\(\int \frac {(a+b x^2)^{5/2} (A+B x^2)}{x^{11}} \, dx\) [232]

Optimal result

Integrand size = 22, antiderivative size = 179 \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^{11}} \, dx=-\frac {b (3 A b+22 a B) \sqrt {a+b x^2}}{96 x^6}-\frac {b^2 (3 A b+118 a B) \sqrt {a+b x^2}}{384 a x^4}+\frac {b^3 (3 A b-10 a B) \sqrt {a+b x^2}}{256 a^2 x^2}-\frac {(A b+2 a B) \left (a+b x^2\right )^{3/2}}{16 x^8}-\frac {A \left (a+b x^2\right )^{5/2}}{10 x^{10}}-\frac {b^4 (3 A b-10 a B) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{256 a^{5/2}} \] Output:

-1/96*b*(3*A*b+22*B*a)*(b*x^2+a)^(1/2)/x^6-1/384*b^2*(3*A*b+118*B*a)*(b*x^ 
2+a)^(1/2)/a/x^4+1/256*b^3*(3*A*b-10*B*a)*(b*x^2+a)^(1/2)/a^2/x^2-1/16*(A* 
b+2*B*a)*(b*x^2+a)^(3/2)/x^8-1/10*A*(b*x^2+a)^(5/2)/x^10-1/256*b^4*(3*A*b- 
10*B*a)*arctanh((b*x^2+a)^(1/2)/a^(1/2))/a^(5/2)
 

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.80 \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^{11}} \, dx=-\frac {\sqrt {a+b x^2} \left (-45 A b^4 x^8+30 a b^3 x^6 \left (A+5 B x^2\right )+96 a^4 \left (4 A+5 B x^2\right )+16 a^3 b x^2 \left (63 A+85 B x^2\right )+4 a^2 b^2 x^4 \left (186 A+295 B x^2\right )\right )}{3840 a^2 x^{10}}+\frac {b^4 (-3 A b+10 a B) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{256 a^{5/2}} \] Input:

Integrate[((a + b*x^2)^(5/2)*(A + B*x^2))/x^11,x]
 

Output:

-1/3840*(Sqrt[a + b*x^2]*(-45*A*b^4*x^8 + 30*a*b^3*x^6*(A + 5*B*x^2) + 96* 
a^4*(4*A + 5*B*x^2) + 16*a^3*b*x^2*(63*A + 85*B*x^2) + 4*a^2*b^2*x^4*(186* 
A + 295*B*x^2)))/(a^2*x^10) + (b^4*(-3*A*b + 10*a*B)*ArcTanh[Sqrt[a + b*x^ 
2]/Sqrt[a]])/(256*a^(5/2))
 

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.89, 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, 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.

\(\displaystyle \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^{11}} \, dx\)

\(\Big \downarrow \) 354

\(\displaystyle \frac {1}{2} \int \frac {\left (b x^2+a\right )^{5/2} \left (B x^2+A\right )}{x^{12}}dx^2\)

\(\Big \downarrow \) 87

\(\displaystyle \frac {1}{2} \left (-\frac {(3 A b-10 a B) \int \frac {\left (b x^2+a\right )^{5/2}}{x^{10}}dx^2}{10 a}-\frac {A \left (a+b x^2\right )^{7/2}}{5 a x^{10}}\right )\)

\(\Big \downarrow \) 51

\(\displaystyle \frac {1}{2} \left (-\frac {(3 A b-10 a B) \left (\frac {5}{8} b \int \frac {\left (b x^2+a\right )^{3/2}}{x^8}dx^2-\frac {\left (a+b x^2\right )^{5/2}}{4 x^8}\right )}{10 a}-\frac {A \left (a+b x^2\right )^{7/2}}{5 a x^{10}}\right )\)

\(\Big \downarrow \) 51

\(\displaystyle \frac {1}{2} \left (-\frac {(3 A b-10 a B) \left (\frac {5}{8} b \left (\frac {1}{2} b \int \frac {\sqrt {b x^2+a}}{x^6}dx^2-\frac {\left (a+b x^2\right )^{3/2}}{3 x^6}\right )-\frac {\left (a+b x^2\right )^{5/2}}{4 x^8}\right )}{10 a}-\frac {A \left (a+b x^2\right )^{7/2}}{5 a x^{10}}\right )\)

