Integrand size = 15, antiderivative size = 123 \[ \int \frac {\left (a+b x^2\right )^{9/2}}{x^2} \, dx=\frac {315}{128} a^3 b x \sqrt {a+b x^2}+\frac {105}{64} a^2 b x \left (a+b x^2\right )^{3/2}+\frac {21}{16} a b x \left (a+b x^2\right )^{5/2}+\frac {9}{8} b x \left (a+b x^2\right )^{7/2}-\frac {\left (a+b x^2\right )^{9/2}}{x}+\frac {315}{128} a^4 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \]
105/64*a^2*b*x*(b*x^2+a)^(3/2)+21/16*a*b*x*(b*x^2+a)^(5/2)+9/8*b*x*(b*x^2+ a)^(7/2)-(b*x^2+a)^(9/2)/x+315/128*a^4*arctanh(x*b^(1/2)/(b*x^2+a)^(1/2))* b^(1/2)+315/128*a^3*b*x*(b*x^2+a)^(1/2)
Time = 0.12 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.77 \[ \int \frac {\left (a+b x^2\right )^{9/2}}{x^2} \, dx=\frac {\sqrt {a+b x^2} \left (-128 a^4+325 a^3 b x^2+210 a^2 b^2 x^4+88 a b^3 x^6+16 b^4 x^8\right )}{128 x}-\frac {315}{128} a^4 \sqrt {b} \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right ) \]
(Sqrt[a + b*x^2]*(-128*a^4 + 325*a^3*b*x^2 + 210*a^2*b^2*x^4 + 88*a*b^3*x^ 6 + 16*b^4*x^8))/(128*x) - (315*a^4*Sqrt[b]*Log[-(Sqrt[b]*x) + Sqrt[a + b* x^2]])/128
Time = 0.20 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.07, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {247, 211, 211, 211, 211, 224, 219}
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.
\(\displaystyle \int \frac {\left (a+b x^2\right )^{9/2}}{x^2} \, dx\) |
\(\Big \downarrow \) 247 |
\(\displaystyle 9 b \int \left (b x^2+a\right )^{7/2}dx-\frac {\left (a+b x^2\right )^{9/2}}{x}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle 9 b \left (\frac {7}{8} a \int \left (b x^2+a\right )^{5/2}dx+\frac {1}{8} x \left (a+b x^2\right )^{7/2}\right )-\frac {\left (a+b x^2\right )^{9/2}}{x}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle 9 b \left (\frac {7}{8} a \left (\frac {5}{6} a \int \left (b x^2+a\right )^{3/2}dx+\frac {1}{6} x \left (a+b x^2\right )^{5/2}\right )+\frac {1}{8} x \left (a+b x^2\right )^{7/2}\right )-\frac {\left (a+b x^2\right )^{9/2}}{x}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle 9 b \left (\frac {7}{8} a \left (\frac {5}{6} a \left (\frac {3}{4} a \int \sqrt {b x^2+a}dx+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )+\frac {1}{6} x \left (a+b x^2\right )^{5/2}\right )+\frac {1}{8} x \left (a+b x^2\right )^{7/2}\right )-\frac {\left (a+b x^2\right )^{9/2}}{x}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle 9 b \left (\frac {7}{8} a \left (\frac {5}{6} a \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{\sqrt {b x^2+a}}dx+\frac {1}{2} x \sqrt {a+b x^2}\right )+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )+\frac {1}{6} x \left (a+b x^2\right )^{5/2}\right )+\frac {1}{8} x \left (a+b x^2\right )^{7/2}\right )-\frac {\left (a+b x^2\right )^{9/2}}{x}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle 9 b \left (\frac {7}{8} a \left (\frac {5}{6} a \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}+\frac {1}{2} x \sqrt {a+b x^2}\right )+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )+\frac {1}{6} x \left (a+b x^2\right )^{5/2}\right )+\frac {1}{8} x \left (a+b x^2\right )^{7/2}\right )-\frac {\left (a+b x^2\right )^{9/2}}{x}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle 9 b \left (\frac {7}{8} a \left (\frac {5}{6} a \left (\frac {3}{4} a \left (\frac {a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 \sqrt {b}}+\frac {1}{2} x \sqrt {a+b x^2}\right )+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )+\frac {1}{6} x \left (a+b x^2\right )^{5/2}\right )+\frac {1}{8} x \left (a+b x^2\right )^{7/2}\right )-\frac {\left (a+b x^2\right )^{9/2}}{x}\) |
-((a + b*x^2)^(9/2)/x) + 9*b*((x*(a + b*x^2)^(7/2))/8 + (7*a*((x*(a + b*x^ 2)^(5/2))/6 + (5*a*((x*(a + b*x^2)^(3/2))/4 + (3*a*((x*Sqrt[a + b*x^2])/2 + (a*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2*Sqrt[b])))/4))/6))/8)
3.5.29.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^p/(c*(m + 1))), x] - Simp[2*b*(p/(c^2*(m + 1))) Int[ (c*x)^(m + 2)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && LtQ[m, -1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomialQ[a, b, c, 2, m, p, x]
Time = 1.95 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.67
method | result | size |
risch | \(-\frac {\sqrt {b \,x^{2}+a}\, \left (-16 x^{8} b^{4}-88 a \,b^{3} x^{6}-210 a^{2} x^{4} b^{2}-325 a^{3} b \,x^{2}+128 a^{4}\right )}{128 x}+\frac {315 a^{4} \sqrt {b}\, \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{128}\) | \(83\) |
pseudoelliptic | \(-\frac {-\frac {315 \,\operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right ) a^{4} b x}{128}+\sqrt {b \,x^{2}+a}\, \left (-\frac {b^{\frac {9}{2}} x^{8}}{8}-\frac {11 a \,b^{\frac {7}{2}} x^{6}}{16}-\frac {105 a^{2} b^{\frac {5}{2}} x^{4}}{64}-\frac {325 a^{3} b^{\frac {3}{2}} x^{2}}{128}+a^{4} \sqrt {b}\right )}{\sqrt {b}\, x}\) | \(91\) |
default | \(-\frac {\left (b \,x^{2}+a \right )^{\frac {11}{2}}}{a x}+\frac {10 b \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {9}{2}}}{10}+\frac {9 a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {7}{2}}}{8}+\frac {7 a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6}+\frac {5 a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6}\right )}{8}\right )}{10}\right )}{a}\) | \(124\) |
-1/128*(b*x^2+a)^(1/2)*(-16*b^4*x^8-88*a*b^3*x^6-210*a^2*b^2*x^4-325*a^3*b *x^2+128*a^4)/x+315/128*a^4*b^(1/2)*ln(x*b^(1/2)+(b*x^2+a)^(1/2))
Time = 0.30 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.50 \[ \int \frac {\left (a+b x^2\right )^{9/2}}{x^2} \, dx=\left [\frac {315 \, a^{4} \sqrt {b} x \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (16 \, b^{4} x^{8} + 88 \, a b^{3} x^{6} + 210 \, a^{2} b^{2} x^{4} + 325 \, a^{3} b x^{2} - 128 \, a^{4}\right )} \sqrt {b x^{2} + a}}{256 \, x}, -\frac {315 \, a^{4} \sqrt {-b} x \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (16 \, b^{4} x^{8} + 88 \, a b^{3} x^{6} + 210 \, a^{2} b^{2} x^{4} + 325 \, a^{3} b x^{2} - 128 \, a^{4}\right )} \sqrt {b x^{2} + a}}{128 \, x}\right ] \]
