Integrand size = 17, antiderivative size = 129 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2} x^{11/2}} \, dx=\frac {2}{3 b \left (a+\frac {b}{x}\right )^{3/2} x^{7/2}}+\frac {14}{3 b^2 \sqrt {a+\frac {b}{x}} x^{5/2}}-\frac {35 \sqrt {a+\frac {b}{x}}}{6 b^3 x^{3/2}}+\frac {35 a \sqrt {a+\frac {b}{x}}}{4 b^4 \sqrt {x}}-\frac {35 a^2 \text {arctanh}\left (\frac {\sqrt {b}}{\sqrt {a+\frac {b}{x}} \sqrt {x}}\right )}{4 b^{9/2}} \]
2/3/b/(a+b/x)^(3/2)/x^(7/2)-35/4*a^2*arctanh(b^(1/2)/(a+b/x)^(1/2)/x^(1/2) )/b^(9/2)+14/3/b^2/x^(5/2)/(a+b/x)^(1/2)-35/6*(a+b/x)^(1/2)/b^3/x^(3/2)+35 /4*a*(a+b/x)^(1/2)/b^4/x^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 10.02 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.43 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2} x^{11/2}} \, dx=-\frac {2 \sqrt {1+\frac {b}{a x}} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {9}{2},\frac {11}{2},-\frac {b}{a x}\right )}{9 a^2 \sqrt {a+\frac {b}{x}} x^{9/2}} \]
(-2*Sqrt[1 + b/(a*x)]*Hypergeometric2F1[5/2, 9/2, 11/2, -(b/(a*x))])/(9*a^ 2*Sqrt[a + b/x]*x^(9/2))
Time = 0.24 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.16, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {860, 252, 252, 262, 262, 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.
\(\displaystyle \int \frac {1}{x^{11/2} \left (a+\frac {b}{x}\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 860 |
\(\displaystyle -2 \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2} x^4}d\frac {1}{\sqrt {x}}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle -2 \left (\frac {7 \int \frac {1}{\left (a+\frac {b}{x}\right )^{3/2} x^3}d\frac {1}{\sqrt {x}}}{3 b}-\frac {1}{3 b x^{7/2} \left (a+\frac {b}{x}\right )^{3/2}}\right )\) |
\(\Big \downarrow \) 252 |
\(\displaystyle -2 \left (\frac {7 \left (\frac {5 \int \frac {1}{\sqrt {a+\frac {b}{x}} x^2}d\frac {1}{\sqrt {x}}}{b}-\frac {1}{b x^{5/2} \sqrt {a+\frac {b}{x}}}\right )}{3 b}-\frac {1}{3 b x^{7/2} \left (a+\frac {b}{x}\right )^{3/2}}\right )\) |
\(\Big \downarrow \) 262 |
\(\displaystyle -2 \left (\frac {7 \left (\frac {5 \left (\frac {\sqrt {a+\frac {b}{x}}}{4 b x^{3/2}}-\frac {3 a \int \frac {1}{\sqrt {a+\frac {b}{x}} x}d\frac {1}{\sqrt {x}}}{4 b}\right )}{b}-\frac {1}{b x^{5/2} \sqrt {a+\frac {b}{x}}}\right )}{3 b}-\frac {1}{3 b x^{7/2} \left (a+\frac {b}{x}\right )^{3/2}}\right )\) |
\(\Big \downarrow \) 262 |
\(\displaystyle -2 \left (\frac {7 \left (\frac {5 \left (\frac {\sqrt {a+\frac {b}{x}}}{4 b x^{3/2}}-\frac {3 a \left (\frac {\sqrt {a+\frac {b}{x}}}{2 b \sqrt {x}}-\frac {a \int \frac {1}{\sqrt {a+\frac {b}{x}}}d\frac {1}{\sqrt {x}}}{2 b}\right )}{4 b}\right )}{b}-\frac {1}{b x^{5/2} \sqrt {a+\frac {b}{x}}}\right )}{3 b}-\frac {1}{3 b x^{7/2} \left (a+\frac {b}{x}\right )^{3/2}}\right )\) |
\(\Big \downarrow \) 224 |
