\(\int \frac {1}{x^3 (a+\frac {b}{(c+d x)^2})^{3/2}} \, dx\) [93]

Optimal result

Integrand size = 19, antiderivative size = 234 \[ \int \frac {1}{x^3 \left (a+\frac {b}{(c+d x)^2}\right )^{3/2}} \, dx=-\frac {c^2 d^2 \sqrt {a+\frac {b}{(c+d x)^2}}}{2 \left (b+a c^2\right )^2 \left (1-\frac {c}{c+d x}\right )^2}-\frac {c^2 \left (5 b-2 a c^2\right ) d^2 \sqrt {a+\frac {b}{(c+d x)^2}}}{2 \left (b+a c^2\right )^3 \left (1-\frac {c}{c+d x}\right )}+\frac {b d^2 \left (b-3 a c^2+\frac {c \left (3 b-a c^2\right )}{c+d x}\right )}{\left (b+a c^2\right )^3 \sqrt {a+\frac {b}{(c+d x)^2}}}-\frac {3 b c \left (2 b-3 a c^2\right ) d^2 \text {arctanh}\left (\frac {a c+\frac {b}{c+d x}}{\sqrt {b+a c^2} \sqrt {a+\frac {b}{(c+d x)^2}}}\right )}{2 \left (b+a c^2\right )^{7/2}} \] Output:

-1/2*c^2*d^2*(a+b/(d*x+c)^2)^(1/2)/(a*c^2+b)^2/(1-c/(d*x+c))^2-1/2*c^2*(-2 
*a*c^2+5*b)*d^2*(a+b/(d*x+c)^2)^(1/2)/(a*c^2+b)^3/(1-c/(d*x+c))+b*d^2*(b-3 
*a*c^2+c*(-a*c^2+3*b)/(d*x+c))/(a*c^2+b)^3/(a+b/(d*x+c)^2)^(1/2)-3/2*b*c*( 
-3*a*c^2+2*b)*d^2*arctanh((a*c+b/(d*x+c))/(a*c^2+b)^(1/2)/(a+b/(d*x+c)^2)^ 
(1/2))/(a*c^2+b)^(7/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 10.53 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.04 \[ \int \frac {1}{x^3 \left (a+\frac {b}{(c+d x)^2}\right )^{3/2}} \, dx=\frac {1}{2} \left (-\frac {a^2 c^4 (c-d x) (c+d x)^2+b^2 \left (c^3+6 c^2 d x-8 c d^2 x^2-2 d^3 x^3\right )+a b c^2 \left (2 c^3+7 c^2 d x+21 c d^2 x^2+12 d^3 x^3\right )}{\left (b+a c^2\right )^3 x^2 (c+d x) \sqrt {\frac {b+a (c+d x)^2}{(c+d x)^2}}}+\frac {3 b c \left (2 b-3 a c^2\right ) d^2 \log (x)}{\left (b+a c^2\right )^{7/2}}-\frac {3 b c \left (2 b-3 a c^2\right ) d^2 \log \left (b+(c+d x) \left (a c+\sqrt {b+a c^2} \sqrt {\frac {b+a (c+d x)^2}{(c+d x)^2}}\right )\right )}{\left (b+a c^2\right )^{7/2}}\right ) \] Input:

Integrate[1/(x^3*(a + b/(c + d*x)^2)^(3/2)),x]
 

Output:

(-((a^2*c^4*(c - d*x)*(c + d*x)^2 + b^2*(c^3 + 6*c^2*d*x - 8*c*d^2*x^2 - 2 
*d^3*x^3) + a*b*c^2*(2*c^3 + 7*c^2*d*x + 21*c*d^2*x^2 + 12*d^3*x^3))/((b + 
 a*c^2)^3*x^2*(c + d*x)*Sqrt[(b + a*(c + d*x)^2)/(c + d*x)^2])) + (3*b*c*( 
2*b - 3*a*c^2)*d^2*Log[x])/(b + a*c^2)^(7/2) - (3*b*c*(2*b - 3*a*c^2)*d^2* 
Log[b + (c + d*x)*(a*c + Sqrt[b + a*c^2]*Sqrt[(b + a*(c + d*x)^2)/(c + d*x 
)^2])])/(b + a*c^2)^(7/2))/2
 

Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.03, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.632, Rules used = {896, 25, 1774, 1803, 25, 593, 25, 688, 25, 679, 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.

