\(\int \frac {x^{3/2} (A+B x)}{(a+b x)^{5/2}} \, dx\) [337]

Optimal result

Integrand size = 20, antiderivative size = 112 \[ \int \frac {x^{3/2} (A+B x)}{(a+b x)^{5/2}} \, dx=-\frac {2 (A b-a B) x^{3/2}}{3 b^2 (a+b x)^{3/2}}-\frac {2 (A b-2 a B) \sqrt {x}}{b^3 \sqrt {a+b x}}+\frac {B \sqrt {x} \sqrt {a+b x}}{b^3}+\frac {(2 A b-5 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{b^{7/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.92 \[ \int \frac {x^{3/2} (A+B x)}{(a+b x)^{5/2}} \, dx=\frac {\sqrt {x} \left (-6 a A b+15 a^2 B-8 A b^2 x+20 a b B x+3 b^2 B x^2\right )}{3 b^3 (a+b x)^{3/2}}+\frac {2 (2 A b-5 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{-\sqrt {a}+\sqrt {a+b x}}\right )}{b^{7/2}} \] Input:

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

Output:

(Sqrt[x]*(-6*a*A*b + 15*a^2*B - 8*A*b^2*x + 20*a*b*B*x + 3*b^2*B*x^2))/(3* 
b^3*(a + b*x)^(3/2)) + (2*(2*A*b - 5*a*B)*ArcTanh[(Sqrt[b]*Sqrt[x])/(-Sqrt 
[a] + Sqrt[a + b*x])])/b^(7/2)
 

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.12, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {87, 57, 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.

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 57

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 65

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

(2*(A*b - a*B)*x^(5/2))/(3*a*b*(a + b*x)^(3/2)) - ((2*A*b - 5*a*B)*((-2*x^ 
(3/2))/(b*Sqrt[a + b*x]) + (3*((Sqrt[x]*Sqrt[a + b*x])/b - (a*ArcTanh[(Sqr 
t[b]*Sqrt[x])/Sqrt[a + b*x]])/b^(3/2)))/b))/(3*a*b)
 

Defintions of rubi rules used

rule 57

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}, x] & 
& GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m 
+ n + 2, 0] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c 
, d, m, n, 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 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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(243\) vs. \(2(90)=180\).

Time = 0.18 (sec) , antiderivative size = 244, normalized size of antiderivative = 2.18

method	result	size
risch	\(\frac {B \sqrt {x}\, \sqrt {b x +a}}{b^{3}}+\frac {\left (2 A \sqrt {b}\, \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )-\frac {5 B a \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{\sqrt {b}}-\frac {4 \left (2 A b -3 B a \right ) \sqrt {b \left (x +\frac {a}{b}\right )^{2}-\left (x +\frac {a}{b}\right ) a}}{b \left (x +\frac {a}{b}\right )}+\frac {2 a^{2} \left (A b -B a \right ) \left (\frac {2 \sqrt {b \left (x +\frac {a}{b}\right )^{2}-\left (x +\frac {a}{b}\right ) a}}{3 a \left (x +\frac {a}{b}\right )^{2}}+\frac {4 b \sqrt {b \left (x +\frac {a}{b}\right )^{2}-\left (x +\frac {a}{b}\right ) a}}{3 a^{2} \left (x +\frac {a}{b}\right )}\right )}{b^{2}}\right ) \sqrt {x \left (b x +a \right )}}{2 b^{3} \sqrt {x}\, \sqrt {b x +a}}\)	\(244\)
default	\(\frac {\left (6 A \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) b^{3} x^{2}-15 B \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a \,b^{2} x^{2}+6 B \,b^{\frac {5}{2}} x^{2} \sqrt {x \left (b x +a \right )}+12 A \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a \,b^{2} x -16 A \sqrt {x \left (b x +a \right )}\, b^{\frac {5}{2}} x -30 B \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{2} b x +40 B \sqrt {x \left (b x +a \right )}\, b^{\frac {3}{2}} a x +6 A \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{2} b -12 A \sqrt {x \left (b x +a \right )}\, b^{\frac {3}{2}} a -15 B \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{3}+30 B \sqrt {x \left (b x +a \right )}\, \sqrt {b}\, a^{2}\right ) \sqrt {x}}{6 \sqrt {x \left (b x +a \right )}\, b^{\frac {7}{2}} \left (b x +a \right )^{\frac {3}{2}}}\)	\(315\)

Input:

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

Output:

B*x^(1/2)*(b*x+a)^(1/2)/b^3+1/2/b^3*(2*A*b^(1/2)*ln((1/2*a+b*x)/b^(1/2)+(b 
*x^2+a*x)^(1/2))-5*B*a*ln((1/2*a+b*x)/b^(1/2)+(b*x^2+a*x)^(1/2))/b^(1/2)-4 
*(2*A*b-3*B*a)/b/(x+a/b)*(b*(x+a/b)^2-(x+a/b)*a)^(1/2)+2*a^2*(A*b-B*a)/b^2 
*(2/3/a/(x+a/b)^2*(b*(x+a/b)^2-(x+a/b)*a)^(1/2)+4/3*b/a^2/(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 = 0.10 (sec) , antiderivative size = 311, normalized size of antiderivative = 2.78 \[ \int \frac {x^{3/2} (A+B x)}{(a+b x)^{5/2}} \, dx=\left [-\frac {3 \, {\left (5 \, B a^{3} - 2 \, A a^{2} b + {\left (5 \, B a b^{2} - 2 \, A b^{3}\right )} x^{2} + 2 \, {\left (5 \, B a^{2} b - 2 \, A a b^{2}\right )} x\right )} \sqrt {b} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) - 2 \, {\left (3 \, B b^{3} x^{2} + 15 \, B a^{2} b - 6 \, A a b^{2} + 4 \, {\left (5 \, B a b^{2} - 2 \, A b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{6 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}}, \frac {3 \, {\left (5 \, B a^{3} - 2 \, A a^{2} b + {\left (5 \, B a b^{2} - 2 \, A b^{3}\right )} x^{2} + 2 \, {\left (5 \, B a^{2} b - 2 \, A a b^{2}\right )} x\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {b x + a}}\right ) + {\left (3 \, B b^{3} x^{2} + 15 \, B a^{2} b - 6 \, A a b^{2} + 4 \, {\left (5 \, B a b^{2} - 2 \, A b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{3 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}}\right ] \] Input:

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

Output:

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 729 vs. \(2 (110) = 220\).

Time = 62.53 (sec) , antiderivative size = 729, normalized size of antiderivative = 6.51 \[ \int \frac {x^{3/2} (A+B x)}{(a+b x)^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

A*(6*a**(39/2)*b**11*x**(27/2)*sqrt(1 + b*x/a)*asinh(sqrt(b)*sqrt(x)/sqrt( 
a))/(3*a**(39/2)*b**(27/2)*x**(27/2)*sqrt(1 + b*x/a) + 3*a**(37/2)*b**(29/ 
2)*x**(29/2)*sqrt(1 + b*x/a)) + 6*a**(37/2)*b**12*x**(29/2)*sqrt(1 + b*x/a 
)*asinh(sqrt(b)*sqrt(x)/sqrt(a))/(3*a**(39/2)*b**(27/2)*x**(27/2)*sqrt(1 + 
 b*x/a) + 3*a**(37/2)*b**(29/2)*x**(29/2)*sqrt(1 + b*x/a)) - 6*a**19*b**(2 
3/2)*x**14/(3*a**(39/2)*b**(27/2)*x**(27/2)*sqrt(1 + b*x/a) + 3*a**(37/2)* 
b**(29/2)*x**(29/2)*sqrt(1 + b*x/a)) - 8*a**18*b**(25/2)*x**15/(3*a**(39/2 
)*b**(27/2)*x**(27/2)*sqrt(1 + b*x/a) + 3*a**(37/2)*b**(29/2)*x**(29/2)*sq 
rt(1 + b*x/a))) + B*(-15*a**(81/2)*b**22*x**(51/2)*sqrt(1 + b*x/a)*asinh(s 
qrt(b)*sqrt(x)/sqrt(a))/(3*a**(79/2)*b**(51/2)*x**(51/2)*sqrt(1 + b*x/a) + 
 3*a**(77/2)*b**(53/2)*x**(53/2)*sqrt(1 + b*x/a)) - 15*a**(79/2)*b**23*x** 
(53/2)*sqrt(1 + b*x/a)*asinh(sqrt(b)*sqrt(x)/sqrt(a))/(3*a**(79/2)*b**(51/ 
2)*x**(51/2)*sqrt(1 + b*x/a) + 3*a**(77/2)*b**(53/2)*x**(53/2)*sqrt(1 + b* 
x/a)) + 15*a**40*b**(45/2)*x**26/(3*a**(79/2)*b**(51/2)*x**(51/2)*sqrt(1 + 
 b*x/a) + 3*a**(77/2)*b**(53/2)*x**(53/2)*sqrt(1 + b*x/a)) + 20*a**39*b**( 
47/2)*x**27/(3*a**(79/2)*b**(51/2)*x**(51/2)*sqrt(1 + b*x/a) + 3*a**(77/2) 
*b**(53/2)*x**(53/2)*sqrt(1 + b*x/a)) + 3*a**38*b**(49/2)*x**28/(3*a**(79/ 
2)*b**(51/2)*x**(51/2)*sqrt(1 + b*x/a) + 3*a**(77/2)*b**(53/2)*x**(53/2)*s 
qrt(1 + b*x/a)))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 334 vs. \(2 (92) = 184\).

