\(\int \frac {x^{5/2} (A+B x)}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx\) [439]

Optimal result

Integrand size = 31, antiderivative size = 238 \[ \int \frac {x^{5/2} (A+B x)}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {2 a^2 (A b-a B) \sqrt {x} (a+b x)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 a (A b-a B) x^{3/2} (a+b x)}{3 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (A b-a B) x^{5/2} (a+b x)}{5 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 B x^{7/2} (a+b x)}{7 b \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 a^{5/2} (A b-a B) (a+b x) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.50 \[ \int \frac {x^{5/2} (A+B x)}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {2 (a+b x) \left (\sqrt {b} \sqrt {x} \left (-105 a^3 B+35 a^2 b (3 A+B x)-7 a b^2 x (5 A+3 B x)+3 b^3 x^2 (7 A+5 B x)\right )+105 a^{5/2} (-A b+a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )\right )}{105 b^{9/2} \sqrt {(a+b x)^2}} \] Input:

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

Output:

(2*(a + b*x)*(Sqrt[b]*Sqrt[x]*(-105*a^3*B + 35*a^2*b*(3*A + B*x) - 7*a*b^2 
*x*(5*A + 3*B*x) + 3*b^3*x^2*(7*A + 5*B*x)) + 105*a^(5/2)*(-(A*b) + a*B)*A 
rcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]]))/(105*b^(9/2)*Sqrt[(a + b*x)^2])
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.55, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.258, Rules used = {1187, 27, 90, 60, 60, 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.

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

\(\Big \downarrow \) 1187

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 90

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

\(\displaystyle \frac {(a+b x) \left (\frac {(A b-a B) \left (\frac {2 x^{5/2}}{5 b}-\frac {a \left (\frac {2 x^{3/2}}{3 b}-\frac {a \left (\frac {2 \sqrt {x}}{b}-\frac {a \int \frac {1}{\sqrt {x} (a+b x)}dx}{b}\right )}{b}\right )}{b}\right )}{b}+\frac {2 B x^{7/2}}{7 b}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {(a+b x) \left (\frac {(A b-a B) \left (\frac {2 x^{5/2}}{5 b}-\frac {a \left (\frac {2 x^{3/2}}{3 b}-\frac {a \left (\frac {2 \sqrt {x}}{b}-\frac {2 a \int \frac {1}{a+b x}d\sqrt {x}}{b}\right )}{b}\right )}{b}\right )}{b}+\frac {2 B x^{7/2}}{7 b}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 218

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

Input:

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

Output:

((a + b*x)*((2*B*x^(7/2))/(7*b) + ((A*b - a*B)*((2*x^(5/2))/(5*b) - (a*((2 
*x^(3/2))/(3*b) - (a*((2*Sqrt[x])/b - (2*Sqrt[a]*ArcTan[(Sqrt[b]*Sqrt[x])/ 
Sqrt[a]])/b^(3/2)))/b))/b))/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 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 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 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 = 0.75 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.55

method	result	size
risch	\(\frac {2 \left (15 x^{3} B \,b^{3}+21 A \,b^{3} x^{2}-21 B a \,b^{2} x^{2}-35 A a \,b^{2} x +35 B \,a^{2} b x +105 A \,a^{2} b -105 B \,a^{3}\right ) \sqrt {x}\, \sqrt {\left (b x +a \right )^{2}}}{105 b^{4} \left (b x +a \right )}-\frac {2 a^{3} \left (A b -B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) \sqrt {\left (b x +a \right )^{2}}}{b^{4} \sqrt {a b}\, \left (b x +a \right )}\)	\(132\)
default	\(\frac {2 \left (b x +a \right ) \left (15 B \sqrt {a b}\, x^{\frac {7}{2}} b^{3}+21 A \sqrt {a b}\, x^{\frac {5}{2}} b^{3}-21 B \sqrt {a b}\, x^{\frac {5}{2}} a \,b^{2}-35 A \sqrt {a b}\, x^{\frac {3}{2}} a \,b^{2}+35 B \sqrt {a b}\, x^{\frac {3}{2}} a^{2} b +105 A \sqrt {a b}\, \sqrt {x}\, a^{2} b -105 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a^{3} b -105 B \sqrt {a b}\, \sqrt {x}\, a^{3}+105 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a^{4}\right )}{105 \sqrt {\left (b x +a \right )^{2}}\, b^{4} \sqrt {a b}}\)	\(163\)

