3.8.69 \(\int \frac {x (c+d x)^{5/2}}{(a+b x)^{3/2}} \, dx\) [769]

3.8.69.1 Optimal result
3.8.69.2 Mathematica [A] (verified)
3.8.69.3 Rubi [A] (verified)
3.8.69.4 Maple [B] (verified)
3.8.69.5 Fricas [A] (verification not implemented)
3.8.69.6 Sympy [F]
3.8.69.7 Maxima [F(-2)]
3.8.69.8 Giac [A] (verification not implemented)
3.8.69.9 Mupad [F(-1)]

3.8.69.1 Optimal result

Integrand size = 20, antiderivative size = 214 \[ \int \frac {x (c+d x)^{5/2}}{(a+b x)^{3/2}} \, dx=\frac {5 (b c-7 a d) (b c-a d) \sqrt {a+b x} \sqrt {c+d x}}{8 b^4}+\frac {5 (b c-7 a d) \sqrt {a+b x} (c+d x)^{3/2}}{12 b^3}+\frac {(b c-7 a d) \sqrt {a+b x} (c+d x)^{5/2}}{3 b^2 (b c-a d)}+\frac {2 a (c+d x)^{7/2}}{b (b c-a d) \sqrt {a+b x}}+\frac {5 (b c-7 a d) (b c-a d)^2 \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 b^{9/2} \sqrt {d}} \]

output
5/8*(-7*a*d+b*c)*(-a*d+b*c)^2*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c 
)^(1/2))/b^(9/2)/d^(1/2)+2*a*(d*x+c)^(7/2)/b/(-a*d+b*c)/(b*x+a)^(1/2)+5/12 
*(-7*a*d+b*c)*(d*x+c)^(3/2)*(b*x+a)^(1/2)/b^3+1/3*(-7*a*d+b*c)*(d*x+c)^(5/ 
2)*(b*x+a)^(1/2)/b^2/(-a*d+b*c)+5/8*(-7*a*d+b*c)*(-a*d+b*c)*(b*x+a)^(1/2)* 
(d*x+c)^(1/2)/b^4
 
3.8.69.2 Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.75 \[ \int \frac {x (c+d x)^{5/2}}{(a+b x)^{3/2}} \, dx=\frac {\sqrt {c+d x} \left (105 a^3 d^2+5 a^2 b d (-38 c+7 d x)+a b^2 \left (81 c^2-68 c d x-14 d^2 x^2\right )+b^3 x \left (33 c^2+26 c d x+8 d^2 x^2\right )\right )}{24 b^4 \sqrt {a+b x}}+\frac {5 (b c-7 a d) (b c-a d)^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{8 b^{9/2} \sqrt {d}} \]

input
Integrate[(x*(c + d*x)^(5/2))/(a + b*x)^(3/2),x]
 
output
(Sqrt[c + d*x]*(105*a^3*d^2 + 5*a^2*b*d*(-38*c + 7*d*x) + a*b^2*(81*c^2 - 
68*c*d*x - 14*d^2*x^2) + b^3*x*(33*c^2 + 26*c*d*x + 8*d^2*x^2)))/(24*b^4*S 
qrt[a + b*x]) + (5*(b*c - 7*a*d)*(b*c - a*d)^2*ArcTanh[(Sqrt[b]*Sqrt[c + d 
*x])/(Sqrt[d]*Sqrt[a + b*x])])/(8*b^(9/2)*Sqrt[d])
 
3.8.69.3 Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 211, normalized size of antiderivative = 0.99, 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, 66, 221}

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.

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 66

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

\(\Big \downarrow \) 221

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

input
Int[(x*(c + d*x)^(5/2))/(a + b*x)^(3/2),x]
 
output
(2*a*(c + d*x)^(7/2))/(b*(b*c - a*d)*Sqrt[a + b*x]) + ((b*c - 7*a*d)*((Sqr 
t[a + b*x]*(c + d*x)^(5/2))/(3*b) + (5*(b*c - a*d)*((Sqrt[a + b*x]*(c + d* 
x)^(3/2))/(2*b) + (3*(b*c - a*d)*((Sqrt[a + b*x]*Sqrt[c + d*x])/b + ((b*c 
- a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(b^(3/2)* 
Sqrt[d])))/(4*b)))/(6*b)))/(b*(b*c - a*d))
 

3.8.69.3.1 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 66
Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 
2   Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre 
eQ[{a, b, c, d}, x] &&  !GtQ[c - a*(d/b), 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 221
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 
3.8.69.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(688\) vs. \(2(178)=356\).

