Integrand size = 13, antiderivative size = 93 \[ \int \frac {x}{\left (a+\frac {b}{x}\right )^{3/2}} \, dx=-\frac {15 b^2}{4 a^3 \sqrt {a+\frac {b}{x}}}-\frac {5 b x}{4 a^2 \sqrt {a+\frac {b}{x}}}+\frac {x^2}{2 a \sqrt {a+\frac {b}{x}}}+\frac {15 b^2 \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{4 a^{7/2}} \]
15/4*b^2*arctanh((a+b/x)^(1/2)/a^(1/2))/a^(7/2)-15/4*b^2/a^3/(a+b/x)^(1/2) -5/4*b*x/a^2/(a+b/x)^(1/2)+1/2*x^2/a/(a+b/x)^(1/2)
Time = 0.17 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.82 \[ \int \frac {x}{\left (a+\frac {b}{x}\right )^{3/2}} \, dx=\frac {\sqrt {a+\frac {b}{x}} x \left (-15 b^2-5 a b x+2 a^2 x^2\right )}{4 a^3 (b+a x)}+\frac {15 b^2 \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{4 a^{7/2}} \]
(Sqrt[a + b/x]*x*(-15*b^2 - 5*a*b*x + 2*a^2*x^2))/(4*a^3*(b + a*x)) + (15* b^2*ArcTanh[Sqrt[a + b/x]/Sqrt[a]])/(4*a^(7/2))
Time = 0.21 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {798, 52, 52, 61, 73, 221}
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.
\(\displaystyle \int \frac {x}{\left (a+\frac {b}{x}\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle -\int \frac {x^3}{\left (a+\frac {b}{x}\right )^{3/2}}d\frac {1}{x}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {5 b \int \frac {x^2}{\left (a+\frac {b}{x}\right )^{3/2}}d\frac {1}{x}}{4 a}+\frac {x^2}{2 a \sqrt {a+\frac {b}{x}}}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {5 b \left (-\frac {3 b \int \frac {x}{\left (a+\frac {b}{x}\right )^{3/2}}d\frac {1}{x}}{2 a}-\frac {x}{a \sqrt {a+\frac {b}{x}}}\right )}{4 a}+\frac {x^2}{2 a \sqrt {a+\frac {b}{x}}}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {5 b \left (-\frac {3 b \left (\frac {\int \frac {x}{\sqrt {a+\frac {b}{x}}}d\frac {1}{x}}{a}+\frac {2}{a \sqrt {a+\frac {b}{x}}}\right )}{2 a}-\frac {x}{a \sqrt {a+\frac {b}{x}}}\right )}{4 a}+\frac {x^2}{2 a \sqrt {a+\frac {b}{x}}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {5 b \left (-\frac {3 b \left (\frac {2 \int \frac {1}{\frac {1}{b x^2}-\frac {a}{b}}d\sqrt {a+\frac {b}{x}}}{a b}+\frac {2}{a \sqrt {a+\frac {b}{x}}}\right )}{2 a}-\frac {x}{a \sqrt {a+\frac {b}{x}}}\right )}{4 a}+\frac {x^2}{2 a \sqrt {a+\frac {b}{x}}}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {5 b \left (-\frac {3 b \left (\frac {2}{a \sqrt {a+\frac {b}{x}}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{a^{3/2}}\right )}{2 a}-\frac {x}{a \sqrt {a+\frac {b}{x}}}\right )}{4 a}+\frac {x^2}{2 a \sqrt {a+\frac {b}{x}}}\) |
x^2/(2*a*Sqrt[a + b/x]) + (5*b*(-(x/(a*Sqrt[a + b/x])) - (3*b*(2/(a*Sqrt[a + b/x]) - (2*ArcTanh[Sqrt[a + b/x]/Sqrt[a]])/a^(3/2)))/(2*a)))/(4*a)
3.18.33.3.1 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)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/n Subst [Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
Time = 0.06 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.39
method | result | size |
