\(\int \frac {A+B x}{x^{3/2} (a^2+2 a b x+b^2 x^2)^2} \, dx\) [400]

Optimal result

Integrand size = 29, antiderivative size = 135 \[ \int \frac {A+B x}{x^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=-\frac {2 A}{a^4 \sqrt {x}}-\frac {(A b-a B) \sqrt {x}}{3 a^2 (a+b x)^3}-\frac {(11 A b-5 a B) \sqrt {x}}{12 a^3 (a+b x)^2}-\frac {(19 A b-5 a B) \sqrt {x}}{8 a^4 (a+b x)}-\frac {5 (7 A b-a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{8 a^{9/2} \sqrt {b}} \] Output:

-2*A/a^4/x^(1/2)-1/3*(A*b-B*a)*x^(1/2)/a^2/(b*x+a)^3-1/12*(11*A*b-5*B*a)*x 
^(1/2)/a^3/(b*x+a)^2-1/8*(19*A*b-5*B*a)*x^(1/2)/a^4/(b*x+a)-5/8*(7*A*b-B*a 
)*arctan(b^(1/2)*x^(1/2)/a^(1/2))/a^(9/2)/b^(1/2)
 

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.83 \[ \int \frac {A+B x}{x^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {-105 A b^3 x^3+5 a b^2 x^2 (-56 A+3 B x)+a^3 (-48 A+33 B x)+a^2 b x (-231 A+40 B x)}{24 a^4 \sqrt {x} (a+b x)^3}+\frac {5 (-7 A b+a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{8 a^{9/2} \sqrt {b}} \] Input:

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

Output:

(-105*A*b^3*x^3 + 5*a*b^2*x^2*(-56*A + 3*B*x) + a^3*(-48*A + 33*B*x) + a^2 
*b*x*(-231*A + 40*B*x))/(24*a^4*Sqrt[x]*(a + b*x)^3) + (5*(-7*A*b + a*B)*A 
rcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(8*a^(9/2)*Sqrt[b])
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.04, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {1184, 27, 87, 52, 52, 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.

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

\(\Big \downarrow \) 1184

\(\displaystyle b^4 \int \frac {A+B x}{b^4 x^{3/2} (a+b x)^4}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \int \frac {A+B x}{x^{3/2} (a+b x)^4}dx\)

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 52

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

\(\Big \downarrow \) 52

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

(A*b - a*B)/(3*a*b*Sqrt[x]*(a + b*x)^3) + ((7*A*b - a*B)*(1/(2*a*Sqrt[x]*( 
a + b*x)^2) + (5*(1/(a*Sqrt[x]*(a + b*x)) + (3*(-2/(a*Sqrt[x]) - (2*Sqrt[b 
]*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/a^(3/2)))/(2*a)))/(4*a)))/(6*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 = 1.15 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.78

method	result	size
derivativedivides	\(-\frac {2 \left (\frac {\left (\frac {19}{16} A \,b^{3}-\frac {5}{16} B a \,b^{2}\right ) x^{\frac {5}{2}}+\frac {a b \left (17 A b -5 B a \right ) x^{\frac {3}{2}}}{6}+\left (\frac {29}{16} A \,a^{2} b -\frac {11}{16} B \,a^{3}\right ) \sqrt {x}}{\left (b x +a \right )^{3}}+\frac {5 \left (7 A b -B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{16 \sqrt {a b}}\right )}{a^{4}}-\frac {2 A}{a^{4} \sqrt {x}}\)	\(105\)
default	\(-\frac {2 \left (\frac {\left (\frac {19}{16} A \,b^{3}-\frac {5}{16} B a \,b^{2}\right ) x^{\frac {5}{2}}+\frac {a b \left (17 A b -5 B a \right ) x^{\frac {3}{2}}}{6}+\left (\frac {29}{16} A \,a^{2} b -\frac {11}{16} B \,a^{3}\right ) \sqrt {x}}{\left (b x +a \right )^{3}}+\frac {5 \left (7 A b -B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{16 \sqrt {a b}}\right )}{a^{4}}-\frac {2 A}{a^{4} \sqrt {x}}\)	\(105\)
risch	\(-\frac {2 A}{a^{4} \sqrt {x}}-\frac {\frac {2 \left (\frac {19}{16} A \,b^{3}-\frac {5}{16} B a \,b^{2}\right ) x^{\frac {5}{2}}+\frac {a b \left (17 A b -5 B a \right ) x^{\frac {3}{2}}}{3}+2 \left (\frac {29}{16} A \,a^{2} b -\frac {11}{16} B \,a^{3}\right ) \sqrt {x}}{\left (b x +a \right )^{3}}+\frac {5 \left (7 A b -B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}}}{a^{4}}\)	\(106\)