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

\(\Big \downarrow \) 896

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1774

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

\(\Big \downarrow \) 1803

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 593

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 688

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 679

\(\displaystyle d^2 \left (\frac {\frac {c}{c+d x}+1}{\left (a c^2+b\right ) \left (1-\frac {c}{c+d x}\right )^2 \sqrt {a+\frac {b}{(c+d x)^2}}}-\frac {c \left (\frac {\frac {3 b \left (2 b-3 a c^2\right ) \int \frac {1}{\sqrt {a+\frac {b}{(c+d x)^2}} \left (1-\frac {c}{c+d x}\right )}d\frac {1}{c+d x}}{a c^2+b}+\frac {c \left (11 b-4 a c^2\right ) \sqrt {a+\frac {b}{(c+d x)^2}}}{\left (a c^2+b\right ) \left (1-\frac {c}{c+d x}\right )}}{2 \left (a c^2+b\right )}+\frac {5 c \sqrt {a+\frac {b}{(c+d x)^2}}}{2 \left (a c^2+b\right ) \left (1-\frac {c}{c+d x}\right )^2}\right )}{a c^2+b}\right )\)

\(\Big \downarrow \) 488

\(\displaystyle d^2 \left (\frac {\frac {c}{c+d x}+1}{\left (a c^2+b\right ) \left (1-\frac {c}{c+d x}\right )^2 \sqrt {a+\frac {b}{(c+d x)^2}}}-\frac {c \left (\frac {\frac {c \left (11 b-4 a c^2\right ) \sqrt {a+\frac {b}{(c+d x)^2}}}{\left (a c^2+b\right ) \left (1-\frac {c}{c+d x}\right )}-\frac {3 b \left (2 b-3 a c^2\right ) \int \frac {1}{a c^2+b-\frac {1}{(c+d x)^2}}d\frac {-\frac {b}{c+d x}-a c}{\sqrt {a+\frac {b}{(c+d x)^2}}}}{a c^2+b}}{2 \left (a c^2+b\right )}+\frac {5 c \sqrt {a+\frac {b}{(c+d x)^2}}}{2 \left (a c^2+b\right ) \left (1-\frac {c}{c+d x}\right )^2}\right )}{a c^2+b}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle d^2 \left (\frac {\frac {c}{c+d x}+1}{\left (a c^2+b\right ) \left (1-\frac {c}{c+d x}\right )^2 \sqrt {a+\frac {b}{(c+d x)^2}}}-\frac {c \left (\frac {\frac {c \left (11 b-4 a c^2\right ) \sqrt {a+\frac {b}{(c+d x)^2}}}{\left (a c^2+b\right ) \left (1-\frac {c}{c+d x}\right )}-\frac {3 b \left (2 b-3 a c^2\right ) \text {arctanh}\left (\frac {-a c-\frac {b}{c+d x}}{\sqrt {a c^2+b} \sqrt {a+\frac {b}{(c+d x)^2}}}\right )}{\left (a c^2+b\right )^{3/2}}}{2 \left (a c^2+b\right )}+\frac {5 c \sqrt {a+\frac {b}{(c+d x)^2}}}{2 \left (a c^2+b\right ) \left (1-\frac {c}{c+d x}\right )^2}\right )}{a c^2+b}\right )\)

Input:

Int[1/(x^3*(a + b/(c + d*x)^2)^(3/2)),x]
 

