\(\int \frac {x^4 (A+B x)}{(a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\) [343]

Optimal result

Integrand size = 29, antiderivative size = 249 \[ \int \frac {x^4 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {a^3 (4 A b-5 a B)}{b^6 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a^4 (A b-a B)}{2 b^6 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 a (A b-2 a B) x (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(A b-3 a B) x^2 (a+b x)}{2 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B x^3 (a+b x)}{3 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 a^2 (3 A b-5 a B) (a+b x) \log (a+b x)}{b^6 \sqrt {a^2+2 a b x+b^2 x^2}} \] Output:

a^3*(4*A*b-5*B*a)/b^6/((b*x+a)^2)^(1/2)-1/2*a^4*(A*b-B*a)/b^6/(b*x+a)/((b* 
x+a)^2)^(1/2)-3*a*(A*b-2*B*a)*x*(b*x+a)/b^5/((b*x+a)^2)^(1/2)+1/2*(A*b-3*B 
*a)*x^2*(b*x+a)/b^4/((b*x+a)^2)^(1/2)+1/3*B*x^3*(b*x+a)/b^3/((b*x+a)^2)^(1 
/2)+2*a^2*(3*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 = 140, normalized size of antiderivative = 0.56 \[ \int \frac {x^4 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {-27 a^5 B+b^5 x^4 (3 A+2 B x)+3 a^4 b (7 A+2 B x)-a b^4 x^3 (12 A+5 B x)+a^2 b^3 x^2 (-33 A+20 B x)+3 a^3 b^2 x (2 A+21 B x)-12 a^2 (-3 A b+5 a B) (a+b x)^2 \log (a+b x)}{6 b^6 (a+b x) \sqrt {(a+b x)^2}} \] Input:

Integrate[(x^4*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]
 

Output:

(-27*a^5*B + b^5*x^4*(3*A + 2*B*x) + 3*a^4*b*(7*A + 2*B*x) - a*b^4*x^3*(12 
*A + 5*B*x) + a^2*b^3*x^2*(-33*A + 20*B*x) + 3*a^3*b^2*x*(2*A + 21*B*x) - 
12*a^2*(-3*A*b + 5*a*B)*(a + b*x)^2*Log[a + b*x])/(6*b^6*(a + b*x)*Sqrt[(a 
 + b*x)^2])
 

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 142, 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.

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

\(\Big \downarrow \) 1187

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 86

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

\(\Big \downarrow \) 2009

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

Input:

Int[(x^4*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]
 

Output:

((a + b*x)*((-3*a*(A*b - 2*a*B)*x)/b^5 + ((A*b - 3*a*B)*x^2)/(2*b^4) + (B* 
x^3)/(3*b^3) - (a^4*(A*b - a*B))/(2*b^6*(a + b*x)^2) + (a^3*(4*A*b - 5*a*B 
))/(b^6*(a + b*x)) + (2*a^2*(3*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 = 1.20 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.62

method	result	size
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\frac {1}{3} x^{3} B \,b^{2}+\frac {1}{2} x^{2} b^{2} A -\frac {3}{2} B a \,x^{2} b -3 a b A x +6 a^{2} B x \right )}{\left (b x +a \right ) b^{5}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\left (4 A \,a^{3} b -5 a^{4} B \right ) x +\frac {a^{4} \left (7 A b -9 B a \right )}{2 b}\right )}{\left (b x +a \right )^{3} b^{5}}+\frac {2 \sqrt {\left (b x +a \right )^{2}}\, a^{2} \left (3 A b -5 B a \right ) \ln \left (b x +a \right )}{\left (b x +a \right ) b^{6}}\)	\(155\)
default	\(\frac {\left (2 B \,x^{5} b^{5}+3 A \,b^{5} x^{4}-5 B a \,b^{4} x^{4}+36 A \ln \left (b x +a \right ) x^{2} a^{2} b^{3}-12 A a \,b^{4} x^{3}-60 B \ln \left (b x +a \right ) x^{2} a^{3} b^{2}+20 B \,a^{2} b^{3} x^{3}+72 A \ln \left (b x +a \right ) x \,a^{3} b^{2}-33 A \,a^{2} b^{3} x^{2}-120 B \ln \left (b x +a \right ) x \,a^{4} b +63 B \,a^{3} b^{2} x^{2}+36 A \ln \left (b x +a \right ) a^{4} b +6 A \,a^{3} b^{2} x -60 B \ln \left (b x +a \right ) a^{5}+6 B \,a^{4} b x +21 A \,a^{4} b -27 B \,a^{5}\right ) \left (b x +a \right )}{6 b^{6} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\)	\(217\)

Input:

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

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.79 \[ \int \frac {x^4 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {2 \, B b^{5} x^{5} - 27 \, B a^{5} + 21 \, A a^{4} b - {\left (5 \, B a b^{4} - 3 \, A b^{5}\right )} x^{4} + 4 \, {\left (5 \, B a^{2} b^{3} - 3 \, A a b^{4}\right )} x^{3} + 3 \, {\left (21 \, B a^{3} b^{2} - 11 \, A a^{2} b^{3}\right )} x^{2} + 6 \, {\left (B a^{4} b + A a^{3} b^{2}\right )} x - 12 \, {\left (5 \, B a^{5} - 3 \, A a^{4} b + {\left (5 \, B a^{3} b^{2} - 3 \, A a^{2} b^{3}\right )} x^{2} + 2 \, {\left (5 \, B a^{4} b - 3 \, A a^{3} b^{2}\right )} x\right )} \log \left (b x + a\right )}{6 \, {\left (b^{8} x^{2} + 2 \, a b^{7} x + a^{2} b^{6}\right )}} \] Input:

integrate(x^4*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas")
 

