\(\int \frac {(A+B x) (a^2+2 a b x+b^2 x^2)^{5/2}}{x} \, dx\) [319]

Optimal result

Integrand size = 29, antiderivative size = 262 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x} \, dx=\frac {5 a^4 A b x \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {5 a^3 A b^2 x^2 \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {10 a^2 A b^3 x^3 \sqrt {a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac {5 a A b^4 x^4 \sqrt {a^2+2 a b x+b^2 x^2}}{4 (a+b x)}+\frac {A b^5 x^5 \sqrt {a^2+2 a b x+b^2 x^2}}{5 (a+b x)}+\frac {B (a+b x)^5 \sqrt {a^2+2 a b x+b^2 x^2}}{6 b}+\frac {a^5 A \sqrt {a^2+2 a b x+b^2 x^2} \log (x)}{a+b x} \] Output:

5*a^4*A*b*x*((b*x+a)^2)^(1/2)/(b*x+a)+5*a^3*A*b^2*x^2*((b*x+a)^2)^(1/2)/(b 
*x+a)+10*a^2*A*b^3*x^3*((b*x+a)^2)^(1/2)/(3*b*x+3*a)+5*a*A*b^4*x^4*((b*x+a 
)^2)^(1/2)/(4*b*x+4*a)+A*b^5*x^5*((b*x+a)^2)^(1/2)/(5*b*x+5*a)+1/6*B*(b*x+ 
a)^5*((b*x+a)^2)^(1/2)/b+a^5*A*((b*x+a)^2)^(1/2)*ln(x)/(b*x+a)
 

Mathematica [A] (verified)

Time = 1.07 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.47 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x} \, dx=\frac {\sqrt {(a+b x)^2} \left (x \left (60 a^5 B+150 a^4 b (2 A+B x)+100 a^3 b^2 x (3 A+2 B x)+50 a^2 b^3 x^2 (4 A+3 B x)+15 a b^4 x^3 (5 A+4 B x)+2 b^5 x^4 (6 A+5 B x)\right )+60 a^5 A \log (x)\right )}{60 (a+b x)} \] Input:

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

Output:

(Sqrt[(a + b*x)^2]*(x*(60*a^5*B + 150*a^4*b*(2*A + B*x) + 100*a^3*b^2*x*(3 
*A + 2*B*x) + 50*a^2*b^3*x^2*(4*A + 3*B*x) + 15*a*b^4*x^3*(5*A + 4*B*x) + 
2*b^5*x^4*(6*A + 5*B*x)) + 60*a^5*A*Log[x]))/(60*(a + b*x))
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.40, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {1187, 27, 90, 49, 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.

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

\(\Big \downarrow \) 1187

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 90

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

\(\Big \downarrow \) 49

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (A \left (a^5 \log (x)+5 a^4 b x+5 a^3 b^2 x^2+\frac {10}{3} a^2 b^3 x^3+\frac {5}{4} a b^4 x^4+\frac {b^5 x^5}{5}\right )+\frac {B (a+b x)^6}{6 b}\right )}{a+b x}\)

Input:

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

Output:

(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*((B*(a + b*x)^6)/(6*b) + A*(5*a^4*b*x + 5*a 
^3*b^2*x^2 + (10*a^2*b^3*x^3)/3 + (5*a*b^4*x^4)/4 + (b^5*x^5)/5 + a^5*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 49

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int 
[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] 
&& IGtQ[m, 0] && IGtQ[m + n + 2, 0]
 

rule 90

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), 
 x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p 
+ 2))   Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, 
p}, x] && NeQ[n + p + 2, 0]
 

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 = 0.99 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.53

method	result	size
default	\(\frac {\left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} \left (10 B \,b^{5} x^{6}+12 A \,b^{5} x^{5}+60 B a \,b^{4} x^{5}+75 A a \,b^{4} x^{4}+150 B \,a^{2} b^{3} x^{4}+200 A \,a^{2} b^{3} x^{3}+200 B \,a^{3} b^{2} x^{3}+300 A \,a^{3} b^{2} x^{2}+150 B \,a^{4} b \,x^{2}+60 A \ln \left (x \right ) a^{5}+300 A \,a^{4} b x +60 B \,a^{5} x \right )}{60 \left (b x +a \right )^{5}}\)	\(139\)
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, B \,b^{5} x^{6}}{6 b x +6 a}+\frac {A \,b^{5} x^{5} \sqrt {\left (b x +a \right )^{2}}}{5 b x +5 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, B a \,b^{4} x^{5}}{b x +a}+\frac {5 \sqrt {\left (b x +a \right )^{2}}\, A a \,b^{4} x^{4}}{4 \left (b x +a \right )}+\frac {5 \sqrt {\left (b x +a \right )^{2}}\, B \,a^{2} b^{3} x^{4}}{2 \left (b x +a \right )}+\frac {10 \sqrt {\left (b x +a \right )^{2}}\, A \,a^{2} b^{3} x^{3}}{3 \left (b x +a \right )}+\frac {10 \sqrt {\left (b x +a \right )^{2}}\, B \,a^{3} b^{2} x^{3}}{3 \left (b x +a \right )}+\frac {5 a^{3} A \,b^{2} x^{2} \sqrt {\left (b x +a \right )^{2}}}{b x +a}+\frac {5 \sqrt {\left (b x +a \right )^{2}}\, B \,a^{4} b \,x^{2}}{2 \left (b x +a \right )}+\frac {5 a^{4} A b x \sqrt {\left (b x +a \right )^{2}}}{b x +a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, B \,a^{5} x}{b x +a}+\frac {a^{5} A \sqrt {\left (b x +a \right )^{2}}\, \ln \left (x \right )}{b x +a}\)	\(310\)

