Integrand size = 17, antiderivative size = 94 \[ \int \left (a+\frac {b}{x}\right )^{5/2} \sqrt {x} \, dx=-\frac {5 b^2 \sqrt {a+\frac {b}{x}}}{\sqrt {x}}+\frac {10}{3} b \left (a+\frac {b}{x}\right )^{3/2} \sqrt {x}+\frac {2}{3} \left (a+\frac {b}{x}\right )^{5/2} x^{3/2}-5 a b^{3/2} \text {arctanh}\left (\frac {\sqrt {b}}{\sqrt {a+\frac {b}{x}} \sqrt {x}}\right ) \]
2/3*(a+b/x)^(5/2)*x^(3/2)-5*a*b^(3/2)*arctanh(b^(1/2)/(a+b/x)^(1/2)/x^(1/2 ))-5*b^2*(a+b/x)^(1/2)/x^(1/2)+10/3*b*(a+b/x)^(3/2)*x^(1/2)
Time = 6.90 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.91 \[ \int \left (a+\frac {b}{x}\right )^{5/2} \sqrt {x} \, dx=\frac {\sqrt {a+\frac {b}{x}} \sqrt {x} \left (\frac {\sqrt {b+a x} \left (-3 b^2+14 a b x+2 a^2 x^2\right )}{3 x}-5 a b^{3/2} \text {arctanh}\left (\frac {\sqrt {b+a x}}{\sqrt {b}}\right )\right )}{\sqrt {b+a x}} \]
(Sqrt[a + b/x]*Sqrt[x]*((Sqrt[b + a*x]*(-3*b^2 + 14*a*b*x + 2*a^2*x^2))/(3 *x) - 5*a*b^(3/2)*ArcTanh[Sqrt[b + a*x]/Sqrt[b]]))/Sqrt[b + a*x]
Time = 0.20 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.11, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {860, 247, 247, 211, 224, 219}
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.
\(\displaystyle \int \sqrt {x} \left (a+\frac {b}{x}\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 860 |
\(\displaystyle -2 \int \left (a+\frac {b}{x}\right )^{5/2} x^2d\frac {1}{\sqrt {x}}\) |
\(\Big \downarrow \) 247 |
\(\displaystyle -2 \left (\frac {5}{3} b \int \left (a+\frac {b}{x}\right )^{3/2} xd\frac {1}{\sqrt {x}}-\frac {1}{3} x^{3/2} \left (a+\frac {b}{x}\right )^{5/2}\right )\) |
\(\Big \downarrow \) 247 |
\(\displaystyle -2 \left (\frac {5}{3} b \left (3 b \int \sqrt {a+\frac {b}{x}}d\frac {1}{\sqrt {x}}-\sqrt {x} \left (a+\frac {b}{x}\right )^{3/2}\right )-\frac {1}{3} x^{3/2} \left (a+\frac {b}{x}\right )^{5/2}\right )\) |
\(\Big \downarrow \) 211 |
\(\displaystyle -2 \left (\frac {5}{3} b \left (3 b \left (\frac {1}{2} a \int \frac {1}{\sqrt {a+\frac {b}{x}}}d\frac {1}{\sqrt {x}}+\frac {\sqrt {a+\frac {b}{x}}}{2 \sqrt {x}}\right )-\sqrt {x} \left (a+\frac {b}{x}\right )^{3/2}\right )-\frac {1}{3} x^{3/2} \left (a+\frac {b}{x}\right )^{5/2}\right )\) |
\(\Big \downarrow \) 224 |
