\(\int \frac {\sqrt {a+\frac {b}{(c+d x)^2}}}{x^2} \, dx\) [68]

Optimal result

Integrand size = 19, antiderivative size = 134 \[ \int \frac {\sqrt {a+\frac {b}{(c+d x)^2}}}{x^2} \, dx=-\frac {d \sqrt {a+\frac {b}{(c+d x)^2}}}{c \left (1-\frac {c}{c+d x}\right )}-\frac {\sqrt {b} d \text {arctanh}\left (\frac {\sqrt {b}}{(c+d x) \sqrt {a+\frac {b}{(c+d x)^2}}}\right )}{c^2}+\frac {b d \text {arctanh}\left (\frac {a c+\frac {b}{c+d x}}{\sqrt {b+a c^2} \sqrt {a+\frac {b}{(c+d x)^2}}}\right )}{c^2 \sqrt {b+a c^2}} \] Output:

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

Mathematica [A] (verified)

Time = 10.47 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.54 \[ \int \frac {\sqrt {a+\frac {b}{(c+d x)^2}}}{x^2} \, dx=-\frac {d \sqrt {\frac {b+a (c+d x)^2}{(c+d x)^2}}}{c}-\frac {\sqrt {\frac {b+a (c+d x)^2}{(c+d x)^2}}}{x}-\frac {b d \log (x)}{c^2 \sqrt {b+a c^2}}+\frac {\sqrt {b} d \log (c+d x)}{c^2}-\frac {\sqrt {b} d \log \left (b+\sqrt {b} (c+d x) \sqrt {\frac {b+a (c+d x)^2}{(c+d x)^2}}\right )}{c^2}+\frac {b d \log \left (b+(c+d x) \left (a c+\sqrt {b+a c^2} \sqrt {\frac {b+a (c+d x)^2}{(c+d x)^2}}\right )\right )}{c^2 \sqrt {b+a c^2}} \] Input:

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

Output:

-((d*Sqrt[(b + a*(c + d*x)^2)/(c + d*x)^2])/c) - Sqrt[(b + a*(c + d*x)^2)/ 
(c + d*x)^2]/x - (b*d*Log[x])/(c^2*Sqrt[b + a*c^2]) + (Sqrt[b]*d*Log[c + d 
*x])/c^2 - (Sqrt[b]*d*Log[b + Sqrt[b]*(c + d*x)*Sqrt[(b + a*(c + d*x)^2)/( 
c + d*x)^2]])/c^2 + (b*d*Log[b + (c + d*x)*(a*c + Sqrt[b + a*c^2]*Sqrt[(b 
+ a*(c + d*x)^2)/(c + d*x)^2])])/(c^2*Sqrt[b + a*c^2])
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.06, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.474, Rules used = {896, 1774, 1799, 492, 605, 224, 219, 488, 219}

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.

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

\(\Big \downarrow \) 896

\(\displaystyle d \int \frac {\sqrt {a+\frac {b}{(c+d x)^2}}}{d^2 x^2}d(c+d x)\)

\(\Big \downarrow \) 1774

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

\(\Big \downarrow \) 1799

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

\(\Big \downarrow \) 492

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

\(\Big \downarrow \) 605

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 488

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

\(\Big \downarrow \) 219

\(\displaystyle -d \left (\frac {\sqrt {a+\frac {b}{(c+d x)^2}}}{c \left (1-\frac {c}{c+d x}\right )}-\frac {b \left (-\frac {\text {arctanh}\left (\frac {-a c-\frac {b}{c+d x}}{\sqrt {a c^2+b} \sqrt {a+\frac {b}{(c+d x)^2}}}\right )}{c \sqrt {a c^2+b}}-\frac {\text {arctanh}\left (\frac {\sqrt {b}}{(c+d x) \sqrt {a+\frac {b}{(c+d x)^2}}}\right )}{\sqrt {b} c}\right )}{c}\right )\)

Input:

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

Output:

-(d*(Sqrt[a + b/(c + d*x)^2]/(c*(1 - c/(c + d*x))) - (b*(-(ArcTanh[Sqrt[b] 
/((c + d*x)*Sqrt[a + b/(c + d*x)^2])]/(Sqrt[b]*c)) - ArcTanh[(-(a*c) - b/( 
c + d*x))/(Sqrt[b + a*c^2]*Sqrt[a + b/(c + d*x)^2])]/(c*Sqrt[b + a*c^2]))) 
/c))
 

Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 0]
 

rule 488

Int[1/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> -Subst[ 
Int[1/(b*c^2 + a*d^2 - x^2), x], x, (a*d - b*c*x)/Sqrt[a + b*x^2]] /; FreeQ 
[{a, b, c, d}, x]
 

rule 492

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
(c + d*x)^(n + 1)*((a + b*x^2)^p/(d*(n + 1))), x] - Simp[2*b*(p/(d*(n + 1)) 
)   Int[x*(c + d*x)^(n + 1)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, 
d, n}, x] && GtQ[p, 0] && (IntegerQ[p] || LtQ[n, -1]) && NeQ[n, -1] &&  !IL 
tQ[n + 2*p + 1, 0] && IntQuadraticQ[a, 0, b, c, d, n, p, x]
 

rule 605

Int[((x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_))/((c_) + (d_.)*(x_)), x_Symbol] 
:> Simp[1/d   Int[x^(m - 1)*(a + b*x^2)^p, x], x] - Simp[c/d   Int[x^(m - 1 
)*((a + b*x^2)^p/(c + d*x)), x], x] /; FreeQ[{a, b, c, d, p}, x] && IGtQ[m, 
 0] && LtQ[-1, p, 0]
 

rule 896

Int[((a_) + (b_.)*(v_)^(n_))^(p_.)*(x_)^(m_.), x_Symbol] :> With[{c = Coeff 
icient[v, x, 0], d = Coefficient[v, x, 1]}, Simp[1/d^(m + 1)   Subst[Int[Si 
mplifyIntegrand[(x - c)^m*(a + b*x^n)^p, x], x], x, v], x] /; NeQ[c, 0]] /; 
 FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && IntegerQ[m]
 

rule 1774

Int[((d_) + (e_.)*(x_)^(mn_.))^(q_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.), x_Sy 
mbol] :> Int[x^(mn*q)*(e + d/x^mn)^q*(a + c*x^n2)^p, x] /; FreeQ[{a, c, d, 
e, mn, p}, x] && EqQ[n2, -2*mn] && IntegerQ[q] && (PosQ[n2] ||  !IntegerQ[p 
])
 

rule 1799

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q 
_.), x_Symbol] :> Simp[1/n   Subst[Int[(d + e*x)^q*(a + c*x^2)^p, x], x, x^ 
n], x] /; FreeQ[{a, c, d, e, m, n, p, q}, x] && EqQ[n2, 2*n] && EqQ[Simplif 
y[m - n + 1], 0]
 

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.67

method	result	size
risch	\(-\frac {\sqrt {\frac {a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b}{\left (d x +c \right )^{2}}}\, \left (d x +c \right )}{c x}+\frac {\left (\frac {d b \ln \left (\frac {2 a \,c^{2}+2 b +2 a d x c +2 \sqrt {a \,c^{2}+b}\, \sqrt {a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b}}{x}\right )}{c^{2} \sqrt {a \,c^{2}+b}}-\frac {d \sqrt {b}\, \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {a \,d^{2} \left (x +\frac {c}{d}\right )^{2}+b}}{x +\frac {c}{d}}\right )}{c^{2}}\right ) \sqrt {\frac {a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b}{\left (d x +c \right )^{2}}}\, \left (d x +c \right )}{\sqrt {a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b}}\)	\(224\)
default	\(-\frac {\sqrt {\frac {a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b}{\left (d x +c \right )^{2}}}\, \left (d x +c \right ) \left (-\sqrt {a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b}\, a c \,d^{2} x^{2}+\sqrt {b}\, \ln \left (\frac {2 \left (\sqrt {b}\, \sqrt {a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b}+b \right ) d}{d x +c}\right ) a \,c^{2} d x -2 \sqrt {a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b}\, a \,c^{2} d x -\sqrt {a \,c^{2}+b}\, \ln \left (\frac {2 a \,c^{2}+2 b +2 a d x c +2 \sqrt {a \,c^{2}+b}\, \sqrt {a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b}}{x}\right ) b d x +b^{\frac {3}{2}} \ln \left (\frac {2 \left (\sqrt {b}\, \sqrt {a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b}+b \right ) d}{d x +c}\right ) d x +\left (a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b \right )^{\frac {3}{2}} c \right )}{\sqrt {a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b}\, c^{2} \left (a \,c^{2}+b \right ) x}\)	\(328\)

