\(\int \frac {(A+B x) (a^2+2 a b x+b^2 x^2)^{5/2}}{x^{12}} \, dx\) [330]

Optimal result

Integrand size = 29, antiderivative size = 241 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{12}} \, dx=\frac {b^4 (A b-a B) (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{6 a^6 x^6}-\frac {A \left (a^2+2 a b x+b^2 x^2\right )^{7/2}}{11 a^2 x^{11}}+\frac {(15 A b-11 a B) \left (a^2+2 a b x+b^2 x^2\right )^{7/2}}{110 a^3 x^{10}}-\frac {b (155 A b-143 a B) \left (a^2+2 a b x+b^2 x^2\right )^{7/2}}{990 a^4 x^9}+\frac {b^2 (325 A b-319 a B) \left (a^2+2 a b x+b^2 x^2\right )^{7/2}}{1980 a^5 x^8}-\frac {b^3 (2305 A b-2299 a B) \left (a^2+2 a b x+b^2 x^2\right )^{7/2}}{13860 a^6 x^7} \] Output:

1/6*b^4*(A*b-B*a)*(b*x+a)*(b^2*x^2+2*a*b*x+a^2)^(5/2)/a^6/x^6-1/11*A*(b^2* 
x^2+2*a*b*x+a^2)^(7/2)/a^2/x^11+1/110*(15*A*b-11*B*a)*(b^2*x^2+2*a*b*x+a^2 
)^(7/2)/a^3/x^10-1/990*b*(155*A*b-143*B*a)*(b^2*x^2+2*a*b*x+a^2)^(7/2)/a^4 
/x^9+1/1980*b^2*(325*A*b-319*B*a)*(b^2*x^2+2*a*b*x+a^2)^(7/2)/a^5/x^8-1/13 
860*b^3*(2305*A*b-2299*B*a)*(b^2*x^2+2*a*b*x+a^2)^(7/2)/a^6/x^7
 

Mathematica [A] (verified)

Time = 1.06 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.52 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{12}} \, dx=-\frac {\sqrt {(a+b x)^2} \left (462 b^5 x^5 (5 A+6 B x)+1650 a b^4 x^4 (6 A+7 B x)+2475 a^2 b^3 x^3 (7 A+8 B x)+1925 a^3 b^2 x^2 (8 A+9 B x)+770 a^4 b x (9 A+10 B x)+126 a^5 (10 A+11 B x)\right )}{13860 x^{11} (a+b x)} \] Input:

Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2))/x^12,x]
 

Output:

-1/13860*(Sqrt[(a + b*x)^2]*(462*b^5*x^5*(5*A + 6*B*x) + 1650*a*b^4*x^4*(6 
*A + 7*B*x) + 2475*a^2*b^3*x^3*(7*A + 8*B*x) + 1925*a^3*b^2*x^2*(8*A + 9*B 
*x) + 770*a^4*b*x*(9*A + 10*B*x) + 126*a^5*(10*A + 11*B*x)))/(x^11*(a + b* 
x))
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 145, 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, 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.

\(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2} (A+B x)}{x^{12}} \, dx\)

\(\Big \downarrow \) 1187

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {b^5 (a+b x)^5 (A+B x)}{x^{12}}dx}{b^5 (a+b x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x)^5 (A+B x)}{x^{12}}dx}{a+b x}\)

\(\Big \downarrow \) 85

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {A a^5}{x^{12}}+\frac {(5 A b+a B) a^4}{x^{11}}+\frac {5 b (2 A b+a B) a^3}{x^{10}}+\frac {10 b^2 (A b+a B) a^2}{x^9}+\frac {5 b^3 (A b+2 a B) a}{x^8}+\frac {b^5 B}{x^6}+\frac {b^4 (A b+5 a B)}{x^7}\right )dx}{a+b x}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (-\frac {a^5 A}{11 x^{11}}-\frac {a^4 (a B+5 A b)}{10 x^{10}}-\frac {5 a^3 b (a B+2 A b)}{9 x^9}-\frac {5 a^2 b^2 (a B+A b)}{4 x^8}-\frac {b^4 (5 a B+A b)}{6 x^6}-\frac {5 a b^3 (2 a B+A b)}{7 x^7}-\frac {b^5 B}{5 x^5}\right )}{a+b x}\)

