\(\int \frac {A+B x}{x^{5/2} (a^2+2 a b x+b^2 x^2)^2} \, dx\) [401]

Optimal result

Integrand size = 29, antiderivative size = 159 \[ \int \frac {A+B x}{x^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=-\frac {2 A}{3 a^4 x^{3/2}}+\frac {2 (4 A b-a B)}{a^5 \sqrt {x}}+\frac {b (A b-a B) \sqrt {x}}{3 a^3 (a+b x)^3}+\frac {b (17 A b-11 a B) \sqrt {x}}{12 a^4 (a+b x)^2}+\frac {b (41 A b-19 a B) \sqrt {x}}{8 a^5 (a+b x)}+\frac {35 \sqrt {b} (3 A b-a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{8 a^{11/2}} \] Output:

-2/3*A/a^4/x^(3/2)+2*(4*A*b-B*a)/a^5/x^(1/2)+1/3*b*(A*b-B*a)*x^(1/2)/a^3/( 
b*x+a)^3+1/12*b*(17*A*b-11*B*a)*x^(1/2)/a^4/(b*x+a)^2+1/8*b*(41*A*b-19*B*a 
)*x^(1/2)/a^5/(b*x+a)+35/8*b^(1/2)*(3*A*b-B*a)*arctan(b^(1/2)*x^(1/2)/a^(1 
/2))/a^(11/2)
 

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.82 \[ \int \frac {A+B x}{x^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {315 A b^4 x^4+3 a^3 b x (48 A-77 B x)+7 a^2 b^2 x^2 (99 A-40 B x)-105 a b^3 x^3 (-8 A+B x)-16 a^4 (A+3 B x)}{24 a^5 x^{3/2} (a+b x)^3}-\frac {35 \sqrt {b} (-3 A b+a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{8 a^{11/2}} \] Input:

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

Output:

(315*A*b^4*x^4 + 3*a^3*b*x*(48*A - 77*B*x) + 7*a^2*b^2*x^2*(99*A - 40*B*x) 
 - 105*a*b^3*x^3*(-8*A + B*x) - 16*a^4*(A + 3*B*x))/(24*a^5*x^(3/2)*(a + b 
*x)^3) - (35*Sqrt[b]*(-3*A*b + a*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(8* 
a^(11/2))
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.01, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {1184, 27, 87, 52, 52, 61, 61, 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.

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

\(\Big \downarrow \) 1184

\(\displaystyle b^4 \int \frac {A+B x}{b^4 x^{5/2} (a+b x)^4}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \int \frac {A+B x}{x^{5/2} (a+b x)^4}dx\)

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 52

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

\(\Big \downarrow \) 52

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

(A*b - a*B)/(3*a*b*x^(3/2)*(a + b*x)^3) + ((3*A*b - a*B)*(1/(2*a*x^(3/2)*( 
a + b*x)^2) + (7*(1/(a*x^(3/2)*(a + b*x)) + (5*(-2/(3*a*x^(3/2)) - (b*(-2/ 
(a*Sqrt[x]) - (2*Sqrt[b]*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/a^(3/2)))/a))/ 
(2*a)))/(4*a)))/(2*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 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 61

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] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0 
] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[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 = 1.07 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.75