Input:

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

Output:

-2/a^4*(((19/16*A*b^3-5/16*B*a*b^2)*x^(5/2)+1/6*a*b*(17*A*b-5*B*a)*x^(3/2) 
+(29/16*A*a^2*b-11/16*B*a^3)*x^(1/2))/(b*x+a)^3+5/16*(7*A*b-B*a)/(a*b)^(1/ 
2)*arctan(b*x^(1/2)/(a*b)^(1/2)))-2*A/a^4/x^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 445, normalized size of antiderivative = 3.30 \[ \int \frac {A+B x}{x^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\left [\frac {15 \, {\left ({\left (B a b^{3} - 7 \, A b^{4}\right )} x^{4} + 3 \, {\left (B a^{2} b^{2} - 7 \, A a b^{3}\right )} x^{3} + 3 \, {\left (B a^{3} b - 7 \, A a^{2} b^{2}\right )} x^{2} + {\left (B a^{4} - 7 \, A a^{3} b\right )} x\right )} \sqrt {-a b} \log \left (\frac {b x - a + 2 \, \sqrt {-a b} \sqrt {x}}{b x + a}\right ) - 2 \, {\left (48 \, A a^{4} b - 15 \, {\left (B a^{2} b^{3} - 7 \, A a b^{4}\right )} x^{3} - 40 \, {\left (B a^{3} b^{2} - 7 \, A a^{2} b^{3}\right )} x^{2} - 33 \, {\left (B a^{4} b - 7 \, A a^{3} b^{2}\right )} x\right )} \sqrt {x}}{48 \, {\left (a^{5} b^{4} x^{4} + 3 \, a^{6} b^{3} x^{3} + 3 \, a^{7} b^{2} x^{2} + a^{8} b x\right )}}, -\frac {15 \, {\left ({\left (B a b^{3} - 7 \, A b^{4}\right )} x^{4} + 3 \, {\left (B a^{2} b^{2} - 7 \, A a b^{3}\right )} x^{3} + 3 \, {\left (B a^{3} b - 7 \, A a^{2} b^{2}\right )} x^{2} + {\left (B a^{4} - 7 \, A a^{3} b\right )} x\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b}}{b \sqrt {x}}\right ) + {\left (48 \, A a^{4} b - 15 \, {\left (B a^{2} b^{3} - 7 \, A a b^{4}\right )} x^{3} - 40 \, {\left (B a^{3} b^{2} - 7 \, A a^{2} b^{3}\right )} x^{2} - 33 \, {\left (B a^{4} b - 7 \, A a^{3} b^{2}\right )} x\right )} \sqrt {x}}{24 \, {\left (a^{5} b^{4} x^{4} + 3 \, a^{6} b^{3} x^{3} + 3 \, a^{7} b^{2} x^{2} + a^{8} b x\right )}}\right ] \] Input:

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

Output:

[1/48*(15*((B*a*b^3 - 7*A*b^4)*x^4 + 3*(B*a^2*b^2 - 7*A*a*b^3)*x^3 + 3*(B* 
a^3*b - 7*A*a^2*b^2)*x^2 + (B*a^4 - 7*A*a^3*b)*x)*sqrt(-a*b)*log((b*x - a 
+ 2*sqrt(-a*b)*sqrt(x))/(b*x + a)) - 2*(48*A*a^4*b - 15*(B*a^2*b^3 - 7*A*a 
*b^4)*x^3 - 40*(B*a^3*b^2 - 7*A*a^2*b^3)*x^2 - 33*(B*a^4*b - 7*A*a^3*b^2)* 
x)*sqrt(x))/(a^5*b^4*x^4 + 3*a^6*b^3*x^3 + 3*a^7*b^2*x^2 + a^8*b*x), -1/24 
*(15*((B*a*b^3 - 7*A*b^4)*x^4 + 3*(B*a^2*b^2 - 7*A*a*b^3)*x^3 + 3*(B*a^3*b 
 - 7*A*a^2*b^2)*x^2 + (B*a^4 - 7*A*a^3*b)*x)*sqrt(a*b)*arctan(sqrt(a*b)/(b 
*sqrt(x))) + (48*A*a^4*b - 15*(B*a^2*b^3 - 7*A*a*b^4)*x^3 - 40*(B*a^3*b^2 
- 7*A*a^2*b^3)*x^2 - 33*(B*a^4*b - 7*A*a^3*b^2)*x)*sqrt(x))/(a^5*b^4*x^4 + 
 3*a^6*b^3*x^3 + 3*a^7*b^2*x^2 + a^8*b*x)]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2660 vs. \(2 (134) = 268\).

