Integrand size = 15, antiderivative size = 103 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{11/2}} \, dx=-\frac {2}{3 b^3 x^{3/2}}+\frac {6 a}{b^4 \sqrt {x}}-\frac {a^3}{2 b^4 \left (a+\frac {b}{x}\right )^2 \sqrt {x}}+\frac {13 a^2}{4 b^4 \left (a+\frac {b}{x}\right ) \sqrt {x}}-\frac {35 a^{3/2} \arctan \left (\frac {\sqrt {b}}{\sqrt {a} \sqrt {x}}\right )}{4 b^{9/2}} \] Output:
-2/3/b^3/x^(3/2)+6*a/b^4/x^(1/2)-1/2*a^3/b^4/(a+b/x)^2/x^(1/2)+13/4*a^2/b^ 4/(a+b/x)/x^(1/2)-35/4*a^(3/2)*arctan(1/a^(1/2)/x^(1/2)*b^(1/2))/b^(9/2)
Time = 0.12 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.79 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{11/2}} \, dx=\frac {-8 b^3+56 a b^2 x+175 a^2 b x^2+105 a^3 x^3}{12 b^4 x^{3/2} (b+a x)^2}+\frac {35 a^{3/2} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{4 b^{9/2}} \] Input:
Integrate[1/((a + b/x)^3*x^(11/2)),x]
Output:
(-8*b^3 + 56*a*b^2*x + 175*a^2*b*x^2 + 105*a^3*x^3)/(12*b^4*x^(3/2)*(b + a *x)^2) + (35*a^(3/2)*ArcTan[(Sqrt[a]*Sqrt[x])/Sqrt[b]])/(4*b^(9/2))
Time = 0.32 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.07, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {795, 52, 52, 61, 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.
\(\displaystyle \int \frac {1}{x^{11/2} \left (a+\frac {b}{x}\right )^3} \, dx\) |
\(\Big \downarrow \) 795 |
\(\displaystyle \int \frac {1}{x^{5/2} (a x+b)^3}dx\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {7 \int \frac {1}{x^{5/2} (b+a x)^2}dx}{4 b}+\frac {1}{2 b x^{3/2} (a x+b)^2}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {7 \left (\frac {5 \int \frac {1}{x^{5/2} (b+a x)}dx}{2 b}+\frac {1}{b x^{3/2} (a x+b)}\right )}{4 b}+\frac {1}{2 b x^{3/2} (a x+b)^2}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {7 \left (\frac {5 \left (-\frac {a \int \frac {1}{x^{3/2} (b+a x)}dx}{b}-\frac {2}{3 b x^{3/2}}\right )}{2 b}+\frac {1}{b x^{3/2} (a x+b)}\right )}{4 b}+\frac {1}{2 b x^{3/2} (a x+b)^2}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {7 \left (\frac {5 \left (-\frac {a \left (-\frac {a \int \frac {1}{\sqrt {x} (b+a x)}dx}{b}-\frac {2}{b \sqrt {x}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{2 b}+\frac {1}{b x^{3/2} (a x+b)}\right )}{4 b}+\frac {1}{2 b x^{3/2} (a x+b)^2}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {7 \left (\frac {5 \left (-\frac {a \left (-\frac {2 a \int \frac {1}{b+a x}d\sqrt {x}}{b}-\frac {2}{b \sqrt {x}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{2 b}+\frac {1}{b x^{3/2} (a x+b)}\right )}{4 b}+\frac {1}{2 b x^{3/2} (a x+b)^2}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {7 \left (\frac {5 \left (-\frac {a \left (-\frac {2 \sqrt {a} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{b^{3/2}}-\frac {2}{b \sqrt {x}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{2 b}+\frac {1}{b x^{3/2} (a x+b)}\right )}{4 b}+\frac {1}{2 b x^{3/2} (a x+b)^2}\) |
Input:
Int[1/((a + b/x)^3*x^(11/2)),x]
Output:
