-b/a) + 8*a**2*b**3*sqrt(-b/a)) + b**2*log(sqrt(x ) - sqrt(-b/a))/(8*a**4*b*x**2*sqrt(-b/a) + 16*a**3*b**2*x*sqrt(-b/a) + 8* a**2*b**3*sqrt(-b/a)) - b**2*log(sqrt(x) + sqrt(-b/a))/(8*a**4*b*x**2*sqrt (-b/a) + 16*a**3*b**2*x*sqrt(-b/a) + 8*a**2*b**3*sqrt(-b/a)), True))
Time = 0.30 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.90 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{5/2}} \, dx=\frac {\frac {a}{\sqrt {x}} - \frac {b}{x^{\frac {3}{2}}}}{4 \, {\left (a^{3} b + \frac {2 \, a^{2} b^{2}}{x} + \frac {a b^{3}}{x^{2}}\right )}} - \frac {\arctan \left (\frac {b}{\sqrt {a b} \sqrt {x}}\right )}{4 \, \sqrt {a b} a b} \]
1/4*(a/sqrt(x) - b/x^(3/2))/(a^3*b + 2*a^2*b^2/x + a*b^3/x^2) - 1/4*arctan (b/(sqrt(a*b)*sqrt(x)))/(sqrt(a*b)*a*b)
Time = 0.29 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.71 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{5/2}} \, dx=\frac {\arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a b} + \frac {a x^{\frac {3}{2}} - b \sqrt {x}}{4 \, {\left (a x + b\right )}^{2} a b} \]
1/4*arctan(a*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a*b) + 1/4*(a*x^(3/2) - b*sqrt( x))/((a*x + b)^2*a*b)
Time = 5.87 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.78 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{5/2}} \, dx=\frac {\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {x}}{\sqrt {b}}\right )}{4\,a^{3/2}\,b^{3/2}}-\frac {\frac {\sqrt {x}}{4\,a}-\frac {x^{3/2}}{4\,b}}{a^2\,x^2+2\,a\,b\,x+b^2} \]