Input:

int(x^(5/2)*(B*x+A)/((b*x+a)^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

2/105*(15*B*b^3*x^3+21*A*b^3*x^2-21*B*a*b^2*x^2-35*A*a*b^2*x+35*B*a^2*b*x+ 
105*A*a^2*b-105*B*a^3)*x^(1/2)/b^4*((b*x+a)^2)^(1/2)/(b*x+a)-2*a^3*(A*b-B* 
a)/b^4/(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 = 229, normalized size of antiderivative = 0.96 \[ \int \frac {x^{5/2} (A+B x)}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\left [-\frac {105 \, {\left (B a^{3} - A a^{2} b\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x - 2 \, b \sqrt {x} \sqrt {-\frac {a}{b}} - a}{b x + a}\right ) - 2 \, {\left (15 \, B b^{3} x^{3} - 105 \, B a^{3} + 105 \, A a^{2} b - 21 \, {\left (B a b^{2} - A b^{3}\right )} x^{2} + 35 \, {\left (B a^{2} b - A a b^{2}\right )} x\right )} \sqrt {x}}{105 \, b^{4}}, \frac {2 \, {\left (105 \, {\left (B a^{3} - A a^{2} b\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {\frac {a}{b}}}{a}\right ) + {\left (15 \, B b^{3} x^{3} - 105 \, B a^{3} + 105 \, A a^{2} b - 21 \, {\left (B a b^{2} - A b^{3}\right )} x^{2} + 35 \, {\left (B a^{2} b - A a b^{2}\right )} x\right )} \sqrt {x}\right )}}{105 \, b^{4}}\right ] \] Input:

integrate(x^(5/2)*(B*x+A)/((b*x+a)^2)^(1/2),x, algorithm="fricas")
 

Output:

[-1/105*(105*(B*a^3 - A*a^2*b)*sqrt(-a/b)*log((b*x - 2*b*sqrt(x)*sqrt(-a/b 
) - a)/(b*x + a)) - 2*(15*B*b^3*x^3 - 105*B*a^3 + 105*A*a^2*b - 21*(B*a*b^ 
2 - A*b^3)*x^2 + 35*(B*a^2*b - A*a*b^2)*x)*sqrt(x))/b^4, 2/105*(105*(B*a^3 
 - A*a^2*b)*sqrt(a/b)*arctan(b*sqrt(x)*sqrt(a/b)/a) + (15*B*b^3*x^3 - 105* 
B*a^3 + 105*A*a^2*b - 21*(B*a*b^2 - A*b^3)*x^2 + 35*(B*a^2*b - A*a*b^2)*x) 
*sqrt(x))/b^4]
 

Sympy [F(-1)]

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

integrate(x**(5/2)*(B*x+A)/((b*x+a)**2)**(1/2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 203, normalized size of antiderivative = 0.85 \[ \int \frac {x^{5/2} (A+B x)}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {6 \, {\left (5 \, B b^{3} x^{2} + 7 \, B a b^{2} x\right )} x^{\frac {5}{2}} - 2 \, {\left (3 \, {\left (9 \, B a b^{2} - 7 \, A b^{3}\right )} x^{2} + 7 \, {\left (7 \, B a^{2} b - 5 \, A a b^{2}\right )} x\right )} x^{\frac {3}{2}} - 7 \, {\left (3 \, {\left (9 \, B a^{2} b - 7 \, A a b^{2}\right )} x^{2} + 5 \, {\left (7 \, B a^{3} - 5 \, A a^{2} b\right )} x\right )} \sqrt {x}}{105 \, {\left (b^{4} x + a b^{3}\right )}} + \frac {2 \, {\left (B a^{4} - A a^{3} b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{4}} + \frac {{\left (9 \, B a^{2} b - 7 \, A a b^{2}\right )} x^{\frac {3}{2}} - 6 \, {\left (B a^{3} - A a^{2} b\right )} \sqrt {x}}{3 \, b^{4}} \] Input:

integrate(x^(5/2)*(B*x+A)/((b*x+a)^2)^(1/2),x, algorithm="maxima")
 

