\(\int \frac {(A+B x) (a^2+2 a b x+b^2 x^2)^{5/2}}{x^{3/2}} \, dx\) [434]

Optimal result

Integrand size = 31, antiderivative size = 314 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{3/2}} \, dx=-\frac {2 a^5 A \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {x} (a+b x)}+\frac {2 a^4 (5 A b+a B) \sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {10 a^3 b (2 A b+a B) x^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac {4 a^2 b^2 (A b+a B) x^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {10 a b^3 (A b+2 a B) x^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}{7 (a+b x)}+\frac {2 b^4 (A b+5 a B) x^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}}{9 (a+b x)}+\frac {2 b^5 B x^{11/2} \sqrt {a^2+2 a b x+b^2 x^2}}{11 (a+b x)} \] Output:

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

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.39 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{3/2}} \, dx=\frac {2 \sqrt {(a+b x)^2} \left (-693 a^5 (A-B x)+1155 a^4 b x (3 A+B x)+462 a^3 b^2 x^2 (5 A+3 B x)+198 a^2 b^3 x^3 (7 A+5 B x)+55 a b^4 x^4 (9 A+7 B x)+7 b^5 x^5 (11 A+9 B x)\right )}{693 \sqrt {x} (a+b x)} \] Input:

Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2))/x^(3/2),x]
 

Output:

(2*Sqrt[(a + b*x)^2]*(-693*a^5*(A - B*x) + 1155*a^4*b*x*(3*A + B*x) + 462* 
a^3*b^2*x^2*(5*A + 3*B*x) + 198*a^2*b^3*x^3*(7*A + 5*B*x) + 55*a*b^4*x^4*( 
9*A + 7*B*x) + 7*b^5*x^5*(11*A + 9*B*x)))/(693*Sqrt[x]*(a + b*x))
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 153, 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.

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

\(\Big \downarrow \) 1187

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 85

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

\(\Big \downarrow \) 2009

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

Input:

Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2))/x^(3/2),x]
 

Output:

(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*((-2*a^5*A)/Sqrt[x] + 2*a^4*(5*A*b + a*B)*S 
qrt[x] + (10*a^3*b*(2*A*b + a*B)*x^(3/2))/3 + 4*a^2*b^2*(A*b + a*B)*x^(5/2 
) + (10*a*b^3*(A*b + 2*a*B)*x^(7/2))/7 + (2*b^4*(A*b + 5*a*B)*x^(9/2))/9 + 
 (2*b^5*B*x^(11/2))/11))/(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 = 140, normalized size of antiderivative = 0.45

method	result	size
gosper	\(-\frac {2 \left (-63 B \,b^{5} x^{6}-77 A \,b^{5} x^{5}-385 B a \,b^{4} x^{5}-495 A a \,b^{4} x^{4}-990 B \,a^{2} b^{3} x^{4}-1386 A \,a^{2} b^{3} x^{3}-1386 B \,a^{3} b^{2} x^{3}-2310 A \,a^{3} b^{2} x^{2}-1155 B \,a^{4} b \,x^{2}-3465 A \,a^{4} b x -693 B \,a^{5} x +693 A \,a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{693 \sqrt {x}\, \left (b x +a \right )^{5}}\)	\(140\)
default	\(-\frac {2 \left (-63 B \,b^{5} x^{6}-77 A \,b^{5} x^{5}-385 B a \,b^{4} x^{5}-495 A a \,b^{4} x^{4}-990 B \,a^{2} b^{3} x^{4}-1386 A \,a^{2} b^{3} x^{3}-1386 B \,a^{3} b^{2} x^{3}-2310 A \,a^{3} b^{2} x^{2}-1155 B \,a^{4} b \,x^{2}-3465 A \,a^{4} b x -693 B \,a^{5} x +693 A \,a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{693 \sqrt {x}\, \left (b x +a \right )^{5}}\)	\(140\)
risch	\(-\frac {2 \sqrt {\left (b x +a \right )^{2}}\, \left (-63 B \,b^{5} x^{6}-77 A \,b^{5} x^{5}-385 B a \,b^{4} x^{5}-495 A a \,b^{4} x^{4}-990 B \,a^{2} b^{3} x^{4}-1386 A \,a^{2} b^{3} x^{3}-1386 B \,a^{3} b^{2} x^{3}-2310 A \,a^{3} b^{2} x^{2}-1155 B \,a^{4} b \,x^{2}-3465 A \,a^{4} b x -693 B \,a^{5} x +693 A \,a^{5}\right )}{693 \left (b x +a \right ) \sqrt {x}}\)	\(140\)
orering	\(-\frac {2 \left (-63 B \,b^{5} x^{6}-77 A \,b^{5} x^{5}-385 B a \,b^{4} x^{5}-495 A a \,b^{4} x^{4}-990 B \,a^{2} b^{3} x^{4}-1386 A \,a^{2} b^{3} x^{3}-1386 B \,a^{3} b^{2} x^{3}-2310 A \,a^{3} b^{2} x^{2}-1155 B \,a^{4} b \,x^{2}-3465 A \,a^{4} b x -693 B \,a^{5} x +693 A \,a^{5}\right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{\frac {5}{2}}}{693 \sqrt {x}\, \left (b x +a \right )^{5}}\)	\(149\)

