\(\int \frac {x^4 (A+B x)}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\) [351]

Optimal result

Integrand size = 29, antiderivative size = 245 \[ \int \frac {x^4 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {2 a (2 A b-5 a B)}{b^6 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a^4 (A b-a B)}{4 b^6 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {a^3 (4 A b-5 a B)}{3 b^6 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a^2 (3 A b-5 a B)}{b^6 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B x (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(A b-5 a B) (a+b x) \log (a+b x)}{b^6 \sqrt {a^2+2 a b x+b^2 x^2}} \] Output:

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

Mathematica [A] (verified)

Time = 1.08 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.52 \[ \int \frac {x^4 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {-77 a^5 B+12 b^5 B x^5+a^4 b (25 A-248 B x)+4 a^3 b^2 x (22 A-63 B x)+12 a^2 b^3 x^2 (9 A-4 B x)+48 a b^4 x^3 (A+B x)+12 (A b-5 a B) (a+b x)^4 \log (a+b x)}{12 b^6 (a+b x)^3 \sqrt {(a+b x)^2}} \] Input:

Integrate[(x^4*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]
 

Output:

(-77*a^5*B + 12*b^5*B*x^5 + a^4*b*(25*A - 248*B*x) + 4*a^3*b^2*x*(22*A - 6 
3*B*x) + 12*a^2*b^3*x^2*(9*A - 4*B*x) + 48*a*b^4*x^3*(A + B*x) + 12*(A*b - 
 5*a*B)*(a + b*x)^4*Log[a + b*x])/(12*b^6*(a + b*x)^3*Sqrt[(a + b*x)^2])
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.60, 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.

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

\(\Big \downarrow \) 1187

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 86

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

\(\Big \downarrow \) 2009

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

Input:

Int[(x^4*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]
 

Output:

((a + b*x)*((B*x)/b^5 - (a^4*(A*b - a*B))/(4*b^6*(a + b*x)^4) + (a^3*(4*A* 
b - 5*a*B))/(3*b^6*(a + b*x)^3) - (a^2*(3*A*b - 5*a*B))/(b^6*(a + b*x)^2) 
+ (2*a*(2*A*b - 5*a*B))/(b^6*(a + b*x)) + ((A*b - 5*a*B)*Log[a + b*x])/b^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 = 1.31 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.61

method	result	size
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, B x}{\left (b x +a \right ) b^{5}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\left (4 A a \,b^{3}-10 B \,a^{2} b^{2}\right ) x^{3}+a^{2} b \left (9 A b -25 B a \right ) x^{2}+\left (\frac {22}{3} A \,a^{3} b -\frac {65}{3} a^{4} B \right ) x +\frac {a^{4} \left (25 A b -77 B a \right )}{12 b}\right )}{\left (b x +a \right )^{5} b^{5}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (A b -5 B a \right ) \ln \left (b x +a \right )}{\left (b x +a \right ) b^{6}}\)	\(150\)
default	\(\frac {\left (12 A \ln \left (b x +a \right ) b^{5} x^{4}-60 B \ln \left (b x +a \right ) a \,b^{4} x^{4}+12 B \,x^{5} b^{5}+48 A \ln \left (b x +a \right ) x^{3} a \,b^{4}-240 B \ln \left (b x +a \right ) x^{3} a^{2} b^{3}+48 B a \,b^{4} x^{4}+72 A \ln \left (b x +a \right ) x^{2} a^{2} b^{3}+48 A a \,b^{4} x^{3}-360 B \ln \left (b x +a \right ) x^{2} a^{3} b^{2}-48 B \,a^{2} b^{3} x^{3}+48 A \ln \left (b x +a \right ) x \,a^{3} b^{2}+108 A \,a^{2} b^{3} x^{2}-240 B \ln \left (b x +a \right ) x \,a^{4} b -252 B \,a^{3} b^{2} x^{2}+12 A \ln \left (b x +a \right ) a^{4} b +88 A \,a^{3} b^{2} x -60 B \ln \left (b x +a \right ) a^{5}-248 B \,a^{4} b x +25 A \,a^{4} b -77 B \,a^{5}\right ) \left (b x +a \right )}{12 b^{6} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\)	\(273\)

