\(\int \frac {x^{11/2} (A+B x)}{(a^2+2 a b x+b^2 x^2)^3} \, dx\) [402]

Optimal result

Integrand size = 29, antiderivative size = 214 \[ \int \frac {x^{11/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=\frac {231 (3 A b-13 a B) \sqrt {x}}{128 b^7}+\frac {2 B x^{3/2}}{3 b^6}-\frac {(A b-a B) x^{11/2}}{5 b^2 (a+b x)^5}-\frac {(11 A b-21 a B) x^{9/2}}{40 b^3 (a+b x)^4}-\frac {(99 A b-269 a B) x^{7/2}}{240 b^4 (a+b x)^3}-\frac {(693 A b-2363 a B) x^{5/2}}{960 b^5 (a+b x)^2}-\frac {(693 A b-2747 a B) x^{3/2}}{384 b^6 (a+b x)}-\frac {231 \sqrt {a} (3 A b-13 a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{128 b^{15/2}} \] Output:

231/128*(3*A*b-13*B*a)*x^(1/2)/b^7+2/3*B*x^(3/2)/b^6-1/5*(A*b-B*a)*x^(11/2 
)/b^2/(b*x+a)^5-1/40*(11*A*b-21*B*a)*x^(9/2)/b^3/(b*x+a)^4-1/240*(99*A*b-2 
69*B*a)*x^(7/2)/b^4/(b*x+a)^3-1/960*(693*A*b-2363*B*a)*x^(5/2)/b^5/(b*x+a) 
^2-1/384*(693*A*b-2747*B*a)*x^(3/2)/b^6/(b*x+a)-231/128*a^(1/2)*(3*A*b-13* 
B*a)*arctan(b^(1/2)*x^(1/2)/a^(1/2))/b^(15/2)
 

Mathematica [A] (verified)

Time = 0.53 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.79 \[ \int \frac {x^{11/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=\frac {\frac {\sqrt {b} \sqrt {x} \left (-45045 a^6 B+66 a^3 b^3 x^2 (1344 A-5135 B x)+5 a b^5 x^4 (6369 A-3328 B x)+55 a^2 b^4 x^3 (1422 A-2509 B x)+462 a^4 b^2 x (105 A-832 B x)+1155 a^5 b (9 A-182 B x)+1280 b^6 x^5 (3 A+B x)\right )}{(a+b x)^5}+3465 \sqrt {a} (-3 A b+13 a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{1920 b^{15/2}} \] Input:

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

Output:

((Sqrt[b]*Sqrt[x]*(-45045*a^6*B + 66*a^3*b^3*x^2*(1344*A - 5135*B*x) + 5*a 
*b^5*x^4*(6369*A - 3328*B*x) + 55*a^2*b^4*x^3*(1422*A - 2509*B*x) + 462*a^ 
4*b^2*x*(105*A - 832*B*x) + 1155*a^5*b*(9*A - 182*B*x) + 1280*b^6*x^5*(3*A 
 + B*x)))/(a + b*x)^5 + 3465*Sqrt[a]*(-3*A*b + 13*a*B)*ArcTan[(Sqrt[b]*Sqr 
t[x])/Sqrt[a]])/(1920*b^(15/2))
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.379, Rules used = {1184, 27, 87, 51, 51, 51, 51, 60, 60, 73, 218}

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.

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

\(\Big \downarrow \) 1184

\(\displaystyle b^6 \int \frac {x^{11/2} (A+B x)}{b^6 (a+b x)^6}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \int \frac {x^{11/2} (A+B x)}{(a+b x)^6}dx\)

\(\Big \downarrow \) 87

\(\displaystyle \frac {x^{13/2} (A b-a B)}{5 a b (a+b x)^5}-\frac {(3 A b-13 a B) \int \frac {x^{11/2}}{(a+b x)^5}dx}{10 a b}\)

\(\Big \downarrow \) 51

\(\displaystyle \frac {x^{13/2} (A b-a B)}{5 a b (a+b x)^5}-\frac {(3 A b-13 a B) \left (\frac {11 \int \frac {x^{9/2}}{(a+b x)^4}dx}{8 b}-\frac {x^{11/2}}{4 b (a+b x)^4}\right )}{10 a b}\)