\(\Big \downarrow \) 51

\(\displaystyle \frac {1}{2} \left (-\frac {(3 A b-10 a B) \left (\frac {5}{8} b \left (\frac {1}{2} b \left (\frac {1}{4} b \int \frac {1}{x^4 \sqrt {b x^2+a}}dx^2-\frac {\sqrt {a+b x^2}}{2 x^4}\right )-\frac {\left (a+b x^2\right )^{3/2}}{3 x^6}\right )-\frac {\left (a+b x^2\right )^{5/2}}{4 x^8}\right )}{10 a}-\frac {A \left (a+b x^2\right )^{7/2}}{5 a x^{10}}\right )\)

\(\Big \downarrow \) 52

\(\displaystyle \frac {1}{2} \left (-\frac {(3 A b-10 a B) \left (\frac {5}{8} b \left (\frac {1}{2} b \left (\frac {1}{4} b \left (-\frac {b \int \frac {1}{x^2 \sqrt {b x^2+a}}dx^2}{2 a}-\frac {\sqrt {a+b x^2}}{a x^2}\right )-\frac {\sqrt {a+b x^2}}{2 x^4}\right )-\frac {\left (a+b x^2\right )^{3/2}}{3 x^6}\right )-\frac {\left (a+b x^2\right )^{5/2}}{4 x^8}\right )}{10 a}-\frac {A \left (a+b x^2\right )^{7/2}}{5 a x^{10}}\right )\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{2} \left (-\frac {(3 A b-10 a B) \left (\frac {5}{8} b \left (\frac {1}{2} b \left (\frac {1}{4} b \left (-\frac {\int \frac {1}{\frac {x^4}{b}-\frac {a}{b}}d\sqrt {b x^2+a}}{a}-\frac {\sqrt {a+b x^2}}{a x^2}\right )-\frac {\sqrt {a+b x^2}}{2 x^4}\right )-\frac {\left (a+b x^2\right )^{3/2}}{3 x^6}\right )-\frac {\left (a+b x^2\right )^{5/2}}{4 x^8}\right )}{10 a}-\frac {A \left (a+b x^2\right )^{7/2}}{5 a x^{10}}\right )\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {1}{2} \left (-\frac {(3 A b-10 a B) \left (\frac {5}{8} b \left (\frac {1}{2} b \left (\frac {1}{4} b \left (\frac {b \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {\sqrt {a+b x^2}}{a x^2}\right )-\frac {\sqrt {a+b x^2}}{2 x^4}\right )-\frac {\left (a+b x^2\right )^{3/2}}{3 x^6}\right )-\frac {\left (a+b x^2\right )^{5/2}}{4 x^8}\right )}{10 a}-\frac {A \left (a+b x^2\right )^{7/2}}{5 a x^{10}}\right )\)

Input:

Int[((a + b*x^2)^(5/2)*(A + B*x^2))/x^11,x]
 

Output:

(-1/5*(A*(a + b*x^2)^(7/2))/(a*x^10) - ((3*A*b - 10*a*B)*(-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))/(10*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 = 0.53 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.74

method	result	size
pseudoelliptic	\(-\frac {31 \left (\frac {15 b^{4} \left (A b -\frac {10 B a}{3}\right ) x^{10} \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{\sqrt {a}}\right )}{248}+\left (\frac {5 b^{3} x^{6} \left (5 x^{2} B +A \right ) a^{\frac {3}{2}}}{124}+b^{2} x^{4} \left (\frac {295 x^{2} B}{186}+A \right ) a^{\frac {5}{2}}+\frac {42 b \,x^{2} \left (\frac {85 x^{2} B}{63}+A \right ) a^{\frac {7}{2}}}{31}+\frac {4 \left (5 x^{2} B +4 A \right ) a^{\frac {9}{2}}}{31}-\frac {15 A \sqrt {a}\, b^{4} x^{8}}{248}\right ) \sqrt {b \,x^{2}+a}\right )}{160 a^{\frac {5}{2}} x^{10}}\)	\(132\)
risch	\(-\frac {\sqrt {b \,x^{2}+a}\, \left (-45 A \,x^{8} b^{4}+150 B \,x^{8} a \,b^{3}+30 A \,x^{6} a \,b^{3}+1180 B \,x^{6} a^{2} b^{2}+744 A \,x^{4} a^{2} b^{2}+1360 B \,x^{4} a^{3} b +1008 A \,x^{2} a^{3} b +480 B \,x^{2} a^{4}+384 A \,a^{4}\right )}{3840 x^{10} a^{2}}-\frac {\left (3 A b -10 B a \right ) b^{4} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{256 a^{\frac {5}{2}}}\)	\(148\)
default	\(A \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{10 a \,x^{10}}-\frac {3 b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{8 a \,x^{8}}-\frac {b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{6 a \,x^{6}}+\frac {b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{4 a \,x^{4}}+\frac {3 b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{2 a \,x^{2}}+\frac {5 b \left (\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{5}+a \left (\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3}+a \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )\right )\right )}{2 a}\right )}{4 a}\right )}{6 a}\right )}{8 a}\right )}{10 a}\right )+B \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{8 a \,x^{8}}-\frac {b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{6 a \,x^{6}}+\frac {b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{4 a \,x^{4}}+\frac {3 b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{2 a \,x^{2}}+\frac {5 b \left (\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{5}+a \left (\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3}+a \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )\right )\right )}{2 a}\right )}{4 a}\right )}{6 a}\right )}{8 a}\right )\)	\(354\)

