\(\int \frac {(A+B x) (a^2+2 a b x+b^2 x^2)^{5/2}}{x^{14}} \, dx\) [332]

Optimal result

Integrand size = 29, antiderivative size = 304 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{14}} \, dx=-\frac {a^5 A \sqrt {a^2+2 a b x+b^2 x^2}}{13 x^{13} (a+b x)}-\frac {a^4 (5 A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{12 x^{12} (a+b x)}-\frac {5 a^3 b (2 A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{11 x^{11} (a+b x)}-\frac {a^2 b^2 (A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{x^{10} (a+b x)}-\frac {5 a b^3 (A b+2 a B) \sqrt {a^2+2 a b x+b^2 x^2}}{9 x^9 (a+b x)}-\frac {b^4 (A b+5 a B) \sqrt {a^2+2 a b x+b^2 x^2}}{8 x^8 (a+b x)}-\frac {b^5 B \sqrt {a^2+2 a b x+b^2 x^2}}{7 x^7 (a+b x)} \] Output:

-1/13*a^5*A*((b*x+a)^2)^(1/2)/x^13/(b*x+a)-1/12*a^4*(5*A*b+B*a)*((b*x+a)^2 
)^(1/2)/x^12/(b*x+a)-5/11*a^3*b*(2*A*b+B*a)*((b*x+a)^2)^(1/2)/x^11/(b*x+a) 
-a^2*b^2*(A*b+B*a)*((b*x+a)^2)^(1/2)/x^10/(b*x+a)-5/9*a*b^3*(A*b+2*B*a)*(( 
b*x+a)^2)^(1/2)/x^9/(b*x+a)-1/8*b^4*(A*b+5*B*a)*((b*x+a)^2)^(1/2)/x^8/(b*x 
+a)-1/7*b^5*B*((b*x+a)^2)^(1/2)/x^7/(b*x+a)
 

Mathematica [A] (verified)

Time = 1.05 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.41 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{14}} \, dx=-\frac {\sqrt {(a+b x)^2} \left (1287 b^5 x^5 (7 A+8 B x)+5005 a b^4 x^4 (8 A+9 B x)+8008 a^2 b^3 x^3 (9 A+10 B x)+6552 a^3 b^2 x^2 (10 A+11 B x)+2730 a^4 b x (11 A+12 B x)+462 a^5 (12 A+13 B x)\right )}{72072 x^{13} (a+b x)} \] Input:

Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2))/x^14,x]
 

Output:

-1/72072*(Sqrt[(a + b*x)^2]*(1287*b^5*x^5*(7*A + 8*B*x) + 5005*a*b^4*x^4*( 
8*A + 9*B*x) + 8008*a^2*b^3*x^3*(9*A + 10*B*x) + 6552*a^3*b^2*x^2*(10*A + 
11*B*x) + 2730*a^4*b*x*(11*A + 12*B*x) + 462*a^5*(12*A + 13*B*x)))/(x^13*( 
a + b*x))
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.47, 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.

\(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2} (A+B x)}{x^{14}} \, dx\)

\(\Big \downarrow \) 1187

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {b^5 (a+b x)^5 (A+B x)}{x^{14}}dx}{b^5 (a+b x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x)^5 (A+B x)}{x^{14}}dx}{a+b x}\)

\(\Big \downarrow \) 85

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {A a^5}{x^{14}}+\frac {(5 A b+a B) a^4}{x^{13}}+\frac {5 b (2 A b+a B) a^3}{x^{12}}+\frac {10 b^2 (A b+a B) a^2}{x^{11}}+\frac {5 b^3 (A b+2 a B) a}{x^{10}}+\frac {b^5 B}{x^8}+\frac {b^4 (A b+5 a B)}{x^9}\right )dx}{a+b x}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (-\frac {a^5 A}{13 x^{13}}-\frac {a^4 (a B+5 A b)}{12 x^{12}}-\frac {5 a^3 b (a B+2 A b)}{11 x^{11}}-\frac {a^2 b^2 (a B+A b)}{x^{10}}-\frac {b^4 (5 a B+A b)}{8 x^8}-\frac {5 a b^3 (2 a B+A b)}{9 x^9}-\frac {b^5 B}{7 x^7}\right )}{a+b x}\)

