Integrand size = 15, antiderivative size = 128 \[ \int \frac {\left (a+b x^2\right )^{9/2}}{x^9} \, dx=\frac {315}{128} b^4 \sqrt {a+b x^2}-\frac {105 b^3 \left (a+b x^2\right )^{3/2}}{128 x^2}-\frac {21 b^2 \left (a+b x^2\right )^{5/2}}{64 x^4}-\frac {3 b \left (a+b x^2\right )^{7/2}}{16 x^6}-\frac {\left (a+b x^2\right )^{9/2}}{8 x^8}-\frac {315}{128} \sqrt {a} b^4 \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \]
-105/128*b^3*(b*x^2+a)^(3/2)/x^2-21/64*b^2*(b*x^2+a)^(5/2)/x^4-3/16*b*(b*x ^2+a)^(7/2)/x^6-1/8*(b*x^2+a)^(9/2)/x^8-315/128*b^4*arctanh((b*x^2+a)^(1/2 )/a^(1/2))*a^(1/2)+315/128*b^4*(b*x^2+a)^(1/2)
Time = 0.13 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.72 \[ \int \frac {\left (a+b x^2\right )^{9/2}}{x^9} \, dx=\frac {\sqrt {a+b x^2} \left (-16 a^4-88 a^3 b x^2-210 a^2 b^2 x^4-325 a b^3 x^6+128 b^4 x^8\right )}{128 x^8}-\frac {315}{128} \sqrt {a} b^4 \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \]
(Sqrt[a + b*x^2]*(-16*a^4 - 88*a^3*b*x^2 - 210*a^2*b^2*x^4 - 325*a*b^3*x^6 + 128*b^4*x^8))/(128*x^8) - (315*Sqrt[a]*b^4*ArcTanh[Sqrt[a + b*x^2]/Sqrt [a]])/128
Time = 0.21 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.07, 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, 51, 51, 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.
| \(\displaystyle \int \frac {\left (a+b x^2\right )^{9/2}}{x^9} \, dx\) |
| \(\Big \downarrow \) 243 |
| \(\displaystyle \frac {1}{2} \int \frac {\left (b x^2+a\right )^{9/2}}{x^{10}}dx^2\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {1}{2} \left (\frac {9}{8} b \int \frac {\left (b x^2+a\right )^{7/2}}{x^8}dx^2-\frac {\left (a+b x^2\right )^{9/2}}{4 x^8}\right )\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {1}{2} \left (\frac {9}{8} b \left (\frac {7}{6} b \int \frac {\left (b x^2+a\right )^{5/2}}{x^6}dx^2-\frac {\left (a+b x^2\right )^{7/2}}{3 x^6}\right )-\frac {\left (a+b x^2\right )^{9/2}}{4 x^8}\right )\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {1}{2} \left (\frac {9}{8} b \left (\frac {7}{6} b \left (\frac {5}{4} b \int \frac {\left (b x^2+a\right )^{3/2}}{x^4}dx^2-\frac {\left (a+b x^2\right )^{5/2}}{2 x^4}\right )-\frac {\left (a+b x^2\right )^{7/2}}{3 x^6}\right )-\frac {\left (a+b x^2\right )^{9/2}}{4 x^8}\right )\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {1}{2} \left (\frac {9}{8} b \left (\frac {7}{6} b \left (\frac {5}{4} b \left (\frac {3}{2} b \int \frac {\sqrt {b x^2+a}}{x^2}dx^2-\frac {\left (a+b x^2\right )^{3/2}}{x^2}\right )-\frac {\left (a+b x^2\right )^{5/2}}{2 x^4}\right )-\frac {\left (a+b x^2\right )^{7/2}}{3 x^6}\right )-\frac {\left (a+b x^2\right )^{9/2}}{4 x^8}\right )\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {1}{2} \left (\frac {9}{8} b \left (\frac {7}{6} b \left (\frac {5}{4} b \left (\frac {3}{2} b \left (a \int \frac {1}{x^2 \sqrt {b x^2+a}}dx^2+2 \sqrt {a+b x^2}\right )-\frac {\left (a+b x^2\right )^{3/2}}{x^2}\right )-\frac {\left (a+b x^2\right )^{5/2}}{2 x^4}\right )-\frac {\left (a+b x^2\right )^{7/2}}{3 x^6}\right )-\frac {\left (a+b x^2\right )^{9/2}}{4 x^8}\right )\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {1}{2} \left (\frac {9}{8} b \left (\frac {7}{6} b \left (\frac {5}{4} b \left (\frac {3}{2} b \left (\frac {2 a \int \frac {1}{\frac {x^4}{b}-\frac {a}{b}}d\sqrt {b x^2+a}}{b}+2 \sqrt {a+b x^2}\right )-\frac {\left (a+b x^2\right )^{3/2}}{x^2}\right )-\frac {\left (a+b x^2\right )^{5/2}}{2 x^4}\right )-\frac {\left (a+b x^2\right )^{7/2}}{3 x^6}\right )-\frac {\left (a+b x^2\right )^{9/2}}{4 x^8}\right )\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {1}{2} \left (\frac {9}{8} b \left (\frac {7}{6} b \left (\frac {5}{4} b \left (\frac {3}{2} b \left (2 \sqrt {a+b x^2}-2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )\right )-\frac {\left (a+b x^2\right )^{3/2}}{x^2}\right )-\frac {\left (a+b x^2\right )^{5/2}}{2 x^4}\right )-\frac {\left (a+b x^2\right )^{7/2}}{3 x^6}\right )-\frac {\left (a+b x^2\right )^{9/2}}{4 x^8}\right )\) |
