3.7.83 \(\int \frac {x (a+b x)^{5/2}}{(c+d x)^{3/2}} \, dx\) [683]

3.7.83.1 Optimal result
3.7.83.2 Mathematica [A] (verified)
3.7.83.3 Rubi [A] (verified)
3.7.83.4 Maple [B] (verified)
3.7.83.5 Fricas [A] (verification not implemented)
3.7.83.6 Sympy [F]
3.7.83.7 Maxima [F(-2)]
3.7.83.8 Giac [A] (verification not implemented)
3.7.83.9 Mupad [F(-1)]

3.7.83.1 Optimal result

Integrand size = 20, antiderivative size = 218 \[ \int \frac {x (a+b x)^{5/2}}{(c+d x)^{3/2}} \, dx=-\frac {2 c (a+b x)^{7/2}}{d (b c-a d) \sqrt {c+d x}}+\frac {5 (b c-a d) (7 b c-a d) \sqrt {a+b x} \sqrt {c+d x}}{8 d^4}-\frac {5 (7 b c-a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 d^3}+\frac {(7 b c-a d) (a+b x)^{5/2} \sqrt {c+d x}}{3 d^2 (b c-a d)}-\frac {5 (b c-a d)^2 (7 b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 \sqrt {b} d^{9/2}} \]

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

Time = 0.52 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.75 \[ \int \frac {x (a+b x)^{5/2}}{(c+d x)^{3/2}} \, dx=\frac {\sqrt {a+b x} \left (3 a^2 d^2 (27 c+11 d x)+2 a b d \left (-95 c^2-34 c d x+13 d^2 x^2\right )+b^2 \left (105 c^3+35 c^2 d x-14 c d^2 x^2+8 d^3 x^3\right )\right )}{24 d^4 \sqrt {c+d x}}-\frac {5 (b c-a d)^2 (7 b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 \sqrt {b} d^{9/2}} \]

input
Integrate[(x*(a + b*x)^(5/2))/(c + d*x)^(3/2),x]
 
output
(Sqrt[a + b*x]*(3*a^2*d^2*(27*c + 11*d*x) + 2*a*b*d*(-95*c^2 - 34*c*d*x + 
13*d^2*x^2) + b^2*(105*c^3 + 35*c^2*d*x - 14*c*d^2*x^2 + 8*d^3*x^3)))/(24* 
d^4*Sqrt[c + d*x]) - (5*(b*c - a*d)^2*(7*b*c - a*d)*ArcTanh[(Sqrt[d]*Sqrt[ 
a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(8*Sqrt[b]*d^(9/2))
 
3.7.83.3 Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 213, normalized size of antiderivative = 0.98, 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.

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 66

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

\(\Big \downarrow \) 221

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

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

3.7.83.3.1 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]
 
3.7.83.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(688\) vs. \(2(182)=364\).

Time = 0.60 (sec) , antiderivative size = 689, normalized size of antiderivative = 3.16