Output:

1/6*(2*B*b^5*x^5 - 27*B*a^5 + 21*A*a^4*b - (5*B*a*b^4 - 3*A*b^5)*x^4 + 4*( 
5*B*a^2*b^3 - 3*A*a*b^4)*x^3 + 3*(21*B*a^3*b^2 - 11*A*a^2*b^3)*x^2 + 6*(B* 
a^4*b + A*a^3*b^2)*x - 12*(5*B*a^5 - 3*A*a^4*b + (5*B*a^3*b^2 - 3*A*a^2*b^ 
3)*x^2 + 2*(5*B*a^4*b - 3*A*a^3*b^2)*x)*log(b*x + a))/(b^8*x^2 + 2*a*b^7*x 
 + a^2*b^6)
 

Sympy [F]

\[ \int \frac {x^4 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {x^{4} \left (A + B x\right )}{\left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(x**4*(B*x+A)/(b**2*x**2+2*a*b*x+a**2)**(3/2),x)
 

Output:

Integral(x**4*(A + B*x)/((a + b*x)**2)**(3/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.22 \[ \int \frac {x^4 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {B x^{4}}{3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} - \frac {7 \, B a x^{3}}{6 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{3}} + \frac {A x^{3}}{2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} + \frac {9 \, B a^{2} x^{2}}{2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{4}} - \frac {5 \, A a x^{2}}{2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{3}} - \frac {10 \, B a^{3} \log \left (x + \frac {a}{b}\right )}{b^{6}} + \frac {6 \, A a^{2} \log \left (x + \frac {a}{b}\right )}{b^{5}} + \frac {9 \, B a^{4}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{6}} - \frac {5 \, A a^{3}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{5}} - \frac {20 \, B a^{4} x}{b^{7} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {12 \, A a^{3} x}{b^{6} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {39 \, B a^{5}}{2 \, b^{8} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {23 \, A a^{4}}{2 \, b^{7} {\left (x + \frac {a}{b}\right )}^{2}} \] Input:

integrate(x^4*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima")
 

Output:

1/3*B*x^4/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^2) - 7/6*B*a*x^3/(sqrt(b^2*x^2 
+ 2*a*b*x + a^2)*b^3) + 1/2*A*x^3/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^2) + 9/ 
2*B*a^2*x^2/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^4) - 5/2*A*a*x^2/(sqrt(b^2*x^ 
2 + 2*a*b*x + a^2)*b^3) - 10*B*a^3*log(x + a/b)/b^6 + 6*A*a^2*log(x + a/b) 
/b^5 + 9*B*a^4/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^6) - 5*A*a^3/(sqrt(b^2*x^2 
 + 2*a*b*x + a^2)*b^5) - 20*B*a^4*x/(b^7*(x + a/b)^2) + 12*A*a^3*x/(b^6*(x 
 + a/b)^2) - 39/2*B*a^5/(b^8*(x + a/b)^2) + 23/2*A*a^4/(b^7*(x + a/b)^2)
 

Giac [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.60 \[ \int \frac {x^4 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {2 \, {\left (5 \, B a^{3} - 3 \, A a^{2} b\right )} \log \left ({\left | b x + a \right |}\right )}{b^{6} \mathrm {sgn}\left (b x + a\right )} - \frac {9 \, B a^{5} - 7 \, A a^{4} b + 2 \, {\left (5 \, B a^{4} b - 4 \, A a^{3} b^{2}\right )} x}{2 \, {\left (b x + a\right )}^{2} b^{6} \mathrm {sgn}\left (b x + a\right )} + \frac {2 \, B b^{6} x^{3} - 9 \, B a b^{5} x^{2} + 3 \, A b^{6} x^{2} + 36 \, B a^{2} b^{4} x - 18 \, A a b^{5} x}{6 \, b^{9} \mathrm {sgn}\left (b x + a\right )} \] Input:

integrate(x^4*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")
 

Output:

-2*(5*B*a^3 - 3*A*a^2*b)*log(abs(b*x + a))/(b^6*sgn(b*x + a)) - 1/2*(9*B*a 
^5 - 7*A*a^4*b + 2*(5*B*a^4*b - 4*A*a^3*b^2)*x)/((b*x + a)^2*b^6*sgn(b*x + 
 a)) + 1/6*(2*B*b^6*x^3 - 9*B*a*b^5*x^2 + 3*A*b^6*x^2 + 36*B*a^2*b^4*x - 1 
8*A*a*b^5*x)/(b^9*sgn(b*x + a))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {x^4\,\left (A+B\,x\right )}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \] Input:

int((x^4*(A + B*x))/(a^2 + b^2*x^2 + 2*a*b*x)^(3/2),x)
 

Output:

int((x^4*(A + B*x))/(a^2 + b^2*x^2 + 2*a*b*x)^(3/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.29 \[ \int \frac {x^4 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {-12 \,\mathrm {log}\left (b x +a \right ) a^{4}-12 \,\mathrm {log}\left (b x +a \right ) a^{3} b x +12 a^{3} b x +6 a^{2} b^{2} x^{2}-2 a \,b^{3} x^{3}+b^{4} x^{4}}{3 b^{5} \left (b x +a \right )} \] Input:

int(x^4*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x)
 

Output:

( - 12*log(a + b*x)*a**4 - 12*log(a + b*x)*a**3*b*x + 12*a**3*b*x + 6*a**2 
*b**2*x**2 - 2*a*b**3*x**3 + b**4*x**4)/(3*b**5*(a + b*x))