(-1/4*(a + b*x^2)^(9/2)/x^8 + (9*b*(-1/3*(a + b*x^2)^(7/2)/x^6 + (7*b*(-1/ 2*(a + b*x^2)^(5/2)/x^4 + (5*b*(-((a + b*x^2)^(3/2)/x^2) + (3*b*(2*Sqrt[a + b*x^2] - 2*Sqrt[a]*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]]))/2))/4))/6))/8)/2
3.5.21.3.1 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 = 1.96 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.73
| method | result | size |
| risch | \(-\frac {a \sqrt {b \,x^{2}+a}\, \left (325 b^{3} x^{6}+210 a \,b^{2} x^{4}+88 a^{2} b \,x^{2}+16 a^{3}\right )}{128 x^{8}}-\frac {315 \sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right ) b^{4}}{128}+b^{4} \sqrt {b \,x^{2}+a}\) | \(93\) |
| pseudoelliptic | \(\frac {-315 \,\operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{\sqrt {a}}\right ) a \,b^{4} x^{8}+128 b^{4} x^{8} \sqrt {b \,x^{2}+a}\, \sqrt {a}-325 a^{\frac {3}{2}} b^{3} x^{6} \sqrt {b \,x^{2}+a}-210 a^{\frac {5}{2}} b^{2} x^{4} \sqrt {b \,x^{2}+a}-88 a^{\frac {7}{2}} b \,x^{2} \sqrt {b \,x^{2}+a}-16 \sqrt {b \,x^{2}+a}\, a^{\frac {9}{2}}}{128 x^{8} \sqrt {a}}\) | \(125\) |
| default | \(-\frac {\left (b \,x^{2}+a \right )^{\frac {11}{2}}}{8 a \,x^{8}}+\frac {3 b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {11}{2}}}{6 a \,x^{6}}+\frac {5 b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {11}{2}}}{4 a \,x^{4}}+\frac {7 b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {11}{2}}}{2 a \,x^{2}}+\frac {9 b \left (\frac {\left (b \,x^{2}+a \right )^{\frac {9}{2}}}{9}+a \left (\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{7}+a \left (\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{5}+a \left (\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3}+a \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )\right )\right )\right )\right )}{2 a}\right )}{4 a}\right )}{6 a}\right )}{8 a}\) | \(191\) |
-1/128*a*(b*x^2+a)^(1/2)*(325*b^3*x^6+210*a*b^2*x^4+88*a^2*b*x^2+16*a^3)/x ^8-315/128*a^(1/2)*ln((2*a+2*a^(1/2)*(b*x^2+a)^(1/2))/x)*b^4+b^4*(b*x^2+a) ^(1/2)
Time = 0.27 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.48 \[ \int \frac {\left (a+b x^2\right )^{9/2}}{x^9} \, dx=\left [\frac {315 \, \sqrt {a} b^{4} x^{8} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (128 \, b^{4} x^{8} - 325 \, a b^{3} x^{6} - 210 \, a^{2} b^{2} x^{4} - 88 \, a^{3} b x^{2} - 16 \, a^{4}\right )} \sqrt {b x^{2} + a}}{256 \, x^{8}}, \frac {315 \, \sqrt {-a} b^{4} x^{8} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (128 \, b^{4} x^{8} - 325 \, a b^{3} x^{6} - 210 \, a^{2} b^{2} x^{4} - 88 \, a^{3} b x^{2} - 16 \, a^{4}\right )} \sqrt {b x^{2} + a}}{128 \, x^{8}}\right ] \]
[1/256*(315*sqrt(a)*b^4*x^8*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a) /x^2) + 2*(128*b^4*x^8 - 325*a*b^3*x^6 - 210*a^2*b^2*x^4 - 88*a^3*b*x^2 - 16*a^4)*sqrt(b*x^2 + a))/x^8, 1/128*(315*sqrt(-a)*b^4*x^8*arctan(sqrt(-a)/ sqrt(b*x^2 + a)) + (128*b^4*x^8 - 325*a*b^3*x^6 - 210*a^2*b^2*x^4 - 88*a^3 *b*x^2 - 16*a^4)*sqrt(b*x^2 + a))/x^8]
