Integrand size = 15, antiderivative size = 82 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 \sqrt {x}} \, dx=\frac {15 \sqrt {x}}{4 a^3}-\frac {x^{5/2}}{2 a (b+a x)^2}-\frac {5 x^{3/2}}{4 a^2 (b+a x)}-\frac {15 \sqrt {b} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{4 a^{7/2}} \]
-1/2*x^(5/2)/a/(a*x+b)^2-5/4*x^(3/2)/a^2/(a*x+b)-15/4*arctan(a^(1/2)*x^(1/ 2)/b^(1/2))*b^(1/2)/a^(7/2)+15/4*x^(1/2)/a^3
Time = 0.20 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.85 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 \sqrt {x}} \, dx=\frac {\sqrt {x} \left (15 b^2+25 a b x+8 a^2 x^2\right )}{4 a^3 (b+a x)^2}-\frac {15 \sqrt {b} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{4 a^{7/2}} \]
(Sqrt[x]*(15*b^2 + 25*a*b*x + 8*a^2*x^2))/(4*a^3*(b + a*x)^2) - (15*Sqrt[b ]*ArcTan[(Sqrt[a]*Sqrt[x])/Sqrt[b]])/(4*a^(7/2))
Time = 0.19 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.12, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {795, 51, 51, 60, 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.
\(\displaystyle \int \frac {1}{\sqrt {x} \left (a+\frac {b}{x}\right )^3} \, dx\) |
\(\Big \downarrow \) 795 |
\(\displaystyle \int \frac {x^{5/2}}{(a x+b)^3}dx\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {5 \int \frac {x^{3/2}}{(b+a x)^2}dx}{4 a}-\frac {x^{5/2}}{2 a (a x+b)^2}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {5 \left (\frac {3 \int \frac {\sqrt {x}}{b+a x}dx}{2 a}-\frac {x^{3/2}}{a (a x+b)}\right )}{4 a}-\frac {x^{5/2}}{2 a (a x+b)^2}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {5 \left (\frac {3 \left (\frac {2 \sqrt {x}}{a}-\frac {b \int \frac {1}{\sqrt {x} (b+a x)}dx}{a}\right )}{2 a}-\frac {x^{3/2}}{a (a x+b)}\right )}{4 a}-\frac {x^{5/2}}{2 a (a x+b)^2}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {5 \left (\frac {3 \left (\frac {2 \sqrt {x}}{a}-\frac {2 b \int \frac {1}{b+a x}d\sqrt {x}}{a}\right )}{2 a}-\frac {x^{3/2}}{a (a x+b)}\right )}{4 a}-\frac {x^{5/2}}{2 a (a x+b)^2}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {5 \left (\frac {3 \left (\frac {2 \sqrt {x}}{a}-\frac {2 \sqrt {b} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{a^{3/2}}\right )}{2 a}-\frac {x^{3/2}}{a (a x+b)}\right )}{4 a}-\frac {x^{5/2}}{2 a (a x+b)^2}\) |
-1/2*x^(5/2)/(a*(b + a*x)^2) + (5*(-(x^(3/2)/(a*(b + a*x))) + (3*((2*Sqrt[ x])/a - (2*Sqrt[b]*ArcTan[(Sqrt[a]*Sqrt[x])/Sqrt[b]])/a^(3/2)))/(2*a)))/(4 *a)
3.17.84.3.1 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/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[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 = 0.08 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.68
method | result | size |
derivativedivides | \(\frac {2 \sqrt {x}}{a^{3}}-\frac {2 b \left (\frac {-\frac {9 a \,x^{\frac {3}{2}}}{8}-\frac {7 b \sqrt {x}}{8}}{\left (a x +b \right )^{2}}+\frac {15 \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{a^{3}}\) | \(56\) |
default | \(\frac {2 \sqrt {x}}{a^{3}}-\frac {2 b \left (\frac {-\frac {9 a \,x^{\frac {3}{2}}}{8}-\frac {7 b \sqrt {x}}{8}}{\left (a x +b \right )^{2}}+\frac {15 \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{a^{3}}\) | \(56\) |