Input:

Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2))/x^14,x]
 

Output:

((-1/13*(a^5*A)/x^13 - (a^4*(5*A*b + a*B))/(12*x^12) - (5*a^3*b*(2*A*b + a 
*B))/(11*x^11) - (a^2*b^2*(A*b + a*B))/x^10 - (5*a*b^3*(A*b + 2*a*B))/(9*x 
^9) - (b^4*(A*b + 5*a*B))/(8*x^8) - (b^5*B)/(7*x^7))*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 = 2.49 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.45

method	result	size
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {B \,b^{5} x^{6}}{7}+\left (-\frac {1}{8} A \,b^{5}-\frac {5}{8} B a \,b^{4}\right ) x^{5}+\left (-\frac {5}{9} A a \,b^{4}-\frac {10}{9} B \,a^{2} b^{3}\right ) x^{4}+\left (-A \,a^{2} b^{3}-B \,a^{3} b^{2}\right ) x^{3}+\left (-\frac {10}{11} a^{3} A \,b^{2}-\frac {5}{11} B \,a^{4} b \right ) x^{2}+\left (-\frac {5}{12} A \,a^{4} b -\frac {1}{12} B \,a^{5}\right ) x -\frac {A \,a^{5}}{13}\right )}{\left (b x +a \right ) x^{13}}\)	\(136\)
gosper	\(-\frac {\left (10296 B \,b^{5} x^{6}+9009 A \,b^{5} x^{5}+45045 B a \,b^{4} x^{5}+40040 A a \,b^{4} x^{4}+80080 B \,a^{2} b^{3} x^{4}+72072 A \,a^{2} b^{3} x^{3}+72072 B \,a^{3} b^{2} x^{3}+65520 A \,a^{3} b^{2} x^{2}+32760 B \,a^{4} b \,x^{2}+30030 A \,a^{4} b x +6006 B \,a^{5} x +5544 A \,a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{72072 x^{13} \left (b x +a \right )^{5}}\)	\(140\)
default	\(-\frac {\left (10296 B \,b^{5} x^{6}+9009 A \,b^{5} x^{5}+45045 B a \,b^{4} x^{5}+40040 A a \,b^{4} x^{4}+80080 B \,a^{2} b^{3} x^{4}+72072 A \,a^{2} b^{3} x^{3}+72072 B \,a^{3} b^{2} x^{3}+65520 A \,a^{3} b^{2} x^{2}+32760 B \,a^{4} b \,x^{2}+30030 A \,a^{4} b x +6006 B \,a^{5} x +5544 A \,a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{72072 x^{13} \left (b x +a \right )^{5}}\)	\(140\)
orering	\(-\frac {\left (10296 B \,b^{5} x^{6}+9009 A \,b^{5} x^{5}+45045 B a \,b^{4} x^{5}+40040 A a \,b^{4} x^{4}+80080 B \,a^{2} b^{3} x^{4}+72072 A \,a^{2} b^{3} x^{3}+72072 B \,a^{3} b^{2} x^{3}+65520 A \,a^{3} b^{2} x^{2}+32760 B \,a^{4} b \,x^{2}+30030 A \,a^{4} b x +6006 B \,a^{5} x +5544 A \,a^{5}\right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{\frac {5}{2}}}{72072 x^{13} \left (b x +a \right )^{5}}\)	\(149\)

Input:

int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(5/2)/x^14,x,method=_RETURNVERBOSE)
 

Output:

((b*x+a)^2)^(1/2)/(b*x+a)*(-1/7*B*b^5*x^6+(-1/8*A*b^5-5/8*B*a*b^4)*x^5+(-5 
/9*A*a*b^4-10/9*B*a^2*b^3)*x^4+(-A*a^2*b^3-B*a^3*b^2)*x^3+(-10/11*a^3*A*b^ 
2-5/11*B*a^4*b)*x^2+(-5/12*A*a^4*b-1/12*B*a^5)*x-1/13*A*a^5)/x^13
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.39 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{14}} \, dx=-\frac {10296 \, B b^{5} x^{6} + 5544 \, A a^{5} + 9009 \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{5} + 40040 \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{4} + 72072 \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{3} + 32760 \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{2} + 6006 \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x}{72072 \, x^{13}} \] Input:

integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(5/2)/x^14,x, algorithm="fricas")
 