Time = 9.42 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.35 \[ \int \frac {\left (a+b x^2\right )^{9/2}}{x^9} \, dx=- \frac {315 \sqrt {a} b^{4} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{128} - \frac {a^{5}}{8 \sqrt {b} x^{9} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {13 a^{4} \sqrt {b}}{16 x^{7} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {149 a^{3} b^{\frac {3}{2}}}{64 x^{5} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {535 a^{2} b^{\frac {5}{2}}}{128 x^{3} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {197 a b^{\frac {7}{2}}}{128 x \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {b^{\frac {9}{2}} x}{\sqrt {\frac {a}{b x^{2}} + 1}} \]
-315*sqrt(a)*b**4*asinh(sqrt(a)/(sqrt(b)*x))/128 - a**5/(8*sqrt(b)*x**9*sq rt(a/(b*x**2) + 1)) - 13*a**4*sqrt(b)/(16*x**7*sqrt(a/(b*x**2) + 1)) - 149 *a**3*b**(3/2)/(64*x**5*sqrt(a/(b*x**2) + 1)) - 535*a**2*b**(5/2)/(128*x** 3*sqrt(a/(b*x**2) + 1)) - 197*a*b**(7/2)/(128*x*sqrt(a/(b*x**2) + 1)) + b* *(9/2)*x/sqrt(a/(b*x**2) + 1)
Time = 0.21 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.39 \[ \int \frac {\left (a+b x^2\right )^{9/2}}{x^9} \, dx=-\frac {315}{128} \, \sqrt {a} b^{4} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right ) + \frac {315}{128} \, \sqrt {b x^{2} + a} b^{4} + \frac {35 \, {\left (b x^{2} + a\right )}^{\frac {9}{2}} b^{4}}{128 \, a^{4}} + \frac {45 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{4}}{128 \, a^{3}} + \frac {63 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{4}}{128 \, a^{2}} + \frac {105 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{4}}{128 \, a} - \frac {35 \, {\left (b x^{2} + a\right )}^{\frac {11}{2}} b^{3}}{128 \, a^{4} x^{2}} - \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {11}{2}} b^{2}}{64 \, a^{3} x^{4}} - \frac {{\left (b x^{2} + a\right )}^{\frac {11}{2}} b}{16 \, a^{2} x^{6}} - \frac {{\left (b x^{2} + a\right )}^{\frac {11}{2}}}{8 \, a x^{8}} \]
-315/128*sqrt(a)*b^4*arcsinh(a/(sqrt(a*b)*abs(x))) + 315/128*sqrt(b*x^2 + a)*b^4 + 35/128*(b*x^2 + a)^(9/2)*b^4/a^4 + 45/128*(b*x^2 + a)^(7/2)*b^4/a ^3 + 63/128*(b*x^2 + a)^(5/2)*b^4/a^2 + 105/128*(b*x^2 + a)^(3/2)*b^4/a - 35/128*(b*x^2 + a)^(11/2)*b^3/(a^4*x^2) - 5/64*(b*x^2 + a)^(11/2)*b^2/(a^3 *x^4) - 1/16*(b*x^2 + a)^(11/2)*b/(a^2*x^6) - 1/8*(b*x^2 + a)^(11/2)/(a*x^ 8)
Time = 0.30 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.95 \[ \int \frac {\left (a+b x^2\right )^{9/2}}{x^9} \, dx=\frac {\frac {315 \, a b^{5} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + 128 \, \sqrt {b x^{2} + a} b^{5} - \frac {325 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} a b^{5} - 765 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a^{2} b^{5} + 643 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{3} b^{5} - 187 \, \sqrt {b x^{2} + a} a^{4} b^{5}}{b^{4} x^{8}}}{128 \, b} \]
1/128*(315*a*b^5*arctan(sqrt(b*x^2 + a)/sqrt(-a))/sqrt(-a) + 128*sqrt(b*x^ 2 + a)*b^5 - (325*(b*x^2 + a)^(7/2)*a*b^5 - 765*(b*x^2 + a)^(5/2)*a^2*b^5 + 643*(b*x^2 + a)^(3/2)*a^3*b^5 - 187*sqrt(b*x^2 + a)*a^4*b^5)/(b^4*x^8))/ b
Time = 5.78 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.82 \[ \int \frac {\left (a+b x^2\right )^{9/2}}{x^9} \, dx=b^4\,\sqrt {b\,x^2+a}-\frac {325\,a\,{\left (b\,x^2+a\right )}^{7/2}}{128\,x^8}+\frac {187\,a^4\,\sqrt {b\,x^2+a}}{128\,x^8}-\frac {643\,a^3\,{\left (b\,x^2+a\right )}^{3/2}}{128\,x^8}+\frac {765\,a^2\,{\left (b\,x^2+a\right )}^{5/2}}{128\,x^8}+\frac {\sqrt {a}\,b^4\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,315{}\mathrm {i}}{128} \]