\(\int \frac {(a+b x^2)^{3/2} (A+B x^2)}{x^9} \, dx\) [211]

Optimal result

Integrand size = 22, antiderivative size = 150 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^9} \, dx=-\frac {(3 A b+8 a B) \sqrt {a+b x^2}}{48 x^6}-\frac {b (3 A b+56 a B) \sqrt {a+b x^2}}{192 a x^4}+\frac {b^2 (3 A b-8 a B) \sqrt {a+b x^2}}{128 a^2 x^2}-\frac {A \left (a+b x^2\right )^{3/2}}{8 x^8}-\frac {b^3 (3 A b-8 a B) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{128 a^{5/2}} \] Output:

-1/48*(3*A*b+8*B*a)*(b*x^2+a)^(1/2)/x^6-1/192*b*(3*A*b+56*B*a)*(b*x^2+a)^( 
1/2)/a/x^4+1/128*b^2*(3*A*b-8*B*a)*(b*x^2+a)^(1/2)/a^2/x^2-1/8*A*(b*x^2+a) 
^(3/2)/x^8-1/128*b^3*(3*A*b-8*B*a)*arctanh((b*x^2+a)^(1/2)/a^(1/2))/a^(5/2 
)
 

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.84 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^9} \, dx=\frac {\sqrt {a+b x^2} \left (-48 a^3 A-72 a^2 A b x^2-64 a^3 B x^2-6 a A b^2 x^4-112 a^2 b B x^4+9 A b^3 x^6-24 a b^2 B x^6\right )}{384 a^2 x^8}+\frac {b^3 (-3 A b+8 a B) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{128 a^{5/2}} \] Input:

Integrate[((a + b*x^2)^(3/2)*(A + B*x^2))/x^9,x]
 

Output:

(Sqrt[a + b*x^2]*(-48*a^3*A - 72*a^2*A*b*x^2 - 64*a^3*B*x^2 - 6*a*A*b^2*x^ 
4 - 112*a^2*b*B*x^4 + 9*A*b^3*x^6 - 24*a*b^2*B*x^6))/(384*a^2*x^8) + (b^3* 
(-3*A*b + 8*a*B)*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/(128*a^(5/2))
 

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.91, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {354, 87, 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.

\(\displaystyle \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^9} \, dx\)

\(\Big \downarrow \) 354

\(\displaystyle \frac {1}{2} \int \frac {\left (b x^2+a\right )^{3/2} \left (B x^2+A\right )}{x^{10}}dx^2\)

\(\Big \downarrow \) 87

\(\displaystyle \frac {1}{2} \left (-\frac {(3 A b-8 a B) \int \frac {\left (b x^2+a\right )^{3/2}}{x^8}dx^2}{8 a}-\frac {A \left (a+b x^2\right )^{5/2}}{4 a x^8}\right )\)

\(\Big \downarrow \) 51

\(\displaystyle \frac {1}{2} \left (-\frac {(3 A b-8 a B) \left (\frac {1}{2} b \int \frac {\sqrt {b x^2+a}}{x^6}dx^2-\frac {\left (a+b x^2\right )^{3/2}}{3 x^6}\right )}{8 a}-\frac {A \left (a+b x^2\right )^{5/2}}{4 a x^8}\right )\)

\(\Big \downarrow \) 51

\(\displaystyle \frac {1}{2} \left (-\frac {(3 A b-8 a B) \left (\frac {1}{2} b \left (\frac {1}{4} b \int \frac {1}{x^4 \sqrt {b x^2+a}}dx^2-\frac {\sqrt {a+b x^2}}{2 x^4}\right )-\frac {\left (a+b x^2\right )^{3/2}}{3 x^6}\right )}{8 a}-\frac {A \left (a+b x^2\right )^{5/2}}{4 a x^8}\right )\)