risch | \(\frac {\left (2 a x -7 b \right ) \left (a x +b \right )}{4 a^{3} \sqrt {\frac {a x +b}{x}}}+\frac {\left (\frac {15 b^{2} \ln \left (\frac {\frac {b}{2}+a x}{\sqrt {a}}+\sqrt {a \,x^{2}+b x}\right )}{8 a^{\frac {7}{2}}}-\frac {2 b^{2} \sqrt {a \left (x +\frac {b}{a}\right )^{2}-b \left (x +\frac {b}{a}\right )}}{a^{4} \left (x +\frac {b}{a}\right )}\right ) \sqrt {x \left (a x +b \right )}}{x \sqrt {\frac {a x +b}{x}}}\) | \(129\) |
default | \(\frac {\sqrt {\frac {a x +b}{x}}\, x \left (4 \sqrt {a \,x^{2}+b x}\, a^{\frac {9}{2}} x^{3}-32 \sqrt {x \left (a x +b \right )}\, a^{\frac {7}{2}} b \,x^{2}+10 \sqrt {a \,x^{2}+b x}\, a^{\frac {7}{2}} b \,x^{2}+16 a^{3} \ln \left (\frac {2 \sqrt {x \left (a x +b \right )}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) b^{2} x^{2}+16 \left (x \left (a x +b \right )\right )^{\frac {3}{2}} a^{\frac {5}{2}} b -64 \sqrt {x \left (a x +b \right )}\, a^{\frac {5}{2}} b^{2} x +8 a^{\frac {5}{2}} \sqrt {a \,x^{2}+b x}\, b^{2} x +32 a^{2} \ln \left (\frac {2 \sqrt {x \left (a x +b \right )}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) b^{3} x -\ln \left (\frac {2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) a^{3} b^{2} x^{2}-32 a^{\frac {3}{2}} \sqrt {x \left (a x +b \right )}\, b^{3}+2 a^{\frac {3}{2}} \sqrt {a \,x^{2}+b x}\, b^{3}+16 a \ln \left (\frac {2 \sqrt {x \left (a x +b \right )}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) b^{4}-2 \ln \left (\frac {2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) a^{2} b^{3} x -\ln \left (\frac {2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) a \,b^{4}\right )}{8 a^{\frac {9}{2}} \sqrt {x \left (a x +b \right )}\, \left (a x +b \right )^{2}}\) | \(395\) |
1/4*(2*a*x-7*b)*(a*x+b)/a^3/((a*x+b)/x)^(1/2)+(15/8*b^2/a^(7/2)*ln((1/2*b+ a*x)/a^(1/2)+(a*x^2+b*x)^(1/2))-2*b^2/a^4/(x+b/a)*(a*(x+b/a)^2-b*(x+b/a))^ (1/2))/x/((a*x+b)/x)^(1/2)*(x*(a*x+b))^(1/2)
Time = 0.31 (sec) , antiderivative size = 186, normalized size of antiderivative = 2.00 \[ \int \frac {x}{\left (a+\frac {b}{x}\right )^{3/2}} \, dx=\left [\frac {15 \, {\left (a b^{2} x + b^{3}\right )} \sqrt {a} \log \left (2 \, a x + 2 \, \sqrt {a} x \sqrt {\frac {a x + b}{x}} + b\right ) + 2 \, {\left (2 \, a^{3} x^{3} - 5 \, a^{2} b x^{2} - 15 \, a b^{2} x\right )} \sqrt {\frac {a x + b}{x}}}{8 \, {\left (a^{5} x + a^{4} b\right )}}, -\frac {15 \, {\left (a b^{2} x + b^{3}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {a x + b}{x}}}{a}\right ) - {\left (2 \, a^{3} x^{3} - 5 \, a^{2} b x^{2} - 15 \, a b^{2} x\right )} \sqrt {\frac {a x + b}{x}}}{4 \, {\left (a^{5} x + a^{4} b\right )}}\right ] \]
[1/8*(15*(a*b^2*x + b^3)*sqrt(a)*log(2*a*x + 2*sqrt(a)*x*sqrt((a*x + b)/x) + b) + 2*(2*a^3*x^3 - 5*a^2*b*x^2 - 15*a*b^2*x)*sqrt((a*x + b)/x))/(a^5*x + a^4*b), -1/4*(15*(a*b^2*x + b^3)*sqrt(-a)*arctan(sqrt(-a)*sqrt((a*x + b )/x)/a) - (2*a^3*x^3 - 5*a^2*b*x^2 - 15*a*b^2*x)*sqrt((a*x + b)/x))/(a^5*x + a^4*b)]
