\(\int \frac {(a+\frac {b}{x})^{5/2}}{(c+\frac {d}{x})^2} \, dx\) [23]

Optimal result

Integrand size = 21, antiderivative size = 166 \[ \int \frac {\left (a+\frac {b}{x}\right )^{5/2}}{\left (c+\frac {d}{x}\right )^2} \, dx=\frac {(b c-2 a d) (b c-a d) \sqrt {a+\frac {b}{x}}}{c^2 d \left (c+\frac {d}{x}\right )}+\frac {a \left (a+\frac {b}{x}\right )^{3/2} x}{c \left (c+\frac {d}{x}\right )}-\frac {(b c-a d)^{3/2} (b c+4 a d) \arctan \left (\frac {\sqrt {d} \sqrt {a+\frac {b}{x}}}{\sqrt {b c-a d}}\right )}{c^3 d^{3/2}}+\frac {a^{3/2} (5 b c-4 a d) \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{c^3} \] Output:

(-2*a*d+b*c)*(-a*d+b*c)*(a+b/x)^(1/2)/c^2/d/(c+d/x)+a*(a+b/x)^(3/2)*x/c/(c 
+d/x)-(-a*d+b*c)^(3/2)*(4*a*d+b*c)*arctan(d^(1/2)*(a+b/x)^(1/2)/(-a*d+b*c) 
^(1/2))/c^3/d^(3/2)+a^(3/2)*(-4*a*d+5*b*c)*arctanh((a+b/x)^(1/2)/a^(1/2))/ 
c^3
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.65 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.88 \[ \int \frac {\left (a+\frac {b}{x}\right )^{5/2}}{\left (c+\frac {d}{x}\right )^2} \, dx=\frac {\frac {c \sqrt {a+\frac {b}{x}} x \left (b^2 c^2-2 a b c d+a^2 d (2 d+c x)\right )}{d (d+c x)}-\frac {(b c-a d)^{3/2} (b c+4 a d) \arctan \left (\frac {\sqrt {d} \sqrt {a+\frac {b}{x}}}{\sqrt {b c-a d}}\right )}{d^{3/2}}-a^{3/2} (-5 b c+4 a d) \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{c^3} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.12, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {899, 109, 27, 166, 25, 174, 73, 218, 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.

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

\(\Big \downarrow \) 899

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

\(\Big \downarrow \) 109

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 166

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 174

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, 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 166

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

rule 174

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

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

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]
 

rule 899

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol 
] :> -Subst[Int[(a + b/x^n)^p*((c + d/x^n)^q/x^2), x], x, 1/x] /; FreeQ[{a, 
 b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && ILtQ[n, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(516\) vs. \(2(146)=292\).

Time = 0.53 (sec) , antiderivative size = 517, normalized size of antiderivative = 3.11

