\(\int \frac {(A+B x) (a^2+2 a b x+b^2 x^2)^{3/2}}{x^9} \, dx\) [308]

Optimal result

Integrand size = 29, antiderivative size = 210 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^9} \, dx=-\frac {a^3 A \sqrt {a^2+2 a b x+b^2 x^2}}{8 x^8 (a+b x)}-\frac {a^2 (3 A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{7 x^7 (a+b x)}-\frac {a b (A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{2 x^6 (a+b x)}-\frac {b^2 (A b+3 a B) \sqrt {a^2+2 a b x+b^2 x^2}}{5 x^5 (a+b x)}-\frac {b^3 B \sqrt {a^2+2 a b x+b^2 x^2}}{4 x^4 (a+b x)} \] Output:

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

Mathematica [A] (verified)

Time = 1.01 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.41 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^9} \, dx=-\frac {\sqrt {(a+b x)^2} \left (14 b^3 x^3 (4 A+5 B x)+28 a b^2 x^2 (5 A+6 B x)+20 a^2 b x (6 A+7 B x)+5 a^3 (7 A+8 B x)\right )}{280 x^8 (a+b x)} \] Input:

Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/x^9,x]
 

Output:

-1/280*(Sqrt[(a + b*x)^2]*(14*b^3*x^3*(4*A + 5*B*x) + 28*a*b^2*x^2*(5*A + 
6*B*x) + 20*a^2*b*x*(6*A + 7*B*x) + 5*a^3*(7*A + 8*B*x)))/(x^8*(a + b*x))
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.49, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {1187, 27, 85, 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.

\(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2} (A+B x)}{x^9} \, dx\)

\(\Big \downarrow \) 1187

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {b^3 (a+b x)^3 (A+B x)}{x^9}dx}{b^3 (a+b x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x)^3 (A+B x)}{x^9}dx}{a+b x}\)

\(\Big \downarrow \) 85

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {A a^3}{x^9}+\frac {(3 A b+a B) a^2}{x^8}+\frac {3 b (A b+a B) a}{x^7}+\frac {b^3 B}{x^5}+\frac {b^2 (A b+3 a B)}{x^6}\right )dx}{a+b x}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (-\frac {a^3 A}{8 x^8}-\frac {a^2 (a B+3 A b)}{7 x^7}-\frac {b^2 (3 a B+A b)}{5 x^5}-\frac {a b (a B+A b)}{2 x^6}-\frac {b^3 B}{4 x^4}\right )}{a+b x}\)

Input:

Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/x^9,x]
 

Output:

((-1/8*(a^3*A)/x^8 - (a^2*(3*A*b + a*B))/(7*x^7) - (a*b*(A*b + a*B))/(2*x^ 
6) - (b^2*(A*b + 3*a*B))/(5*x^5) - (b^3*B)/(4*x^4))*Sqrt[a^2 + 2*a*b*x + b 
^2*x^2])/(a + b*x)
 

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 85

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_] : 
> Int[ExpandIntegrand[(a + b*x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, 
 d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && NeQ[b*e + a* 
f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n 
 + p + 2, 0] && RationalQ[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 
1])
 

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 = 1.45 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.43

method	result	size
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {B \,b^{3} x^{4}}{4}+\left (-\frac {1}{5} A \,b^{3}-\frac {3}{5} B a \,b^{2}\right ) x^{3}+\left (-\frac {1}{2} A a \,b^{2}-\frac {1}{2} B \,a^{2} b \right ) x^{2}+\left (-\frac {3}{7} A \,a^{2} b -\frac {1}{7} B \,a^{3}\right ) x -\frac {a^{3} A}{8}\right )}{\left (b x +a \right ) x^{8}}\)	\(90\)
gosper	\(-\frac {\left (70 B \,b^{3} x^{4}+56 A \,b^{3} x^{3}+168 B a \,b^{2} x^{3}+140 A a \,b^{2} x^{2}+140 B \,a^{2} b \,x^{2}+120 A \,a^{2} b x +40 B \,a^{3} x +35 a^{3} A \right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{280 x^{8} \left (b x +a \right )^{3}}\)	\(92\)
default	\(-\frac {\left (70 B \,b^{3} x^{4}+56 A \,b^{3} x^{3}+168 B a \,b^{2} x^{3}+140 A a \,b^{2} x^{2}+140 B \,a^{2} b \,x^{2}+120 A \,a^{2} b x +40 B \,a^{3} x +35 a^{3} A \right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{280 x^{8} \left (b x +a \right )^{3}}\)	\(92\)
orering	\(-\frac {\left (70 B \,b^{3} x^{4}+56 A \,b^{3} x^{3}+168 B a \,b^{2} x^{3}+140 A a \,b^{2} x^{2}+140 B \,a^{2} b \,x^{2}+120 A \,a^{2} b x +40 B \,a^{3} x +35 a^{3} A \right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{\frac {3}{2}}}{280 x^{8} \left (b x +a \right )^{3}}\)	\(101\)

Input:

int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/x^9,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.35 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^9} \, dx=-\frac {70 \, B b^{3} x^{4} + 35 \, A a^{3} + 56 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{3} + 140 \, {\left (B a^{2} b + A a b^{2}\right )} x^{2} + 40 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{280 \, x^{8}} \] Input:

integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/x^9,x, algorithm="fricas")
 

Output:

-1/280*(70*B*b^3*x^4 + 35*A*a^3 + 56*(3*B*a*b^2 + A*b^3)*x^3 + 140*(B*a^2* 
b + A*a*b^2)*x^2 + 40*(B*a^3 + 3*A*a^2*b)*x)/x^8
 

Sympy [F]

\[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^9} \, dx=\int \frac {\left (A + B x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{x^{9}}\, dx \] Input:

integrate((B*x+A)*(b**2*x**2+2*a*b*x+a**2)**(3/2)/x**9,x)
 

Output:

Integral((A + B*x)*((a + b*x)**2)**(3/2)/x**9, x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 495 vs. \(2 (145) = 290\).