\(\Big \downarrow \) 52

\(\displaystyle \frac {1}{2} \left (-\frac {(3 A b-8 a B) \left (\frac {1}{2} b \left (\frac {1}{4} b \left (-\frac {b \int \frac {1}{x^2 \sqrt {b x^2+a}}dx^2}{2 a}-\frac {\sqrt {a+b x^2}}{a x^2}\right )-\frac {\sqrt {a+b x^2}}{2 x^4}\right )-\frac {\left (a+b x^2\right )^{3/2}}{3 x^6}\right )}{8 a}-\frac {A \left (a+b x^2\right )^{5/2}}{4 a x^8}\right )\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{2} \left (-\frac {(3 A b-8 a B) \left (\frac {1}{2} b \left (\frac {1}{4} b \left (-\frac {\int \frac {1}{\frac {x^4}{b}-\frac {a}{b}}d\sqrt {b x^2+a}}{a}-\frac {\sqrt {a+b x^2}}{a x^2}\right )-\frac {\sqrt {a+b x^2}}{2 x^4}\right )-\frac {\left (a+b x^2\right )^{3/2}}{3 x^6}\right )}{8 a}-\frac {A \left (a+b x^2\right )^{5/2}}{4 a x^8}\right )\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {1}{2} \left (-\frac {(3 A b-8 a B) \left (\frac {1}{2} b \left (\frac {1}{4} b \left (\frac {b \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {\sqrt {a+b x^2}}{a x^2}\right )-\frac {\sqrt {a+b x^2}}{2 x^4}\right )-\frac {\left (a+b x^2\right )^{3/2}}{3 x^6}\right )}{8 a}-\frac {A \left (a+b x^2\right )^{5/2}}{4 a x^8}\right )\)

Input:

Int[((a + b*x^2)^(3/2)*(A + B*x^2))/x^9,x]
 

Output:

(-1/4*(A*(a + b*x^2)^(5/2))/(a*x^8) - ((3*A*b - 8*a*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*a))/2
 

Defintions of rubi rules used

rule 51

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]
 

rule 52

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]
 

rule 73

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]
 

rule 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

rule 221

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]
 

rule 354

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S 
ymbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x 
, x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ 
[(m - 1)/2]
 

Maple [A] (verified)

Time = 0.56 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.75

method	result	size
pseudoelliptic	\(-\frac {3 \left (\frac {\left (A b -\frac {8 B a}{3}\right ) b^{3} x^{8} \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{\sqrt {a}}\right )}{8}+\sqrt {b \,x^{2}+a}\, \left (\frac {b^{2} x^{4} \left (4 x^{2} B +A \right ) a^{\frac {3}{2}}}{12}+b \,x^{2} \left (\frac {14 x^{2} B}{9}+A \right ) a^{\frac {5}{2}}+\left (\frac {8 x^{2} B}{9}+\frac {2 A}{3}\right ) a^{\frac {7}{2}}-\frac {A \sqrt {a}\, b^{3} x^{6}}{8}\right )\right )}{16 a^{\frac {5}{2}} x^{8}}\)	\(112\)
risch	\(-\frac {\sqrt {b \,x^{2}+a}\, \left (-9 A \,b^{3} x^{6}+24 B a \,b^{2} x^{6}+6 a A \,b^{2} x^{4}+112 B \,a^{2} b \,x^{4}+72 a^{2} A b \,x^{2}+64 B \,a^{3} x^{2}+48 a^{3} A \right )}{384 x^{8} a^{2}}-\frac {\left (3 A b -8 B a \right ) b^{3} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{128 a^{\frac {5}{2}}}\)	\(124\)
default	\(A \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{8 a \,x^{8}}-\frac {3 b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6 a \,x^{6}}-\frac {b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{4 a \,x^{4}}+\frac {b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{2 a \,x^{2}}+\frac {3 b \left (\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3}+a \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )\right )}{2 a}\right )}{4 a}\right )}{6 a}\right )}{8 a}\right )+B \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6 a \,x^{6}}-\frac {b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{4 a \,x^{4}}+\frac {b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{2 a \,x^{2}}+\frac {3 b \left (\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3}+a \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )\right )}{2 a}\right )}{4 a}\right )}{6 a}\right )\)	\(278\)

