3.7.39 \(\int \frac {(a+b x)^{3/2}}{x^3 (c+d x)^{3/2}} \, dx\) [639]

3.7.39.1 Optimal result
3.7.39.2 Mathematica [A] (verified)
3.7.39.3 Rubi [A] (verified)
3.7.39.4 Maple [B] (verified)
3.7.39.5 Fricas [A] (verification not implemented)
3.7.39.6 Sympy [F]
3.7.39.7 Maxima [F(-2)]
3.7.39.8 Giac [B] (verification not implemented)
3.7.39.9 Mupad [F(-1)]

3.7.39.1 Optimal result

Integrand size = 22, antiderivative size = 175 \[ \int \frac {(a+b x)^{3/2}}{x^3 (c+d x)^{3/2}} \, dx=\frac {3 (b c-5 a d) (b c-a d) \sqrt {a+b x}}{4 a c^3 \sqrt {c+d x}}-\frac {(b c-5 a d) (a+b x)^{3/2}}{4 a c^2 x \sqrt {c+d x}}-\frac {(a+b x)^{5/2}}{2 a c x^2 \sqrt {c+d x}}-\frac {3 (b c-5 a d) (b c-a d) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 \sqrt {a} c^{7/2}} \]

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

Time = 10.07 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.74 \[ \int \frac {(a+b x)^{3/2}}{x^3 (c+d x)^{3/2}} \, dx=\frac {\sqrt {a+b x} \left (-b c x (5 c+13 d x)+a \left (-2 c^2+5 c d x+15 d^2 x^2\right )\right )}{4 c^3 x^2 \sqrt {c+d x}}-\frac {3 \left (b^2 c^2-6 a b c d+5 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 \sqrt {a} c^{7/2}} \]

input
Integrate[(a + b*x)^(3/2)/(x^3*(c + d*x)^(3/2)),x]
 
output
(Sqrt[a + b*x]*(-(b*c*x*(5*c + 13*d*x)) + a*(-2*c^2 + 5*c*d*x + 15*d^2*x^2 
)))/(4*c^3*x^2*Sqrt[c + d*x]) - (3*(b^2*c^2 - 6*a*b*c*d + 5*a^2*d^2)*ArcTa 
nh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(4*Sqrt[a]*c^(7/2))
 
3.7.39.3 Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.90, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {107, 105, 105, 104, 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.

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

\(\Big \downarrow \) 107

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

\(\Big \downarrow \) 105

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

\(\Big \downarrow \) 105

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

\(\Big \downarrow \) 104

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

\(\Big \downarrow \) 221

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

input
Int[(a + b*x)^(3/2)/(x^3*(c + d*x)^(3/2)),x]
 
output
-1/2*(a + b*x)^(5/2)/(a*c*x^2*Sqrt[c + d*x]) + ((b*c - 5*a*d)*(-((a + b*x) 
^(3/2)/(c*x*Sqrt[c + d*x])) + (3*(b*c - a*d)*((2*Sqrt[a + b*x])/(c*Sqrt[c 
+ d*x]) - (2*Sqrt[a]*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x 
])])/c^(3/2)))/(2*c)))/(4*a*c)
 

3.7.39.3.1 Defintions of rubi rules used

rule 104
Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x 
_)), x_] :> With[{q = Denominator[m]}, Simp[q   Subst[Int[x^(q*(m + 1) - 1) 
/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] 
] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L 
tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
 

rule 105
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((m + 
1)*(b*e - a*f))), x] - Simp[n*((d*e - c*f)/((m + 1)*(b*e - a*f)))   Int[(a 
+ b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, 
e, f, m, p}, x] && EqQ[m + n + p + 2, 0] && GtQ[n, 0] && (SumSimplerQ[m, 1] 
 ||  !SumSimplerQ[p, 1]) && NeQ[m, -1]
 

rule 107
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[b*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 
)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[(a*d*f*(m + 1) + b*c*f*(n + 
 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f))   Int[(a + b*x)^(m + 
 1)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x 
] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || SumSimplerQ[m, 1])
 

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]
 
3.7.39.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(463\) vs. \(2(143)=286\).

