Integrand size = 15, antiderivative size = 117 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{x^9} \, dx=-\frac {a \sqrt {a+b x^2}}{8 x^8}-\frac {3 b \sqrt {a+b x^2}}{16 x^6}-\frac {b^2 \sqrt {a+b x^2}}{64 a x^4}+\frac {3 b^3 \sqrt {a+b x^2}}{128 a^2 x^2}-\frac {3 b^4 \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{128 a^{5/2}} \] Output:
-1/8*a*(b*x^2+a)^(1/2)/x^8-3/16*b*(b*x^2+a)^(1/2)/x^6-1/64*b^2*(b*x^2+a)^( 1/2)/a/x^4+3/128*b^3*(b*x^2+a)^(1/2)/a^2/x^2-3/128*b^4*arctanh((b*x^2+a)^( 1/2)/a^(1/2))/a^(5/2)
Time = 0.12 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.72 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{x^9} \, dx=\frac {\sqrt {a+b x^2} \left (-16 a^3-24 a^2 b x^2-2 a b^2 x^4+3 b^3 x^6\right )}{128 a^2 x^8}-\frac {3 b^4 \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{128 a^{5/2}} \] Input:
Integrate[(a + b*x^2)^(3/2)/x^9,x]
Output:
(Sqrt[a + b*x^2]*(-16*a^3 - 24*a^2*b*x^2 - 2*a*b^2*x^4 + 3*b^3*x^6))/(128* a^2*x^8) - (3*b^4*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/(128*a^(5/2))
Time = 0.19 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.09, 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, 52, 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.
\(\displaystyle \int \frac {\left (a+b x^2\right )^{3/2}}{x^9} \, dx\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \int \frac {\left (b x^2+a\right )^{3/2}}{x^{10}}dx^2\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {1}{2} \left (\frac {3}{8} b \int \frac {\sqrt {b x^2+a}}{x^8}dx^2-\frac {\left (a+b x^2\right )^{3/2}}{4 x^8}\right )\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {1}{2} \left (\frac {3}{8} b \left (\frac {1}{6} b \int \frac {1}{x^6 \sqrt {b x^2+a}}dx^2-\frac {\sqrt {a+b x^2}}{3 x^6}\right )-\frac {\left (a+b x^2\right )^{3/2}}{4 x^8}\right )\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{2} \left (\frac {3}{8} b \left (\frac {1}{6} b \left (-\frac {3 b \int \frac {1}{x^4 \sqrt {b x^2+a}}dx^2}{4 a}-\frac {\sqrt {a+b x^2}}{2 a x^4}\right )-\frac {\sqrt {a+b x^2}}{3 x^6}\right )-\frac {\left (a+b x^2\right )^{3/2}}{4 x^8}\right )\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{2} \left (\frac {3}{8} b \left (\frac {1}{6} b \left (-\frac {3 b \left (-\frac {b \int \frac {1}{x^2 \sqrt {b x^2+a}}dx^2}{2 a}-\frac {\sqrt {a+b x^2}}{a x^2}\right )}{4 a}-\frac {\sqrt {a+b x^2}}{2 a x^4}\right )-\frac {\sqrt {a+b x^2}}{3 x^6}\right )-\frac {\left (a+b x^2\right )^{3/2}}{4 x^8}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} \left (\frac {3}{8} b \left (\frac {1}{6} b \left (-\frac {3 b \left (-\frac {\int \frac {1}{\frac {x^4}{b}-\frac {a}{b}}d\sqrt {b x^2+a}}{a}-\frac {\sqrt {a+b x^2}}{a x^2}\right )}{4 a}-\frac {\sqrt {a+b x^2}}{2 a x^4}\right )-\frac {\sqrt {a+b x^2}}{3 x^6}\right )-\frac {\left (a+b x^2\right )^{3/2}}{4 x^8}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{2} \left (\frac {3}{8} b \left (\frac {1}{6} b \left (-\frac {3 b \left (\frac {b \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {\sqrt {a+b x^2}}{a x^2}\right )}{4 a}-\frac {\sqrt {a+b x^2}}{2 a x^4}\right )-\frac {\sqrt {a+b x^2}}{3 x^6}\right )-\frac {\left (a+b x^2\right )^{3/2}}{4 x^8}\right )\) |
Input:
Int[(a + b*x^2)^(3/2)/x^9,x]
Output:
(-1/4*(a + b*x^2)^(3/2)/x^8 + (3*b*(-1/3*Sqrt[a + b*x^2]/x^6 + (b*(-1/2*Sq rt[a + b*x^2]/(a*x^4) - (3*b*(-(Sqrt[a + b*x^2]/(a*x^2)) + (b*ArcTanh[Sqrt [a + b*x^2]/Sqrt[a]])/a^(3/2)))/(4*a)))/6))/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 = 0.35 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.67
method | result | size |
pseudoelliptic | \(\frac {-\frac {3 \,\operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{\sqrt {a}}\right ) b^{4} x^{8}}{128}+\frac {3 \sqrt {b \,x^{2}+a}\, \left (\sqrt {a}\, b^{3} x^{6}-\frac {2 a^{\frac {3}{2}} b^{2} x^{4}}{3}-8 a^{\frac {5}{2}} b \,x^{2}-\frac {16 a^{\frac {7}{2}}}{3}\right )}{128}}{a^{\frac {5}{2}} x^{8}}\) | \(78\) |
risch | \(-\frac {\sqrt {b \,x^{2}+a}\, \left (-3 b^{3} x^{6}+2 a \,b^{2} x^{4}+24 a^{2} b \,x^{2}+16 a^{3}\right )}{128 x^{8} a^{2}}-\frac {3 b^{4} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{128 a^{\frac {5}{2}}}\) | \(82\) |
default | \(-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{8 a \,x^{8}}-\frac {3 b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6 a \,x^{6}}-\frac {b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{4 a \,x^{4}}+\frac {b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{2 a \,x^{2}}+\frac {3 b \left (\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3}+a \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )\right )}{2 a}\right )}{4 a}\right )}{6 a}\right )}{8 a}\) | \(149\) |
Input:
int((b*x^2+a)^(3/2)/x^9,x,method=_RETURNVERBOSE)
Output:
3/128/a^(5/2)*(-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^6-2/3*a^(3/2)*b^2*x^4-8*a^(5/2)*b*x^2-16/3*a^(7/2)))/x^8
Time = 0.09 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.56 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{x^9} \, dx=\left [\frac {3 \, \sqrt {a} b^{4} x^{8} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (3 \, a b^{3} x^{6} - 2 \, a^{2} b^{2} x^{4} - 24 \, a^{3} b x^{2} - 16 \, a^{4}\right )} \sqrt {b x^{2} + a}}{256 \, a^{3} x^{8}}, \frac {3 \, \sqrt {-a} b^{4} x^{8} \arctan \left (\frac {\sqrt {b x^{2} + a} \sqrt {-a}}{a}\right ) + {\left (3 \, a b^{3} x^{6} - 2 \, a^{2} b^{2} x^{4} - 24 \, a^{3} b x^{2} - 16 \, a^{4}\right )} \sqrt {b x^{2} + a}}{128 \, a^{3} x^{8}}\right ] \] Input:
integrate((b*x^2+a)^(3/2)/x^9,x, algorithm="fricas")
Output:
[1/256*(3*sqrt(a)*b^4*x^8*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x ^2) + 2*(3*a*b^3*x^6 - 2*a^2*b^2*x^4 - 24*a^3*b*x^2 - 16*a^4)*sqrt(b*x^2 + a))/(a^3*x^8), 1/128*(3*sqrt(-a)*b^4*x^8*arctan(sqrt(b*x^2 + a)*sqrt(-a)/ a) + (3*a*b^3*x^6 - 2*a^2*b^2*x^4 - 24*a^3*b*x^2 - 16*a^4)*sqrt(b*x^2 + a) )/(a^3*x^8)]
Time = 11.82 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.26 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{x^9} \, dx=- \frac {a^{2}}{8 \sqrt {b} x^{9} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {5 a \sqrt {b}}{16 x^{7} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {13 b^{\frac {3}{2}}}{64 x^{5} \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {b^{\frac {5}{2}}}{128 a x^{3} \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {3 b^{\frac {7}{2}}}{128 a^{2} x \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {3 b^{4} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{128 a^{\frac {5}{2}}} \] Input:
integrate((b*x**2+a)**(3/2)/x**9,x)
Output:
-a**2/(8*sqrt(b)*x**9*sqrt(a/(b*x**2) + 1)) - 5*a*sqrt(b)/(16*x**7*sqrt(a/ (b*x**2) + 1)) - 13*b**(3/2)/(64*x**5*sqrt(a/(b*x**2) + 1)) + b**(5/2)/(12 8*a*x**3*sqrt(a/(b*x**2) + 1)) + 3*b**(7/2)/(128*a**2*x*sqrt(a/(b*x**2) + 1)) - 3*b**4*asinh(sqrt(a)/(sqrt(b)*x))/(128*a**(5/2))
