\(\int \frac {(A+B x) (a^2+2 a b x+b^2 x^2)^{3/2}}{x^{10}} \, dx\) [309]

Optimal result

Integrand size = 29, antiderivative size = 210 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^{10}} \, dx=-\frac {a^3 A \sqrt {a^2+2 a b x+b^2 x^2}}{9 x^9 (a+b x)}-\frac {a^2 (3 A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{8 x^8 (a+b x)}-\frac {3 a b (A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{7 x^7 (a+b x)}-\frac {b^2 (A b+3 a B) \sqrt {a^2+2 a b x+b^2 x^2}}{6 x^6 (a+b x)}-\frac {b^3 B \sqrt {a^2+2 a b x+b^2 x^2}}{5 x^5 (a+b x)} \] Output:

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

Mathematica [A] (verified)

Time = 1.04 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.41 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^{10}} \, dx=-\frac {\sqrt {(a+b x)^2} \left (84 b^3 x^3 (5 A+6 B x)+180 a b^2 x^2 (6 A+7 B x)+135 a^2 b x (7 A+8 B x)+35 a^3 (8 A+9 B x)\right )}{2520 x^9 (a+b x)} \] Input:

Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/x^10,x]
 

Output:

-1/2520*(Sqrt[(a + b*x)^2]*(84*b^3*x^3*(5*A + 6*B*x) + 180*a*b^2*x^2*(6*A 
+ 7*B*x) + 135*a^2*b*x*(7*A + 8*B*x) + 35*a^3*(8*A + 9*B*x)))/(x^9*(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.

\(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2} (A+B x)}{x^{10}} \, dx\)

\(\Big \downarrow \) 1187

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {b^3 (a+b x)^3 (A+B x)}{x^{10}}dx}{b^3 (a+b x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x)^3 (A+B x)}{x^{10}}dx}{a+b x}\)

\(\Big \downarrow \) 85

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {A a^3}{x^{10}}+\frac {(3 A b+a B) a^2}{x^9}+\frac {3 b (A b+a B) a}{x^8}+\frac {b^3 B}{x^6}+\frac {b^2 (A b+3 a B)}{x^7}\right )dx}{a+b x}\)

\(\Big \downarrow \) 2009

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

Input:

Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/x^10,x]
 

Output:

((-1/9*(a^3*A)/x^9 - (a^2*(3*A*b + a*B))/(8*x^8) - (3*a*b*(A*b + a*B))/(7* 
x^7) - (b^2*(A*b + 3*a*B))/(6*x^6) - (b^3*B)/(5*x^5))*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 = 1.56 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.43

method	result	size
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {B \,b^{3} x^{4}}{5}+\left (-\frac {1}{6} A \,b^{3}-\frac {1}{2} B a \,b^{2}\right ) x^{3}+\left (-\frac {3}{7} A a \,b^{2}-\frac {3}{7} B \,a^{2} b \right ) x^{2}+\left (-\frac {3}{8} A \,a^{2} b -\frac {1}{8} B \,a^{3}\right ) x -\frac {a^{3} A}{9}\right )}{\left (b x +a \right ) x^{9}}\)	\(90\)
gosper	\(-\frac {\left (504 B \,b^{3} x^{4}+420 A \,b^{3} x^{3}+1260 B a \,b^{2} x^{3}+1080 A a \,b^{2} x^{2}+1080 B \,a^{2} b \,x^{2}+945 A \,a^{2} b x +315 B \,a^{3} x +280 a^{3} A \right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{2520 x^{9} \left (b x +a \right )^{3}}\)	\(92\)
default	\(-\frac {\left (504 B \,b^{3} x^{4}+420 A \,b^{3} x^{3}+1260 B a \,b^{2} x^{3}+1080 A a \,b^{2} x^{2}+1080 B \,a^{2} b \,x^{2}+945 A \,a^{2} b x +315 B \,a^{3} x +280 a^{3} A \right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{2520 x^{9} \left (b x +a \right )^{3}}\)	\(92\)
orering	\(-\frac {\left (504 B \,b^{3} x^{4}+420 A \,b^{3} x^{3}+1260 B a \,b^{2} x^{3}+1080 A a \,b^{2} x^{2}+1080 B \,a^{2} b \,x^{2}+945 A \,a^{2} b x +315 B \,a^{3} x +280 a^{3} A \right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{\frac {3}{2}}}{2520 x^{9} \left (b x +a \right )^{3}}\)	\(101\)

Input:

int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/x^10,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.35 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^{10}} \, dx=-\frac {504 \, B b^{3} x^{4} + 280 \, A a^{3} + 420 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{3} + 1080 \, {\left (B a^{2} b + A a b^{2}\right )} x^{2} + 315 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{2520 \, x^{9}} \] Input:

integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/x^10,x, algorithm="fricas")
 

Output:

-1/2520*(504*B*b^3*x^4 + 280*A*a^3 + 420*(3*B*a*b^2 + A*b^3)*x^3 + 1080*(B 
*a^2*b + A*a*b^2)*x^2 + 315*(B*a^3 + 3*A*a^2*b)*x)/x^9
 

Sympy [F]

\[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^{10}} \, dx=\int \frac {\left (A + B x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{x^{10}}\, dx \] Input:

integrate((B*x+A)*(b**2*x**2+2*a*b*x+a**2)**(3/2)/x**10,x)
 

Output:

Integral((A + B*x)*((a + b*x)**2)**(3/2)/x**10, x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 555 vs. \(2 (145) = 290\).

