\(\int \frac {x^{5/2} (A+B x)}{(a+b x)^{3/2}} \, dx\) [327]

Optimal result

Integrand size = 20, antiderivative size = 153 \[ \int \frac {x^{5/2} (A+B x)}{(a+b x)^{3/2}} \, dx=-\frac {2 (A b-a B) x^{5/2}}{b^2 \sqrt {a+b x}}-\frac {5 a (6 A b-7 a B) \sqrt {x} \sqrt {a+b x}}{8 b^4}+\frac {5 (6 A b-7 a B) x^{3/2} \sqrt {a+b x}}{12 b^3}+\frac {B x^{5/2} \sqrt {a+b x}}{3 b^2}+\frac {5 a^2 (6 A b-7 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{8 b^{9/2}} \] Output:

-2*(A*b-B*a)*x^(5/2)/b^2/(b*x+a)^(1/2)-5/8*a*(6*A*b-7*B*a)*x^(1/2)*(b*x+a) 
^(1/2)/b^4+5/12*(6*A*b-7*B*a)*x^(3/2)*(b*x+a)^(1/2)/b^3+1/3*B*x^(5/2)*(b*x 
+a)^(1/2)/b^2+5/8*a^2*(6*A*b-7*B*a)*arctanh(b^(1/2)*x^(1/2)/(b*x+a)^(1/2)) 
/b^(9/2)
 

Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.82 \[ \int \frac {x^{5/2} (A+B x)}{(a+b x)^{3/2}} \, dx=\frac {\sqrt {x} \left (105 a^3 B+4 b^3 x^2 (3 A+2 B x)-2 a b^2 x (15 A+7 B x)+a^2 (-90 A b+35 b B x)\right )}{24 b^4 \sqrt {a+b x}}+\frac {5 a^2 (-6 A b+7 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}-\sqrt {a+b x}}\right )}{4 b^{9/2}} \] Input:

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

Output:

(Sqrt[x]*(105*a^3*B + 4*b^3*x^2*(3*A + 2*B*x) - 2*a*b^2*x*(15*A + 7*B*x) + 
 a^2*(-90*A*b + 35*b*B*x)))/(24*b^4*Sqrt[a + b*x]) + (5*a^2*(-6*A*b + 7*a* 
B)*ArcTanh[(Sqrt[b]*Sqrt[x])/(Sqrt[a] - Sqrt[a + b*x])])/(4*b^(9/2))
 

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {87, 60, 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.

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

\(\Big \downarrow \) 87

\(\displaystyle \frac {2 x^{7/2} (A b-a B)}{a b \sqrt {a+b x}}-\frac {(6 A b-7 a B) \int \frac {x^{5/2}}{\sqrt {a+b x}}dx}{a b}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {2 x^{7/2} (A b-a B)}{a b \sqrt {a+b x}}-\frac {(6 A b-7 a B) \left (\frac {x^{5/2} \sqrt {a+b x}}{3 b}-\frac {5 a \int \frac {x^{3/2}}{\sqrt {a+b x}}dx}{6 b}\right )}{a b}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {2 x^{7/2} (A b-a B)}{a b \sqrt {a+b x}}-\frac {(6 A b-7 a B) \left (\frac {x^{5/2} \sqrt {a+b x}}{3 b}-\frac {5 a \left (\frac {x^{3/2} \sqrt {a+b x}}{2 b}-\frac {3 a \int \frac {\sqrt {x}}{\sqrt {a+b x}}dx}{4 b}\right )}{6 b}\right )}{a b}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {2 x^{7/2} (A b-a B)}{a b \sqrt {a+b x}}-\frac {(6 A b-7 a B) \left (\frac {x^{5/2} \sqrt {a+b x}}{3 b}-\frac {5 a \left (\frac {x^{3/2} \sqrt {a+b x}}{2 b}-\frac {3 a \left (\frac {\sqrt {x} \sqrt {a+b x}}{b}-\frac {a \int \frac {1}{\sqrt {x} \sqrt {a+b x}}dx}{2 b}\right )}{4 b}\right )}{6 b}\right )}{a b}\)