Input:

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

Output:

1/60*((b*x+a)^2)^(5/2)*(10*B*b^5*x^6+12*A*b^5*x^5+60*B*a*b^4*x^5+75*A*a*b^ 
4*x^4+150*B*a^2*b^3*x^4+200*A*a^2*b^3*x^3+200*B*a^3*b^2*x^3+300*A*a^3*b^2* 
x^2+150*B*a^4*b*x^2+60*A*ln(x)*a^5+300*A*a^4*b*x+60*B*a^5*x)/(b*x+a)^5
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.44 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x} \, dx=\frac {1}{6} \, B b^{5} x^{6} + A a^{5} \log \left (x\right ) + \frac {1}{5} \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{5} + \frac {5}{4} \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{4} + \frac {10}{3} \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{3} + \frac {5}{2} \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{2} + {\left (B a^{5} + 5 \, A a^{4} b\right )} x \] Input:

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

Output:

1/6*B*b^5*x^6 + A*a^5*log(x) + 1/5*(5*B*a*b^4 + A*b^5)*x^5 + 5/4*(2*B*a^2* 
b^3 + A*a*b^4)*x^4 + 10/3*(B*a^3*b^2 + A*a^2*b^3)*x^3 + 5/2*(B*a^4*b + 2*A 
*a^3*b^2)*x^2 + (B*a^5 + 5*A*a^4*b)*x
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 236, normalized size of antiderivative = 0.90 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x} \, dx=\left (-1\right )^{2 \, b^{2} x + 2 \, a b} A a^{5} \log \left (2 \, b^{2} x + 2 \, a b\right ) - \left (-1\right )^{2 \, a b x + 2 \, a^{2}} A a^{5} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right ) + \frac {1}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A a^{3} b x + \frac {3}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A a^{4} + \frac {1}{4} \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A a b x + \frac {7}{12} \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A a^{2} + \frac {1}{6} \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B x + \frac {1}{5} \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B a}{6 \, b} \] Input:

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

Output:

(-1)^(2*b^2*x + 2*a*b)*A*a^5*log(2*b^2*x + 2*a*b) - (-1)^(2*a*b*x + 2*a^2) 
*A*a^5*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)*A*a^3*b*x + 3/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*A*a^4 + 1/4*(b^2*x^2 + 2 
*a*b*x + a^2)^(3/2)*A*a*b*x + 7/12*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*a^2 + 
 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*x + 1/5*(b^2*x^2 + 2*a*b*x + a^2)^( 
5/2)*A + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*a/b
 

Giac [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.73 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x} \, dx=\frac {1}{6} \, B b^{5} x^{6} \mathrm {sgn}\left (b x + a\right ) + B a b^{4} x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{5} \, A b^{5} x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{2} \, B a^{2} b^{3} x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{4} \, A a b^{4} x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {10}{3} \, B a^{3} b^{2} x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {10}{3} \, A a^{2} b^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{2} \, B a^{4} b x^{2} \mathrm {sgn}\left (b x + a\right ) + 5 \, A a^{3} b^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + B a^{5} x \mathrm {sgn}\left (b x + a\right ) + 5 \, A a^{4} b x \mathrm {sgn}\left (b x + a\right ) + A a^{5} \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (b x + a\right ) \] Input:

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

Output:

1/6*B*b^5*x^6*sgn(b*x + a) + B*a*b^4*x^5*sgn(b*x + a) + 1/5*A*b^5*x^5*sgn( 
b*x + a) + 5/2*B*a^2*b^3*x^4*sgn(b*x + a) + 5/4*A*a*b^4*x^4*sgn(b*x + a) + 
 10/3*B*a^3*b^2*x^3*sgn(b*x + a) + 10/3*A*a^2*b^3*x^3*sgn(b*x + a) + 5/2*B 
*a^4*b*x^2*sgn(b*x + a) + 5*A*a^3*b^2*x^2*sgn(b*x + a) + B*a^5*x*sgn(b*x + 
 a) + 5*A*a^4*b*x*sgn(b*x + a) + A*a^5*log(abs(x))*sgn(b*x + a)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.24 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x} \, dx=\mathrm {log}\left (x \right ) a^{6}+6 a^{5} b x +\frac {15 a^{4} b^{2} x^{2}}{2}+\frac {20 a^{3} b^{3} x^{3}}{3}+\frac {15 a^{2} b^{4} x^{4}}{4}+\frac {6 a \,b^{5} x^{5}}{5}+\frac {b^{6} x^{6}}{6} \] Input:

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

Output:

(60*log(x)*a**6 + 360*a**5*b*x + 450*a**4*b**2*x**2 + 400*a**3*b**3*x**3 + 
 225*a**2*b**4*x**4 + 72*a*b**5*x**5 + 10*b**6*x**6)/60