Time = 0.04 (sec) , antiderivative size = 495, normalized size of antiderivative = 2.36 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^9} \, dx=-\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B b^{7}}{4 \, a^{7}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A b^{8}}{4 \, a^{8}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B b^{6}}{4 \, a^{6} x} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A b^{7}}{4 \, a^{7} x} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{5}}{4 \, a^{7} x^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{6}}{4 \, a^{8} x^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{4}}{4 \, a^{6} x^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{5}}{4 \, a^{7} x^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{3}}{4 \, a^{5} x^{4}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{4}}{4 \, a^{6} x^{4}} - \frac {17 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{2}}{70 \, a^{4} x^{5}} + \frac {69 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{3}}{280 \, a^{5} x^{5}} + \frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b}{14 \, a^{3} x^{6}} - \frac {13 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{2}}{56 \, a^{4} x^{6}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B}{7 \, a^{2} x^{7}} + \frac {11 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b}{56 \, a^{3} x^{7}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A}{8 \, a^{2} x^{8}} \] Input:

integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/x^9,x, algorithm="maxima")
 

Output:

-1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*b^7/a^7 + 1/4*(b^2*x^2 + 2*a*b*x + 
a^2)^(3/2)*A*b^8/a^8 - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*b^6/(a^6*x) + 
 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*b^7/(a^7*x) + 1/4*(b^2*x^2 + 2*a*b* 
x + a^2)^(5/2)*B*b^5/(a^7*x^2) - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^6 
/(a^8*x^2) - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*b^4/(a^6*x^3) + 1/4*(b^ 
2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^5/(a^7*x^3) + 1/4*(b^2*x^2 + 2*a*b*x + a^ 
2)^(5/2)*B*b^3/(a^5*x^4) - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^4/(a^6* 
x^4) - 17/70*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*b^2/(a^4*x^5) + 69/280*(b^2 
*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^3/(a^5*x^5) + 3/14*(b^2*x^2 + 2*a*b*x + a^ 
2)^(5/2)*B*b/(a^3*x^6) - 13/56*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^2/(a^4* 
x^6) - 1/7*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B/(a^2*x^7) + 11/56*(b^2*x^2 + 
2*a*b*x + a^2)^(5/2)*A*b/(a^3*x^7) - 1/8*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A 
/(a^2*x^8)
 

Giac [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.71 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^9} \, dx=\frac {{\left (2 \, B a b^{7} - A b^{8}\right )} \mathrm {sgn}\left (b x + a\right )}{280 \, a^{5}} - \frac {70 \, B b^{3} x^{4} \mathrm {sgn}\left (b x + a\right ) + 168 \, B a b^{2} x^{3} \mathrm {sgn}\left (b x + a\right ) + 56 \, A b^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + 140 \, B a^{2} b x^{2} \mathrm {sgn}\left (b x + a\right ) + 140 \, A a b^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 40 \, B a^{3} x \mathrm {sgn}\left (b x + a\right ) + 120 \, A a^{2} b x \mathrm {sgn}\left (b x + a\right ) + 35 \, A a^{3} \mathrm {sgn}\left (b x + a\right )}{280 \, x^{8}} \] Input:

integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/x^9,x, algorithm="giac")
 

Output:

1/280*(2*B*a*b^7 - A*b^8)*sgn(b*x + a)/a^5 - 1/280*(70*B*b^3*x^4*sgn(b*x + 
 a) + 168*B*a*b^2*x^3*sgn(b*x + a) + 56*A*b^3*x^3*sgn(b*x + a) + 140*B*a^2 
*b*x^2*sgn(b*x + a) + 140*A*a*b^2*x^2*sgn(b*x + a) + 40*B*a^3*x*sgn(b*x + 
a) + 120*A*a^2*b*x*sgn(b*x + a) + 35*A*a^3*sgn(b*x + a))/x^8
 

Mupad [B] (verification not implemented)

Time = 10.76 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.93 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^9} \, dx=-\frac {\left (\frac {B\,a^3}{7}+\frac {3\,A\,b\,a^2}{7}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{x^7\,\left (a+b\,x\right )}-\frac {\left (\frac {A\,b^3}{5}+\frac {3\,B\,a\,b^2}{5}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{x^5\,\left (a+b\,x\right )}-\frac {A\,a^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{8\,x^8\,\left (a+b\,x\right )}-\frac {B\,b^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{4\,x^4\,\left (a+b\,x\right )}-\frac {a\,b\,\left (A\,b+B\,a\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{2\,x^6\,\left (a+b\,x\right )} \] Input:

int(((A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/x^9,x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.22 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^9} \, dx=\frac {-70 b^{4} x^{4}-224 a \,b^{3} x^{3}-280 a^{2} b^{2} x^{2}-160 a^{3} b x -35 a^{4}}{280 x^{8}} \] Input:

int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/x^9,x)
 

Output:

( - 35*a**4 - 160*a**3*b*x - 280*a**2*b**2*x**2 - 224*a*b**3*x**3 - 70*b** 
4*x**4)/(280*x**8)