method	result	size
risch	\(-\frac {2 \left (-12 A b x +3 B a x +A a \right )}{3 a^{5} x^{\frac {3}{2}}}+\frac {b \left (\frac {2 \left (\frac {41}{16} A \,b^{3}-\frac {19}{16} B a \,b^{2}\right ) x^{\frac {5}{2}}+\frac {a b \left (35 A b -17 B a \right ) x^{\frac {3}{2}}}{3}+2 \left (\frac {55}{16} A \,a^{2} b -\frac {29}{16} B \,a^{3}\right ) \sqrt {x}}{\left (b x +a \right )^{3}}+\frac {35 \left (3 A b -B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{a^{5}}\)	\(119\)
derivativedivides	\(\frac {2 b \left (\frac {\left (\frac {41}{16} A \,b^{3}-\frac {19}{16} B a \,b^{2}\right ) x^{\frac {5}{2}}+\frac {a b \left (35 A b -17 B a \right ) x^{\frac {3}{2}}}{6}+\left (\frac {55}{16} A \,a^{2} b -\frac {29}{16} B \,a^{3}\right ) \sqrt {x}}{\left (b x +a \right )^{3}}+\frac {35 \left (3 A b -B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{16 \sqrt {a b}}\right )}{a^{5}}-\frac {2 A}{3 a^{4} x^{\frac {3}{2}}}-\frac {2 \left (-4 A b +B a \right )}{a^{5} \sqrt {x}}\)	\(122\)
default	\(\frac {2 b \left (\frac {\left (\frac {41}{16} A \,b^{3}-\frac {19}{16} B a \,b^{2}\right ) x^{\frac {5}{2}}+\frac {a b \left (35 A b -17 B a \right ) x^{\frac {3}{2}}}{6}+\left (\frac {55}{16} A \,a^{2} b -\frac {29}{16} B \,a^{3}\right ) \sqrt {x}}{\left (b x +a \right )^{3}}+\frac {35 \left (3 A b -B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{16 \sqrt {a b}}\right )}{a^{5}}-\frac {2 A}{3 a^{4} x^{\frac {3}{2}}}-\frac {2 \left (-4 A b +B a \right )}{a^{5} \sqrt {x}}\)	\(122\)

Input:

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

Output:

-2/3*(-12*A*b*x+3*B*a*x+A*a)/a^5/x^(3/2)+1/a^5*b*(2*((41/16*A*b^3-19/16*B* 
a*b^2)*x^(5/2)+1/6*a*b*(35*A*b-17*B*a)*x^(3/2)+(55/16*A*a^2*b-29/16*B*a^3) 
*x^(1/2))/(b*x+a)^3+35/8*(3*A*b-B*a)/(a*b)^(1/2)*arctan(b*x^(1/2)/(a*b)^(1 
/2)))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 477, normalized size of antiderivative = 3.00 \[ \int \frac {A+B x}{x^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\left [-\frac {105 \, {\left ({\left (B a b^{3} - 3 \, A b^{4}\right )} x^{5} + 3 \, {\left (B a^{2} b^{2} - 3 \, A a b^{3}\right )} x^{4} + 3 \, {\left (B a^{3} b - 3 \, A a^{2} b^{2}\right )} x^{3} + {\left (B a^{4} - 3 \, A a^{3} b\right )} x^{2}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x + 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - a}{b x + a}\right ) + 2 \, {\left (16 \, A a^{4} + 105 \, {\left (B a b^{3} - 3 \, A b^{4}\right )} x^{4} + 280 \, {\left (B a^{2} b^{2} - 3 \, A a b^{3}\right )} x^{3} + 231 \, {\left (B a^{3} b - 3 \, A a^{2} b^{2}\right )} x^{2} + 48 \, {\left (B a^{4} - 3 \, A a^{3} b\right )} x\right )} \sqrt {x}}{48 \, {\left (a^{5} b^{3} x^{5} + 3 \, a^{6} b^{2} x^{4} + 3 \, a^{7} b x^{3} + a^{8} x^{2}\right )}}, -\frac {105 \, {\left ({\left (B a b^{3} - 3 \, A b^{4}\right )} x^{5} + 3 \, {\left (B a^{2} b^{2} - 3 \, A a b^{3}\right )} x^{4} + 3 \, {\left (B a^{3} b - 3 \, A a^{2} b^{2}\right )} x^{3} + {\left (B a^{4} - 3 \, A a^{3} b\right )} x^{2}\right )} \sqrt {\frac {b}{a}} \arctan \left (\sqrt {x} \sqrt {\frac {b}{a}}\right ) + {\left (16 \, A a^{4} + 105 \, {\left (B a b^{3} - 3 \, A b^{4}\right )} x^{4} + 280 \, {\left (B a^{2} b^{2} - 3 \, A a b^{3}\right )} x^{3} + 231 \, {\left (B a^{3} b - 3 \, A a^{2} b^{2}\right )} x^{2} + 48 \, {\left (B a^{4} - 3 \, A a^{3} b\right )} x\right )} \sqrt {x}}{24 \, {\left (a^{5} b^{3} x^{5} + 3 \, a^{6} b^{2} x^{4} + 3 \, a^{7} b x^{3} + a^{8} x^{2}\right )}}\right ] \] Input:

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

Output:

[-1/48*(105*((B*a*b^3 - 3*A*b^4)*x^5 + 3*(B*a^2*b^2 - 3*A*a*b^3)*x^4 + 3*( 
B*a^3*b - 3*A*a^2*b^2)*x^3 + (B*a^4 - 3*A*a^3*b)*x^2)*sqrt(-b/a)*log((b*x 
+ 2*a*sqrt(x)*sqrt(-b/a) - a)/(b*x + a)) + 2*(16*A*a^4 + 105*(B*a*b^3 - 3* 
A*b^4)*x^4 + 280*(B*a^2*b^2 - 3*A*a*b^3)*x^3 + 231*(B*a^3*b - 3*A*a^2*b^2) 
*x^2 + 48*(B*a^4 - 3*A*a^3*b)*x)*sqrt(x))/(a^5*b^3*x^5 + 3*a^6*b^2*x^4 + 3 
*a^7*b*x^3 + a^8*x^2), -1/24*(105*((B*a*b^3 - 3*A*b^4)*x^5 + 3*(B*a^2*b^2 
- 3*A*a*b^3)*x^4 + 3*(B*a^3*b - 3*A*a^2*b^2)*x^3 + (B*a^4 - 3*A*a^3*b)*x^2 
)*sqrt(b/a)*arctan(sqrt(x)*sqrt(b/a)) + (16*A*a^4 + 105*(B*a*b^3 - 3*A*b^4 
)*x^4 + 280*(B*a^2*b^2 - 3*A*a*b^3)*x^3 + 231*(B*a^3*b - 3*A*a^2*b^2)*x^2 
+ 48*(B*a^4 - 3*A*a^3*b)*x)*sqrt(x))/(a^5*b^3*x^5 + 3*a^6*b^2*x^4 + 3*a^7* 
b*x^3 + a^8*x^2)]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.99 \[ \int \frac {A+B x}{x^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=-\frac {16 \, A a^{4} + 105 \, {\left (B a b^{3} - 3 \, A b^{4}\right )} x^{4} + 280 \, {\left (B a^{2} b^{2} - 3 \, A a b^{3}\right )} x^{3} + 231 \, {\left (B a^{3} b - 3 \, A a^{2} b^{2}\right )} x^{2} + 48 \, {\left (B a^{4} - 3 \, A a^{3} b\right )} x}{24 \, {\left (a^{5} b^{3} x^{\frac {9}{2}} + 3 \, a^{6} b^{2} x^{\frac {7}{2}} + 3 \, a^{7} b x^{\frac {5}{2}} + a^{8} x^{\frac {3}{2}}\right )}} - \frac {35 \, {\left (B a b - 3 \, A b^{2}\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{5}} \] Input:

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

Output:

-1/24*(16*A*a^4 + 105*(B*a*b^3 - 3*A*b^4)*x^4 + 280*(B*a^2*b^2 - 3*A*a*b^3 
)*x^3 + 231*(B*a^3*b - 3*A*a^2*b^2)*x^2 + 48*(B*a^4 - 3*A*a^3*b)*x)/(a^5*b 
^3*x^(9/2) + 3*a^6*b^2*x^(7/2) + 3*a^7*b*x^(5/2) + a^8*x^(3/2)) - 35/8*(B* 
a*b - 3*A*b^2)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^5)
 

