Integrand size = 15, antiderivative size = 118 \[ \int \frac {\left (a+b x^2\right )^{9/2}}{x^3} \, dx=\frac {9}{2} a^3 b \sqrt {a+b x^2}+\frac {3}{2} a^2 b \left (a+b x^2\right )^{3/2}+\frac {9}{10} a b \left (a+b x^2\right )^{5/2}+\frac {9}{14} b \left (a+b x^2\right )^{7/2}-\frac {\left (a+b x^2\right )^{9/2}}{2 x^2}-\frac {9}{2} a^{7/2} b \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \]
3/2*a^2*b*(b*x^2+a)^(3/2)+9/10*a*b*(b*x^2+a)^(5/2)+9/14*b*(b*x^2+a)^(7/2)- 1/2*(b*x^2+a)^(9/2)/x^2-9/2*a^(7/2)*b*arctanh((b*x^2+a)^(1/2)/a^(1/2))+9/2 *a^3*b*(b*x^2+a)^(1/2)
Time = 0.08 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.76 \[ \int \frac {\left (a+b x^2\right )^{9/2}}{x^3} \, dx=\frac {\sqrt {a+b x^2} \left (-35 a^4+388 a^3 b x^2+156 a^2 b^2 x^4+58 a b^3 x^6+10 b^4 x^8\right )}{70 x^2}-\frac {9}{2} a^{7/2} b \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \]
(Sqrt[a + b*x^2]*(-35*a^4 + 388*a^3*b*x^2 + 156*a^2*b^2*x^4 + 58*a*b^3*x^6 + 10*b^4*x^8))/(70*x^2) - (9*a^(7/2)*b*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/ 2
Time = 0.21 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.01, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {243, 51, 60, 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.
\(\displaystyle \int \frac {\left (a+b x^2\right )^{9/2}}{x^3} \, dx\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \int \frac {\left (b x^2+a\right )^{9/2}}{x^4}dx^2\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {1}{2} \left (\frac {9}{2} b \int \frac {\left (b x^2+a\right )^{7/2}}{x^2}dx^2-\frac {\left (a+b x^2\right )^{9/2}}{x^2}\right )\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{2} \left (\frac {9}{2} b \left (a \int \frac {\left (b x^2+a\right )^{5/2}}{x^2}dx^2+\frac {2}{7} \left (a+b x^2\right )^{7/2}\right )-\frac {\left (a+b x^2\right )^{9/2}}{x^2}\right )\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{2} \left (\frac {9}{2} b \left (a \left (a \int \frac {\left (b x^2+a\right )^{3/2}}{x^2}dx^2+\frac {2}{5} \left (a+b x^2\right )^{5/2}\right )+\frac {2}{7} \left (a+b x^2\right )^{7/2}\right )-\frac {\left (a+b x^2\right )^{9/2}}{x^2}\right )\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{2} \left (\frac {9}{2} b \left (a \left (a \left (a \int \frac {\sqrt {b x^2+a}}{x^2}dx^2+\frac {2}{3} \left (a+b x^2\right )^{3/2}\right )+\frac {2}{5} \left (a+b x^2\right )^{5/2}\right )+\frac {2}{7} \left (a+b x^2\right )^{7/2}\right )-\frac {\left (a+b x^2\right )^{9/2}}{x^2}\right )\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{2} \left (\frac {9}{2} b \left (a \left (a \left (a \left (a \int \frac {1}{x^2 \sqrt {b x^2+a}}dx^2+2 \sqrt {a+b x^2}\right )+\frac {2}{3} \left (a+b x^2\right )^{3/2}\right )+\frac {2}{5} \left (a+b x^2\right )^{5/2}\right )+\frac {2}{7} \left (a+b x^2\right )^{7/2}\right )-\frac {\left (a+b x^2\right )^{9/2}}{x^2}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} \left (\frac {9}{2} b \left (a \left (a \left (a \left (\frac {2 a \int \frac {1}{\frac {x^4}{b}-\frac {a}{b}}d\sqrt {b x^2+a}}{b}+2 \sqrt {a+b x^2}\right )+\frac {2}{3} \left (a+b x^2\right )^{3/2}\right )+\frac {2}{5} \left (a+b x^2\right )^{5/2}\right )+\frac {2}{7} \left (a+b x^2\right )^{7/2}\right )-\frac {\left (a+b x^2\right )^{9/2}}{x^2}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{2} \left (\frac {9}{2} b \left (a \left (a \left (a \left (2 \sqrt {a+b x^2}-2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )\right )+\frac {2}{3} \left (a+b x^2\right )^{3/2}\right )+\frac {2}{5} \left (a+b x^2\right )^{5/2}\right )+\frac {2}{7} \left (a+b x^2\right )^{7/2}\right )-\frac {\left (a+b x^2\right )^{9/2}}{x^2}\right )\) |
(-((a + b*x^2)^(9/2)/x^2) + (9*b*((2*(a + b*x^2)^(7/2))/7 + a*((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)/2
3.5.18.3.1 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 = 1.93 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.76
method | result | size |
pseudoelliptic | \(-\frac {9 \left (\operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{\sqrt {a}}\right ) a^{4} b \,x^{2}-\frac {2 \sqrt {b \,x^{2}+a}\, \left (\sqrt {a}\, b^{4} x^{8}+\frac {29 a^{\frac {3}{2}} b^{3} x^{6}}{5}+\frac {78 a^{\frac {5}{2}} b^{2} x^{4}}{5}+\frac {194 a^{\frac {7}{2}} b \,x^{2}}{5}-\frac {7 a^{\frac {9}{2}}}{2}\right )}{63}\right )}{2 \sqrt {a}\, x^{2}}\) | \(90\) |
risch | \(-\frac {a^{4} \sqrt {b \,x^{2}+a}}{2 x^{2}}+\frac {b^{4} x^{6} \sqrt {b \,x^{2}+a}}{7}+\frac {29 b^{3} a \,x^{4} \sqrt {b \,x^{2}+a}}{35}+\frac {78 b^{2} a^{2} x^{2} \sqrt {b \,x^{2}+a}}{35}+\frac {194 a^{3} b \sqrt {b \,x^{2}+a}}{35}-\frac {9 b \,a^{\frac {7}{2}} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{2}\) | \(118\) |
default | \(-\frac {\left (b \,x^{2}+a \right )^{\frac {11}{2}}}{2 a \,x^{2}}+\frac {9 b \left (\frac {\left (b \,x^{2}+a \right )^{\frac {9}{2}}}{9}+a \left (\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{7}+a \left (\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{5}+a \left (\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3}+a \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )\right )\right )\right )\right )}{2 a}\) | \(119\) |
-9/2*(arctanh((b*x^2+a)^(1/2)/a^(1/2))*a^4*b*x^2-2/63*(b*x^2+a)^(1/2)*(a^( 1/2)*b^4*x^8+29/5*a^(3/2)*b^3*x^6+78/5*a^(5/2)*b^2*x^4+194/5*a^(7/2)*b*x^2 -7/2*a^(9/2)))/a^(1/2)/x^2
Time = 0.26 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.59 \[ \int \frac {\left (a+b x^2\right )^{9/2}}{x^3} \, dx=\left [\frac {315 \, a^{\frac {7}{2}} b x^{2} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (10 \, b^{4} x^{8} + 58 \, a b^{3} x^{6} + 156 \, a^{2} b^{2} x^{4} + 388 \, a^{3} b x^{2} - 35 \, a^{4}\right )} \sqrt {b x^{2} + a}}{140 \, x^{2}}, \frac {315 \, \sqrt {-a} a^{3} b x^{2} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (10 \, b^{4} x^{8} + 58 \, a b^{3} x^{6} + 156 \, a^{2} b^{2} x^{4} + 388 \, a^{3} b x^{2} - 35 \, a^{4}\right )} \sqrt {b x^{2} + a}}{70 \, x^{2}}\right ] \]
[1/140*(315*a^(7/2)*b*x^2*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x ^2) + 2*(10*b^4*x^8 + 58*a*b^3*x^6 + 156*a^2*b^2*x^4 + 388*a^3*b*x^2 - 35* a^4)*sqrt(b*x^2 + a))/x^2, 1/70*(315*sqrt(-a)*a^3*b*x^2*arctan(sqrt(-a)/sq rt(b*x^2 + a)) + (10*b^4*x^8 + 58*a*b^3*x^6 + 156*a^2*b^2*x^4 + 388*a^3*b* x^2 - 35*a^4)*sqrt(b*x^2 + a))/x^2]