Time = 0.05 (sec) , antiderivative size = 555, normalized size of antiderivative = 2.64 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^{10}} \, dx=\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B b^{8}}{4 \, a^{8}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A b^{9}}{4 \, a^{9}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B b^{7}}{4 \, a^{7} x} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A b^{8}}{4 \, a^{8} x} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{6}}{4 \, a^{8} x^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{7}}{4 \, a^{9} x^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{5}}{4 \, a^{7} x^{3}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{6}}{4 \, a^{8} x^{3}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{4}}{4 \, a^{6} x^{4}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{5}}{4 \, a^{7} x^{4}} + \frac {69 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{3}}{280 \, a^{5} x^{5}} - \frac {125 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{4}}{504 \, a^{6} x^{5}} - \frac {13 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{2}}{56 \, a^{4} x^{6}} + \frac {121 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{3}}{504 \, a^{5} x^{6}} + \frac {11 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b}{56 \, a^{3} x^{7}} - \frac {37 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{2}}{168 \, a^{4} x^{7}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B}{8 \, a^{2} x^{8}} + \frac {13 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b}{72 \, a^{3} x^{8}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A}{9 \, a^{2} x^{9}} \] Input:

integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/x^10,x, algorithm="maxima")
 

Output:

1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*b^8/a^8 - 1/4*(b^2*x^2 + 2*a*b*x + a 
^2)^(3/2)*A*b^9/a^9 + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*b^7/(a^7*x) - 
1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*b^8/(a^8*x) - 1/4*(b^2*x^2 + 2*a*b*x 
 + a^2)^(5/2)*B*b^6/(a^8*x^2) + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^7/ 
(a^9*x^2) + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*b^5/(a^7*x^3) - 1/4*(b^2 
*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^6/(a^8*x^3) - 1/4*(b^2*x^2 + 2*a*b*x + a^2 
)^(5/2)*B*b^4/(a^6*x^4) + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^5/(a^7*x 
^4) + 69/280*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*b^3/(a^5*x^5) - 125/504*(b^ 
2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^4/(a^6*x^5) - 13/56*(b^2*x^2 + 2*a*b*x + 
a^2)^(5/2)*B*b^2/(a^4*x^6) + 121/504*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^3 
/(a^5*x^6) + 11/56*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*b/(a^3*x^7) - 37/168* 
(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^2/(a^4*x^7) - 1/8*(b^2*x^2 + 2*a*b*x + 
 a^2)^(5/2)*B/(a^2*x^8) + 13/72*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b/(a^3*x 
^8) - 1/9*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A/(a^2*x^9)
 

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.71 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^{10}} \, dx=-\frac {{\left (9 \, B a b^{8} - 5 \, A b^{9}\right )} \mathrm {sgn}\left (b x + a\right )}{2520 \, a^{6}} - \frac {504 \, B b^{3} x^{4} \mathrm {sgn}\left (b x + a\right ) + 1260 \, B a b^{2} x^{3} \mathrm {sgn}\left (b x + a\right ) + 420 \, A b^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + 1080 \, B a^{2} b x^{2} \mathrm {sgn}\left (b x + a\right ) + 1080 \, A a b^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 315 \, B a^{3} x \mathrm {sgn}\left (b x + a\right ) + 945 \, A a^{2} b x \mathrm {sgn}\left (b x + a\right ) + 280 \, A a^{3} \mathrm {sgn}\left (b x + a\right )}{2520 \, x^{9}} \] Input:

integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/x^10,x, algorithm="giac")
 

Output:

-1/2520*(9*B*a*b^8 - 5*A*b^9)*sgn(b*x + a)/a^6 - 1/2520*(504*B*b^3*x^4*sgn 
(b*x + a) + 1260*B*a*b^2*x^3*sgn(b*x + a) + 420*A*b^3*x^3*sgn(b*x + a) + 1 
080*B*a^2*b*x^2*sgn(b*x + a) + 1080*A*a*b^2*x^2*sgn(b*x + a) + 315*B*a^3*x 
*sgn(b*x + a) + 945*A*a^2*b*x*sgn(b*x + a) + 280*A*a^3*sgn(b*x + a))/x^9
 

Mupad [B] (verification not implemented)

Time = 10.81 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.93 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^{10}} \, dx=-\frac {\left (\frac {B\,a^3}{8}+\frac {3\,A\,b\,a^2}{8}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{x^8\,\left (a+b\,x\right )}-\frac {\left (\frac {A\,b^3}{6}+\frac {B\,a\,b^2}{2}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{x^6\,\left (a+b\,x\right )}-\frac {A\,a^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{9\,x^9\,\left (a+b\,x\right )}-\frac {B\,b^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{5\,x^5\,\left (a+b\,x\right )}-\frac {3\,a\,b\,\left (A\,b+B\,a\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{7\,x^7\,\left (a+b\,x\right )} \] Input:

int(((A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/x^10,x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.22 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^{10}} \, dx=\frac {-126 b^{4} x^{4}-420 a \,b^{3} x^{3}-540 a^{2} b^{2} x^{2}-315 a^{3} b x -70 a^{4}}{630 x^{9}} \] Input:

int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/x^10,x)
 

Output:

( - 70*a**4 - 315*a**3*b*x - 540*a**2*b**2*x**2 - 420*a*b**3*x**3 - 126*b* 
*4*x**4)/(630*x**9)