[1/256*(315*a^4*sqrt(b)*x*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(16*b^4*x^8 + 88*a*b^3*x^6 + 210*a^2*b^2*x^4 + 325*a^3*b*x^2 - 128*a^4 )*sqrt(b*x^2 + a))/x, -1/128*(315*a^4*sqrt(-b)*x*arctan(sqrt(-b)*x/sqrt(b* x^2 + a)) - (16*b^4*x^8 + 88*a*b^3*x^6 + 210*a^2*b^2*x^4 + 325*a^3*b*x^2 - 128*a^4)*sqrt(b*x^2 + a))/x]
Time = 9.72 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.41 \[ \int \frac {\left (a+b x^2\right )^{9/2}}{x^2} \, dx=- \frac {a^{\frac {9}{2}}}{x \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {197 a^{\frac {7}{2}} b x}{128 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {535 a^{\frac {5}{2}} b^{2} x^{3}}{128 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {149 a^{\frac {3}{2}} b^{3} x^{5}}{64 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {13 \sqrt {a} b^{4} x^{7}}{16 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {315 a^{4} \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{128} + \frac {b^{5} x^{9}}{8 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} \]
-a**(9/2)/(x*sqrt(1 + b*x**2/a)) + 197*a**(7/2)*b*x/(128*sqrt(1 + b*x**2/a )) + 535*a**(5/2)*b**2*x**3/(128*sqrt(1 + b*x**2/a)) + 149*a**(3/2)*b**3*x **5/(64*sqrt(1 + b*x**2/a)) + 13*sqrt(a)*b**4*x**7/(16*sqrt(1 + b*x**2/a)) + 315*a**4*sqrt(b)*asinh(sqrt(b)*x/sqrt(a))/128 + b**5*x**9/(8*sqrt(a)*sq rt(1 + b*x**2/a))
Time = 0.22 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.74 \[ \int \frac {\left (a+b x^2\right )^{9/2}}{x^2} \, dx=\frac {9}{8} \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b x + \frac {21}{16} \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a b x + \frac {105}{64} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2} b x + \frac {315}{128} \, \sqrt {b x^{2} + a} a^{3} b x + \frac {315}{128} \, a^{4} \sqrt {b} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right ) - \frac {{\left (b x^{2} + a\right )}^{\frac {9}{2}}}{x} \]
9/8*(b*x^2 + a)^(7/2)*b*x + 21/16*(b*x^2 + a)^(5/2)*a*b*x + 105/64*(b*x^2 + a)^(3/2)*a^2*b*x + 315/128*sqrt(b*x^2 + a)*a^3*b*x + 315/128*a^4*sqrt(b) *arcsinh(b*x/sqrt(a*b)) - (b*x^2 + a)^(9/2)/x
Time = 0.30 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.93 \[ \int \frac {\left (a+b x^2\right )^{9/2}}{x^2} \, dx=-\frac {315}{256} \, a^{4} \sqrt {b} \log \left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2}\right ) + \frac {2 \, a^{5} \sqrt {b}}{{\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a} + \frac {1}{128} \, {\left (325 \, a^{3} b + 2 \, {\left (105 \, a^{2} b^{2} + 4 \, {\left (2 \, b^{4} x^{2} + 11 \, a b^{3}\right )} x^{2}\right )} x^{2}\right )} \sqrt {b x^{2} + a} x \]
-315/256*a^4*sqrt(b)*log((sqrt(b)*x - sqrt(b*x^2 + a))^2) + 2*a^5*sqrt(b)/ ((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a) + 1/128*(325*a^3*b + 2*(105*a^2*b^2 + 4*(2*b^4*x^2 + 11*a*b^3)*x^2)*x^2)*sqrt(b*x^2 + a)*x
Time = 6.00 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.33 \[ \int \frac {\left (a+b x^2\right )^{9/2}}{x^2} \, dx=-\frac {{\left (b\,x^2+a\right )}^{9/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {9}{2},-\frac {1}{2};\ \frac {1}{2};\ -\frac {b\,x^2}{a}\right )}{x\,{\left (\frac {b\,x^2}{a}+1\right )}^{9/2}} \]