method	result	size
risch	\(\frac {a^{2} x \sqrt {\frac {a x +b}{x}}}{c^{2}}-\frac {\left (\frac {a^{\frac {3}{2}} \left (4 a d -5 b c \right ) \ln \left (\frac {\frac {b}{2}+a x}{\sqrt {a}}+\sqrt {a \,x^{2}+b x}\right )}{c}+\frac {2 \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \left (-\frac {c^{2} \sqrt {a \left (x +\frac {d}{c}\right )^{2}-\frac {\left (2 a d -b c \right ) \left (x +\frac {d}{c}\right )}{c}+\frac {\left (a d -b c \right ) d}{c^{2}}}}{\left (a d -b c \right ) d \left (x +\frac {d}{c}\right )}-\frac {\left (2 a d -b c \right ) c \ln \left (\frac {\frac {2 \left (a d -b c \right ) d}{c^{2}}-\frac {\left (2 a d -b c \right ) \left (x +\frac {d}{c}\right )}{c}+2 \sqrt {\frac {\left (a d -b c \right ) d}{c^{2}}}\, \sqrt {a \left (x +\frac {d}{c}\right )^{2}-\frac {\left (2 a d -b c \right ) \left (x +\frac {d}{c}\right )}{c}+\frac {\left (a d -b c \right ) d}{c^{2}}}}{x +\frac {d}{c}}\right )}{2 \left (a d -b c \right ) d \sqrt {\frac {\left (a d -b c \right ) d}{c^{2}}}}\right )}{c^{3}}+\frac {6 a \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \ln \left (\frac {\frac {2 \left (a d -b c \right ) d}{c^{2}}-\frac {\left (2 a d -b c \right ) \left (x +\frac {d}{c}\right )}{c}+2 \sqrt {\frac {\left (a d -b c \right ) d}{c^{2}}}\, \sqrt {a \left (x +\frac {d}{c}\right )^{2}-\frac {\left (2 a d -b c \right ) \left (x +\frac {d}{c}\right )}{c}+\frac {\left (a d -b c \right ) d}{c^{2}}}}{x +\frac {d}{c}}\right )}{c^{2} \sqrt {\frac {\left (a d -b c \right ) d}{c^{2}}}}\right ) \sqrt {\frac {a x +b}{x}}\, \sqrt {x \left (a x +b \right )}}{2 c^{2} \left (a x +b \right )}\)	\(517\)
default	\(\text {Expression too large to display}\)	\(1325\)

Input:

int((a+b/x)^(5/2)/(c+1/x*d)^2,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 1013, normalized size of antiderivative = 6.10 \[ \int \frac {\left (a+\frac {b}{x}\right )^{5/2}}{\left (c+\frac {d}{x}\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

[-1/2*((5*a*b*c*d^2 - 4*a^2*d^3 + (5*a*b*c^2*d - 4*a^2*c*d^2)*x)*sqrt(a)*l 
og(2*a*x - 2*sqrt(a)*x*sqrt((a*x + b)/x) + b) + (b^2*c^2*d + 3*a*b*c*d^2 - 
 4*a^2*d^3 + (b^2*c^3 + 3*a*b*c^2*d - 4*a^2*c*d^2)*x)*sqrt(-(b*c - a*d)/d) 
*log((2*d*x*sqrt(-(b*c - a*d)/d)*sqrt((a*x + b)/x) + b*d - (b*c - 2*a*d)*x 
)/(c*x + d)) - 2*(a^2*c^2*d*x^2 + (b^2*c^3 - 2*a*b*c^2*d + 2*a^2*c*d^2)*x) 
*sqrt((a*x + b)/x))/(c^4*d*x + c^3*d^2), -1/2*(2*(5*a*b*c*d^2 - 4*a^2*d^3 
+ (5*a*b*c^2*d - 4*a^2*c*d^2)*x)*sqrt(-a)*arctan(sqrt(-a)*x*sqrt((a*x + b) 
/x)/(a*x + b)) + (b^2*c^2*d + 3*a*b*c*d^2 - 4*a^2*d^3 + (b^2*c^3 + 3*a*b*c 
^2*d - 4*a^2*c*d^2)*x)*sqrt(-(b*c - a*d)/d)*log((2*d*x*sqrt(-(b*c - a*d)/d 
)*sqrt((a*x + b)/x) + b*d - (b*c - 2*a*d)*x)/(c*x + d)) - 2*(a^2*c^2*d*x^2 
 + (b^2*c^3 - 2*a*b*c^2*d + 2*a^2*c*d^2)*x)*sqrt((a*x + b)/x))/(c^4*d*x + 
c^3*d^2), 1/2*(2*(b^2*c^2*d + 3*a*b*c*d^2 - 4*a^2*d^3 + (b^2*c^3 + 3*a*b*c 
^2*d - 4*a^2*c*d^2)*x)*sqrt((b*c - a*d)/d)*arctan(-d*sqrt((b*c - a*d)/d)*s 
qrt((a*x + b)/x)/(b*c - a*d)) - (5*a*b*c*d^2 - 4*a^2*d^3 + (5*a*b*c^2*d - 
4*a^2*c*d^2)*x)*sqrt(a)*log(2*a*x - 2*sqrt(a)*x*sqrt((a*x + b)/x) + b) + 2 
*(a^2*c^2*d*x^2 + (b^2*c^3 - 2*a*b*c^2*d + 2*a^2*c*d^2)*x)*sqrt((a*x + b)/ 
x))/(c^4*d*x + c^3*d^2), -((5*a*b*c*d^2 - 4*a^2*d^3 + (5*a*b*c^2*d - 4*a^2 
*c*d^2)*x)*sqrt(-a)*arctan(sqrt(-a)*x*sqrt((a*x + b)/x)/(a*x + b)) - (b^2* 
c^2*d + 3*a*b*c*d^2 - 4*a^2*d^3 + (b^2*c^3 + 3*a*b*c^2*d - 4*a^2*c*d^2)*x) 
*sqrt((b*c - a*d)/d)*arctan(-d*sqrt((b*c - a*d)/d)*sqrt((a*x + b)/x)/(b...
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 667 vs. \(2 (146) = 292\).

