Integrand size = 15, antiderivative size = 64 \[ \int \frac {x^2}{\left (a+\frac {b}{x^3}\right )^{3/2}} \, dx=\frac {b}{a^2 \sqrt {a+\frac {b}{x^3}}}+\frac {x^3}{3 a \sqrt {a+\frac {b}{x^3}}}-\frac {b \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x^3}}}{\sqrt {a}}\right )}{a^{5/2}} \]
Time = 0.41 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.31 \[ \int \frac {x^2}{\left (a+\frac {b}{x^3}\right )^{3/2}} \, dx=\frac {\sqrt {a} x^{3/2} \left (3 b+a x^3\right )-3 b \sqrt {b+a x^3} \log \left (\sqrt {a} x^{3/2}+\sqrt {b+a x^3}\right )}{3 a^{5/2} \sqrt {a+\frac {b}{x^3}} x^{3/2}} \]
(Sqrt[a]*x^(3/2)*(3*b + a*x^3) - 3*b*Sqrt[b + a*x^3]*Log[Sqrt[a]*x^(3/2) + Sqrt[b + a*x^3]])/(3*a^(5/2)*Sqrt[a + b/x^3]*x^(3/2))
Time = 0.20 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.14, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, 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.
\(\displaystyle \int \frac {x^2}{\left (a+\frac {b}{x^3}\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle -\frac {1}{3} \int \frac {x^6}{\left (a+\frac {b}{x^3}\right )^{3/2}}d\frac {1}{x^3}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{3} \left (\frac {3 b \int \frac {x^3}{\left (a+\frac {b}{x^3}\right )^{3/2}}d\frac {1}{x^3}}{2 a}+\frac {x^3}{a \sqrt {a+\frac {b}{x^3}}}\right )\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {1}{3} \left (\frac {3 b \left (\frac {\int \frac {x^3}{\sqrt {a+\frac {b}{x^3}}}d\frac {1}{x^3}}{a}+\frac {2}{a \sqrt {a+\frac {b}{x^3}}}\right )}{2 a}+\frac {x^3}{a \sqrt {a+\frac {b}{x^3}}}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{3} \left (\frac {3 b \left (\frac {2 \int \frac {1}{\frac {1}{b x^6}-\frac {a}{b}}d\sqrt {a+\frac {b}{x^3}}}{a b}+\frac {2}{a \sqrt {a+\frac {b}{x^3}}}\right )}{2 a}+\frac {x^3}{a \sqrt {a+\frac {b}{x^3}}}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{3} \left (\frac {3 b \left (\frac {2}{a \sqrt {a+\frac {b}{x^3}}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x^3}}}{\sqrt {a}}\right )}{a^{3/2}}\right )}{2 a}+\frac {x^3}{a \sqrt {a+\frac {b}{x^3}}}\right )\) |
(x^3/(a*Sqrt[a + b/x^3]) + (3*b*(2/(a*Sqrt[a + b/x^3]) - (2*ArcTanh[Sqrt[a + b/x^3]/Sqrt[a]])/a^(3/2)))/(2*a))/3
3.21.38.3.1 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 = 0.42 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.30
method | result | size |
default | \(-\frac {\left (a \,x^{3}+b \right ) \left (-x^{5} a^{\frac {7}{2}}-3 a^{\frac {5}{2}} b \,x^{2}+3 \,\operatorname {arctanh}\left (\frac {\sqrt {x \left (a \,x^{3}+b \right )}}{x^{2} \sqrt {a}}\right ) b \,a^{2} \sqrt {x \left (a \,x^{3}+b \right )}\right )}{3 \left (\frac {a \,x^{3}+b}{x^{3}}\right )^{\frac {3}{2}} x^{5} a^{\frac {9}{2}}}\) | \(83\) |
risch | \(\frac {a \,x^{3}+b}{3 a^{2} \sqrt {\frac {a \,x^{3}+b}{x^{3}}}}-\frac {b \left (\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {x \left (a \,x^{3}+b \right )}}{x^{2} \sqrt {a}}\right )}{\sqrt {a}}-\frac {4 x^{2}}{3 \sqrt {\left (x^{3}+\frac {b}{a}\right ) a x}}\right ) \sqrt {x \left (a \,x^{3}+b \right )}}{2 a^{2} x^{2} \sqrt {\frac {a \,x^{3}+b}{x^{3}}}}\) | \(104\) |
-1/3/((a*x^3+b)/x^3)^(3/2)/x^5*(a*x^3+b)*(-x^5*a^(7/2)-3*a^(5/2)*b*x^2+3*a rctanh((x*(a*x^3+b))^(1/2)/x^2/a^(1/2))*b*a^2*(x*(a*x^3+b))^(1/2))/a^(9/2)