Input:

Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2))/x^12,x]
 

Output:

((-1/11*(a^5*A)/x^11 - (a^4*(5*A*b + a*B))/(10*x^10) - (5*a^3*b*(2*A*b + a 
*B))/(9*x^9) - (5*a^2*b^2*(A*b + a*B))/(4*x^8) - (5*a*b^3*(A*b + 2*a*B))/( 
7*x^7) - (b^4*(A*b + 5*a*B))/(6*x^6) - (b^5*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 = 2.05 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.56

method	result	size
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {B \,b^{5} x^{6}}{5}+\left (-\frac {1}{6} A \,b^{5}-\frac {5}{6} B a \,b^{4}\right ) x^{5}+\left (-\frac {5}{7} A a \,b^{4}-\frac {10}{7} B \,a^{2} b^{3}\right ) x^{4}+\left (-\frac {5}{4} A \,a^{2} b^{3}-\frac {5}{4} B \,a^{3} b^{2}\right ) x^{3}+\left (-\frac {10}{9} a^{3} A \,b^{2}-\frac {5}{9} B \,a^{4} b \right ) x^{2}+\left (-\frac {1}{2} A \,a^{4} b -\frac {1}{10} B \,a^{5}\right ) x -\frac {A \,a^{5}}{11}\right )}{\left (b x +a \right ) x^{11}}\)	\(136\)
gosper	\(-\frac {\left (2772 B \,b^{5} x^{6}+2310 A \,b^{5} x^{5}+11550 B a \,b^{4} x^{5}+9900 A a \,b^{4} x^{4}+19800 B \,a^{2} b^{3} x^{4}+17325 A \,a^{2} b^{3} x^{3}+17325 B \,a^{3} b^{2} x^{3}+15400 A \,a^{3} b^{2} x^{2}+7700 B \,a^{4} b \,x^{2}+6930 A \,a^{4} b x +1386 B \,a^{5} x +1260 A \,a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{13860 x^{11} \left (b x +a \right )^{5}}\)	\(140\)
default	\(-\frac {\left (2772 B \,b^{5} x^{6}+2310 A \,b^{5} x^{5}+11550 B a \,b^{4} x^{5}+9900 A a \,b^{4} x^{4}+19800 B \,a^{2} b^{3} x^{4}+17325 A \,a^{2} b^{3} x^{3}+17325 B \,a^{3} b^{2} x^{3}+15400 A \,a^{3} b^{2} x^{2}+7700 B \,a^{4} b \,x^{2}+6930 A \,a^{4} b x +1386 B \,a^{5} x +1260 A \,a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{13860 x^{11} \left (b x +a \right )^{5}}\)	\(140\)
orering	\(-\frac {\left (2772 B \,b^{5} x^{6}+2310 A \,b^{5} x^{5}+11550 B a \,b^{4} x^{5}+9900 A a \,b^{4} x^{4}+19800 B \,a^{2} b^{3} x^{4}+17325 A \,a^{2} b^{3} x^{3}+17325 B \,a^{3} b^{2} x^{3}+15400 A \,a^{3} b^{2} x^{2}+7700 B \,a^{4} b \,x^{2}+6930 A \,a^{4} b x +1386 B \,a^{5} x +1260 A \,a^{5}\right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{\frac {5}{2}}}{13860 x^{11} \left (b x +a \right )^{5}}\)	\(149\)

Input:

int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(5/2)/x^12,x,method=_RETURNVERBOSE)
 

Output:

((b*x+a)^2)^(1/2)/(b*x+a)*(-1/5*B*b^5*x^6+(-1/6*A*b^5-5/6*B*a*b^4)*x^5+(-5 
/7*A*a*b^4-10/7*B*a^2*b^3)*x^4+(-5/4*A*a^2*b^3-5/4*B*a^3*b^2)*x^3+(-10/9*a 
^3*A*b^2-5/9*B*a^4*b)*x^2+(-1/2*A*a^4*b-1/10*B*a^5)*x-1/11*A*a^5)/x^11
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.49 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{12}} \, dx=-\frac {2772 \, B b^{5} x^{6} + 1260 \, A a^{5} + 2310 \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{5} + 9900 \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{4} + 17325 \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{3} + 7700 \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{2} + 1386 \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x}{13860 \, x^{11}} \] Input:

integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(5/2)/x^12,x, algorithm="fricas")
 