\(\Big \downarrow \) 51

\(\displaystyle \frac {x^{13/2} (A b-a B)}{5 a b (a+b x)^5}-\frac {(3 A b-13 a B) \left (\frac {11 \left (\frac {3 \int \frac {x^{7/2}}{(a+b x)^3}dx}{2 b}-\frac {x^{9/2}}{3 b (a+b x)^3}\right )}{8 b}-\frac {x^{11/2}}{4 b (a+b x)^4}\right )}{10 a b}\)

\(\Big \downarrow \) 51

\(\displaystyle \frac {x^{13/2} (A b-a B)}{5 a b (a+b x)^5}-\frac {(3 A b-13 a B) \left (\frac {11 \left (\frac {3 \left (\frac {7 \int \frac {x^{5/2}}{(a+b x)^2}dx}{4 b}-\frac {x^{7/2}}{2 b (a+b x)^2}\right )}{2 b}-\frac {x^{9/2}}{3 b (a+b x)^3}\right )}{8 b}-\frac {x^{11/2}}{4 b (a+b x)^4}\right )}{10 a b}\)

\(\Big \downarrow \) 51

\(\displaystyle \frac {x^{13/2} (A b-a B)}{5 a b (a+b x)^5}-\frac {(3 A b-13 a B) \left (\frac {11 \left (\frac {3 \left (\frac {7 \left (\frac {5 \int \frac {x^{3/2}}{a+b x}dx}{2 b}-\frac {x^{5/2}}{b (a+b x)}\right )}{4 b}-\frac {x^{7/2}}{2 b (a+b x)^2}\right )}{2 b}-\frac {x^{9/2}}{3 b (a+b x)^3}\right )}{8 b}-\frac {x^{11/2}}{4 b (a+b x)^4}\right )}{10 a b}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {x^{13/2} (A b-a B)}{5 a b (a+b x)^5}-\frac {(3 A b-13 a B) \left (\frac {11 \left (\frac {3 \left (\frac {7 \left (\frac {5 \left (\frac {2 x^{3/2}}{3 b}-\frac {a \int \frac {\sqrt {x}}{a+b x}dx}{b}\right )}{2 b}-\frac {x^{5/2}}{b (a+b x)}\right )}{4 b}-\frac {x^{7/2}}{2 b (a+b x)^2}\right )}{2 b}-\frac {x^{9/2}}{3 b (a+b x)^3}\right )}{8 b}-\frac {x^{11/2}}{4 b (a+b x)^4}\right )}{10 a b}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {x^{13/2} (A b-a B)}{5 a b (a+b x)^5}-\frac {(3 A b-13 a B) \left (\frac {11 \left (\frac {3 \left (\frac {7 \left (\frac {5 \left (\frac {2 x^{3/2}}{3 b}-\frac {a \left (\frac {2 \sqrt {x}}{b}-\frac {a \int \frac {1}{\sqrt {x} (a+b x)}dx}{b}\right )}{b}\right )}{2 b}-\frac {x^{5/2}}{b (a+b x)}\right )}{4 b}-\frac {x^{7/2}}{2 b (a+b x)^2}\right )}{2 b}-\frac {x^{9/2}}{3 b (a+b x)^3}\right )}{8 b}-\frac {x^{11/2}}{4 b (a+b x)^4}\right )}{10 a b}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {x^{13/2} (A b-a B)}{5 a b (a+b x)^5}-\frac {(3 A b-13 a B) \left (\frac {11 \left (\frac {3 \left (\frac {7 \left (\frac {5 \left (\frac {2 x^{3/2}}{3 b}-\frac {a \left (\frac {2 \sqrt {x}}{b}-\frac {2 a \int \frac {1}{a+b x}d\sqrt {x}}{b}\right )}{b}\right )}{2 b}-\frac {x^{5/2}}{b (a+b x)}\right )}{4 b}-\frac {x^{7/2}}{2 b (a+b x)^2}\right )}{2 b}-\frac {x^{9/2}}{3 b (a+b x)^3}\right )}{8 b}-\frac {x^{11/2}}{4 b (a+b x)^4}\right )}{10 a b}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {x^{13/2} (A b-a B)}{5 a b (a+b x)^5}-\frac {(3 A b-13 a B) \left (\frac {11 \left (\frac {3 \left (\frac {7 \left (\frac {5 \left (\frac {2 x^{3/2}}{3 b}-\frac {a \left (\frac {2 \sqrt {x}}{b}-\frac {2 \sqrt {a} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{3/2}}\right )}{b}\right )}{2 b}-\frac {x^{5/2}}{b (a+b x)}\right )}{4 b}-\frac {x^{7/2}}{2 b (a+b x)^2}\right )}{2 b}-\frac {x^{9/2}}{3 b (a+b x)^3}\right )}{8 b}-\frac {x^{11/2}}{4 b (a+b x)^4}\right )}{10 a b}\)