Time = 0.58 (sec) , antiderivative size = 689, normalized size of antiderivative = 3.22

method result size
default \(-\frac {\sqrt {d x +c}\, \left (-16 b^{3} d^{2} x^{3} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+105 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{3} b \,d^{3} x -225 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} b^{2} c \,d^{2} x +135 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a \,b^{3} c^{2} d x -15 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{4} c^{3} x +28 a \,b^{2} d^{2} x^{2} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}-52 b^{3} c d \,x^{2} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+105 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{4} d^{3}-225 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{3} b c \,d^{2}+135 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} b^{2} c^{2} d -15 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a \,b^{3} c^{3}-70 a^{2} b \,d^{2} x \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+136 a \,b^{2} c d x \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}-66 b^{3} c^{2} x \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}-210 a^{3} d^{2} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+380 a^{2} b c d \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}-162 a \,b^{2} c^{2} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\right )}{48 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {b x +a}\, b^{4}}\) \(689\)

input
int(x*(d*x+c)^(5/2)/(b*x+a)^(3/2),x,method=_RETURNVERBOSE)
 
output
-1/48*(d*x+c)^(1/2)*(-16*b^3*d^2*x^3*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+1 
05*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1 
/2))*a^3*b*d^3*x-225*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2) 
+a*d+b*c)/(b*d)^(1/2))*a^2*b^2*c*d^2*x+135*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x 
+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a*b^3*c^2*d*x-15*ln(1/2*(2*b* 
d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*b^4*c^3*x+ 
28*a*b^2*d^2*x^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-52*b^3*c*d*x^2*((b*x+ 
a)*(d*x+c))^(1/2)*(b*d)^(1/2)+105*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2 
)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a^4*d^3-225*ln(1/2*(2*b*d*x+2*((b*x+a) 
*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a^3*b*c*d^2+135*ln(1/2*( 
2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a^2*b^ 
2*c^2*d-15*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/ 
(b*d)^(1/2))*a*b^3*c^3-70*a^2*b*d^2*x*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+ 
136*a*b^2*c*d*x*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-66*b^3*c^2*x*((b*x+a)* 
(d*x+c))^(1/2)*(b*d)^(1/2)-210*a^3*d^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2) 
+380*a^2*b*c*d*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-162*a*b^2*c^2*((b*x+a)* 
(d*x+c))^(1/2)*(b*d)^(1/2))/((b*x+a)*(d*x+c))^(1/2)/(b*d)^(1/2)/(b*x+a)^(1 
/2)/b^4
 
3.8.69.5 Fricas [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 594, normalized size of antiderivative = 2.78 \[ \int \frac {x (c+d x)^{5/2}}{(a+b x)^{3/2}} \, dx=\left [-\frac {15 \, {\left (a b^{3} c^{3} - 9 \, a^{2} b^{2} c^{2} d + 15 \, a^{3} b c d^{2} - 7 \, a^{4} d^{3} + {\left (b^{4} c^{3} - 9 \, a b^{3} c^{2} d + 15 \, a^{2} b^{2} c d^{2} - 7 \, a^{3} b d^{3}\right )} x\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} - 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) - 4 \, {\left (8 \, b^{4} d^{3} x^{3} + 81 \, a b^{3} c^{2} d - 190 \, a^{2} b^{2} c d^{2} + 105 \, a^{3} b d^{3} + 2 \, {\left (13 \, b^{4} c d^{2} - 7 \, a b^{3} d^{3}\right )} x^{2} + {\left (33 \, b^{4} c^{2} d - 68 \, a b^{3} c d^{2} + 35 \, a^{2} b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{96 \, {\left (b^{6} d x + a b^{5} d\right )}}, -\frac {15 \, {\left (a b^{3} c^{3} - 9 \, a^{2} b^{2} c^{2} d + 15 \, a^{3} b c d^{2} - 7 \, a^{4} d^{3} + {\left (b^{4} c^{3} - 9 \, a b^{3} c^{2} d + 15 \, a^{2} b^{2} c d^{2} - 7 \, a^{3} b d^{3}\right )} x\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) - 2 \, {\left (8 \, b^{4} d^{3} x^{3} + 81 \, a b^{3} c^{2} d - 190 \, a^{2} b^{2} c d^{2} + 105 \, a^{3} b d^{3} + 2 \, {\left (13 \, b^{4} c d^{2} - 7 \, a b^{3} d^{3}\right )} x^{2} + {\left (33 \, b^{4} c^{2} d - 68 \, a b^{3} c d^{2} + 35 \, a^{2} b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, {\left (b^{6} d x + a b^{5} d\right )}}\right ] \]

input
integrate(x*(d*x+c)^(5/2)/(b*x+a)^(3/2),x, algorithm="fricas")
 
