\(\int \frac {x^{3/2} (A+B x)}{(a+b x)^{3/2}} \, dx\) [328]

Optimal result

Integrand size = 20, antiderivative size = 120 \[ \int \frac {x^{3/2} (A+B x)}{(a+b x)^{3/2}} \, dx=-\frac {2 (A b-a B) x^{3/2}}{b^2 \sqrt {a+b x}}+\frac {3 (4 A b-5 a B) \sqrt {x} \sqrt {a+b x}}{4 b^3}+\frac {B x^{3/2} \sqrt {a+b x}}{2 b^2}-\frac {3 a (4 A b-5 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{4 b^{7/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.86 \[ \int \frac {x^{3/2} (A+B x)}{(a+b x)^{3/2}} \, dx=\frac {\sqrt {x} \left (-15 a^2 B+a b (12 A-5 B x)+2 b^2 x (2 A+B x)\right )}{4 b^3 \sqrt {a+b x}}+\frac {3 a (-4 A b+5 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{-\sqrt {a}+\sqrt {a+b x}}\right )}{2 b^{7/2}} \] Input:

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

Output:

(Sqrt[x]*(-15*a^2*B + a*b*(12*A - 5*B*x) + 2*b^2*x*(2*A + B*x)))/(4*b^3*Sq 
rt[a + b*x]) + (3*a*(-4*A*b + 5*a*B)*ArcTanh[(Sqrt[b]*Sqrt[x])/(-Sqrt[a] + 
 Sqrt[a + b*x])])/(2*b^(7/2))
 

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {87, 60, 60, 65, 219}

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.

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 65

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

(2*(A*b - a*B)*x^(5/2))/(a*b*Sqrt[a + b*x]) - ((4*A*b - 5*a*B)*((x^(3/2)*S 
qrt[a + b*x])/(2*b) - (3*a*((Sqrt[x]*Sqrt[a + b*x])/b - (a*ArcTanh[(Sqrt[b 
]*Sqrt[x])/Sqrt[a + b*x]])/b^(3/2)))/(4*b)))/(a*b)
 

Defintions of rubi rules used

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 65

Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[2   Sub 
st[Int[1/(b - d*x^2), x], x, Sqrt[b*x]/Sqrt[c + d*x]], x] /; FreeQ[{b, c, d 
}, x] &&  !GtQ[c, 0]
 

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 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.37

method	result	size
risch	\(\frac {\left (2 b B x +4 A b -7 B a \right ) \sqrt {x}\, \sqrt {b x +a}}{4 b^{3}}-\frac {a \left (12 A \sqrt {b}\, \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )-\frac {15 B a \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{\sqrt {b}}-\frac {16 \left (A b -B a \right ) \sqrt {b \left (x +\frac {a}{b}\right )^{2}-\left (x +\frac {a}{b}\right ) a}}{b \left (x +\frac {a}{b}\right )}\right ) \sqrt {x \left (b x +a \right )}}{8 b^{3} \sqrt {x}\, \sqrt {b x +a}}\)	\(164\)
default	\(-\frac {\left (-4 B \,b^{\frac {5}{2}} x^{2} \sqrt {x \left (b x +a \right )}+12 A \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a \,b^{2} x -8 A \sqrt {x \left (b x +a \right )}\, b^{\frac {5}{2}} x -15 B \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{2} b x +10 B \sqrt {x \left (b x +a \right )}\, b^{\frac {3}{2}} a x +12 A \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{2} b -24 A \sqrt {x \left (b x +a \right )}\, b^{\frac {3}{2}} a -15 B \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{3}+30 B \sqrt {x \left (b x +a \right )}\, \sqrt {b}\, a^{2}\right ) \sqrt {x}}{8 b^{\frac {7}{2}} \sqrt {x \left (b x +a \right )}\, \sqrt {b x +a}}\)	\(244\)

Input:

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

Output:

1/4*(2*B*b*x+4*A*b-7*B*a)*x^(1/2)*(b*x+a)^(1/2)/b^3-1/8/b^3*a*(12*A*b^(1/2 
)*ln((1/2*a+b*x)/b^(1/2)+(b*x^2+a*x)^(1/2))-15*B*a*ln((1/2*a+b*x)/b^(1/2)+ 
(b*x^2+a*x)^(1/2))/b^(1/2)-16*(A*b-B*a)/b/(x+a/b)*(b*(x+a/b)^2-(x+a/b)*a)^ 
(1/2))*(x*(b*x+a))^(1/2)/x^(1/2)/(b*x+a)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 252, normalized size of antiderivative = 2.10 \[ \int \frac {x^{3/2} (A+B x)}{(a+b x)^{3/2}} \, dx=\left [-\frac {3 \, {\left (5 \, B a^{3} - 4 \, A a^{2} b + {\left (5 \, B a^{2} b - 4 \, A a b^{2}\right )} x\right )} \sqrt {b} \log \left (2 \, b x - 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) - 2 \, {\left (2 \, B b^{3} x^{2} - 15 \, B a^{2} b + 12 \, A a b^{2} - {\left (5 \, B a b^{2} - 4 \, A b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{8 \, {\left (b^{5} x + a b^{4}\right )}}, -\frac {3 \, {\left (5 \, B a^{3} - 4 \, A a^{2} b + {\left (5 \, B a^{2} b - 4 \, A a b^{2}\right )} x\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {b x + a}}\right ) - {\left (2 \, B b^{3} x^{2} - 15 \, B a^{2} b + 12 \, A a b^{2} - {\left (5 \, B a b^{2} - 4 \, A b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{4 \, {\left (b^{5} x + a b^{4}\right )}}\right ] \] Input:

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

Output:

[-1/8*(3*(5*B*a^3 - 4*A*a^2*b + (5*B*a^2*b - 4*A*a*b^2)*x)*sqrt(b)*log(2*b 
*x - 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) - 2*(2*B*b^3*x^2 - 15*B*a^2*b + 
12*A*a*b^2 - (5*B*a*b^2 - 4*A*b^3)*x)*sqrt(b*x + a)*sqrt(x))/(b^5*x + a*b^ 
4), -1/4*(3*(5*B*a^3 - 4*A*a^2*b + (5*B*a^2*b - 4*A*a*b^2)*x)*sqrt(-b)*arc 
tan(sqrt(-b)*sqrt(x)/sqrt(b*x + a)) - (2*B*b^3*x^2 - 15*B*a^2*b + 12*A*a*b 
^2 - (5*B*a*b^2 - 4*A*b^3)*x)*sqrt(b*x + a)*sqrt(x))/(b^5*x + a*b^4)]
 

Sympy [A] (verification not implemented)

Time = 14.91 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.52 \[ \int \frac {x^{3/2} (A+B x)}{(a+b x)^{3/2}} \, dx=A \left (\frac {3 \sqrt {a} \sqrt {x}}{b^{2} \sqrt {1 + \frac {b x}{a}}} - \frac {3 a \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{b^{\frac {5}{2}}} + \frac {x^{\frac {3}{2}}}{\sqrt {a} b \sqrt {1 + \frac {b x}{a}}}\right ) + B \left (- \frac {15 a^{\frac {3}{2}} \sqrt {x}}{4 b^{3} \sqrt {1 + \frac {b x}{a}}} - \frac {5 \sqrt {a} x^{\frac {3}{2}}}{4 b^{2} \sqrt {1 + \frac {b x}{a}}} + \frac {15 a^{2} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{4 b^{\frac {7}{2}}} + \frac {x^{\frac {5}{2}}}{2 \sqrt {a} b \sqrt {1 + \frac {b x}{a}}}\right ) \] Input:

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

Output:

A*(3*sqrt(a)*sqrt(x)/(b**2*sqrt(1 + b*x/a)) - 3*a*asinh(sqrt(b)*sqrt(x)/sq 
rt(a))/b**(5/2) + x**(3/2)/(sqrt(a)*b*sqrt(1 + b*x/a))) + B*(-15*a**(3/2)* 
sqrt(x)/(4*b**3*sqrt(1 + b*x/a)) - 5*sqrt(a)*x**(3/2)/(4*b**2*sqrt(1 + b*x 
/a)) + 15*a**2*asinh(sqrt(b)*sqrt(x)/sqrt(a))/(4*b**(7/2)) + x**(5/2)/(2*s 
qrt(a)*b*sqrt(1 + b*x/a)))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 246 vs. \(2 (94) = 188\).

