\(\int \frac {A+B x}{x^2 (a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\) [357]

Optimal result

Integrand size = 29, antiderivative size = 282 \[ \int \frac {A+B x}{x^2 \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {4 A b-a B}{a^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {A b-a B}{4 a^2 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 A b-a B}{3 a^3 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 A b-a B}{2 a^4 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {A (a+b x)}{a^5 x \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(5 A b-a B) (a+b x) \log (x)}{a^6 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(5 A b-a B) (a+b x) \log (a+b x)}{a^6 \sqrt {a^2+2 a b x+b^2 x^2}} \] Output:

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

Mathematica [A] (verified)

Time = 1.11 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.52 \[ \int \frac {A+B x}{x^2 \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {a \left (-60 A b^4 x^4+6 a b^3 x^3 (-35 A+2 B x)+2 a^2 b^2 x^2 (-130 A+21 B x)+a^4 (-12 A+25 B x)+a^3 b x (-125 A+52 B x)\right )+12 (-5 A b+a B) x (a+b x)^4 \log (x)+12 (5 A b-a B) x (a+b x)^4 \log (a+b x)}{12 a^6 x (a+b x)^3 \sqrt {(a+b x)^2}} \] Input:

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

Output:

(a*(-60*A*b^4*x^4 + 6*a*b^3*x^3*(-35*A + 2*B*x) + 2*a^2*b^2*x^2*(-130*A + 
21*B*x) + a^4*(-12*A + 25*B*x) + a^3*b*x*(-125*A + 52*B*x)) + 12*(-5*A*b + 
 a*B)*x*(a + b*x)^4*Log[x] + 12*(5*A*b - a*B)*x*(a + b*x)^4*Log[a + b*x])/ 
(12*a^6*x*(a + b*x)^3*Sqrt[(a + b*x)^2])
 

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.57, 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.

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

\(\Big \downarrow \) 1187

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 86

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

((a + b*x)*(-(A/(a^5*x)) - (A*b - a*B)/(4*a^2*(a + b*x)^4) - (2*A*b - a*B) 
/(3*a^3*(a + b*x)^3) - (3*A*b - a*B)/(2*a^4*(a + b*x)^2) - (4*A*b - a*B)/( 
a^5*(a + b*x)) - ((5*A*b - a*B)*Log[x])/a^6 + ((5*A*b - a*B)*Log[a + b*x]) 
/a^6))/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 = 1.17 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.61

method	result	size
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {\left (5 A b -B a \right ) b^{3} x^{4}}{a^{5}}-\frac {7 b^{2} \left (5 A b -B a \right ) x^{3}}{2 a^{4}}-\frac {13 \left (5 A b -B a \right ) b \,x^{2}}{3 a^{3}}-\frac {25 \left (5 A b -B a \right ) x}{12 a^{2}}-\frac {A}{a}\right )}{\left (b x +a \right )^{5} x}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (5 A b -B a \right ) \ln \left (-b x -a \right )}{\left (b x +a \right ) a^{6}}-\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (5 A b -B a \right ) \ln \left (x \right )}{\left (b x +a \right ) a^{6}}\)	\(172\)
default	\(\frac {\left (-60 A a \,b^{4} x^{4}+72 B \ln \left (x \right ) a^{3} b^{2} x^{3}+12 B \ln \left (x \right ) a^{5} x +52 B \,a^{4} b \,x^{2}-125 A \,a^{4} b x +48 B \ln \left (x \right ) a^{2} b^{3} x^{4}-12 B \ln \left (b x +a \right ) a \,b^{4} x^{5}-48 B \ln \left (b x +a \right ) a^{4} b \,x^{2}-72 B \ln \left (b x +a \right ) a^{3} b^{2} x^{3}-48 B \ln \left (b x +a \right ) a^{2} b^{3} x^{4}+12 B \ln \left (x \right ) a \,b^{4} x^{5}+48 B \ln \left (x \right ) a^{4} b \,x^{2}-12 B \ln \left (b x +a \right ) a^{5} x +60 A \ln \left (b x +a \right ) x \,a^{4} b -60 A \ln \left (x \right ) x \,a^{4} b -240 A \ln \left (x \right ) x^{4} a \,b^{4}+240 A \ln \left (b x +a \right ) x^{4} a \,b^{4}-360 A \ln \left (x \right ) x^{3} a^{2} b^{3}+360 A \ln \left (b x +a \right ) x^{3} a^{2} b^{3}-240 A \ln \left (x \right ) x^{2} a^{3} b^{2}+240 A \ln \left (b x +a \right ) x^{2} a^{3} b^{2}-60 A \ln \left (x \right ) x^{5} b^{5}+60 A \ln \left (b x +a \right ) x^{5} b^{5}+12 B \,a^{2} b^{3} x^{4}-210 A \,a^{2} b^{3} x^{3}-12 A \,a^{5}+42 B \,a^{3} b^{2} x^{3}-260 A \,a^{3} b^{2} x^{2}+25 B \,a^{5} x \right ) \left (b x +a \right )}{12 x \,a^{6} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\)	\(397\)

