Integrand size = 15, antiderivative size = 126 \[ \int \frac {\left (a+b x^2\right )^{9/2}}{x^5} \, dx=\frac {63}{8} a^2 b^2 \sqrt {a+b x^2}+\frac {21}{8} a b^2 \left (a+b x^2\right )^{3/2}+\frac {63}{40} b^2 \left (a+b x^2\right )^{5/2}-\frac {9 b \left (a+b x^2\right )^{7/2}}{8 x^2}-\frac {\left (a+b x^2\right )^{9/2}}{4 x^4}-\frac {63}{8} a^{5/2} b^2 \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \]
21/8*a*b^2*(b*x^2+a)^(3/2)+63/40*b^2*(b*x^2+a)^(5/2)-9/8*b*(b*x^2+a)^(7/2) /x^2-1/4*(b*x^2+a)^(9/2)/x^4-63/8*a^(5/2)*b^2*arctanh((b*x^2+a)^(1/2)/a^(1 /2))+63/8*a^2*b^2*(b*x^2+a)^(1/2)
Time = 0.12 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.73 \[ \int \frac {\left (a+b x^2\right )^{9/2}}{x^5} \, dx=\frac {\sqrt {a+b x^2} \left (-10 a^4-85 a^3 b x^2+288 a^2 b^2 x^4+56 a b^3 x^6+8 b^4 x^8\right )}{40 x^4}-\frac {63}{8} a^{5/2} b^2 \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \]
(Sqrt[a + b*x^2]*(-10*a^4 - 85*a^3*b*x^2 + 288*a^2*b^2*x^4 + 56*a*b^3*x^6 + 8*b^4*x^8))/(40*x^4) - (63*a^(5/2)*b^2*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]]) /8
Time = 0.21 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.99, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {243, 51, 51, 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.
| \(\displaystyle \int \frac {\left (a+b x^2\right )^{9/2}}{x^5} \, dx\) |
| \(\Big \downarrow \) 243 |
| \(\displaystyle \frac {1}{2} \int \frac {\left (b x^2+a\right )^{9/2}}{x^6}dx^2\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {1}{2} \left (\frac {9}{4} b \int \frac {\left (b x^2+a\right )^{7/2}}{x^4}dx^2-\frac {\left (a+b x^2\right )^{9/2}}{2 x^4}\right )\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {1}{2} \left (\frac {9}{4} b \left (\frac {7}{2} b \int \frac {\left (b x^2+a\right )^{5/2}}{x^2}dx^2-\frac {\left (a+b x^2\right )^{7/2}}{x^2}\right )-\frac {\left (a+b x^2\right )^{9/2}}{2 x^4}\right )\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {1}{2} \left (\frac {9}{4} b \left (\frac {7}{2} b \left (a \int \frac {\left (b x^2+a\right )^{3/2}}{x^2}dx^2+\frac {2}{5} \left (a+b x^2\right )^{5/2}\right )-\frac {\left (a+b x^2\right )^{7/2}}{x^2}\right )-\frac {\left (a+b x^2\right )^{9/2}}{2 x^4}\right )\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {1}{2} \left (\frac {9}{4} b \left (\frac {7}{2} b \left (a \left (a \int \frac {\sqrt {b x^2+a}}{x^2}dx^2+\frac {2}{3} \left (a+b x^2\right )^{3/2}\right )+\frac {2}{5} \left (a+b x^2\right )^{5/2}\right )-\frac {\left (a+b x^2\right )^{7/2}}{x^2}\right )-\frac {\left (a+b x^2\right )^{9/2}}{2 x^4}\right )\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {1}{2} \left (\frac {9}{4} b \left (\frac {7}{2} b \left (a \left (a \left (a \int \frac {1}{x^2 \sqrt {b x^2+a}}dx^2+2 \sqrt {a+b x^2}\right )+\frac {2}{3} \left (a+b x^2\right )^{3/2}\right )+\frac {2}{5} \left (a+b x^2\right )^{5/2}\right )-\frac {\left (a+b x^2\right )^{7/2}}{x^2}\right )-\frac {\left (a+b x^2\right )^{9/2}}{2 x^4}\right )\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {1}{2} \left (\frac {9}{4} b \left (\frac {7}{2} b \left (a \left (a \left (\frac {2 a \int \frac {1}{\frac {x^4}{b}-\frac {a}{b}}d\sqrt {b x^2+a}}{b}+2 \sqrt {a+b x^2}\right )+\frac {2}{3} \left (a+b x^2\right )^{3/2}\right )+\frac {2}{5} \left (a+b x^2\right )^{5/2}\right )-\frac {\left (a+b x^2\right )^{7/2}}{x^2}\right )-\frac {\left (a+b x^2\right )^{9/2}}{2 x^4}\right )\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {1}{2} \left (\frac {9}{4} b \left (\frac {7}{2} b \left (a \left (a \left (2 \sqrt {a+b x^2}-2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )\right )+\frac {2}{3} \left (a+b x^2\right )^{3/2}\right )+\frac {2}{5} \left (a+b x^2\right )^{5/2}\right )-\frac {\left (a+b x^2\right )^{7/2}}{x^2}\right )-\frac {\left (a+b x^2\right )^{9/2}}{2 x^4}\right )\) |