Time = 3.64 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.13 \[ \int \frac {x}{\left (a+\frac {b}{x}\right )^{3/2}} \, dx=\frac {x^{\frac {5}{2}}}{2 a \sqrt {b} \sqrt {\frac {a x}{b} + 1}} - \frac {5 \sqrt {b} x^{\frac {3}{2}}}{4 a^{2} \sqrt {\frac {a x}{b} + 1}} - \frac {15 b^{\frac {3}{2}} \sqrt {x}}{4 a^{3} \sqrt {\frac {a x}{b} + 1}} + \frac {15 b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}} \right )}}{4 a^{\frac {7}{2}}} \]
x**(5/2)/(2*a*sqrt(b)*sqrt(a*x/b + 1)) - 5*sqrt(b)*x**(3/2)/(4*a**2*sqrt(a *x/b + 1)) - 15*b**(3/2)*sqrt(x)/(4*a**3*sqrt(a*x/b + 1)) + 15*b**2*asinh( sqrt(a)*sqrt(x)/sqrt(b))/(4*a**(7/2))
Time = 0.28 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.31 \[ \int \frac {x}{\left (a+\frac {b}{x}\right )^{3/2}} \, dx=-\frac {15 \, {\left (a + \frac {b}{x}\right )}^{2} b^{2} - 25 \, {\left (a + \frac {b}{x}\right )} a b^{2} + 8 \, a^{2} b^{2}}{4 \, {\left ({\left (a + \frac {b}{x}\right )}^{\frac {5}{2}} a^{3} - 2 \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} a^{4} + \sqrt {a + \frac {b}{x}} a^{5}\right )}} - \frac {15 \, b^{2} \log \left (\frac {\sqrt {a + \frac {b}{x}} - \sqrt {a}}{\sqrt {a + \frac {b}{x}} + \sqrt {a}}\right )}{8 \, a^{\frac {7}{2}}} \]
-1/4*(15*(a + b/x)^2*b^2 - 25*(a + b/x)*a*b^2 + 8*a^2*b^2)/((a + b/x)^(5/2 )*a^3 - 2*(a + b/x)^(3/2)*a^4 + sqrt(a + b/x)*a^5) - 15/8*b^2*log((sqrt(a + b/x) - sqrt(a))/(sqrt(a + b/x) + sqrt(a)))/a^(7/2)
Time = 0.29 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.45 \[ \int \frac {x}{\left (a+\frac {b}{x}\right )^{3/2}} \, dx=\frac {1}{4} \, \sqrt {a x^{2} + b x} {\left (\frac {2 \, x}{a^{2} \mathrm {sgn}\left (x\right )} - \frac {7 \, b}{a^{3} \mathrm {sgn}\left (x\right )}\right )} - \frac {15 \, b^{2} \log \left ({\left | 2 \, {\left (\sqrt {a} x - \sqrt {a x^{2} + b x}\right )} \sqrt {a} + b \right |}\right )}{8 \, a^{\frac {7}{2}} \mathrm {sgn}\left (x\right )} + \frac {{\left (15 \, b^{2} \log \left ({\left | b \right |}\right ) + 16 \, b^{2}\right )} \mathrm {sgn}\left (x\right )}{8 \, a^{\frac {7}{2}}} - \frac {2 \, b^{3}}{{\left ({\left (\sqrt {a} x - \sqrt {a x^{2} + b x}\right )} \sqrt {a} + b\right )} a^{\frac {7}{2}} \mathrm {sgn}\left (x\right )} \]
1/4*sqrt(a*x^2 + b*x)*(2*x/(a^2*sgn(x)) - 7*b/(a^3*sgn(x))) - 15/8*b^2*log (abs(2*(sqrt(a)*x - sqrt(a*x^2 + b*x))*sqrt(a) + b))/(a^(7/2)*sgn(x)) + 1/ 8*(15*b^2*log(abs(b)) + 16*b^2)*sgn(x)/a^(7/2) - 2*b^3/(((sqrt(a)*x - sqrt (a*x^2 + b*x))*sqrt(a) + b)*a^(7/2)*sgn(x))
Time = 5.96 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.78 \[ \int \frac {x}{\left (a+\frac {b}{x}\right )^{3/2}} \, dx=\frac {15\,b^2\,\mathrm {atanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{4\,a^{7/2}}-\frac {15\,b^2}{4\,a^3\,\sqrt {a+\frac {b}{x}}}+\frac {x^2}{2\,a\,\sqrt {a+\frac {b}{x}}}-\frac {5\,b\,x}{4\,a^2\,\sqrt {a+\frac {b}{x}}} \]