Integrand size = 15, antiderivative size = 70 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{7/2}} \, dx=\frac {\sqrt {x}}{2 b (b+a x)^2}+\frac {3 \sqrt {x}}{4 b^2 (b+a x)}+\frac {3 \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{4 \sqrt {a} b^{5/2}} \]
3/4*arctan(a^(1/2)*x^(1/2)/b^(1/2))/b^(5/2)/a^(1/2)+1/2*x^(1/2)/b/(a*x+b)^ 2+3/4*x^(1/2)/b^2/(a*x+b)
Time = 0.11 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.84 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{7/2}} \, dx=\frac {\sqrt {x} (5 b+3 a x)}{4 b^2 (b+a x)^2}+\frac {3 \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{4 \sqrt {a} b^{5/2}} \]
(Sqrt[x]*(5*b + 3*a*x))/(4*b^2*(b + a*x)^2) + (3*ArcTan[(Sqrt[a]*Sqrt[x])/ Sqrt[b]])/(4*Sqrt[a]*b^(5/2))
Time = 0.18 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {795, 52, 52, 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.
\(\displaystyle \int \frac {1}{x^{7/2} \left (a+\frac {b}{x}\right )^3} \, dx\) |
\(\Big \downarrow \) 795 |
\(\displaystyle \int \frac {1}{\sqrt {x} (a x+b)^3}dx\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {3 \int \frac {1}{\sqrt {x} (b+a x)^2}dx}{4 b}+\frac {\sqrt {x}}{2 b (a x+b)^2}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {3 \left (\frac {\int \frac {1}{\sqrt {x} (b+a x)}dx}{2 b}+\frac {\sqrt {x}}{b (a x+b)}\right )}{4 b}+\frac {\sqrt {x}}{2 b (a x+b)^2}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {3 \left (\frac {\int \frac {1}{b+a x}d\sqrt {x}}{b}+\frac {\sqrt {x}}{b (a x+b)}\right )}{4 b}+\frac {\sqrt {x}}{2 b (a x+b)^2}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {3 \left (\frac {\arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{\sqrt {a} b^{3/2}}+\frac {\sqrt {x}}{b (a x+b)}\right )}{4 b}+\frac {\sqrt {x}}{2 b (a x+b)^2}\) |
Sqrt[x]/(2*b*(b + a*x)^2) + (3*(Sqrt[x]/(b*(b + a*x)) + ArcTan[(Sqrt[a]*Sq rt[x])/Sqrt[b]]/(Sqrt[a]*b^(3/2))))/(4*b)
3.17.87.3.1 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] :> 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 = 0.04 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.84
method | result | size |
derivativedivides | \(\frac {\sqrt {x}}{2 b \left (a x +b \right )^{2}}+\frac {\frac {3 \sqrt {x}}{4 b \left (a x +b \right )}+\frac {3 \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{4 b \sqrt {a b}}}{b}\) | \(59\) |
default | \(\frac {\sqrt {x}}{2 b \left (a x +b \right )^{2}}+\frac {\frac {3 \sqrt {x}}{4 b \left (a x +b \right )}+\frac {3 \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{4 b \sqrt {a b}}}{b}\) | \(59\) |
1/2*x^(1/2)/b/(a*x+b)^2+3/2/b*(1/2*x^(1/2)/b/(a*x+b)+1/2/b/(a*b)^(1/2)*arc tan(a*x^(1/2)/(a*b)^(1/2)))
Time = 0.30 (sec) , antiderivative size = 186, normalized size of antiderivative = 2.66 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{7/2}} \, dx=\left [-\frac {3 \, {\left (a^{2} x^{2} + 2 \, a b x + b^{2}\right )} \sqrt {-a b} \log \left (\frac {a x - b - 2 \, \sqrt {-a b} \sqrt {x}}{a x + b}\right ) - 2 \, {\left (3 \, a^{2} b x + 5 \, a b^{2}\right )} \sqrt {x}}{8 \, {\left (a^{3} b^{3} x^{2} + 2 \, a^{2} b^{4} x + a b^{5}\right )}}, -\frac {3 \, {\left (a^{2} x^{2} + 2 \, a b x + b^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b}}{a \sqrt {x}}\right ) - {\left (3 \, a^{2} b x + 5 \, a b^{2}\right )} \sqrt {x}}{4 \, {\left (a^{3} b^{3} x^{2} + 2 \, a^{2} b^{4} x + a b^{5}\right )}}\right ] \]
[-1/8*(3*(a^2*x^2 + 2*a*b*x + b^2)*sqrt(-a*b)*log((a*x - b - 2*sqrt(-a*b)* sqrt(x))/(a*x + b)) - 2*(3*a^2*b*x + 5*a*b^2)*sqrt(x))/(a^3*b^3*x^2 + 2*a^ 2*b^4*x + a*b^5), -1/4*(3*(a^2*x^2 + 2*a*b*x + b^2)*sqrt(a*b)*arctan(sqrt( a*b)/(a*sqrt(x))) - (3*a^2*b*x + 5*a*b^2)*sqrt(x))/(a^3*b^3*x^2 + 2*a^2*b^ 4*x + a*b^5)]
Leaf count of result is larger than twice the leaf count of optimal. 632 vs. \(2 (61) = 122\).
