\(\int \frac {(c+d x)^{5/2}}{x (a+b x)^{3/2}} \, dx\) [379]

Optimal result

Integrand size = 22, antiderivative size = 163 \[ \int \frac {(c+d x)^{5/2}}{x (a+b x)^{3/2}} \, dx=-\frac {d (2 b c-3 a d) \sqrt {a+b x} \sqrt {c+d x}}{a b^2}+\frac {2 (b c-a d) (c+d x)^{3/2}}{a b \sqrt {a+b x}}-\frac {2 c^{5/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{a^{3/2}}+\frac {d^{3/2} (5 b c-3 a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{5/2}} \] Output:

-d*(-3*a*d+2*b*c)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/a/b^2+2*(-a*d+b*c)*(d*x+c)^( 
3/2)/a/b/(b*x+a)^(1/2)-2*c^(5/2)*arctanh(c^(1/2)*(b*x+a)^(1/2)/a^(1/2)/(d* 
x+c)^(1/2))/a^(3/2)+d^(3/2)*(-3*a*d+5*b*c)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b 
^(1/2)/(d*x+c)^(1/2))/b^(5/2)
 

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.90 \[ \int \frac {(c+d x)^{5/2}}{x (a+b x)^{3/2}} \, dx=\frac {\sqrt {c+d x} \left (2 b^2 c^2+3 a^2 d^2+a b d (-4 c+d x)\right )}{a b^2 \sqrt {a+b x}}-\frac {2 c^{5/2} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c+d x}}{\sqrt {c} \sqrt {a+b x}}\right )}{a^{3/2}}+\frac {d^{3/2} (5 b c-3 a d) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{b^{5/2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.11, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {109, 27, 171, 27, 175, 66, 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.

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

\(\Big \downarrow \) 109

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 171

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 175

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

\(\Big \downarrow \) 66

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

\(\Big \downarrow \) 104

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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 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 109

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

rule 171

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[h*(a + b*x)^m*(c + d*x)^(n + 1)*(( 
e + f*x)^(p + 1)/(d*f*(m + n + p + 2))), x] + Simp[1/(d*f*(m + n + p + 2)) 
  Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2 
) - h*(b*c*e*m + a*(d*e*(n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) 
+ h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x], x], x] /; Fre 
eQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] 
 && IntegersQ[2*m, 2*n, 2*p]
 

rule 175

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_ 
)))/((a_.) + (b_.)*(x_)), x_] :> Simp[h/b   Int[(c + d*x)^n*(e + f*x)^p, x] 
, x] + Simp[(b*g - a*h)/b   Int[(c + d*x)^n*((e + f*x)^p/(a + b*x)), x], x] 
 /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]
 

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(491\) vs. \(2(131)=262\).

Time = 0.24 (sec) , antiderivative size = 492, normalized size of antiderivative = 3.02

method	result	size
default	\(-\frac {\sqrt {x d +c}\, \left (2 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}+2 a c}{x}\right ) b^{3} c^{3} x \sqrt {d b}+3 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) a^{2} b \,d^{3} x \sqrt {a c}-5 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) a \,b^{2} c \,d^{2} x \sqrt {a c}+2 \sqrt {d b}\, \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}+2 a c}{x}\right ) a \,b^{2} c^{3}+3 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) \sqrt {a c}\, a^{3} d^{3}-5 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) \sqrt {a c}\, a^{2} b c \,d^{2}-2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, \sqrt {a c}\, a b \,d^{2} x -6 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, \sqrt {a c}\, a^{2} d^{2}+8 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, \sqrt {a c}\, a b c d -4 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, \sqrt {a c}\, b^{2} c^{2}\right )}{2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, \sqrt {a c}\, \sqrt {b x +a}\, a \,b^{2}}\)	\(492\)

Input:

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

Output:

-1/2*(d*x+c)^(1/2)*(2*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^3*c^3*x*(d*b)^(1/2)+3*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1 
/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*a^2*b*d^3*x*(a*c)^(1/2)-5*ln(1/2*(2* 
b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*a*b^2*c* 
d^2*x*(a*c)^(1/2)+2*(d*b)^(1/2)*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^2*c^3+3*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2 
)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*(a*c)^(1/2)*a^3*d^3-5*ln(1/2*(2*b*d*x+ 
2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*(a*c)^(1/2)*a^ 
2*b*c*d^2-2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)*(a*c)^(1/2)*a*b*d^2*x-6*(( 
b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)*(a*c)^(1/2)*a^2*d^2+8*((b*x+a)*(d*x+c))^ 
(1/2)*(d*b)^(1/2)*(a*c)^(1/2)*a*b*c*d-4*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2 
)*(a*c)^(1/2)*b^2*c^2)/((b*x+a)*(d*x+c))^(1/2)/(d*b)^(1/2)/(a*c)^(1/2)/(b* 
x+a)^(1/2)/a/b^2
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 269 vs. \(2 (131) = 262\).