\(\displaystyle -2 \left (\frac {5}{3} b \left (3 b \left (\frac {1}{2} a \int \frac {1}{1-\frac {b}{x}}d\frac {1}{\sqrt {a+\frac {b}{x}} \sqrt {x}}+\frac {\sqrt {a+\frac {b}{x}}}{2 \sqrt {x}}\right )-\sqrt {x} \left (a+\frac {b}{x}\right )^{3/2}\right )-\frac {1}{3} x^{3/2} \left (a+\frac {b}{x}\right )^{5/2}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -2 \left (\frac {5}{3} b \left (3 b \left (\frac {a \text {arctanh}\left (\frac {\sqrt {b}}{\sqrt {x} \sqrt {a+\frac {b}{x}}}\right )}{2 \sqrt {b}}+\frac {\sqrt {a+\frac {b}{x}}}{2 \sqrt {x}}\right )-\sqrt {x} \left (a+\frac {b}{x}\right )^{3/2}\right )-\frac {1}{3} x^{3/2} \left (a+\frac {b}{x}\right )^{5/2}\right )\) |
-2*(-1/3*((a + b/x)^(5/2)*x^(3/2)) + (5*b*(-((a + b/x)^(3/2)*Sqrt[x]) + 3* b*(Sqrt[a + b/x]/(2*Sqrt[x]) + (a*ArcTanh[Sqrt[b]/(Sqrt[a + b/x]*Sqrt[x])] )/(2*Sqrt[b]))))/3)
3.18.73.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^p/(c*(m + 1))), x] - Simp[2*b*(p/(c^2*(m + 1))) Int[ (c*x)^(m + 2)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && LtQ[m, -1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Simp[-k/c Subst[Int[(a + b/(c^n*x^(k*n)))^p/x^(k*(m + 1 ) + 1), x], x, 1/(c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && ILtQ[n, 0] && FractionQ[m]
Time = 0.05 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.87
method | result | size |
risch | \(-\frac {b^{2} \sqrt {\frac {a x +b}{x}}}{\sqrt {x}}+\frac {a \left (\frac {4 \left (a x +b \right )^{\frac {3}{2}}}{3}+8 b \sqrt {a x +b}-10 b^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {a x +b}}{\sqrt {b}}\right )\right ) \sqrt {\frac {a x +b}{x}}\, \sqrt {x}}{2 \sqrt {a x +b}}\) | \(82\) |
default | \(-\frac {\sqrt {\frac {a x +b}{x}}\, \left (-2 a^{2} x^{2} \sqrt {b}\, \sqrt {a x +b}+15 \,\operatorname {arctanh}\left (\frac {\sqrt {a x +b}}{\sqrt {b}}\right ) a \,b^{2} x -14 a \,b^{\frac {3}{2}} x \sqrt {a x +b}+3 b^{\frac {5}{2}} \sqrt {a x +b}\right )}{3 \sqrt {x}\, \sqrt {a x +b}\, \sqrt {b}}\) | \(91\) |
-b^2/x^(1/2)*((a*x+b)/x)^(1/2)+1/2*a*(4/3*(a*x+b)^(3/2)+8*b*(a*x+b)^(1/2)- 10*b^(3/2)*arctanh((a*x+b)^(1/2)/b^(1/2)))*((a*x+b)/x)^(1/2)/(a*x+b)^(1/2) *x^(1/2)
Time = 0.30 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.64 \[ \int \left (a+\frac {b}{x}\right )^{5/2} \sqrt {x} \, dx=\left [\frac {15 \, a b^{\frac {3}{2}} x \log \left (\frac {a x - 2 \, \sqrt {b} \sqrt {x} \sqrt {\frac {a x + b}{x}} + 2 \, b}{x}\right ) + 2 \, {\left (2 \, a^{2} x^{2} + 14 \, a b x - 3 \, b^{2}\right )} \sqrt {x} \sqrt {\frac {a x + b}{x}}}{6 \, x}, \frac {15 \, a \sqrt {-b} b x \arctan \left (\frac {\sqrt {-b} \sqrt {x} \sqrt {\frac {a x + b}{x}}}{b}\right ) + {\left (2 \, a^{2} x^{2} + 14 \, a b x - 3 \, b^{2}\right )} \sqrt {x} \sqrt {\frac {a x + b}{x}}}{3 \, x}\right ] \]
