Integrand size = 15, antiderivative size = 70 \[ \int \frac {\sqrt {x}}{\left (a+\frac {b}{x}\right )^2} \, dx=-\frac {5 b \sqrt {x}}{a^3}+\frac {5 x^{3/2}}{3 a^2}-\frac {x^{5/2}}{a (b+a x)}+\frac {5 b^{3/2} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{a^{7/2}} \]
5/3*x^(3/2)/a^2-x^(5/2)/a/(a*x+b)+5*b^(3/2)*arctan(a^(1/2)*x^(1/2)/b^(1/2) )/a^(7/2)-5*b*x^(1/2)/a^3
Time = 0.13 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.97 \[ \int \frac {\sqrt {x}}{\left (a+\frac {b}{x}\right )^2} \, dx=\frac {\sqrt {x} \left (-15 b^2-10 a b x+2 a^2 x^2\right )}{3 a^3 (b+a x)}+\frac {5 b^{3/2} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{a^{7/2}} \]
(Sqrt[x]*(-15*b^2 - 10*a*b*x + 2*a^2*x^2))/(3*a^3*(b + a*x)) + (5*b^(3/2)* ArcTan[(Sqrt[a]*Sqrt[x])/Sqrt[b]])/a^(7/2)
Time = 0.18 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.20, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {795, 51, 60, 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.
\(\displaystyle \int \frac {\sqrt {x}}{\left (a+\frac {b}{x}\right )^2} \, dx\) |
\(\Big \downarrow \) 795 |
\(\displaystyle \int \frac {x^{5/2}}{(a x+b)^2}dx\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {5 \int \frac {x^{3/2}}{b+a x}dx}{2 a}-\frac {x^{5/2}}{a (a x+b)}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {5 \left (\frac {2 x^{3/2}}{3 a}-\frac {b \int \frac {\sqrt {x}}{b+a x}dx}{a}\right )}{2 a}-\frac {x^{5/2}}{a (a x+b)}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {5 \left (\frac {2 x^{3/2}}{3 a}-\frac {b \left (\frac {2 \sqrt {x}}{a}-\frac {b \int \frac {1}{\sqrt {x} (b+a x)}dx}{a}\right )}{a}\right )}{2 a}-\frac {x^{5/2}}{a (a x+b)}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {5 \left (\frac {2 x^{3/2}}{3 a}-\frac {b \left (\frac {2 \sqrt {x}}{a}-\frac {2 b \int \frac {1}{b+a x}d\sqrt {x}}{a}\right )}{a}\right )}{2 a}-\frac {x^{5/2}}{a (a x+b)}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {5 \left (\frac {2 x^{3/2}}{3 a}-\frac {b \left (\frac {2 \sqrt {x}}{a}-\frac {2 \sqrt {b} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{a^{3/2}}\right )}{a}\right )}{2 a}-\frac {x^{5/2}}{a (a x+b)}\) |
-(x^(5/2)/(a*(b + a*x))) + (5*((2*x^(3/2))/(3*a) - (b*((2*Sqrt[x])/a - (2* Sqrt[b]*ArcTan[(Sqrt[a]*Sqrt[x])/Sqrt[b]])/a^(3/2)))/a))/(2*a)
3.17.75.3.1 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 = 0.06 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.79
method | result | size |
risch | \(\frac {2 \left (a x -6 b \right ) \sqrt {x}}{3 a^{3}}+\frac {b^{2} \left (-\frac {\sqrt {x}}{a x +b}+\frac {5 \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}}\right )}{a^{3}}\) | \(55\) |