Input:

int((b*x^2+a)^(3/2)*(B*x^2+A)/x^9,x,method=_RETURNVERBOSE)
 

Output:

-3/16*(1/8*(A*b-8/3*B*a)*b^3*x^8*arctanh((b*x^2+a)^(1/2)/a^(1/2))+(b*x^2+a 
)^(1/2)*(1/12*b^2*x^4*(4*B*x^2+A)*a^(3/2)+b*x^2*(14/9*x^2*B+A)*a^(5/2)+(8/ 
9*x^2*B+2/3*A)*a^(7/2)-1/8*A*a^(1/2)*b^3*x^6))/a^(5/2)/x^8
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.83 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^9} \, dx=\left [-\frac {3 \, {\left (8 \, B a b^{3} - 3 \, A b^{4}\right )} \sqrt {a} x^{8} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (3 \, {\left (8 \, B a^{2} b^{2} - 3 \, A a b^{3}\right )} x^{6} + 48 \, A a^{4} + 2 \, {\left (56 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{4} + 8 \, {\left (8 \, B a^{4} + 9 \, A a^{3} b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{768 \, a^{3} x^{8}}, -\frac {3 \, {\left (8 \, B a b^{3} - 3 \, A b^{4}\right )} \sqrt {-a} x^{8} \arctan \left (\frac {\sqrt {b x^{2} + a} \sqrt {-a}}{a}\right ) + {\left (3 \, {\left (8 \, B a^{2} b^{2} - 3 \, A a b^{3}\right )} x^{6} + 48 \, A a^{4} + 2 \, {\left (56 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{4} + 8 \, {\left (8 \, B a^{4} + 9 \, A a^{3} b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{384 \, a^{3} x^{8}}\right ] \] Input:

integrate((b*x^2+a)^(3/2)*(B*x^2+A)/x^9,x, algorithm="fricas")
 

Output:

[-1/768*(3*(8*B*a*b^3 - 3*A*b^4)*sqrt(a)*x^8*log(-(b*x^2 - 2*sqrt(b*x^2 + 
a)*sqrt(a) + 2*a)/x^2) + 2*(3*(8*B*a^2*b^2 - 3*A*a*b^3)*x^6 + 48*A*a^4 + 2 
*(56*B*a^3*b + 3*A*a^2*b^2)*x^4 + 8*(8*B*a^4 + 9*A*a^3*b)*x^2)*sqrt(b*x^2 
+ a))/(a^3*x^8), -1/384*(3*(8*B*a*b^3 - 3*A*b^4)*sqrt(-a)*x^8*arctan(sqrt( 
b*x^2 + a)*sqrt(-a)/a) + (3*(8*B*a^2*b^2 - 3*A*a*b^3)*x^6 + 48*A*a^4 + 2*( 
56*B*a^3*b + 3*A*a^2*b^2)*x^4 + 8*(8*B*a^4 + 9*A*a^3*b)*x^2)*sqrt(b*x^2 + 
a))/(a^3*x^8)]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 287 vs. \(2 (139) = 278\).

Time = 113.00 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.91 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^9} \, dx=- \frac {A a^{2}}{8 \sqrt {b} x^{9} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {5 A a \sqrt {b}}{16 x^{7} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {13 A b^{\frac {3}{2}}}{64 x^{5} \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {A b^{\frac {5}{2}}}{128 a x^{3} \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {3 A b^{\frac {7}{2}}}{128 a^{2} x \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {3 A b^{4} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{128 a^{\frac {5}{2}}} - \frac {B a^{2}}{6 \sqrt {b} x^{7} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {11 B a \sqrt {b}}{24 x^{5} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {17 B b^{\frac {3}{2}}}{48 x^{3} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {B b^{\frac {5}{2}}}{16 a x \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {B b^{3} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{16 a^{\frac {3}{2}}} \] Input:

integrate((b*x**2+a)**(3/2)*(B*x**2+A)/x**9,x)
 

