\(\int \frac {(a+b x^2)^{3/2} (A+B x^2)}{x^{11}} \, dx\) [212]

Optimal result

Integrand size = 22, antiderivative size = 179 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^{11}} \, dx=-\frac {(3 A b+10 a B) \sqrt {a+b x^2}}{80 x^8}-\frac {b (A b+30 a B) \sqrt {a+b x^2}}{160 a x^6}+\frac {b^2 (A b-2 a B) \sqrt {a+b x^2}}{128 a^2 x^4}-\frac {3 b^3 (A b-2 a B) \sqrt {a+b x^2}}{256 a^3 x^2}-\frac {A \left (a+b x^2\right )^{3/2}}{10 x^{10}}+\frac {3 b^4 (A b-2 a B) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{256 a^{7/2}} \] Output:

-1/80*(3*A*b+10*B*a)*(b*x^2+a)^(1/2)/x^8-1/160*b*(A*b+30*B*a)*(b*x^2+a)^(1 
/2)/a/x^6+1/128*b^2*(A*b-2*B*a)*(b*x^2+a)^(1/2)/a^2/x^4-3/256*b^3*(A*b-2*B 
*a)*(b*x^2+a)^(1/2)/a^3/x^2-1/10*A*(b*x^2+a)^(3/2)/x^10+3/256*b^4*(A*b-2*B 
*a)*arctanh((b*x^2+a)^(1/2)/a^(1/2))/a^(7/2)
 

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.79 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^{11}} \, dx=-\frac {\sqrt {a+b x^2} \left (15 A b^4 x^8-10 a b^3 x^6 \left (A+3 B x^2\right )+4 a^2 b^2 x^4 \left (2 A+5 B x^2\right )+32 a^4 \left (4 A+5 B x^2\right )+16 a^3 b x^2 \left (11 A+15 B x^2\right )\right )}{1280 a^3 x^{10}}+\frac {3 b^4 (A b-2 a B) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{256 a^{7/2}} \] Input:

Integrate[((a + b*x^2)^(3/2)*(A + B*x^2))/x^11,x]
 

Output:

-1/1280*(Sqrt[a + b*x^2]*(15*A*b^4*x^8 - 10*a*b^3*x^6*(A + 3*B*x^2) + 4*a^ 
2*b^2*x^4*(2*A + 5*B*x^2) + 32*a^4*(4*A + 5*B*x^2) + 16*a^3*b*x^2*(11*A + 
15*B*x^2)))/(a^3*x^10) + (3*b^4*(A*b - 2*a*B)*ArcTanh[Sqrt[a + b*x^2]/Sqrt 
[a]])/(256*a^(7/2))
 

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.92, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {354, 87, 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.

\(\displaystyle \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^{11}} \, dx\)

\(\Big \downarrow \) 354

\(\displaystyle \frac {1}{2} \int \frac {\left (b x^2+a\right )^{3/2} \left (B x^2+A\right )}{x^{12}}dx^2\)

\(\Big \downarrow \) 87

\(\displaystyle \frac {1}{2} \left (-\frac {(A b-2 a B) \int \frac {\left (b x^2+a\right )^{3/2}}{x^{10}}dx^2}{2 a}-\frac {A \left (a+b x^2\right )^{5/2}}{5 a x^{10}}\right )\)

\(\Big \downarrow \) 51

\(\displaystyle \frac {1}{2} \left (-\frac {(A b-2 a B) \left (\frac {3}{8} b \int \frac {\sqrt {b x^2+a}}{x^8}dx^2-\frac {\left (a+b x^2\right )^{3/2}}{4 x^8}\right )}{2 a}-\frac {A \left (a+b x^2\right )^{5/2}}{5 a x^{10}}\right )\)

\(\Big \downarrow \) 51

\(\displaystyle \frac {1}{2} \left (-\frac {(A b-2 a B) \left (\frac {3}{8} b \left (\frac {1}{6} b \int \frac {1}{x^6 \sqrt {b x^2+a}}dx^2-\frac {\sqrt {a+b x^2}}{3 x^6}\right )-\frac {\left (a+b x^2\right )^{3/2}}{4 x^8}\right )}{2 a}-\frac {A \left (a+b x^2\right )^{5/2}}{5 a x^{10}}\right )\)

