Integrand size = 15, antiderivative size = 82 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{9/2}} \, dx=-\frac {15}{4 b^3 \sqrt {x}}+\frac {1}{2 b \sqrt {x} (b+a x)^2}+\frac {5}{4 b^2 \sqrt {x} (b+a x)}-\frac {15 \sqrt {a} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{4 b^{7/2}} \]
-15/4*arctan(a^(1/2)*x^(1/2)/b^(1/2))*a^(1/2)/b^(7/2)-15/4/b^3/x^(1/2)+1/2 /b/(a*x+b)^2/x^(1/2)+5/4/b^2/(a*x+b)/x^(1/2)
Time = 0.16 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.85 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{9/2}} \, dx=\frac {-8 b^2-25 a b x-15 a^2 x^2}{4 b^3 \sqrt {x} (b+a x)^2}-\frac {15 \sqrt {a} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{4 b^{7/2}} \]
(-8*b^2 - 25*a*b*x - 15*a^2*x^2)/(4*b^3*Sqrt[x]*(b + a*x)^2) - (15*Sqrt[a] *ArcTan[(Sqrt[a]*Sqrt[x])/Sqrt[b]])/(4*b^(7/2))
Time = 0.19 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.11, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {795, 52, 52, 61, 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.
\(\displaystyle \int \frac {1}{x^{9/2} \left (a+\frac {b}{x}\right )^3} \, dx\) |
\(\Big \downarrow \) 795 |
\(\displaystyle \int \frac {1}{x^{3/2} (a x+b)^3}dx\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {5 \int \frac {1}{x^{3/2} (b+a x)^2}dx}{4 b}+\frac {1}{2 b \sqrt {x} (a x+b)^2}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {5 \left (\frac {3 \int \frac {1}{x^{3/2} (b+a x)}dx}{2 b}+\frac {1}{b \sqrt {x} (a x+b)}\right )}{4 b}+\frac {1}{2 b \sqrt {x} (a x+b)^2}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {5 \left (\frac {3 \left (-\frac {a \int \frac {1}{\sqrt {x} (b+a x)}dx}{b}-\frac {2}{b \sqrt {x}}\right )}{2 b}+\frac {1}{b \sqrt {x} (a x+b)}\right )}{4 b}+\frac {1}{2 b \sqrt {x} (a x+b)^2}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {5 \left (\frac {3 \left (-\frac {2 a \int \frac {1}{b+a x}d\sqrt {x}}{b}-\frac {2}{b \sqrt {x}}\right )}{2 b}+\frac {1}{b \sqrt {x} (a x+b)}\right )}{4 b}+\frac {1}{2 b \sqrt {x} (a x+b)^2}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {5 \left (\frac {3 \left (-\frac {2 \sqrt {a} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{b^{3/2}}-\frac {2}{b \sqrt {x}}\right )}{2 b}+\frac {1}{b \sqrt {x} (a x+b)}\right )}{4 b}+\frac {1}{2 b \sqrt {x} (a x+b)^2}\) |
1/(2*b*Sqrt[x]*(b + a*x)^2) + (5*(1/(b*Sqrt[x]*(b + a*x)) + (3*(-2/(b*Sqrt [x]) - (2*Sqrt[a]*ArcTan[(Sqrt[a]*Sqrt[x])/Sqrt[b]])/b^(3/2)))/(2*b)))/(4* b)
3.17.88.3.1 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 + 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] :> 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] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[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 = 56, normalized size of antiderivative = 0.68
method | result | size |
derivativedivides | \(-\frac {2 a \left (\frac {\frac {7 a \,x^{\frac {3}{2}}}{8}+\frac {9 b \sqrt {x}}{8}}{\left (a x +b \right )^{2}}+\frac {15 \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{b^{3}}-\frac {2}{\sqrt {x}\, b^{3}}\) | \(56\) |
default | \(-\frac {2 a \left (\frac {\frac {7 a \,x^{\frac {3}{2}}}{8}+\frac {9 b \sqrt {x}}{8}}{\left (a x +b \right )^{2}}+\frac {15 \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{b^{3}}-\frac {2}{\sqrt {x}\, b^{3}}\) | \(56\) |
