\(\int \frac {x^{5/2} (A+B x)}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\) [458]

Optimal result

Integrand size = 31, antiderivative size = 252 \[ \int \frac {x^{5/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {(5 A b-93 a B) \sqrt {x}}{64 a b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(A b-a B) x^{5/2}}{4 b^2 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(5 A b-13 a B) x^{3/2}}{24 b^3 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(5 A b-29 a B) \sqrt {x}}{32 b^4 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 (A b+7 a B) (a+b x) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{64 a^{3/2} b^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}} \] Output:

1/64*(5*A*b-93*B*a)*x^(1/2)/a/b^4/((b*x+a)^2)^(1/2)-1/4*(A*b-B*a)*x^(5/2)/ 
b^2/(b*x+a)^3/((b*x+a)^2)^(1/2)-1/24*(5*A*b-13*B*a)*x^(3/2)/b^3/(b*x+a)^2/ 
((b*x+a)^2)^(1/2)-1/32*(5*A*b-29*B*a)*x^(1/2)/b^4/(b*x+a)/((b*x+a)^2)^(1/2 
)+5/64*(A*b+7*B*a)*(b*x+a)*arctan(b^(1/2)*x^(1/2)/a^(1/2))/a^(3/2)/b^(9/2) 
/((b*x+a)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.58 \[ \int \frac {x^{5/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {-\sqrt {a} \sqrt {b} \sqrt {x} \left (105 a^4 B-15 A b^4 x^3+5 a^3 b (3 A+77 B x)+a b^3 x^2 (73 A+279 B x)+a^2 b^2 x (55 A+511 B x)\right )+15 (A b+7 a B) (a+b x)^4 \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{192 a^{3/2} b^{9/2} (a+b x)^3 \sqrt {(a+b x)^2}} \] Input:

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

Output:

(-(Sqrt[a]*Sqrt[b]*Sqrt[x]*(105*a^4*B - 15*A*b^4*x^3 + 5*a^3*b*(3*A + 77*B 
*x) + a*b^3*x^2*(73*A + 279*B*x) + a^2*b^2*x*(55*A + 511*B*x))) + 15*(A*b 
+ 7*a*B)*(a + b*x)^4*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(192*a^(3/2)*b^(9/ 
2)*(a + b*x)^3*Sqrt[(a + b*x)^2])
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.69, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.258, Rules used = {1187, 27, 87, 51, 51, 51, 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.

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

\(\Big \downarrow \) 1187

\(\displaystyle \frac {b^5 (a+b x) \int \frac {x^{5/2} (A+B x)}{b^5 (a+b x)^5}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(a+b x) \int \frac {x^{5/2} (A+B x)}{(a+b x)^5}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 51

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

\(\Big \downarrow \) 51

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

\(\Big \downarrow \) 51

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

((a + b*x)*(((A*b - a*B)*x^(7/2))/(4*a*b*(a + b*x)^4) + ((A*b + 7*a*B)*(-1 
/3*x^(5/2)/(b*(a + b*x)^3) + (5*(-1/2*x^(3/2)/(b*(a + b*x)^2) + (3*(-(Sqrt 
[x]/(b*(a + b*x))) + ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]]/(Sqrt[a]*b^(3/2)))) 
/(4*b)))/(6*b)))/(8*a*b)))/Sqrt[a^2 + 2*a*b*x + b^2*x^2]
 

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 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 1187

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_ 
) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(a + b*x + c*x^2)^FracPart[p]/(c^ 
IntPart[p]*(b/2 + c*x)^(2*FracPart[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, p}, x] && EqQ[b^2 
 - 4*a*c, 0] &&  !IntegerQ[p]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(356\) vs. \(2(169)=338\).

Time = 1.24 (sec) , antiderivative size = 357, normalized size of antiderivative = 1.42

method	result	size
default	\(\frac {\left (15 A \sqrt {a b}\, x^{\frac {7}{2}} b^{4}-279 B \sqrt {a b}\, x^{\frac {7}{2}} a \,b^{3}-73 A \sqrt {a b}\, x^{\frac {5}{2}} a \,b^{3}+15 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) b^{5} x^{4}-511 B \sqrt {a b}\, x^{\frac {5}{2}} a^{2} b^{2}+105 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a \,b^{4} x^{4}+60 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a \,b^{4} x^{3}+420 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a^{2} b^{3} x^{3}-55 A \sqrt {a b}\, x^{\frac {3}{2}} a^{2} b^{2}+90 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a^{2} b^{3} x^{2}-385 B \sqrt {a b}\, x^{\frac {3}{2}} a^{3} b +630 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a^{3} b^{2} x^{2}+60 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a^{3} b^{2} x +420 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a^{4} b x -15 A \sqrt {a b}\, \sqrt {x}\, a^{3} b +15 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a^{4} b -105 B \sqrt {a b}\, \sqrt {x}\, a^{4}+105 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a^{5}\right ) \left (b x +a \right )}{192 \sqrt {a b}\, b^{4} a \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\)	\(357\)