Input:

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

Output:

((A*b - a*B)*x^(13/2))/(5*a*b*(a + b*x)^5) - ((3*A*b - 13*a*B)*(-1/4*x^(11 
/2)/(b*(a + b*x)^4) + (11*(-1/3*x^(9/2)/(b*(a + b*x)^3) + (3*(-1/2*x^(7/2) 
/(b*(a + b*x)^2) + (7*(-(x^(5/2)/(b*(a + b*x))) + (5*((2*x^(3/2))/(3*b) - 
(a*((2*Sqrt[x])/b - (2*Sqrt[a]*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/b^(3/2)) 
)/b))/(2*b)))/(4*b)))/(2*b)))/(8*b)))/(10*a*b)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

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 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 1184

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_ 
) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[1/c^p   Int[(d + e*x)^m*(f + g*x 
)^n*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x] && E 
qQ[b^2 - 4*a*c, 0] && IntegerQ[p]
 

Maple [A] (verified)

Time = 2.15 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.77

method	result	size
risch	\(\frac {2 \left (B b x +3 A b -18 B a \right ) \sqrt {x}}{3 b^{7}}-\frac {a \left (\frac {2 \left (-\frac {843}{256} A \,b^{5}+\frac {2373}{256} B a \,b^{4}\right ) x^{\frac {9}{2}}+2 \left (-\frac {1327}{128} A a \,b^{4}+\frac {12131}{384} B \,a^{2} b^{3}\right ) x^{\frac {7}{2}}+2 \left (-\frac {131}{10} A \,a^{2} b^{3}+\frac {1253}{30} B \,a^{3} b^{2}\right ) x^{\frac {5}{2}}-\frac {a^{3} b \left (2931 A b -9629 B a \right ) x^{\frac {3}{2}}}{192}+2 \left (-\frac {437}{256} A \,a^{4} b +\frac {1467}{256} B \,a^{5}\right ) \sqrt {x}}{\left (b x +a \right )^{5}}+\frac {231 \left (3 A b -13 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{128 \sqrt {a b}}\right )}{b^{7}}\)	\(165\)
derivativedivides	\(\frac {\frac {2 B b \,x^{\frac {3}{2}}}{3}+2 A b \sqrt {x}-12 B a \sqrt {x}}{b^{7}}-\frac {2 a \left (\frac {\left (-\frac {843}{256} A \,b^{5}+\frac {2373}{256} B a \,b^{4}\right ) x^{\frac {9}{2}}+\left (-\frac {1327}{128} A a \,b^{4}+\frac {12131}{384} B \,a^{2} b^{3}\right ) x^{\frac {7}{2}}+\left (-\frac {131}{10} A \,a^{2} b^{3}+\frac {1253}{30} B \,a^{3} b^{2}\right ) x^{\frac {5}{2}}-\frac {a^{3} b \left (2931 A b -9629 B a \right ) x^{\frac {3}{2}}}{384}+\left (-\frac {437}{256} A \,a^{4} b +\frac {1467}{256} B \,a^{5}\right ) \sqrt {x}}{\left (b x +a \right )^{5}}+\frac {231 \left (3 A b -13 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{256 \sqrt {a b}}\right )}{b^{7}}\)	\(169\)
default	\(\frac {\frac {2 B b \,x^{\frac {3}{2}}}{3}+2 A b \sqrt {x}-12 B a \sqrt {x}}{b^{7}}-\frac {2 a \left (\frac {\left (-\frac {843}{256} A \,b^{5}+\frac {2373}{256} B a \,b^{4}\right ) x^{\frac {9}{2}}+\left (-\frac {1327}{128} A a \,b^{4}+\frac {12131}{384} B \,a^{2} b^{3}\right ) x^{\frac {7}{2}}+\left (-\frac {131}{10} A \,a^{2} b^{3}+\frac {1253}{30} B \,a^{3} b^{2}\right ) x^{\frac {5}{2}}-\frac {a^{3} b \left (2931 A b -9629 B a \right ) x^{\frac {3}{2}}}{384}+\left (-\frac {437}{256} A \,a^{4} b +\frac {1467}{256} B \,a^{5}\right ) \sqrt {x}}{\left (b x +a \right )^{5}}+\frac {231 \left (3 A b -13 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{256 \sqrt {a b}}\right )}{b^{7}}\)	\(169\)