Time = 1.54 (sec) , antiderivative size = 464, normalized size of antiderivative = 2.65

method result size
default \(-\frac {\sqrt {b x +a}\, \left (15 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) a^{2} d^{3} x^{3}-18 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) a b c \,d^{2} x^{3}+3 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) b^{2} c^{2} d \,x^{3}+15 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) a^{2} c \,d^{2} x^{2}-18 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) a b \,c^{2} d \,x^{2}+3 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) b^{2} c^{3} x^{2}-30 a \,d^{2} x^{2} \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+26 b c d \,x^{2} \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}-10 a c d x \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+10 b \,c^{2} x \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+4 a \,c^{2} \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\right )}{8 c^{3} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, x^{2} \sqrt {a c}\, \sqrt {d x +c}}\) \(464\)

input
int((b*x+a)^(3/2)/x^3/(d*x+c)^(3/2),x,method=_RETURNVERBOSE)
 
output
-1/8*(b*x+a)^(1/2)*(15*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/ 
2)+2*a*c)/x)*a^2*d^3*x^3-18*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c) 
)^(1/2)+2*a*c)/x)*a*b*c*d^2*x^3+3*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*( 
d*x+c))^(1/2)+2*a*c)/x)*b^2*c^2*d*x^3+15*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((b 
*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*a^2*c*d^2*x^2-18*ln((a*d*x+b*c*x+2*(a*c)^(1 
/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*a*b*c^2*d*x^2+3*ln((a*d*x+b*c*x+2*(a 
*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*b^2*c^3*x^2-30*a*d^2*x^2*(a*c) 
^(1/2)*((b*x+a)*(d*x+c))^(1/2)+26*b*c*d*x^2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^ 
(1/2)-10*a*c*d*x*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+10*b*c^2*x*(a*c)^(1/2 
)*((b*x+a)*(d*x+c))^(1/2)+4*a*c^2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/c^3 
/((b*x+a)*(d*x+c))^(1/2)/x^2/(a*c)^(1/2)/(d*x+c)^(1/2)
 
3.7.39.5 Fricas [A] (verification not implemented)

Time = 0.61 (sec) , antiderivative size = 472, normalized size of antiderivative = 2.70 \[ \int \frac {(a+b x)^{3/2}}{x^3 (c+d x)^{3/2}} \, dx=\left [\frac {3 \, {\left ({\left (b^{2} c^{2} d - 6 \, a b c d^{2} + 5 \, a^{2} d^{3}\right )} x^{3} + {\left (b^{2} c^{3} - 6 \, a b c^{2} d + 5 \, a^{2} c d^{2}\right )} x^{2}\right )} \sqrt {a c} \log \left (\frac {8 \, a^{2} c^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{2} - 4 \, {\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {a c} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) - 4 \, {\left (2 \, a^{2} c^{3} + {\left (13 \, a b c^{2} d - 15 \, a^{2} c d^{2}\right )} x^{2} + 5 \, {\left (a b c^{3} - a^{2} c^{2} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{16 \, {\left (a c^{4} d x^{3} + a c^{5} x^{2}\right )}}, \frac {3 \, {\left ({\left (b^{2} c^{2} d - 6 \, a b c d^{2} + 5 \, a^{2} d^{3}\right )} x^{3} + {\left (b^{2} c^{3} - 6 \, a b c^{2} d + 5 \, a^{2} c d^{2}\right )} x^{2}\right )} \sqrt {-a c} \arctan \left (\frac {{\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {-a c} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (a b c d x^{2} + a^{2} c^{2} + {\left (a b c^{2} + a^{2} c d\right )} x\right )}}\right ) - 2 \, {\left (2 \, a^{2} c^{3} + {\left (13 \, a b c^{2} d - 15 \, a^{2} c d^{2}\right )} x^{2} + 5 \, {\left (a b c^{3} - a^{2} c^{2} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{8 \, {\left (a c^{4} d x^{3} + a c^{5} x^{2}\right )}}\right ] \]

input
integrate((b*x+a)^(3/2)/x^3/(d*x+c)^(3/2),x, algorithm="fricas")
 
output
[1/16*(3*((b^2*c^2*d - 6*a*b*c*d^2 + 5*a^2*d^3)*x^3 + (b^2*c^3 - 6*a*b*c^2 
*d + 5*a^2*c*d^2)*x^2)*sqrt(a*c)*log((8*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a 
^2*d^2)*x^2 - 4*(2*a*c + (b*c + a*d)*x)*sqrt(a*c)*sqrt(b*x + a)*sqrt(d*x + 
 c) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) - 4*(2*a^2*c^3 + (13*a*b*c^2*d - 15*a^ 
2*c*d^2)*x^2 + 5*(a*b*c^3 - a^2*c^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a* 
c^4*d*x^3 + a*c^5*x^2), 1/8*(3*((b^2*c^2*d - 6*a*b*c*d^2 + 5*a^2*d^3)*x^3 
+ (b^2*c^3 - 6*a*b*c^2*d + 5*a^2*c*d^2)*x^2)*sqrt(-a*c)*arctan(1/2*(2*a*c 
+ (b*c + a*d)*x)*sqrt(-a*c)*sqrt(b*x + a)*sqrt(d*x + c)/(a*b*c*d*x^2 + a^2 
*c^2 + (a*b*c^2 + a^2*c*d)*x)) - 2*(2*a^2*c^3 + (13*a*b*c^2*d - 15*a^2*c*d 
^2)*x^2 + 5*(a*b*c^3 - a^2*c^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a*c^4*d 
*x^3 + a*c^5*x^2)]
 
3.7.39.6 Sympy [F]

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

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

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

input
integrate((b*x+a)^(3/2)/x^3/(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
 
3.7.39.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1096 vs. \(2 (143) = 286\).