output
[-1/96*(15*(a*b^3*c^3 - 9*a^2*b^2*c^2*d + 15*a^3*b*c*d^2 - 7*a^4*d^3 + (b^ 
4*c^3 - 9*a*b^3*c^2*d + 15*a^2*b^2*c*d^2 - 7*a^3*b*d^3)*x)*sqrt(b*d)*log(8 
*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 - 4*(2*b*d*x + b*c + a*d)*sqr 
t(b*d)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(b^2*c*d + a*b*d^2)*x) - 4*(8*b^4*d 
^3*x^3 + 81*a*b^3*c^2*d - 190*a^2*b^2*c*d^2 + 105*a^3*b*d^3 + 2*(13*b^4*c* 
d^2 - 7*a*b^3*d^3)*x^2 + (33*b^4*c^2*d - 68*a*b^3*c*d^2 + 35*a^2*b^2*d^3)* 
x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^6*d*x + a*b^5*d), -1/48*(15*(a*b^3*c^3 
- 9*a^2*b^2*c^2*d + 15*a^3*b*c*d^2 - 7*a^4*d^3 + (b^4*c^3 - 9*a*b^3*c^2*d 
+ 15*a^2*b^2*c*d^2 - 7*a^3*b*d^3)*x)*sqrt(-b*d)*arctan(1/2*(2*b*d*x + b*c 
+ a*d)*sqrt(-b*d)*sqrt(b*x + a)*sqrt(d*x + c)/(b^2*d^2*x^2 + a*b*c*d + (b^ 
2*c*d + a*b*d^2)*x)) - 2*(8*b^4*d^3*x^3 + 81*a*b^3*c^2*d - 190*a^2*b^2*c*d 
^2 + 105*a^3*b*d^3 + 2*(13*b^4*c*d^2 - 7*a*b^3*d^3)*x^2 + (33*b^4*c^2*d - 
68*a*b^3*c*d^2 + 35*a^2*b^2*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^6*d*x 
+ a*b^5*d)]
 
3.8.69.6 Sympy [F]

\[ \int \frac {x (c+d x)^{5/2}}{(a+b x)^{3/2}} \, dx=\int \frac {x \left (c + d x\right )^{\frac {5}{2}}}{\left (a + b x\right )^{\frac {3}{2}}}\, dx \]

input
integrate(x*(d*x+c)**(5/2)/(b*x+a)**(3/2),x)
 
output
Integral(x*(c + d*x)**(5/2)/(a + b*x)**(3/2), x)
 
3.8.69.7 Maxima [F(-2)]

Exception generated. \[ \int \frac {x (c+d x)^{5/2}}{(a+b x)^{3/2}} \, dx=\text {Exception raised: ValueError} \]

input
integrate(x*(d*x+c)^(5/2)/(b*x+a)^(3/2),x, algorithm="maxima")
 
output
Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for m 
ore detail
 
3.8.69.8 Giac [A] (verification not implemented)

Time = 0.47 (sec) , antiderivative size = 340, normalized size of antiderivative = 1.59 \[ \int \frac {x (c+d x)^{5/2}}{(a+b x)^{3/2}} \, dx=\frac {1}{24} \, \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \sqrt {b x + a} {\left (2 \, {\left (b x + a\right )} {\left (\frac {4 \, {\left (b x + a\right )} d^{2} {\left | b \right |}}{b^{6}} + \frac {13 \, b^{18} c d^{5} {\left | b \right |} - 19 \, a b^{17} d^{6} {\left | b \right |}}{b^{23} d^{4}}\right )} + \frac {3 \, {\left (11 \, b^{19} c^{2} d^{4} {\left | b \right |} - 40 \, a b^{18} c d^{5} {\left | b \right |} + 29 \, a^{2} b^{17} d^{6} {\left | b \right |}\right )}}{b^{23} d^{4}}\right )} - \frac {5 \, {\left (b^{3} c^{3} {\left | b \right |} - 9 \, a b^{2} c^{2} d {\left | b \right |} + 15 \, a^{2} b c d^{2} {\left | b \right |} - 7 \, a^{3} d^{3} {\left | b \right |}\right )} \log \left ({\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}{16 \, \sqrt {b d} b^{5}} + \frac {4 \, {\left (a b^{3} c^{3} d {\left | b \right |} - 3 \, a^{2} b^{2} c^{2} d^{2} {\left | b \right |} + 3 \, a^{3} b c d^{3} {\left | b \right |} - a^{4} d^{4} {\left | b \right |}\right )}}{{\left (b^{2} c - a b d - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )} \sqrt {b d} b^{4}} \]

input
integrate(x*(d*x+c)^(5/2)/(b*x+a)^(3/2),x, algorithm="giac")
 
output
1/24*sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a)*(2*(b*x + a)*(4*(b* 
x + a)*d^2*abs(b)/b^6 + (13*b^18*c*d^5*abs(b) - 19*a*b^17*d^6*abs(b))/(b^2 
3*d^4)) + 3*(11*b^19*c^2*d^4*abs(b) - 40*a*b^18*c*d^5*abs(b) + 29*a^2*b^17 
*d^6*abs(b))/(b^23*d^4)) - 5/16*(b^3*c^3*abs(b) - 9*a*b^2*c^2*d*abs(b) + 1 
5*a^2*b*c*d^2*abs(b) - 7*a^3*d^3*abs(b))*log((sqrt(b*d)*sqrt(b*x + a) - sq 
rt(b^2*c + (b*x + a)*b*d - a*b*d))^2)/(sqrt(b*d)*b^5) + 4*(a*b^3*c^3*d*abs 
(b) - 3*a^2*b^2*c^2*d^2*abs(b) + 3*a^3*b*c*d^3*abs(b) - a^4*d^4*abs(b))/(( 
b^2*c - a*b*d - (sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a* 
b*d))^2)*sqrt(b*d)*b^4)
 
3.8.69.9 Mupad [F(-1)]

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

input
int((x*(c + d*x)^(5/2))/(a + b*x)^(3/2),x)
 
output
int((x*(c + d*x)^(5/2))/(a + b*x)^(3/2), x)