Output:

-1/13860*(2772*B*b^5*x^6 + 1260*A*a^5 + 2310*(5*B*a*b^4 + A*b^5)*x^5 + 990 
0*(2*B*a^2*b^3 + A*a*b^4)*x^4 + 17325*(B*a^3*b^2 + A*a^2*b^3)*x^3 + 7700*( 
B*a^4*b + 2*A*a^3*b^2)*x^2 + 1386*(B*a^5 + 5*A*a^4*b)*x)/x^11
 

Sympy [F]

\[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{12}} \, dx=\int \frac {\left (A + B x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}{x^{12}}\, dx \] Input:

integrate((B*x+A)*(b**2*x**2+2*a*b*x+a**2)**(5/2)/x**12,x)
 

Output:

Integral((A + B*x)*((a + b*x)**2)**(5/2)/x**12, x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 675 vs. \(2 (217) = 434\).

Time = 0.06 (sec) , antiderivative size = 675, normalized size of antiderivative = 2.80 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{12}} \, dx =\text {Too large to display} \] Input:

integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(5/2)/x^12,x, algorithm="maxima")
 

Output:

1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*b^10/a^10 - 1/6*(b^2*x^2 + 2*a*b*x + 
 a^2)^(5/2)*A*b^11/a^11 + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*b^9/(a^9*x 
) - 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^10/(a^10*x) - 1/6*(b^2*x^2 + 2 
*a*b*x + a^2)^(7/2)*B*b^8/(a^10*x^2) + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(7/2) 
*A*b^9/(a^11*x^2) + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*B*b^7/(a^9*x^3) - 
1/6*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*A*b^8/(a^10*x^3) - 1/6*(b^2*x^2 + 2*a* 
b*x + a^2)^(7/2)*B*b^6/(a^8*x^4) + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*A*b 
^7/(a^9*x^4) + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*B*b^5/(a^7*x^5) - 1/6*( 
b^2*x^2 + 2*a*b*x + a^2)^(7/2)*A*b^6/(a^8*x^5) - 1/6*(b^2*x^2 + 2*a*b*x + 
a^2)^(7/2)*B*b^4/(a^6*x^6) + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*A*b^5/(a^ 
7*x^6) + 209/1260*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*B*b^3/(a^5*x^7) - 461/27 
72*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*A*b^4/(a^6*x^7) - 29/180*(b^2*x^2 + 2*a 
*b*x + a^2)^(7/2)*B*b^2/(a^4*x^8) + 65/396*(b^2*x^2 + 2*a*b*x + a^2)^(7/2) 
*A*b^3/(a^5*x^8) + 13/90*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*B*b/(a^3*x^9) - 3 
1/198*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*A*b^2/(a^4*x^9) - 1/10*(b^2*x^2 + 2* 
a*b*x + a^2)^(7/2)*B/(a^2*x^10) + 3/22*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*A*b 
/(a^3*x^10) - 1/11*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*A/(a^2*x^11)
 

Giac [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 221, normalized size of antiderivative = 0.92 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{12}} \, dx=-\frac {{\left (11 \, B a b^{10} - 5 \, A b^{11}\right )} \mathrm {sgn}\left (b x + a\right )}{13860 \, a^{6}} - \frac {2772 \, B b^{5} x^{6} \mathrm {sgn}\left (b x + a\right ) + 11550 \, B a b^{4} x^{5} \mathrm {sgn}\left (b x + a\right ) + 2310 \, A b^{5} x^{5} \mathrm {sgn}\left (b x + a\right ) + 19800 \, B a^{2} b^{3} x^{4} \mathrm {sgn}\left (b x + a\right ) + 9900 \, A a b^{4} x^{4} \mathrm {sgn}\left (b x + a\right ) + 17325 \, B a^{3} b^{2} x^{3} \mathrm {sgn}\left (b x + a\right ) + 17325 \, A a^{2} b^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + 7700 \, B a^{4} b x^{2} \mathrm {sgn}\left (b x + a\right ) + 15400 \, A a^{3} b^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 1386 \, B a^{5} x \mathrm {sgn}\left (b x + a\right ) + 6930 \, A a^{4} b x \mathrm {sgn}\left (b x + a\right ) + 1260 \, A a^{5} \mathrm {sgn}\left (b x + a\right )}{13860 \, x^{11}} \] Input:

integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(5/2)/x^12,x, algorithm="giac")
 

Output:

-1/13860*(11*B*a*b^10 - 5*A*b^11)*sgn(b*x + a)/a^6 - 1/13860*(2772*B*b^5*x 
^6*sgn(b*x + a) + 11550*B*a*b^4*x^5*sgn(b*x + a) + 2310*A*b^5*x^5*sgn(b*x 
+ a) + 19800*B*a^2*b^3*x^4*sgn(b*x + a) + 9900*A*a*b^4*x^4*sgn(b*x + a) + 
17325*B*a^3*b^2*x^3*sgn(b*x + a) + 17325*A*a^2*b^3*x^3*sgn(b*x + a) + 7700 
*B*a^4*b*x^2*sgn(b*x + a) + 15400*A*a^3*b^2*x^2*sgn(b*x + a) + 1386*B*a^5* 
x*sgn(b*x + a) + 6930*A*a^4*b*x*sgn(b*x + a) + 1260*A*a^5*sgn(b*x + a))/x^ 
11
 

Mupad [B] (verification not implemented)

Time = 10.92 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.18 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{12}} \, dx=-\frac {\left (\frac {B\,a^5}{10}+\frac {A\,b\,a^4}{2}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{x^{10}\,\left (a+b\,x\right )}-\frac {\left (\frac {A\,b^5}{6}+\frac {5\,B\,a\,b^4}{6}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{x^6\,\left (a+b\,x\right )}-\frac {A\,a^5\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{11\,x^{11}\,\left (a+b\,x\right )}-\frac {B\,b^5\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{5\,x^5\,\left (a+b\,x\right )}-\frac {5\,a\,b^3\,\left (A\,b+2\,B\,a\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{7\,x^7\,\left (a+b\,x\right )}-\frac {5\,a^3\,b\,\left (2\,A\,b+B\,a\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{9\,x^9\,\left (a+b\,x\right )}-\frac {5\,a^2\,b^2\,\left (A\,b+B\,a\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{4\,x^8\,\left (a+b\,x\right )} \] Input:

int(((A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(5/2))/x^12,x)
 

Output:

- (((B*a^5)/10 + (A*a^4*b)/2)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(x^10*(a + 
b*x)) - (((A*b^5)/6 + (5*B*a*b^4)/6)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(x^6 
*(a + b*x)) - (A*a^5*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(11*x^11*(a + b*x)) 
- (B*b^5*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(5*x^5*(a + b*x)) - (5*a*b^3*(A* 
b + 2*B*a)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(7*x^7*(a + b*x)) - (5*a^3*b*( 
2*A*b + B*a)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(9*x^9*(a + b*x)) - (5*a^2*b 
^2*(A*b + B*a)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(4*x^8*(a + b*x))
 

Reduce [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.28 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{12}} \, dx=\frac {-462 b^{6} x^{6}-2310 a \,b^{5} x^{5}-4950 a^{2} b^{4} x^{4}-5775 a^{3} b^{3} x^{3}-3850 a^{4} b^{2} x^{2}-1386 a^{5} b x -210 a^{6}}{2310 x^{11}} \] Input:

int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(5/2)/x^12,x)
 

Output:

( - 210*a**6 - 1386*a**5*b*x - 3850*a**4*b**2*x**2 - 5775*a**3*b**3*x**3 - 
 4950*a**2*b**4*x**4 - 2310*a*b**5*x**5 - 462*b**6*x**6)/(2310*x**11)