Input:

int((b*x^2+a)^(5/2)*(B*x^2+A)/x^11,x,method=_RETURNVERBOSE)
 

Output:

-31/160/a^(5/2)*(15/248*b^4*(A*b-10/3*B*a)*x^10*arctanh((b*x^2+a)^(1/2)/a^ 
(1/2))+(5/124*b^3*x^6*(5*B*x^2+A)*a^(3/2)+b^2*x^4*(295/186*x^2*B+A)*a^(5/2 
)+42/31*b*x^2*(85/63*x^2*B+A)*a^(7/2)+4/31*(5*B*x^2+4*A)*a^(9/2)-15/248*A* 
a^(1/2)*b^4*x^8)*(b*x^2+a)^(1/2))/x^10
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 322, normalized size of antiderivative = 1.80 \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^{11}} \, dx=\left [-\frac {15 \, {\left (10 \, B a b^{4} - 3 \, A b^{5}\right )} \sqrt {a} x^{10} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (15 \, {\left (10 \, B a^{2} b^{3} - 3 \, A a b^{4}\right )} x^{8} + 10 \, {\left (118 \, B a^{3} b^{2} + 3 \, A a^{2} b^{3}\right )} x^{6} + 384 \, A a^{5} + 8 \, {\left (170 \, B a^{4} b + 93 \, A a^{3} b^{2}\right )} x^{4} + 48 \, {\left (10 \, B a^{5} + 21 \, A a^{4} b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{7680 \, a^{3} x^{10}}, -\frac {15 \, {\left (10 \, B a b^{4} - 3 \, A b^{5}\right )} \sqrt {-a} x^{10} \arctan \left (\frac {\sqrt {b x^{2} + a} \sqrt {-a}}{a}\right ) + {\left (15 \, {\left (10 \, B a^{2} b^{3} - 3 \, A a b^{4}\right )} x^{8} + 10 \, {\left (118 \, B a^{3} b^{2} + 3 \, A a^{2} b^{3}\right )} x^{6} + 384 \, A a^{5} + 8 \, {\left (170 \, B a^{4} b + 93 \, A a^{3} b^{2}\right )} x^{4} + 48 \, {\left (10 \, B a^{5} + 21 \, A a^{4} b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{3840 \, a^{3} x^{10}}\right ] \] Input:

integrate((b*x^2+a)^(5/2)*(B*x^2+A)/x^11,x, algorithm="fricas")
 

Output:

[-1/7680*(15*(10*B*a*b^4 - 3*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*(10*B*a^2*b^3 - 3*A*a*b^4)*x^8 + 10*(11 
8*B*a^3*b^2 + 3*A*a^2*b^3)*x^6 + 384*A*a^5 + 8*(170*B*a^4*b + 93*A*a^3*b^2 
)*x^4 + 48*(10*B*a^5 + 21*A*a^4*b)*x^2)*sqrt(b*x^2 + a))/(a^3*x^10), -1/38 
40*(15*(10*B*a*b^4 - 3*A*b^5)*sqrt(-a)*x^10*arctan(sqrt(b*x^2 + a)*sqrt(-a 
)/a) + (15*(10*B*a^2*b^3 - 3*A*a*b^4)*x^8 + 10*(118*B*a^3*b^2 + 3*A*a^2*b^ 
3)*x^6 + 384*A*a^5 + 8*(170*B*a^4*b + 93*A*a^3*b^2)*x^4 + 48*(10*B*a^5 + 2 
1*A*a^4*b)*x^2)*sqrt(b*x^2 + a))/(a^3*x^10)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^{11}} \, dx=\text {Timed out} \] Input:

integrate((b*x**2+a)**(5/2)*(B*x**2+A)/x**11,x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 330 vs. \(2 (151) = 302\).