\(\displaystyle -2 \left (\frac {7 \left (\frac {5 \left (\frac {\sqrt {a+\frac {b}{x}}}{4 b x^{3/2}}-\frac {3 a \left (\frac {\sqrt {a+\frac {b}{x}}}{2 b \sqrt {x}}-\frac {a \int \frac {1}{1-\frac {b}{x}}d\frac {1}{\sqrt {a+\frac {b}{x}} \sqrt {x}}}{2 b}\right )}{4 b}\right )}{b}-\frac {1}{b x^{5/2} \sqrt {a+\frac {b}{x}}}\right )}{3 b}-\frac {1}{3 b x^{7/2} \left (a+\frac {b}{x}\right )^{3/2}}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -2 \left (\frac {7 \left (\frac {5 \left (\frac {\sqrt {a+\frac {b}{x}}}{4 b x^{3/2}}-\frac {3 a \left (\frac {\sqrt {a+\frac {b}{x}}}{2 b \sqrt {x}}-\frac {a \text {arctanh}\left (\frac {\sqrt {b}}{\sqrt {x} \sqrt {a+\frac {b}{x}}}\right )}{2 b^{3/2}}\right )}{4 b}\right )}{b}-\frac {1}{b x^{5/2} \sqrt {a+\frac {b}{x}}}\right )}{3 b}-\frac {1}{3 b x^{7/2} \left (a+\frac {b}{x}\right )^{3/2}}\right )\) |
-2*(-1/3*1/(b*(a + b/x)^(3/2)*x^(7/2)) + (7*(-(1/(b*Sqrt[a + b/x]*x^(5/2)) ) + (5*(Sqrt[a + b/x]/(4*b*x^(3/2)) - (3*a*(Sqrt[a + b/x]/(2*b*Sqrt[x]) - (a*ArcTanh[Sqrt[b]/(Sqrt[a + b/x]*Sqrt[x])])/(2*b^(3/2))))/(4*b)))/b))/(3* b))
3.19.3.3.1 Defintions of rubi rules used
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*(c*x )^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* (p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c }, x] && LtQ[p, -1] && GtQ[m, 1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi alQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 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 = 0.07 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.78
method | result | size |
risch | \(\frac {\left (a x +b \right ) \left (11 a x -2 b \right )}{4 b^{4} x^{\frac {5}{2}} \sqrt {\frac {a x +b}{x}}}+\frac {a^{2} \left (-\frac {70 \,\operatorname {arctanh}\left (\frac {\sqrt {a x +b}}{\sqrt {b}}\right )}{\sqrt {b}}+\frac {48}{\sqrt {a x +b}}+\frac {16 b}{3 \left (a x +b \right )^{\frac {3}{2}}}\right ) \sqrt {a x +b}}{8 b^{4} \sqrt {x}\, \sqrt {\frac {a x +b}{x}}}\) | \(100\) |
default | \(-\frac {\sqrt {\frac {a x +b}{x}}\, \left (105 \,\operatorname {arctanh}\left (\frac {\sqrt {a x +b}}{\sqrt {b}}\right ) \sqrt {a x +b}\, a^{3} x^{3}+105 \,\operatorname {arctanh}\left (\frac {\sqrt {a x +b}}{\sqrt {b}}\right ) a^{2} b \,x^{2} \sqrt {a x +b}-105 a^{3} x^{3} \sqrt {b}-140 a^{2} b^{\frac {3}{2}} x^{2}-21 b^{\frac {5}{2}} a x +6 b^{\frac {7}{2}}\right )}{12 x^{\frac {3}{2}} \left (a x +b \right )^{2} b^{\frac {9}{2}}}\) | \(117\) |
1/4*(a*x+b)*(11*a*x-2*b)/b^4/x^(5/2)/((a*x+b)/x)^(1/2)+1/8/b^4*a^2*(-70/b^ (1/2)*arctanh((a*x+b)^(1/2)/b^(1/2))+48/(a*x+b)^(1/2)+16/3*b/(a*x+b)^(3/2) )/x^(1/2)/((a*x+b)/x)^(1/2)*(a*x+b)^(1/2)
Time = 0.32 (sec) , antiderivative size = 283, normalized size of antiderivative = 2.19 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2} x^{11/2}} \, dx=\left [\frac {105 \, {\left (a^{4} x^{4} + 2 \, a^{3} b x^{3} + a^{2} b^{2} x^{2}\right )} \sqrt {b} \log \left (\frac {a x - 2 \, \sqrt {b} \sqrt {x} \sqrt {\frac {a x + b}{x}} + 2 \, b}{x}\right ) + 2 \, {\left (105 \, a^{3} b x^{3} + 140 \, a^{2} b^{2} x^{2} + 21 \, a b^{3} x - 6 \, b^{4}\right )} \sqrt {x} \sqrt {\frac {a x + b}{x}}}{24 \, {\left (a^{2} b^{5} x^{4} + 2 \, a b^{6} x^{3} + b^{7} x^{2}\right )}}, \frac {105 \, {\left (a^{4} x^{4} + 2 \, a^{3} b x^{3} + a^{2} b^{2} x^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {x} \sqrt {\frac {a x + b}{x}}}{b}\right ) + {\left (105 \, a^{3} b x^{3} + 140 \, a^{2} b^{2} x^{2} + 21 \, a b^{3} x - 6 \, b^{4}\right )} \sqrt {x} \sqrt {\frac {a x + b}{x}}}{12 \, {\left (a^{2} b^{5} x^{4} + 2 \, a b^{6} x^{3} + b^{7} x^{2}\right )}}\right ] \]
[1/24*(105*(a^4*x^4 + 2*a^3*b*x^3 + a^2*b^2*x^2)*sqrt(b)*log((a*x - 2*sqrt (b)*sqrt(x)*sqrt((a*x + b)/x) + 2*b)/x) + 2*(105*a^3*b*x^3 + 140*a^2*b^2*x ^2 + 21*a*b^3*x - 6*b^4)*sqrt(x)*sqrt((a*x + b)/x))/(a^2*b^5*x^4 + 2*a*b^6 *x^3 + b^7*x^2), 1/12*(105*(a^4*x^4 + 2*a^3*b*x^3 + a^2*b^2*x^2)*sqrt(-b)* arctan(sqrt(-b)*sqrt(x)*sqrt((a*x + b)/x)/b) + (105*a^3*b*x^3 + 140*a^2*b^ 2*x^2 + 21*a*b^3*x - 6*b^4)*sqrt(x)*sqrt((a*x + b)/x))/(a^2*b^5*x^4 + 2*a* b^6*x^3 + b^7*x^2)]
Timed out. \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2} x^{11/2}} \, dx=\text {Timed out} \]
Time = 0.29 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.26 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2} x^{11/2}} \, dx=\frac {105 \, {\left (a + \frac {b}{x}\right )}^{3} a^{2} x^{3} - 175 \, {\left (a + \frac {b}{x}\right )}^{2} a^{2} b x^{2} + 56 \, {\left (a + \frac {b}{x}\right )} a^{2} b^{2} x + 8 \, a^{2} b^{3}}{12 \, {\left ({\left (a + \frac {b}{x}\right )}^{\frac {7}{2}} b^{4} x^{\frac {7}{2}} - 2 \, {\left (a + \frac {b}{x}\right )}^{\frac {5}{2}} b^{5} x^{\frac {5}{2}} + {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} b^{6} x^{\frac {3}{2}}\right )}} + \frac {35 \, a^{2} \log \left (\frac {\sqrt {a + \frac {b}{x}} \sqrt {x} - \sqrt {b}}{\sqrt {a + \frac {b}{x}} \sqrt {x} + \sqrt {b}}\right )}{8 \, b^{\frac {9}{2}}} \]
1/12*(105*(a + b/x)^3*a^2*x^3 - 175*(a + b/x)^2*a^2*b*x^2 + 56*(a + b/x)*a ^2*b^2*x + 8*a^2*b^3)/((a + b/x)^(7/2)*b^4*x^(7/2) - 2*(a + b/x)^(5/2)*b^5 *x^(5/2) + (a + b/x)^(3/2)*b^6*x^(3/2)) + 35/8*a^2*log((sqrt(a + b/x)*sqrt (x) - sqrt(b))/(sqrt(a + b/x)*sqrt(x) + sqrt(b)))/b^(9/2)
Time = 0.36 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.72 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2} x^{11/2}} \, dx=\frac {35 \, a^{2} \arctan \left (\frac {\sqrt {a x + b}}{\sqrt {-b}}\right )}{4 \, \sqrt {-b} b^{4}} + \frac {2 \, {\left (9 \, {\left (a x + b\right )} a^{2} + a^{2} b\right )}}{3 \, {\left (a x + b\right )}^{\frac {3}{2}} b^{4}} + \frac {11 \, {\left (a x + b\right )}^{\frac {3}{2}} a^{2} - 13 \, \sqrt {a x + b} a^{2} b}{4 \, a^{2} b^{4} x^{2}} \]
35/4*a^2*arctan(sqrt(a*x + b)/sqrt(-b))/(sqrt(-b)*b^4) + 2/3*(9*(a*x + b)* a^2 + a^2*b)/((a*x + b)^(3/2)*b^4) + 1/4*(11*(a*x + b)^(3/2)*a^2 - 13*sqrt (a*x + b)*a^2*b)/(a^2*b^4*x^2)
Timed out. \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2} x^{11/2}} \, dx=\int \frac {1}{x^{11/2}\,{\left (a+\frac {b}{x}\right )}^{5/2}} \,d x \]