\(\int \frac {\sqrt {x} (A+B x)}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\) [460]

Optimal result

Integrand size = 31, antiderivative size = 258 \[ \int \frac {\sqrt {x} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {(5 A b+3 a B) \sqrt {x}}{64 a^3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(A b-a B) \sqrt {x}}{4 b^2 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(A b-9 a B) \sqrt {x}}{24 a b^2 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(5 A b+3 a B) \sqrt {x}}{96 a^2 b^2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(5 A b+3 a B) (a+b x) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{64 a^{7/2} b^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}} \] Output:

1/64*(5*A*b+3*B*a)*x^(1/2)/a^3/b^2/((b*x+a)^2)^(1/2)-1/4*(A*b-B*a)*x^(1/2) 
/b^2/(b*x+a)^3/((b*x+a)^2)^(1/2)+1/24*(A*b-9*B*a)*x^(1/2)/a/b^2/(b*x+a)^2/ 
((b*x+a)^2)^(1/2)+1/96*(5*A*b+3*B*a)*x^(1/2)/a^2/b^2/(b*x+a)/((b*x+a)^2)^( 
1/2)+1/64*(5*A*b+3*B*a)*(b*x+a)*arctan(b^(1/2)*x^(1/2)/a^(1/2))/a^(7/2)/b^ 
(5/2)/((b*x+a)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.57 \[ \int \frac {\sqrt {x} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {\sqrt {a} \sqrt {b} \sqrt {x} \left (-9 a^4 B+15 A b^4 x^3+a b^3 x^2 (55 A+9 B x)-3 a^3 b (5 A+11 B x)+a^2 b^2 x (73 A+33 B x)\right )+3 (5 A b+3 a B) (a+b x)^4 \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{192 a^{7/2} b^{5/2} (a+b x)^3 \sqrt {(a+b x)^2}} \] Input:

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

Output:

(Sqrt[a]*Sqrt[b]*Sqrt[x]*(-9*a^4*B + 15*A*b^4*x^3 + a*b^3*x^2*(55*A + 9*B* 
x) - 3*a^3*b*(5*A + 11*B*x) + a^2*b^2*x*(73*A + 33*B*x)) + 3*(5*A*b + 3*a* 
B)*(a + b*x)^4*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(192*a^(7/2)*b^(5/2)*(a 
+ b*x)^3*Sqrt[(a + b*x)^2])
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.68, 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, 52, 52, 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.

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

\(\Big \downarrow \) 1187

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 51

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

\(\Big \downarrow \) 52

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

\(\Big \downarrow \) 52

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 218

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

Input:

Int[(Sqrt[x]*(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^(3/2))/(4*a*b*(a + b*x)^4) + ((5*A*b + 3*a*B)*( 
-1/3*Sqrt[x]/(b*(a + b*x)^3) + (Sqrt[x]/(2*a*(a + b*x)^2) + (3*(Sqrt[x]/(a 
*(a + b*x)) + ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]]/(a^(3/2)*Sqrt[b])))/(4*a)) 
/(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 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 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. \(2(175)=350\).

Time = 1.17 (sec) , antiderivative size = 357, normalized size of antiderivative = 1.38

method	result	size
default	\(\frac {\left (15 A \sqrt {a b}\, x^{\frac {7}{2}} b^{4}+9 B \sqrt {a b}\, x^{\frac {7}{2}} a \,b^{3}+55 A \sqrt {a b}\, x^{\frac {5}{2}} a \,b^{3}+15 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) b^{5} x^{4}+33 B \sqrt {a b}\, x^{\frac {5}{2}} a^{2} b^{2}+9 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a \,b^{4} x^{4}+60 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a \,b^{4} x^{3}+36 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a^{2} b^{3} x^{3}+73 A \sqrt {a b}\, x^{\frac {3}{2}} a^{2} b^{2}+90 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a^{2} b^{3} x^{2}-33 B \sqrt {a b}\, x^{\frac {3}{2}} a^{3} b +54 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a^{3} b^{2} x^{2}+60 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a^{3} b^{2} x +36 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a^{4} b x -15 A \sqrt {a b}\, \sqrt {x}\, a^{3} b +15 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a^{4} b -9 B \sqrt {a b}\, \sqrt {x}\, a^{4}+9 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a^{5}\right ) \left (b x +a \right )}{192 \sqrt {a b}\, b^{2} a^{3} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\)	\(357\)

