\(\int \frac {x^{5/2} (A+B x)}{(a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\) [448]

Optimal result

Integrand size = 31, antiderivative size = 238 \[ \int \frac {x^{5/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {(5 A b-9 a B) x^{3/2}}{4 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(A b-a B) x^{5/2}}{2 b^2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 (3 A b-7 a B) \sqrt {x} (a+b x)}{4 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 B x^{3/2} (a+b x)}{3 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 \sqrt {a} (3 A b-7 a B) (a+b x) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 b^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.53 \[ \int \frac {x^{5/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {(a+b x) \left (\sqrt {b} \sqrt {x} \left (-105 a^3 B+a b^2 x (75 A-56 B x)+5 a^2 b (9 A-35 B x)+8 b^3 x^2 (3 A+B x)\right )+15 \sqrt {a} (-3 A b+7 a B) (a+b x)^2 \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )\right )}{12 b^{9/2} \left ((a+b x)^2\right )^{3/2}} \] Input:

Integrate[(x^(5/2)*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]
 

Output:

((a + b*x)*(Sqrt[b]*Sqrt[x]*(-105*a^3*B + a*b^2*x*(75*A - 56*B*x) + 5*a^2* 
b*(9*A - 35*B*x) + 8*b^3*x^2*(3*A + B*x)) + 15*Sqrt[a]*(-3*A*b + 7*a*B)*(a 
 + b*x)^2*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]]))/(12*b^(9/2)*((a + b*x)^2)^(3 
/2))
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.67, 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, 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.

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

\(\Big \downarrow \) 1187

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 51

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 218

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

Input:

Int[(x^(5/2)*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]
 

Output:

((a + b*x)*(((A*b - a*B)*x^(7/2))/(2*a*b*(a + b*x)^2) - ((3*A*b - 7*a*B)*( 
-(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*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 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 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 [A] (verified)

Time = 1.11 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.55

method	result	size
risch	\(\frac {2 \left (B b x +3 A b -9 B a \right ) \sqrt {x}\, \sqrt {\left (b x +a \right )^{2}}}{3 b^{4} \left (b x +a \right )}-\frac {a \left (\frac {2 \left (-\frac {9}{8} b^{2} A +\frac {13}{8} a b B \right ) x^{\frac {3}{2}}-\frac {a \left (7 A b -11 B a \right ) \sqrt {x}}{4}}{\left (b x +a \right )^{2}}+\frac {5 \left (3 A b -7 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \sqrt {a b}}\right ) \sqrt {\left (b x +a \right )^{2}}}{b^{4} \left (b x +a \right )}\)	\(130\)
default	\(\frac {\left (8 B \sqrt {a b}\, x^{\frac {7}{2}} b^{3}-56 B \sqrt {a b}\, x^{\frac {5}{2}} a \,b^{2}+75 A \sqrt {a b}\, x^{\frac {3}{2}} a \,b^{2}+24 A \sqrt {a b}\, x^{\frac {5}{2}} b^{3}-45 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a \,b^{3} x^{2}-175 B \sqrt {a b}\, x^{\frac {3}{2}} a^{2} b +105 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a^{2} b^{2} x^{2}-90 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a^{2} b^{2} x +210 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a^{3} b x +45 A \sqrt {a b}\, \sqrt {x}\, a^{2} b -45 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a^{3} b -105 B \sqrt {a b}\, \sqrt {x}\, a^{3}+105 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a^{4}\right ) \left (b x +a \right )}{12 \sqrt {a b}\, b^{4} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\)	\(247\)