Giac [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.86 \[ \int \frac {A+B x}{x^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=-\frac {35 \, {\left (B a b - 3 \, A b^{2}\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{5}} - \frac {105 \, B a b^{3} x^{4} - 315 \, A b^{4} x^{4} + 280 \, B a^{2} b^{2} x^{3} - 840 \, A a b^{3} x^{3} + 231 \, B a^{3} b x^{2} - 693 \, A a^{2} b^{2} x^{2} + 48 \, B a^{4} x - 144 \, A a^{3} b x + 16 \, A a^{4}}{24 \, {\left (b x^{\frac {3}{2}} + a \sqrt {x}\right )}^{3} a^{5}} \] Input:

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

Output:

-35/8*(B*a*b - 3*A*b^2)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^5) - 1/24 
*(105*B*a*b^3*x^4 - 315*A*b^4*x^4 + 280*B*a^2*b^2*x^3 - 840*A*a*b^3*x^3 + 
231*B*a^3*b*x^2 - 693*A*a^2*b^2*x^2 + 48*B*a^4*x - 144*A*a^3*b*x + 16*A*a^ 
4)/((b*x^(3/2) + a*sqrt(x))^3*a^5)
 

Mupad [B] (verification not implemented)

Time = 10.68 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.91 \[ \int \frac {A+B x}{x^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {\frac {2\,x\,\left (3\,A\,b-B\,a\right )}{a^2}-\frac {2\,A}{3\,a}+\frac {35\,b^2\,x^3\,\left (3\,A\,b-B\,a\right )}{3\,a^4}+\frac {35\,b^3\,x^4\,\left (3\,A\,b-B\,a\right )}{8\,a^5}+\frac {77\,b\,x^2\,\left (3\,A\,b-B\,a\right )}{8\,a^3}}{a^3\,x^{3/2}+b^3\,x^{9/2}+3\,a^2\,b\,x^{5/2}+3\,a\,b^2\,x^{7/2}}+\frac {35\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )\,\left (3\,A\,b-B\,a\right )}{8\,a^{11/2}} \] Input:

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

Output:

((2*x*(3*A*b - B*a))/a^2 - (2*A)/(3*a) + (35*b^2*x^3*(3*A*b - B*a))/(3*a^4 
) + (35*b^3*x^4*(3*A*b - B*a))/(8*a^5) + (77*b*x^2*(3*A*b - B*a))/(8*a^3)) 
/(a^3*x^(3/2) + b^3*x^(9/2) + 3*a^2*b*x^(5/2) + 3*a*b^2*x^(7/2)) + (35*b^( 
1/2)*atan((b^(1/2)*x^(1/2))/a^(1/2))*(3*A*b - B*a))/(8*a^(11/2))
 

Reduce [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.91 \[ \int \frac {A+B x}{x^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {105 \sqrt {x}\, \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b x +210 \sqrt {x}\, \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) a \,b^{2} x^{2}+105 \sqrt {x}\, \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) b^{3} x^{3}-8 a^{4}+56 a^{3} b x +175 a^{2} b^{2} x^{2}+105 a \,b^{3} x^{3}}{12 \sqrt {x}\, a^{5} x \left (b^{2} x^{2}+2 a b x +a^{2}\right )} \] Input:

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

Output:

(105*sqrt(x)*sqrt(b)*sqrt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*a**2*b*x 
+ 210*sqrt(x)*sqrt(b)*sqrt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*a*b**2*x 
**2 + 105*sqrt(x)*sqrt(b)*sqrt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*b**3 
*x**3 - 8*a**4 + 56*a**3*b*x + 175*a**2*b**2*x**2 + 105*a*b**3*x**3)/(12*s 
qrt(x)*a**5*x*(a**2 + 2*a*b*x + b**2*x**2))