Input:

int(x^(1/2)*(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+9*B*(a*b)^(1/2)*x^(7/2)*a*b^3+55*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+33*B*(a* 
b)^(1/2)*x^(5/2)*a^2*b^2+9*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+36*B*arctan(b*x^(1/2)/(a*b)^(1/2)) 
*a^2*b^3*x^3+73*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^2-33*B*(a*b)^(1/2)*x^(3/2)*a^3*b+54*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+36*B*a 
rctan(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*a 
rctan(b*x^(1/2)/(a*b)^(1/2))*a^4*b-9*B*(a*b)^(1/2)*x^(1/2)*a^4+9*B*arctan( 
b*x^(1/2)/(a*b)^(1/2))*a^5)*(b*x+a)/(a*b)^(1/2)/b^2/a^3/((b*x+a)^2)^(5/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 537, normalized size of antiderivative = 2.08 \[ \int \frac {\sqrt {x} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\left [-\frac {3 \, {\left (3 \, B a^{5} + 5 \, A a^{4} b + {\left (3 \, B a b^{4} + 5 \, A b^{5}\right )} x^{4} + 4 \, {\left (3 \, B a^{2} b^{3} + 5 \, A a b^{4}\right )} x^{3} + 6 \, {\left (3 \, B a^{3} b^{2} + 5 \, A a^{2} b^{3}\right )} x^{2} + 4 \, {\left (3 \, B a^{4} b + 5 \, A a^{3} b^{2}\right )} x\right )} \sqrt {-a b} \log \left (\frac {b x - a - 2 \, \sqrt {-a b} \sqrt {x}}{b x + a}\right ) + 2 \, {\left (9 \, B a^{5} b + 15 \, A a^{4} b^{2} - 3 \, {\left (3 \, B a^{2} b^{4} + 5 \, A a b^{5}\right )} x^{3} - 11 \, {\left (3 \, B a^{3} b^{3} + 5 \, A a^{2} b^{4}\right )} x^{2} + {\left (33 \, B a^{4} b^{2} - 73 \, A a^{3} b^{3}\right )} x\right )} \sqrt {x}}{384 \, {\left (a^{4} b^{7} x^{4} + 4 \, a^{5} b^{6} x^{3} + 6 \, a^{6} b^{5} x^{2} + 4 \, a^{7} b^{4} x + a^{8} b^{3}\right )}}, -\frac {3 \, {\left (3 \, B a^{5} + 5 \, A a^{4} b + {\left (3 \, B a b^{4} + 5 \, A b^{5}\right )} x^{4} + 4 \, {\left (3 \, B a^{2} b^{3} + 5 \, A a b^{4}\right )} x^{3} + 6 \, {\left (3 \, B a^{3} b^{2} + 5 \, A a^{2} b^{3}\right )} x^{2} + 4 \, {\left (3 \, B a^{4} b + 5 \, A a^{3} b^{2}\right )} x\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b}}{b \sqrt {x}}\right ) + {\left (9 \, B a^{5} b + 15 \, A a^{4} b^{2} - 3 \, {\left (3 \, B a^{2} b^{4} + 5 \, A a b^{5}\right )} x^{3} - 11 \, {\left (3 \, B a^{3} b^{3} + 5 \, A a^{2} b^{4}\right )} x^{2} + {\left (33 \, B a^{4} b^{2} - 73 \, A a^{3} b^{3}\right )} x\right )} \sqrt {x}}{192 \, {\left (a^{4} b^{7} x^{4} + 4 \, a^{5} b^{6} x^{3} + 6 \, a^{6} b^{5} x^{2} + 4 \, a^{7} b^{4} x + a^{8} b^{3}\right )}}\right ] \] Input:

integrate(x^(1/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fricas 
")
 

Output:

[-1/384*(3*(3*B*a^5 + 5*A*a^4*b + (3*B*a*b^4 + 5*A*b^5)*x^4 + 4*(3*B*a^2*b 
^3 + 5*A*a*b^4)*x^3 + 6*(3*B*a^3*b^2 + 5*A*a^2*b^3)*x^2 + 4*(3*B*a^4*b + 5 
*A*a^3*b^2)*x)*sqrt(-a*b)*log((b*x - a - 2*sqrt(-a*b)*sqrt(x))/(b*x + a)) 
+ 2*(9*B*a^5*b + 15*A*a^4*b^2 - 3*(3*B*a^2*b^4 + 5*A*a*b^5)*x^3 - 11*(3*B* 
a^3*b^3 + 5*A*a^2*b^4)*x^2 + (33*B*a^4*b^2 - 73*A*a^3*b^3)*x)*sqrt(x))/(a^ 
4*b^7*x^4 + 4*a^5*b^6*x^3 + 6*a^6*b^5*x^2 + 4*a^7*b^4*x + a^8*b^3), -1/192 
*(3*(3*B*a^5 + 5*A*a^4*b + (3*B*a*b^4 + 5*A*b^5)*x^4 + 4*(3*B*a^2*b^3 + 5* 
A*a*b^4)*x^3 + 6*(3*B*a^3*b^2 + 5*A*a^2*b^3)*x^2 + 4*(3*B*a^4*b + 5*A*a^3* 
b^2)*x)*sqrt(a*b)*arctan(sqrt(a*b)/(b*sqrt(x))) + (9*B*a^5*b + 15*A*a^4*b^ 
2 - 3*(3*B*a^2*b^4 + 5*A*a*b^5)*x^3 - 11*(3*B*a^3*b^3 + 5*A*a^2*b^4)*x^2 + 
 (33*B*a^4*b^2 - 73*A*a^3*b^3)*x)*sqrt(x))/(a^4*b^7*x^4 + 4*a^5*b^6*x^3 + 
6*a^6*b^5*x^2 + 4*a^7*b^4*x + a^8*b^3)]
 

Sympy [F(-1)]

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

integrate(x**(1/2)*(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. 368 vs. \(2 (176) = 352\).

Time = 0.19 (sec) , antiderivative size = 368, normalized size of antiderivative = 1.43 \[ \int \frac {\sqrt {x} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {15 \, {\left ({\left (B a b^{5} + A b^{6}\right )} x^{2} - {\left (3 \, B a^{2} b^{4} + 7 \, A a b^{5}\right )} x\right )} x^{\frac {9}{2}} + 30 \, {\left ({\left (B a^{2} b^{4} + A a b^{5}\right )} x^{2} - 3 \, {\left (3 \, B a^{3} b^{3} + 7 \, A a^{2} b^{4}\right )} x\right )} x^{\frac {7}{2}} - 20 \, {\left (6 \, {\left (B a^{3} b^{3} + A a^{2} b^{4}\right )} x^{2} + 11 \, {\left (3 \, B a^{4} b^{2} + 7 \, A a^{3} b^{3}\right )} x\right )} x^{\frac {5}{2}} - 2 \, {\left (255 \, {\left (B a^{4} b^{2} + A a^{3} b^{3}\right )} x^{2} + 139 \, {\left (3 \, B a^{5} b + 7 \, A a^{4} b^{2}\right )} x\right )} x^{\frac {3}{2}} + {\left (3 \, {\left (3 \, B a^{5} b - 253 \, A a^{4} b^{2}\right )} x^{2} - 5 \, {\left (3 \, B a^{6} + 263 \, A a^{5} b\right )} x\right )} \sqrt {x}}{1920 \, {\left (a^{5} b^{6} x^{5} + 5 \, a^{6} b^{5} x^{4} + 10 \, a^{7} b^{4} x^{3} + 10 \, a^{8} b^{3} x^{2} + 5 \, a^{9} b^{2} x + a^{10} b\right )}} + \frac {{\left (3 \, B a + 5 \, A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{64 \, \sqrt {a b} a^{3} b^{2}} + \frac {{\left (B a b + A b^{2}\right )} x^{\frac {3}{2}} - 2 \, {\left (3 \, B a^{2} + 5 \, A a b\right )} \sqrt {x}}{128 \, a^{5} b^{2}} \] Input:

integrate(x^(1/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="maxima 
")
 

Output:

-1/1920*(15*((B*a*b^5 + A*b^6)*x^2 - (3*B*a^2*b^4 + 7*A*a*b^5)*x)*x^(9/2) 
+ 30*((B*a^2*b^4 + A*a*b^5)*x^2 - 3*(3*B*a^3*b^3 + 7*A*a^2*b^4)*x)*x^(7/2) 
 - 20*(6*(B*a^3*b^3 + A*a^2*b^4)*x^2 + 11*(3*B*a^4*b^2 + 7*A*a^3*b^3)*x)*x 
^(5/2) - 2*(255*(B*a^4*b^2 + A*a^3*b^3)*x^2 + 139*(3*B*a^5*b + 7*A*a^4*b^2 
)*x)*x^(3/2) + (3*(3*B*a^5*b - 253*A*a^4*b^2)*x^2 - 5*(3*B*a^6 + 263*A*a^5 
*b)*x)*sqrt(x))/(a^5*b^6*x^5 + 5*a^6*b^5*x^4 + 10*a^7*b^4*x^3 + 10*a^8*b^3 
*x^2 + 5*a^9*b^2*x + a^10*b) + 1/64*(3*B*a + 5*A*b)*arctan(b*sqrt(x)/sqrt( 
a*b))/(sqrt(a*b)*a^3*b^2) + 1/128*((B*a*b + A*b^2)*x^(3/2) - 2*(3*B*a^2 + 
5*A*a*b)*sqrt(x))/(a^5*b^2)
 

Giac [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.57 \[ \int \frac {\sqrt {x} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {{\left (3 \, B a + 5 \, A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{64 \, \sqrt {a b} a^{3} b^{2} \mathrm {sgn}\left (b x + a\right )} + \frac {9 \, B a b^{3} x^{\frac {7}{2}} + 15 \, A b^{4} x^{\frac {7}{2}} + 33 \, B a^{2} b^{2} x^{\frac {5}{2}} + 55 \, A a b^{3} x^{\frac {5}{2}} - 33 \, B a^{3} b x^{\frac {3}{2}} + 73 \, A a^{2} b^{2} x^{\frac {3}{2}} - 9 \, B a^{4} \sqrt {x} - 15 \, A a^{3} b \sqrt {x}}{192 \, {\left (b x + a\right )}^{4} a^{3} b^{2} \mathrm {sgn}\left (b x + a\right )} \] Input:

integrate(x^(1/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac")
 

Output:

1/64*(3*B*a + 5*A*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^3*b^2*sgn(b* 
x + a)) + 1/192*(9*B*a*b^3*x^(7/2) + 15*A*b^4*x^(7/2) + 33*B*a^2*b^2*x^(5/ 
2) + 55*A*a*b^3*x^(5/2) - 33*B*a^3*b*x^(3/2) + 73*A*a^2*b^2*x^(3/2) - 9*B* 
a^4*sqrt(x) - 15*A*a^3*b*sqrt(x))/((b*x + a)^4*a^3*b^2*sgn(b*x + a))
 

Mupad [F(-1)]

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

int((x^(1/2)*(A + B*x))/(a^2 + b^2*x^2 + 2*a*b*x)^(5/2),x)
 

Output:

int((x^(1/2)*(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 = 0.64 \[ \int \frac {\sqrt {x} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {3 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) a^{3}+9 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b x +9 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) a \,b^{2} x^{2}+3 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) b^{3} x^{3}-3 \sqrt {x}\, a^{3} b +8 \sqrt {x}\, a^{2} b^{2} x +3 \sqrt {x}\, a \,b^{3} x^{2}}{24 a^{3} b^{2} \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 a^{2} b x +a^{3}\right )} \] Input:

int(x^(1/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x)
 

Output:

(3*sqrt(b)*sqrt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*a**3 + 9*sqrt(b)*sq 
rt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*a**2*b*x + 9*sqrt(b)*sqrt(a)*ata 
n((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*a*b**2*x**2 + 3*sqrt(b)*sqrt(a)*atan((sqr 
t(x)*b)/(sqrt(b)*sqrt(a)))*b**3*x**3 - 3*sqrt(x)*a**3*b + 8*sqrt(x)*a**2*b 
**2*x + 3*sqrt(x)*a*b**3*x**2)/(24*a**3*b**2*(a**3 + 3*a**2*b*x + 3*a*b**2 
*x**2 + b**3*x**3))