Integrand size = 13, antiderivative size = 69 \[ \int \frac {x}{\left (a+\frac {b}{x^2}\right )^{3/2}} \, dx=\frac {3 b}{2 a^2 \sqrt {a+\frac {b}{x^2}}}+\frac {x^2}{2 a \sqrt {a+\frac {b}{x^2}}}-\frac {3 b \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x^2}}}{\sqrt {a}}\right )}{2 a^{5/2}} \]
-3/2*b*arctanh((a+b/x^2)^(1/2)/a^(1/2))/a^(5/2)+3/2*b/a^2/(a+b/x^2)^(1/2)+ 1/2*x^2/a/(a+b/x^2)^(1/2)
Time = 0.26 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.20 \[ \int \frac {x}{\left (a+\frac {b}{x^2}\right )^{3/2}} \, dx=\frac {\sqrt {a} x \left (3 b+a x^2\right )+6 b \sqrt {b+a x^2} \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {b}-\sqrt {b+a x^2}}\right )}{2 a^{5/2} \sqrt {a+\frac {b}{x^2}} x} \]
(Sqrt[a]*x*(3*b + a*x^2) + 6*b*Sqrt[b + a*x^2]*ArcTanh[(Sqrt[a]*x)/(Sqrt[b ] - Sqrt[b + a*x^2])])/(2*a^(5/2)*Sqrt[a + b/x^2]*x)
Time = 0.20 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {798, 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.
\(\displaystyle \int \frac {x}{\left (a+\frac {b}{x^2}\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle -\frac {1}{2} \int \frac {x^4}{\left (a+\frac {b}{x^2}\right )^{3/2}}d\frac {1}{x^2}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{2} \left (\frac {3 b \int \frac {x^2}{\left (a+\frac {b}{x^2}\right )^{3/2}}d\frac {1}{x^2}}{2 a}+\frac {x^2}{a \sqrt {a+\frac {b}{x^2}}}\right )\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {1}{2} \left (\frac {3 b \left (\frac {\int \frac {x^2}{\sqrt {a+\frac {b}{x^2}}}d\frac {1}{x^2}}{a}+\frac {2}{a \sqrt {a+\frac {b}{x^2}}}\right )}{2 a}+\frac {x^2}{a \sqrt {a+\frac {b}{x^2}}}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} \left (\frac {3 b \left (\frac {2 \int \frac {1}{\frac {1}{b x^4}-\frac {a}{b}}d\sqrt {a+\frac {b}{x^2}}}{a b}+\frac {2}{a \sqrt {a+\frac {b}{x^2}}}\right )}{2 a}+\frac {x^2}{a \sqrt {a+\frac {b}{x^2}}}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{2} \left (\frac {3 b \left (\frac {2}{a \sqrt {a+\frac {b}{x^2}}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x^2}}}{\sqrt {a}}\right )}{a^{3/2}}\right )}{2 a}+\frac {x^2}{a \sqrt {a+\frac {b}{x^2}}}\right )\) |
(x^2/(a*Sqrt[a + b/x^2]) + (3*b*(2/(a*Sqrt[a + b/x^2]) - (2*ArcTanh[Sqrt[a + b/x^2]/Sqrt[a]])/a^(3/2)))/(2*a))/2
3.20.31.3.1 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 = 0.04 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.06
method | result | size |
default | \(\frac {\left (a \,x^{2}+b \right ) \left (x^{3} a^{\frac {5}{2}}+3 a^{\frac {3}{2}} b x -3 \ln \left (\sqrt {a}\, x +\sqrt {a \,x^{2}+b}\right ) \sqrt {a \,x^{2}+b}\, a b \right )}{2 \left (\frac {a \,x^{2}+b}{x^{2}}\right )^{\frac {3}{2}} x^{3} a^{\frac {7}{2}}}\) | \(73\) |
risch | \(\frac {a \,x^{2}+b}{2 a^{2} \sqrt {\frac {a \,x^{2}+b}{x^{2}}}}+\frac {\left (\frac {b x}{a^{2} \sqrt {a \,x^{2}+b}}-\frac {3 b \ln \left (\sqrt {a}\, x +\sqrt {a \,x^{2}+b}\right )}{2 a^{\frac {5}{2}}}\right ) \sqrt {a \,x^{2}+b}}{\sqrt {\frac {a \,x^{2}+b}{x^{2}}}\, x}\) | \(91\) |
1/2*(a*x^2+b)*(x^3*a^(5/2)+3*a^(3/2)*b*x-3*ln(a^(1/2)*x+(a*x^2+b)^(1/2))*( a*x^2+b)^(1/2)*a*b)/((a*x^2+b)/x^2)^(3/2)/x^3/a^(7/2)