Time = 75.86 (sec) , antiderivative size = 2660, normalized size of antiderivative = 19.70 \[ \int \frac {A+B x}{x^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

Piecewise((zoo*(-2*A/(9*x**(9/2)) - 2*B/(7*x**(7/2))), Eq(a, 0) & Eq(b, 0) 
), ((-2*A/sqrt(x) + 2*B*sqrt(x))/a**4, Eq(b, 0)), ((-2*A/(9*x**(9/2)) - 2* 
B/(7*x**(7/2)))/b**4, Eq(a, 0)), (-105*A*a**3*b*sqrt(x)*log(sqrt(x) - sqrt 
(-a/b))/(48*a**7*b*sqrt(x)*sqrt(-a/b) + 144*a**6*b**2*x**(3/2)*sqrt(-a/b) 
+ 144*a**5*b**3*x**(5/2)*sqrt(-a/b) + 48*a**4*b**4*x**(7/2)*sqrt(-a/b)) + 
105*A*a**3*b*sqrt(x)*log(sqrt(x) + sqrt(-a/b))/(48*a**7*b*sqrt(x)*sqrt(-a/ 
b) + 144*a**6*b**2*x**(3/2)*sqrt(-a/b) + 144*a**5*b**3*x**(5/2)*sqrt(-a/b) 
 + 48*a**4*b**4*x**(7/2)*sqrt(-a/b)) - 96*A*a**3*b*sqrt(-a/b)/(48*a**7*b*s 
qrt(x)*sqrt(-a/b) + 144*a**6*b**2*x**(3/2)*sqrt(-a/b) + 144*a**5*b**3*x**( 
5/2)*sqrt(-a/b) + 48*a**4*b**4*x**(7/2)*sqrt(-a/b)) - 315*A*a**2*b**2*x**( 
3/2)*log(sqrt(x) - sqrt(-a/b))/(48*a**7*b*sqrt(x)*sqrt(-a/b) + 144*a**6*b* 
*2*x**(3/2)*sqrt(-a/b) + 144*a**5*b**3*x**(5/2)*sqrt(-a/b) + 48*a**4*b**4* 
x**(7/2)*sqrt(-a/b)) + 315*A*a**2*b**2*x**(3/2)*log(sqrt(x) + sqrt(-a/b))/ 
(48*a**7*b*sqrt(x)*sqrt(-a/b) + 144*a**6*b**2*x**(3/2)*sqrt(-a/b) + 144*a* 
*5*b**3*x**(5/2)*sqrt(-a/b) + 48*a**4*b**4*x**(7/2)*sqrt(-a/b)) - 462*A*a* 
*2*b**2*x*sqrt(-a/b)/(48*a**7*b*sqrt(x)*sqrt(-a/b) + 144*a**6*b**2*x**(3/2 
)*sqrt(-a/b) + 144*a**5*b**3*x**(5/2)*sqrt(-a/b) + 48*a**4*b**4*x**(7/2)*s 
qrt(-a/b)) - 315*A*a*b**3*x**(5/2)*log(sqrt(x) - sqrt(-a/b))/(48*a**7*b*sq 
rt(x)*sqrt(-a/b) + 144*a**6*b**2*x**(3/2)*sqrt(-a/b) + 144*a**5*b**3*x**(5 
/2)*sqrt(-a/b) + 48*a**4*b**4*x**(7/2)*sqrt(-a/b)) + 315*A*a*b**3*x**(5...
 

