\(\int \frac {x^{3/2} (A+B x)}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx\) [440]

Optimal result

Integrand size = 31, antiderivative size = 190 \[ \int \frac {x^{3/2} (A+B x)}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-\frac {2 a (A b-a B) \sqrt {x} (a+b x)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (A b-a B) x^{3/2} (a+b x)}{3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 B x^{5/2} (a+b x)}{5 b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 a^{3/2} (A b-a B) (a+b x) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.53 \[ \int \frac {x^{3/2} (A+B x)}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {2 (a+b x) \left (\sqrt {b} \sqrt {x} \left (15 a^2 B-5 a b (3 A+B x)+b^2 x (5 A+3 B x)\right )-15 a^{3/2} (-A b+a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )\right )}{15 b^{7/2} \sqrt {(a+b x)^2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.58, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {1187, 27, 90, 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.

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

\(\Big \downarrow \) 1187

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 90

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

((a + b*x)*((2*B*x^(5/2))/(5*b) + ((A*b - a*B)*((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))/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 = 0.83 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.57

method	result	size
risch	\(-\frac {2 \left (-3 x^{2} B \,b^{2}-5 x \,b^{2} A +5 x a b B +15 a b A -15 a^{2} B \right ) \sqrt {x}\, \sqrt {\left (b x +a \right )^{2}}}{15 b^{3} \left (b x +a \right )}+\frac {2 a^{2} \left (A b -B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) \sqrt {\left (b x +a \right )^{2}}}{b^{3} \sqrt {a b}\, \left (b x +a \right )}\)	\(108\)
default	\(\frac {2 \left (b x +a \right ) \left (3 B \,x^{\frac {5}{2}} \sqrt {a b}\, b^{2}+5 A \,x^{\frac {3}{2}} \sqrt {a b}\, b^{2}-5 B \,x^{\frac {3}{2}} \sqrt {a b}\, a b -15 A \sqrt {x}\, \sqrt {a b}\, a b +15 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a^{2} b +15 B \sqrt {x}\, \sqrt {a b}\, a^{2}-15 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a^{3}\right )}{15 \sqrt {\left (b x +a \right )^{2}}\, b^{3} \sqrt {a b}}\)	\(129\)

Input:

int(x^(3/2)*(B*x+A)/((b*x+a)^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-2/15*(-3*B*b^2*x^2-5*A*b^2*x+5*B*a*b*x+15*A*a*b-15*B*a^2)*x^(1/2)/b^3*((b 
*x+a)^2)^(1/2)/(b*x+a)+2*a^2*(A*b-B*a)/b^3/(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 = 0.08 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.95 \[ \int \frac {x^{3/2} (A+B x)}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\left [-\frac {15 \, {\left (B a^{2} - A a b\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x + 2 \, b \sqrt {x} \sqrt {-\frac {a}{b}} - a}{b x + a}\right ) - 2 \, {\left (3 \, B b^{2} x^{2} + 15 \, B a^{2} - 15 \, A a b - 5 \, {\left (B a b - A b^{2}\right )} x\right )} \sqrt {x}}{15 \, b^{3}}, -\frac {2 \, {\left (15 \, {\left (B a^{2} - A a b\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {\frac {a}{b}}}{a}\right ) - {\left (3 \, B b^{2} x^{2} + 15 \, B a^{2} - 15 \, A a b - 5 \, {\left (B a b - A b^{2}\right )} x\right )} \sqrt {x}\right )}}{15 \, b^{3}}\right ] \] Input:

integrate(x^(3/2)*(B*x+A)/((b*x+a)^2)^(1/2),x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

integrate(x**(3/2)*(B*x+A)/((b*x+a)**2)**(1/2),x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.77 \[ \int \frac {x^{3/2} (A+B x)}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {2 \, {\left (3 \, B b^{2} x^{2} + 5 \, B a b x\right )} x^{\frac {3}{2}} + {\left (3 \, {\left (7 \, B a b - 5 \, A b^{2}\right )} x^{2} + 5 \, {\left (5 \, B a^{2} - 3 \, A a b\right )} x\right )} \sqrt {x}}{15 \, {\left (b^{3} x + a b^{2}\right )}} - \frac {2 \, {\left (B a^{3} - A a^{2} b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{3}} - \frac {{\left (7 \, B a b - 5 \, A b^{2}\right )} x^{\frac {3}{2}} - 6 \, {\left (B a^{2} - A a b\right )} \sqrt {x}}{3 \, b^{3}} \] Input:

integrate(x^(3/2)*(B*x+A)/((b*x+a)^2)^(1/2),x, algorithm="maxima")
 

Output:

1/15*(2*(3*B*b^2*x^2 + 5*B*a*b*x)*x^(3/2) + (3*(7*B*a*b - 5*A*b^2)*x^2 + 5 
*(5*B*a^2 - 3*A*a*b)*x)*sqrt(x))/(b^3*x + a*b^2) - 2*(B*a^3 - A*a^2*b)*arc 
tan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^3) - 1/3*((7*B*a*b - 5*A*b^2)*x^(3/2 
) - 6*(B*a^2 - A*a*b)*sqrt(x))/b^3
 

Giac [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.70 \[ \int \frac {x^{3/2} (A+B x)}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-\frac {2 \, {\left (B a^{3} \mathrm {sgn}\left (b x + a\right ) - A a^{2} b \mathrm {sgn}\left (b x + a\right )\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{3}} + \frac {2 \, {\left (3 \, B b^{4} x^{\frac {5}{2}} \mathrm {sgn}\left (b x + a\right ) - 5 \, B a b^{3} x^{\frac {3}{2}} \mathrm {sgn}\left (b x + a\right ) + 5 \, A b^{4} x^{\frac {3}{2}} \mathrm {sgn}\left (b x + a\right ) + 15 \, B a^{2} b^{2} \sqrt {x} \mathrm {sgn}\left (b x + a\right ) - 15 \, A a b^{3} \sqrt {x} \mathrm {sgn}\left (b x + a\right )\right )}}{15 \, b^{5}} \] Input:

integrate(x^(3/2)*(B*x+A)/((b*x+a)^2)^(1/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

int((x^(3/2)*(A + B*x))/((a + b*x)^2)^(1/2),x)
 

Output:

int((x^(3/2)*(A + B*x))/((a + b*x)^2)^(1/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.04 \[ \int \frac {x^{3/2} (A+B x)}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {2 \sqrt {x}\, x^{2}}{5} \] Input:

int(x^(3/2)*(B*x+A)/((b*x+a)^2)^(1/2),x)
 

Output:

(2*sqrt(x)*x**2)/5