\(\int \frac {(A+B x) (a^2+2 a b x+b^2 x^2)^{5/2}}{x^{9/2}} \, dx\) [437]

Optimal result

Integrand size = 31, antiderivative size = 316 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{9/2}} \, dx=-\frac {2 a^5 A \sqrt {a^2+2 a b x+b^2 x^2}}{7 x^{7/2} (a+b x)}-\frac {2 a^4 (5 A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{5 x^{5/2} (a+b x)}-\frac {10 a^3 b (2 A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{3 x^{3/2} (a+b x)}-\frac {20 a^2 b^2 (A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {x} (a+b x)}+\frac {10 a b^3 (A b+2 a B) \sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {2 b^4 (A b+5 a B) x^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac {2 b^5 B x^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}{5 (a+b x)} \] Output:

-2/7*a^5*A*((b*x+a)^2)^(1/2)/x^(7/2)/(b*x+a)-2/5*a^4*(5*A*b+B*a)*((b*x+a)^ 
2)^(1/2)/x^(5/2)/(b*x+a)-10/3*a^3*b*(2*A*b+B*a)*((b*x+a)^2)^(1/2)/x^(3/2)/ 
(b*x+a)-20*a^2*b^2*(A*b+B*a)*((b*x+a)^2)^(1/2)/x^(1/2)/(b*x+a)+10*a*b^3*(A 
*b+2*B*a)*x^(1/2)*((b*x+a)^2)^(1/2)/(b*x+a)+2*b^4*(A*b+5*B*a)*x^(3/2)*((b* 
x+a)^2)^(1/2)/(3*b*x+3*a)+2*b^5*B*x^(5/2)*((b*x+a)^2)^(1/2)/(5*b*x+5*a)
 

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.39 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{9/2}} \, dx=-\frac {2 \sqrt {(a+b x)^2} \left (1050 a^2 b^3 x^3 (A-B x)-175 a b^4 x^4 (3 A+B x)+350 a^3 b^2 x^2 (A+3 B x)-7 b^5 x^5 (5 A+3 B x)+35 a^4 b x (3 A+5 B x)+3 a^5 (5 A+7 B x)\right )}{105 x^{7/2} (a+b x)} \] Input:

Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2))/x^(9/2),x]
 

Output:

(-2*Sqrt[(a + b*x)^2]*(1050*a^2*b^3*x^3*(A - B*x) - 175*a*b^4*x^4*(3*A + B 
*x) + 350*a^3*b^2*x^2*(A + 3*B*x) - 7*b^5*x^5*(5*A + 3*B*x) + 35*a^4*b*x*( 
3*A + 5*B*x) + 3*a^5*(5*A + 7*B*x)))/(105*x^(7/2)*(a + b*x))
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.49, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, 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.

\(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2} (A+B x)}{x^{9/2}} \, dx\)

\(\Big \downarrow \) 1187

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {b^5 (a+b x)^5 (A+B x)}{x^{9/2}}dx}{b^5 (a+b x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x)^5 (A+B x)}{x^{9/2}}dx}{a+b x}\)

\(\Big \downarrow \) 85

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (-\frac {2 a^5 A}{7 x^{7/2}}-\frac {2 a^4 (a B+5 A b)}{5 x^{5/2}}-\frac {10 a^3 b (a B+2 A b)}{3 x^{3/2}}-\frac {20 a^2 b^2 (a B+A b)}{\sqrt {x}}+\frac {2}{3} b^4 x^{3/2} (5 a B+A b)+10 a b^3 \sqrt {x} (2 a B+A b)+\frac {2}{5} b^5 B x^{5/2}\right )}{a+b x}\)

Input:

Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2))/x^(9/2),x]
 

Output:

(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*((-2*a^5*A)/(7*x^(7/2)) - (2*a^4*(5*A*b + a 
*B))/(5*x^(5/2)) - (10*a^3*b*(2*A*b + a*B))/(3*x^(3/2)) - (20*a^2*b^2*(A*b 
 + a*B))/Sqrt[x] + 10*a*b^3*(A*b + 2*a*B)*Sqrt[x] + (2*b^4*(A*b + 5*a*B)*x 
^(3/2))/3 + (2*b^5*B*x^(5/2))/5))/(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 = 0.94 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.44