Output:

d^2*((1 + c/(c + d*x))/((b + a*c^2)*Sqrt[a + b/(c + d*x)^2]*(1 - c/(c + d* 
x))^2) - (c*((5*c*Sqrt[a + b/(c + d*x)^2])/(2*(b + a*c^2)*(1 - c/(c + d*x) 
)^2) + ((c*(11*b - 4*a*c^2)*Sqrt[a + b/(c + d*x)^2])/((b + a*c^2)*(1 - c/( 
c + d*x))) - (3*b*(2*b - 3*a*c^2)*ArcTanh[(-(a*c) - b/(c + d*x))/(Sqrt[b + 
 a*c^2]*Sqrt[a + b/(c + d*x)^2])])/(b + a*c^2)^(3/2))/(2*(b + a*c^2))))/(b 
 + a*c^2))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 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 593

Int[(x_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
 Simp[(c + d*x)^(n + 1)*(c - d*x)*((a + b*x^2)^(p + 1)/(2*(p + 1)*(b*c^2 + 
a*d^2))), x] - Simp[d/(2*(p + 1)*(b*c^2 + a*d^2))   Int[(c + d*x)^n*(a + b* 
x^2)^(p + 1)*(c*n - d*(n + 2*p + 4)*x), x], x] /; FreeQ[{a, b, c, d, n}, x] 
 && LtQ[p, -1] && NeQ[b*c^2 + a*d^2, 0]
 

rule 679

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p 
_.), x_Symbol] :> Simp[(-(e*f - d*g))*(d + e*x)^(m + 1)*((a + c*x^2)^(p + 1 
)/(2*(p + 1)*(c*d^2 + a*e^2))), x] + Simp[(c*d*f + a*e*g)/(c*d^2 + a*e^2) 
 Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, 
 p}, x] && EqQ[Simplify[m + 2*p + 3], 0]
 

rule 688

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p 
_.), x_Symbol] :> Simp[(e*f - d*g)*(d + e*x)^(m + 1)*((a + c*x^2)^(p + 1)/( 
(m + 1)*(c*d^2 + a*e^2))), x] + Simp[1/((m + 1)*(c*d^2 + a*e^2))   Int[(d + 
 e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m 
 + 2*p + 3)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] && LtQ[m, -1] 
&& (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
 

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. \(573\) vs. \(2(216)=432\).

Time = 0.13 (sec) , antiderivative size = 574, normalized size of antiderivative = 2.45