Time = 0.05 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.11 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{x^9} \, dx=-\frac {3 \, b^{4} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{128 \, a^{\frac {5}{2}}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{4}}{128 \, a^{4}} + \frac {3 \, \sqrt {b x^{2} + a} b^{4}}{128 \, a^{3}} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{3}}{128 \, a^{4} x^{2}} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{2}}{64 \, a^{3} x^{4}} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} b}{16 \, a^{2} x^{6}} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}}}{8 \, a x^{8}} \] Input:
integrate((b*x^2+a)^(3/2)/x^9,x, algorithm="maxima")
Output:
-3/128*b^4*arcsinh(a/(sqrt(a*b)*abs(x)))/a^(5/2) + 1/128*(b*x^2 + a)^(3/2) *b^4/a^4 + 3/128*sqrt(b*x^2 + a)*b^4/a^3 - 1/128*(b*x^2 + a)^(5/2)*b^3/(a^ 4*x^2) - 1/64*(b*x^2 + a)^(5/2)*b^2/(a^3*x^4) + 1/16*(b*x^2 + a)^(5/2)*b/( a^2*x^6) - 1/8*(b*x^2 + a)^(5/2)/(a*x^8)
Time = 0.13 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.93 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{x^9} \, dx=\frac {\frac {3 \, b^{5} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} + \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{5} - 11 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a b^{5} - 11 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2} b^{5} + 3 \, \sqrt {b x^{2} + a} a^{3} b^{5}}{a^{2} b^{4} x^{8}}}{128 \, b} \] Input:
integrate((b*x^2+a)^(3/2)/x^9,x, algorithm="giac")
Output:
1/128*(3*b^5*arctan(sqrt(b*x^2 + a)/sqrt(-a))/(sqrt(-a)*a^2) + (3*(b*x^2 + a)^(7/2)*b^5 - 11*(b*x^2 + a)^(5/2)*a*b^5 - 11*(b*x^2 + a)^(3/2)*a^2*b^5 + 3*sqrt(b*x^2 + a)*a^3*b^5)/(a^2*b^4*x^8))/b
Time = 0.94 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.76 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{x^9} \, dx=\frac {3\,a\,\sqrt {b\,x^2+a}}{128\,x^8}-\frac {11\,{\left (b\,x^2+a\right )}^{3/2}}{128\,x^8}-\frac {11\,{\left (b\,x^2+a\right )}^{5/2}}{128\,a\,x^8}+\frac {3\,{\left (b\,x^2+a\right )}^{7/2}}{128\,a^2\,x^8}+\frac {b^4\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,3{}\mathrm {i}}{128\,a^{5/2}} \] Input:
int((a + b*x^2)^(3/2)/x^9,x)
Output:
(b^4*atan(((a + b*x^2)^(1/2)*1i)/a^(1/2))*3i)/(128*a^(5/2)) - (11*(a + b*x ^2)^(3/2))/(128*x^8) + (3*a*(a + b*x^2)^(1/2))/(128*x^8) - (11*(a + b*x^2) ^(5/2))/(128*a*x^8) + (3*(a + b*x^2)^(7/2))/(128*a^2*x^8)
Time = 0.22 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.19 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{x^9} \, dx=\frac {-16 \sqrt {b \,x^{2}+a}\, a^{4}-24 \sqrt {b \,x^{2}+a}\, a^{3} b \,x^{2}-2 \sqrt {b \,x^{2}+a}\, a^{2} b^{2} x^{4}+3 \sqrt {b \,x^{2}+a}\, a \,b^{3} x^{6}+3 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}-\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{4} x^{8}-3 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{4} x^{8}}{128 a^{3} x^{8}} \] Input:
int((b*x^2+a)^(3/2)/x^9,x)
Output:
( - 16*sqrt(a + b*x**2)*a**4 - 24*sqrt(a + b*x**2)*a**3*b*x**2 - 2*sqrt(a + b*x**2)*a**2*b**2*x**4 + 3*sqrt(a + b*x**2)*a*b**3*x**6 + 3*sqrt(a)*log( (sqrt(a + b*x**2) - sqrt(a) + sqrt(b)*x)/sqrt(a))*b**4*x**8 - 3*sqrt(a)*lo g((sqrt(a + b*x**2) + sqrt(a) + sqrt(b)*x)/sqrt(a))*b**4*x**8)/(128*a**3*x **8)