\(\int \frac {1}{x^3 (a+b \sqrt {c+d x})^2} \, dx\) [124]

Optimal result

Integrand size = 19, antiderivative size = 299 \[ \int \frac {1}{x^3 \left (a+b \sqrt {c+d x}\right )^2} \, dx=\frac {a b^2 \left (a^2+11 b^2 c\right ) d^2}{2 c \left (a^2-b^2 c\right )^3 \left (a+b \sqrt {c+d x}\right )}-\frac {a^2+b^2 c-2 a b \sqrt {c+d x}}{2 \left (a^2-b^2 c\right )^2 x^2}-\frac {b d \left (5 a b c-\left (a^2+2 b^2 c\right ) \sqrt {c+d x}\right )}{2 c \left (a^2-b^2 c\right )^2 x \left (a+b \sqrt {c+d x}\right )}-\frac {a b \left (a^4-10 a^2 b^2 c-15 b^4 c^2\right ) d^2 \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{2 c^{3/2} \left (a^2-b^2 c\right )^4}+\frac {b^4 \left (5 a^2+b^2 c\right ) d^2 \log (x)}{\left (a^2-b^2 c\right )^4}-\frac {2 b^4 \left (5 a^2+b^2 c\right ) d^2 \log \left (a+b \sqrt {c+d x}\right )}{\left (a^2-b^2 c\right )^4} \] Output:

1/2*a*b^2*(11*b^2*c+a^2)*d^2/c/(-b^2*c+a^2)^3/(a+b*(d*x+c)^(1/2))-1/2*(a^2 
+b^2*c-2*a*b*(d*x+c)^(1/2))/(-b^2*c+a^2)^2/x^2-1/2*b*d*(5*a*b*c-(2*b^2*c+a 
^2)*(d*x+c)^(1/2))/c/(-b^2*c+a^2)^2/x/(a+b*(d*x+c)^(1/2))-1/2*a*b*(-15*b^4 
*c^2-10*a^2*b^2*c+a^4)*d^2*arctanh((d*x+c)^(1/2)/c^(1/2))/c^(3/2)/(-b^2*c+ 
a^2)^4+b^4*(b^2*c+5*a^2)*d^2*ln(x)/(-b^2*c+a^2)^4-2*b^4*(b^2*c+5*a^2)*d^2* 
ln(a+b*(d*x+c)^(1/2))/(-b^2*c+a^2)^4
 

Mathematica [A] (verified)

Time = 0.84 (sec) , antiderivative size = 301, normalized size of antiderivative = 1.01 \[ \int \frac {1}{x^3 \left (a+b \sqrt {c+d x}\right )^2} \, dx=\frac {1}{2} \left (\frac {a^5 c-b^5 c^2 (c-2 d x) \sqrt {c+d x}+a^2 b^3 c (2 c-d x) \sqrt {c+d x}-a^4 b (c+d x)^{3/2}+a b^4 c \left (c^2-3 c d x-11 d^2 x^2\right )-a^3 b^2 \left (2 c^2-3 c d x+d^2 x^2\right )}{c \left (-a^2+b^2 c\right )^3 x^2 \left (a+b \sqrt {c+d x}\right )}+\frac {\left (-a^5 b+10 a^3 b^3 c+15 a b^5 c^2\right ) d^2 \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{c^{3/2} \left (a^2-b^2 c\right )^4}+\frac {2 b^4 \left (5 a^2+b^2 c\right ) d^2 \log (-d x)}{\left (a^2-b^2 c\right )^4}-\frac {4 b^4 \left (5 a^2+b^2 c\right ) d^2 \log \left (a+b \sqrt {c+d x}\right )}{\left (a^2-b^2 c\right )^4}\right ) \] Input:

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

Output:

((a^5*c - b^5*c^2*(c - 2*d*x)*Sqrt[c + d*x] + a^2*b^3*c*(2*c - d*x)*Sqrt[c 
 + d*x] - a^4*b*(c + d*x)^(3/2) + a*b^4*c*(c^2 - 3*c*d*x - 11*d^2*x^2) - a 
^3*b^2*(2*c^2 - 3*c*d*x + d^2*x^2))/(c*(-a^2 + b^2*c)^3*x^2*(a + b*Sqrt[c 
+ d*x])) + ((-(a^5*b) + 10*a^3*b^3*c + 15*a*b^5*c^2)*d^2*ArcTanh[Sqrt[c + 
d*x]/Sqrt[c]])/(c^(3/2)*(a^2 - b^2*c)^4) + (2*b^4*(5*a^2 + b^2*c)*d^2*Log[ 
-(d*x)])/(a^2 - b^2*c)^4 - (4*b^4*(5*a^2 + b^2*c)*d^2*Log[a + b*Sqrt[c + d 
*x]])/(a^2 - b^2*c)^4)/2
 

Rubi [A] (verified)

Time = 0.95 (sec) , antiderivative size = 336, normalized size of antiderivative = 1.12, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.474, Rules used = {896, 25, 1732, 593, 27, 686, 25, 657, 2009}

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.