Input:

int(x^4*(B*x+A)/(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)*B/b^5*x+((b*x+a)^2)^(1/2)/(b*x+a)^5*((4*A*a*b^3- 
10*B*a^2*b^2)*x^3+a^2*b*(9*A*b-25*B*a)*x^2+(22/3*A*a^3*b-65/3*a^4*B)*x+1/1 
2*a^4*(25*A*b-77*B*a)/b)/b^5+((b*x+a)^2)^(1/2)/(b*x+a)/b^6*(A*b-5*B*a)*ln( 
b*x+a)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.03 \[ \int \frac {x^4 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {12 \, B b^{5} x^{5} + 48 \, B a b^{4} x^{4} - 77 \, B a^{5} + 25 \, A a^{4} b - 48 \, {\left (B a^{2} b^{3} - A a b^{4}\right )} x^{3} - 36 \, {\left (7 \, B a^{3} b^{2} - 3 \, A a^{2} b^{3}\right )} x^{2} - 8 \, {\left (31 \, B a^{4} b - 11 \, A a^{3} b^{2}\right )} x - 12 \, {\left (5 \, B a^{5} - A a^{4} b + {\left (5 \, B a b^{4} - A b^{5}\right )} x^{4} + 4 \, {\left (5 \, B a^{2} b^{3} - A a b^{4}\right )} x^{3} + 6 \, {\left (5 \, B a^{3} b^{2} - A a^{2} b^{3}\right )} x^{2} + 4 \, {\left (5 \, B a^{4} b - A a^{3} b^{2}\right )} x\right )} \log \left (b x + a\right )}{12 \, {\left (b^{10} x^{4} + 4 \, a b^{9} x^{3} + 6 \, a^{2} b^{8} x^{2} + 4 \, a^{3} b^{7} x + a^{4} b^{6}\right )}} \] Input:

integrate(x^4*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fricas")
 

Output:

1/12*(12*B*b^5*x^5 + 48*B*a*b^4*x^4 - 77*B*a^5 + 25*A*a^4*b - 48*(B*a^2*b^ 
3 - A*a*b^4)*x^3 - 36*(7*B*a^3*b^2 - 3*A*a^2*b^3)*x^2 - 8*(31*B*a^4*b - 11 
*A*a^3*b^2)*x - 12*(5*B*a^5 - A*a^4*b + (5*B*a*b^4 - A*b^5)*x^4 + 4*(5*B*a 
^2*b^3 - A*a*b^4)*x^3 + 6*(5*B*a^3*b^2 - A*a^2*b^3)*x^2 + 4*(5*B*a^4*b - A 
*a^3*b^2)*x)*log(b*x + a))/(b^10*x^4 + 4*a*b^9*x^3 + 6*a^2*b^8*x^2 + 4*a^3 
*b^7*x + a^4*b^6)
 

Sympy [F]

\[ \int \frac {x^4 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int \frac {x^{4} \left (A + B x\right )}{\left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(x**4*(B*x+A)/(b**2*x**2+2*a*b*x+a**2)**(5/2),x)
 

Output:

Integral(x**4*(A + B*x)/((a + b*x)**2)**(5/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 211, normalized size of antiderivative = 0.86 \[ \int \frac {x^4 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {1}{12} \, B {\left (\frac {12 \, b^{5} x^{5} + 48 \, a b^{4} x^{4} - 48 \, a^{2} b^{3} x^{3} - 252 \, a^{3} b^{2} x^{2} - 248 \, a^{4} b x - 77 \, a^{5}}{b^{10} x^{4} + 4 \, a b^{9} x^{3} + 6 \, a^{2} b^{8} x^{2} + 4 \, a^{3} b^{7} x + a^{4} b^{6}} - \frac {60 \, a \log \left (b x + a\right )}{b^{6}}\right )} + \frac {1}{12} \, A {\left (\frac {48 \, a b^{3} x^{3} + 108 \, a^{2} b^{2} x^{2} + 88 \, a^{3} b x + 25 \, a^{4}}{b^{9} x^{4} + 4 \, a b^{8} x^{3} + 6 \, a^{2} b^{7} x^{2} + 4 \, a^{3} b^{6} x + a^{4} b^{5}} + \frac {12 \, \log \left (b x + a\right )}{b^{5}}\right )} \] Input:

integrate(x^4*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="maxima")
 