(-1/2*(a + b*x^2)^(9/2)/x^4 + (9*b*(-((a + b*x^2)^(7/2)/x^2) + (7*b*((2*(a + b*x^2)^(5/2))/5 + a*((2*(a + b*x^2)^(3/2))/3 + a*(2*Sqrt[a + b*x^2] - 2 *Sqrt[a]*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]]))))/2))/4)/2
3.5.19.3.1 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/(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_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Time = 1.94 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.73
| method | result | size |
| pseudoelliptic | \(-\frac {63 \left (\operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{\sqrt {a}}\right ) a^{3} b^{2} x^{4}-\frac {8 \left (\sqrt {a}\, b^{4} x^{8}+7 a^{\frac {3}{2}} b^{3} x^{6}+36 a^{\frac {5}{2}} b^{2} x^{4}-\frac {85 a^{\frac {7}{2}} b \,x^{2}}{8}-\frac {5 a^{\frac {9}{2}}}{4}\right ) \sqrt {b \,x^{2}+a}}{315}\right )}{8 \sqrt {a}\, x^{4}}\) | \(92\) |
| risch | \(-\frac {a^{3} \sqrt {b \,x^{2}+a}\, \left (17 b \,x^{2}+2 a \right )}{8 x^{4}}+\frac {b^{4} x^{4} \sqrt {b \,x^{2}+a}}{5}+\frac {7 b^{3} a \,x^{2} \sqrt {b \,x^{2}+a}}{5}+\frac {36 a^{2} b^{2} \sqrt {b \,x^{2}+a}}{5}-\frac {63 b^{2} a^{\frac {5}{2}} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{8}\) | \(112\) |
| default | \(-\frac {\left (b \,x^{2}+a \right )^{\frac {11}{2}}}{4 a \,x^{4}}+\frac {7 b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {11}{2}}}{2 a \,x^{2}}+\frac {9 b \left (\frac {\left (b \,x^{2}+a \right )^{\frac {9}{2}}}{9}+a \left (\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{7}+a \left (\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{5}+a \left (\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3}+a \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )\right )\right )\right )\right )}{2 a}\right )}{4 a}\) | \(143\) |
-63/8*(arctanh((b*x^2+a)^(1/2)/a^(1/2))*a^3*b^2*x^4-8/315*(a^(1/2)*b^4*x^8 +7*a^(3/2)*b^3*x^6+36*a^(5/2)*b^2*x^4-85/8*a^(7/2)*b*x^2-5/4*a^(9/2))*(b*x ^2+a)^(1/2))/a^(1/2)/x^4
Time = 0.28 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.52 \[ \int \frac {\left (a+b x^2\right )^{9/2}}{x^5} \, dx=\left [\frac {315 \, a^{\frac {5}{2}} b^{2} x^{4} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (8 \, b^{4} x^{8} + 56 \, a b^{3} x^{6} + 288 \, a^{2} b^{2} x^{4} - 85 \, a^{3} b x^{2} - 10 \, a^{4}\right )} \sqrt {b x^{2} + a}}{80 \, x^{4}}, \frac {315 \, \sqrt {-a} a^{2} b^{2} x^{4} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (8 \, b^{4} x^{8} + 56 \, a b^{3} x^{6} + 288 \, a^{2} b^{2} x^{4} - 85 \, a^{3} b x^{2} - 10 \, a^{4}\right )} \sqrt {b x^{2} + a}}{40 \, x^{4}}\right ] \]
[1/80*(315*a^(5/2)*b^2*x^4*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/ x^2) + 2*(8*b^4*x^8 + 56*a*b^3*x^6 + 288*a^2*b^2*x^4 - 85*a^3*b*x^2 - 10*a ^4)*sqrt(b*x^2 + a))/x^4, 1/40*(315*sqrt(-a)*a^2*b^2*x^4*arctan(sqrt(-a)/s qrt(b*x^2 + a)) + (8*b^4*x^8 + 56*a*b^3*x^6 + 288*a^2*b^2*x^4 - 85*a^3*b*x ^2 - 10*a^4)*sqrt(b*x^2 + a))/x^4]
