Integrand size = 15, antiderivative size = 88 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{9/2}} \, dx=-\frac {2}{b^3 \sqrt {x}}+\frac {a^2}{2 b^3 \left (a+\frac {b}{x}\right )^2 \sqrt {x}}-\frac {9 a}{4 b^3 \left (a+\frac {b}{x}\right ) \sqrt {x}}+\frac {15 \sqrt {a} \arctan \left (\frac {\sqrt {b}}{\sqrt {a} \sqrt {x}}\right )}{4 b^{7/2}} \] Output:
-2/b^3/x^(1/2)+1/2*a^2/b^3/(a+b/x)^2/x^(1/2)-9/4*a/b^3/(a+b/x)/x^(1/2)+15/ 4*a^(1/2)*arctan(1/a^(1/2)/x^(1/2)*b^(1/2))/b^(7/2)
Time = 0.16 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.80 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{9/2}} \, dx=\frac {-8 b^2-25 a b x-15 a^2 x^2}{4 b^3 \sqrt {x} (b+a x)^2}-\frac {15 \sqrt {a} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{4 b^{7/2}} \] Input:
Integrate[1/((a + b/x)^3*x^(9/2)),x]
Output:
(-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 = 0.30 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.03, 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 |
\(\displaystyle \frac {5 \left (\frac {3 \left (-\frac {2 \sqrt {a} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{b^{3/2}}-\frac {2}{b \sqrt {x}}\right )}{2 b}+\frac {1}{b \sqrt {x} (a x+b)}\right )}{4 b}+\frac {1}{2 b \sqrt {x} (a x+b)^2}\) |
Input:
Int[1/((a + b/x)^3*x^(9/2)),x]
Output:
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)
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 = 0.21 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.64
method | result | size |
derivativedivides | \(-\frac {2}{b^{3} \sqrt {x}}-\frac {2 a \left (\frac {\frac {7 a \,x^{\frac {3}{2}}}{8}+\frac {9 b \sqrt {x}}{8}}{\left (a x +b \right )^{2}}+\frac {15 \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{b^{3}}\) | \(56\) |
default | \(-\frac {2}{b^{3} \sqrt {x}}-\frac {2 a \left (\frac {\frac {7 a \,x^{\frac {3}{2}}}{8}+\frac {9 b \sqrt {x}}{8}}{\left (a x +b \right )^{2}}+\frac {15 \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{b^{3}}\) | \(56\) |
risch | \(-\frac {2}{b^{3} \sqrt {x}}-\frac {a \left (\frac {\frac {7 a \,x^{\frac {3}{2}}}{4}+\frac {9 b \sqrt {x}}{4}}{\left (a x +b \right )^{2}}+\frac {15 \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{4 \sqrt {a b}}\right )}{b^{3}}\) | \(57\) |
Input:
int(1/(a+b/x)^3/x^(9/2),x,method=_RETURNVERBOSE)
Output:
-2/b^3/x^(1/2)-2/b^3*a*((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)))
Time = 0.08 (sec) , antiderivative size = 209, normalized size of antiderivative = 2.38 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{9/2}} \, dx=\left [\frac {15 \, {\left (a^{2} x^{3} + 2 \, a b x^{2} + b^{2} x\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {a x - 2 \, b \sqrt {x} \sqrt {-\frac {a}{b}} - b}{a x + b}\right ) - 2 \, {\left (15 \, a^{2} x^{2} + 25 \, a b x + 8 \, b^{2}\right )} \sqrt {x}}{8 \, {\left (a^{2} b^{3} x^{3} + 2 \, a b^{4} x^{2} + b^{5} x\right )}}, -\frac {15 \, {\left (a^{2} x^{3} + 2 \, a b x^{2} + b^{2} x\right )} \sqrt {\frac {a}{b}} \arctan \left (\sqrt {x} \sqrt {\frac {a}{b}}\right ) + {\left (15 \, a^{2} x^{2} + 25 \, a b x + 8 \, b^{2}\right )} \sqrt {x}}{4 \, {\left (a^{2} b^{3} x^{3} + 2 \, a b^{4} x^{2} + b^{5} x\right )}}\right ] \] Input:
integrate(1/(a+b/x)^3/x^(9/2),x, algorithm="fricas")
Output:
[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)*s qrt(a/b)*arctan(sqrt(x)*sqrt(a/b)) + (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. \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{9/2}} \, dx=\text {Timed out} \] Input:
integrate(1/(a+b/x)**3/x**(9/2),x)
Output:
Timed out
Time = 0.11 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.85 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{9/2}} \, dx=-\frac {\frac {7 \, a^{2}}{\sqrt {x}} + \frac {9 \, a b}{x^{\frac {3}{2}}}}{4 \, {\left (a^{2} b^{3} + \frac {2 \, a b^{4}}{x} + \frac {b^{5}}{x^{2}}\right )}} + \frac {15 \, a \arctan \left (\frac {b}{\sqrt {a b} \sqrt {x}}\right )}{4 \, \sqrt {a b} b^{3}} - \frac {2}{b^{3} \sqrt {x}} \] Input:
integrate(1/(a+b/x)^3/x^(9/2),x, algorithm="maxima")
Output:
-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 = 0.13 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.67 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{9/2}} \, dx=-\frac {15 \, a \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} b^{3}} - \frac {2}{b^{3} \sqrt {x}} - \frac {7 \, a^{2} x^{\frac {3}{2}} + 9 \, a b \sqrt {x}}{4 \, {\left (a x + b\right )}^{2} b^{3}} \] Input:
integrate(1/(a+b/x)^3/x^(9/2),x, algorithm="giac")
Output:
-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 = 0.32 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.80 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{9/2}} \, dx=-\frac {\frac {2}{b}+\frac {15\,a^2\,x^2}{4\,b^3}+\frac {25\,a\,x}{4\,b^2}}{a^2\,x^{5/2}+b^2\,\sqrt {x}+2\,a\,b\,x^{3/2}}-\frac {15\,\sqrt {a}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {x}}{\sqrt {b}}\right )}{4\,b^{7/2}} \] Input:
int(1/(x^(9/2)*(a + b/x)^3),x)
Output:
- (2/b + (15*a^2*x^2)/(4*b^3) + (25*a*x)/(4*b^2))/(a^2*x^(5/2) + b^2*x^(1/ 2) + 2*a*b*x^(3/2)) - (15*a^(1/2)*atan((a^(1/2)*x^(1/2))/b^(1/2)))/(4*b^(7 /2))
Time = 0.21 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.41 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{9/2}} \, dx=\frac {-15 \sqrt {x}\, \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, a}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} x^{2}-30 \sqrt {x}\, \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, a}{\sqrt {b}\, \sqrt {a}}\right ) a b x -15 \sqrt {x}\, \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, a}{\sqrt {b}\, \sqrt {a}}\right ) b^{2}-15 a^{2} b \,x^{2}-25 a \,b^{2} x -8 b^{3}}{4 \sqrt {x}\, b^{4} \left (a^{2} x^{2}+2 a b x +b^{2}\right )} \] Input:
int(1/(a+b/x)^3/x^(9/2),x)
Output:
( - 15*sqrt(x)*sqrt(b)*sqrt(a)*atan((sqrt(x)*a)/(sqrt(b)*sqrt(a)))*a**2*x* *2 - 30*sqrt(x)*sqrt(b)*sqrt(a)*atan((sqrt(x)*a)/(sqrt(b)*sqrt(a)))*a*b*x - 15*sqrt(x)*sqrt(b)*sqrt(a)*atan((sqrt(x)*a)/(sqrt(b)*sqrt(a)))*b**2 - 15 *a**2*b*x**2 - 25*a*b**2*x - 8*b**3)/(4*sqrt(x)*b**4*(a**2*x**2 + 2*a*b*x + b**2))