Output:

1/12*B*((12*b^5*x^5 + 48*a*b^4*x^4 - 48*a^2*b^3*x^3 - 252*a^3*b^2*x^2 - 24 
8*a^4*b*x - 77*a^5)/(b^10*x^4 + 4*a*b^9*x^3 + 6*a^2*b^8*x^2 + 4*a^3*b^7*x 
+ a^4*b^6) - 60*a*log(b*x + a)/b^6) + 1/12*A*((48*a*b^3*x^3 + 108*a^2*b^2* 
x^2 + 88*a^3*b*x + 25*a^4)/(b^9*x^4 + 4*a*b^8*x^3 + 6*a^2*b^7*x^2 + 4*a^3* 
b^6*x + a^4*b^5) + 12*log(b*x + a)/b^5)
 

Giac [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.59 \[ \int \frac {x^4 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {B x}{b^{5} \mathrm {sgn}\left (b x + a\right )} - \frac {{\left (5 \, B a - A b\right )} \log \left ({\left | b x + a \right |}\right )}{b^{6} \mathrm {sgn}\left (b x + a\right )} - \frac {77 \, B a^{5} - 25 \, A a^{4} b + 24 \, {\left (5 \, B a^{2} b^{3} - 2 \, A a b^{4}\right )} x^{3} + 12 \, {\left (25 \, B a^{3} b^{2} - 9 \, A a^{2} b^{3}\right )} x^{2} + 4 \, {\left (65 \, B a^{4} b - 22 \, A a^{3} b^{2}\right )} x}{12 \, {\left (b x + a\right )}^{4} b^{6} \mathrm {sgn}\left (b x + a\right )} \] Input:

integrate(x^4*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac")
 

Output:

B*x/(b^5*sgn(b*x + a)) - (5*B*a - A*b)*log(abs(b*x + a))/(b^6*sgn(b*x + a) 
) - 1/12*(77*B*a^5 - 25*A*a^4*b + 24*(5*B*a^2*b^3 - 2*A*a*b^4)*x^3 + 12*(2 
5*B*a^3*b^2 - 9*A*a^2*b^3)*x^2 + 4*(65*B*a^4*b - 22*A*a^3*b^2)*x)/((b*x + 
a)^4*b^6*sgn(b*x + a))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int \frac {x^4\,\left (A+B\,x\right )}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}} \,d x \] Input:

int((x^4*(A + B*x))/(a^2 + b^2*x^2 + 2*a*b*x)^(5/2),x)
 

Output:

int((x^4*(A + B*x))/(a^2 + b^2*x^2 + 2*a*b*x)^(5/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.49 \[ \int \frac {x^4 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {-12 \,\mathrm {log}\left (b x +a \right ) a^{4}-36 \,\mathrm {log}\left (b x +a \right ) a^{3} b x -36 \,\mathrm {log}\left (b x +a \right ) a^{2} b^{2} x^{2}-12 \,\mathrm {log}\left (b x +a \right ) a \,b^{3} x^{3}-10 a^{4}-18 a^{3} b x +12 a \,b^{3} x^{3}+3 b^{4} x^{4}}{3 b^{5} \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 a^{2} b x +a^{3}\right )} \] Input:

int(x^4*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x)
 

Output:

( - 12*log(a + b*x)*a**4 - 36*log(a + b*x)*a**3*b*x - 36*log(a + b*x)*a**2 
*b**2*x**2 - 12*log(a + b*x)*a*b**3*x**3 - 10*a**4 - 18*a**3*b*x + 12*a*b* 
*3*x**3 + 3*b**4*x**4)/(3*b**5*(a**3 + 3*a**2*b*x + 3*a*b**2*x**2 + b**3*x 
**3))