Input:

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

Output:

2/3*(B*b*x+3*A*b-18*B*a)*x^(1/2)/b^7-a/b^7*(2*((-843/256*A*b^5+2373/256*B* 
a*b^4)*x^(9/2)+(-1327/128*A*a*b^4+12131/384*B*a^2*b^3)*x^(7/2)+(-131/10*A* 
a^2*b^3+1253/30*B*a^3*b^2)*x^(5/2)-1/384*a^3*b*(2931*A*b-9629*B*a)*x^(3/2) 
+(-437/256*A*a^4*b+1467/256*B*a^5)*x^(1/2))/(b*x+a)^5+231/128*(3*A*b-13*B* 
a)/(a*b)^(1/2)*arctan(b*x^(1/2)/(a*b)^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 703, normalized size of antiderivative = 3.29 \[ \int \frac {x^{11/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx =\text {Too large to display} \] Input:

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

Output:

[-1/3840*(3465*(13*B*a^6 - 3*A*a^5*b + (13*B*a*b^5 - 3*A*b^6)*x^5 + 5*(13* 
B*a^2*b^4 - 3*A*a*b^5)*x^4 + 10*(13*B*a^3*b^3 - 3*A*a^2*b^4)*x^3 + 10*(13* 
B*a^4*b^2 - 3*A*a^3*b^3)*x^2 + 5*(13*B*a^5*b - 3*A*a^4*b^2)*x)*sqrt(-a/b)* 
log((b*x - 2*b*sqrt(x)*sqrt(-a/b) - a)/(b*x + a)) - 2*(1280*B*b^6*x^6 - 45 
045*B*a^6 + 10395*A*a^5*b - 1280*(13*B*a*b^5 - 3*A*b^6)*x^5 - 10615*(13*B* 
a^2*b^4 - 3*A*a*b^5)*x^4 - 26070*(13*B*a^3*b^3 - 3*A*a^2*b^4)*x^3 - 29568* 
(13*B*a^4*b^2 - 3*A*a^3*b^3)*x^2 - 16170*(13*B*a^5*b - 3*A*a^4*b^2)*x)*sqr 
t(x))/(b^12*x^5 + 5*a*b^11*x^4 + 10*a^2*b^10*x^3 + 10*a^3*b^9*x^2 + 5*a^4* 
b^8*x + a^5*b^7), 1/1920*(3465*(13*B*a^6 - 3*A*a^5*b + (13*B*a*b^5 - 3*A*b 
^6)*x^5 + 5*(13*B*a^2*b^4 - 3*A*a*b^5)*x^4 + 10*(13*B*a^3*b^3 - 3*A*a^2*b^ 
4)*x^3 + 10*(13*B*a^4*b^2 - 3*A*a^3*b^3)*x^2 + 5*(13*B*a^5*b - 3*A*a^4*b^2 
)*x)*sqrt(a/b)*arctan(b*sqrt(x)*sqrt(a/b)/a) + (1280*B*b^6*x^6 - 45045*B*a 
^6 + 10395*A*a^5*b - 1280*(13*B*a*b^5 - 3*A*b^6)*x^5 - 10615*(13*B*a^2*b^4 
 - 3*A*a*b^5)*x^4 - 26070*(13*B*a^3*b^3 - 3*A*a^2*b^4)*x^3 - 29568*(13*B*a 
^4*b^2 - 3*A*a^3*b^3)*x^2 - 16170*(13*B*a^5*b - 3*A*a^4*b^2)*x)*sqrt(x))/( 
b^12*x^5 + 5*a*b^11*x^4 + 10*a^2*b^10*x^3 + 10*a^3*b^9*x^2 + 5*a^4*b^8*x + 
 a^5*b^7)]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.08 \[ \int \frac {x^{11/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=-\frac {45 \, {\left (791 \, B a^{2} b^{4} - 281 \, A a b^{5}\right )} x^{\frac {9}{2}} + 10 \, {\left (12131 \, B a^{3} b^{3} - 3981 \, A a^{2} b^{4}\right )} x^{\frac {7}{2}} + 128 \, {\left (1253 \, B a^{4} b^{2} - 393 \, A a^{3} b^{3}\right )} x^{\frac {5}{2}} + 10 \, {\left (9629 \, B a^{5} b - 2931 \, A a^{4} b^{2}\right )} x^{\frac {3}{2}} + 15 \, {\left (1467 \, B a^{6} - 437 \, A a^{5} b\right )} \sqrt {x}}{1920 \, {\left (b^{12} x^{5} + 5 \, a b^{11} x^{4} + 10 \, a^{2} b^{10} x^{3} + 10 \, a^{3} b^{9} x^{2} + 5 \, a^{4} b^{8} x + a^{5} b^{7}\right )}} + \frac {231 \, {\left (13 \, B a^{2} - 3 \, A a b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{128 \, \sqrt {a b} b^{7}} + \frac {2 \, {\left (B b x^{\frac {3}{2}} - 3 \, {\left (6 \, B a - A b\right )} \sqrt {x}\right )}}{3 \, b^{7}} \] Input:

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

Output:

-1/1920*(45*(791*B*a^2*b^4 - 281*A*a*b^5)*x^(9/2) + 10*(12131*B*a^3*b^3 - 
3981*A*a^2*b^4)*x^(7/2) + 128*(1253*B*a^4*b^2 - 393*A*a^3*b^3)*x^(5/2) + 1 
0*(9629*B*a^5*b - 2931*A*a^4*b^2)*x^(3/2) + 15*(1467*B*a^6 - 437*A*a^5*b)* 
sqrt(x))/(b^12*x^5 + 5*a*b^11*x^4 + 10*a^2*b^10*x^3 + 10*a^3*b^9*x^2 + 5*a 
^4*b^8*x + a^5*b^7) + 231/128*(13*B*a^2 - 3*A*a*b)*arctan(b*sqrt(x)/sqrt(a 
*b))/(sqrt(a*b)*b^7) + 2/3*(B*b*x^(3/2) - 3*(6*B*a - A*b)*sqrt(x))/b^7
 

Giac [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.89 \[ \int \frac {x^{11/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=\frac {231 \, {\left (13 \, B a^{2} - 3 \, A a b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{128 \, \sqrt {a b} b^{7}} - \frac {35595 \, B a^{2} b^{4} x^{\frac {9}{2}} - 12645 \, A a b^{5} x^{\frac {9}{2}} + 121310 \, B a^{3} b^{3} x^{\frac {7}{2}} - 39810 \, A a^{2} b^{4} x^{\frac {7}{2}} + 160384 \, B a^{4} b^{2} x^{\frac {5}{2}} - 50304 \, A a^{3} b^{3} x^{\frac {5}{2}} + 96290 \, B a^{5} b x^{\frac {3}{2}} - 29310 \, A a^{4} b^{2} x^{\frac {3}{2}} + 22005 \, B a^{6} \sqrt {x} - 6555 \, A a^{5} b \sqrt {x}}{1920 \, {\left (b x + a\right )}^{5} b^{7}} + \frac {2 \, {\left (B b^{12} x^{\frac {3}{2}} - 18 \, B a b^{11} \sqrt {x} + 3 \, A b^{12} \sqrt {x}\right )}}{3 \, b^{18}} \] Input:

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

Output:

231/128*(13*B*a^2 - 3*A*a*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^7) - 
 1/1920*(35595*B*a^2*b^4*x^(9/2) - 12645*A*a*b^5*x^(9/2) + 121310*B*a^3*b^ 
3*x^(7/2) - 39810*A*a^2*b^4*x^(7/2) + 160384*B*a^4*b^2*x^(5/2) - 50304*A*a 
^3*b^3*x^(5/2) + 96290*B*a^5*b*x^(3/2) - 29310*A*a^4*b^2*x^(3/2) + 22005*B 
*a^6*sqrt(x) - 6555*A*a^5*b*sqrt(x))/((b*x + a)^5*b^7) + 2/3*(B*b^12*x^(3/ 
2) - 18*B*a*b^11*sqrt(x) + 3*A*b^12*sqrt(x))/b^18
 