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

\(\Big \downarrow \) 896

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1732

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

\(\Big \downarrow \) 593

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 686

\(\displaystyle -2 d^2 \left (\frac {b \left (\frac {3 a b c-\left (a^2+2 b^2 c\right ) \sqrt {c+d x}}{2 c d x \left (a^2-b^2 c\right ) \left (a+b \sqrt {c+d x}\right )}-\frac {\int \frac {a \left (a^2-7 b^2 c\right )+2 b \left (a^2+2 b^2 c\right ) \sqrt {c+d x}}{d x \left (a+b \sqrt {c+d x}\right )^2}d\sqrt {c+d x}}{2 c \left (a^2-b^2 c\right )}\right )}{2 \left (a^2-b^2 c\right )}+\frac {a-b \sqrt {c+d x}}{4 d^2 x^2 \left (a^2-b^2 c\right ) \left (a+b \sqrt {c+d x}\right )}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle -2 d^2 \left (\frac {b \left (\frac {\int -\frac {a \left (a^2-7 b^2 c\right )+2 b \left (a^2+2 b^2 c\right ) \sqrt {c+d x}}{d x \left (a+b \sqrt {c+d x}\right )^2}d\sqrt {c+d x}}{2 c \left (a^2-b^2 c\right )}+\frac {3 a b c-\left (a^2+2 b^2 c\right ) \sqrt {c+d x}}{2 c d x \left (a^2-b^2 c\right ) \left (a+b \sqrt {c+d x}\right )}\right )}{2 \left (a^2-b^2 c\right )}+\frac {a-b \sqrt {c+d x}}{4 d^2 x^2 \left (a^2-b^2 c\right ) \left (a+b \sqrt {c+d x}\right )}\right )\)

\(\Big \downarrow \) 657

\(\displaystyle -2 d^2 \left (\frac {b \left (\frac {\int \left (\frac {a \left (a^2+11 b^2 c\right ) b^2}{\left (a^2-b^2 c\right ) \left (a+b \sqrt {c+d x}\right )^2}-\frac {4 c \left (5 a^2+b^2 c\right ) \sqrt {c+d x} b^3+a \left (a^4-10 b^2 c a^2-15 b^4 c^2\right )}{\left (a^2-b^2 c\right )^2 d x}+\frac {4 c \left (c b^6+5 a^2 b^4\right )}{\left (a^2-b^2 c\right )^2 \left (a+b \sqrt {c+d x}\right )}\right )d\sqrt {c+d x}}{2 c \left (a^2-b^2 c\right )}+\frac {3 a b c-\left (a^2+2 b^2 c\right ) \sqrt {c+d x}}{2 c d x \left (a^2-b^2 c\right ) \left (a+b \sqrt {c+d x}\right )}\right )}{2 \left (a^2-b^2 c\right )}+\frac {a-b \sqrt {c+d x}}{4 d^2 x^2 \left (a^2-b^2 c\right ) \left (a+b \sqrt {c+d x}\right )}\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle -2 d^2 \left (\frac {a-b \sqrt {c+d x}}{4 d^2 x^2 \left (a^2-b^2 c\right ) \left (a+b \sqrt {c+d x}\right )}+\frac {b \left (\frac {3 a b c-\left (a^2+2 b^2 c\right ) \sqrt {c+d x}}{2 c d x \left (a^2-b^2 c\right ) \left (a+b \sqrt {c+d x}\right )}+\frac {-\frac {a b \left (a^2+11 b^2 c\right )}{\left (a^2-b^2 c\right ) \left (a+b \sqrt {c+d x}\right )}-\frac {2 b^3 c \left (5 a^2+b^2 c\right ) \log (-d x)}{\left (a^2-b^2 c\right )^2}+\frac {4 b^3 c \left (5 a^2+b^2 c\right ) \log \left (a+b \sqrt {c+d x}\right )}{\left (a^2-b^2 c\right )^2}+\frac {a \left (a^4-10 a^2 b^2 c-15 b^4 c^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{\sqrt {c} \left (a^2-b^2 c\right )^2}}{2 c \left (a^2-b^2 c\right )}\right )}{2 \left (a^2-b^2 c\right )}\right )\)

Input:

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

Output:

-2*d^2*((a - b*Sqrt[c + d*x])/(4*(a^2 - b^2*c)*d^2*x^2*(a + b*Sqrt[c + d*x 
])) + (b*((3*a*b*c - (a^2 + 2*b^2*c)*Sqrt[c + d*x])/(2*c*(a^2 - b^2*c)*d*x 
*(a + b*Sqrt[c + d*x])) + (-((a*b*(a^2 + 11*b^2*c))/((a^2 - b^2*c)*(a + b* 
Sqrt[c + d*x]))) + (a*(a^4 - 10*a^2*b^2*c - 15*b^4*c^2)*ArcTanh[Sqrt[c + d 
*x]/Sqrt[c]])/(Sqrt[c]*(a^2 - b^2*c)^2) - (2*b^3*c*(5*a^2 + b^2*c)*Log[-(d 
*x)])/(a^2 - b^2*c)^2 + (4*b^3*c*(5*a^2 + b^2*c)*Log[a + b*Sqrt[c + d*x]]) 
/(a^2 - b^2*c)^2)/(2*c*(a^2 - b^2*c))))/(2*(a^2 - b^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 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 657

Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_) + (c_.)*( 
x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g*x)^n/(a + c*x^ 
2)), x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IntegersQ[n]
 

rule 686

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p 
_), x_Symbol] :> Simp[(-(d + e*x)^(m + 1))*(f*a*c*e - a*g*c*d + c*(c*d*f + 
a*e*g)*x)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1)*(c*d^2 + a*e^2))), x] + Simp[ 
1/(2*a*c*(p + 1)*(c*d^2 + a*e^2))   Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Sim 
p[f*(c^2*d^2*(2*p + 3) + a*c*e^2*(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f 
+ a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && LtQ 
[p, -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 1732

Int[((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symb 
ol] :> With[{g = Denominator[n]}, Simp[g   Subst[Int[x^(g - 1)*(d + e*x^(g* 
n))^q*(a + c*x^(2*g*n))^p, x], x, x^(1/g)], x]] /; FreeQ[{a, c, d, e, p, q} 
, x] && EqQ[n2, 2*n] && FractionQ[n]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.01

method	result	size
derivativedivides	\(2 d^{2} \left (\frac {b^{4} a}{\left (-b^{2} c +a^{2}\right )^{3} \left (a +b \sqrt {d x +c}\right )}-\frac {b^{4} \left (b^{2} c +5 a^{2}\right ) \ln \left (a +b \sqrt {d x +c}\right )}{\left (-b^{2} c +a^{2}\right )^{4}}-\frac {\frac {-\frac {a b \left (-7 b^{4} c^{2}+6 a^{2} b^{2} c +a^{4}\right ) \left (d x +c \right )^{\frac {3}{2}}}{4 c}+\left (-\frac {1}{2} b^{6} c^{2}-a^{2} c \,b^{4}+\frac {3}{2} a^{4} b^{2}\right ) \left (d x +c \right )+\left (-\frac {9}{4} a \,b^{5} c^{2}+\frac {5}{2} a^{3} b^{3} c -\frac {1}{4} a^{5} b \right ) \sqrt {d x +c}+\frac {3 b^{6} c^{3}}{4}+\frac {3 a^{2} b^{4} c^{2}}{4}-\frac {7 a^{4} b^{2} c}{4}+\frac {a^{6}}{4}}{d^{2} x^{2}}+\frac {b \left (-\frac {\left (4 b^{5} c^{2}+20 a^{2} b^{3} c \right ) \ln \left (-d x \right )}{2}+\frac {\left (-15 a \,b^{4} c^{2}-10 a^{3} b^{2} c +a^{5}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{\sqrt {c}}\right )}{4 c}}{\left (-b^{2} c +a^{2}\right )^{4}}\right )\)	\(303\)
default	\(2 d^{2} \left (\frac {b^{4} a}{\left (-b^{2} c +a^{2}\right )^{3} \left (a +b \sqrt {d x +c}\right )}-\frac {b^{4} \left (b^{2} c +5 a^{2}\right ) \ln \left (a +b \sqrt {d x +c}\right )}{\left (-b^{2} c +a^{2}\right )^{4}}-\frac {\frac {-\frac {a b \left (-7 b^{4} c^{2}+6 a^{2} b^{2} c +a^{4}\right ) \left (d x +c \right )^{\frac {3}{2}}}{4 c}+\left (-\frac {1}{2} b^{6} c^{2}-a^{2} c \,b^{4}+\frac {3}{2} a^{4} b^{2}\right ) \left (d x +c \right )+\left (-\frac {9}{4} a \,b^{5} c^{2}+\frac {5}{2} a^{3} b^{3} c -\frac {1}{4} a^{5} b \right ) \sqrt {d x +c}+\frac {3 b^{6} c^{3}}{4}+\frac {3 a^{2} b^{4} c^{2}}{4}-\frac {7 a^{4} b^{2} c}{4}+\frac {a^{6}}{4}}{d^{2} x^{2}}+\frac {b \left (-\frac {\left (4 b^{5} c^{2}+20 a^{2} b^{3} c \right ) \ln \left (-d x \right )}{2}+\frac {\left (-15 a \,b^{4} c^{2}-10 a^{3} b^{2} c +a^{5}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{\sqrt {c}}\right )}{4 c}}{\left (-b^{2} c +a^{2}\right )^{4}}\right )\)	\(303\)