Maxima [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.98 \[ \int \frac {A+B x}{x^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=-\frac {48 \, A a^{3} - 15 \, {\left (B a b^{2} - 7 \, A b^{3}\right )} x^{3} - 40 \, {\left (B a^{2} b - 7 \, A a b^{2}\right )} x^{2} - 33 \, {\left (B a^{3} - 7 \, A a^{2} b\right )} x}{24 \, {\left (a^{4} b^{3} x^{\frac {7}{2}} + 3 \, a^{5} b^{2} x^{\frac {5}{2}} + 3 \, a^{6} b x^{\frac {3}{2}} + a^{7} \sqrt {x}\right )}} + \frac {5 \, {\left (B a - 7 \, A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{4}} \] Input:

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

Output:

-1/24*(48*A*a^3 - 15*(B*a*b^2 - 7*A*b^3)*x^3 - 40*(B*a^2*b - 7*A*a*b^2)*x^ 
2 - 33*(B*a^3 - 7*A*a^2*b)*x)/(a^4*b^3*x^(7/2) + 3*a^5*b^2*x^(5/2) + 3*a^6 
*b*x^(3/2) + a^7*sqrt(x)) + 5/8*(B*a - 7*A*b)*arctan(b*sqrt(x)/sqrt(a*b))/ 
(sqrt(a*b)*a^4)
 

Giac [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.81 \[ \int \frac {A+B x}{x^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {5 \, {\left (B a - 7 \, A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{4}} - \frac {2 \, A}{a^{4} \sqrt {x}} + \frac {15 \, B a b^{2} x^{\frac {5}{2}} - 57 \, A b^{3} x^{\frac {5}{2}} + 40 \, B a^{2} b x^{\frac {3}{2}} - 136 \, A a b^{2} x^{\frac {3}{2}} + 33 \, B a^{3} \sqrt {x} - 87 \, A a^{2} b \sqrt {x}}{24 \, {\left (b x + a\right )}^{3} a^{4}} \] Input:

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

Output:

5/8*(B*a - 7*A*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^4) - 2*A/(a^4*s 
qrt(x)) + 1/24*(15*B*a*b^2*x^(5/2) - 57*A*b^3*x^(5/2) + 40*B*a^2*b*x^(3/2) 
 - 136*A*a*b^2*x^(3/2) + 33*B*a^3*sqrt(x) - 87*A*a^2*b*sqrt(x))/((b*x + a) 
^3*a^4)
 

Mupad [B] (verification not implemented)

Time = 10.71 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.09 \[ \int \frac {A+B x}{x^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=-\frac {\frac {2\,A}{a}+\frac {11\,x\,\left (7\,A\,b-B\,a\right )}{8\,a^2}+\frac {5\,b^2\,x^3\,\left (7\,A\,b-B\,a\right )}{8\,a^4}+\frac {5\,b\,x^2\,\left (7\,A\,b-B\,a\right )}{3\,a^3}}{a^3\,\sqrt {x}+b^3\,x^{7/2}+3\,a^2\,b\,x^{3/2}+3\,a\,b^2\,x^{5/2}}-\frac {5\,\mathrm {atan}\left (\frac {5\,\sqrt {b}\,\sqrt {x}\,\left (7\,A\,b-B\,a\right )}{\sqrt {a}\,\left (35\,A\,b-5\,B\,a\right )}\right )\,\left (7\,A\,b-B\,a\right )}{8\,a^{9/2}\,\sqrt {b}} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.92 \[ \int \frac {A+B x}{x^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {-15 \sqrt {x}\, \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) a^{2}-30 \sqrt {x}\, \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) a b x -15 \sqrt {x}\, \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) b^{2} x^{2}-8 a^{3}-25 a^{2} b x -15 a \,b^{2} x^{2}}{4 \sqrt {x}\, a^{4} \left (b^{2} x^{2}+2 a b x +a^{2}\right )} \] Input:

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

Output:

( - 15*sqrt(x)*sqrt(b)*sqrt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*a**2 - 
30*sqrt(x)*sqrt(b)*sqrt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*a*b*x - 15* 
sqrt(x)*sqrt(b)*sqrt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*b**2*x**2 - 8* 
a**3 - 25*a**2*b*x - 15*a*b**2*x**2)/(4*sqrt(x)*a**4*(a**2 + 2*a*b*x + b** 
2*x**2))