Time = 0.38 (sec) , antiderivative size = 215, normalized size of antiderivative = 3.36 \[ \int \frac {x^2}{\left (a+\frac {b}{x^3}\right )^{3/2}} \, dx=\left [\frac {3 \, {\left (a b x^{3} + b^{2}\right )} \sqrt {a} \log \left (-8 \, a^{2} x^{6} - 8 \, a b x^{3} - b^{2} + 4 \, {\left (2 \, a x^{6} + b x^{3}\right )} \sqrt {a} \sqrt {\frac {a x^{3} + b}{x^{3}}}\right ) + 4 \, {\left (a^{2} x^{6} + 3 \, a b x^{3}\right )} \sqrt {\frac {a x^{3} + b}{x^{3}}}}{12 \, {\left (a^{4} x^{3} + a^{3} b\right )}}, \frac {3 \, {\left (a b x^{3} + b^{2}\right )} \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} x^{3} \sqrt {\frac {a x^{3} + b}{x^{3}}}}{2 \, a x^{3} + b}\right ) + 2 \, {\left (a^{2} x^{6} + 3 \, a b x^{3}\right )} \sqrt {\frac {a x^{3} + b}{x^{3}}}}{6 \, {\left (a^{4} x^{3} + a^{3} b\right )}}\right ] \]
[1/12*(3*(a*b*x^3 + b^2)*sqrt(a)*log(-8*a^2*x^6 - 8*a*b*x^3 - b^2 + 4*(2*a *x^6 + b*x^3)*sqrt(a)*sqrt((a*x^3 + b)/x^3)) + 4*(a^2*x^6 + 3*a*b*x^3)*sqr t((a*x^3 + b)/x^3))/(a^4*x^3 + a^3*b), 1/6*(3*(a*b*x^3 + b^2)*sqrt(-a)*arc tan(2*sqrt(-a)*x^3*sqrt((a*x^3 + b)/x^3)/(2*a*x^3 + b)) + 2*(a^2*x^6 + 3*a *b*x^3)*sqrt((a*x^3 + b)/x^3))/(a^4*x^3 + a^3*b)]
Time = 1.75 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.14 \[ \int \frac {x^2}{\left (a+\frac {b}{x^3}\right )^{3/2}} \, dx=\frac {x^{\frac {9}{2}}}{3 a \sqrt {b} \sqrt {\frac {a x^{3}}{b} + 1}} + \frac {\sqrt {b} x^{\frac {3}{2}}}{a^{2} \sqrt {\frac {a x^{3}}{b} + 1}} - \frac {b \operatorname {asinh}{\left (\frac {\sqrt {a} x^{\frac {3}{2}}}{\sqrt {b}} \right )}}{a^{\frac {5}{2}}} \]
x**(9/2)/(3*a*sqrt(b)*sqrt(a*x**3/b + 1)) + sqrt(b)*x**(3/2)/(a**2*sqrt(a* x**3/b + 1)) - b*asinh(sqrt(a)*x**(3/2)/sqrt(b))/a**(5/2)
Time = 0.28 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.34 \[ \int \frac {x^2}{\left (a+\frac {b}{x^3}\right )^{3/2}} \, dx=\frac {3 \, {\left (a + \frac {b}{x^{3}}\right )} b - 2 \, a b}{3 \, {\left ({\left (a + \frac {b}{x^{3}}\right )}^{\frac {3}{2}} a^{2} - \sqrt {a + \frac {b}{x^{3}}} a^{3}\right )}} + \frac {b \log \left (\frac {\sqrt {a + \frac {b}{x^{3}}} - \sqrt {a}}{\sqrt {a + \frac {b}{x^{3}}} + \sqrt {a}}\right )}{2 \, a^{\frac {5}{2}}} \]
1/3*(3*(a + b/x^3)*b - 2*a*b)/((a + b/x^3)^(3/2)*a^2 - sqrt(a + b/x^3)*a^3 ) + 1/2*b*log((sqrt(a + b/x^3) - sqrt(a))/(sqrt(a + b/x^3) + sqrt(a)))/a^( 5/2)
Time = 0.37 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.92 \[ \int \frac {x^2}{\left (a+\frac {b}{x^3}\right )^{3/2}} \, dx=\frac {\sqrt {a x^{4} + b x} x}{3 \, a^{2}} + \frac {b \arctan \left (\frac {\sqrt {a + \frac {b}{x^{3}}}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} + \frac {2 \, b}{3 \, \sqrt {a + \frac {b}{x^{3}}} a^{2}} \]
1/3*sqrt(a*x^4 + b*x)*x/a^2 + b*arctan(sqrt(a + b/x^3)/sqrt(-a))/(sqrt(-a) *a^2) + 2/3*b/(sqrt(a + b/x^3)*a^2)
Time = 6.28 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.14 \[ \int \frac {x^2}{\left (a+\frac {b}{x^3}\right )^{3/2}} \, dx=\frac {b}{a^2\,\sqrt {a+\frac {b}{x^3}}}+\frac {b\,\ln \left (x^6\,{\left (\sqrt {a+\frac {b}{x^3}}-\sqrt {a}\right )}^3\,\left (\sqrt {a+\frac {b}{x^3}}+\sqrt {a}\right )\right )}{2\,a^{5/2}}+\frac {x^3}{3\,a\,\sqrt {a+\frac {b}{x^3}}} \]