method	result	size
default	\(\frac {\left (a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b \right ) \left (\left (a \,c^{2}+b \right )^{\frac {3}{2}} a^{2} c^{4} d^{3} x^{3}+9 \sqrt {a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b}\, \ln \left (\frac {2 a \,c^{2}+2 b +2 a d x c +2 \sqrt {a \,c^{2}+b}\, \sqrt {a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b}}{x}\right ) a^{2} b \,c^{5} d^{2} x^{2}+\left (a \,c^{2}+b \right )^{\frac {3}{2}} a^{2} c^{5} d^{2} x^{2}-\left (a \,c^{2}+b \right )^{\frac {3}{2}} a^{2} c^{6} d x -12 \left (a \,c^{2}+b \right )^{\frac {3}{2}} a b \,c^{2} d^{3} x^{3}+3 \sqrt {a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b}\, \ln \left (\frac {2 a \,c^{2}+2 b +2 a d x c +2 \sqrt {a \,c^{2}+b}\, \sqrt {a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b}}{x}\right ) a \,b^{2} c^{3} d^{2} x^{2}-\left (a \,c^{2}+b \right )^{\frac {3}{2}} a^{2} c^{7}-21 \left (a \,c^{2}+b \right )^{\frac {3}{2}} a b \,c^{3} d^{2} x^{2}-7 \left (a \,c^{2}+b \right )^{\frac {3}{2}} a b \,c^{4} d x +2 \left (a \,c^{2}+b \right )^{\frac {3}{2}} b^{2} d^{3} x^{3}-6 \sqrt {a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b}\, \ln \left (\frac {2 a \,c^{2}+2 b +2 a d x c +2 \sqrt {a \,c^{2}+b}\, \sqrt {a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b}}{x}\right ) b^{3} c \,d^{2} x^{2}-2 \left (a \,c^{2}+b \right )^{\frac {3}{2}} a b \,c^{5}+8 \left (a \,c^{2}+b \right )^{\frac {3}{2}} b^{2} c \,d^{2} x^{2}-6 \left (a \,c^{2}+b \right )^{\frac {3}{2}} b^{2} c^{2} d x -\left (a \,c^{2}+b \right )^{\frac {3}{2}} b^{2} c^{3}\right )}{2 \left (\frac {a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b}{\left (d x +c \right )^{2}}\right )^{\frac {3}{2}} \left (d x +c \right )^{3} \left (a \,c^{2}+b \right )^{\frac {9}{2}} x^{2}}\)	\(574\)
risch	\(-\frac {c^{2} \left (a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b \right ) \left (-a d x \,c^{2}+c^{3} a +6 b d x +b c \right )}{2 \left (a \,c^{2}+b \right )^{3} x^{2} \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 )}+\frac {\left (\frac {d^{3} b^{2} x}{\left (a \,c^{2}+b \right )^{3} \sqrt {a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b}}+\frac {d^{2} b^{2} c}{\left (a \,c^{2}+b \right )^{3} \sqrt {a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b}}-\frac {9 d^{3} c^{4} a^{2} x}{2 \left (a \,c^{2}+b \right )^{3} \sqrt {a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b}}-\frac {9 d^{2} c^{5} a^{2}}{2 \left (a \,c^{2}+b \right )^{3} \sqrt {a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b}}-\frac {3 d^{2} b \,a^{2} c^{5}}{\left (a \,c^{2}+b \right )^{4} \sqrt {a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b}}-\frac {9 d^{2} b^{2} a \,c^{3}}{2 \left (a \,c^{2}+b \right )^{4} \sqrt {a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b}}+\frac {3 d^{2} b^{3} c}{\left (a \,c^{2}+b \right )^{4} \sqrt {a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b}}+\frac {9 d^{3} c^{6} a^{3} x}{2 \left (a \,c^{2}+b \right )^{4} \sqrt {a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b}}+\frac {3 d^{3} b \,c^{4} a^{2} x}{2 \left (a \,c^{2}+b \right )^{4} \sqrt {a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b}}-\frac {3 d^{3} b^{2} c^{2} a x}{\left (a \,c^{2}+b \right )^{4} \sqrt {a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b}}+\frac {9 d^{2} c^{7} a^{3}}{2 \left (a \,c^{2}+b \right )^{4} \sqrt {a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b}}+\frac {9 d^{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 ) a^{2} c^{5}}{2 \left (a \,c^{2}+b \right )^{\frac {9}{2}}}+\frac {3 d^{2} b^{2} \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 ) a \,c^{3}}{2 \left (a \,c^{2}+b \right )^{\frac {9}{2}}}-\frac {3 d^{2} b^{3} c \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 )}{\left (a \,c^{2}+b \right )^{\frac {9}{2}}}+\frac {7 d^{2} b \,c^{3} a}{2 \left (a \,c^{2}+b \right )^{3} \sqrt {a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b}}\right ) \sqrt {a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b}}{\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 )}\)	\(903\)

Input:

int(1/x^3/(a+b/(d*x+c)^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/2*(a*d^2*x^2+2*a*c*d*x+a*c^2+b)*((a*c^2+b)^(3/2)*a^2*c^4*d^3*x^3+9*(a*d^ 
2*x^2+2*a*c*d*x+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^2*b*c^5*d^2*x^2+(a*c^2+b)^(3/2)*a^2*c 
^5*d^2*x^2-(a*c^2+b)^(3/2)*a^2*c^6*d*x-12*(a*c^2+b)^(3/2)*a*b*c^2*d^3*x^3+ 
3*(a*d^2*x^2+2*a*c*d*x+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^2*c^3*d^2*x^2-(a*c^2+b)^(3/2 
)*a^2*c^7-21*(a*c^2+b)^(3/2)*a*b*c^3*d^2*x^2-7*(a*c^2+b)^(3/2)*a*b*c^4*d*x 
+2*(a*c^2+b)^(3/2)*b^2*d^3*x^3-6*(a*d^2*x^2+2*a*c*d*x+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 
^3*c*d^2*x^2-2*(a*c^2+b)^(3/2)*a*b*c^5+8*(a*c^2+b)^(3/2)*b^2*c*d^2*x^2-6*( 
a*c^2+b)^(3/2)*b^2*c^2*d*x-(a*c^2+b)^(3/2)*b^2*c^3)/((a*d^2*x^2+2*a*c*d*x+ 
a*c^2+b)/(d*x+c)^2)^(3/2)/(d*x+c)^3/(a*c^2+b)^(9/2)/x^2
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 628 vs. \(2 (216) = 432\).

Time = 2.17 (sec) , antiderivative size = 1260, normalized size of antiderivative = 5.38 \[ \int \frac {1}{x^3 \left (a+\frac {b}{(c+d x)^2}\right )^{3/2}} \, dx=\text {Too large to display} \] Input:

integrate(1/x^3/(a+b/(d*x+c)^2)^(3/2),x, algorithm="fricas")
 

Output:

[1/4*(3*((3*a^2*b*c^3 - 2*a*b^2*c)*d^4*x^4 + 2*(3*a^2*b*c^4 - 2*a*b^2*c^2) 
*d^3*x^3 + (3*a^2*b*c^5 + a*b^2*c^3 - 2*b^3*c)*d^2*x^2)*sqrt(a*c^2 + b)*lo 
g(-(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^3*c^10 + 3*a^2*b*c^8 + 3*a*b^2*c^6 - (a^3*c^6 - 11*a^2*b*c^ 
4 - 10*a*b^2*c^2 + 2*b^3)*d^4*x^4 + b^3*c^4 - (2*a^3*c^7 - 31*a^2*b*c^5 - 
23*a*b^2*c^3 + 10*b^3*c)*d^3*x^3 + 2*(14*a^2*b*c^6 + 13*a*b^2*c^4 - b^3*c^ 
2)*d^2*x^2 + (2*a^3*c^9 + 11*a^2*b*c^7 + 16*a*b^2*c^5 + 7*b^3*c^3)*d*x)*sq 
rt((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^5*c 
^8 + 4*a^4*b*c^6 + 6*a^3*b^2*c^4 + 4*a^2*b^3*c^2 + a*b^4)*d^2*x^4 + 2*(a^5 
*c^9 + 4*a^4*b*c^7 + 6*a^3*b^2*c^5 + 4*a^2*b^3*c^3 + a*b^4*c)*d*x^3 + (a^5 
*c^10 + 5*a^4*b*c^8 + 10*a^3*b^2*c^6 + 10*a^2*b^3*c^4 + 5*a*b^4*c^2 + b^5) 
*x^2), -1/2*(3*((3*a^2*b*c^3 - 2*a*b^2*c)*d^4*x^4 + 2*(3*a^2*b*c^4 - 2*a*b 
^2*c^2)*d^3*x^3 + (3*a^2*b*c^5 + a*b^2*c^3 - 2*b^3*c)*d^2*x^2)*sqrt(-a*c^2 
 - b)*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^3*c^10 + 3*a^2*b*c^8 + 3*a*b^2*c^6 - (a^3*c^6 - 11*a^2*b*c^4 - 1 
0*a*b^2*c^2 + 2*b^3)*d^4*x^4 + b^3*c^4 - (2*a^3*c^7 - 31*a^2*b*c^5 - 23...
 

