Integrand size = 15, antiderivative size = 85 \[ \int \frac {\left (a+b x^n\right )^{5/2}}{x} \, dx=\frac {2 a^2 \sqrt {a+b x^n}}{n}+\frac {2 a \left (a+b x^n\right )^{3/2}}{3 n}+\frac {2 \left (a+b x^n\right )^{5/2}}{5 n}-\frac {2 a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+b x^n}}{\sqrt {a}}\right )}{n} \]
2/3*a*(a+b*x^n)^(3/2)/n+2/5*(a+b*x^n)^(5/2)/n-2*a^(5/2)*arctanh((a+b*x^n)^ (1/2)/a^(1/2))/n+2*a^2*(a+b*x^n)^(1/2)/n
Time = 0.10 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.81 \[ \int \frac {\left (a+b x^n\right )^{5/2}}{x} \, dx=\frac {2 \sqrt {a+b x^n} \left (23 a^2+11 a b x^n+3 b^2 x^{2 n}\right )-30 a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+b x^n}}{\sqrt {a}}\right )}{15 n} \]
(2*Sqrt[a + b*x^n]*(23*a^2 + 11*a*b*x^n + 3*b^2*x^(2*n)) - 30*a^(5/2)*ArcT anh[Sqrt[a + b*x^n]/Sqrt[a]])/(15*n)
Time = 0.19 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.93, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {798, 60, 60, 60, 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.
\(\displaystyle \int \frac {\left (a+b x^n\right )^{5/2}}{x} \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {\int x^{-n} \left (b x^n+a\right )^{5/2}dx^n}{n}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {a \int x^{-n} \left (b x^n+a\right )^{3/2}dx^n+\frac {2}{5} \left (a+b x^n\right )^{5/2}}{n}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {a \left (a \int x^{-n} \sqrt {b x^n+a}dx^n+\frac {2}{3} \left (a+b x^n\right )^{3/2}\right )+\frac {2}{5} \left (a+b x^n\right )^{5/2}}{n}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {a \left (a \left (a \int \frac {x^{-n}}{\sqrt {b x^n+a}}dx^n+2 \sqrt {a+b x^n}\right )+\frac {2}{3} \left (a+b x^n\right )^{3/2}\right )+\frac {2}{5} \left (a+b x^n\right )^{5/2}}{n}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {a \left (a \left (\frac {2 a \int \frac {1}{\frac {x^{2 n}}{b}-\frac {a}{b}}d\sqrt {b x^n+a}}{b}+2 \sqrt {a+b x^n}\right )+\frac {2}{3} \left (a+b x^n\right )^{3/2}\right )+\frac {2}{5} \left (a+b x^n\right )^{5/2}}{n}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {a \left (a \left (2 \sqrt {a+b x^n}-2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b x^n}}{\sqrt {a}}\right )\right )+\frac {2}{3} \left (a+b x^n\right )^{3/2}\right )+\frac {2}{5} \left (a+b x^n\right )^{5/2}}{n}\) |
((2*(a + b*x^n)^(5/2))/5 + a*((2*(a + b*x^n)^(3/2))/3 + a*(2*Sqrt[a + b*x^ n] - 2*Sqrt[a]*ArcTanh[Sqrt[a + b*x^n]/Sqrt[a]])))/n
3.25.99.3.1 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 + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[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 = 3.74 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.73
method | result | size |
derivativedivides | \(\frac {\frac {2 \left (a +b \,x^{n}\right )^{\frac {5}{2}}}{5}+\frac {2 a \left (a +b \,x^{n}\right )^{\frac {3}{2}}}{3}+2 a^{2} \sqrt {a +b \,x^{n}}-2 a^{\frac {5}{2}} \operatorname {arctanh}\left (\frac {\sqrt {a +b \,x^{n}}}{\sqrt {a}}\right )}{n}\) | \(62\) |
