\(\int \frac {(A+B x) (a^2+2 a b x+b^2 x^2)^{5/2}}{x^2} \, dx\) [320]

Optimal result

Integrand size = 29, antiderivative size = 294 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^2} \, dx=-\frac {a^5 A \sqrt {a^2+2 a b x+b^2 x^2}}{x (a+b x)}+\frac {5 a^3 b (2 A b+a B) x \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {5 a^2 b^2 (A b+a B) x^2 \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {5 a b^3 (A b+2 a B) x^3 \sqrt {a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac {b^4 (A b+5 a B) x^4 \sqrt {a^2+2 a b x+b^2 x^2}}{4 (a+b x)}+\frac {b^5 B x^5 \sqrt {a^2+2 a b x+b^2 x^2}}{5 (a+b x)}+\frac {a^4 (5 A b+a B) \sqrt {a^2+2 a b x+b^2 x^2} \log (x)}{a+b x} \] Output:

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

Mathematica [A] (verified)

Time = 1.06 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.44 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^2} \, dx=\frac {\sqrt {(a+b x)^2} \left (-60 a^5 A+300 a^4 b B x^2+300 a^3 b^2 x^2 (2 A+B x)+100 a^2 b^3 x^3 (3 A+2 B x)+25 a b^4 x^4 (4 A+3 B x)+3 b^5 x^5 (5 A+4 B x)+60 a^4 (5 A b+a B) x \log (x)\right )}{60 x (a+b x)} \] Input:

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

Output:

(Sqrt[(a + b*x)^2]*(-60*a^5*A + 300*a^4*b*B*x^2 + 300*a^3*b^2*x^2*(2*A + B 
*x) + 100*a^2*b^3*x^3*(3*A + 2*B*x) + 25*a*b^4*x^4*(4*A + 3*B*x) + 3*b^5*x 
^5*(5*A + 4*B*x) + 60*a^4*(5*A*b + a*B)*x*Log[x]))/(60*x*(a + b*x))
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.45, 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.

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

\(\Big \downarrow \) 1187

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 85

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (-\frac {a^5 A}{x}+a^4 \log (x) (a B+5 A b)+5 a^3 b x (a B+2 A b)+5 a^2 b^2 x^2 (a B+A b)+\frac {1}{4} b^4 x^4 (5 a B+A b)+\frac {5}{3} a b^3 x^3 (2 a B+A b)+\frac {1}{5} b^5 B x^5\right )}{a+b x}\)

Input:

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

Output:

(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-((a^5*A)/x) + 5*a^3*b*(2*A*b + a*B)*x + 5 
*a^2*b^2*(A*b + a*B)*x^2 + (5*a*b^3*(A*b + 2*a*B)*x^3)/3 + (b^4*(A*b + 5*a 
*B)*x^4)/4 + (b^5*B*x^5)/5 + a^4*(5*A*b + a*B)*Log[x]))/(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 = 1.03 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.49

method	result	size
default	\(\frac {\left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} \left (12 B \,b^{5} x^{6}+15 A \,b^{5} x^{5}+75 B a \,b^{4} x^{5}+100 A a \,b^{4} x^{4}+200 B \,a^{2} b^{3} x^{4}+300 A \,a^{2} b^{3} x^{3}+300 B \,a^{3} b^{2} x^{3}+300 A \ln \left (x \right ) x \,a^{4} b +600 A \,a^{3} b^{2} x^{2}+60 B \ln \left (x \right ) a^{5} x +300 B \,a^{4} b \,x^{2}-60 A \,a^{5}\right )}{60 x \left (b x +a \right )^{5}}\)	\(144\)
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, b \left (\frac {1}{5} b^{4} B \,x^{5}+\frac {1}{4} A \,b^{4} x^{4}+\frac {5}{4} B a \,b^{3} x^{4}+\frac {5}{3} A a \,b^{3} x^{3}+\frac {10}{3} B \,a^{2} b^{2} x^{3}+5 A \,a^{2} b^{2} x^{2}+5 B \,a^{3} b \,x^{2}+10 A \,a^{3} b x +5 a^{4} B x \right )}{b x +a}-\frac {a^{5} A \sqrt {\left (b x +a \right )^{2}}}{x \left (b x +a \right )}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (5 A \,a^{4} b +B \,a^{5}\right ) \ln \left (x \right )}{b x +a}\)	\(165\)

Input:

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

Output:

1/60*((b*x+a)^2)^(5/2)*(12*B*b^5*x^6+15*A*b^5*x^5+75*B*a*b^4*x^5+100*A*a*b 
^4*x^4+200*B*a^2*b^3*x^4+300*A*a^2*b^3*x^3+300*B*a^3*b^2*x^3+300*A*ln(x)*x 
*a^4*b+600*A*a^3*b^2*x^2+60*B*ln(x)*a^5*x+300*B*a^4*b*x^2-60*A*a^5)/x/(b*x 
+a)^5
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.41 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^2} \, dx=\frac {12 \, B b^{5} x^{6} - 60 \, A a^{5} + 15 \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{5} + 100 \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{4} + 300 \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{3} + 300 \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{2} + 60 \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x \log \left (x\right )}{60 \, x} \] Input:

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

Output:

1/60*(12*B*b^5*x^6 - 60*A*a^5 + 15*(5*B*a*b^4 + A*b^5)*x^5 + 100*(2*B*a^2* 
b^3 + A*a*b^4)*x^4 + 300*(B*a^3*b^2 + A*a^2*b^3)*x^3 + 300*(B*a^4*b + 2*A* 
a^3*b^2)*x^2 + 60*(B*a^5 + 5*A*a^4*b)*x*log(x))/x
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 386, normalized size of antiderivative = 1.31 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^2} \, dx=\left (-1\right )^{2 \, b^{2} x + 2 \, a b} B a^{5} \log \left (2 \, b^{2} x + 2 \, a b\right ) + 5 \, \left (-1\right )^{2 \, b^{2} x + 2 \, a b} A a^{4} b \log \left (2 \, b^{2} x + 2 \, a b\right ) - \left (-1\right )^{2 \, a b x + 2 \, a^{2}} B a^{5} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right ) - 5 \, \left (-1\right )^{2 \, a b x + 2 \, a^{2}} A a^{4} b \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right ) + \frac {1}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B a^{3} b x + \frac {5}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A a^{2} b^{2} x + \frac {3}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B a^{4} + \frac {15}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A a^{3} b + \frac {1}{4} \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B a b x + \frac {5}{4} \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A b^{2} x + \frac {7}{12} \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B a^{2} + \frac {35}{12} \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A a b + \frac {1}{5} \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A}{x} \] Input:

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

Output:

(-1)^(2*b^2*x + 2*a*b)*B*a^5*log(2*b^2*x + 2*a*b) + 5*(-1)^(2*b^2*x + 2*a* 
b)*A*a^4*b*log(2*b^2*x + 2*a*b) - (-1)^(2*a*b*x + 2*a^2)*B*a^5*log(2*a*b*x 
/abs(x) + 2*a^2/abs(x)) - 5*(-1)^(2*a*b*x + 2*a^2)*A*a^4*b*log(2*a*b*x/abs 
(x) + 2*a^2/abs(x)) + 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B*a^3*b*x + 5/2*sq 
rt(b^2*x^2 + 2*a*b*x + a^2)*A*a^2*b^2*x + 3/2*sqrt(b^2*x^2 + 2*a*b*x + a^2 
)*B*a^4 + 15/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*A*a^3*b + 1/4*(b^2*x^2 + 2*a* 
b*x + a^2)^(3/2)*B*a*b*x + 5/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*b^2*x + 7 
/12*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*a^2 + 35/12*(b^2*x^2 + 2*a*b*x + a^2 
)^(3/2)*A*a*b + 1/5*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B - (b^2*x^2 + 2*a*b*x 
 + a^2)^(5/2)*A/x
 

Giac [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.65 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^2} \, dx=\frac {1}{5} \, B b^{5} x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{4} \, B a b^{4} x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{4} \, A b^{5} x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {10}{3} \, B a^{2} b^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{3} \, A a b^{4} x^{3} \mathrm {sgn}\left (b x + a\right ) + 5 \, B a^{3} b^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 5 \, A a^{2} b^{3} x^{2} \mathrm {sgn}\left (b x + a\right ) + 5 \, B a^{4} b x \mathrm {sgn}\left (b x + a\right ) + 10 \, A a^{3} b^{2} x \mathrm {sgn}\left (b x + a\right ) - \frac {A a^{5} \mathrm {sgn}\left (b x + a\right )}{x} + {\left (B a^{5} \mathrm {sgn}\left (b x + a\right ) + 5 \, A a^{4} b \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | x \right |}\right ) \] Input:

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

Output:

1/5*B*b^5*x^5*sgn(b*x + a) + 5/4*B*a*b^4*x^4*sgn(b*x + a) + 1/4*A*b^5*x^4* 
sgn(b*x + a) + 10/3*B*a^2*b^3*x^3*sgn(b*x + a) + 5/3*A*a*b^4*x^3*sgn(b*x + 
 a) + 5*B*a^3*b^2*x^2*sgn(b*x + a) + 5*A*a^2*b^3*x^2*sgn(b*x + a) + 5*B*a^ 
4*b*x*sgn(b*x + a) + 10*A*a^3*b^2*x*sgn(b*x + a) - A*a^5*sgn(b*x + a)/x + 
(B*a^5*sgn(b*x + a) + 5*A*a^4*b*sgn(b*x + a))*log(abs(x))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.24 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^2} \, dx=\frac {60 \,\mathrm {log}\left (x \right ) a^{5} b x -10 a^{6}+150 a^{4} b^{2} x^{2}+100 a^{3} b^{3} x^{3}+50 a^{2} b^{4} x^{4}+15 a \,b^{5} x^{5}+2 b^{6} x^{6}}{10 x} \] Input:

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

Output:

(60*log(x)*a**5*b*x - 10*a**6 + 150*a**4*b**2*x**2 + 100*a**3*b**3*x**3 + 
50*a**2*b**4*x**4 + 15*a*b**5*x**5 + 2*b**6*x**6)/(10*x)