risch | \(\frac {2 \sqrt {x}}{a^{3}}-\frac {b \left (\frac {-\frac {9 a \,x^{\frac {3}{2}}}{4}-\frac {7 b \sqrt {x}}{4}}{\left (a x +b \right )^{2}}+\frac {15 \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{4 \sqrt {a b}}\right )}{a^{3}}\) | \(57\) |
2*x^(1/2)/a^3-2/a^3*b*((-9/8*a*x^(3/2)-7/8*b*x^(1/2))/(a*x+b)^2+15/8/(a*b) ^(1/2)*arctan(a*x^(1/2)/(a*b)^(1/2)))
Time = 0.26 (sec) , antiderivative size = 200, normalized size of antiderivative = 2.44 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 \sqrt {x}} \, dx=\left [\frac {15 \, {\left (a^{2} x^{2} + 2 \, a b x + b^{2}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {a x - 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - b}{a x + b}\right ) + 2 \, {\left (8 \, a^{2} x^{2} + 25 \, a b x + 15 \, b^{2}\right )} \sqrt {x}}{8 \, {\left (a^{5} x^{2} + 2 \, a^{4} b x + a^{3} b^{2}\right )}}, -\frac {15 \, {\left (a^{2} x^{2} + 2 \, a b x + b^{2}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {x} \sqrt {\frac {b}{a}}}{b}\right ) - {\left (8 \, a^{2} x^{2} + 25 \, a b x + 15 \, b^{2}\right )} \sqrt {x}}{4 \, {\left (a^{5} x^{2} + 2 \, a^{4} b x + a^{3} b^{2}\right )}}\right ] \]
[1/8*(15*(a^2*x^2 + 2*a*b*x + b^2)*sqrt(-b/a)*log((a*x - 2*a*sqrt(x)*sqrt( -b/a) - b)/(a*x + b)) + 2*(8*a^2*x^2 + 25*a*b*x + 15*b^2)*sqrt(x))/(a^5*x^ 2 + 2*a^4*b*x + a^3*b^2), -1/4*(15*(a^2*x^2 + 2*a*b*x + b^2)*sqrt(b/a)*arc tan(a*sqrt(x)*sqrt(b/a)/b) - (8*a^2*x^2 + 25*a*b*x + 15*b^2)*sqrt(x))/(a^5 *x^2 + 2*a^4*b*x + a^3*b^2)]
Leaf count of result is larger than twice the leaf count of optimal. 683 vs. \(2 (73) = 146\).
Time = 7.88 (sec) , antiderivative size = 683, normalized size of antiderivative = 8.33 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 \sqrt {x}} \, dx=\begin {cases} \tilde {\infty } x^{\frac {7}{2}} & \text {for}\: a = 0 \wedge b = 0 \\\frac {2 x^{\frac {7}{2}}}{7 b^{3}} & \text {for}\: a = 0 \\\frac {2 \sqrt {x}}{a^{3}} & \text {for}\: b = 0 \\\frac {16 a^{3} x^{\frac {5}{2}} \sqrt {- \frac {b}{a}}}{8 a^{6} x^{2} \sqrt {- \frac {b}{a}} + 16 a^{5} b x \sqrt {- \frac {b}{a}} + 8 a^{4} b^{2} \sqrt {- \frac {b}{a}}} + \frac {50 a^{2} b x^{\frac {3}{2}} \sqrt {- \frac {b}{a}}}{8 a^{6} x^{2} \sqrt {- \frac {b}{a}} + 16 a^{5} b x \sqrt {- \frac {b}{a}} + 8 a^{4} b^{2} \sqrt {- \frac {b}{a}}} - \frac {15 a^{2} b x^{2} \log {\left (\sqrt {x} - \sqrt {- \frac {b}{a}} \right )}}{8 a^{6} x^{2} \sqrt {- \frac {b}{a}} + 16 a^{5} b x \sqrt {- \frac {b}{a}} + 8 a^{4} b^{2} \sqrt {- \frac {b}{a}}} + \frac {15 a^{2} b x^{2} \log {\left (\sqrt {x} + \sqrt {- \frac {b}{a}} \right )}}{8 a^{6} x^{2} \sqrt {- \frac {b}{a}} + 16 a^{5} b x \sqrt {- \frac {b}{a}} + 8 a^{4} b^{2} \sqrt {- \frac {b}{a}}} + \frac {30 a b^{2} \sqrt {x} \sqrt {- \frac {b}{a}}}{8 a^{6} x^{2} \sqrt {- \frac {b}{a}} + 16 a^{5} b x \sqrt {- \frac {b}{a}} + 8 a^{4} b^{2} \sqrt {- \frac {b}{a}}} - \frac {30 a b^{2} x \log {\left (\sqrt {x} - \sqrt {- \frac {b}{a}} \right )}}{8 a^{6} x^{2} \sqrt {- \frac {b}{a}} + 16 a^{5} b x \sqrt {- \frac {b}{a}} + 8 a^{4} b^{2} \sqrt {- \frac {b}{a}}} + \frac {30 a b^{2} x \log {\left (\sqrt {x} + \sqrt {- \frac {b}{a}} \right )}}{8 a^{6} x^{2} \sqrt {- \frac {b}{a}} + 16 a^{5} b x \sqrt {- \frac {b}{a}} + 8 a^{4} b^{2} \sqrt {- \frac {b}{a}}} - \frac {15 b^{3} \log {\left (\sqrt {x} - \sqrt {- \frac {b}{a}} \right )}}{8 a^{6} x^{2} \sqrt {- \frac {b}{a}} + 16 a^{5} b x \sqrt {- \frac {b}{a}} + 8 a^{4} b^{2} \sqrt {- \frac {b}{a}}} + \frac {15 b^{3} \log {\left (\sqrt {x} + \sqrt {- \frac {b}{a}} \right )}}{8 a^{6} x^{2} \sqrt {- \frac {b}{a}} + 16 a^{5} b x \sqrt {- \frac {b}{a}} + 8 a^{4} b^{2} \sqrt {- \frac {b}{a}}} & \text {otherwise} \end {cases} \]
Piecewise((zoo*x**(7/2), Eq(a, 0) & Eq(b, 0)), (2*x**(7/2)/(7*b**3), Eq(a, 0)), (2*sqrt(x)/a**3, Eq(b, 0)), (16*a**3*x**(5/2)*sqrt(-b/a)/(8*a**6*x** 2*sqrt(-b/a) + 16*a**5*b*x*sqrt(-b/a) + 8*a**4*b**2*sqrt(-b/a)) + 50*a**2* b*x**(3/2)*sqrt(-b/a)/(8*a**6*x**2*sqrt(-b/a) + 16*a**5*b*x*sqrt(-b/a) + 8 *a**4*b**2*sqrt(-b/a)) - 15*a**2*b*x**2*log(sqrt(x) - sqrt(-b/a))/(8*a**6* x**2*sqrt(-b/a) + 16*a**5*b*x*sqrt(-b/a) + 8*a**4*b**2*sqrt(-b/a)) + 15*a* *2*b*x**2*log(sqrt(x) + sqrt(-b/a))/(8*a**6*x**2*sqrt(-b/a) + 16*a**5*b*x* sqrt(-b/a) + 8*a**4*b**2*sqrt(-b/a)) + 30*a*b**2*sqrt(x)*sqrt(-b/a)/(8*a** 6*x**2*sqrt(-b/a) + 16*a**5*b*x*sqrt(-b/a) + 8*a**4*b**2*sqrt(-b/a)) - 30* a*b**2*x*log(sqrt(x) - sqrt(-b/a))/(8*a**6*x**2*sqrt(-b/a) + 16*a**5*b*x*s qrt(-b/a) + 8*a**4*b**2*sqrt(-b/a)) + 30*a*b**2*x*log(sqrt(x) + sqrt(-b/a) )/(8*a**6*x**2*sqrt(-b/a) + 16*a**5*b*x*sqrt(-b/a) + 8*a**4*b**2*sqrt(-b/a )) - 15*b**3*log(sqrt(x) - sqrt(-b/a))/(8*a**6*x**2*sqrt(-b/a) + 16*a**5*b *x*sqrt(-b/a) + 8*a**4*b**2*sqrt(-b/a)) + 15*b**3*log(sqrt(x) + sqrt(-b/a) )/(8*a**6*x**2*sqrt(-b/a) + 16*a**5*b*x*sqrt(-b/a) + 8*a**4*b**2*sqrt(-b/a )), True))
Time = 0.28 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.91 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 \sqrt {x}} \, dx=\frac {8 \, a^{2} + \frac {25 \, a b}{x} + \frac {15 \, b^{2}}{x^{2}}}{4 \, {\left (\frac {a^{5}}{\sqrt {x}} + \frac {2 \, a^{4} b}{x^{\frac {3}{2}}} + \frac {a^{3} b^{2}}{x^{\frac {5}{2}}}\right )}} + \frac {15 \, b \arctan \left (\frac {b}{\sqrt {a b} \sqrt {x}}\right )}{4 \, \sqrt {a b} a^{3}} \]
1/4*(8*a^2 + 25*a*b/x + 15*b^2/x^2)/(a^5/sqrt(x) + 2*a^4*b/x^(3/2) + a^3*b ^2/x^(5/2)) + 15/4*b*arctan(b/(sqrt(a*b)*sqrt(x)))/(sqrt(a*b)*a^3)
Time = 0.27 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.72 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 \sqrt {x}} \, dx=-\frac {15 \, b \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{3}} + \frac {2 \, \sqrt {x}}{a^{3}} + \frac {9 \, a b x^{\frac {3}{2}} + 7 \, b^{2} \sqrt {x}}{4 \, {\left (a x + b\right )}^{2} a^{3}} \]
-15/4*b*arctan(a*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^3) + 2*sqrt(x)/a^3 + 1/4* (9*a*b*x^(3/2) + 7*b^2*sqrt(x))/((a*x + b)^2*a^3)
Time = 5.83 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.84 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 \sqrt {x}} \, dx=\frac {\frac {7\,b^2\,\sqrt {x}}{4}+\frac {9\,a\,b\,x^{3/2}}{4}}{a^5\,x^2+2\,a^4\,b\,x+a^3\,b^2}+\frac {2\,\sqrt {x}}{a^3}-\frac {15\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {x}}{\sqrt {b}}\right )}{4\,a^{7/2}} \]