Input:

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

Output:

((b*x+a)^2)^(1/2)/(b*x+a)^5*(-(5*A*b-B*a)/a^5*b^3*x^4-7/2/a^4*b^2*(5*A*b-B 
*a)*x^3-13/3/a^3*(5*A*b-B*a)*b*x^2-25/12/a^2*(5*A*b-B*a)*x-A/a)/x+((b*x+a) 
^2)^(1/2)/(b*x+a)*(5*A*b-B*a)/a^6*ln(-b*x-a)-((b*x+a)^2)^(1/2)/(b*x+a)*(5* 
A*b-B*a)/a^6*ln(x)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 347, normalized size of antiderivative = 1.23 \[ \int \frac {A+B x}{x^2 \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {12 \, A a^{5} - 12 \, {\left (B a^{2} b^{3} - 5 \, A a b^{4}\right )} x^{4} - 42 \, {\left (B a^{3} b^{2} - 5 \, A a^{2} b^{3}\right )} x^{3} - 52 \, {\left (B a^{4} b - 5 \, A a^{3} b^{2}\right )} x^{2} - 25 \, {\left (B a^{5} - 5 \, A a^{4} b\right )} x + 12 \, {\left ({\left (B a b^{4} - 5 \, A b^{5}\right )} x^{5} + 4 \, {\left (B a^{2} b^{3} - 5 \, A a b^{4}\right )} x^{4} + 6 \, {\left (B a^{3} b^{2} - 5 \, A a^{2} b^{3}\right )} x^{3} + 4 \, {\left (B a^{4} b - 5 \, A a^{3} b^{2}\right )} x^{2} + {\left (B a^{5} - 5 \, A a^{4} b\right )} x\right )} \log \left (b x + a\right ) - 12 \, {\left ({\left (B a b^{4} - 5 \, A b^{5}\right )} x^{5} + 4 \, {\left (B a^{2} b^{3} - 5 \, A a b^{4}\right )} x^{4} + 6 \, {\left (B a^{3} b^{2} - 5 \, A a^{2} b^{3}\right )} x^{3} + 4 \, {\left (B a^{4} b - 5 \, A a^{3} b^{2}\right )} x^{2} + {\left (B a^{5} - 5 \, A a^{4} b\right )} x\right )} \log \left (x\right )}{12 \, {\left (a^{6} b^{4} x^{5} + 4 \, a^{7} b^{3} x^{4} + 6 \, a^{8} b^{2} x^{3} + 4 \, a^{9} b x^{2} + a^{10} x\right )}} \] Input:

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

Output:

-1/12*(12*A*a^5 - 12*(B*a^2*b^3 - 5*A*a*b^4)*x^4 - 42*(B*a^3*b^2 - 5*A*a^2 
*b^3)*x^3 - 52*(B*a^4*b - 5*A*a^3*b^2)*x^2 - 25*(B*a^5 - 5*A*a^4*b)*x + 12 
*((B*a*b^4 - 5*A*b^5)*x^5 + 4*(B*a^2*b^3 - 5*A*a*b^4)*x^4 + 6*(B*a^3*b^2 - 
 5*A*a^2*b^3)*x^3 + 4*(B*a^4*b - 5*A*a^3*b^2)*x^2 + (B*a^5 - 5*A*a^4*b)*x) 
*log(b*x + a) - 12*((B*a*b^4 - 5*A*b^5)*x^5 + 4*(B*a^2*b^3 - 5*A*a*b^4)*x^ 
4 + 6*(B*a^3*b^2 - 5*A*a^2*b^3)*x^3 + 4*(B*a^4*b - 5*A*a^3*b^2)*x^2 + (B*a 
^5 - 5*A*a^4*b)*x)*log(x))/(a^6*b^4*x^5 + 4*a^7*b^3*x^4 + 6*a^8*b^2*x^3 + 
4*a^9*b*x^2 + a^10*x)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 276, normalized size of antiderivative = 0.98 \[ \int \frac {A+B x}{x^2 \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {\left (-1\right )^{2 \, a b x + 2 \, a^{2}} B \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right )}{a^{5}} + \frac {5 \, \left (-1\right )^{2 \, a b x + 2 \, a^{2}} A b \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right )}{a^{6}} + \frac {B}{3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{2}} - \frac {5 \, A b}{3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{3}} + \frac {B}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{4}} - \frac {5 \, A b}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{5}} - \frac {A}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{2} x} + \frac {B}{2 \, a^{3} b^{2} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {5 \, A}{2 \, a^{4} b {\left (x + \frac {a}{b}\right )}^{2}} + \frac {B}{4 \, a b^{4} {\left (x + \frac {a}{b}\right )}^{4}} - \frac {A}{4 \, a^{2} b^{3} {\left (x + \frac {a}{b}\right )}^{4}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.60 \[ \int \frac {A+B x}{x^2 \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {{\left (B a - 5 \, A b\right )} \log \left ({\left | x \right |}\right )}{a^{6} \mathrm {sgn}\left (b x + a\right )} - \frac {{\left (B a b - 5 \, A b^{2}\right )} \log \left ({\left | b x + a \right |}\right )}{a^{6} b \mathrm {sgn}\left (b x + a\right )} - \frac {12 \, A a^{5} - 12 \, {\left (B a^{2} b^{3} - 5 \, A a b^{4}\right )} x^{4} - 42 \, {\left (B a^{3} b^{2} - 5 \, A a^{2} b^{3}\right )} x^{3} - 52 \, {\left (B a^{4} b - 5 \, A a^{3} b^{2}\right )} x^{2} - 25 \, {\left (B a^{5} - 5 \, A a^{4} b\right )} x}{12 \, {\left (b x + a\right )}^{4} a^{6} x \mathrm {sgn}\left (b x + a\right )} \] Input:

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

Output:

(B*a - 5*A*b)*log(abs(x))/(a^6*sgn(b*x + a)) - (B*a*b - 5*A*b^2)*log(abs(b 
*x + a))/(a^6*b*sgn(b*x + a)) - 1/12*(12*A*a^5 - 12*(B*a^2*b^3 - 5*A*a*b^4 
)*x^4 - 42*(B*a^3*b^2 - 5*A*a^2*b^3)*x^3 - 52*(B*a^4*b - 5*A*a^3*b^2)*x^2 
- 25*(B*a^5 - 5*A*a^4*b)*x)/((b*x + a)^4*a^6*x*sgn(b*x + a))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.61 \[ \int \frac {A+B x}{x^2 \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {12 \,\mathrm {log}\left (b x +a \right ) a^{3} b x +36 \,\mathrm {log}\left (b x +a \right ) a^{2} b^{2} x^{2}+36 \,\mathrm {log}\left (b x +a \right ) a \,b^{3} x^{3}+12 \,\mathrm {log}\left (b x +a \right ) b^{4} x^{4}-12 \,\mathrm {log}\left (x \right ) a^{3} b x -36 \,\mathrm {log}\left (x \right ) a^{2} b^{2} x^{2}-36 \,\mathrm {log}\left (x \right ) a \,b^{3} x^{3}-12 \,\mathrm {log}\left (x \right ) b^{4} x^{4}-3 a^{4}-18 a^{3} b x -18 a^{2} b^{2} x^{2}+4 b^{4} x^{4}}{3 a^{5} x \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 a^{2} b x +a^{3}\right )} \] Input:

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

Output:

(12*log(a + b*x)*a**3*b*x + 36*log(a + b*x)*a**2*b**2*x**2 + 36*log(a + b* 
x)*a*b**3*x**3 + 12*log(a + b*x)*b**4*x**4 - 12*log(x)*a**3*b*x - 36*log(x 
)*a**2*b**2*x**2 - 36*log(x)*a*b**3*x**3 - 12*log(x)*b**4*x**4 - 3*a**4 - 
18*a**3*b*x - 18*a**2*b**2*x**2 + 4*b**4*x**4)/(3*a**5*x*(a**3 + 3*a**2*b* 
x + 3*a*b**2*x**2 + b**3*x**3))