[1/6*(15*a*b^(3/2)*x*log((a*x - 2*sqrt(b)*sqrt(x)*sqrt((a*x + b)/x) + 2*b) /x) + 2*(2*a^2*x^2 + 14*a*b*x - 3*b^2)*sqrt(x)*sqrt((a*x + b)/x))/x, 1/3*( 15*a*sqrt(-b)*b*x*arctan(sqrt(-b)*sqrt(x)*sqrt((a*x + b)/x)/b) + (2*a^2*x^ 2 + 14*a*b*x - 3*b^2)*sqrt(x)*sqrt((a*x + b)/x))/x]
Time = 7.35 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.05 \[ \int \left (a+\frac {b}{x}\right )^{5/2} \sqrt {x} \, dx=\frac {2 a^{2} \sqrt {b} x \sqrt {\frac {a x}{b} + 1}}{3} + \frac {14 a b^{\frac {3}{2}} \sqrt {\frac {a x}{b} + 1}}{3} + \frac {5 a b^{\frac {3}{2}} \log {\left (\frac {a x}{b} \right )}}{2} - 5 a b^{\frac {3}{2}} \log {\left (\sqrt {\frac {a x}{b} + 1} + 1 \right )} - \frac {b^{\frac {5}{2}} \sqrt {\frac {a x}{b} + 1}}{x} \]
2*a**2*sqrt(b)*x*sqrt(a*x/b + 1)/3 + 14*a*b**(3/2)*sqrt(a*x/b + 1)/3 + 5*a *b**(3/2)*log(a*x/b)/2 - 5*a*b**(3/2)*log(sqrt(a*x/b + 1) + 1) - b**(5/2)* sqrt(a*x/b + 1)/x
Time = 0.27 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.18 \[ \int \left (a+\frac {b}{x}\right )^{5/2} \sqrt {x} \, dx=\frac {2}{3} \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} a x^{\frac {3}{2}} + \frac {5}{2} \, a b^{\frac {3}{2}} \log \left (\frac {\sqrt {a + \frac {b}{x}} \sqrt {x} - \sqrt {b}}{\sqrt {a + \frac {b}{x}} \sqrt {x} + \sqrt {b}}\right ) + 4 \, \sqrt {a + \frac {b}{x}} a b \sqrt {x} - \frac {\sqrt {a + \frac {b}{x}} a b^{2} \sqrt {x}}{{\left (a + \frac {b}{x}\right )} x - b} \]
2/3*(a + b/x)^(3/2)*a*x^(3/2) + 5/2*a*b^(3/2)*log((sqrt(a + b/x)*sqrt(x) - sqrt(b))/(sqrt(a + b/x)*sqrt(x) + sqrt(b))) + 4*sqrt(a + b/x)*a*b*sqrt(x) - sqrt(a + b/x)*a*b^2*sqrt(x)/((a + b/x)*x - b)
Time = 0.33 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.79 \[ \int \left (a+\frac {b}{x}\right )^{5/2} \sqrt {x} \, dx=\frac {\frac {15 \, a^{2} b^{2} \arctan \left (\frac {\sqrt {a x + b}}{\sqrt {-b}}\right )}{\sqrt {-b}} + 2 \, {\left (a x + b\right )}^{\frac {3}{2}} a^{2} + 12 \, \sqrt {a x + b} a^{2} b - \frac {3 \, \sqrt {a x + b} a b^{2}}{x}}{3 \, a} \]
1/3*(15*a^2*b^2*arctan(sqrt(a*x + b)/sqrt(-b))/sqrt(-b) + 2*(a*x + b)^(3/2 )*a^2 + 12*sqrt(a*x + b)*a^2*b - 3*sqrt(a*x + b)*a*b^2/x)/a
Timed out. \[ \int \left (a+\frac {b}{x}\right )^{5/2} \sqrt {x} \, dx=\int \sqrt {x}\,{\left (a+\frac {b}{x}\right )}^{5/2} \,d x \]