Time = 1.13 (sec) , antiderivative size = 1183, normalized size of antiderivative = 7.26 \[ \int \frac {(c+d x)^{5/2}}{x (a+b x)^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

[-1/4*((5*a^2*b*c*d - 3*a^3*d^2 + (5*a*b^2*c*d - 3*a^2*b*d^2)*x)*sqrt(d/b) 
*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 - 4*(2*b^2*d*x + b^2*c 
+ a*b*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(d/b) + 8*(b^2*c*d + a*b*d^2)*x) 
- 2*(b^3*c^2*x + a*b^2*c^2)*sqrt(c/a)*log((8*a^2*c^2 + (b^2*c^2 + 6*a*b*c* 
d + a^2*d^2)*x^2 - 4*(2*a^2*c + (a*b*c + a^2*d)*x)*sqrt(b*x + a)*sqrt(d*x 
+ c)*sqrt(c/a) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) - 4*(a*b*d^2*x + 2*b^2*c^2 
- 4*a*b*c*d + 3*a^2*d^2)*sqrt(b*x + a)*sqrt(d*x + c))/(a*b^3*x + a^2*b^2), 
 -1/2*((5*a^2*b*c*d - 3*a^3*d^2 + (5*a*b^2*c*d - 3*a^2*b*d^2)*x)*sqrt(-d/b 
)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-d/b)/ 
(b*d^2*x^2 + a*c*d + (b*c*d + a*d^2)*x)) - (b^3*c^2*x + a*b^2*c^2)*sqrt(c/ 
a)*log((8*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^2 - 4*(2*a^2*c + (a* 
b*c + a^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(c/a) + 8*(a*b*c^2 + a^2*c 
*d)*x)/x^2) - 2*(a*b*d^2*x + 2*b^2*c^2 - 4*a*b*c*d + 3*a^2*d^2)*sqrt(b*x + 
 a)*sqrt(d*x + c))/(a*b^3*x + a^2*b^2), 1/4*(4*(b^3*c^2*x + a*b^2*c^2)*sqr 
t(-c/a)*arctan(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*sqr 
t(-c/a)/(b*c*d*x^2 + a*c^2 + (b*c^2 + a*c*d)*x)) - (5*a^2*b*c*d - 3*a^3*d^ 
2 + (5*a*b^2*c*d - 3*a^2*b*d^2)*x)*sqrt(d/b)*log(8*b^2*d^2*x^2 + b^2*c^2 + 
 6*a*b*c*d + a^2*d^2 - 4*(2*b^2*d*x + b^2*c + a*b*d)*sqrt(b*x + a)*sqrt(d* 
x + c)*sqrt(d/b) + 8*(b^2*c*d + a*b*d^2)*x) + 4*(a*b*d^2*x + 2*b^2*c^2 - 4 
*a*b*c*d + 3*a^2*d^2)*sqrt(b*x + a)*sqrt(d*x + c))/(a*b^3*x + a^2*b^2),...
 

Sympy [F]

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

integrate((d*x+c)**(5/2)/x/(b*x+a)**(3/2),x)
 

Output:

Integral((c + d*x)**(5/2)/(x*(a + b*x)**(3/2)), x)
 

Maxima [F(-2)]

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

integrate((d*x+c)^(5/2)/x/(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
 

Giac [F(-2)]

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

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:index.cc index_m i_lex_is_greater E 
rror: Bad Argument Value
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 403, normalized size of antiderivative = 2.47 \[ \int \frac {(c+d x)^{5/2}}{x (a+b x)^{3/2}} \, dx=\frac {4 \sqrt {c}\, \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (-\sqrt {2 \sqrt {d}\, \sqrt {c}\, \sqrt {b}\, \sqrt {a}+a d +b c}+\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}\right ) b^{3} c^{2}+4 \sqrt {c}\, \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {2 \sqrt {d}\, \sqrt {c}\, \sqrt {b}\, \sqrt {a}+a d +b c}+\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}\right ) b^{3} c^{2}-4 \sqrt {c}\, \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (2 \sqrt {d}\, \sqrt {b}\, \sqrt {d x +c}\, \sqrt {b x +a}+2 \sqrt {d}\, \sqrt {c}\, \sqrt {b}\, \sqrt {a}+2 b d x \right ) b^{3} c^{2}-12 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a^{3} d^{2}+20 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a^{2} b c d +9 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, a^{3} d^{2}-17 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, a^{2} b c d +8 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, a \,b^{2} c^{2}+12 \sqrt {d x +c}\, a^{3} b \,d^{2}-16 \sqrt {d x +c}\, a^{2} b^{2} c d +4 \sqrt {d x +c}\, a^{2} b^{2} d^{2} x +8 \sqrt {d x +c}\, a \,b^{3} c^{2}}{4 \sqrt {b x +a}\, a^{2} b^{3}} \] Input:

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

Output:

(4*sqrt(c)*sqrt(a)*sqrt(a + b*x)*log( - sqrt(2*sqrt(d)*sqrt(c)*sqrt(b)*sqr 
t(a) + a*d + b*c) + sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))*b**3*c* 
*2 + 4*sqrt(c)*sqrt(a)*sqrt(a + b*x)*log(sqrt(2*sqrt(d)*sqrt(c)*sqrt(b)*sq 
rt(a) + a*d + b*c) + sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))*b**3*c 
**2 - 4*sqrt(c)*sqrt(a)*sqrt(a + b*x)*log(2*sqrt(d)*sqrt(b)*sqrt(c + d*x)* 
sqrt(a + b*x) + 2*sqrt(d)*sqrt(c)*sqrt(b)*sqrt(a) + 2*b*d*x)*b**3*c**2 - 1 
2*sqrt(d)*sqrt(b)*sqrt(a + b*x)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt( 
c + d*x))/sqrt(a*d - b*c))*a**3*d**2 + 20*sqrt(d)*sqrt(b)*sqrt(a + b*x)*lo 
g((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a**2*b* 
c*d + 9*sqrt(d)*sqrt(b)*sqrt(a + b*x)*a**3*d**2 - 17*sqrt(d)*sqrt(b)*sqrt( 
a + b*x)*a**2*b*c*d + 8*sqrt(d)*sqrt(b)*sqrt(a + b*x)*a*b**2*c**2 + 12*sqr 
t(c + d*x)*a**3*b*d**2 - 16*sqrt(c + d*x)*a**2*b**2*c*d + 4*sqrt(c + d*x)* 
a**2*b**2*d**2*x + 8*sqrt(c + d*x)*a*b**3*c**2)/(4*sqrt(a + b*x)*a**2*b**3 
)