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

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

Time = 0.39 (sec) , antiderivative size = 598, normalized size of antiderivative = 2.74 \[ \int \frac {x (a+b x)^{5/2}}{(c+d x)^{3/2}} \, dx=\left [-\frac {15 \, {\left (7 \, b^{3} c^{4} - 15 \, a b^{2} c^{3} d + 9 \, a^{2} b c^{2} d^{2} - a^{3} c d^{3} + {\left (7 \, b^{3} c^{3} d - 15 \, a b^{2} c^{2} d^{2} + 9 \, a^{2} b c d^{3} - a^{3} d^{4}\right )} x\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) - 4 \, {\left (8 \, b^{3} d^{4} x^{3} + 105 \, b^{3} c^{3} d - 190 \, a b^{2} c^{2} d^{2} + 81 \, a^{2} b c d^{3} - 2 \, {\left (7 \, b^{3} c d^{3} - 13 \, a b^{2} d^{4}\right )} x^{2} + {\left (35 \, b^{3} c^{2} d^{2} - 68 \, a b^{2} c d^{3} + 33 \, a^{2} b d^{4}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{96 \, {\left (b d^{6} x + b c d^{5}\right )}}, \frac {15 \, {\left (7 \, b^{3} c^{4} - 15 \, a b^{2} c^{3} d + 9 \, a^{2} b c^{2} d^{2} - a^{3} c d^{3} + {\left (7 \, b^{3} c^{3} d - 15 \, a b^{2} c^{2} d^{2} + 9 \, a^{2} b c d^{3} - a^{3} d^{4}\right )} x\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) + 2 \, {\left (8 \, b^{3} d^{4} x^{3} + 105 \, b^{3} c^{3} d - 190 \, a b^{2} c^{2} d^{2} + 81 \, a^{2} b c d^{3} - 2 \, {\left (7 \, b^{3} c d^{3} - 13 \, a b^{2} d^{4}\right )} x^{2} + {\left (35 \, b^{3} c^{2} d^{2} - 68 \, a b^{2} c d^{3} + 33 \, a^{2} b d^{4}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, {\left (b d^{6} x + b c d^{5}\right )}}\right ] \]

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

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

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

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

input
integrate(x*(b*x+a)^(5/2)/(d*x+c)^(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
 
3.7.83.8 Giac [A] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.34 \[ \int \frac {x (a+b x)^{5/2}}{(c+d x)^{3/2}} \, dx=\frac {{\left ({\left (b x + a\right )} {\left (2 \, {\left (b x + a\right )} {\left (\frac {4 \, {\left (b x + a\right )} {\left | b \right |}}{b d} - \frac {7 \, b^{2} c d^{5} {\left | b \right |} - a b d^{6} {\left | b \right |}}{b^{2} d^{7}}\right )} + \frac {5 \, {\left (7 \, b^{3} c^{2} d^{4} {\left | b \right |} - 8 \, a b^{2} c d^{5} {\left | b \right |} + a^{2} b d^{6} {\left | b \right |}\right )}}{b^{2} d^{7}}\right )} + \frac {15 \, {\left (7 \, b^{4} c^{3} d^{3} {\left | b \right |} - 15 \, a b^{3} c^{2} d^{4} {\left | b \right |} + 9 \, a^{2} b^{2} c d^{5} {\left | b \right |} - a^{3} b d^{6} {\left | b \right |}\right )}}{b^{2} d^{7}}\right )} \sqrt {b x + a}}{24 \, \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}} + \frac {5 \, {\left (7 \, b^{3} c^{3} {\left | b \right |} - 15 \, a b^{2} c^{2} d {\left | b \right |} + 9 \, a^{2} b c d^{2} {\left | b \right |} - a^{3} d^{3} {\left | b \right |}\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{8 \, \sqrt {b d} b d^{4}} \]

input
integrate(x*(b*x+a)^(5/2)/(d*x+c)^(3/2),x, algorithm="giac")
 
output
1/24*((b*x + a)*(2*(b*x + a)*(4*(b*x + a)*abs(b)/(b*d) - (7*b^2*c*d^5*abs( 
b) - a*b*d^6*abs(b))/(b^2*d^7)) + 5*(7*b^3*c^2*d^4*abs(b) - 8*a*b^2*c*d^5* 
abs(b) + a^2*b*d^6*abs(b))/(b^2*d^7)) + 15*(7*b^4*c^3*d^3*abs(b) - 15*a*b^ 
3*c^2*d^4*abs(b) + 9*a^2*b^2*c*d^5*abs(b) - a^3*b*d^6*abs(b))/(b^2*d^7))*s 
qrt(b*x + a)/sqrt(b^2*c + (b*x + a)*b*d - a*b*d) + 5/8*(7*b^3*c^3*abs(b) - 
 15*a*b^2*c^2*d*abs(b) + 9*a^2*b*c*d^2*abs(b) - a^3*d^3*abs(b))*log(abs(-s 
qrt(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(sqrt(b*d)* 
b*d^4)
 
3.7.83.9 Mupad [F(-1)]

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

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