Time = 0.05 (sec) , antiderivative size = 246, normalized size of antiderivative = 2.05 \[ \int \frac {x^{3/2} (A+B x)}{(a+b x)^{3/2}} \, dx=-\frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} B a}{b^{4} x^{2} + 2 \, a b^{3} x + a^{2} b^{2}} - \frac {3 \, \sqrt {b x^{2} + a x} B a^{2}}{b^{4} x + a b^{3}} + \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} A}{b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b} + \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} B}{2 \, {\left (b^{3} x + a b^{2}\right )}} + \frac {3 \, \sqrt {b x^{2} + a x} A a}{b^{3} x + a b^{2}} + \frac {15 \, B a^{2} \log \left (2 \, x + \frac {a}{b} + \frac {2 \, \sqrt {b x^{2} + a x}}{\sqrt {b}}\right )}{8 \, b^{\frac {7}{2}}} - \frac {3 \, A a \log \left (2 \, x + \frac {a}{b} + \frac {2 \, \sqrt {b x^{2} + a x}}{\sqrt {b}}\right )}{2 \, b^{\frac {5}{2}}} - \frac {3 \, \sqrt {b x^{2} + a x} B a}{4 \, b^{3}} \] Input:

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

Output:

-(b*x^2 + a*x)^(3/2)*B*a/(b^4*x^2 + 2*a*b^3*x + a^2*b^2) - 3*sqrt(b*x^2 + 
a*x)*B*a^2/(b^4*x + a*b^3) + (b*x^2 + a*x)^(3/2)*A/(b^3*x^2 + 2*a*b^2*x + 
a^2*b) + 1/2*(b*x^2 + a*x)^(3/2)*B/(b^3*x + a*b^2) + 3*sqrt(b*x^2 + a*x)*A 
*a/(b^3*x + a*b^2) + 15/8*B*a^2*log(2*x + a/b + 2*sqrt(b*x^2 + a*x)/sqrt(b 
))/b^(7/2) - 3/2*A*a*log(2*x + a/b + 2*sqrt(b*x^2 + a*x)/sqrt(b))/b^(5/2) 
- 3/4*sqrt(b*x^2 + a*x)*B*a/b^3
 

Giac [A] (verification not implemented)

Time = 15.50 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.42 \[ \int \frac {x^{3/2} (A+B x)}{(a+b x)^{3/2}} \, dx=\frac {1}{4} \, \sqrt {{\left (b x + a\right )} b - a b} \sqrt {b x + a} {\left (\frac {2 \, {\left (b x + a\right )} B {\left | b \right |}}{b^{5}} - \frac {9 \, B a b^{9} {\left | b \right |} - 4 \, A b^{10} {\left | b \right |}}{b^{14}}\right )} - \frac {3 \, {\left (5 \, B a^{2} {\left | b \right |} - 4 \, A a b {\left | b \right |}\right )} \log \left ({\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2}\right )}{8 \, b^{\frac {9}{2}}} - \frac {4 \, {\left (B a^{3} {\left | b \right |} - A a^{2} b {\left | b \right |}\right )}}{{\left ({\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} + a b\right )} b^{\frac {7}{2}}} \] Input:

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

Output:

1/4*sqrt((b*x + a)*b - a*b)*sqrt(b*x + a)*(2*(b*x + a)*B*abs(b)/b^5 - (9*B 
*a*b^9*abs(b) - 4*A*b^10*abs(b))/b^14) - 3/8*(5*B*a^2*abs(b) - 4*A*a*b*abs 
(b))*log((sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2)/b^(9/2) - 4* 
(B*a^3*abs(b) - A*a^2*b*abs(b))/(((sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)* 
b - a*b))^2 + a*b)*b^(7/2))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.48 \[ \int \frac {x^{3/2} (A+B x)}{(a+b x)^{3/2}} \, dx=\frac {-3 \sqrt {x}\, \sqrt {b x +a}\, a b +2 \sqrt {x}\, \sqrt {b x +a}\, b^{2} x +3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) a^{2}}{4 b^{3}} \] Input:

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

Output:

( - 3*sqrt(x)*sqrt(a + b*x)*a*b + 2*sqrt(x)*sqrt(a + b*x)*b**2*x + 3*sqrt( 
b)*log((sqrt(a + b*x) + sqrt(x)*sqrt(b))/sqrt(a))*a**2)/(4*b**3)