default | \(\frac {\frac {2 \left (a +b \,x^{n}\right )^{\frac {5}{2}}}{5}+\frac {2 a \left (a +b \,x^{n}\right )^{\frac {3}{2}}}{3}+2 a^{2} \sqrt {a +b \,x^{n}}-2 a^{\frac {5}{2}} \operatorname {arctanh}\left (\frac {\sqrt {a +b \,x^{n}}}{\sqrt {a}}\right )}{n}\) | \(62\) |
risch | \(\frac {2 \left (3 b^{2} {\mathrm e}^{2 n \ln \left (x \right )}+11 a \,{\mathrm e}^{n \ln \left (x \right )} b +23 a^{2}\right ) \sqrt {a +b \,{\mathrm e}^{n \ln \left (x \right )}}}{15 n}-\frac {2 a^{\frac {5}{2}} \operatorname {arctanh}\left (\frac {\sqrt {a +b \,{\mathrm e}^{n \ln \left (x \right )}}}{\sqrt {a}}\right )}{n}\) | \(69\) |
1/n*(2/5*(a+b*x^n)^(5/2)+2/3*a*(a+b*x^n)^(3/2)+2*a^2*(a+b*x^n)^(1/2)-2*a^( 5/2)*arctanh((a+b*x^n)^(1/2)/a^(1/2)))
Time = 0.29 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.69 \[ \int \frac {\left (a+b x^n\right )^{5/2}}{x} \, dx=\left [\frac {15 \, a^{\frac {5}{2}} \log \left (\frac {b x^{n} - 2 \, \sqrt {b x^{n} + a} \sqrt {a} + 2 \, a}{x^{n}}\right ) + 2 \, {\left (3 \, b^{2} x^{2 \, n} + 11 \, a b x^{n} + 23 \, a^{2}\right )} \sqrt {b x^{n} + a}}{15 \, n}, \frac {2 \, {\left (15 \, \sqrt {-a} a^{2} \arctan \left (\frac {\sqrt {b x^{n} + a} \sqrt {-a}}{a}\right ) + {\left (3 \, b^{2} x^{2 \, n} + 11 \, a b x^{n} + 23 \, a^{2}\right )} \sqrt {b x^{n} + a}\right )}}{15 \, n}\right ] \]
[1/15*(15*a^(5/2)*log((b*x^n - 2*sqrt(b*x^n + a)*sqrt(a) + 2*a)/x^n) + 2*( 3*b^2*x^(2*n) + 11*a*b*x^n + 23*a^2)*sqrt(b*x^n + a))/n, 2/15*(15*sqrt(-a) *a^2*arctan(sqrt(b*x^n + a)*sqrt(-a)/a) + (3*b^2*x^(2*n) + 11*a*b*x^n + 23 *a^2)*sqrt(b*x^n + a))/n]
Time = 3.80 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.38 \[ \int \frac {\left (a+b x^n\right )^{5/2}}{x} \, dx=\frac {46 a^{\frac {5}{2}} \sqrt {1 + \frac {b x^{n}}{a}}}{15 n} + \frac {a^{\frac {5}{2}} \log {\left (\frac {b x^{n}}{a} \right )}}{n} - \frac {2 a^{\frac {5}{2}} \log {\left (\sqrt {1 + \frac {b x^{n}}{a}} + 1 \right )}}{n} + \frac {22 a^{\frac {3}{2}} b x^{n} \sqrt {1 + \frac {b x^{n}}{a}}}{15 n} + \frac {2 \sqrt {a} b^{2} x^{2 n} \sqrt {1 + \frac {b x^{n}}{a}}}{5 n} \]
46*a**(5/2)*sqrt(1 + b*x**n/a)/(15*n) + a**(5/2)*log(b*x**n/a)/n - 2*a**(5 /2)*log(sqrt(1 + b*x**n/a) + 1)/n + 22*a**(3/2)*b*x**n*sqrt(1 + b*x**n/a)/ (15*n) + 2*sqrt(a)*b**2*x**(2*n)*sqrt(1 + b*x**n/a)/(5*n)
Time = 0.30 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.98 \[ \int \frac {\left (a+b x^n\right )^{5/2}}{x} \, dx=\frac {a^{\frac {5}{2}} \log \left (\frac {\sqrt {b x^{n} + a} - \sqrt {a}}{\sqrt {b x^{n} + a} + \sqrt {a}}\right )}{n} + \frac {2 \, {\left (3 \, {\left (b x^{n} + a\right )}^{\frac {5}{2}} + 5 \, {\left (b x^{n} + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b x^{n} + a} a^{2}\right )}}{15 \, n} \]
a^(5/2)*log((sqrt(b*x^n + a) - sqrt(a))/(sqrt(b*x^n + a) + sqrt(a)))/n + 2 /15*(3*(b*x^n + a)^(5/2) + 5*(b*x^n + a)^(3/2)*a + 15*sqrt(b*x^n + a)*a^2) /n
\[ \int \frac {\left (a+b x^n\right )^{5/2}}{x} \, dx=\int { \frac {{\left (b x^{n} + a\right )}^{\frac {5}{2}}}{x} \,d x } \]
Timed out. \[ \int \frac {\left (a+b x^n\right )^{5/2}}{x} \, dx=\int \frac {{\left (a+b\,x^n\right )}^{5/2}}{x} \,d x \]