Input:

int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(5/2)/x^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-2/693*(-63*B*b^5*x^6-77*A*b^5*x^5-385*B*a*b^4*x^5-495*A*a*b^4*x^4-990*B*a 
^2*b^3*x^4-1386*A*a^2*b^3*x^3-1386*B*a^3*b^2*x^3-2310*A*a^3*b^2*x^2-1155*B 
*a^4*b*x^2-3465*A*a^4*b*x-693*B*a^5*x+693*A*a^5)*((b*x+a)^2)^(5/2)/x^(1/2) 
/(b*x+a)^5
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.38 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{3/2}} \, dx=\frac {2 \, {\left (63 \, B b^{5} x^{6} - 693 \, A a^{5} + 77 \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{5} + 495 \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{4} + 1386 \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{3} + 1155 \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{2} + 693 \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x\right )}}{693 \, \sqrt {x}} \] Input:

integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(5/2)/x^(3/2),x, algorithm="fricas 
")
 

Output:

2/693*(63*B*b^5*x^6 - 693*A*a^5 + 77*(5*B*a*b^4 + A*b^5)*x^5 + 495*(2*B*a^ 
2*b^3 + A*a*b^4)*x^4 + 1386*(B*a^3*b^2 + A*a^2*b^3)*x^3 + 1155*(B*a^4*b + 
2*A*a^3*b^2)*x^2 + 693*(B*a^5 + 5*A*a^4*b)*x)/sqrt(x)
 

Sympy [F]

\[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{3/2}} \, dx=\int \frac {\left (A + B x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}{x^{\frac {3}{2}}}\, dx \] Input:

integrate((B*x+A)*(b**2*x**2+2*a*b*x+a**2)**(5/2)/x**(3/2),x)
 

Output:

Integral((A + B*x)*((a + b*x)**2)**(5/2)/x**(3/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 238, normalized size of antiderivative = 0.76 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{3/2}} \, dx=\frac {2}{315} \, {\left (5 \, {\left (7 \, b^{5} x^{2} + 9 \, a b^{4} x\right )} x^{\frac {5}{2}} + 36 \, {\left (5 \, a b^{4} x^{2} + 7 \, a^{2} b^{3} x\right )} x^{\frac {3}{2}} + 126 \, {\left (3 \, a^{2} b^{3} x^{2} + 5 \, a^{3} b^{2} x\right )} \sqrt {x} + \frac {420 \, {\left (a^{3} b^{2} x^{2} + 3 \, a^{4} b x\right )}}{\sqrt {x}} + \frac {315 \, {\left (a^{4} b x^{2} - a^{5} x\right )}}{x^{\frac {3}{2}}}\right )} A + \frac {2}{3465} \, {\left (35 \, {\left (9 \, b^{5} x^{2} + 11 \, a b^{4} x\right )} x^{\frac {7}{2}} + 220 \, {\left (7 \, a b^{4} x^{2} + 9 \, a^{2} b^{3} x\right )} x^{\frac {5}{2}} + 594 \, {\left (5 \, a^{2} b^{3} x^{2} + 7 \, a^{3} b^{2} x\right )} x^{\frac {3}{2}} + 924 \, {\left (3 \, a^{3} b^{2} x^{2} + 5 \, a^{4} b x\right )} \sqrt {x} + \frac {1155 \, {\left (a^{4} b x^{2} + 3 \, a^{5} x\right )}}{\sqrt {x}}\right )} B \] Input:

integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(5/2)/x^(3/2),x, algorithm="maxima 
")
 

Output:

2/315*(5*(7*b^5*x^2 + 9*a*b^4*x)*x^(5/2) + 36*(5*a*b^4*x^2 + 7*a^2*b^3*x)* 
x^(3/2) + 126*(3*a^2*b^3*x^2 + 5*a^3*b^2*x)*sqrt(x) + 420*(a^3*b^2*x^2 + 3 
*a^4*b*x)/sqrt(x) + 315*(a^4*b*x^2 - a^5*x)/x^(3/2))*A + 2/3465*(35*(9*b^5 
*x^2 + 11*a*b^4*x)*x^(7/2) + 220*(7*a*b^4*x^2 + 9*a^2*b^3*x)*x^(5/2) + 594 
*(5*a^2*b^3*x^2 + 7*a^3*b^2*x)*x^(3/2) + 924*(3*a^3*b^2*x^2 + 5*a^4*b*x)*s 
qrt(x) + 1155*(a^4*b*x^2 + 3*a^5*x)/sqrt(x))*B
 