Time = 0.32 (sec) , antiderivative size = 192, normalized size of antiderivative = 2.78 \[ \int \frac {x}{\left (a+\frac {b}{x^2}\right )^{3/2}} \, dx=\left [\frac {3 \, {\left (a b x^{2} + b^{2}\right )} \sqrt {a} \log \left (-2 \, a x^{2} + 2 \, \sqrt {a} x^{2} \sqrt {\frac {a x^{2} + b}{x^{2}}} - b\right ) + 2 \, {\left (a^{2} x^{4} + 3 \, a b x^{2}\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{4 \, {\left (a^{4} x^{2} + a^{3} b\right )}}, \frac {3 \, {\left (a b x^{2} + b^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a} x^{2} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{a x^{2} + b}\right ) + {\left (a^{2} x^{4} + 3 \, a b x^{2}\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{2 \, {\left (a^{4} x^{2} + a^{3} b\right )}}\right ] \]
[1/4*(3*(a*b*x^2 + b^2)*sqrt(a)*log(-2*a*x^2 + 2*sqrt(a)*x^2*sqrt((a*x^2 + b)/x^2) - b) + 2*(a^2*x^4 + 3*a*b*x^2)*sqrt((a*x^2 + b)/x^2))/(a^4*x^2 + a^3*b), 1/2*(3*(a*b*x^2 + b^2)*sqrt(-a)*arctan(sqrt(-a)*x^2*sqrt((a*x^2 + b)/x^2)/(a*x^2 + b)) + (a^2*x^4 + 3*a*b*x^2)*sqrt((a*x^2 + b)/x^2))/(a^4*x ^2 + a^3*b)]
Time = 1.69 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.03 \[ \int \frac {x}{\left (a+\frac {b}{x^2}\right )^{3/2}} \, dx=\frac {x^{3}}{2 a \sqrt {b} \sqrt {\frac {a x^{2}}{b} + 1}} + \frac {3 \sqrt {b} x}{2 a^{2} \sqrt {\frac {a x^{2}}{b} + 1}} - \frac {3 b \operatorname {asinh}{\left (\frac {\sqrt {a} x}{\sqrt {b}} \right )}}{2 a^{\frac {5}{2}}} \]
x**3/(2*a*sqrt(b)*sqrt(a*x**2/b + 1)) + 3*sqrt(b)*x/(2*a**2*sqrt(a*x**2/b + 1)) - 3*b*asinh(sqrt(a)*x/sqrt(b))/(2*a**(5/2))
Time = 0.28 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.25 \[ \int \frac {x}{\left (a+\frac {b}{x^2}\right )^{3/2}} \, dx=\frac {3 \, {\left (a + \frac {b}{x^{2}}\right )} b - 2 \, a b}{2 \, {\left ({\left (a + \frac {b}{x^{2}}\right )}^{\frac {3}{2}} a^{2} - \sqrt {a + \frac {b}{x^{2}}} a^{3}\right )}} + \frac {3 \, b \log \left (\frac {\sqrt {a + \frac {b}{x^{2}}} - \sqrt {a}}{\sqrt {a + \frac {b}{x^{2}}} + \sqrt {a}}\right )}{4 \, a^{\frac {5}{2}}} \]
1/2*(3*(a + b/x^2)*b - 2*a*b)/((a + b/x^2)^(3/2)*a^2 - sqrt(a + b/x^2)*a^3 ) + 3/4*b*log((sqrt(a + b/x^2) - sqrt(a))/(sqrt(a + b/x^2) + sqrt(a)))/a^( 5/2)
Time = 0.32 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.07 \[ \int \frac {x}{\left (a+\frac {b}{x^2}\right )^{3/2}} \, dx=\frac {x {\left (\frac {x^{2}}{a \mathrm {sgn}\left (x\right )} + \frac {3 \, b}{a^{2} \mathrm {sgn}\left (x\right )}\right )}}{2 \, \sqrt {a x^{2} + b}} - \frac {3 \, b \log \left ({\left | b \right |}\right ) \mathrm {sgn}\left (x\right )}{4 \, a^{\frac {5}{2}}} + \frac {3 \, b \log \left ({\left | -\sqrt {a} x + \sqrt {a x^{2} + b} \right |}\right )}{2 \, a^{\frac {5}{2}} \mathrm {sgn}\left (x\right )} \]
1/2*x*(x^2/(a*sgn(x)) + 3*b/(a^2*sgn(x)))/sqrt(a*x^2 + b) - 3/4*b*log(abs( b))*sgn(x)/a^(5/2) + 3/2*b*log(abs(-sqrt(a)*x + sqrt(a*x^2 + b)))/(a^(5/2) *sgn(x))
Time = 6.33 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.77 \[ \int \frac {x}{\left (a+\frac {b}{x^2}\right )^{3/2}} \, dx=\frac {3\,b}{2\,a^2\,\sqrt {a+\frac {b}{x^2}}}+\frac {x^2}{2\,a\,\sqrt {a+\frac {b}{x^2}}}-\frac {3\,b\,\mathrm {atanh}\left (\frac {\sqrt {a+\frac {b}{x^2}}}{\sqrt {a}}\right )}{2\,a^{5/2}} \]