Time = 10.11 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.42 \[ \int \frac {\left (a+b x^2\right )^{9/2}}{x^3} \, dx=- \frac {a^{\frac {9}{2}} \sqrt {1 + \frac {b x^{2}}{a}}}{2 x^{2}} + \frac {194 a^{\frac {7}{2}} b \sqrt {1 + \frac {b x^{2}}{a}}}{35} + \frac {9 a^{\frac {7}{2}} b \log {\left (\frac {b x^{2}}{a} \right )}}{4} - \frac {9 a^{\frac {7}{2}} b \log {\left (\sqrt {1 + \frac {b x^{2}}{a}} + 1 \right )}}{2} + \frac {78 a^{\frac {5}{2}} b^{2} x^{2} \sqrt {1 + \frac {b x^{2}}{a}}}{35} + \frac {29 a^{\frac {3}{2}} b^{3} x^{4} \sqrt {1 + \frac {b x^{2}}{a}}}{35} + \frac {\sqrt {a} b^{4} x^{6} \sqrt {1 + \frac {b x^{2}}{a}}}{7} \]
-a**(9/2)*sqrt(1 + b*x**2/a)/(2*x**2) + 194*a**(7/2)*b*sqrt(1 + b*x**2/a)/ 35 + 9*a**(7/2)*b*log(b*x**2/a)/4 - 9*a**(7/2)*b*log(sqrt(1 + b*x**2/a) + 1)/2 + 78*a**(5/2)*b**2*x**2*sqrt(1 + b*x**2/a)/35 + 29*a**(3/2)*b**3*x**4 *sqrt(1 + b*x**2/a)/35 + sqrt(a)*b**4*x**6*sqrt(1 + b*x**2/a)/7
Time = 0.29 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.90 \[ \int \frac {\left (a+b x^2\right )^{9/2}}{x^3} \, dx=-\frac {9}{2} \, a^{\frac {7}{2}} b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right ) + \frac {9}{14} \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b + \frac {{\left (b x^{2} + a\right )}^{\frac {9}{2}} b}{2 \, a} + \frac {9}{10} \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a b + \frac {3}{2} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2} b + \frac {9}{2} \, \sqrt {b x^{2} + a} a^{3} b - \frac {{\left (b x^{2} + a\right )}^{\frac {11}{2}}}{2 \, a x^{2}} \]
-9/2*a^(7/2)*b*arcsinh(a/(sqrt(a*b)*abs(x))) + 9/14*(b*x^2 + a)^(7/2)*b + 1/2*(b*x^2 + a)^(9/2)*b/a + 9/10*(b*x^2 + a)^(5/2)*a*b + 3/2*(b*x^2 + a)^( 3/2)*a^2*b + 9/2*sqrt(b*x^2 + a)*a^3*b - 1/2*(b*x^2 + a)^(11/2)/(a*x^2)
Time = 0.30 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.98 \[ \int \frac {\left (a+b x^2\right )^{9/2}}{x^3} \, dx=\frac {\frac {315 \, a^{4} b^{2} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + 10 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{2} + 28 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a b^{2} + 70 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2} b^{2} + 280 \, \sqrt {b x^{2} + a} a^{3} b^{2} - \frac {35 \, \sqrt {b x^{2} + a} a^{4} b}{x^{2}}}{70 \, b} \]
1/70*(315*a^4*b^2*arctan(sqrt(b*x^2 + a)/sqrt(-a))/sqrt(-a) + 10*(b*x^2 + a)^(7/2)*b^2 + 28*(b*x^2 + a)^(5/2)*a*b^2 + 70*(b*x^2 + a)^(3/2)*a^2*b^2 + 280*sqrt(b*x^2 + a)*a^3*b^2 - 35*sqrt(b*x^2 + a)*a^4*b/x^2)/b
Time = 4.90 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.81 \[ \int \frac {\left (a+b x^2\right )^{9/2}}{x^3} \, dx=\frac {b\,{\left (b\,x^2+a\right )}^{7/2}}{7}+4\,a^3\,b\,\sqrt {b\,x^2+a}+a^2\,b\,{\left (b\,x^2+a\right )}^{3/2}-\frac {a^4\,\sqrt {b\,x^2+a}}{2\,x^2}+\frac {2\,a\,b\,{\left (b\,x^2+a\right )}^{5/2}}{5}+\frac {a^{7/2}\,b\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,9{}\mathrm {i}}{2} \]