Time = 114.36 (sec) , antiderivative size = 632, normalized size of antiderivative = 9.03 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{7/2}} \, dx=\begin {cases} \tilde {\infty } \sqrt {x} & \text {for}\: a = 0 \wedge b = 0 \\\frac {2 \sqrt {x}}{b^{3}} & \text {for}\: a = 0 \\- \frac {2}{5 a^{3} x^{\frac {5}{2}}} & \text {for}\: b = 0 \\\frac {6 a^{2} x^{\frac {3}{2}} \sqrt {- \frac {b}{a}}}{8 a^{3} b^{2} x^{2} \sqrt {- \frac {b}{a}} + 16 a^{2} b^{3} x \sqrt {- \frac {b}{a}} + 8 a b^{4} \sqrt {- \frac {b}{a}}} + \frac {3 a^{2} x^{2} \log {\left (\sqrt {x} - \sqrt {- \frac {b}{a}} \right )}}{8 a^{3} b^{2} x^{2} \sqrt {- \frac {b}{a}} + 16 a^{2} b^{3} x \sqrt {- \frac {b}{a}} + 8 a b^{4} \sqrt {- \frac {b}{a}}} - \frac {3 a^{2} x^{2} \log {\left (\sqrt {x} + \sqrt {- \frac {b}{a}} \right )}}{8 a^{3} b^{2} x^{2} \sqrt {- \frac {b}{a}} + 16 a^{2} b^{3} x \sqrt {- \frac {b}{a}} + 8 a b^{4} \sqrt {- \frac {b}{a}}} + \frac {10 a b \sqrt {x} \sqrt {- \frac {b}{a}}}{8 a^{3} b^{2} x^{2} \sqrt {- \frac {b}{a}} + 16 a^{2} b^{3} x \sqrt {- \frac {b}{a}} + 8 a b^{4} \sqrt {- \frac {b}{a}}} + \frac {6 a b x \log {\left (\sqrt {x} - \sqrt {- \frac {b}{a}} \right )}}{8 a^{3} b^{2} x^{2} \sqrt {- \frac {b}{a}} + 16 a^{2} b^{3} x \sqrt {- \frac {b}{a}} + 8 a b^{4} \sqrt {- \frac {b}{a}}} - \frac {6 a b x \log {\left (\sqrt {x} + \sqrt {- \frac {b}{a}} \right )}}{8 a^{3} b^{2} x^{2} \sqrt {- \frac {b}{a}} + 16 a^{2} b^{3} x \sqrt {- \frac {b}{a}} + 8 a b^{4} \sqrt {- \frac {b}{a}}} + \frac {3 b^{2} \log {\left (\sqrt {x} - \sqrt {- \frac {b}{a}} \right )}}{8 a^{3} b^{2} x^{2} \sqrt {- \frac {b}{a}} + 16 a^{2} b^{3} x \sqrt {- \frac {b}{a}} + 8 a b^{4} \sqrt {- \frac {b}{a}}} - \frac {3 b^{2} \log {\left (\sqrt {x} + \sqrt {- \frac {b}{a}} \right )}}{8 a^{3} b^{2} x^{2} \sqrt {- \frac {b}{a}} + 16 a^{2} b^{3} x \sqrt {- \frac {b}{a}} + 8 a b^{4} \sqrt {- \frac {b}{a}}} & \text {otherwise} \end {cases} \]