Time = 9.62 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.39 \[ \int \frac {\left (a+b x^2\right )^{9/2}}{x^5} \, dx=- \frac {63 a^{\frac {5}{2}} b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{8} - \frac {a^{5}}{4 \sqrt {b} x^{5} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {19 a^{4} \sqrt {b}}{8 x^{3} \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {203 a^{3} b^{\frac {3}{2}}}{40 x \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {43 a^{2} b^{\frac {5}{2}} x}{5 \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {8 a b^{\frac {7}{2}} x^{3}}{5 \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {b^{\frac {9}{2}} x^{5}}{5 \sqrt {\frac {a}{b x^{2}} + 1}} \]
-63*a**(5/2)*b**2*asinh(sqrt(a)/(sqrt(b)*x))/8 - a**5/(4*sqrt(b)*x**5*sqrt (a/(b*x**2) + 1)) - 19*a**4*sqrt(b)/(8*x**3*sqrt(a/(b*x**2) + 1)) + 203*a* *3*b**(3/2)/(40*x*sqrt(a/(b*x**2) + 1)) + 43*a**2*b**(5/2)*x/(5*sqrt(a/(b* x**2) + 1)) + 8*a*b**(7/2)*x**3/(5*sqrt(a/(b*x**2) + 1)) + b**(9/2)*x**5/( 5*sqrt(a/(b*x**2) + 1))
Time = 0.28 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.08 \[ \int \frac {\left (a+b x^2\right )^{9/2}}{x^5} \, dx=-\frac {63}{8} \, a^{\frac {5}{2}} b^{2} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right ) + \frac {63}{40} \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{2} + \frac {7 \, {\left (b x^{2} + a\right )}^{\frac {9}{2}} b^{2}}{8 \, a^{2}} + \frac {9 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{2}}{8 \, a} + \frac {21}{8} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a b^{2} + \frac {63}{8} \, \sqrt {b x^{2} + a} a^{2} b^{2} - \frac {7 \, {\left (b x^{2} + a\right )}^{\frac {11}{2}} b}{8 \, a^{2} x^{2}} - \frac {{\left (b x^{2} + a\right )}^{\frac {11}{2}}}{4 \, a x^{4}} \]
-63/8*a^(5/2)*b^2*arcsinh(a/(sqrt(a*b)*abs(x))) + 63/40*(b*x^2 + a)^(5/2)* b^2 + 7/8*(b*x^2 + a)^(9/2)*b^2/a^2 + 9/8*(b*x^2 + a)^(7/2)*b^2/a + 21/8*( b*x^2 + a)^(3/2)*a*b^2 + 63/8*sqrt(b*x^2 + a)*a^2*b^2 - 7/8*(b*x^2 + a)^(1 1/2)*b/(a^2*x^2) - 1/4*(b*x^2 + a)^(11/2)/(a*x^4)
Time = 0.30 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.98 \[ \int \frac {\left (a+b x^2\right )^{9/2}}{x^5} \, dx=\frac {\frac {315 \, a^{3} b^{3} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + 8 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{3} + 40 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a b^{3} + 240 \, \sqrt {b x^{2} + a} a^{2} b^{3} - \frac {5 \, {\left (17 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{3} b^{3} - 15 \, \sqrt {b x^{2} + a} a^{4} b^{3}\right )}}{b^{2} x^{4}}}{40 \, b} \]
1/40*(315*a^3*b^3*arctan(sqrt(b*x^2 + a)/sqrt(-a))/sqrt(-a) + 8*(b*x^2 + a )^(5/2)*b^3 + 40*(b*x^2 + a)^(3/2)*a*b^3 + 240*sqrt(b*x^2 + a)*a^2*b^3 - 5 *(17*(b*x^2 + a)^(3/2)*a^3*b^3 - 15*sqrt(b*x^2 + a)*a^4*b^3)/(b^2*x^4))/b
Time = 5.13 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.05 \[ \int \frac {\left (a+b x^2\right )^{9/2}}{x^5} \, dx=\frac {\frac {15\,a^4\,b^2\,\sqrt {b\,x^2+a}}{8}-\frac {17\,a^3\,b^2\,{\left (b\,x^2+a\right )}^{3/2}}{8}}{{\left (b\,x^2+a\right )}^2-2\,a\,\left (b\,x^2+a\right )+a^2}+\frac {b^2\,{\left (b\,x^2+a\right )}^{5/2}}{5}+a\,b^2\,{\left (b\,x^2+a\right )}^{3/2}+6\,a^2\,b^2\,\sqrt {b\,x^2+a}+\frac {a^{5/2}\,b^2\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,63{}\mathrm {i}}{8} \]