Giac [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.63 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{3/2}} \, dx=\frac {2}{11} \, B b^{5} x^{\frac {11}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {10}{9} \, B a b^{4} x^{\frac {9}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{9} \, A b^{5} x^{\frac {9}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {20}{7} \, B a^{2} b^{3} x^{\frac {7}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {10}{7} \, A a b^{4} x^{\frac {7}{2}} \mathrm {sgn}\left (b x + a\right ) + 4 \, B a^{3} b^{2} x^{\frac {5}{2}} \mathrm {sgn}\left (b x + a\right ) + 4 \, A a^{2} b^{3} x^{\frac {5}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {10}{3} \, B a^{4} b x^{\frac {3}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {20}{3} \, A a^{3} b^{2} x^{\frac {3}{2}} \mathrm {sgn}\left (b x + a\right ) + 2 \, B a^{5} \sqrt {x} \mathrm {sgn}\left (b x + a\right ) + 10 \, A a^{4} b \sqrt {x} \mathrm {sgn}\left (b x + a\right ) - \frac {2 \, A a^{5} \mathrm {sgn}\left (b x + a\right )}{\sqrt {x}} \] Input:

integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(5/2)/x^(3/2),x, algorithm="giac")
 

Output:

2/11*B*b^5*x^(11/2)*sgn(b*x + a) + 10/9*B*a*b^4*x^(9/2)*sgn(b*x + a) + 2/9 
*A*b^5*x^(9/2)*sgn(b*x + a) + 20/7*B*a^2*b^3*x^(7/2)*sgn(b*x + a) + 10/7*A 
*a*b^4*x^(7/2)*sgn(b*x + a) + 4*B*a^3*b^2*x^(5/2)*sgn(b*x + a) + 4*A*a^2*b 
^3*x^(5/2)*sgn(b*x + a) + 10/3*B*a^4*b*x^(3/2)*sgn(b*x + a) + 20/3*A*a^3*b 
^2*x^(3/2)*sgn(b*x + a) + 2*B*a^5*sqrt(x)*sgn(b*x + a) + 10*A*a^4*b*sqrt(x 
)*sgn(b*x + a) - 2*A*a^5*sgn(b*x + a)/sqrt(x)
 

Mupad [B] (verification not implemented)

Time = 11.19 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.45 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{3/2}} \, dx=\frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (\frac {2\,B\,b^4\,x^6}{11}-\frac {2\,A\,a^5}{b}+\frac {10\,a^3\,x^2\,\left (2\,A\,b+B\,a\right )}{3}+\frac {x^5\,\left (154\,A\,b^5+770\,B\,a\,b^4\right )}{693\,b}+4\,a^2\,b\,x^3\,\left (A\,b+B\,a\right )+\frac {10\,a\,b^2\,x^4\,\left (A\,b+2\,B\,a\right )}{7}+\frac {2\,a^4\,x\,\left (5\,A\,b+B\,a\right )}{b}\right )}{x^{3/2}+\frac {a\,\sqrt {x}}{b}} \] Input:

int(((A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(5/2))/x^(3/2),x)
 

Output:

((a^2 + b^2*x^2 + 2*a*b*x)^(1/2)*((2*B*b^4*x^6)/11 - (2*A*a^5)/b + (10*a^3 
*x^2*(2*A*b + B*a))/3 + (x^5*(154*A*b^5 + 770*B*a*b^4))/(693*b) + 4*a^2*b* 
x^3*(A*b + B*a) + (10*a*b^2*x^4*(A*b + 2*B*a))/7 + (2*a^4*x*(5*A*b + B*a)) 
/b))/(x^(3/2) + (a*x^(1/2))/b)
 

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.22 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{3/2}} \, dx=\frac {\frac {2}{11} b^{6} x^{6}+\frac {4}{3} a \,b^{5} x^{5}+\frac {30}{7} a^{2} b^{4} x^{4}+8 a^{3} b^{3} x^{3}+10 a^{4} b^{2} x^{2}+12 a^{5} b x -2 a^{6}}{\sqrt {x}} \] Input:

int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(5/2)/x^(3/2),x)
 

Output:

(2*( - 231*a**6 + 1386*a**5*b*x + 1155*a**4*b**2*x**2 + 924*a**3*b**3*x**3 
 + 495*a**2*b**4*x**4 + 154*a*b**5*x**5 + 21*b**6*x**6))/(231*sqrt(x))