risch | \(-\frac {2}{\sqrt {x}\, b^{3}}-\frac {a \left (\frac {\frac {7 a \,x^{\frac {3}{2}}}{4}+\frac {9 b \sqrt {x}}{4}}{\left (a x +b \right )^{2}}+\frac {15 \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{4 \sqrt {a b}}\right )}{b^{3}}\) | \(57\) |
-2*a/b^3*((7/8*a*x^(3/2)+9/8*b*x^(1/2))/(a*x+b)^2+15/8/(a*b)^(1/2)*arctan( a*x^(1/2)/(a*b)^(1/2)))-2/x^(1/2)/b^3
Time = 0.30 (sec) , antiderivative size = 214, normalized size of antiderivative = 2.61 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{9/2}} \, dx=\left [\frac {15 \, {\left (a^{2} x^{3} + 2 \, a b x^{2} + b^{2} x\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {a x - 2 \, b \sqrt {x} \sqrt {-\frac {a}{b}} - b}{a x + b}\right ) - 2 \, {\left (15 \, a^{2} x^{2} + 25 \, a b x + 8 \, b^{2}\right )} \sqrt {x}}{8 \, {\left (a^{2} b^{3} x^{3} + 2 \, a b^{4} x^{2} + b^{5} x\right )}}, \frac {15 \, {\left (a^{2} x^{3} + 2 \, a b x^{2} + b^{2} x\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {\frac {a}{b}}}{a \sqrt {x}}\right ) - {\left (15 \, a^{2} x^{2} + 25 \, a b x + 8 \, b^{2}\right )} \sqrt {x}}{4 \, {\left (a^{2} b^{3} x^{3} + 2 \, a b^{4} x^{2} + b^{5} x\right )}}\right ] \]
[1/8*(15*(a^2*x^3 + 2*a*b*x^2 + b^2*x)*sqrt(-a/b)*log((a*x - 2*b*sqrt(x)*s qrt(-a/b) - b)/(a*x + b)) - 2*(15*a^2*x^2 + 25*a*b*x + 8*b^2)*sqrt(x))/(a^ 2*b^3*x^3 + 2*a*b^4*x^2 + b^5*x), 1/4*(15*(a^2*x^3 + 2*a*b*x^2 + b^2*x)*sq rt(a/b)*arctan(b*sqrt(a/b)/(a*sqrt(x))) - (15*a^2*x^2 + 25*a*b*x + 8*b^2)* sqrt(x))/(a^2*b^3*x^3 + 2*a*b^4*x^2 + b^5*x)]
Timed out. \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{9/2}} \, dx=\text {Timed out} \]
Time = 0.28 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.91 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{9/2}} \, dx=-\frac {\frac {7 \, a^{2}}{\sqrt {x}} + \frac {9 \, a b}{x^{\frac {3}{2}}}}{4 \, {\left (a^{2} b^{3} + \frac {2 \, a b^{4}}{x} + \frac {b^{5}}{x^{2}}\right )}} + \frac {15 \, a \arctan \left (\frac {b}{\sqrt {a b} \sqrt {x}}\right )}{4 \, \sqrt {a b} b^{3}} - \frac {2}{b^{3} \sqrt {x}} \]
-1/4*(7*a^2/sqrt(x) + 9*a*b/x^(3/2))/(a^2*b^3 + 2*a*b^4/x + b^5/x^2) + 15/ 4*a*arctan(b/(sqrt(a*b)*sqrt(x)))/(sqrt(a*b)*b^3) - 2/(b^3*sqrt(x))
Time = 0.27 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.72 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{9/2}} \, dx=-\frac {15 \, a \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} b^{3}} - \frac {2}{b^{3} \sqrt {x}} - \frac {7 \, a^{2} x^{\frac {3}{2}} + 9 \, a b \sqrt {x}}{4 \, {\left (a x + b\right )}^{2} b^{3}} \]
-15/4*a*arctan(a*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^3) - 2/(b^3*sqrt(x)) - 1/ 4*(7*a^2*x^(3/2) + 9*a*b*sqrt(x))/((a*x + b)^2*b^3)
Time = 5.93 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.85 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{9/2}} \, dx=-\frac {\frac {2}{b}+\frac {15\,a^2\,x^2}{4\,b^3}+\frac {25\,a\,x}{4\,b^2}}{a^2\,x^{5/2}+b^2\,\sqrt {x}+2\,a\,b\,x^{3/2}}-\frac {15\,\sqrt {a}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {x}}{\sqrt {b}}\right )}{4\,b^{7/2}} \]