1/(2*b*x^(3/2)*(b + a*x)^2) + (7*(1/(b*x^(3/2)*(b + a*x)) + (5*(-2/(3*b*x^ (3/2)) - (a*(-2/(b*Sqrt[x]) - (2*Sqrt[a]*ArcTan[(Sqrt[a]*Sqrt[x])/Sqrt[b]] )/b^(3/2)))/b))/(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 = 0.23 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.64
method | result | size |
risch | \(\frac {6 a x -\frac {2 b}{3}}{b^{4} x^{\frac {3}{2}}}+\frac {a^{2} \left (\frac {\frac {11 a \,x^{\frac {3}{2}}}{4}+\frac {13 b \sqrt {x}}{4}}{\left (a x +b \right )^{2}}+\frac {35 \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{4 \sqrt {a b}}\right )}{b^{4}}\) | \(66\) |
derivativedivides | \(-\frac {2}{3 b^{3} x^{\frac {3}{2}}}+\frac {6 a}{b^{4} \sqrt {x}}+\frac {2 a^{2} \left (\frac {\frac {11 a \,x^{\frac {3}{2}}}{8}+\frac {13 b \sqrt {x}}{8}}{\left (a x +b \right )^{2}}+\frac {35 \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{b^{4}}\) | \(67\) |
default | \(-\frac {2}{3 b^{3} x^{\frac {3}{2}}}+\frac {6 a}{b^{4} \sqrt {x}}+\frac {2 a^{2} \left (\frac {\frac {11 a \,x^{\frac {3}{2}}}{8}+\frac {13 b \sqrt {x}}{8}}{\left (a x +b \right )^{2}}+\frac {35 \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{b^{4}}\) | \(67\) |
Input:
int(1/(a+b/x)^3/x^(11/2),x,method=_RETURNVERBOSE)
Output:
2/3*(9*a*x-b)/b^4/x^(3/2)+a^2/b^4*(2*(11/8*a*x^(3/2)+13/8*b*x^(1/2))/(a*x+ b)^2+35/4/(a*b)^(1/2)*arctan(a*x^(1/2)/(a*b)^(1/2)))
Time = 0.09 (sec) , antiderivative size = 245, normalized size of antiderivative = 2.38 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{11/2}} \, dx=\left [\frac {105 \, {\left (a^{3} x^{4} + 2 \, a^{2} b x^{3} + a b^{2} x^{2}\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {a x + 2 \, b \sqrt {x} \sqrt {-\frac {a}{b}} - b}{a x + b}\right ) + 2 \, {\left (105 \, a^{3} x^{3} + 175 \, a^{2} b x^{2} + 56 \, a b^{2} x - 8 \, b^{3}\right )} \sqrt {x}}{24 \, {\left (a^{2} b^{4} x^{4} + 2 \, a b^{5} x^{3} + b^{6} x^{2}\right )}}, \frac {105 \, {\left (a^{3} x^{4} + 2 \, a^{2} b x^{3} + a b^{2} x^{2}\right )} \sqrt {\frac {a}{b}} \arctan \left (\sqrt {x} \sqrt {\frac {a}{b}}\right ) + {\left (105 \, a^{3} x^{3} + 175 \, a^{2} b x^{2} + 56 \, a b^{2} x - 8 \, b^{3}\right )} \sqrt {x}}{12 \, {\left (a^{2} b^{4} x^{4} + 2 \, a b^{5} x^{3} + b^{6} x^{2}\right )}}\right ] \] Input:
integrate(1/(a+b/x)^3/x^(11/2),x, algorithm="fricas")
Output:
[1/24*(105*(a^3*x^4 + 2*a^2*b*x^3 + a*b^2*x^2)*sqrt(-a/b)*log((a*x + 2*b*s qrt(x)*sqrt(-a/b) - b)/(a*x + b)) + 2*(105*a^3*x^3 + 175*a^2*b*x^2 + 56*a* b^2*x - 8*b^3)*sqrt(x))/(a^2*b^4*x^4 + 2*a*b^5*x^3 + b^6*x^2), 1/12*(105*( a^3*x^4 + 2*a^2*b*x^3 + a*b^2*x^2)*sqrt(a/b)*arctan(sqrt(x)*sqrt(a/b)) + ( 105*a^3*x^3 + 175*a^2*b*x^2 + 56*a*b^2*x - 8*b^3)*sqrt(x))/(a^2*b^4*x^4 + 2*a*b^5*x^3 + b^6*x^2)]
Timed out. \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{11/2}} \, dx=\text {Timed out} \] Input:
integrate(1/(a+b/x)**3/x**(11/2),x)
Output:
Timed out