Input:

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

Output:

2*d^2*(b^4/(-b^2*c+a^2)^3*a/(a+b*(d*x+c)^(1/2))-b^4*(b^2*c+5*a^2)/(-b^2*c+ 
a^2)^4*ln(a+b*(d*x+c)^(1/2))-1/(-b^2*c+a^2)^4*((-1/4*a*b*(-7*b^4*c^2+6*a^2 
*b^2*c+a^4)/c*(d*x+c)^(3/2)+(-1/2*b^6*c^2-a^2*c*b^4+3/2*a^4*b^2)*(d*x+c)+( 
-9/4*a*b^5*c^2+5/2*a^3*b^3*c-1/4*a^5*b)*(d*x+c)^(1/2)+3/4*b^6*c^3+3/4*a^2* 
b^4*c^2-7/4*a^4*b^2*c+1/4*a^6)/d^2/x^2+1/4*b/c*(-1/2*(4*b^5*c^2+20*a^2*b^3 
*c)*ln(-d*x)+(-15*a*b^4*c^2-10*a^3*b^2*c+a^5)/c^(1/2)*arctanh((d*x+c)^(1/2 
)/c^(1/2)))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 619 vs. \(2 (283) = 566\).

Time = 0.95 (sec) , antiderivative size = 1249, normalized size of antiderivative = 4.18 \[ \int \frac {1}{x^3 \left (a+b \sqrt {c+d x}\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

[-1/4*(2*b^8*c^6 - 4*a^2*b^6*c^5 + 4*a^6*b^2*c^3 - 2*a^8*c^2 - 4*(b^8*c^4 
+ 4*a^2*b^6*c^3 - 5*a^4*b^4*c^2)*d^2*x^2 - 2*(b^8*c^5 + 3*a^2*b^6*c^4 - 9* 
a^4*b^4*c^3 + 5*a^6*b^2*c^2)*d*x - ((15*a*b^7*c^2 + 10*a^3*b^5*c - a^5*b^3 
)*d^3*x^3 + (15*a*b^7*c^3 - 5*a^3*b^5*c^2 - 11*a^5*b^3*c + a^7*b)*d^2*x^2) 
*sqrt(c)*log((d*x + 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x) + 8*((b^8*c^3 + 5*a^ 
2*b^6*c^2)*d^3*x^3 + (b^8*c^4 + 4*a^2*b^6*c^3 - 5*a^4*b^4*c^2)*d^2*x^2)*lo 
g(sqrt(d*x + c)*b + a) - 4*((b^8*c^3 + 5*a^2*b^6*c^2)*d^3*x^3 + (b^8*c^4 + 
 4*a^2*b^6*c^3 - 5*a^4*b^4*c^2)*d^2*x^2)*log(x) - 2*(2*a*b^7*c^5 - 6*a^3*b 
^5*c^4 + 6*a^5*b^3*c^3 - 2*a^7*b*c^2 - (11*a*b^7*c^3 - 10*a^3*b^5*c^2 - a^ 
5*b^3*c)*d^2*x^2 - (5*a*b^7*c^4 - 9*a^3*b^5*c^3 + 3*a^5*b^3*c^2 + a^7*b*c) 
*d*x)*sqrt(d*x + c))/((b^10*c^6 - 4*a^2*b^8*c^5 + 6*a^4*b^6*c^4 - 4*a^6*b^ 
4*c^3 + a^8*b^2*c^2)*d*x^3 + (b^10*c^7 - 5*a^2*b^8*c^6 + 10*a^4*b^6*c^5 - 
10*a^6*b^4*c^4 + 5*a^8*b^2*c^3 - a^10*c^2)*x^2), -1/2*(b^8*c^6 - 2*a^2*b^6 
*c^5 + 2*a^6*b^2*c^3 - a^8*c^2 - 2*(b^8*c^4 + 4*a^2*b^6*c^3 - 5*a^4*b^4*c^ 
2)*d^2*x^2 - (b^8*c^5 + 3*a^2*b^6*c^4 - 9*a^4*b^4*c^3 + 5*a^6*b^2*c^2)*d*x 
 + ((15*a*b^7*c^2 + 10*a^3*b^5*c - a^5*b^3)*d^3*x^3 + (15*a*b^7*c^3 - 5*a^ 
3*b^5*c^2 - 11*a^5*b^3*c + a^7*b)*d^2*x^2)*sqrt(-c)*arctan(sqrt(-c)/sqrt(d 
*x + c)) + 4*((b^8*c^3 + 5*a^2*b^6*c^2)*d^3*x^3 + (b^8*c^4 + 4*a^2*b^6*c^3 
 - 5*a^4*b^4*c^2)*d^2*x^2)*log(sqrt(d*x + c)*b + a) - 2*((b^8*c^3 + 5*a^2* 
b^6*c^2)*d^3*x^3 + (b^8*c^4 + 4*a^2*b^6*c^3 - 5*a^4*b^4*c^2)*d^2*x^2)*l...
 

