\(\int \frac {A+B x}{x^3 (a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\) [350]

Optimal result

Integrand size = 29, antiderivative size = 243 \[ \int \frac {A+B x}{x^3 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {b (3 A b-2 a B)}{a^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b (A b-a B)}{2 a^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {A (a+b x)}{2 a^3 x^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(3 A b-a B) (a+b x)}{a^4 x \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 b (2 A b-a B) (a+b x) \log (x)}{a^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 b (2 A b-a B) (a+b x) \log (a+b x)}{a^5 \sqrt {a^2+2 a b x+b^2 x^2}} \] Output:

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

Mathematica [A] (verified)

Time = 1.09 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.55 \[ \int \frac {A+B x}{x^3 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {-a \left (-12 A b^3 x^3+6 a b^2 x^2 (-3 A+B x)+a^3 (A+2 B x)+a^2 b x (-4 A+9 B x)\right )+6 b (2 A b-a B) x^2 (a+b x)^2 \log (x)+6 b (-2 A b+a B) x^2 (a+b x)^2 \log (a+b x)}{2 a^5 x^2 (a+b x) \sqrt {(a+b x)^2}} \] Input:

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

Output:

(-(a*(-12*A*b^3*x^3 + 6*a*b^2*x^2*(-3*A + B*x) + a^3*(A + 2*B*x) + a^2*b*x 
*(-4*A + 9*B*x))) + 6*b*(2*A*b - a*B)*x^2*(a + b*x)^2*Log[x] + 6*b*(-2*A*b 
 + a*B)*x^2*(a + b*x)^2*Log[a + b*x])/(2*a^5*x^2*(a + b*x)*Sqrt[(a + b*x)^ 
2])
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.56, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {1187, 27, 86, 2009}

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.

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

\(\Big \downarrow \) 1187

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 86

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

((a + b*x)*(-1/2*A/(a^3*x^2) + (3*A*b - a*B)/(a^4*x) + (b*(A*b - a*B))/(2* 
a^3*(a + b*x)^2) + (b*(3*A*b - 2*a*B))/(a^4*(a + b*x)) + (3*b*(2*A*b - a*B 
)*Log[x])/a^5 - (3*b*(2*A*b - a*B)*Log[a + b*x])/a^5))/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 86

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ 
.), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; 
 FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 
] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p 
+ 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
 

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]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [A] (verified)

Time = 3.13 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.63

method	result	size
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\frac {3 b^{2} \left (2 A b -B a \right ) x^{3}}{a^{4}}+\frac {9 b \left (2 A b -B a \right ) x^{2}}{2 a^{3}}+\frac {\left (2 A b -B a \right ) x}{a^{2}}-\frac {A}{2 a}\right )}{\left (b x +a \right )^{3} x^{2}}-\frac {3 \sqrt {\left (b x +a \right )^{2}}\, b \left (2 A b -B a \right ) \ln \left (b x +a \right )}{\left (b x +a \right ) a^{5}}+\frac {3 \sqrt {\left (b x +a \right )^{2}}\, b \left (2 A b -B a \right ) \ln \left (-x \right )}{\left (b x +a \right ) a^{5}}\)	\(153\)
default	\(-\frac {\left (12 A \ln \left (b x +a \right ) x^{4} b^{4}-12 A \ln \left (x \right ) x^{4} b^{4}-6 B \ln \left (b x +a \right ) x^{4} a \,b^{3}+6 B \ln \left (x \right ) x^{4} a \,b^{3}+24 A \ln \left (b x +a \right ) x^{3} a \,b^{3}-24 A \ln \left (x \right ) x^{3} a \,b^{3}-12 B \ln \left (b x +a \right ) x^{3} a^{2} b^{2}+12 B \ln \left (x \right ) x^{3} a^{2} b^{2}+12 A \ln \left (b x +a \right ) x^{2} a^{2} b^{2}-12 A \ln \left (x \right ) x^{2} a^{2} b^{2}-12 A a \,b^{3} x^{3}-6 B \ln \left (b x +a \right ) x^{2} a^{3} b +6 B \ln \left (x \right ) x^{2} a^{3} b +6 B \,a^{2} b^{2} x^{3}-18 A \,a^{2} b^{2} x^{2}+9 B \,a^{3} b \,x^{2}-4 A \,a^{3} b x +2 a^{4} B x +a^{4} A \right ) \left (b x +a \right )}{2 x^{2} a^{5} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\)	\(262\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 225, normalized size of antiderivative = 0.93 \[ \int \frac {A+B x}{x^3 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {A a^{4} + 6 \, {\left (B a^{2} b^{2} - 2 \, A a b^{3}\right )} x^{3} + 9 \, {\left (B a^{3} b - 2 \, A a^{2} b^{2}\right )} x^{2} + 2 \, {\left (B a^{4} - 2 \, A a^{3} b\right )} x - 6 \, {\left ({\left (B a b^{3} - 2 \, A b^{4}\right )} x^{4} + 2 \, {\left (B a^{2} b^{2} - 2 \, A a b^{3}\right )} x^{3} + {\left (B a^{3} b - 2 \, A a^{2} b^{2}\right )} x^{2}\right )} \log \left (b x + a\right ) + 6 \, {\left ({\left (B a b^{3} - 2 \, A b^{4}\right )} x^{4} + 2 \, {\left (B a^{2} b^{2} - 2 \, A a b^{3}\right )} x^{3} + {\left (B a^{3} b - 2 \, A a^{2} b^{2}\right )} x^{2}\right )} \log \left (x\right )}{2 \, {\left (a^{5} b^{2} x^{4} + 2 \, a^{6} b x^{3} + a^{7} x^{2}\right )}} \] Input:

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

Output:

-1/2*(A*a^4 + 6*(B*a^2*b^2 - 2*A*a*b^3)*x^3 + 9*(B*a^3*b - 2*A*a^2*b^2)*x^ 
2 + 2*(B*a^4 - 2*A*a^3*b)*x - 6*((B*a*b^3 - 2*A*b^4)*x^4 + 2*(B*a^2*b^2 - 
2*A*a*b^3)*x^3 + (B*a^3*b - 2*A*a^2*b^2)*x^2)*log(b*x + a) + 6*((B*a*b^3 - 
 2*A*b^4)*x^4 + 2*(B*a^2*b^2 - 2*A*a*b^3)*x^3 + (B*a^3*b - 2*A*a^2*b^2)*x^ 
2)*log(x))/(a^5*b^2*x^4 + 2*a^6*b*x^3 + a^7*x^2)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.03 \[ \int \frac {A+B x}{x^3 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {3 \, \left (-1\right )^{2 \, a b x + 2 \, a^{2}} B b \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right )}{a^{4}} - \frac {6 \, \left (-1\right )^{2 \, a b x + 2 \, a^{2}} A b^{2} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right )}{a^{5}} - \frac {3 \, B b}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{3}} + \frac {6 \, A b^{2}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{4}} - \frac {B}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{2} x} + \frac {5 \, A b}{2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{3} x} - \frac {A}{2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{2} x^{2}} + \frac {A}{2 \, a^{3} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {B}{2 \, a^{2} b {\left (x + \frac {a}{b}\right )}^{2}} \] Input:

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

Output:

3*(-1)^(2*a*b*x + 2*a^2)*B*b*log(2*a*b*x/abs(x) + 2*a^2/abs(x))/a^4 - 6*(- 
1)^(2*a*b*x + 2*a^2)*A*b^2*log(2*a*b*x/abs(x) + 2*a^2/abs(x))/a^5 - 3*B*b/ 
(sqrt(b^2*x^2 + 2*a*b*x + a^2)*a^3) + 6*A*b^2/(sqrt(b^2*x^2 + 2*a*b*x + a^ 
2)*a^4) - B/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*a^2*x) + 5/2*A*b/(sqrt(b^2*x^2 
+ 2*a*b*x + a^2)*a^3*x) - 1/2*A/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*a^2*x^2) + 
1/2*A/(a^3*(x + a/b)^2) - 1/2*B/(a^2*b*(x + a/b)^2)
 

Giac [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.61 \[ \int \frac {A+B x}{x^3 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {3 \, {\left (B a b - 2 \, A b^{2}\right )} \log \left ({\left | x \right |}\right )}{a^{5} \mathrm {sgn}\left (b x + a\right )} + \frac {3 \, {\left (B a b^{2} - 2 \, A b^{3}\right )} \log \left ({\left | b x + a \right |}\right )}{a^{5} b \mathrm {sgn}\left (b x + a\right )} - \frac {6 \, B a b^{2} x^{3} - 12 \, A b^{3} x^{3} + 9 \, B a^{2} b x^{2} - 18 \, A a b^{2} x^{2} + 2 \, B a^{3} x - 4 \, A a^{2} b x + A a^{3}}{2 \, {\left (b x^{2} + a x\right )}^{2} a^{4} \mathrm {sgn}\left (b x + a\right )} \] Input:

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

Output:

-3*(B*a*b - 2*A*b^2)*log(abs(x))/(a^5*sgn(b*x + a)) + 3*(B*a*b^2 - 2*A*b^3 
)*log(abs(b*x + a))/(a^5*b*sgn(b*x + a)) - 1/2*(6*B*a*b^2*x^3 - 12*A*b^3*x 
^3 + 9*B*a^2*b*x^2 - 18*A*a*b^2*x^2 + 2*B*a^3*x - 4*A*a^2*b*x + A*a^3)/((b 
*x^2 + a*x)^2*a^4*sgn(b*x + a))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.35 \[ \int \frac {A+B x}{x^3 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {-6 \,\mathrm {log}\left (b x +a \right ) a \,b^{2} x^{2}-6 \,\mathrm {log}\left (b x +a \right ) b^{3} x^{3}+6 \,\mathrm {log}\left (x \right ) a \,b^{2} x^{2}+6 \,\mathrm {log}\left (x \right ) b^{3} x^{3}-a^{3}+3 a^{2} b x -6 b^{3} x^{3}}{2 a^{4} x^{2} \left (b x +a \right )} \] Input:

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

Output:

( - 6*log(a + b*x)*a*b**2*x**2 - 6*log(a + b*x)*b**3*x**3 + 6*log(x)*a*b** 
2*x**2 + 6*log(x)*b**3*x**3 - a**3 + 3*a**2*b*x - 6*b**3*x**3)/(2*a**4*x** 
2*(a + b*x))