Output:

-A*a**2/(8*sqrt(b)*x**9*sqrt(a/(b*x**2) + 1)) - 5*A*a*sqrt(b)/(16*x**7*sqr 
t(a/(b*x**2) + 1)) - 13*A*b**(3/2)/(64*x**5*sqrt(a/(b*x**2) + 1)) + A*b**( 
5/2)/(128*a*x**3*sqrt(a/(b*x**2) + 1)) + 3*A*b**(7/2)/(128*a**2*x*sqrt(a/( 
b*x**2) + 1)) - 3*A*b**4*asinh(sqrt(a)/(sqrt(b)*x))/(128*a**(5/2)) - B*a** 
2/(6*sqrt(b)*x**7*sqrt(a/(b*x**2) + 1)) - 11*B*a*sqrt(b)/(24*x**5*sqrt(a/( 
b*x**2) + 1)) - 17*B*b**(3/2)/(48*x**3*sqrt(a/(b*x**2) + 1)) - B*b**(5/2)/ 
(16*a*x*sqrt(a/(b*x**2) + 1)) + B*b**3*asinh(sqrt(a)/(sqrt(b)*x))/(16*a**( 
3/2))
 

Maxima [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.68 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^9} \, dx=\frac {B b^{3} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{16 \, a^{\frac {3}{2}}} - \frac {3 \, A b^{4} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{128 \, a^{\frac {5}{2}}} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B b^{3}}{48 \, a^{3}} - \frac {\sqrt {b x^{2} + a} B b^{3}}{16 \, a^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A b^{4}}{128 \, a^{4}} + \frac {3 \, \sqrt {b x^{2} + a} A b^{4}}{128 \, a^{3}} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B b^{2}}{48 \, a^{3} x^{2}} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A b^{3}}{128 \, a^{4} x^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B b}{24 \, a^{2} x^{4}} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A b^{2}}{64 \, a^{3} x^{4}} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B}{6 \, a x^{6}} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A b}{16 \, a^{2} x^{6}} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A}{8 \, a x^{8}} \] Input:

integrate((b*x^2+a)^(3/2)*(B*x^2+A)/x^9,x, algorithm="maxima")
                                                                                    
                                                                                    
 

Output:

1/16*B*b^3*arcsinh(a/(sqrt(a*b)*abs(x)))/a^(3/2) - 3/128*A*b^4*arcsinh(a/( 
sqrt(a*b)*abs(x)))/a^(5/2) - 1/48*(b*x^2 + a)^(3/2)*B*b^3/a^3 - 1/16*sqrt( 
b*x^2 + a)*B*b^3/a^2 + 1/128*(b*x^2 + a)^(3/2)*A*b^4/a^4 + 3/128*sqrt(b*x^ 
2 + a)*A*b^4/a^3 + 1/48*(b*x^2 + a)^(5/2)*B*b^2/(a^3*x^2) - 1/128*(b*x^2 + 
 a)^(5/2)*A*b^3/(a^4*x^2) + 1/24*(b*x^2 + a)^(5/2)*B*b/(a^2*x^4) - 1/64*(b 
*x^2 + a)^(5/2)*A*b^2/(a^3*x^4) - 1/6*(b*x^2 + a)^(5/2)*B/(a*x^6) + 1/16*( 
b*x^2 + a)^(5/2)*A*b/(a^2*x^6) - 1/8*(b*x^2 + a)^(5/2)*A/(a*x^8)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.29 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^9} \, dx=-\frac {\frac {3 \, {\left (8 \, B a b^{4} - 3 \, A b^{5}\right )} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} + \frac {24 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} B a b^{4} + 40 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} B a^{2} b^{4} - 88 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a^{3} b^{4} + 24 \, \sqrt {b x^{2} + a} B a^{4} b^{4} - 9 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} A b^{5} + 33 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} A a b^{5} + 33 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A a^{2} b^{5} - 9 \, \sqrt {b x^{2} + a} A a^{3} b^{5}}{a^{2} b^{4} x^{8}}}{384 \, b} \] Input:

integrate((b*x^2+a)^(3/2)*(B*x^2+A)/x^9,x, algorithm="giac")
 