Input:

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

Output:

-1/c/x*((a*d^2*x^2+2*a*c*d*x+a*c^2+b)/(d*x+c)^2)^(1/2)*(d*x+c)+(1/c^2*d*b/ 
(a*c^2+b)^(1/2)*ln((2*a*c^2+2*b+2*a*d*x*c+2*(a*c^2+b)^(1/2)*(a*d^2*x^2+2*a 
*c*d*x+a*c^2+b)^(1/2))/x)-1/c^2*d*b^(1/2)*ln((2*b+2*b^(1/2)*(a*d^2*(x+c/d) 
^2+b)^(1/2))/(x+c/d)))*((a*d^2*x^2+2*a*c*d*x+a*c^2+b)/(d*x+c)^2)^(1/2)*(d* 
x+c)/(a*d^2*x^2+2*a*c*d*x+a*c^2+b)^(1/2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 313 vs. \(2 (118) = 236\).

Time = 0.19 (sec) , antiderivative size = 1334, normalized size of antiderivative = 9.96 \[ \int \frac {\sqrt {a+\frac {b}{(c+d x)^2}}}{x^2} \, dx=\text {Too large to display} \] Input:

integrate((a+b/(d*x+c)^2)^(1/2)/x^2,x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

integrate((a+b/(d*x+c)**2)**(1/2)/x**2,x)
 

Output:

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

Maxima [F]

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

integrate((a+b/(d*x+c)^2)^(1/2)/x^2,x, algorithm="maxima")
 

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 299 vs. \(2 (118) = 236\).

Time = 0.20 (sec) , antiderivative size = 299, normalized size of antiderivative = 2.23 \[ \int \frac {\sqrt {a+\frac {b}{(c+d x)^2}}}{x^2} \, dx=-\frac {2 \, b d \arctan \left (-\frac {\sqrt {a d^{2}} x - \sqrt {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}}{\sqrt {-a c^{2} - b}}\right ) \mathrm {sgn}\left (d x + c\right )}{\sqrt {-a c^{2} - b} c^{2}} + \frac {2 \, b d \arctan \left (-\frac {\sqrt {a} c {\left | d \right |} + {\left (\sqrt {a d^{2}} x - \sqrt {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}\right )} d}{\sqrt {-b} d}\right ) \mathrm {sgn}\left (d x + c\right )}{\sqrt {-b} c^{2}} - \frac {2 \, {\left (a^{\frac {3}{2}} b^{2} c^{2} d^{4} {\left | d \right |} \mathrm {sgn}\left (d x + c\right ) + {\left (\sqrt {a d^{2}} x - \sqrt {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}\right )} a b^{2} c d^{5} \mathrm {sgn}\left (d x + c\right ) + \sqrt {a} b^{3} d^{4} {\left | d \right |} \mathrm {sgn}\left (d x + c\right )\right )}}{{\left (a c^{2} - {\left (\sqrt {a d^{2}} x - \sqrt {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}\right )}^{2} + b\right )} b^{2} c d^{4}} \] Input:

integrate((a+b/(d*x+c)^2)^(1/2)/x^2,x, algorithm="giac")
 

Output:

-2*b*d*arctan(-(sqrt(a*d^2)*x - sqrt(a*d^2*x^2 + 2*a*c*d*x + a*c^2 + b))/s 
qrt(-a*c^2 - b))*sgn(d*x + c)/(sqrt(-a*c^2 - b)*c^2) + 2*b*d*arctan(-(sqrt 
(a)*c*abs(d) + (sqrt(a*d^2)*x - sqrt(a*d^2*x^2 + 2*a*c*d*x + a*c^2 + b))*d 
)/(sqrt(-b)*d))*sgn(d*x + c)/(sqrt(-b)*c^2) - 2*(a^(3/2)*b^2*c^2*d^4*abs(d 
)*sgn(d*x + c) + (sqrt(a*d^2)*x - sqrt(a*d^2*x^2 + 2*a*c*d*x + a*c^2 + b)) 
*a*b^2*c*d^5*sgn(d*x + c) + sqrt(a)*b^3*d^4*abs(d)*sgn(d*x + c))/((a*c^2 - 
 (sqrt(a*d^2)*x - sqrt(a*d^2*x^2 + 2*a*c*d*x + a*c^2 + b))^2 + b)*b^2*c*d^ 
4)
 

Mupad [F(-1)]

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

int((a + b/(c + d*x)^2)^(1/2)/x^2,x)
 

Output:

int((a + b/(c + d*x)^2)^(1/2)/x^2, x)
 

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 289, normalized size of antiderivative = 2.16 \[ \int \frac {\sqrt {a+\frac {b}{(c+d x)^2}}}{x^2} \, dx=\frac {-2 \sqrt {a \,d^{2} x^{2}+2 a c d x +a \,c^{2}+b}\, a \,c^{3}-2 \sqrt {a \,d^{2} x^{2}+2 a c d x +a \,c^{2}+b}\, b c +2 \sqrt {a \,c^{2}+b}\, \mathrm {log}\left (-\sqrt {a \,c^{2}+b}\, \sqrt {a \,d^{2} x^{2}+2 a c d x +a \,c^{2}+b}-a \,c^{2}-a c d x -b \right ) b d x -2 \sqrt {a \,c^{2}+b}\, \mathrm {log}\left (x \right ) b d x +\sqrt {b}\, \mathrm {log}\left (\sqrt {a \,d^{2} x^{2}+2 a c d x +a \,c^{2}+b}-\sqrt {b}\right ) a \,c^{2} d x +\sqrt {b}\, \mathrm {log}\left (\sqrt {a \,d^{2} x^{2}+2 a c d x +a \,c^{2}+b}-\sqrt {b}\right ) b d x -\sqrt {b}\, \mathrm {log}\left (\sqrt {a \,d^{2} x^{2}+2 a c d x +a \,c^{2}+b}+\sqrt {b}\right ) a \,c^{2} d x -\sqrt {b}\, \mathrm {log}\left (\sqrt {a \,d^{2} x^{2}+2 a c d x +a \,c^{2}+b}+\sqrt {b}\right ) b d x}{2 c^{2} x \left (a \,c^{2}+b \right )} \] Input:

int((a+b/(d*x+c)^2)^(1/2)/x^2,x)
 

Output:

( - 2*sqrt(a*c**2 + 2*a*c*d*x + a*d**2*x**2 + b)*a*c**3 - 2*sqrt(a*c**2 + 
2*a*c*d*x + a*d**2*x**2 + b)*b*c + 2*sqrt(a*c**2 + b)*log( - sqrt(a*c**2 + 
 b)*sqrt(a*c**2 + 2*a*c*d*x + a*d**2*x**2 + b) - a*c**2 - a*c*d*x - b)*b*d 
*x - 2*sqrt(a*c**2 + b)*log(x)*b*d*x + sqrt(b)*log(sqrt(a*c**2 + 2*a*c*d*x 
 + a*d**2*x**2 + b) - sqrt(b))*a*c**2*d*x + sqrt(b)*log(sqrt(a*c**2 + 2*a* 
c*d*x + a*d**2*x**2 + b) - sqrt(b))*b*d*x - sqrt(b)*log(sqrt(a*c**2 + 2*a* 
c*d*x + a*d**2*x**2 + b) + sqrt(b))*a*c**2*d*x - sqrt(b)*log(sqrt(a*c**2 + 
 2*a*c*d*x + a*d**2*x**2 + b) + sqrt(b))*b*d*x)/(2*c**2*x*(a*c**2 + b))