\(\int \frac {(a+\frac {b}{(c+d x)^2})^{3/2}}{x} \, dx\) [75]

Optimal result

Integrand size = 19, antiderivative size = 187 \[ \int \frac {\left (a+\frac {b}{(c+d x)^2}\right )^{3/2}}{x} \, dx=\frac {b \sqrt {a+\frac {b}{(c+d x)^2}}}{c^2}+\frac {b \sqrt {a+\frac {b}{(c+d x)^2}}}{2 c (c+d x)}+\frac {\sqrt {b} \left (2 b+3 a c^2\right ) \text {arctanh}\left (\frac {\sqrt {b}}{(c+d x) \sqrt {a+\frac {b}{(c+d x)^2}}}\right )}{2 c^3}+a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{(c+d x)^2}}}{\sqrt {a}}\right )-\frac {\left (b+a c^2\right )^{3/2} \text {arctanh}\left (\frac {a c+\frac {b}{c+d x}}{\sqrt {b+a c^2} \sqrt {a+\frac {b}{(c+d x)^2}}}\right )}{c^3} \] Output:

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

Mathematica [A] (verified)

Time = 11.21 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.28 \[ \int \frac {\left (a+\frac {b}{(c+d x)^2}\right )^{3/2}}{x} \, dx=\frac {(c+d x) \left (a+\frac {b}{(c+d x)^2}\right )^{3/2} \left (b c (3 c+2 d x) \sqrt {b+a (c+d x)^2}+4 \left (-b-a c^2\right )^{3/2} (c+d x)^2 \arctan \left (\frac {-\sqrt {a} d x+\sqrt {b+a (c+d x)^2}}{\sqrt {-b-a c^2}}\right )-2 \sqrt {b} \left (2 b+3 a c^2\right ) (c+d x)^2 \text {arctanh}\left (\frac {\sqrt {a} (c+d x)-\sqrt {b+a (c+d x)^2}}{\sqrt {b}}\right )-2 a^{3/2} c^3 (c+d x)^2 \log \left (-\sqrt {a} (c+d x)+\sqrt {b+a (c+d x)^2}\right )\right )}{2 c^3 \left (b+a (c+d x)^2\right )^{3/2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.87 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.13, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.947, Rules used = {896, 25, 1774, 1803, 25, 606, 25, 243, 60, 73, 221, 682, 27, 719, 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.

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

\(\Big \downarrow \) 896

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1774

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

\(\Big \downarrow \) 1803

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 606

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

\(\Big \downarrow \) 682

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 719

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 488

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

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

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 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 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 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 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 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 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

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 606

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

rule 682

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p 
_.), x_Symbol] :> Simp[(d + e*x)^(m + 1)*(c*e*f*(m + 2*p + 2) - g*c*d*(2*p 
+ 1) + g*c*e*(m + 2*p + 1)*x)*((a + c*x^2)^p/(c*e^2*(m + 2*p + 1)*(m + 2*p 
+ 2))), x] + Simp[2*(p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)))   Int[(d + e*x) 
^m*(a + c*x^2)^(p - 1)*Simp[f*a*c*e^2*(m + 2*p + 2) + a*c*d*e*g*m - (c^2*f* 
d*e*(m + 2*p + 2) - g*(c^2*d^2*(2*p + 1) + a*c*e^2*(m + 2*p + 1)))*x, x], x 
], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && GtQ[p, 0] && (IntegerQ[p] ||  ! 
RationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])) &&  !ILtQ[m + 2*p, 0] && (Intege 
rQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
 

rule 719

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p 
_.), x_Symbol] :> Simp[g/e   Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] + 
Simp[(e*f - d*g)/e   Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, 
d, e, f, g, m, p}, x] &&  !IGtQ[m, 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 1803

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q 
_.), x_Symbol] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(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] && IntegerQ[Simplify[(m + 1)/n]]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1454\) vs. \(2(161)=322\).

Time = 0.16 (sec) , antiderivative size = 1455, normalized size of antiderivative = 7.78

method	result	size
default	\(\text {Expression too large to display}\)	\(1455\)

Input:

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

Output:

1/2*(2*ln((a*d^2*x+a*c*d+(a*d^2*x^2+2*a*c*d*x+a*c^2+b)^(1/2)*(a*d^2)^(1/2) 
)/(a*d^2)^(1/2))*a^2*b*c^5*d-2*(a*d^2)^(1/2)*(a*c^2+b)^(1/2)*ln(2*(a*d*x*c 
+a*c^2+(a*c^2+b)^(1/2)*(a*d^2*x^2+2*a*c*d*x+a*c^2+b)^(1/2)+b)/x)*b^2*c^2+6 
*ln(2*(b^(1/2)*(a*d^2*x^2+2*a*c*d*x+a*c^2+b)^(1/2)+b)*d/(d*x+c))*(a*d^2)^( 
1/2)*b^(3/2)*a*c^3*d*x+3*ln(2*(b^(1/2)*(a*d^2*x^2+2*a*c*d*x+a*c^2+b)^(1/2) 
+b)*d/(d*x+c))*(a*d^2)^(1/2)*b^(3/2)*a*c^2*d^2*x^2-4*(a*d^2)^(1/2)*(a*c^2+ 
b)^(1/2)*ln(2*(a*d*x*c+a*c^2+(a*c^2+b)^(1/2)*(a*d^2*x^2+2*a*c*d*x+a*c^2+b) 
^(1/2)+b)/x)*b^2*c*d*x-7*(a*d^2*x^2+2*a*c*d*x+a*c^2+b)^(1/2)*(a*d^2)^(1/2) 
*a*b*c^2*d^2*x^2-8*(a*d^2*x^2+2*a*c*d*x+a*c^2+b)^(1/2)*(a*d^2)^(1/2)*a*b*c 
^3*d*x+4*ln(2*(b^(1/2)*(a*d^2*x^2+2*a*c*d*x+a*c^2+b)^(1/2)+b)*d/(d*x+c))*( 
a*d^2)^(1/2)*b^(5/2)*c*d*x+3*ln(2*(b^(1/2)*(a*d^2*x^2+2*a*c*d*x+a*c^2+b)^( 
1/2)+b)*d/(d*x+c))*(a*d^2)^(1/2)*b^(3/2)*a*c^4+2*ln(2*(b^(1/2)*(a*d^2*x^2+ 
2*a*c*d*x+a*c^2+b)^(1/2)+b)*d/(d*x+c))*(a*d^2)^(1/2)*b^(5/2)*d^2*x^2-7*(a* 
d^2*x^2+2*a*c*d*x+a*c^2+b)^(3/2)*(a*d^2)^(1/2)*a*c^2*d^2*x^2-4*(a*d^2)^(1/ 
2)*(a*c^2+b)^(1/2)*ln(2*(a*d*x*c+a*c^2+(a*c^2+b)^(1/2)*(a*d^2*x^2+2*a*c*d* 
x+a*c^2+b)^(1/2)+b)/x)*a*b*c^3*d*x-2*(a*d^2)^(1/2)*(a*c^2+b)^(1/2)*ln(2*(a 
*d*x*c+a*c^2+(a*c^2+b)^(1/2)*(a*d^2*x^2+2*a*c*d*x+a*c^2+b)^(1/2)+b)/x)*a*b 
*c^2*d^2*x^2-2*(a*d^2*x^2+2*a*c*d*x+a*c^2+b)^(1/2)*(a*d^2)^(1/2)*a*b*c*d^3 
*x^3-2*(a*d^2)^(1/2)*(a*c^2+b)^(1/2)*ln(2*(a*d*x*c+a*c^2+(a*c^2+b)^(1/2)*( 
a*d^2*x^2+2*a*c*d*x+a*c^2+b)^(1/2)+b)/x)*a*b*c^4-2*(a*d^2)^(1/2)*(a*c^2...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 439 vs. \(2 (161) = 322\).

Time = 9.35 (sec) , antiderivative size = 3665, normalized size of antiderivative = 19.60 \[ \int \frac {\left (a+\frac {b}{(c+d x)^2}\right )^{3/2}}{x} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 1155, normalized size of antiderivative = 6.18 \[ \int \frac {\left (a+\frac {b}{(c+d x)^2}\right )^{3/2}}{x} \, dx =\text {Too large to display} \] Input:

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

Output:

(6*sqrt(a*c**2 + 2*a*c*d*x + a*d**2*x**2 + b)*b*c**2 + 4*sqrt(a*c**2 + 2*a 
*c*d*x + a*d**2*x**2 + b)*b*c*d*x + 4*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)*a*c** 
4 + 8*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)*a*c**3*d*x + 4*sqrt(a*c**2 + b)*log(s 
qrt(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)*a*c**2*d**2*x**2 + 4*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*c**2 + 8*sq 
rt(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*c*d*x + 4*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**2*x**2 - 4*sqrt(a*c**2 + b)*log(x)*a*c**4 - 8*sqrt(a*c**2 + b)*log(x)*a 
*c**3*d*x - 4*sqrt(a*c**2 + b)*log(x)*a*c**2*d**2*x**2 - 4*sqrt(a*c**2 + b 
)*log(x)*b*c**2 - 8*sqrt(a*c**2 + b)*log(x)*b*c*d*x - 4*sqrt(a*c**2 + b)*l 
og(x)*b*d**2*x**2 + 4*sqrt(a)*log( - sqrt(a)*sqrt(a*c**2 + 2*a*c*d*x + a*d 
**2*x**2 + b) - a*c - a*d*x)*a*c**5 + 8*sqrt(a)*log( - sqrt(a)*sqrt(a*c**2 
 + 2*a*c*d*x + a*d**2*x**2 + b) - a*c - a*d*x)*a*c**4*d*x + 4*sqrt(a)*log( 
 - sqrt(a)*sqrt(a*c**2 + 2*a*c*d*x + a*d**2*x**2 + b) - a*c - a*d*x)*a*c** 
3*d**2*x**2 - 3*sqrt(b)*log(sqrt(a*c**2 + 2*a*c*d*x + a*d**2*x**2 + b) - s 
qrt(b))*a*c**4 - 6*sqrt(b)*log(sqrt(a*c**2 + 2*a*c*d*x + a*d**2*x**2 + ...