Mupad [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.15 \[ \int \frac {x^{11/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=\sqrt {x}\,\left (\frac {2\,A}{b^6}-\frac {12\,B\,a}{b^7}\right )+\frac {x^{3/2}\,\left (\frac {977\,A\,a^4\,b^2}{64}-\frac {9629\,B\,a^5\,b}{192}\right )-x^{9/2}\,\left (\frac {2373\,B\,a^2\,b^4}{128}-\frac {843\,A\,a\,b^5}{128}\right )-\sqrt {x}\,\left (\frac {1467\,B\,a^6}{128}-\frac {437\,A\,a^5\,b}{128}\right )+x^{5/2}\,\left (\frac {131\,A\,a^3\,b^3}{5}-\frac {1253\,B\,a^4\,b^2}{15}\right )+x^{7/2}\,\left (\frac {1327\,A\,a^2\,b^4}{64}-\frac {12131\,B\,a^3\,b^3}{192}\right )}{a^5\,b^7+5\,a^4\,b^8\,x+10\,a^3\,b^9\,x^2+10\,a^2\,b^{10}\,x^3+5\,a\,b^{11}\,x^4+b^{12}\,x^5}+\frac {2\,B\,x^{3/2}}{3\,b^6}+\frac {231\,\sqrt {a}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {b}\,\sqrt {x}\,\left (3\,A\,b-13\,B\,a\right )}{13\,B\,a^2-3\,A\,a\,b}\right )\,\left (3\,A\,b-13\,B\,a\right )}{128\,b^{15/2}} \] Input:

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

Output:

x^(1/2)*((2*A)/b^6 - (12*B*a)/b^7) + (x^(3/2)*((977*A*a^4*b^2)/64 - (9629* 
B*a^5*b)/192) - x^(9/2)*((2373*B*a^2*b^4)/128 - (843*A*a*b^5)/128) - x^(1/ 
2)*((1467*B*a^6)/128 - (437*A*a^5*b)/128) + x^(5/2)*((131*A*a^3*b^3)/5 - ( 
1253*B*a^4*b^2)/15) + x^(7/2)*((1327*A*a^2*b^4)/64 - (12131*B*a^3*b^3)/192 
))/(a^5*b^7 + b^12*x^5 + 5*a^4*b^8*x + 5*a*b^11*x^4 + 10*a^3*b^9*x^2 + 10* 
a^2*b^10*x^3) + (2*B*x^(3/2))/(3*b^6) + (231*a^(1/2)*atan((a^(1/2)*b^(1/2) 
*x^(1/2)*(3*A*b - 13*B*a))/(13*B*a^2 - 3*A*a*b))*(3*A*b - 13*B*a))/(128*b^ 
(15/2))
 

Reduce [B] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.12 \[ \int \frac {x^{11/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=\frac {3465 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) a^{5}+13860 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) a^{4} b x +20790 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) a^{3} b^{2} x^{2}+13860 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b^{3} x^{3}+3465 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) a \,b^{4} x^{4}-3465 \sqrt {x}\, a^{5} b -12705 \sqrt {x}\, a^{4} b^{2} x -16863 \sqrt {x}\, a^{3} b^{3} x^{2}-9207 \sqrt {x}\, a^{2} b^{4} x^{3}-1408 \sqrt {x}\, a \,b^{5} x^{4}+128 \sqrt {x}\, b^{6} x^{5}}{192 b^{7} \left (b^{4} x^{4}+4 a \,b^{3} x^{3}+6 a^{2} b^{2} x^{2}+4 a^{3} b x +a^{4}\right )} \] Input:

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

Output:

(3465*sqrt(b)*sqrt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*a**5 + 13860*sqr 
t(b)*sqrt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*a**4*b*x + 20790*sqrt(b)* 
sqrt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*a**3*b**2*x**2 + 13860*sqrt(b) 
*sqrt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*a**2*b**3*x**3 + 3465*sqrt(b) 
*sqrt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*a*b**4*x**4 - 3465*sqrt(x)*a* 
*5*b - 12705*sqrt(x)*a**4*b**2*x - 16863*sqrt(x)*a**3*b**3*x**2 - 9207*sqr 
t(x)*a**2*b**4*x**3 - 1408*sqrt(x)*a*b**5*x**4 + 128*sqrt(x)*b**6*x**5)/(1 
92*b**7*(a**4 + 4*a**3*b*x + 6*a**2*b**2*x**2 + 4*a*b**3*x**3 + b**4*x**4) 
)