\(\Big \downarrow \) 65

\(\displaystyle \frac {2 x^{7/2} (A b-a B)}{a b \sqrt {a+b x}}-\frac {(6 A b-7 a B) \left (\frac {x^{5/2} \sqrt {a+b x}}{3 b}-\frac {5 a \left (\frac {x^{3/2} \sqrt {a+b x}}{2 b}-\frac {3 a \left (\frac {\sqrt {x} \sqrt {a+b x}}{b}-\frac {a \int \frac {1}{1-\frac {b x}{a+b x}}d\frac {\sqrt {x}}{\sqrt {a+b x}}}{b}\right )}{4 b}\right )}{6 b}\right )}{a b}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {2 x^{7/2} (A b-a B)}{a b \sqrt {a+b x}}-\frac {(6 A b-7 a B) \left (\frac {x^{5/2} \sqrt {a+b x}}{3 b}-\frac {5 a \left (\frac {x^{3/2} \sqrt {a+b x}}{2 b}-\frac {3 a \left (\frac {\sqrt {x} \sqrt {a+b x}}{b}-\frac {a \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{b^{3/2}}\right )}{4 b}\right )}{6 b}\right )}{a b}\)

Input:

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

Output:

(2*(A*b - a*B)*x^(7/2))/(a*b*Sqrt[a + b*x]) - ((6*A*b - 7*a*B)*((x^(5/2)*S 
qrt[a + b*x])/(3*b) - (5*a*((x^(3/2)*Sqrt[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)))/(6*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 = 186, normalized size of antiderivative = 1.22

method	result	size
risch	\(-\frac {\left (-8 b^{2} B \,x^{2}-12 A \,b^{2} x +22 B a b x +42 a b A -57 a^{2} B \right ) \sqrt {x}\, \sqrt {b x +a}}{24 b^{4}}+\frac {a^{2} \left (30 A \sqrt {b}\, \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )-\frac {35 B a \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{\sqrt {b}}-\frac {32 \left (A b -B a \right ) \sqrt {b \left (x +\frac {a}{b}\right )^{2}-\left (x +\frac {a}{b}\right ) a}}{b \left (x +\frac {a}{b}\right )}\right ) \sqrt {x \left (b x +a \right )}}{16 b^{4} \sqrt {x}\, \sqrt {b x +a}}\)	\(186\)
default	\(\frac {\left (16 B \,b^{\frac {7}{2}} x^{3} \sqrt {x \left (b x +a \right )}+24 A \,b^{\frac {7}{2}} x^{2} \sqrt {x \left (b x +a \right )}-28 B a \,b^{\frac {5}{2}} x^{2} \sqrt {x \left (b x +a \right )}+90 A \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{2} b^{2} x -60 A \sqrt {x \left (b x +a \right )}\, b^{\frac {5}{2}} a x -105 B \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{3} b x +70 B \sqrt {x \left (b x +a \right )}\, b^{\frac {3}{2}} a^{2} x +90 A \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{3} b -180 A \sqrt {x \left (b x +a \right )}\, b^{\frac {3}{2}} a^{2}-105 B \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{4}+210 B \sqrt {x \left (b x +a \right )}\, \sqrt {b}\, a^{3}\right ) \sqrt {x}}{48 b^{\frac {9}{2}} \sqrt {x \left (b x +a \right )}\, \sqrt {b x +a}}\)	\(288\)

Input:

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

Output:

-1/24*(-8*B*b^2*x^2-12*A*b^2*x+22*B*a*b*x+42*A*a*b-57*B*a^2)*x^(1/2)*(b*x+ 
a)^(1/2)/b^4+1/16*a^2/b^4*(30*A*b^(1/2)*ln((1/2*a+b*x)/b^(1/2)+(b*x^2+a*x) 
^(1/2))-35*B*a*ln((1/2*a+b*x)/b^(1/2)+(b*x^2+a*x)^(1/2))/b^(1/2)-32*(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 = 0.09 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.98 \[ \int \frac {x^{5/2} (A+B x)}{(a+b x)^{3/2}} \, dx=\left [-\frac {15 \, {\left (7 \, B a^{4} - 6 \, A a^{3} b + {\left (7 \, B a^{3} b - 6 \, A a^{2} b^{2}\right )} x\right )} \sqrt {b} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) - 2 \, {\left (8 \, B b^{4} x^{3} + 105 \, B a^{3} b - 90 \, A a^{2} b^{2} - 2 \, {\left (7 \, B a b^{3} - 6 \, A b^{4}\right )} x^{2} + 5 \, {\left (7 \, B a^{2} b^{2} - 6 \, A a b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{48 \, {\left (b^{6} x + a b^{5}\right )}}, \frac {15 \, {\left (7 \, B a^{4} - 6 \, A a^{3} b + {\left (7 \, B a^{3} b - 6 \, A a^{2} b^{2}\right )} x\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {b x + a}}\right ) + {\left (8 \, B b^{4} x^{3} + 105 \, B a^{3} b - 90 \, A a^{2} b^{2} - 2 \, {\left (7 \, B a b^{3} - 6 \, A b^{4}\right )} x^{2} + 5 \, {\left (7 \, B a^{2} b^{2} - 6 \, A a b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{24 \, {\left (b^{6} x + a b^{5}\right )}}\right ] \] Input:

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

Output:

[-1/48*(15*(7*B*a^4 - 6*A*a^3*b + (7*B*a^3*b - 6*A*a^2*b^2)*x)*sqrt(b)*log 
(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) - 2*(8*B*b^4*x^3 + 105*B*a^3 
*b - 90*A*a^2*b^2 - 2*(7*B*a*b^3 - 6*A*b^4)*x^2 + 5*(7*B*a^2*b^2 - 6*A*a*b 
^3)*x)*sqrt(b*x + a)*sqrt(x))/(b^6*x + a*b^5), 1/24*(15*(7*B*a^4 - 6*A*a^3 
*b + (7*B*a^3*b - 6*A*a^2*b^2)*x)*sqrt(-b)*arctan(sqrt(-b)*sqrt(x)/sqrt(b* 
x + a)) + (8*B*b^4*x^3 + 105*B*a^3*b - 90*A*a^2*b^2 - 2*(7*B*a*b^3 - 6*A*b 
^4)*x^2 + 5*(7*B*a^2*b^2 - 6*A*a*b^3)*x)*sqrt(b*x + a)*sqrt(x))/(b^6*x + a 
*b^5)]
 

Sympy [A] (verification not implemented)

Time = 72.56 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.59 \[ \int \frac {x^{5/2} (A+B x)}{(a+b x)^{3/2}} \, dx=A \left (- \frac {15 a^{\frac {3}{2}} \sqrt {x}}{4 b^{3} \sqrt {1 + \frac {b x}{a}}} - \frac {5 \sqrt {a} x^{\frac {3}{2}}}{4 b^{2} \sqrt {1 + \frac {b x}{a}}} + \frac {15 a^{2} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{4 b^{\frac {7}{2}}} + \frac {x^{\frac {5}{2}}}{2 \sqrt {a} b \sqrt {1 + \frac {b x}{a}}}\right ) + B \left (\frac {35 a^{\frac {5}{2}} \sqrt {x}}{8 b^{4} \sqrt {1 + \frac {b x}{a}}} + \frac {35 a^{\frac {3}{2}} x^{\frac {3}{2}}}{24 b^{3} \sqrt {1 + \frac {b x}{a}}} - \frac {7 \sqrt {a} x^{\frac {5}{2}}}{12 b^{2} \sqrt {1 + \frac {b x}{a}}} - \frac {35 a^{3} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{8 b^{\frac {9}{2}}} + \frac {x^{\frac {7}{2}}}{3 \sqrt {a} b \sqrt {1 + \frac {b x}{a}}}\right ) \] Input:

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

Output:

A*(-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*sqrt(a)*b*sqrt(1 + b*x/a))) + B*(35*a**(5/2)*sqrt(x)/(8*b** 
4*sqrt(1 + b*x/a)) + 35*a**(3/2)*x**(3/2)/(24*b**3*sqrt(1 + b*x/a)) - 7*sq 
rt(a)*x**(5/2)/(12*b**2*sqrt(1 + b*x/a)) - 35*a**3*asinh(sqrt(b)*sqrt(x)/s 
qrt(a))/(8*b**(9/2)) + x**(7/2)/(3*sqrt(a)*b*sqrt(1 + b*x/a)))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.39 \[ \int \frac {x^{5/2} (A+B x)}{(a+b x)^{3/2}} \, dx=\frac {B x^{4}}{3 \, \sqrt {b x^{2} + a x} b} - \frac {7 \, B a x^{3}}{12 \, \sqrt {b x^{2} + a x} b^{2}} + \frac {A x^{3}}{2 \, \sqrt {b x^{2} + a x} b} + \frac {35 \, B a^{2} x^{2}}{24 \, \sqrt {b x^{2} + a x} b^{3}} - \frac {5 \, A a x^{2}}{4 \, \sqrt {b x^{2} + a x} b^{2}} + \frac {35 \, B a^{3} x}{8 \, \sqrt {b x^{2} + a x} b^{4}} - \frac {15 \, A a^{2} x}{4 \, \sqrt {b x^{2} + a x} b^{3}} - \frac {35 \, B a^{3} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{16 \, b^{\frac {9}{2}}} + \frac {15 \, A a^{2} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{8 \, b^{\frac {7}{2}}} \] Input:

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

Output:

1/3*B*x^4/(sqrt(b*x^2 + a*x)*b) - 7/12*B*a*x^3/(sqrt(b*x^2 + a*x)*b^2) + 1 
/2*A*x^3/(sqrt(b*x^2 + a*x)*b) + 35/24*B*a^2*x^2/(sqrt(b*x^2 + a*x)*b^3) - 
 5/4*A*a*x^2/(sqrt(b*x^2 + a*x)*b^2) + 35/8*B*a^3*x/(sqrt(b*x^2 + a*x)*b^4 
) - 15/4*A*a^2*x/(sqrt(b*x^2 + a*x)*b^3) - 35/16*B*a^3*log(2*b*x + a + 2*s 
qrt(b*x^2 + a*x)*sqrt(b))/b^(9/2) + 15/8*A*a^2*log(2*b*x + a + 2*sqrt(b*x^ 
2 + a*x)*sqrt(b))/b^(7/2)
 

Giac [A] (verification not implemented)

Time = 15.54 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.35 \[ \int \frac {x^{5/2} (A+B x)}{(a+b x)^{3/2}} \, dx=\frac {1}{24} \, \sqrt {{\left (b x + a\right )} b - a b} \sqrt {b x + a} {\left (2 \, {\left (b x + a\right )} {\left (\frac {4 \, {\left (b x + a\right )} B {\left | b \right |}}{b^{6}} - \frac {19 \, B a b^{17} {\left | b \right |} - 6 \, A b^{18} {\left | b \right |}}{b^{23}}\right )} + \frac {3 \, {\left (29 \, B a^{2} b^{17} {\left | b \right |} - 18 \, A a b^{18} {\left | b \right |}\right )}}{b^{23}}\right )} + \frac {5 \, {\left (7 \, B a^{3} {\left | b \right |} - 6 \, A a^{2} b {\left | b \right |}\right )} \log \left ({\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2}\right )}{16 \, b^{\frac {11}{2}}} + \frac {4 \, {\left (B a^{4} {\left | b \right |} - A a^{3} b {\left | b \right |}\right )}}{{\left ({\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} + a b\right )} b^{\frac {9}{2}}} \] Input:

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

Output:

1/24*sqrt((b*x + a)*b - a*b)*sqrt(b*x + a)*(2*(b*x + a)*(4*(b*x + a)*B*abs 
(b)/b^6 - (19*B*a*b^17*abs(b) - 6*A*b^18*abs(b))/b^23) + 3*(29*B*a^2*b^17* 
abs(b) - 18*A*a*b^18*abs(b))/b^23) + 5/16*(7*B*a^3*abs(b) - 6*A*a^2*b*abs( 
b))*log((sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2)/b^(11/2) + 4* 
(B*a^4*abs(b) - A*a^3*b*abs(b))/(((sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)* 
b - a*b))^2 + a*b)*b^(9/2))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.50 \[ \int \frac {x^{5/2} (A+B x)}{(a+b x)^{3/2}} \, dx=\frac {15 \sqrt {x}\, \sqrt {b x +a}\, a^{2} b -10 \sqrt {x}\, \sqrt {b x +a}\, a \,b^{2} x +8 \sqrt {x}\, \sqrt {b x +a}\, b^{3} x^{2}-15 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) a^{3}}{24 b^{4}} \] Input:

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

Output:

(15*sqrt(x)*sqrt(a + b*x)*a**2*b - 10*sqrt(x)*sqrt(a + b*x)*a*b**2*x + 8*s 
qrt(x)*sqrt(a + b*x)*b**3*x**2 - 15*sqrt(b)*log((sqrt(a + b*x) + sqrt(x)*s 
qrt(b))/sqrt(a))*a**3)/(24*b**4)