Integrand size = 15, antiderivative size = 71 \[ \int \sqrt {a+\frac {b}{x^3}} x^5 \, dx=\frac {b \sqrt {a+\frac {b}{x^3}} x^3}{12 a}+\frac {1}{6} \sqrt {a+\frac {b}{x^3}} x^6-\frac {b^2 \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x^3}}}{\sqrt {a}}\right )}{12 a^{3/2}} \]
-1/12*b^2*arctanh((a+b/x^3)^(1/2)/a^(1/2))/a^(3/2)+1/12*b*x^3*(a+b/x^3)^(1 /2)/a+1/6*x^6*(a+b/x^3)^(1/2)
Time = 0.57 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.31 \[ \int \sqrt {a+\frac {b}{x^3}} x^5 \, dx=\frac {\sqrt {a+\frac {b}{x^3}} x^3 \left (b+2 a x^3\right )}{12 a}-\frac {b^2 \sqrt {a+\frac {b}{x^3}} x^{3/2} \log \left (\sqrt {a} x^{3/2}+\sqrt {b+a x^3}\right )}{12 a^{3/2} \sqrt {b+a x^3}} \]
(Sqrt[a + b/x^3]*x^3*(b + 2*a*x^3))/(12*a) - (b^2*Sqrt[a + b/x^3]*x^(3/2)* Log[Sqrt[a]*x^(3/2) + Sqrt[b + a*x^3]])/(12*a^(3/2)*Sqrt[b + a*x^3])
Time = 0.18 (sec) , antiderivative size = 73, 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 = {798, 51, 52, 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.
\(\displaystyle \int x^5 \sqrt {a+\frac {b}{x^3}} \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle -\frac {1}{3} \int \sqrt {a+\frac {b}{x^3}} x^9d\frac {1}{x^3}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{2} x^6 \sqrt {a+\frac {b}{x^3}}-\frac {1}{4} b \int \frac {x^6}{\sqrt {a+\frac {b}{x^3}}}d\frac {1}{x^3}\right )\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{2} x^6 \sqrt {a+\frac {b}{x^3}}-\frac {1}{4} b \left (-\frac {b \int \frac {x^3}{\sqrt {a+\frac {b}{x^3}}}d\frac {1}{x^3}}{2 a}-\frac {x^3 \sqrt {a+\frac {b}{x^3}}}{a}\right )\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{2} x^6 \sqrt {a+\frac {b}{x^3}}-\frac {1}{4} b \left (-\frac {\int \frac {1}{\frac {1}{b x^6}-\frac {a}{b}}d\sqrt {a+\frac {b}{x^3}}}{a}-\frac {x^3 \sqrt {a+\frac {b}{x^3}}}{a}\right )\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{2} x^6 \sqrt {a+\frac {b}{x^3}}-\frac {1}{4} b \left (\frac {b \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x^3}}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {x^3 \sqrt {a+\frac {b}{x^3}}}{a}\right )\right )\) |
((Sqrt[a + b/x^3]*x^6)/2 - (b*(-((Sqrt[a + b/x^3]*x^3)/a) + (b*ArcTanh[Sqr t[a + b/x^3]/Sqrt[a]])/a^(3/2)))/4)/3
3.20.91.3.1 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/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x ] && ILtQ[m, -1] && FractionQ[n] && GtQ[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] && 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)*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 = 1.87 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.30
method | result | size |
risch | \(\frac {x^{3} \left (2 a \,x^{3}+b \right ) \sqrt {\frac {a \,x^{3}+b}{x^{3}}}}{12 a}-\frac {\operatorname {arctanh}\left (\frac {\sqrt {x \left (a \,x^{3}+b \right )}}{x^{2} \sqrt {a}}\right ) b^{2} x \sqrt {\frac {a \,x^{3}+b}{x^{3}}}\, \sqrt {x \left (a \,x^{3}+b \right )}}{12 a^{\frac {3}{2}} \left (a \,x^{3}+b \right )}\) | \(92\) |
default | \(\frac {\sqrt {\frac {a \,x^{3}+b}{x^{3}}}\, x^{2} \left (2 \sqrt {x \left (a \,x^{3}+b \right )}\, a^{\frac {3}{2}} x^{4}+b x \sqrt {x \left (a \,x^{3}+b \right )}\, \sqrt {a}-\operatorname {arctanh}\left (\frac {\sqrt {x \left (a \,x^{3}+b \right )}}{x^{2} \sqrt {a}}\right ) b^{2}\right )}{12 \sqrt {x \left (a \,x^{3}+b \right )}\, a^{\frac {3}{2}}}\) | \(94\) |