\(\Big \downarrow \) 52

\(\displaystyle \frac {1}{2} \left (-\frac {(A b-2 a B) \left (\frac {3}{8} b \left (\frac {1}{6} b \left (-\frac {3 b \int \frac {1}{x^4 \sqrt {b x^2+a}}dx^2}{4 a}-\frac {\sqrt {a+b x^2}}{2 a x^4}\right )-\frac {\sqrt {a+b x^2}}{3 x^6}\right )-\frac {\left (a+b x^2\right )^{3/2}}{4 x^8}\right )}{2 a}-\frac {A \left (a+b x^2\right )^{5/2}}{5 a x^{10}}\right )\)

\(\Big \downarrow \) 52

\(\displaystyle \frac {1}{2} \left (-\frac {(A b-2 a B) \left (\frac {3}{8} b \left (\frac {1}{6} b \left (-\frac {3 b \left (-\frac {b \int \frac {1}{x^2 \sqrt {b x^2+a}}dx^2}{2 a}-\frac {\sqrt {a+b x^2}}{a x^2}\right )}{4 a}-\frac {\sqrt {a+b x^2}}{2 a x^4}\right )-\frac {\sqrt {a+b x^2}}{3 x^6}\right )-\frac {\left (a+b x^2\right )^{3/2}}{4 x^8}\right )}{2 a}-\frac {A \left (a+b x^2\right )^{5/2}}{5 a x^{10}}\right )\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{2} \left (-\frac {(A b-2 a B) \left (\frac {3}{8} b \left (\frac {1}{6} b \left (-\frac {3 b \left (-\frac {\int \frac {1}{\frac {x^4}{b}-\frac {a}{b}}d\sqrt {b x^2+a}}{a}-\frac {\sqrt {a+b x^2}}{a x^2}\right )}{4 a}-\frac {\sqrt {a+b x^2}}{2 a x^4}\right )-\frac {\sqrt {a+b x^2}}{3 x^6}\right )-\frac {\left (a+b x^2\right )^{3/2}}{4 x^8}\right )}{2 a}-\frac {A \left (a+b x^2\right )^{5/2}}{5 a x^{10}}\right )\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {1}{2} \left (-\frac {(A b-2 a B) \left (\frac {3}{8} b \left (\frac {1}{6} b \left (-\frac {3 b \left (\frac {b \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {\sqrt {a+b x^2}}{a x^2}\right )}{4 a}-\frac {\sqrt {a+b x^2}}{2 a x^4}\right )-\frac {\sqrt {a+b x^2}}{3 x^6}\right )-\frac {\left (a+b x^2\right )^{3/2}}{4 x^8}\right )}{2 a}-\frac {A \left (a+b x^2\right )^{5/2}}{5 a x^{10}}\right )\)

Input:

Int[((a + b*x^2)^(3/2)*(A + B*x^2))/x^11,x]
 

Output:

(-1/5*(A*(a + b*x^2)^(5/2))/(a*x^10) - ((A*b - 2*a*B)*(-1/4*(a + b*x^2)^(3 
/2)/x^8 + (3*b*(-1/3*Sqrt[a + b*x^2]/x^6 + (b*(-1/2*Sqrt[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*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 = 0.46 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.73

method	result	size
pseudoelliptic	\(-\frac {-\frac {15 b^{4} x^{10} \left (A b -2 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{\sqrt {a}}\right )}{8}+\left (-\frac {5 b^{3} x^{6} \left (3 x^{2} B +A \right ) a^{\frac {3}{2}}}{4}+b^{2} x^{4} \left (\frac {5 x^{2} B}{2}+A \right ) a^{\frac {5}{2}}+22 \left (\frac {15 x^{2} B}{11}+A \right ) b \,x^{2} a^{\frac {7}{2}}+\left (20 x^{2} B +16 A \right ) a^{\frac {9}{2}}+\frac {15 A \sqrt {a}\, b^{4} x^{8}}{8}\right ) \sqrt {b \,x^{2}+a}}{160 a^{\frac {7}{2}} x^{10}}\)	\(131\)
risch	\(-\frac {\sqrt {b \,x^{2}+a}\, \left (15 A \,x^{8} b^{4}-30 B \,x^{8} a \,b^{3}-10 A \,x^{6} a \,b^{3}+20 B \,x^{6} a^{2} b^{2}+8 A \,x^{4} a^{2} b^{2}+240 B \,x^{4} a^{3} b +176 A \,x^{2} a^{3} b +160 B \,x^{2} a^{4}+128 A \,a^{4}\right )}{1280 x^{10} a^{3}}+\frac {3 \left (A b -2 B a \right ) b^{4} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{256 a^{\frac {7}{2}}}\)	\(147\)
default	\(A \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{10 a \,x^{10}}-\frac {b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{8 a \,x^{8}}-\frac {3 b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6 a \,x^{6}}-\frac {b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{4 a \,x^{4}}+\frac {b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{2 a \,x^{2}}+\frac {3 b \left (\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3}+a \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )\right )}{2 a}\right )}{4 a}\right )}{6 a}\right )}{8 a}\right )}{2 a}\right )+B \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{8 a \,x^{8}}-\frac {3 b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6 a \,x^{6}}-\frac {b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{4 a \,x^{4}}+\frac {b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{2 a \,x^{2}}+\frac {3 b \left (\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3}+a \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )\right )}{2 a}\right )}{4 a}\right )}{6 a}\right )}{8 a}\right )\)	\(326\)