Time = 0.17 (sec) , antiderivative size = 667, normalized size of antiderivative = 4.02 \[ \int \frac {\left (a+\frac {b}{x}\right )^{5/2}}{\left (c+\frac {d}{x}\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

sqrt(a*x^2 + b*x)*a^2*sgn(x)/c^2 - 1/2*(5*a^2*b*c*sgn(x) - 4*a^3*d*sgn(x)) 
*log(abs(-2*(sqrt(a)*x - sqrt(a*x^2 + b*x))*sqrt(a) - b))/(sqrt(a)*c^3) + 
(b^3*c^3*sgn(x) + 2*a*b^2*c^2*d*sgn(x) - 7*a^2*b*c*d^2*sgn(x) + 4*a^3*d^3* 
sgn(x))*arctan(-((sqrt(a)*x - sqrt(a*x^2 + b*x))*c + sqrt(a)*d)/sqrt(b*c*d 
 - a*d^2))/(sqrt(b*c*d - a*d^2)*c^3*d) + 1/2*(2*sqrt(a)*b^3*c^3*arctan(sqr 
t(a)*d/sqrt(b*c*d - a*d^2)) + 4*a^(3/2)*b^2*c^2*d*arctan(sqrt(a)*d/sqrt(b* 
c*d - a*d^2)) - 14*a^(5/2)*b*c*d^2*arctan(sqrt(a)*d/sqrt(b*c*d - a*d^2)) + 
 8*a^(7/2)*d^3*arctan(sqrt(a)*d/sqrt(b*c*d - a*d^2)) + 5*sqrt(b*c*d - a*d^ 
2)*a^2*b*c*d*log(abs(b)) - 4*sqrt(b*c*d - a*d^2)*a^3*d^2*log(abs(b)) - 2*s 
qrt(b*c*d - a*d^2)*a*b^2*c^2 + 4*sqrt(b*c*d - a*d^2)*a^2*b*c*d - 2*sqrt(b* 
c*d - a*d^2)*a^3*d^2)*sgn(x)/(sqrt(b*c*d - a*d^2)*sqrt(a)*c^3*d) - ((sqrt( 
a)*x - sqrt(a*x^2 + b*x))*sqrt(a)*b^3*c^3*sgn(x) - 4*(sqrt(a)*x - sqrt(a*x 
^2 + b*x))*a^(3/2)*b^2*c^2*d*sgn(x) + 5*(sqrt(a)*x - sqrt(a*x^2 + b*x))*a^ 
(5/2)*b*c*d^2*sgn(x) - 2*(sqrt(a)*x - sqrt(a*x^2 + b*x))*a^(7/2)*d^3*sgn(x 
) - a*b^3*c^2*d*sgn(x) + 2*a^2*b^2*c*d^2*sgn(x) - a^3*b*d^3*sgn(x))/(((sqr 
t(a)*x - sqrt(a*x^2 + b*x))^2*c + 2*(sqrt(a)*x - sqrt(a*x^2 + b*x))*sqrt(a 
)*d + b*d)*sqrt(a)*c^3*d)
 