1/12*x^3*(2*a*x^3+b)/a*((a*x^3+b)/x^3)^(1/2)-1/12/a^(3/2)*arctanh((x*(a*x^ 3+b))^(1/2)/x^2/a^(1/2))*b^2*x*((a*x^3+b)/x^3)^(1/2)*(x*(a*x^3+b))^(1/2)/( a*x^3+b)
Time = 0.38 (sec) , antiderivative size = 175, normalized size of antiderivative = 2.46 \[ \int \sqrt {a+\frac {b}{x^3}} x^5 \, dx=\left [\frac {\sqrt {a} b^{2} \log \left (-8 \, a^{2} x^{6} - 8 \, a b x^{3} - b^{2} + 4 \, {\left (2 \, a x^{6} + b x^{3}\right )} \sqrt {a} \sqrt {\frac {a x^{3} + b}{x^{3}}}\right ) + 4 \, {\left (2 \, a^{2} x^{6} + a b x^{3}\right )} \sqrt {\frac {a x^{3} + b}{x^{3}}}}{48 \, a^{2}}, \frac {\sqrt {-a} b^{2} \arctan \left (\frac {2 \, \sqrt {-a} x^{3} \sqrt {\frac {a x^{3} + b}{x^{3}}}}{2 \, a x^{3} + b}\right ) + 2 \, {\left (2 \, a^{2} x^{6} + a b x^{3}\right )} \sqrt {\frac {a x^{3} + b}{x^{3}}}}{24 \, a^{2}}\right ] \]
[1/48*(sqrt(a)*b^2*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*(2*a^2*x^6 + a*b*x^3)*sqrt((a*x^3 + b)/ x^3))/a^2, 1/24*(sqrt(-a)*b^2*arctan(2*sqrt(-a)*x^3*sqrt((a*x^3 + b)/x^3)/ (2*a*x^3 + b)) + 2*(2*a^2*x^6 + a*b*x^3)*sqrt((a*x^3 + b)/x^3))/a^2]
Time = 2.08 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.41 \[ \int \sqrt {a+\frac {b}{x^3}} x^5 \, dx=\frac {a x^{\frac {15}{2}}}{6 \sqrt {b} \sqrt {\frac {a x^{3}}{b} + 1}} + \frac {\sqrt {b} x^{\frac {9}{2}}}{4 \sqrt {\frac {a x^{3}}{b} + 1}} + \frac {b^{\frac {3}{2}} x^{\frac {3}{2}}}{12 a \sqrt {\frac {a x^{3}}{b} + 1}} - \frac {b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a} x^{\frac {3}{2}}}{\sqrt {b}} \right )}}{12 a^{\frac {3}{2}}} \]
a*x**(15/2)/(6*sqrt(b)*sqrt(a*x**3/b + 1)) + sqrt(b)*x**(9/2)/(4*sqrt(a*x* *3/b + 1)) + b**(3/2)*x**(3/2)/(12*a*sqrt(a*x**3/b + 1)) - b**2*asinh(sqrt (a)*x**(3/2)/sqrt(b))/(12*a**(3/2))
Time = 0.28 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.41 \[ \int \sqrt {a+\frac {b}{x^3}} x^5 \, dx=\frac {b^{2} \log \left (\frac {\sqrt {a + \frac {b}{x^{3}}} - \sqrt {a}}{\sqrt {a + \frac {b}{x^{3}}} + \sqrt {a}}\right )}{24 \, a^{\frac {3}{2}}} + \frac {{\left (a + \frac {b}{x^{3}}\right )}^{\frac {3}{2}} b^{2} + \sqrt {a + \frac {b}{x^{3}}} a b^{2}}{12 \, {\left ({\left (a + \frac {b}{x^{3}}\right )}^{2} a - 2 \, {\left (a + \frac {b}{x^{3}}\right )} a^{2} + a^{3}\right )}} \]
1/24*b^2*log((sqrt(a + b/x^3) - sqrt(a))/(sqrt(a + b/x^3) + sqrt(a)))/a^(3 /2) + 1/12*((a + b/x^3)^(3/2)*b^2 + sqrt(a + b/x^3)*a*b^2)/((a + b/x^3)^2* a - 2*(a + b/x^3)*a^2 + a^3)
Time = 0.33 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.77 \[ \int \sqrt {a+\frac {b}{x^3}} x^5 \, dx=\frac {1}{12} \, \sqrt {a x^{4} + b x} {\left (2 \, x^{3} + \frac {b}{a}\right )} x + \frac {b^{2} \arctan \left (\frac {\sqrt {a + \frac {b}{x^{3}}}}{\sqrt {-a}}\right )}{12 \, \sqrt {-a} a} \]
1/12*sqrt(a*x^4 + b*x)*(2*x^3 + b/a)*x + 1/12*b^2*arctan(sqrt(a + b/x^3)/s qrt(-a))/(sqrt(-a)*a)
Time = 6.32 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.07 \[ \int \sqrt {a+\frac {b}{x^3}} x^5 \, dx=\frac {x^6\,\sqrt {a+\frac {b}{x^3}}}{6}+\frac {b^2\,\ln \left (x^6\,{\left (\sqrt {a+\frac {b}{x^3}}-\sqrt {a}\right )}^3\,\left (\sqrt {a+\frac {b}{x^3}}+\sqrt {a}\right )\right )}{24\,a^{3/2}}+\frac {b\,x^3\,\sqrt {a+\frac {b}{x^3}}}{12\,a} \]