Time = 0.11 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.86 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{11/2}} \, dx=\frac {\frac {11 \, a^{3}}{\sqrt {x}} + \frac {13 \, a^{2} b}{x^{\frac {3}{2}}}}{4 \, {\left (a^{2} b^{4} + \frac {2 \, a b^{5}}{x} + \frac {b^{6}}{x^{2}}\right )}} - \frac {35 \, a^{2} \arctan \left (\frac {b}{\sqrt {a b} \sqrt {x}}\right )}{4 \, \sqrt {a b} b^{4}} + \frac {2 \, {\left (\frac {9 \, a}{\sqrt {x}} - \frac {b}{x^{\frac {3}{2}}}\right )}}{3 \, b^{4}} \] Input:
integrate(1/(a+b/x)^3/x^(11/2),x, algorithm="maxima")
Output:
1/4*(11*a^3/sqrt(x) + 13*a^2*b/x^(3/2))/(a^2*b^4 + 2*a*b^5/x + b^6/x^2) - 35/4*a^2*arctan(b/(sqrt(a*b)*sqrt(x)))/(sqrt(a*b)*b^4) + 2/3*(9*a/sqrt(x) - b/x^(3/2))/b^4
Time = 0.12 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.69 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{11/2}} \, dx=\frac {35 \, a^{2} \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} b^{4}} + \frac {2 \, {\left (9 \, a x - b\right )}}{3 \, b^{4} x^{\frac {3}{2}}} + \frac {11 \, a^{3} x^{\frac {3}{2}} + 13 \, a^{2} b \sqrt {x}}{4 \, {\left (a x + b\right )}^{2} b^{4}} \] Input:
integrate(1/(a+b/x)^3/x^(11/2),x, algorithm="giac")
Output:
35/4*a^2*arctan(a*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^4) + 2/3*(9*a*x - b)/(b^ 4*x^(3/2)) + 1/4*(11*a^3*x^(3/2) + 13*a^2*b*sqrt(x))/((a*x + b)^2*b^4)
Time = 0.32 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.78 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{11/2}} \, dx=\frac {\frac {175\,a^2\,x^2}{12\,b^3}-\frac {2}{3\,b}+\frac {35\,a^3\,x^3}{4\,b^4}+\frac {14\,a\,x}{3\,b^2}}{a^2\,x^{7/2}+b^2\,x^{3/2}+2\,a\,b\,x^{5/2}}+\frac {35\,a^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {x}}{\sqrt {b}}\right )}{4\,b^{9/2}} \] Input:
int(1/(x^(11/2)*(a + b/x)^3),x)
Output:
((175*a^2*x^2)/(12*b^3) - 2/(3*b) + (35*a^3*x^3)/(4*b^4) + (14*a*x)/(3*b^2 ))/(a^2*x^(7/2) + b^2*x^(3/2) + 2*a*b*x^(5/2)) + (35*a^(3/2)*atan((a^(1/2) *x^(1/2))/b^(1/2)))/(4*b^(9/2))
Time = 0.22 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.40 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{11/2}} \, dx=\frac {105 \sqrt {x}\, \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, a}{\sqrt {b}\, \sqrt {a}}\right ) a^{3} x^{3}+210 \sqrt {x}\, \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, a}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b \,x^{2}+105 \sqrt {x}\, \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, a}{\sqrt {b}\, \sqrt {a}}\right ) a \,b^{2} x +105 a^{3} b \,x^{3}+175 a^{2} b^{2} x^{2}+56 a \,b^{3} x -8 b^{4}}{12 \sqrt {x}\, b^{5} x \left (a^{2} x^{2}+2 a b x +b^{2}\right )} \] Input:
int(1/(a+b/x)^3/x^(11/2),x)
Output:
(105*sqrt(x)*sqrt(b)*sqrt(a)*atan((sqrt(x)*a)/(sqrt(b)*sqrt(a)))*a**3*x**3 + 210*sqrt(x)*sqrt(b)*sqrt(a)*atan((sqrt(x)*a)/(sqrt(b)*sqrt(a)))*a**2*b* x**2 + 105*sqrt(x)*sqrt(b)*sqrt(a)*atan((sqrt(x)*a)/(sqrt(b)*sqrt(a)))*a*b **2*x + 105*a**3*b*x**3 + 175*a**2*b**2*x**2 + 56*a*b**3*x - 8*b**4)/(12*s qrt(x)*b**5*x*(a**2*x**2 + 2*a*b*x + b**2))