Output:

-1/384*(3*(8*B*a*b^4 - 3*A*b^5)*arctan(sqrt(b*x^2 + a)/sqrt(-a))/(sqrt(-a) 
*a^2) + (24*(b*x^2 + a)^(7/2)*B*a*b^4 + 40*(b*x^2 + a)^(5/2)*B*a^2*b^4 - 8 
8*(b*x^2 + a)^(3/2)*B*a^3*b^4 + 24*sqrt(b*x^2 + a)*B*a^4*b^4 - 9*(b*x^2 + 
a)^(7/2)*A*b^5 + 33*(b*x^2 + a)^(5/2)*A*a*b^5 + 33*(b*x^2 + a)^(3/2)*A*a^2 
*b^5 - 9*sqrt(b*x^2 + a)*A*a^3*b^5)/(a^2*b^4*x^8))/b
 

Mupad [B] (verification not implemented)

Time = 3.07 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.13 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^9} \, dx=\frac {3\,A\,a\,\sqrt {b\,x^2+a}}{128\,x^8}-\frac {B\,{\left (b\,x^2+a\right )}^{3/2}}{6\,x^6}-\frac {11\,A\,{\left (b\,x^2+a\right )}^{3/2}}{128\,x^8}+\frac {B\,a\,\sqrt {b\,x^2+a}}{16\,x^6}-\frac {11\,A\,{\left (b\,x^2+a\right )}^{5/2}}{128\,a\,x^8}+\frac {3\,A\,{\left (b\,x^2+a\right )}^{7/2}}{128\,a^2\,x^8}-\frac {B\,{\left (b\,x^2+a\right )}^{5/2}}{16\,a\,x^6}+\frac {A\,b^4\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,3{}\mathrm {i}}{128\,a^{5/2}}-\frac {B\,b^3\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,1{}\mathrm {i}}{16\,a^{3/2}} \] Input:

int(((A + B*x^2)*(a + b*x^2)^(3/2))/x^9,x)
 

Output:

(A*b^4*atan(((a + b*x^2)^(1/2)*1i)/a^(1/2))*3i)/(128*a^(5/2)) - (B*(a + b* 
x^2)^(3/2))/(6*x^6) - (11*A*(a + b*x^2)^(3/2))/(128*x^8) - (B*b^3*atan(((a 
 + b*x^2)^(1/2)*1i)/a^(1/2))*1i)/(16*a^(3/2)) + (3*A*a*(a + b*x^2)^(1/2))/ 
(128*x^8) + (B*a*(a + b*x^2)^(1/2))/(16*x^6) - (11*A*(a + b*x^2)^(5/2))/(1 
28*a*x^8) + (3*A*(a + b*x^2)^(7/2))/(128*a^2*x^8) - (B*(a + b*x^2)^(5/2))/ 
(16*a*x^6)
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.93 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^9} \, dx=\frac {-48 \sqrt {b \,x^{2}+a}\, a^{4}-136 \sqrt {b \,x^{2}+a}\, a^{3} b \,x^{2}-118 \sqrt {b \,x^{2}+a}\, a^{2} b^{2} x^{4}-15 \sqrt {b \,x^{2}+a}\, a \,b^{3} x^{6}-15 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}-\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{4} x^{8}+15 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{4} x^{8}}{384 a^{2} x^{8}} \] Input:

int((b*x^2+a)^(3/2)*(B*x^2+A)/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)