Input:

int((b*x^2+a)^(3/2)*(B*x^2+A)/x^11,x,method=_RETURNVERBOSE)
 

Output:

-1/160*(-15/8*b^4*x^10*(A*b-2*B*a)*arctanh((b*x^2+a)^(1/2)/a^(1/2))+(-5/4* 
b^3*x^6*(3*B*x^2+A)*a^(3/2)+b^2*x^4*(5/2*x^2*B+A)*a^(5/2)+22*(15/11*x^2*B+ 
A)*b*x^2*a^(7/2)+(20*B*x^2+16*A)*a^(9/2)+15/8*A*a^(1/2)*b^4*x^8)*(b*x^2+a) 
^(1/2))/a^(7/2)/x^10
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 320, normalized size of antiderivative = 1.79 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^{11}} \, dx=\left [-\frac {15 \, {\left (2 \, B a b^{4} - A b^{5}\right )} \sqrt {a} x^{10} \log \left (-\frac {b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 2 \, {\left (15 \, {\left (2 \, B a^{2} b^{3} - A a b^{4}\right )} x^{8} - 10 \, {\left (2 \, B a^{3} b^{2} - A a^{2} b^{3}\right )} x^{6} - 128 \, A a^{5} - 8 \, {\left (30 \, B a^{4} b + A a^{3} b^{2}\right )} x^{4} - 16 \, {\left (10 \, B a^{5} + 11 \, A a^{4} b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{2560 \, a^{4} x^{10}}, \frac {15 \, {\left (2 \, B a b^{4} - A b^{5}\right )} \sqrt {-a} x^{10} \arctan \left (\frac {\sqrt {b x^{2} + a} \sqrt {-a}}{a}\right ) + {\left (15 \, {\left (2 \, B a^{2} b^{3} - A a b^{4}\right )} x^{8} - 10 \, {\left (2 \, B a^{3} b^{2} - A a^{2} b^{3}\right )} x^{6} - 128 \, A a^{5} - 8 \, {\left (30 \, B a^{4} b + A a^{3} b^{2}\right )} x^{4} - 16 \, {\left (10 \, B a^{5} + 11 \, A a^{4} b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{1280 \, a^{4} x^{10}}\right ] \] Input:

integrate((b*x^2+a)^(3/2)*(B*x^2+A)/x^11,x, algorithm="fricas")
 

Output:

[-1/2560*(15*(2*B*a*b^4 - A*b^5)*sqrt(a)*x^10*log(-(b*x^2 + 2*sqrt(b*x^2 + 
 a)*sqrt(a) + 2*a)/x^2) - 2*(15*(2*B*a^2*b^3 - A*a*b^4)*x^8 - 10*(2*B*a^3* 
b^2 - A*a^2*b^3)*x^6 - 128*A*a^5 - 8*(30*B*a^4*b + A*a^3*b^2)*x^4 - 16*(10 
*B*a^5 + 11*A*a^4*b)*x^2)*sqrt(b*x^2 + a))/(a^4*x^10), 1/1280*(15*(2*B*a*b 
^4 - A*b^5)*sqrt(-a)*x^10*arctan(sqrt(b*x^2 + a)*sqrt(-a)/a) + (15*(2*B*a^ 
2*b^3 - A*a*b^4)*x^8 - 10*(2*B*a^3*b^2 - A*a^2*b^3)*x^6 - 128*A*a^5 - 8*(3 
0*B*a^4*b + A*a^3*b^2)*x^4 - 16*(10*B*a^5 + 11*A*a^4*b)*x^2)*sqrt(b*x^2 + 
a))/(a^4*x^10)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^{11}} \, dx=\text {Timed out} \] Input:

integrate((b*x**2+a)**(3/2)*(B*x**2+A)/x**11,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.64 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^{11}} \, dx=-\frac {3 \, B b^{4} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{128 \, a^{\frac {5}{2}}} + \frac {3 \, A b^{5} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{256 \, a^{\frac {7}{2}}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B b^{4}}{128 \, a^{4}} + \frac {3 \, \sqrt {b x^{2} + a} B b^{4}}{128 \, a^{3}} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A b^{5}}{256 \, a^{5}} - \frac {3 \, \sqrt {b x^{2} + a} A b^{5}}{256 \, a^{4}} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B b^{3}}{128 \, a^{4} x^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A b^{4}}{256 \, a^{5} x^{2}} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B b^{2}}{64 \, a^{3} x^{4}} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A b^{3}}{128 \, a^{4} x^{4}} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B b}{16 \, a^{2} x^{6}} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A b^{2}}{32 \, a^{3} x^{6}} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B}{8 \, a x^{8}} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A b}{16 \, a^{2} x^{8}} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A}{10 \, a x^{10}} \] Input:

integrate((b*x^2+a)^(3/2)*(B*x^2+A)/x^11,x, algorithm="maxima")
 

Output:

-3/128*B*b^4*arcsinh(a/(sqrt(a*b)*abs(x)))/a^(5/2) + 3/256*A*b^5*arcsinh(a 
/(sqrt(a*b)*abs(x)))/a^(7/2) + 1/128*(b*x^2 + a)^(3/2)*B*b^4/a^4 + 3/128*s 
qrt(b*x^2 + a)*B*b^4/a^3 - 1/256*(b*x^2 + a)^(3/2)*A*b^5/a^5 - 3/256*sqrt( 
b*x^2 + a)*A*b^5/a^4 - 1/128*(b*x^2 + a)^(5/2)*B*b^3/(a^4*x^2) + 1/256*(b* 
x^2 + a)^(5/2)*A*b^4/(a^5*x^2) - 1/64*(b*x^2 + a)^(5/2)*B*b^2/(a^3*x^4) + 
1/128*(b*x^2 + a)^(5/2)*A*b^3/(a^4*x^4) + 1/16*(b*x^2 + a)^(5/2)*B*b/(a^2* 
x^6) - 1/32*(b*x^2 + a)^(5/2)*A*b^2/(a^3*x^6) - 1/8*(b*x^2 + a)^(5/2)*B/(a 
*x^8) + 1/16*(b*x^2 + a)^(5/2)*A*b/(a^2*x^8) - 1/10*(b*x^2 + a)^(5/2)*A/(a 
*x^10)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.05 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^{11}} \, dx=\frac {1}{1280} \, b^{5} {\left (\frac {15 \, {\left (2 \, B a - A b\right )} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{3} b} + \frac {30 \, {\left (b x^{2} + a\right )}^{\frac {9}{2}} B a - 140 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} B a^{2} + 140 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a^{4} - 30 \, \sqrt {b x^{2} + a} B a^{5} - 15 \, {\left (b x^{2} + a\right )}^{\frac {9}{2}} A b + 70 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} A a b - 128 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} A a^{2} b - 70 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A a^{3} b + 15 \, \sqrt {b x^{2} + a} A a^{4} b}{a^{3} b^{6} x^{10}}\right )} \] Input:

integrate((b*x^2+a)^(3/2)*(B*x^2+A)/x^11,x, algorithm="giac")
 

Output:

1/1280*b^5*(15*(2*B*a - A*b)*arctan(sqrt(b*x^2 + a)/sqrt(-a))/(sqrt(-a)*a^ 
3*b) + (30*(b*x^2 + a)^(9/2)*B*a - 140*(b*x^2 + a)^(7/2)*B*a^2 + 140*(b*x^ 
2 + a)^(3/2)*B*a^4 - 30*sqrt(b*x^2 + a)*B*a^5 - 15*(b*x^2 + a)^(9/2)*A*b + 
 70*(b*x^2 + a)^(7/2)*A*a*b - 128*(b*x^2 + a)^(5/2)*A*a^2*b - 70*(b*x^2 + 
a)^(3/2)*A*a^3*b + 15*sqrt(b*x^2 + a)*A*a^4*b)/(a^3*b^6*x^10))
 

Mupad [B] (verification not implemented)

Time = 4.08 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.15 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^{11}} \, dx=\frac {3\,A\,a\,\sqrt {b\,x^2+a}}{256\,x^{10}}-\frac {11\,B\,{\left (b\,x^2+a\right )}^{3/2}}{128\,x^8}-\frac {7\,A\,{\left (b\,x^2+a\right )}^{3/2}}{128\,x^{10}}+\frac {3\,B\,a\,\sqrt {b\,x^2+a}}{128\,x^8}-\frac {A\,{\left (b\,x^2+a\right )}^{5/2}}{10\,a\,x^{10}}+\frac {7\,A\,{\left (b\,x^2+a\right )}^{7/2}}{128\,a^2\,x^{10}}-\frac {3\,A\,{\left (b\,x^2+a\right )}^{9/2}}{256\,a^3\,x^{10}}-\frac {11\,B\,{\left (b\,x^2+a\right )}^{5/2}}{128\,a\,x^8}+\frac {3\,B\,{\left (b\,x^2+a\right )}^{7/2}}{128\,a^2\,x^8}-\frac {A\,b^5\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,3{}\mathrm {i}}{256\,a^{7/2}}+\frac {B\,b^4\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,3{}\mathrm {i}}{128\,a^{5/2}} \] Input:

int(((A + B*x^2)*(a + b*x^2)^(3/2))/x^11,x)
 

Output:

(B*b^4*atan(((a + b*x^2)^(1/2)*1i)/a^(1/2))*3i)/(128*a^(5/2)) - (11*B*(a + 
 b*x^2)^(3/2))/(128*x^8) - (A*b^5*atan(((a + b*x^2)^(1/2)*1i)/a^(1/2))*3i) 
/(256*a^(7/2)) - (7*A*(a + b*x^2)^(3/2))/(128*x^10) + (3*A*a*(a + b*x^2)^( 
1/2))/(256*x^10) + (3*B*a*(a + b*x^2)^(1/2))/(128*x^8) - (A*(a + b*x^2)^(5 
/2))/(10*a*x^10) + (7*A*(a + b*x^2)^(7/2))/(128*a^2*x^10) - (3*A*(a + b*x^ 
2)^(9/2))/(256*a^3*x^10) - (11*B*(a + b*x^2)^(5/2))/(128*a*x^8) + (3*B*(a 
+ b*x^2)^(7/2))/(128*a^2*x^8)
 

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.88 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^{11}} \, dx=\frac {-128 \sqrt {b \,x^{2}+a}\, a^{5}-336 \sqrt {b \,x^{2}+a}\, a^{4} b \,x^{2}-248 \sqrt {b \,x^{2}+a}\, a^{3} b^{2} x^{4}-10 \sqrt {b \,x^{2}+a}\, a^{2} b^{3} x^{6}+15 \sqrt {b \,x^{2}+a}\, a \,b^{4} x^{8}+15 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}-\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{5} x^{10}-15 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{5} x^{10}}{1280 a^{3} x^{10}} \] Input:

int((b*x^2+a)^(3/2)*(B*x^2+A)/x^11,x)
 

Output:

( - 128*sqrt(a + b*x**2)*a**5 - 336*sqrt(a + b*x**2)*a**4*b*x**2 - 248*sqr 
t(a + b*x**2)*a**3*b**2*x**4 - 10*sqrt(a + b*x**2)*a**2*b**3*x**6 + 15*sqr 
t(a + b*x**2)*a*b**4*x**8 + 15*sqrt(a)*log((sqrt(a + b*x**2) - sqrt(a) + s 
qrt(b)*x)/sqrt(a))*b**5*x**10 - 15*sqrt(a)*log((sqrt(a + b*x**2) + sqrt(a) 
 + sqrt(b)*x)/sqrt(a))*b**5*x**10)/(1280*a**3*x**10)