Mupad [B] (verification not implemented)

Time = 1.75 (sec) , antiderivative size = 1153, normalized size of antiderivative = 6.95 \[ \int \frac {\left (a+\frac {b}{x}\right )^{5/2}}{\left (c+\frac {d}{x}\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(((a + b/x)^(1/2)*(a*b^3*c^2 + 2*a^3*b*d^2 - 3*a^2*b^2*c*d))/(c^2*d) - (b* 
(a + b/x)^(3/2)*(2*a^2*d^2 + b^2*c^2 - 2*a*b*c*d))/(c^2*d))/((a + b/x)*(2* 
a*d - b*c) - d*(a + b/x)^2 - a^2*d + a*b*c) - (atanh((10*b^9*(a + b/x)^(1/ 
2)*(a^3)^(1/2))/(10*a^2*b^9 + (32*a^3*b^8*d)/c - (132*a^4*b^7*d^2)/c^2 + ( 
130*a^5*b^6*d^3)/c^3 - (40*a^6*b^5*d^4)/c^4) + (32*a*b^8*(a + b/x)^(1/2)*( 
a^3)^(1/2))/(32*a^3*b^8 + (10*a^2*b^9*c)/d - (132*a^4*b^7*d)/c + (130*a^5* 
b^6*d^2)/c^2 - (40*a^6*b^5*d^3)/c^3) - (132*a^2*b^7*d*(a + b/x)^(1/2)*(a^3 
)^(1/2))/(32*a^3*b^8*c - 132*a^4*b^7*d + (10*a^2*b^9*c^2)/d + (130*a^5*b^6 
*d^2)/c - (40*a^6*b^5*d^3)/c^2) + (130*a^3*b^6*d^2*(a + b/x)^(1/2)*(a^3)^( 
1/2))/(32*a^3*b^8*c^2 + 130*a^5*b^6*d^2 + (10*a^2*b^9*c^3)/d - (40*a^6*b^5 
*d^3)/c - 132*a^4*b^7*c*d) - (40*a^4*b^5*d^3*(a + b/x)^(1/2)*(a^3)^(1/2))/ 
(32*a^3*b^8*c^3 - 40*a^6*b^5*d^3 - 132*a^4*b^7*c^2*d + 130*a^5*b^6*c*d^2 + 
 (10*a^2*b^9*c^4)/d))*(4*a*d - 5*b*c)*(a^3)^(1/2))/c^3 + (atanh((30*a^3*b^ 
6*(a + b/x)^(1/2)*(a^3*d^6 - b^3*c^3*d^3 + 3*a*b^2*c^2*d^4 - 3*a^2*b*c*d^5 
)^(1/2))/(14*a^2*b^9*c^3 + 110*a^5*b^6*d^3 - 4*a^3*b^8*c^2*d - 82*a^4*b^7* 
c*d^2 + (2*a*b^10*c^4)/d - (40*a^6*b^5*d^4)/c) + (18*a^2*b^7*(a + b/x)^(1/ 
2)*(a^3*d^6 - b^3*c^3*d^3 + 3*a*b^2*c^2*d^4 - 3*a^2*b*c*d^5)^(1/2))/(2*a*b 
^10*c^3 - 82*a^4*b^7*d^3 + 14*a^2*b^9*c^2*d - 4*a^3*b^8*c*d^2 + (110*a^5*b 
^6*d^4)/c - (40*a^6*b^5*d^5)/c^2) + (40*a^4*b^5*(a + b/x)^(1/2)*(a^3*d^6 - 
 b^3*c^3*d^3 + 3*a*b^2*c^2*d^4 - 3*a^2*b*c*d^5)^(1/2))/(4*a^3*b^8*c^3 +...
 