Piecewise((zoo*sqrt(x), Eq(a, 0) & Eq(b, 0)), (2*sqrt(x)/b**3, Eq(a, 0)), (-2/(5*a**3*x**(5/2)), Eq(b, 0)), (6*a**2*x**(3/2)*sqrt(-b/a)/(8*a**3*b**2 *x**2*sqrt(-b/a) + 16*a**2*b**3*x*sqrt(-b/a) + 8*a*b**4*sqrt(-b/a)) + 3*a* *2*x**2*log(sqrt(x) - sqrt(-b/a))/(8*a**3*b**2*x**2*sqrt(-b/a) + 16*a**2*b **3*x*sqrt(-b/a) + 8*a*b**4*sqrt(-b/a)) - 3*a**2*x**2*log(sqrt(x) + sqrt(- b/a))/(8*a**3*b**2*x**2*sqrt(-b/a) + 16*a**2*b**3*x*sqrt(-b/a) + 8*a*b**4* sqrt(-b/a)) + 10*a*b*sqrt(x)*sqrt(-b/a)/(8*a**3*b**2*x**2*sqrt(-b/a) + 16* a**2*b**3*x*sqrt(-b/a) + 8*a*b**4*sqrt(-b/a)) + 6*a*b*x*log(sqrt(x) - sqrt (-b/a))/(8*a**3*b**2*x**2*sqrt(-b/a) + 16*a**2*b**3*x*sqrt(-b/a) + 8*a*b** 4*sqrt(-b/a)) - 6*a*b*x*log(sqrt(x) + sqrt(-b/a))/(8*a**3*b**2*x**2*sqrt(- b/a) + 16*a**2*b**3*x*sqrt(-b/a) + 8*a*b**4*sqrt(-b/a)) + 3*b**2*log(sqrt( x) - sqrt(-b/a))/(8*a**3*b**2*x**2*sqrt(-b/a) + 16*a**2*b**3*x*sqrt(-b/a) + 8*a*b**4*sqrt(-b/a)) - 3*b**2*log(sqrt(x) + sqrt(-b/a))/(8*a**3*b**2*x** 2*sqrt(-b/a) + 16*a**2*b**3*x*sqrt(-b/a) + 8*a*b**4*sqrt(-b/a)), True))
Time = 0.31 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.90 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{7/2}} \, dx=\frac {\frac {3 \, a}{\sqrt {x}} + \frac {5 \, b}{x^{\frac {3}{2}}}}{4 \, {\left (a^{2} b^{2} + \frac {2 \, a b^{3}}{x} + \frac {b^{4}}{x^{2}}\right )}} - \frac {3 \, \arctan \left (\frac {b}{\sqrt {a b} \sqrt {x}}\right )}{4 \, \sqrt {a b} b^{2}} \]
1/4*(3*a/sqrt(x) + 5*b/x^(3/2))/(a^2*b^2 + 2*a*b^3/x + b^4/x^2) - 3/4*arct an(b/(sqrt(a*b)*sqrt(x)))/(sqrt(a*b)*b^2)
Time = 0.27 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.67 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{7/2}} \, dx=\frac {3 \, \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} b^{2}} + \frac {3 \, a x^{\frac {3}{2}} + 5 \, b \sqrt {x}}{4 \, {\left (a x + b\right )}^{2} b^{2}} \]
3/4*arctan(a*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^2) + 1/4*(3*a*x^(3/2) + 5*b*s qrt(x))/((a*x + b)^2*b^2)
Time = 5.86 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.81 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{7/2}} \, dx=\frac {\frac {5\,\sqrt {x}}{4\,b}+\frac {3\,a\,x^{3/2}}{4\,b^2}}{a^2\,x^2+2\,a\,b\,x+b^2}+\frac {3\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {x}}{\sqrt {b}}\right )}{4\,\sqrt {a}\,b^{5/2}} \]