derivativedivides | \(\frac {\frac {2 a \,x^{\frac {3}{2}}}{3}-4 b \sqrt {x}}{a^{3}}+\frac {2 b^{2} \left (-\frac {\sqrt {x}}{2 \left (a x +b \right )}+\frac {5 \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{a^{3}}\) | \(59\) |
default | \(\frac {\frac {2 a \,x^{\frac {3}{2}}}{3}-4 b \sqrt {x}}{a^{3}}+\frac {2 b^{2} \left (-\frac {\sqrt {x}}{2 \left (a x +b \right )}+\frac {5 \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{a^{3}}\) | \(59\) |
2/3*(a*x-6*b)*x^(1/2)/a^3+1/a^3*b^2*(-x^(1/2)/(a*x+b)+5/(a*b)^(1/2)*arctan (a*x^(1/2)/(a*b)^(1/2)))
Time = 0.28 (sec) , antiderivative size = 161, normalized size of antiderivative = 2.30 \[ \int \frac {\sqrt {x}}{\left (a+\frac {b}{x}\right )^2} \, dx=\left [\frac {15 \, {\left (a b x + b^{2}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {a x + 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - b}{a x + b}\right ) + 2 \, {\left (2 \, a^{2} x^{2} - 10 \, a b x - 15 \, b^{2}\right )} \sqrt {x}}{6 \, {\left (a^{4} x + a^{3} b\right )}}, \frac {15 \, {\left (a b x + b^{2}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {x} \sqrt {\frac {b}{a}}}{b}\right ) + {\left (2 \, a^{2} x^{2} - 10 \, a b x - 15 \, b^{2}\right )} \sqrt {x}}{3 \, {\left (a^{4} x + a^{3} b\right )}}\right ] \]
[1/6*(15*(a*b*x + b^2)*sqrt(-b/a)*log((a*x + 2*a*sqrt(x)*sqrt(-b/a) - b)/( a*x + b)) + 2*(2*a^2*x^2 - 10*a*b*x - 15*b^2)*sqrt(x))/(a^4*x + a^3*b), 1/ 3*(15*(a*b*x + b^2)*sqrt(b/a)*arctan(a*sqrt(x)*sqrt(b/a)/b) + (2*a^2*x^2 - 10*a*b*x - 15*b^2)*sqrt(x))/(a^4*x + a^3*b)]
Leaf count of result is larger than twice the leaf count of optimal. 389 vs. \(2 (63) = 126\).
Time = 2.11 (sec) , antiderivative size = 389, normalized size of antiderivative = 5.56 \[ \int \frac {\sqrt {x}}{\left (a+\frac {b}{x}\right )^2} \, dx=\begin {cases} \tilde {\infty } x^{\frac {7}{2}} & \text {for}\: a = 0 \wedge b = 0 \\\frac {2 x^{\frac {7}{2}}}{7 b^{2}} & \text {for}\: a = 0 \\\frac {2 x^{\frac {3}{2}}}{3 a^{2}} & \text {for}\: b = 0 \\\frac {4 a^{3} x^{\frac {5}{2}} \sqrt {- \frac {b}{a}}}{6 a^{5} x \sqrt {- \frac {b}{a}} + 6 a^{4} b \sqrt {- \frac {b}{a}}} - \frac {20 a^{2} b x^{\frac {3}{2}} \sqrt {- \frac {b}{a}}}{6 a^{5} x \sqrt {- \frac {b}{a}} + 6 a^{4} b \sqrt {- \frac {b}{a}}} - \frac {30 a b^{2} \sqrt {x} \sqrt {- \frac {b}{a}}}{6 a^{5} x \sqrt {- \frac {b}{a}} + 6 a^{4} b \sqrt {- \frac {b}{a}}} + \frac {15 a b^{2} x \log {\left (\sqrt {x} - \sqrt {- \frac {b}{a}} \right )}}{6 a^{5} x \sqrt {- \frac {b}{a}} + 6 a^{4} b \sqrt {- \frac {b}{a}}} - \frac {15 a b^{2} x \log {\left (\sqrt {x} + \sqrt {- \frac {b}{a}} \right )}}{6 a^{5} x \sqrt {- \frac {b}{a}} + 6 a^{4} b \sqrt {- \frac {b}{a}}} + \frac {15 b^{3} \log {\left (\sqrt {x} - \sqrt {- \frac {b}{a}} \right )}}{6 a^{5} x \sqrt {- \frac {b}{a}} + 6 a^{4} b \sqrt {- \frac {b}{a}}} - \frac {15 b^{3} \log {\left (\sqrt {x} + \sqrt {- \frac {b}{a}} \right )}}{6 a^{5} x \sqrt {- \frac {b}{a}} + 6 a^{4} b \sqrt {- \frac {b}{a}}} & \text {otherwise} \end {cases} \]
Piecewise((zoo*x**(7/2), Eq(a, 0) & Eq(b, 0)), (2*x**(7/2)/(7*b**2), Eq(a, 0)), (2*x**(3/2)/(3*a**2), Eq(b, 0)), (4*a**3*x**(5/2)*sqrt(-b/a)/(6*a**5 *x*sqrt(-b/a) + 6*a**4*b*sqrt(-b/a)) - 20*a**2*b*x**(3/2)*sqrt(-b/a)/(6*a* *5*x*sqrt(-b/a) + 6*a**4*b*sqrt(-b/a)) - 30*a*b**2*sqrt(x)*sqrt(-b/a)/(6*a **5*x*sqrt(-b/a) + 6*a**4*b*sqrt(-b/a)) + 15*a*b**2*x*log(sqrt(x) - sqrt(- b/a))/(6*a**5*x*sqrt(-b/a) + 6*a**4*b*sqrt(-b/a)) - 15*a*b**2*x*log(sqrt(x ) + sqrt(-b/a))/(6*a**5*x*sqrt(-b/a) + 6*a**4*b*sqrt(-b/a)) + 15*b**3*log( sqrt(x) - sqrt(-b/a))/(6*a**5*x*sqrt(-b/a) + 6*a**4*b*sqrt(-b/a)) - 15*b** 3*log(sqrt(x) + sqrt(-b/a))/(6*a**5*x*sqrt(-b/a) + 6*a**4*b*sqrt(-b/a)), T rue))
Time = 0.28 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.94 \[ \int \frac {\sqrt {x}}{\left (a+\frac {b}{x}\right )^2} \, dx=\frac {2 \, a^{2} - \frac {10 \, a b}{x} - \frac {15 \, b^{2}}{x^{2}}}{3 \, {\left (\frac {a^{4}}{x^{\frac {3}{2}}} + \frac {a^{3} b}{x^{\frac {5}{2}}}\right )}} - \frac {5 \, b^{2} \arctan \left (\frac {b}{\sqrt {a b} \sqrt {x}}\right )}{\sqrt {a b} a^{3}} \]
1/3*(2*a^2 - 10*a*b/x - 15*b^2/x^2)/(a^4/x^(3/2) + a^3*b/x^(5/2)) - 5*b^2* arctan(b/(sqrt(a*b)*sqrt(x)))/(sqrt(a*b)*a^3)
Time = 0.26 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.93 \[ \int \frac {\sqrt {x}}{\left (a+\frac {b}{x}\right )^2} \, dx=\frac {5 \, b^{2} \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} - \frac {b^{2} \sqrt {x}}{{\left (a x + b\right )} a^{3}} + \frac {2 \, {\left (a^{4} x^{\frac {3}{2}} - 6 \, a^{3} b \sqrt {x}\right )}}{3 \, a^{6}} \]
5*b^2*arctan(a*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^3) - b^2*sqrt(x)/((a*x + b) *a^3) + 2/3*(a^4*x^(3/2) - 6*a^3*b*sqrt(x))/a^6
Time = 0.07 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.83 \[ \int \frac {\sqrt {x}}{\left (a+\frac {b}{x}\right )^2} \, dx=\frac {2\,x^{3/2}}{3\,a^2}-\frac {4\,b\,\sqrt {x}}{a^3}-\frac {b^2\,\sqrt {x}}{x\,a^4+b\,a^3}+\frac {5\,b^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {x}}{\sqrt {b}}\right )}{a^{7/2}} \]