Integrand size = 15, antiderivative size = 117 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^9} \, dx=-\frac {a^2 \sqrt {a+b x^2}}{8 x^8}-\frac {17 a b \sqrt {a+b x^2}}{48 x^6}-\frac {59 b^2 \sqrt {a+b x^2}}{192 x^4}-\frac {5 b^3 \sqrt {a+b x^2}}{128 a x^2}+\frac {5 b^4 \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{128 a^{3/2}} \] Output:
-1/8*a^2*(b*x^2+a)^(1/2)/x^8-17/48*a*b*(b*x^2+a)^(1/2)/x^6-59/192*b^2*(b*x ^2+a)^(1/2)/x^4-5/128*b^3*(b*x^2+a)^(1/2)/a/x^2+5/128*b^4*arctanh((b*x^2+a )^(1/2)/a^(1/2))/a^(3/2)
Time = 0.11 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.72 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^9} \, dx=\frac {\sqrt {a+b x^2} \left (-48 a^3-136 a^2 b x^2-118 a b^2 x^4-15 b^3 x^6\right )}{384 a x^8}+\frac {5 b^4 \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{128 a^{3/2}} \] Input:
Integrate[(a + b*x^2)^(5/2)/x^9,x]
Output:
(Sqrt[a + b*x^2]*(-48*a^3 - 136*a^2*b*x^2 - 118*a*b^2*x^4 - 15*b^3*x^6))/( 384*a*x^8) + (5*b^4*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/(128*a^(3/2))
Time = 0.19 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.03, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {243, 51, 51, 51, 52, 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.
| \(\displaystyle \int \frac {\left (a+b x^2\right )^{5/2}}{x^9} \, dx\) |
| \(\Big \downarrow \) 243 |
| \(\displaystyle \frac {1}{2} \int \frac {\left (b x^2+a\right )^{5/2}}{x^{10}}dx^2\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {1}{2} \left (\frac {5}{8} b \int \frac {\left (b x^2+a\right )^{3/2}}{x^8}dx^2-\frac {\left (a+b x^2\right )^{5/2}}{4 x^8}\right )\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {1}{2} \left (\frac {5}{8} b \left (\frac {1}{2} b \int \frac {\sqrt {b x^2+a}}{x^6}dx^2-\frac {\left (a+b x^2\right )^{3/2}}{3 x^6}\right )-\frac {\left (a+b x^2\right )^{5/2}}{4 x^8}\right )\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {1}{2} \left (\frac {5}{8} b \left (\frac {1}{2} b \left (\frac {1}{4} b \int \frac {1}{x^4 \sqrt {b x^2+a}}dx^2-\frac {\sqrt {a+b x^2}}{2 x^4}\right )-\frac {\left (a+b x^2\right )^{3/2}}{3 x^6}\right )-\frac {\left (a+b x^2\right )^{5/2}}{4 x^8}\right )\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {1}{2} \left (\frac {5}{8} b \left (\frac {1}{2} b \left (\frac {1}{4} b \left (-\frac {b \int \frac {1}{x^2 \sqrt {b x^2+a}}dx^2}{2 a}-\frac {\sqrt {a+b x^2}}{a x^2}\right )-\frac {\sqrt {a+b x^2}}{2 x^4}\right )-\frac {\left (a+b x^2\right )^{3/2}}{3 x^6}\right )-\frac {\left (a+b x^2\right )^{5/2}}{4 x^8}\right )\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {1}{2} \left (\frac {5}{8} b \left (\frac {1}{2} b \left (\frac {1}{4} b \left (-\frac {\int \frac {1}{\frac {x^4}{b}-\frac {a}{b}}d\sqrt {b x^2+a}}{a}-\frac {\sqrt {a+b x^2}}{a x^2}\right )-\frac {\sqrt {a+b x^2}}{2 x^4}\right )-\frac {\left (a+b x^2\right )^{3/2}}{3 x^6}\right )-\frac {\left (a+b x^2\right )^{5/2}}{4 x^8}\right )\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {1}{2} \left (\frac {5}{8} b \left (\frac {1}{2} b \left (\frac {1}{4} b \left (\frac {b \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {\sqrt {a+b x^2}}{a x^2}\right )-\frac {\sqrt {a+b x^2}}{2 x^4}\right )-\frac {\left (a+b x^2\right )^{3/2}}{3 x^6}\right )-\frac {\left (a+b x^2\right )^{5/2}}{4 x^8}\right )\) |
Input:
Int[(a + b*x^2)^(5/2)/x^9,x]
Output:
(-1/4*(a + b*x^2)^(5/2)/x^8 + (5*b*(-1/3*(a + b*x^2)^(3/2)/x^6 + (b*(-1/2* Sqrt[a + b*x^2]/x^4 + (b*(-(Sqrt[a + b*x^2]/(a*x^2)) + (b*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/a^(3/2)))/4))/2))/8)/2