Input:

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

Output:

1/192*(15*A*(a*b)^(1/2)*x^(7/2)*b^4-279*B*(a*b)^(1/2)*x^(7/2)*a*b^3-73*A*( 
a*b)^(1/2)*x^(5/2)*a*b^3+15*A*arctan(b*x^(1/2)/(a*b)^(1/2))*b^5*x^4-511*B* 
(a*b)^(1/2)*x^(5/2)*a^2*b^2+105*B*arctan(b*x^(1/2)/(a*b)^(1/2))*a*b^4*x^4+ 
60*A*arctan(b*x^(1/2)/(a*b)^(1/2))*a*b^4*x^3+420*B*arctan(b*x^(1/2)/(a*b)^ 
(1/2))*a^2*b^3*x^3-55*A*(a*b)^(1/2)*x^(3/2)*a^2*b^2+90*A*arctan(b*x^(1/2)/ 
(a*b)^(1/2))*a^2*b^3*x^2-385*B*(a*b)^(1/2)*x^(3/2)*a^3*b+630*B*arctan(b*x^ 
(1/2)/(a*b)^(1/2))*a^3*b^2*x^2+60*A*arctan(b*x^(1/2)/(a*b)^(1/2))*a^3*b^2* 
x+420*B*arctan(b*x^(1/2)/(a*b)^(1/2))*a^4*b*x-15*A*(a*b)^(1/2)*x^(1/2)*a^3 
*b+15*A*arctan(b*x^(1/2)/(a*b)^(1/2))*a^4*b-105*B*(a*b)^(1/2)*x^(1/2)*a^4+ 
105*B*arctan(b*x^(1/2)/(a*b)^(1/2))*a^5)*(b*x+a)/(a*b)^(1/2)/b^4/a/((b*x+a 
)^2)^(5/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 525, normalized size of antiderivative = 2.08 \[ \int \frac {x^{5/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\left [-\frac {15 \, {\left (7 \, B a^{5} + A a^{4} b + {\left (7 \, B a b^{4} + A b^{5}\right )} x^{4} + 4 \, {\left (7 \, B a^{2} b^{3} + A a b^{4}\right )} x^{3} + 6 \, {\left (7 \, B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{2} + 4 \, {\left (7 \, B a^{4} b + A a^{3} b^{2}\right )} x\right )} \sqrt {-a b} \log \left (\frac {b x - a - 2 \, \sqrt {-a b} \sqrt {x}}{b x + a}\right ) + 2 \, {\left (105 \, B a^{5} b + 15 \, A a^{4} b^{2} + 3 \, {\left (93 \, B a^{2} b^{4} - 5 \, A a b^{5}\right )} x^{3} + 73 \, {\left (7 \, B a^{3} b^{3} + A a^{2} b^{4}\right )} x^{2} + 55 \, {\left (7 \, B a^{4} b^{2} + A a^{3} b^{3}\right )} x\right )} \sqrt {x}}{384 \, {\left (a^{2} b^{9} x^{4} + 4 \, a^{3} b^{8} x^{3} + 6 \, a^{4} b^{7} x^{2} + 4 \, a^{5} b^{6} x + a^{6} b^{5}\right )}}, -\frac {15 \, {\left (7 \, B a^{5} + A a^{4} b + {\left (7 \, B a b^{4} + A b^{5}\right )} x^{4} + 4 \, {\left (7 \, B a^{2} b^{3} + A a b^{4}\right )} x^{3} + 6 \, {\left (7 \, B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{2} + 4 \, {\left (7 \, B a^{4} b + A a^{3} b^{2}\right )} x\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b}}{b \sqrt {x}}\right ) + {\left (105 \, B a^{5} b + 15 \, A a^{4} b^{2} + 3 \, {\left (93 \, B a^{2} b^{4} - 5 \, A a b^{5}\right )} x^{3} + 73 \, {\left (7 \, B a^{3} b^{3} + A a^{2} b^{4}\right )} x^{2} + 55 \, {\left (7 \, B a^{4} b^{2} + A a^{3} b^{3}\right )} x\right )} \sqrt {x}}{192 \, {\left (a^{2} b^{9} x^{4} + 4 \, a^{3} b^{8} x^{3} + 6 \, a^{4} b^{7} x^{2} + 4 \, a^{5} b^{6} x + a^{6} b^{5}\right )}}\right ] \] Input:

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

Output:

[-1/384*(15*(7*B*a^5 + A*a^4*b + (7*B*a*b^4 + A*b^5)*x^4 + 4*(7*B*a^2*b^3 
+ A*a*b^4)*x^3 + 6*(7*B*a^3*b^2 + A*a^2*b^3)*x^2 + 4*(7*B*a^4*b + A*a^3*b^ 
2)*x)*sqrt(-a*b)*log((b*x - a - 2*sqrt(-a*b)*sqrt(x))/(b*x + a)) + 2*(105* 
B*a^5*b + 15*A*a^4*b^2 + 3*(93*B*a^2*b^4 - 5*A*a*b^5)*x^3 + 73*(7*B*a^3*b^ 
3 + A*a^2*b^4)*x^2 + 55*(7*B*a^4*b^2 + A*a^3*b^3)*x)*sqrt(x))/(a^2*b^9*x^4 
 + 4*a^3*b^8*x^3 + 6*a^4*b^7*x^2 + 4*a^5*b^6*x + a^6*b^5), -1/192*(15*(7*B 
*a^5 + A*a^4*b + (7*B*a*b^4 + A*b^5)*x^4 + 4*(7*B*a^2*b^3 + A*a*b^4)*x^3 + 
 6*(7*B*a^3*b^2 + A*a^2*b^3)*x^2 + 4*(7*B*a^4*b + A*a^3*b^2)*x)*sqrt(a*b)* 
arctan(sqrt(a*b)/(b*sqrt(x))) + (105*B*a^5*b + 15*A*a^4*b^2 + 3*(93*B*a^2* 
b^4 - 5*A*a*b^5)*x^3 + 73*(7*B*a^3*b^3 + A*a^2*b^4)*x^2 + 55*(7*B*a^4*b^2 
+ A*a^3*b^3)*x)*sqrt(x))/(a^2*b^9*x^4 + 4*a^3*b^8*x^3 + 6*a^4*b^7*x^2 + 4* 
a^5*b^6*x + a^6*b^5)]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 376 vs. \(2 (169) = 338\).

Time = 0.18 (sec) , antiderivative size = 376, normalized size of antiderivative = 1.49 \[ \int \frac {x^{5/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {5 \, {\left (7 \, {\left (9 \, B a b^{5} + A b^{6}\right )} x^{2} - 3 \, {\left (7 \, B a^{2} b^{4} + 3 \, A a b^{5}\right )} x\right )} x^{\frac {9}{2}} + 10 \, {\left (7 \, {\left (9 \, B a^{2} b^{4} + A a b^{5}\right )} x^{2} - 9 \, {\left (7 \, B a^{3} b^{3} + 3 \, A a^{2} b^{4}\right )} x\right )} x^{\frac {7}{2}} + 20 \, {\left (2 \, {\left (33 \, B a^{3} b^{3} - 7 \, A a^{2} b^{4}\right )} x^{2} - {\left (13 \, B a^{4} b^{2} + 33 \, A a^{3} b^{3}\right )} x\right )} x^{\frac {5}{2}} + 2 \, {\left (45 \, {\left (9 \, B a^{4} b^{2} + A a^{3} b^{3}\right )} x^{2} - 11 \, {\left (7 \, B a^{5} b + 3 \, A a^{4} b^{2}\right )} x\right )} x^{\frac {3}{2}} + {\left (21 \, {\left (9 \, B a^{5} b + A a^{4} b^{2}\right )} x^{2} - 5 \, {\left (7 \, B a^{6} + 3 \, A a^{5} b\right )} x\right )} \sqrt {x}}{1920 \, {\left (a^{3} b^{8} x^{5} + 5 \, a^{4} b^{7} x^{4} + 10 \, a^{5} b^{6} x^{3} + 10 \, a^{6} b^{5} x^{2} + 5 \, a^{7} b^{4} x + a^{8} b^{3}\right )}} + \frac {5 \, {\left (7 \, B a + A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{64 \, \sqrt {a b} a b^{4}} + \frac {7 \, {\left (9 \, B a b + A b^{2}\right )} x^{\frac {3}{2}} - 30 \, {\left (7 \, B a^{2} + A a b\right )} \sqrt {x}}{384 \, a^{3} b^{4}} \] Input:

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

Output:

-1/1920*(5*(7*(9*B*a*b^5 + A*b^6)*x^2 - 3*(7*B*a^2*b^4 + 3*A*a*b^5)*x)*x^( 
9/2) + 10*(7*(9*B*a^2*b^4 + A*a*b^5)*x^2 - 9*(7*B*a^3*b^3 + 3*A*a^2*b^4)*x 
)*x^(7/2) + 20*(2*(33*B*a^3*b^3 - 7*A*a^2*b^4)*x^2 - (13*B*a^4*b^2 + 33*A* 
a^3*b^3)*x)*x^(5/2) + 2*(45*(9*B*a^4*b^2 + A*a^3*b^3)*x^2 - 11*(7*B*a^5*b 
+ 3*A*a^4*b^2)*x)*x^(3/2) + (21*(9*B*a^5*b + A*a^4*b^2)*x^2 - 5*(7*B*a^6 + 
 3*A*a^5*b)*x)*sqrt(x))/(a^3*b^8*x^5 + 5*a^4*b^7*x^4 + 10*a^5*b^6*x^3 + 10 
*a^6*b^5*x^2 + 5*a^7*b^4*x + a^8*b^3) + 5/64*(7*B*a + A*b)*arctan(b*sqrt(x 
)/sqrt(a*b))/(sqrt(a*b)*a*b^4) + 1/384*(7*(9*B*a*b + A*b^2)*x^(3/2) - 30*( 
7*B*a^2 + A*a*b)*sqrt(x))/(a^3*b^4)
 

Giac [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.58 \[ \int \frac {x^{5/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {5 \, {\left (7 \, B a + A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{64 \, \sqrt {a b} a b^{4} \mathrm {sgn}\left (b x + a\right )} - \frac {279 \, B a b^{3} x^{\frac {7}{2}} - 15 \, A b^{4} x^{\frac {7}{2}} + 511 \, B a^{2} b^{2} x^{\frac {5}{2}} + 73 \, A a b^{3} x^{\frac {5}{2}} + 385 \, B a^{3} b x^{\frac {3}{2}} + 55 \, A a^{2} b^{2} x^{\frac {3}{2}} + 105 \, B a^{4} \sqrt {x} + 15 \, A a^{3} b \sqrt {x}}{192 \, {\left (b x + a\right )}^{4} a b^{4} \mathrm {sgn}\left (b x + a\right )} \] Input:

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

Output:

5/64*(7*B*a + A*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a*b^4*sgn(b*x + 
a)) - 1/192*(279*B*a*b^3*x^(7/2) - 15*A*b^4*x^(7/2) + 511*B*a^2*b^2*x^(5/2 
) + 73*A*a*b^3*x^(5/2) + 385*B*a^3*b*x^(3/2) + 55*A*a^2*b^2*x^(3/2) + 105* 
B*a^4*sqrt(x) + 15*A*a^3*b*sqrt(x))/((b*x + a)^4*a*b^4*sgn(b*x + a))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.65 \[ \int \frac {x^{5/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) a^{3}+45 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b x +45 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) a \,b^{2} x^{2}+15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) b^{3} x^{3}-15 \sqrt {x}\, a^{3} b -40 \sqrt {x}\, a^{2} b^{2} x -33 \sqrt {x}\, a \,b^{3} x^{2}}{24 a \,b^{4} \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 a^{2} b x +a^{3}\right )} \] Input:

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

Output:

(15*sqrt(b)*sqrt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*a**3 + 45*sqrt(b)* 
sqrt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*a**2*b*x + 45*sqrt(b)*sqrt(a)* 
atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*a*b**2*x**2 + 15*sqrt(b)*sqrt(a)*atan( 
(sqrt(x)*b)/(sqrt(b)*sqrt(a)))*b**3*x**3 - 15*sqrt(x)*a**3*b - 40*sqrt(x)* 
a**2*b**2*x - 33*sqrt(x)*a*b**3*x**2)/(24*a*b**4*(a**3 + 3*a**2*b*x + 3*a* 
b**2*x**2 + b**3*x**3))