Input:

int(x^(5/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

2/3*(B*b*x+3*A*b-9*B*a)*x^(1/2)/b^4*((b*x+a)^2)^(1/2)/(b*x+a)-a/b^4*(2*((- 
9/8*b^2*A+13/8*a*b*B)*x^(3/2)-1/8*a*(7*A*b-11*B*a)*x^(1/2))/(b*x+a)^2+5/4* 
(3*A*b-7*B*a)/(a*b)^(1/2)*arctan(b*x^(1/2)/(a*b)^(1/2)))*((b*x+a)^2)^(1/2) 
/(b*x+a)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 349, normalized size of antiderivative = 1.47 \[ \int \frac {x^{5/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\left [-\frac {15 \, {\left (7 \, B a^{3} - 3 \, A a^{2} b + {\left (7 \, B a b^{2} - 3 \, A b^{3}\right )} x^{2} + 2 \, {\left (7 \, B a^{2} b - 3 \, A a b^{2}\right )} x\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x - 2 \, b \sqrt {x} \sqrt {-\frac {a}{b}} - a}{b x + a}\right ) - 2 \, {\left (8 \, B b^{3} x^{3} - 105 \, B a^{3} + 45 \, A a^{2} b - 8 \, {\left (7 \, B a b^{2} - 3 \, A b^{3}\right )} x^{2} - 25 \, {\left (7 \, B a^{2} b - 3 \, A a b^{2}\right )} x\right )} \sqrt {x}}{24 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}}, \frac {15 \, {\left (7 \, B a^{3} - 3 \, A a^{2} b + {\left (7 \, B a b^{2} - 3 \, A b^{3}\right )} x^{2} + 2 \, {\left (7 \, B a^{2} b - 3 \, A a b^{2}\right )} x\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {\frac {a}{b}}}{a}\right ) + {\left (8 \, B b^{3} x^{3} - 105 \, B a^{3} + 45 \, A a^{2} b - 8 \, {\left (7 \, B a b^{2} - 3 \, A b^{3}\right )} x^{2} - 25 \, {\left (7 \, B a^{2} b - 3 \, A a b^{2}\right )} x\right )} \sqrt {x}}{12 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}}\right ] \] Input:

integrate(x^(5/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas 
")
 

Output:

[-1/24*(15*(7*B*a^3 - 3*A*a^2*b + (7*B*a*b^2 - 3*A*b^3)*x^2 + 2*(7*B*a^2*b 
 - 3*A*a*b^2)*x)*sqrt(-a/b)*log((b*x - 2*b*sqrt(x)*sqrt(-a/b) - a)/(b*x + 
a)) - 2*(8*B*b^3*x^3 - 105*B*a^3 + 45*A*a^2*b - 8*(7*B*a*b^2 - 3*A*b^3)*x^ 
2 - 25*(7*B*a^2*b - 3*A*a*b^2)*x)*sqrt(x))/(b^6*x^2 + 2*a*b^5*x + a^2*b^4) 
, 1/12*(15*(7*B*a^3 - 3*A*a^2*b + (7*B*a*b^2 - 3*A*b^3)*x^2 + 2*(7*B*a^2*b 
 - 3*A*a*b^2)*x)*sqrt(a/b)*arctan(b*sqrt(x)*sqrt(a/b)/a) + (8*B*b^3*x^3 - 
105*B*a^3 + 45*A*a^2*b - 8*(7*B*a*b^2 - 3*A*b^3)*x^2 - 25*(7*B*a^2*b - 3*A 
*a*b^2)*x)*sqrt(x))/(b^6*x^2 + 2*a*b^5*x + a^2*b^4)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^{5/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\text {Timed out} \] Input:

integrate(x**(5/2)*(B*x+A)/(b**2*x**2+2*a*b*x+a**2)**(3/2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.06 \[ \int \frac {x^{5/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {{\left ({\left (89 \, B a b^{3} - 35 \, A b^{4}\right )} x^{2} + 3 \, {\left (19 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} x\right )} x^{\frac {5}{2}} + 12 \, {\left (4 \, {\left (3 \, B a^{2} b^{2} - A a b^{3}\right )} x^{2} + {\left (7 \, B a^{3} b - A a^{2} b^{2}\right )} x\right )} x^{\frac {3}{2}} + {\left (21 \, {\left (3 \, B a^{3} b - A a^{2} b^{2}\right )} x^{2} + 5 \, {\left (7 \, B a^{4} - A a^{3} b\right )} x\right )} \sqrt {x}}{24 \, {\left (a b^{6} x^{3} + 3 \, a^{2} b^{5} x^{2} + 3 \, a^{3} b^{4} x + a^{4} b^{3}\right )}} + \frac {5 \, {\left (7 \, B a^{2} - 3 \, A a b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} b^{4}} + \frac {5 \, {\left (7 \, {\left (3 \, B a b - A b^{2}\right )} x^{\frac {3}{2}} - 6 \, {\left (7 \, B a^{2} - 3 \, A a b\right )} \sqrt {x}\right )}}{24 \, a b^{4}} \] Input:

integrate(x^(5/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima 
")
 

Output:

-1/24*(((89*B*a*b^3 - 35*A*b^4)*x^2 + 3*(19*B*a^2*b^2 - 5*A*a*b^3)*x)*x^(5 
/2) + 12*(4*(3*B*a^2*b^2 - A*a*b^3)*x^2 + (7*B*a^3*b - A*a^2*b^2)*x)*x^(3/ 
2) + (21*(3*B*a^3*b - A*a^2*b^2)*x^2 + 5*(7*B*a^4 - A*a^3*b)*x)*sqrt(x))/( 
a*b^6*x^3 + 3*a^2*b^5*x^2 + 3*a^3*b^4*x + a^4*b^3) + 5/4*(7*B*a^2 - 3*A*a* 
b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^4) + 5/24*(7*(3*B*a*b - A*b^2) 
*x^(3/2) - 6*(7*B*a^2 - 3*A*a*b)*sqrt(x))/(a*b^4)
 

Giac [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.60 \[ \int \frac {x^{5/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {5 \, {\left (7 \, B a^{2} - 3 \, A a b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} b^{4} \mathrm {sgn}\left (b x + a\right )} - \frac {13 \, B a^{2} b x^{\frac {3}{2}} - 9 \, A a b^{2} x^{\frac {3}{2}} + 11 \, B a^{3} \sqrt {x} - 7 \, A a^{2} b \sqrt {x}}{4 \, {\left (b x + a\right )}^{2} b^{4} \mathrm {sgn}\left (b x + a\right )} + \frac {2 \, {\left (B b^{6} x^{\frac {3}{2}} - 9 \, B a b^{5} \sqrt {x} + 3 \, A b^{6} \sqrt {x}\right )}}{3 \, b^{9} \mathrm {sgn}\left (b x + a\right )} \] Input:

integrate(x^(5/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")
 

Output:

5/4*(7*B*a^2 - 3*A*a*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^4*sgn(b*x 
 + a)) - 1/4*(13*B*a^2*b*x^(3/2) - 9*A*a*b^2*x^(3/2) + 11*B*a^3*sqrt(x) - 
7*A*a^2*b*sqrt(x))/((b*x + a)^2*b^4*sgn(b*x + a)) + 2/3*(B*b^6*x^(3/2) - 9 
*B*a*b^5*sqrt(x) + 3*A*b^6*sqrt(x))/(b^9*sgn(b*x + a))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^{5/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {x^{5/2}\,\left (A+B\,x\right )}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \] Input:

int((x^(5/2)*(A + B*x))/(a^2 + b^2*x^2 + 2*a*b*x)^(3/2),x)
 

Output:

int((x^(5/2)*(A + B*x))/(a^2 + b^2*x^2 + 2*a*b*x)^(3/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.35 \[ \int \frac {x^{5/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) a^{2}+15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) a b x -15 \sqrt {x}\, a^{2} b -10 \sqrt {x}\, a \,b^{2} x +2 \sqrt {x}\, b^{3} x^{2}}{3 b^{4} \left (b x +a \right )} \] Input:

int(x^(5/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x)
 

Output:

(15*sqrt(b)*sqrt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*a**2 + 15*sqrt(b)* 
sqrt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*a*b*x - 15*sqrt(x)*a**2*b - 10 
*sqrt(x)*a*b**2*x + 2*sqrt(x)*b**3*x**2)/(3*b**4*(a + b*x))