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 + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
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 = 0.44 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.67
| method | result | size |
| pseudoelliptic | \(-\frac {5 \left (-\operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{\sqrt {a}}\right ) b^{4} x^{8}+\sqrt {b \,x^{2}+a}\, \left (\sqrt {a}\, b^{3} x^{6}+\frac {118 a^{\frac {3}{2}} b^{2} x^{4}}{15}+\frac {136 a^{\frac {5}{2}} b \,x^{2}}{15}+\frac {16 a^{\frac {7}{2}}}{5}\right )\right )}{128 a^{\frac {3}{2}} x^{8}}\) | \(78\) |
| risch | \(-\frac {\sqrt {b \,x^{2}+a}\, \left (15 b^{3} x^{6}+118 a \,b^{2} x^{4}+136 a^{2} b \,x^{2}+48 a^{3}\right )}{384 x^{8} a}+\frac {5 b^{4} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{128 a^{\frac {3}{2}}}\) | \(82\) |
| default | \(-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{8 a \,x^{8}}-\frac {b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{6 a \,x^{6}}+\frac {b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{4 a \,x^{4}}+\frac {3 b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{2 a \,x^{2}}+\frac {5 b \left (\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{5}+a \left (\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3}+a \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )\right )\right )}{2 a}\right )}{4 a}\right )}{6 a}\right )}{8 a}\) | \(163\) |
Input:
int((b*x^2+a)^(5/2)/x^9,x,method=_RETURNVERBOSE)
Output:
-5/128/a^(3/2)*(-arctanh((b*x^2+a)^(1/2)/a^(1/2))*b^4*x^8+(b*x^2+a)^(1/2)* (a^(1/2)*b^3*x^6+118/15*a^(3/2)*b^2*x^4+136/15*a^(5/2)*b*x^2+16/5*a^(7/2)) )/x^8
Time = 0.08 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.56 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^9} \, dx=\left [\frac {15 \, \sqrt {a} b^{4} x^{8} \log \left (-\frac {b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 2 \, {\left (15 \, a b^{3} x^{6} + 118 \, a^{2} b^{2} x^{4} + 136 \, a^{3} b x^{2} + 48 \, a^{4}\right )} \sqrt {b x^{2} + a}}{768 \, a^{2} x^{8}}, -\frac {15 \, \sqrt {-a} b^{4} x^{8} \arctan \left (\frac {\sqrt {b x^{2} + a} \sqrt {-a}}{a}\right ) + {\left (15 \, a b^{3} x^{6} + 118 \, a^{2} b^{2} x^{4} + 136 \, a^{3} b x^{2} + 48 \, a^{4}\right )} \sqrt {b x^{2} + a}}{384 \, a^{2} x^{8}}\right ] \] Input:
integrate((b*x^2+a)^(5/2)/x^9,x, algorithm="fricas")
Output:
[1/768*(15*sqrt(a)*b^4*x^8*log(-(b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/ x^2) - 2*(15*a*b^3*x^6 + 118*a^2*b^2*x^4 + 136*a^3*b*x^2 + 48*a^4)*sqrt(b* x^2 + a))/(a^2*x^8), -1/384*(15*sqrt(-a)*b^4*x^8*arctan(sqrt(b*x^2 + a)*sq rt(-a)/a) + (15*a*b^3*x^6 + 118*a^2*b^2*x^4 + 136*a^3*b*x^2 + 48*a^4)*sqrt (b*x^2 + a))/(a^2*x^8)]
Time = 6.99 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.28 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^9} \, dx=- \frac {a^{3}}{8 \sqrt {b} x^{9} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {23 a^{2} \sqrt {b}}{48 x^{7} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {127 a b^{\frac {3}{2}}}{192 x^{5} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {133 b^{\frac {5}{2}}}{384 x^{3} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {5 b^{\frac {7}{2}}}{128 a x \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {5 b^{4} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{128 a^{\frac {3}{2}}} \] Input:
integrate((b*x**2+a)**(5/2)/x**9,x)
Output:
-a**3/(8*sqrt(b)*x**9*sqrt(a/(b*x**2) + 1)) - 23*a**2*sqrt(b)/(48*x**7*sqr t(a/(b*x**2) + 1)) - 127*a*b**(3/2)/(192*x**5*sqrt(a/(b*x**2) + 1)) - 133* b**(5/2)/(384*x**3*sqrt(a/(b*x**2) + 1)) - 5*b**(7/2)/(128*a*x*sqrt(a/(b*x **2) + 1)) + 5*b**4*asinh(sqrt(a)/(sqrt(b)*x))/(128*a**(3/2))
Time = 0.04 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.26 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^9} \, dx=\frac {5 \, b^{4} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{128 \, a^{\frac {3}{2}}} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{4}}{128 \, a^{4}} - \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{4}}{384 \, a^{3}} - \frac {5 \, \sqrt {b x^{2} + a} b^{4}}{128 \, a^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{3}}{128 \, a^{4} x^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{2}}{192 \, a^{3} x^{4}} + \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} b}{48 \, a^{2} x^{6}} - \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}}}{8 \, a x^{8}} \] Input:
integrate((b*x^2+a)^(5/2)/x^9,x, algorithm="maxima")
Output:
5/128*b^4*arcsinh(a/(sqrt(a*b)*abs(x)))/a^(3/2) - 1/128*(b*x^2 + a)^(5/2)* b^4/a^4 - 5/384*(b*x^2 + a)^(3/2)*b^4/a^3 - 5/128*sqrt(b*x^2 + a)*b^4/a^2 + 1/128*(b*x^2 + a)^(7/2)*b^3/(a^4*x^2) + 1/192*(b*x^2 + a)^(7/2)*b^2/(a^3 *x^4) + 1/48*(b*x^2 + a)^(7/2)*b/(a^2*x^6) - 1/8*(b*x^2 + a)^(7/2)/(a*x^8)
Time = 0.12 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.93 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^9} \, dx=-\frac {\frac {15 \, b^{5} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} + \frac {15 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{5} + 73 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a b^{5} - 55 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2} b^{5} + 15 \, \sqrt {b x^{2} + a} a^{3} b^{5}}{a b^{4} x^{8}}}{384 \, b} \] Input:
integrate((b*x^2+a)^(5/2)/x^9,x, algorithm="giac")
Output:
-1/384*(15*b^5*arctan(sqrt(b*x^2 + a)/sqrt(-a))/(sqrt(-a)*a) + (15*(b*x^2 + a)^(7/2)*b^5 + 73*(b*x^2 + a)^(5/2)*a*b^5 - 55*(b*x^2 + a)^(3/2)*a^2*b^5 + 15*sqrt(b*x^2 + a)*a^3*b^5)/(a*b^4*x^8))/b
Time = 1.18 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.76 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^9} \, dx=\frac {55\,a\,{\left (b\,x^2+a\right )}^{3/2}}{384\,x^8}-\frac {73\,{\left (b\,x^2+a\right )}^{5/2}}{384\,x^8}-\frac {5\,a^2\,\sqrt {b\,x^2+a}}{128\,x^8}-\frac {5\,{\left (b\,x^2+a\right )}^{7/2}}{128\,a\,x^8}-\frac {b^4\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,5{}\mathrm {i}}{128\,a^{3/2}} \] Input:
int((a + b*x^2)^(5/2)/x^9,x)
Output:
(55*a*(a + b*x^2)^(3/2))/(384*x^8) - (b^4*atan(((a + b*x^2)^(1/2)*1i)/a^(1 /2))*5i)/(128*a^(3/2)) - (73*(a + b*x^2)^(5/2))/(384*x^8) - (5*a^2*(a + b* x^2)^(1/2))/(128*x^8) - (5*(a + b*x^2)^(7/2))/(128*a*x^8)
Time = 0.21 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.19 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^9} \, dx=\frac {-48 \sqrt {b \,x^{2}+a}\, a^{4}-136 \sqrt {b \,x^{2}+a}\, a^{3} b \,x^{2}-118 \sqrt {b \,x^{2}+a}\, a^{2} b^{2} x^{4}-15 \sqrt {b \,x^{2}+a}\, a \,b^{3} x^{6}-15 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}-\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{4} x^{8}+15 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{4} x^{8}}{384 a^{2} x^{8}} \] Input:
int((b*x^2+a)^(5/2)/x^9,x)
Output:
( - 48*sqrt(a + b*x**2)*a**4 - 136*sqrt(a + b*x**2)*a**3*b*x**2 - 118*sqrt (a + b*x**2)*a**2*b**2*x**4 - 15*sqrt(a + b*x**2)*a*b**3*x**6 - 15*sqrt(a) *log((sqrt(a + b*x**2) - sqrt(a) + sqrt(b)*x)/sqrt(a))*b**4*x**8 + 15*sqrt (a)*log((sqrt(a + b*x**2) + sqrt(a) + sqrt(b)*x)/sqrt(a))*b**4*x**8)/(384* a**2*x**8)