method	result	size
gosper	\(-\frac {2 \left (-21 B \,b^{5} x^{6}-35 A \,b^{5} x^{5}-175 B a \,b^{4} x^{5}-525 A a \,b^{4} x^{4}-1050 B \,a^{2} b^{3} x^{4}+1050 A \,a^{2} b^{3} x^{3}+1050 B \,a^{3} b^{2} x^{3}+350 A \,a^{3} b^{2} x^{2}+175 B \,a^{4} b \,x^{2}+105 A \,a^{4} b x +21 B \,a^{5} x +15 A \,a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{105 x^{\frac {7}{2}} \left (b x +a \right )^{5}}\)	\(140\)
default	\(-\frac {2 \left (-21 B \,b^{5} x^{6}-35 A \,b^{5} x^{5}-175 B a \,b^{4} x^{5}-525 A a \,b^{4} x^{4}-1050 B \,a^{2} b^{3} x^{4}+1050 A \,a^{2} b^{3} x^{3}+1050 B \,a^{3} b^{2} x^{3}+350 A \,a^{3} b^{2} x^{2}+175 B \,a^{4} b \,x^{2}+105 A \,a^{4} b x +21 B \,a^{5} x +15 A \,a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{105 x^{\frac {7}{2}} \left (b x +a \right )^{5}}\)	\(140\)
risch	\(-\frac {2 \sqrt {\left (b x +a \right )^{2}}\, \left (-21 B \,b^{5} x^{6}-35 A \,b^{5} x^{5}-175 B a \,b^{4} x^{5}-525 A a \,b^{4} x^{4}-1050 B \,a^{2} b^{3} x^{4}+1050 A \,a^{2} b^{3} x^{3}+1050 B \,a^{3} b^{2} x^{3}+350 A \,a^{3} b^{2} x^{2}+175 B \,a^{4} b \,x^{2}+105 A \,a^{4} b x +21 B \,a^{5} x +15 A \,a^{5}\right )}{105 \left (b x +a \right ) x^{\frac {7}{2}}}\)	\(140\)
orering	\(-\frac {2 \left (-21 B \,b^{5} x^{6}-35 A \,b^{5} x^{5}-175 B a \,b^{4} x^{5}-525 A a \,b^{4} x^{4}-1050 B \,a^{2} b^{3} x^{4}+1050 A \,a^{2} b^{3} x^{3}+1050 B \,a^{3} b^{2} x^{3}+350 A \,a^{3} b^{2} x^{2}+175 B \,a^{4} b \,x^{2}+105 A \,a^{4} b x +21 B \,a^{5} x +15 A \,a^{5}\right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{\frac {5}{2}}}{105 x^{\frac {7}{2}} \left (b x +a \right )^{5}}\)	\(149\)

Input:

int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(5/2)/x^(9/2),x,method=_RETURNVERBOSE)
 

Output:

-2/105*(-21*B*b^5*x^6-35*A*b^5*x^5-175*B*a*b^4*x^5-525*A*a*b^4*x^4-1050*B* 
a^2*b^3*x^4+1050*A*a^2*b^3*x^3+1050*B*a^3*b^2*x^3+350*A*a^3*b^2*x^2+175*B* 
a^4*b*x^2+105*A*a^4*b*x+21*B*a^5*x+15*A*a^5)*((b*x+a)^2)^(5/2)/x^(7/2)/(b* 
x+a)^5
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.38 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{9/2}} \, dx=\frac {2 \, {\left (21 \, B b^{5} x^{6} - 15 \, A a^{5} + 35 \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{5} + 525 \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{4} - 1050 \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{3} - 175 \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{2} - 21 \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x\right )}}{105 \, x^{\frac {7}{2}}} \] Input:

integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(5/2)/x^(9/2),x, algorithm="fricas 
")
 

Output:

2/105*(21*B*b^5*x^6 - 15*A*a^5 + 35*(5*B*a*b^4 + A*b^5)*x^5 + 525*(2*B*a^2 
*b^3 + A*a*b^4)*x^4 - 1050*(B*a^3*b^2 + A*a^2*b^3)*x^3 - 175*(B*a^4*b + 2* 
A*a^3*b^2)*x^2 - 21*(B*a^5 + 5*A*a^4*b)*x)/x^(7/2)
 

Sympy [F]

\[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{9/2}} \, dx=\int \frac {\left (A + B x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}{x^{\frac {9}{2}}}\, dx \] Input:

integrate((B*x+A)*(b**2*x**2+2*a*b*x+a**2)**(5/2)/x**(9/2),x)
 

Output:

Integral((A + B*x)*((a + b*x)**2)**(5/2)/x**(9/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 234, normalized size of antiderivative = 0.74 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{9/2}} \, dx=\frac {2}{15} \, {\left ({\left (3 \, b^{5} x^{2} + 5 \, a b^{4} x\right )} \sqrt {x} + \frac {20 \, {\left (a b^{4} x^{2} + 3 \, a^{2} b^{3} x\right )}}{\sqrt {x}} + \frac {90 \, {\left (a^{2} b^{3} x^{2} - a^{3} b^{2} x\right )}}{x^{\frac {3}{2}}} - \frac {20 \, {\left (3 \, a^{3} b^{2} x^{2} + a^{4} b x\right )}}{x^{\frac {5}{2}}} - \frac {5 \, a^{4} b x^{2} + 3 \, a^{5} x}{x^{\frac {7}{2}}}\right )} B + \frac {2}{105} \, A {\left (\frac {35 \, {\left (b^{5} x^{2} + 3 \, a b^{4} x\right )}}{\sqrt {x}} + \frac {420 \, {\left (a b^{4} x^{2} - a^{2} b^{3} x\right )}}{x^{\frac {3}{2}}} - \frac {210 \, {\left (3 \, a^{2} b^{3} x^{2} + a^{3} b^{2} x\right )}}{x^{\frac {5}{2}}} - \frac {28 \, {\left (5 \, a^{3} b^{2} x^{2} + 3 \, a^{4} b x\right )}}{x^{\frac {7}{2}}} - \frac {3 \, {\left (7 \, a^{4} b x^{2} + 5 \, a^{5} x\right )}}{x^{\frac {9}{2}}}\right )} \] Input:

integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(5/2)/x^(9/2),x, algorithm="maxima 
")
 

Output:

2/15*((3*b^5*x^2 + 5*a*b^4*x)*sqrt(x) + 20*(a*b^4*x^2 + 3*a^2*b^3*x)/sqrt( 
x) + 90*(a^2*b^3*x^2 - a^3*b^2*x)/x^(3/2) - 20*(3*a^3*b^2*x^2 + a^4*b*x)/x 
^(5/2) - (5*a^4*b*x^2 + 3*a^5*x)/x^(7/2))*B + 2/105*A*(35*(b^5*x^2 + 3*a*b 
^4*x)/sqrt(x) + 420*(a*b^4*x^2 - a^2*b^3*x)/x^(3/2) - 210*(3*a^2*b^3*x^2 + 
 a^3*b^2*x)/x^(5/2) - 28*(5*a^3*b^2*x^2 + 3*a^4*b*x)/x^(7/2) - 3*(7*a^4*b* 
x^2 + 5*a^5*x)/x^(9/2))
 

Giac [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.62 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{9/2}} \, dx=\frac {2}{5} \, B b^{5} x^{\frac {5}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {10}{3} \, B a b^{4} x^{\frac {3}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{3} \, A b^{5} x^{\frac {3}{2}} \mathrm {sgn}\left (b x + a\right ) + 20 \, B a^{2} b^{3} \sqrt {x} \mathrm {sgn}\left (b x + a\right ) + 10 \, A a b^{4} \sqrt {x} \mathrm {sgn}\left (b x + a\right ) - \frac {2 \, {\left (1050 \, B a^{3} b^{2} x^{3} \mathrm {sgn}\left (b x + a\right ) + 1050 \, A a^{2} b^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + 175 \, B a^{4} b x^{2} \mathrm {sgn}\left (b x + a\right ) + 350 \, A a^{3} b^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 21 \, B a^{5} x \mathrm {sgn}\left (b x + a\right ) + 105 \, A a^{4} b x \mathrm {sgn}\left (b x + a\right ) + 15 \, A a^{5} \mathrm {sgn}\left (b x + a\right )\right )}}{105 \, x^{\frac {7}{2}}} \] Input:

integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(5/2)/x^(9/2),x, algorithm="giac")
 

Output:

2/5*B*b^5*x^(5/2)*sgn(b*x + a) + 10/3*B*a*b^4*x^(3/2)*sgn(b*x + a) + 2/3*A 
*b^5*x^(3/2)*sgn(b*x + a) + 20*B*a^2*b^3*sqrt(x)*sgn(b*x + a) + 10*A*a*b^4 
*sqrt(x)*sgn(b*x + a) - 2/105*(1050*B*a^3*b^2*x^3*sgn(b*x + a) + 1050*A*a^ 
2*b^3*x^3*sgn(b*x + a) + 175*B*a^4*b*x^2*sgn(b*x + a) + 350*A*a^3*b^2*x^2* 
sgn(b*x + a) + 21*B*a^5*x*sgn(b*x + a) + 105*A*a^4*b*x*sgn(b*x + a) + 15*A 
*a^5*sgn(b*x + a))/x^(7/2)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{9/2}} \, dx=\int \frac {\left (A+B\,x\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}}{x^{9/2}} \,d x \] Input:

int(((A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(5/2))/x^(9/2),x)
 

Output:

int(((A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(5/2))/x^(9/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.23 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{9/2}} \, dx=\frac {\frac {2}{5} b^{6} x^{6}+4 a \,b^{5} x^{5}+30 a^{2} b^{4} x^{4}-40 a^{3} b^{3} x^{3}-10 a^{4} b^{2} x^{2}-\frac {12}{5} a^{5} b x -\frac {2}{7} a^{6}}{\sqrt {x}\, x^{3}} \] Input:

int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(5/2)/x^(9/2),x)
 

Output:

(2*( - 5*a**6 - 42*a**5*b*x - 175*a**4*b**2*x**2 - 700*a**3*b**3*x**3 + 52 
5*a**2*b**4*x**4 + 70*a*b**5*x**5 + 7*b**6*x**6))/(35*sqrt(x)*x**3)