\(\int \frac {x^{3/2} (A+B x)}{(a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\) [449]

Optimal result

Integrand size = 31, antiderivative size = 189 \[ \int \frac {x^{3/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {(3 A b-7 a B) \sqrt {x}}{4 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(A b-a B) x^{3/2}}{2 b^2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 B \sqrt {x} (a+b x)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 (A b-5 a B) (a+b x) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 \sqrt {a} b^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.57 \[ \int \frac {x^{3/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {(a+b x) \left (\sqrt {b} \sqrt {x} \left (15 a^2 B+b^2 x (-5 A+8 B x)+a (-3 A b+25 b B x)\right )+\frac {3 (A b-5 a B) (a+b x)^2 \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{\sqrt {a}}\right )}{4 b^{7/2} \left ((a+b x)^2\right )^{3/2}} \] Input:

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

Output:

((a + b*x)*(Sqrt[b]*Sqrt[x]*(15*a^2*B + b^2*x*(-5*A + 8*B*x) + a*(-3*A*b + 
 25*b*B*x)) + (3*(A*b - 5*a*B)*(a + b*x)^2*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a 
]])/Sqrt[a]))/(4*b^(7/2)*((a + b*x)^2)^(3/2))
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.74, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {1187, 27, 87, 51, 60, 73, 218}

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.

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

\(\Big \downarrow \) 1187

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 51

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

((a + b*x)*(((A*b - a*B)*x^(5/2))/(2*a*b*(a + b*x)^2) - ((A*b - 5*a*B)*(-( 
x^(3/2)/(b*(a + b*x))) + (3*((2*Sqrt[x])/b - (2*Sqrt[a]*ArcTan[(Sqrt[b]*Sq 
rt[x])/Sqrt[a]])/b^(3/2)))/(2*b)))/(4*a*b)))/Sqrt[a^2 + 2*a*b*x + b^2*x^2]
 

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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) 
Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x 
] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
 

rule 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 87

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

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

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]
 

Maple [A] (verified)

Time = 1.18 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.61

method	result	size
risch	\(\frac {2 B \sqrt {x}\, \sqrt {\left (b x +a \right )^{2}}}{b^{3} \left (b x +a \right )}+\frac {\left (\frac {2 \left (-\frac {5}{8} b^{2} A +\frac {9}{8} a b B \right ) x^{\frac {3}{2}}-\frac {a \left (3 A b -7 B a \right ) \sqrt {x}}{4}}{\left (b x +a \right )^{2}}+\frac {3 \left (A b -5 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \sqrt {a b}}\right ) \sqrt {\left (b x +a \right )^{2}}}{b^{3} \left (b x +a \right )}\)	\(115\)
default	\(-\frac {\left (5 A \,x^{\frac {3}{2}} \sqrt {a b}\, b^{2}-3 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) b^{3} x^{2}-25 B \,x^{\frac {3}{2}} \sqrt {a b}\, a b -8 B \,x^{\frac {5}{2}} \sqrt {a b}\, b^{2}+15 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a \,b^{2} x^{2}-6 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a \,b^{2} x +30 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a^{2} b x +3 A \sqrt {x}\, \sqrt {a b}\, a b -3 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a^{2} b -15 B \sqrt {x}\, \sqrt {a b}\, a^{2}+15 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a^{3}\right ) \left (b x +a \right )}{4 \sqrt {a b}\, b^{3} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\)	\(208\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.69 \[ \int \frac {x^{3/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\left [\frac {3 \, {\left (5 \, B a^{3} - A a^{2} b + {\left (5 \, B a b^{2} - A b^{3}\right )} x^{2} + 2 \, {\left (5 \, B a^{2} b - A a b^{2}\right )} x\right )} \sqrt {-a b} \log \left (\frac {b x - a - 2 \, \sqrt {-a b} \sqrt {x}}{b x + a}\right ) + 2 \, {\left (8 \, B a b^{3} x^{2} + 15 \, B a^{3} b - 3 \, A a^{2} b^{2} + 5 \, {\left (5 \, B a^{2} b^{2} - A a b^{3}\right )} x\right )} \sqrt {x}}{8 \, {\left (a b^{6} x^{2} + 2 \, a^{2} b^{5} x + a^{3} b^{4}\right )}}, \frac {3 \, {\left (5 \, B a^{3} - A a^{2} b + {\left (5 \, B a b^{2} - A b^{3}\right )} x^{2} + 2 \, {\left (5 \, B a^{2} b - A a b^{2}\right )} x\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b}}{b \sqrt {x}}\right ) + {\left (8 \, B a b^{3} x^{2} + 15 \, B a^{3} b - 3 \, A a^{2} b^{2} + 5 \, {\left (5 \, B a^{2} b^{2} - A a b^{3}\right )} x\right )} \sqrt {x}}{4 \, {\left (a b^{6} x^{2} + 2 \, a^{2} b^{5} x + a^{3} b^{4}\right )}}\right ] \] Input:

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

Output:

[1/8*(3*(5*B*a^3 - A*a^2*b + (5*B*a*b^2 - A*b^3)*x^2 + 2*(5*B*a^2*b - A*a* 
b^2)*x)*sqrt(-a*b)*log((b*x - a - 2*sqrt(-a*b)*sqrt(x))/(b*x + a)) + 2*(8* 
B*a*b^3*x^2 + 15*B*a^3*b - 3*A*a^2*b^2 + 5*(5*B*a^2*b^2 - A*a*b^3)*x)*sqrt 
(x))/(a*b^6*x^2 + 2*a^2*b^5*x + a^3*b^4), 1/4*(3*(5*B*a^3 - A*a^2*b + (5*B 
*a*b^2 - A*b^3)*x^2 + 2*(5*B*a^2*b - A*a*b^2)*x)*sqrt(a*b)*arctan(sqrt(a*b 
)/(b*sqrt(x))) + (8*B*a*b^3*x^2 + 15*B*a^3*b - 3*A*a^2*b^2 + 5*(5*B*a^2*b^ 
2 - A*a*b^3)*x)*sqrt(x))/(a*b^6*x^2 + 2*a^2*b^5*x + a^3*b^4)]
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.25 \[ \int \frac {x^{3/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {{\left (5 \, {\left (7 \, B a b^{3} - A b^{4}\right )} x^{2} + 3 \, {\left (5 \, B a^{2} b^{2} + A a b^{3}\right )} x\right )} x^{\frac {5}{2}} + 12 \, {\left (4 \, B a^{2} b^{2} x^{2} + {\left (B a^{3} b + A a^{2} b^{2}\right )} x\right )} x^{\frac {3}{2}} + {\left (3 \, {\left (7 \, B a^{3} b - A a^{2} b^{2}\right )} x^{2} + {\left (5 \, B a^{4} + A a^{3} b\right )} x\right )} \sqrt {x}}{24 \, {\left (a^{2} b^{5} x^{3} + 3 \, a^{3} b^{4} x^{2} + 3 \, a^{4} b^{3} x + a^{5} b^{2}\right )}} - \frac {3 \, {\left (5 \, B a - A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} b^{3}} - \frac {5 \, {\left (7 \, B a b - A b^{2}\right )} x^{\frac {3}{2}} - 18 \, {\left (5 \, B a^{2} - A a b\right )} \sqrt {x}}{24 \, a^{2} b^{3}} \] Input:

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

Output:

1/24*((5*(7*B*a*b^3 - A*b^4)*x^2 + 3*(5*B*a^2*b^2 + A*a*b^3)*x)*x^(5/2) + 
12*(4*B*a^2*b^2*x^2 + (B*a^3*b + A*a^2*b^2)*x)*x^(3/2) + (3*(7*B*a^3*b - A 
*a^2*b^2)*x^2 + (5*B*a^4 + A*a^3*b)*x)*sqrt(x))/(a^2*b^5*x^3 + 3*a^3*b^4*x 
^2 + 3*a^4*b^3*x + a^5*b^2) - 3/4*(5*B*a - A*b)*arctan(b*sqrt(x)/sqrt(a*b) 
)/(sqrt(a*b)*b^3) - 1/24*(5*(7*B*a*b - A*b^2)*x^(3/2) - 18*(5*B*a^2 - A*a* 
b)*sqrt(x))/(a^2*b^3)
 

Giac [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.59 \[ \int \frac {x^{3/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {2 \, B \sqrt {x}}{b^{3} \mathrm {sgn}\left (b x + a\right )} - \frac {3 \, {\left (5 \, B a - A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} b^{3} \mathrm {sgn}\left (b x + a\right )} + \frac {9 \, B a b x^{\frac {3}{2}} - 5 \, A b^{2} x^{\frac {3}{2}} + 7 \, B a^{2} \sqrt {x} - 3 \, A a b \sqrt {x}}{4 \, {\left (b x + a\right )}^{2} b^{3} \mathrm {sgn}\left (b x + a\right )} \] Input:

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

Output:

2*B*sqrt(x)/(b^3*sgn(b*x + a)) - 3/4*(5*B*a - A*b)*arctan(b*sqrt(x)/sqrt(a 
*b))/(sqrt(a*b)*b^3*sgn(b*x + a)) + 1/4*(9*B*a*b*x^(3/2) - 5*A*b^2*x^(3/2) 
 + 7*B*a^2*sqrt(x) - 3*A*a*b*sqrt(x))/((b*x + a)^2*b^3*sgn(b*x + a))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.35 \[ \int \frac {x^{3/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {-3 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) a -3 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) b x +3 \sqrt {x}\, a b +2 \sqrt {x}\, b^{2} x}{b^{3} \left (b x +a \right )} \] Input:

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

Output:

( - 3*sqrt(b)*sqrt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*a - 3*sqrt(b)*sq 
rt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*b*x + 3*sqrt(x)*a*b + 2*sqrt(x)* 
b**2*x)/(b**3*(a + b*x))