Time = 0.05 (sec) , antiderivative size = 330, normalized size of antiderivative = 1.84 \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^{11}} \, dx=\frac {5 \, B b^{4} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{128 \, a^{\frac {3}{2}}} - \frac {3 \, A b^{5} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{256 \, a^{\frac {5}{2}}} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B b^{4}}{128 \, a^{4}} - \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B b^{4}}{384 \, a^{3}} - \frac {5 \, \sqrt {b x^{2} + a} B b^{4}}{128 \, a^{2}} + \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} A b^{5}}{1280 \, a^{5}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A b^{5}}{256 \, a^{4}} + \frac {3 \, \sqrt {b x^{2} + a} A b^{5}}{256 \, a^{3}} + \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} B b^{3}}{128 \, a^{4} x^{2}} - \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} A b^{4}}{1280 \, a^{5} x^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} B b^{2}}{192 \, a^{3} x^{4}} - \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} A b^{3}}{640 \, a^{4} x^{4}} + \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} B b}{48 \, a^{2} x^{6}} - \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} A b^{2}}{160 \, a^{3} x^{6}} - \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} B}{8 \, a x^{8}} + \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} A b}{80 \, a^{2} x^{8}} - \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} A}{10 \, a x^{10}} \] Input:

integrate((b*x^2+a)^(5/2)*(B*x^2+A)/x^11,x, algorithm="maxima")
 

Output:

5/128*B*b^4*arcsinh(a/(sqrt(a*b)*abs(x)))/a^(3/2) - 3/256*A*b^5*arcsinh(a/ 
(sqrt(a*b)*abs(x)))/a^(5/2) - 1/128*(b*x^2 + a)^(5/2)*B*b^4/a^4 - 5/384*(b 
*x^2 + a)^(3/2)*B*b^4/a^3 - 5/128*sqrt(b*x^2 + a)*B*b^4/a^2 + 3/1280*(b*x^ 
2 + a)^(5/2)*A*b^5/a^5 + 1/256*(b*x^2 + a)^(3/2)*A*b^5/a^4 + 3/256*sqrt(b* 
x^2 + a)*A*b^5/a^3 + 1/128*(b*x^2 + a)^(7/2)*B*b^3/(a^4*x^2) - 3/1280*(b*x 
^2 + a)^(7/2)*A*b^4/(a^5*x^2) + 1/192*(b*x^2 + a)^(7/2)*B*b^2/(a^3*x^4) - 
1/640*(b*x^2 + a)^(7/2)*A*b^3/(a^4*x^4) + 1/48*(b*x^2 + a)^(7/2)*B*b/(a^2* 
x^6) - 1/160*(b*x^2 + a)^(7/2)*A*b^2/(a^3*x^6) - 1/8*(b*x^2 + a)^(7/2)*B/( 
a*x^8) + 3/80*(b*x^2 + a)^(7/2)*A*b/(a^2*x^8) - 1/10*(b*x^2 + a)^(7/2)*A/( 
a*x^10)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.13 \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^{11}} \, dx=-\frac {1}{3840} \, b^{5} {\left (\frac {15 \, {\left (10 \, B a - 3 \, A b\right )} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2} b} + \frac {150 \, {\left (b x^{2} + a\right )}^{\frac {9}{2}} B a + 580 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} B a^{2} - 1280 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} B a^{3} + 700 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a^{4} - 150 \, \sqrt {b x^{2} + a} B a^{5} - 45 \, {\left (b x^{2} + a\right )}^{\frac {9}{2}} A b + 210 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} A a b + 384 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} A a^{2} b - 210 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A a^{3} b + 45 \, \sqrt {b x^{2} + a} A a^{4} b}{a^{2} b^{6} x^{10}}\right )} \] Input:

integrate((b*x^2+a)^(5/2)*(B*x^2+A)/x^11,x, algorithm="giac")
 

Output:

-1/3840*b^5*(15*(10*B*a - 3*A*b)*arctan(sqrt(b*x^2 + a)/sqrt(-a))/(sqrt(-a 
)*a^2*b) + (150*(b*x^2 + a)^(9/2)*B*a + 580*(b*x^2 + a)^(7/2)*B*a^2 - 1280 
*(b*x^2 + a)^(5/2)*B*a^3 + 700*(b*x^2 + a)^(3/2)*B*a^4 - 150*sqrt(b*x^2 + 
a)*B*a^5 - 45*(b*x^2 + a)^(9/2)*A*b + 210*(b*x^2 + a)^(7/2)*A*a*b + 384*(b 
*x^2 + a)^(5/2)*A*a^2*b - 210*(b*x^2 + a)^(3/2)*A*a^3*b + 45*sqrt(b*x^2 + 
a)*A*a^4*b)/(a^2*b^6*x^10))
 

Mupad [B] (verification not implemented)

Time = 5.34 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.15 \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^{11}} \, dx=\frac {7\,A\,a\,{\left (b\,x^2+a\right )}^{3/2}}{128\,x^{10}}-\frac {73\,B\,{\left (b\,x^2+a\right )}^{5/2}}{384\,x^8}-\frac {A\,{\left (b\,x^2+a\right )}^{5/2}}{10\,x^{10}}+\frac {55\,B\,a\,{\left (b\,x^2+a\right )}^{3/2}}{384\,x^8}-\frac {3\,A\,a^2\,\sqrt {b\,x^2+a}}{256\,x^{10}}-\frac {7\,A\,{\left (b\,x^2+a\right )}^{7/2}}{128\,a\,x^{10}}+\frac {3\,A\,{\left (b\,x^2+a\right )}^{9/2}}{256\,a^2\,x^{10}}-\frac {5\,B\,a^2\,\sqrt {b\,x^2+a}}{128\,x^8}-\frac {5\,B\,{\left (b\,x^2+a\right )}^{7/2}}{128\,a\,x^8}+\frac {A\,b^5\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,3{}\mathrm {i}}{256\,a^{5/2}}-\frac {B\,b^4\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,5{}\mathrm {i}}{128\,a^{3/2}} \] Input:

int(((A + B*x^2)*(a + b*x^2)^(5/2))/x^11,x)
 

Output:

(A*b^5*atan(((a + b*x^2)^(1/2)*1i)/a^(1/2))*3i)/(256*a^(5/2)) - (73*B*(a + 
 b*x^2)^(5/2))/(384*x^8) - (A*(a + b*x^2)^(5/2))/(10*x^10) - (B*b^4*atan(( 
(a + b*x^2)^(1/2)*1i)/a^(1/2))*5i)/(128*a^(3/2)) + (7*A*a*(a + b*x^2)^(3/2 
))/(128*x^10) + (55*B*a*(a + b*x^2)^(3/2))/(384*x^8) - (3*A*a^2*(a + b*x^2 
)^(1/2))/(256*x^10) - (7*A*(a + b*x^2)^(7/2))/(128*a*x^10) + (3*A*(a + b*x 
^2)^(9/2))/(256*a^2*x^10) - (5*B*a^2*(a + b*x^2)^(1/2))/(128*x^8) - (5*B*( 
a + b*x^2)^(7/2))/(128*a*x^8)
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.88 \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^{11}} \, dx=\frac {-384 \sqrt {b \,x^{2}+a}\, a^{5}-1488 \sqrt {b \,x^{2}+a}\, a^{4} b \,x^{2}-2104 \sqrt {b \,x^{2}+a}\, a^{3} b^{2} x^{4}-1210 \sqrt {b \,x^{2}+a}\, a^{2} b^{3} x^{6}-105 \sqrt {b \,x^{2}+a}\, a \,b^{4} x^{8}-105 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}-\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{5} x^{10}+105 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{5} x^{10}}{3840 a^{2} x^{10}} \] Input:

int((b*x^2+a)^(5/2)*(B*x^2+A)/x^11,x)
 

Output:

( - 384*sqrt(a + b*x**2)*a**5 - 1488*sqrt(a + b*x**2)*a**4*b*x**2 - 2104*s 
qrt(a + b*x**2)*a**3*b**2*x**4 - 1210*sqrt(a + b*x**2)*a**2*b**3*x**6 - 10 
5*sqrt(a + b*x**2)*a*b**4*x**8 - 105*sqrt(a)*log((sqrt(a + b*x**2) - sqrt( 
a) + sqrt(b)*x)/sqrt(a))*b**5*x**10 + 105*sqrt(a)*log((sqrt(a + b*x**2) + 
sqrt(a) + sqrt(b)*x)/sqrt(a))*b**5*x**10)/(3840*a**2*x**10)