Reduce [B] (verification not implemented)

Time = 0.51 (sec) , antiderivative size = 1310, normalized size of antiderivative = 7.89 \[ \int \frac {\left (a+\frac {b}{x}\right )^{5/2}}{\left (c+\frac {d}{x}\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(4*sqrt(d)*sqrt(a*d - b*c)*log(sqrt(c)*sqrt(a*x + b) - sqrt(2*sqrt(d)*sqrt 
(a)*sqrt(a*d - b*c) - 2*a*d + b*c) + sqrt(x)*sqrt(c)*sqrt(a))*a**2*c*d**2* 
x + 4*sqrt(d)*sqrt(a*d - b*c)*log(sqrt(c)*sqrt(a*x + b) - sqrt(2*sqrt(d)*s 
qrt(a)*sqrt(a*d - b*c) - 2*a*d + b*c) + sqrt(x)*sqrt(c)*sqrt(a))*a**2*d**3 
 - 3*sqrt(d)*sqrt(a*d - b*c)*log(sqrt(c)*sqrt(a*x + b) - sqrt(2*sqrt(d)*sq 
rt(a)*sqrt(a*d - b*c) - 2*a*d + b*c) + sqrt(x)*sqrt(c)*sqrt(a))*a*b*c**2*d 
*x - 3*sqrt(d)*sqrt(a*d - b*c)*log(sqrt(c)*sqrt(a*x + b) - sqrt(2*sqrt(d)* 
sqrt(a)*sqrt(a*d - b*c) - 2*a*d + b*c) + sqrt(x)*sqrt(c)*sqrt(a))*a*b*c*d* 
*2 - sqrt(d)*sqrt(a*d - b*c)*log(sqrt(c)*sqrt(a*x + b) - sqrt(2*sqrt(d)*sq 
rt(a)*sqrt(a*d - b*c) - 2*a*d + b*c) + sqrt(x)*sqrt(c)*sqrt(a))*b**2*c**3* 
x - sqrt(d)*sqrt(a*d - b*c)*log(sqrt(c)*sqrt(a*x + b) - sqrt(2*sqrt(d)*sqr 
t(a)*sqrt(a*d - b*c) - 2*a*d + b*c) + sqrt(x)*sqrt(c)*sqrt(a))*b**2*c**2*d 
 + 4*sqrt(d)*sqrt(a*d - b*c)*log(sqrt(c)*sqrt(a*x + b) + sqrt(2*sqrt(d)*sq 
rt(a)*sqrt(a*d - b*c) - 2*a*d + b*c) + sqrt(x)*sqrt(c)*sqrt(a))*a**2*c*d** 
2*x + 4*sqrt(d)*sqrt(a*d - b*c)*log(sqrt(c)*sqrt(a*x + b) + sqrt(2*sqrt(d) 
*sqrt(a)*sqrt(a*d - b*c) - 2*a*d + b*c) + sqrt(x)*sqrt(c)*sqrt(a))*a**2*d* 
*3 - 3*sqrt(d)*sqrt(a*d - b*c)*log(sqrt(c)*sqrt(a*x + b) + sqrt(2*sqrt(d)* 
sqrt(a)*sqrt(a*d - b*c) - 2*a*d + b*c) + sqrt(x)*sqrt(c)*sqrt(a))*a*b*c**2 
*d*x - 3*sqrt(d)*sqrt(a*d - b*c)*log(sqrt(c)*sqrt(a*x + b) + sqrt(2*sqrt(d 
)*sqrt(a)*sqrt(a*d - b*c) - 2*a*d + b*c) + sqrt(x)*sqrt(c)*sqrt(a))*a*b...