Output:

-1/72072*(10296*B*b^5*x^6 + 5544*A*a^5 + 9009*(5*B*a*b^4 + A*b^5)*x^5 + 40 
040*(2*B*a^2*b^3 + A*a*b^4)*x^4 + 72072*(B*a^3*b^2 + A*a^2*b^3)*x^3 + 3276 
0*(B*a^4*b + 2*A*a^3*b^2)*x^2 + 6006*(B*a^5 + 5*A*a^4*b)*x)/x^13
 

Sympy [F]

\[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{14}} \, dx=\int \frac {\left (A + B x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}{x^{14}}\, dx \] Input:

integrate((B*x+A)*(b**2*x**2+2*a*b*x+a**2)**(5/2)/x**14,x)
 

Output:

Integral((A + B*x)*((a + b*x)**2)**(5/2)/x**14, x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 795 vs. \(2 (215) = 430\).

Time = 0.06 (sec) , antiderivative size = 795, normalized size of antiderivative = 2.62 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{14}} \, dx =\text {Too large to display} \] Input:

integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(5/2)/x^14,x, algorithm="maxima")
 

Output:

1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*b^12/a^12 - 1/6*(b^2*x^2 + 2*a*b*x + 
 a^2)^(5/2)*A*b^13/a^13 + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*b^11/(a^11 
*x) - 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^12/(a^12*x) - 1/6*(b^2*x^2 + 
 2*a*b*x + a^2)^(7/2)*B*b^10/(a^12*x^2) + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(7 
/2)*A*b^11/(a^13*x^2) + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*B*b^9/(a^11*x^ 
3) - 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*A*b^10/(a^12*x^3) - 1/6*(b^2*x^2 
+ 2*a*b*x + a^2)^(7/2)*B*b^8/(a^10*x^4) + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(7 
/2)*A*b^9/(a^11*x^4) + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*B*b^7/(a^9*x^5) 
 - 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*A*b^8/(a^10*x^5) - 1/6*(b^2*x^2 + 2 
*a*b*x + a^2)^(7/2)*B*b^6/(a^8*x^6) + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)* 
A*b^7/(a^9*x^6) + 923/5544*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*B*b^5/(a^7*x^7) 
 - 1715/10296*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*A*b^6/(a^8*x^7) - 131/792*(b 
^2*x^2 + 2*a*b*x + a^2)^(7/2)*B*b^4/(a^6*x^8) + 1709/10296*(b^2*x^2 + 2*a* 
b*x + a^2)^(7/2)*A*b^5/(a^7*x^8) + 16/99*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*B 
*b^3/(a^5*x^9) - 211/1287*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*A*b^4/(a^6*x^9) 
- 5/33*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*B*b^2/(a^4*x^10) + 68/429*(b^2*x^2 
+ 2*a*b*x + a^2)^(7/2)*A*b^3/(a^5*x^10) + 17/132*(b^2*x^2 + 2*a*b*x + a^2) 
^(7/2)*B*b/(a^3*x^11) - 251/1716*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*A*b^2/(a^ 
4*x^11) - 1/12*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*B/(a^2*x^12) + 19/156*(b^2* 
x^2 + 2*a*b*x + a^2)^(7/2)*A*b/(a^3*x^12) - 1/13*(b^2*x^2 + 2*a*b*x + a...
 

Giac [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 221, normalized size of antiderivative = 0.73 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{14}} \, dx=-\frac {{\left (13 \, B a b^{12} - 7 \, A b^{13}\right )} \mathrm {sgn}\left (b x + a\right )}{72072 \, a^{8}} - \frac {10296 \, B b^{5} x^{6} \mathrm {sgn}\left (b x + a\right ) + 45045 \, B a b^{4} x^{5} \mathrm {sgn}\left (b x + a\right ) + 9009 \, A b^{5} x^{5} \mathrm {sgn}\left (b x + a\right ) + 80080 \, B a^{2} b^{3} x^{4} \mathrm {sgn}\left (b x + a\right ) + 40040 \, A a b^{4} x^{4} \mathrm {sgn}\left (b x + a\right ) + 72072 \, B a^{3} b^{2} x^{3} \mathrm {sgn}\left (b x + a\right ) + 72072 \, A a^{2} b^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + 32760 \, B a^{4} b x^{2} \mathrm {sgn}\left (b x + a\right ) + 65520 \, A a^{3} b^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 6006 \, B a^{5} x \mathrm {sgn}\left (b x + a\right ) + 30030 \, A a^{4} b x \mathrm {sgn}\left (b x + a\right ) + 5544 \, A a^{5} \mathrm {sgn}\left (b x + a\right )}{72072 \, x^{13}} \] Input:

integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(5/2)/x^14,x, algorithm="giac")
 

Output:

-1/72072*(13*B*a*b^12 - 7*A*b^13)*sgn(b*x + a)/a^8 - 1/72072*(10296*B*b^5* 
x^6*sgn(b*x + a) + 45045*B*a*b^4*x^5*sgn(b*x + a) + 9009*A*b^5*x^5*sgn(b*x 
 + a) + 80080*B*a^2*b^3*x^4*sgn(b*x + a) + 40040*A*a*b^4*x^4*sgn(b*x + a) 
+ 72072*B*a^3*b^2*x^3*sgn(b*x + a) + 72072*A*a^2*b^3*x^3*sgn(b*x + a) + 32 
760*B*a^4*b*x^2*sgn(b*x + a) + 65520*A*a^3*b^2*x^2*sgn(b*x + a) + 6006*B*a 
^5*x*sgn(b*x + a) + 30030*A*a^4*b*x*sgn(b*x + a) + 5544*A*a^5*sgn(b*x + a) 
)/x^13
 

Mupad [B] (verification not implemented)

Time = 10.72 (sec) , antiderivative size = 284, normalized size of antiderivative = 0.93 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{14}} \, dx=-\frac {\left (\frac {B\,a^5}{12}+\frac {5\,A\,b\,a^4}{12}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{x^{12}\,\left (a+b\,x\right )}-\frac {\left (\frac {A\,b^5}{8}+\frac {5\,B\,a\,b^4}{8}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{x^8\,\left (a+b\,x\right )}-\frac {A\,a^5\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{13\,x^{13}\,\left (a+b\,x\right )}-\frac {B\,b^5\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{7\,x^7\,\left (a+b\,x\right )}-\frac {5\,a\,b^3\,\left (A\,b+2\,B\,a\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{9\,x^9\,\left (a+b\,x\right )}-\frac {5\,a^3\,b\,\left (2\,A\,b+B\,a\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{11\,x^{11}\,\left (a+b\,x\right )}-\frac {a^2\,b^2\,\left (A\,b+B\,a\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{x^{10}\,\left (a+b\,x\right )} \] Input:

int(((A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(5/2))/x^14,x)
 

Output:

- (((B*a^5)/12 + (5*A*a^4*b)/12)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(x^12*(a 
 + b*x)) - (((A*b^5)/8 + (5*B*a*b^4)/8)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/( 
x^8*(a + b*x)) - (A*a^5*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(13*x^13*(a + b*x 
)) - (B*b^5*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(7*x^7*(a + b*x)) - (5*a*b^3* 
(A*b + 2*B*a)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(9*x^9*(a + b*x)) - (5*a^3* 
b*(2*A*b + B*a)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(11*x^11*(a + b*x)) - (a^ 
2*b^2*(A*b + B*a)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(x^10*(a + b*x))
 

Reduce [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.22 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{14}} \, dx=\frac {-1716 b^{6} x^{6}-9009 a \,b^{5} x^{5}-20020 a^{2} b^{4} x^{4}-24024 a^{3} b^{3} x^{3}-16380 a^{4} b^{2} x^{2}-6006 a^{5} b x -924 a^{6}}{12012 x^{13}} \] Input:

int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(5/2)/x^14,x)
 

Output:

( - 924*a**6 - 6006*a**5*b*x - 16380*a**4*b**2*x**2 - 24024*a**3*b**3*x**3 
 - 20020*a**2*b**4*x**4 - 9009*a*b**5*x**5 - 1716*b**6*x**6)/(12012*x**13)