Sympy [F]

\[ \int \frac {1}{x^3 \left (a+\frac {b}{(c+d x)^2}\right )^{3/2}} \, dx=\int \frac {1}{x^{3} \left (\frac {a c^{2} + 2 a c d x + a d^{2} x^{2} + b}{c^{2} + 2 c d x + d^{2} x^{2}}\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(1/x**3/(a+b/(d*x+c)**2)**(3/2),x)
 

Output:

Integral(1/(x**3*((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)
                                                                                    
                                                                                    
 

Maxima [F]

\[ \int \frac {1}{x^3 \left (a+\frac {b}{(c+d x)^2}\right )^{3/2}} \, dx=\int { \frac {1}{{\left (a + \frac {b}{{\left (d x + c\right )}^{2}}\right )}^{\frac {3}{2}} x^{3}} \,d x } \] Input:

integrate(1/x^3/(a+b/(d*x+c)^2)^(3/2),x, algorithm="maxima")
 

Output:

integrate(1/((a + b/(d*x + c)^2)^(3/2)*x^3), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 961 vs. \(2 (216) = 432\).

Time = 0.21 (sec) , antiderivative size = 961, normalized size of antiderivative = 4.11 \[ \int \frac {1}{x^3 \left (a+\frac {b}{(c+d x)^2}\right )^{3/2}} \, dx =\text {Too large to display} \] Input:

integrate(1/x^3/(a+b/(d*x+c)^2)^(3/2),x, algorithm="giac")
 

Output:

-((3*a^4*b^2*c^8*d^4*sgn(d*x + c) + 8*a^3*b^3*c^6*d^4*sgn(d*x + c) + 6*a^2 
*b^4*c^4*d^4*sgn(d*x + c) - b^6*d^4*sgn(d*x + c))*x/(a^6*b*c^12*d + 6*a^5* 
b^2*c^10*d + 15*a^4*b^3*c^8*d + 20*a^3*b^4*c^6*d + 15*a^2*b^5*c^4*d + 6*a* 
b^6*c^2*d + b^7*d) + 4*(a^4*b^2*c^9*d^3*sgn(d*x + c) + 2*a^3*b^3*c^7*d^3*s 
gn(d*x + c) - 2*a*b^5*c^3*d^3*sgn(d*x + c) - b^6*c*d^3*sgn(d*x + c))/(a^6* 
b*c^12*d + 6*a^5*b^2*c^10*d + 15*a^4*b^3*c^8*d + 20*a^3*b^4*c^6*d + 15*a^2 
*b^5*c^4*d + 6*a*b^6*c^2*d + b^7*d))/sqrt(a*d^2*x^2 + 2*a*c*d*x + a*c^2 + 
b) - 3*(3*a*b*c^3*d^2 - 2*b^2*c*d^2)*arctan(-(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))/((a^3*c^6*sgn(d*x + c) + 3* 
a^2*b*c^4*sgn(d*x + c) + 3*a*b^2*c^2*sgn(d*x + c) + b^3*sgn(d*x + c))*sqrt 
(-a*c^2 - b)) + (2*a^(7/2)*c^8*d*abs(d) + 4*(sqrt(a*d^2)*x - sqrt(a*d^2*x^ 
2 + 2*a*c*d*x + a*c^2 + b))*a^3*c^7*d^2 + 2*(sqrt(a*d^2)*x - sqrt(a*d^2*x^ 
2 + 2*a*c*d*x + a*c^2 + b))^2*a^(5/2)*c^6*d*abs(d) - 2*a^(5/2)*b*c^6*d*abs 
(d) - (sqrt(a*d^2)*x - sqrt(a*d^2*x^2 + 2*a*c*d*x + a*c^2 + b))*a^2*b*c^5* 
d^2 + 8*(sqrt(a*d^2)*x - sqrt(a*d^2*x^2 + 2*a*c*d*x + a*c^2 + b))^2*a^(3/2 
)*b*c^4*d*abs(d) + 7*(sqrt(a*d^2)*x - sqrt(a*d^2*x^2 + 2*a*c*d*x + a*c^2 + 
 b))^3*a*b*c^3*d^2 - 10*a^(3/2)*b^2*c^4*d*abs(d) - 5*(sqrt(a*d^2)*x - sqrt 
(a*d^2*x^2 + 2*a*c*d*x + a*c^2 + b))*a*b^2*c^3*d^2 + 6*(sqrt(a*d^2)*x - sq 
rt(a*d^2*x^2 + 2*a*c*d*x + a*c^2 + b))^2*sqrt(a)*b^2*c^2*d*abs(d) - 6*sqrt 
(a)*b^3*c^2*d*abs(d))/((a^3*c^6*sgn(d*x + c) + 3*a^2*b*c^4*sgn(d*x + c)...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^3 \left (a+\frac {b}{(c+d x)^2}\right )^{3/2}} \, dx=\int \frac {1}{x^3\,{\left (a+\frac {b}{{\left (c+d\,x\right )}^2}\right )}^{3/2}} \,d x \] Input:

int(1/(x^3*(a + b/(c + d*x)^2)^(3/2)),x)
 

Output:

int(1/(x^3*(a + b/(c + d*x)^2)^(3/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 1394, normalized size of antiderivative = 5.96 \[ \int \frac {1}{x^3 \left (a+\frac {b}{(c+d x)^2}\right )^{3/2}} \, dx =\text {Too large to display} \] Input:

int(1/x^3/(a+b/(d*x+c)^2)^(3/2),x)
 

Output:

( - sqrt(a*c**2 + 2*a*c*d*x + a*d**2*x**2 + b)*a**3*c**9 - sqrt(a*c**2 + 2 
*a*c*d*x + a*d**2*x**2 + b)*a**3*c**8*d*x + sqrt(a*c**2 + 2*a*c*d*x + a*d* 
*2*x**2 + b)*a**3*c**7*d**2*x**2 + sqrt(a*c**2 + 2*a*c*d*x + a*d**2*x**2 + 
 b)*a**3*c**6*d**3*x**3 - 3*sqrt(a*c**2 + 2*a*c*d*x + a*d**2*x**2 + b)*a** 
2*b*c**7 - 8*sqrt(a*c**2 + 2*a*c*d*x + a*d**2*x**2 + b)*a**2*b*c**6*d*x - 
20*sqrt(a*c**2 + 2*a*c*d*x + a*d**2*x**2 + b)*a**2*b*c**5*d**2*x**2 - 11*s 
qrt(a*c**2 + 2*a*c*d*x + a*d**2*x**2 + b)*a**2*b*c**4*d**3*x**3 - 3*sqrt(a 
*c**2 + 2*a*c*d*x + a*d**2*x**2 + b)*a*b**2*c**5 - 13*sqrt(a*c**2 + 2*a*c* 
d*x + a*d**2*x**2 + b)*a*b**2*c**4*d*x - 13*sqrt(a*c**2 + 2*a*c*d*x + a*d* 
*2*x**2 + b)*a*b**2*c**3*d**2*x**2 - 10*sqrt(a*c**2 + 2*a*c*d*x + a*d**2*x 
**2 + b)*a*b**2*c**2*d**3*x**3 - sqrt(a*c**2 + 2*a*c*d*x + a*d**2*x**2 + b 
)*b**3*c**3 - 6*sqrt(a*c**2 + 2*a*c*d*x + a*d**2*x**2 + b)*b**3*c**2*d*x + 
 8*sqrt(a*c**2 + 2*a*c*d*x + a*d**2*x**2 + b)*b**3*c*d**2*x**2 + 2*sqrt(a* 
c**2 + 2*a*c*d*x + a*d**2*x**2 + b)*b**3*d**3*x**3 + 9*sqrt(a*c**2 + b)*lo 
g( - 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**2*b*c**5*d**2*x**2 + 18*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**2*b*c**4*d**3*x**3 + 9*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**2*b*c**3* 
d**4*x**4 + 3*sqrt(a*c**2 + b)*log( - sqrt(a*c**2 + b)*sqrt(a*c**2 + 2*...