Output:

1/105*(6*(5*B*b^3*x^2 + 7*B*a*b^2*x)*x^(5/2) - 2*(3*(9*B*a*b^2 - 7*A*b^3)* 
x^2 + 7*(7*B*a^2*b - 5*A*a*b^2)*x)*x^(3/2) - 7*(3*(9*B*a^2*b - 7*A*a*b^2)* 
x^2 + 5*(7*B*a^3 - 5*A*a^2*b)*x)*sqrt(x))/(b^4*x + a*b^3) + 2*(B*a^4 - A*a 
^3*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^4) + 1/3*((9*B*a^2*b - 7*A* 
a*b^2)*x^(3/2) - 6*(B*a^3 - A*a^2*b)*sqrt(x))/b^4
 

Giac [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.71 \[ \int \frac {x^{5/2} (A+B x)}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {2 \, {\left (B a^{4} \mathrm {sgn}\left (b x + a\right ) - A a^{3} b \mathrm {sgn}\left (b x + a\right )\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{4}} + \frac {2 \, {\left (15 \, B b^{6} x^{\frac {7}{2}} \mathrm {sgn}\left (b x + a\right ) - 21 \, B a b^{5} x^{\frac {5}{2}} \mathrm {sgn}\left (b x + a\right ) + 21 \, A b^{6} x^{\frac {5}{2}} \mathrm {sgn}\left (b x + a\right ) + 35 \, B a^{2} b^{4} x^{\frac {3}{2}} \mathrm {sgn}\left (b x + a\right ) - 35 \, A a b^{5} x^{\frac {3}{2}} \mathrm {sgn}\left (b x + a\right ) - 105 \, B a^{3} b^{3} \sqrt {x} \mathrm {sgn}\left (b x + a\right ) + 105 \, A a^{2} b^{4} \sqrt {x} \mathrm {sgn}\left (b x + a\right )\right )}}{105 \, b^{7}} \] Input:

integrate(x^(5/2)*(B*x+A)/((b*x+a)^2)^(1/2),x, algorithm="giac")
 

Output:

2*(B*a^4*sgn(b*x + a) - A*a^3*b*sgn(b*x + a))*arctan(b*sqrt(x)/sqrt(a*b))/ 
(sqrt(a*b)*b^4) + 2/105*(15*B*b^6*x^(7/2)*sgn(b*x + a) - 21*B*a*b^5*x^(5/2 
)*sgn(b*x + a) + 21*A*b^6*x^(5/2)*sgn(b*x + a) + 35*B*a^2*b^4*x^(3/2)*sgn( 
b*x + a) - 35*A*a*b^5*x^(3/2)*sgn(b*x + a) - 105*B*a^3*b^3*sqrt(x)*sgn(b*x 
 + a) + 105*A*a^2*b^4*sqrt(x)*sgn(b*x + a))/b^7
 

Mupad [F(-1)]

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

int((x^(5/2)*(A + B*x))/((a + b*x)^2)^(1/2),x)
 

Output:

int((x^(5/2)*(A + B*x))/((a + b*x)^2)^(1/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.03 \[ \int \frac {x^{5/2} (A+B x)}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {2 \sqrt {x}\, x^{3}}{7} \] Input:

int(x^(5/2)*(B*x+A)/((b*x+a)^2)^(1/2),x)
 

Output:

(2*sqrt(x)*x**3)/7