Time = 1.71 (sec) , antiderivative size = 1096, normalized size of antiderivative = 6.26 \[ \int \frac {(a+b x)^{3/2}}{x^3 (c+d x)^{3/2}} \, dx=-\frac {2 \, {\left (b^{3} c d - a b^{2} d^{2}\right )} \sqrt {b x + a}}{\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} c^{3} {\left | b \right |}} - \frac {3 \, {\left (\sqrt {b d} b^{4} c^{2} - 6 \, \sqrt {b d} a b^{3} c d + 5 \, \sqrt {b d} a^{2} b^{2} d^{2}\right )} \arctan \left (-\frac {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}}{2 \, \sqrt {-a b c d} b}\right )}{4 \, \sqrt {-a b c d} b c^{3} {\left | b \right |}} - \frac {5 \, \sqrt {b d} b^{10} c^{5} - 27 \, \sqrt {b d} a b^{9} c^{4} d + 58 \, \sqrt {b d} a^{2} b^{8} c^{3} d^{2} - 62 \, \sqrt {b d} a^{3} b^{7} c^{2} d^{3} + 33 \, \sqrt {b d} a^{4} b^{6} c d^{4} - 7 \, \sqrt {b d} a^{5} b^{5} d^{5} - 15 \, \sqrt {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} b^{8} c^{4} + 40 \, \sqrt {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} a b^{7} c^{3} d - 14 \, \sqrt {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} a^{2} b^{6} c^{2} d^{2} - 32 \, \sqrt {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} a^{3} b^{5} c d^{3} + 21 \, \sqrt {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} a^{4} b^{4} d^{4} + 15 \, \sqrt {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 )}^{4} b^{6} c^{3} - 11 \, \sqrt {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 )}^{4} a b^{5} c^{2} d + \sqrt {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 )}^{4} a^{2} b^{4} c d^{2} - 21 \, \sqrt {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 )}^{4} a^{3} b^{3} d^{3} - 5 \, \sqrt {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 )}^{6} b^{4} c^{2} - 2 \, \sqrt {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 )}^{6} a b^{3} c d + 7 \, \sqrt {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 )}^{6} a^{2} b^{2} d^{2}}{2 \, {\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2} - 2 \, {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} b^{2} c - 2 \, {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} 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 )}^{4}\right )}^{2} c^{3} {\left | b \right |}} \]

input
integrate((b*x+a)^(3/2)/x^3/(d*x+c)^(3/2),x, algorithm="giac")
 
output
-2*(b^3*c*d - a*b^2*d^2)*sqrt(b*x + a)/(sqrt(b^2*c + (b*x + a)*b*d - a*b*d 
)*c^3*abs(b)) - 3/4*(sqrt(b*d)*b^4*c^2 - 6*sqrt(b*d)*a*b^3*c*d + 5*sqrt(b* 
d)*a^2*b^2*d^2)*arctan(-1/2*(b^2*c + a*b*d - (sqrt(b*d)*sqrt(b*x + a) - sq 
rt(b^2*c + (b*x + a)*b*d - a*b*d))^2)/(sqrt(-a*b*c*d)*b))/(sqrt(-a*b*c*d)* 
b*c^3*abs(b)) - 1/2*(5*sqrt(b*d)*b^10*c^5 - 27*sqrt(b*d)*a*b^9*c^4*d + 58* 
sqrt(b*d)*a^2*b^8*c^3*d^2 - 62*sqrt(b*d)*a^3*b^7*c^2*d^3 + 33*sqrt(b*d)*a^ 
4*b^6*c*d^4 - 7*sqrt(b*d)*a^5*b^5*d^5 - 15*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + 
 a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^8*c^4 + 40*sqrt(b*d)*(sqrt( 
b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b^7*c^3*d - 
14*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d 
))^2*a^2*b^6*c^2*d^2 - 32*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c 
+ (b*x + a)*b*d - a*b*d))^2*a^3*b^5*c*d^3 + 21*sqrt(b*d)*(sqrt(b*d)*sqrt(b 
*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^4*b^4*d^4 + 15*sqrt(b*d 
)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*b^6*c^ 
3 - 11*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a 
*b*d))^4*a*b^5*c^2*d + sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + ( 
b*x + a)*b*d - a*b*d))^4*a^2*b^4*c*d^2 - 21*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x 
+ a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^3*b^3*d^3 - 5*sqrt(b*d)*(s 
qrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*b^4*c^2 - 
2*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b...
 
3.7.39.9 Mupad [F(-1)]

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

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