Time = 0.04 (sec) , antiderivative size = 334, normalized size of antiderivative = 2.98 \[ \int \frac {x^{3/2} (A+B x)}{(a+b x)^{5/2}} \, dx=\frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} B a}{3 \, {\left (b^{5} x^{3} + 3 \, a b^{4} x^{2} + 3 \, a^{2} b^{3} x + a^{3} b^{2}\right )}} - \frac {\sqrt {b x^{2} + a x} B a^{2}}{3 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}} - \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} A}{3 \, {\left (b^{4} x^{3} + 3 \, a b^{3} x^{2} + 3 \, a^{2} b^{2} x + a^{3} b\right )}} + \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} B}{b^{4} x^{2} + 2 \, a b^{3} x + a^{2} b^{2}} + \frac {\sqrt {b x^{2} + a x} A a}{3 \, {\left (b^{4} x^{2} + 2 \, a b^{3} x + a^{2} b^{2}\right )}} + \frac {16 \, \sqrt {b x^{2} + a x} B a}{3 \, {\left (b^{4} x + a b^{3}\right )}} - \frac {7 \, \sqrt {b x^{2} + a x} A}{3 \, {\left (b^{3} x + a b^{2}\right )}} - \frac {5 \, B a \log \left (2 \, x + \frac {a}{b} + \frac {2 \, \sqrt {b x^{2} + a x}}{\sqrt {b}}\right )}{2 \, b^{\frac {7}{2}}} + \frac {A \log \left (2 \, x + \frac {a}{b} + \frac {2 \, \sqrt {b x^{2} + a x}}{\sqrt {b}}\right )}{b^{\frac {5}{2}}} \] Input:

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

Output:

1/3*(b*x^2 + a*x)^(3/2)*B*a/(b^5*x^3 + 3*a*b^4*x^2 + 3*a^2*b^3*x + a^3*b^2 
) - 1/3*sqrt(b*x^2 + a*x)*B*a^2/(b^5*x^2 + 2*a*b^4*x + a^2*b^3) - 1/3*(b*x 
^2 + a*x)^(3/2)*A/(b^4*x^3 + 3*a*b^3*x^2 + 3*a^2*b^2*x + a^3*b) + (b*x^2 + 
 a*x)^(3/2)*B/(b^4*x^2 + 2*a*b^3*x + a^2*b^2) + 1/3*sqrt(b*x^2 + a*x)*A*a/ 
(b^4*x^2 + 2*a*b^3*x + a^2*b^2) + 16/3*sqrt(b*x^2 + a*x)*B*a/(b^4*x + a*b^ 
3) - 7/3*sqrt(b*x^2 + a*x)*A/(b^3*x + a*b^2) - 5/2*B*a*log(2*x + a/b + 2*s 
qrt(b*x^2 + a*x)/sqrt(b))/b^(7/2) + A*log(2*x + a/b + 2*sqrt(b*x^2 + a*x)/ 
sqrt(b))/b^(5/2)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 297 vs. \(2 (92) = 184\).

Time = 15.52 (sec) , antiderivative size = 297, normalized size of antiderivative = 2.65 \[ \int \frac {x^{3/2} (A+B x)}{(a+b x)^{5/2}} \, dx=\frac {\sqrt {{\left (b x + a\right )} b - a b} \sqrt {b x + a} B {\left | b \right |}}{b^{5}} + \frac {{\left (5 \, B a {\left | b \right |} - 2 \, A b {\left | b \right |}\right )} \log \left ({\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2}\right )}{2 \, b^{\frac {9}{2}}} + \frac {4 \, {\left (9 \, B a^{2} {\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{4} {\left | b \right |} + 12 \, B a^{3} {\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} b {\left | b \right |} - 6 \, A a {\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{4} b {\left | b \right |} + 7 \, B a^{4} b^{2} {\left | b \right |} - 6 \, A a^{2} {\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} b^{2} {\left | b \right |} - 4 \, A a^{3} b^{3} {\left | b \right |}\right )}}{3 \, {\left ({\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} + a b\right )}^{3} b^{\frac {7}{2}}} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.61 \[ \int \frac {x^{3/2} (A+B x)}{(a+b x)^{5/2}} \, dx=\frac {-12 \sqrt {b}\, \sqrt {b x +a}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) a +9 \sqrt {b}\, \sqrt {b x +a}\, a +12 \sqrt {x}\, a b +4 \sqrt {x}\, b^{2} x}{4 \sqrt {b x +a}\, b^{3}} \] Input:

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

Output:

( - 12*sqrt(b)*sqrt(a + b*x)*log((sqrt(a + b*x) + sqrt(x)*sqrt(b))/sqrt(a) 
)*a + 9*sqrt(b)*sqrt(a + b*x)*a + 12*sqrt(x)*a*b + 4*sqrt(x)*b**2*x)/(4*sq 
rt(a + b*x)*b**3)