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 659 vs. \(2 (283) = 566\).

Time = 0.13 (sec) , antiderivative size = 659, normalized size of antiderivative = 2.20 \[ \int \frac {1}{x^3 \left (a+b \sqrt {c+d x}\right )^2} \, dx=\frac {1}{4} \, d^{2} {\left (\frac {4 \, {\left (b^{6} c + 5 \, a^{2} b^{4}\right )} \log \left (d x\right )}{b^{8} c^{4} - 4 \, a^{2} b^{6} c^{3} + 6 \, a^{4} b^{4} c^{2} - 4 \, a^{6} b^{2} c + a^{8}} - \frac {8 \, {\left (b^{6} c + 5 \, a^{2} b^{4}\right )} \log \left (\sqrt {d x + c} b + a\right )}{b^{8} c^{4} - 4 \, a^{2} b^{6} c^{3} + 6 \, a^{4} b^{4} c^{2} - 4 \, a^{6} b^{2} c + a^{8}} - \frac {{\left (15 \, a b^{5} c^{2} + 10 \, a^{3} b^{3} c - a^{5} b\right )} \log \left (\frac {\sqrt {d x + c} - \sqrt {c}}{\sqrt {d x + c} + \sqrt {c}}\right )}{{\left (b^{8} c^{5} - 4 \, a^{2} b^{6} c^{4} + 6 \, a^{4} b^{4} c^{3} - 4 \, a^{6} b^{2} c^{2} + a^{8} c\right )} \sqrt {c}} - \frac {2 \, {\left (7 \, a b^{4} c^{3} + 6 \, a^{3} b^{2} c^{2} - a^{5} c + {\left (11 \, a b^{4} c + a^{3} b^{2}\right )} {\left (d x + c\right )}^{2} - {\left (2 \, b^{5} c^{2} - a^{2} b^{3} c - a^{4} b\right )} {\left (d x + c\right )}^{\frac {3}{2}} - {\left (19 \, a b^{4} c^{2} + 5 \, a^{3} b^{2} c\right )} {\left (d x + c\right )} + 3 \, {\left (b^{5} c^{3} - a^{2} b^{3} c^{2}\right )} \sqrt {d x + c}\right )}}{a b^{6} c^{6} - 3 \, a^{3} b^{4} c^{5} + 3 \, a^{5} b^{2} c^{4} - a^{7} c^{3} + {\left (b^{7} c^{4} - 3 \, a^{2} b^{5} c^{3} + 3 \, a^{4} b^{3} c^{2} - a^{6} b c\right )} {\left (d x + c\right )}^{\frac {5}{2}} + {\left (a b^{6} c^{4} - 3 \, a^{3} b^{4} c^{3} + 3 \, a^{5} b^{2} c^{2} - a^{7} c\right )} {\left (d x + c\right )}^{2} - 2 \, {\left (b^{7} c^{5} - 3 \, a^{2} b^{5} c^{4} + 3 \, a^{4} b^{3} c^{3} - a^{6} b c^{2}\right )} {\left (d x + c\right )}^{\frac {3}{2}} - 2 \, {\left (a b^{6} c^{5} - 3 \, a^{3} b^{4} c^{4} + 3 \, a^{5} b^{2} c^{3} - a^{7} c^{2}\right )} {\left (d x + c\right )} + {\left (b^{7} c^{6} - 3 \, a^{2} b^{5} c^{5} + 3 \, a^{4} b^{3} c^{4} - a^{6} b c^{3}\right )} \sqrt {d x + c}}\right )} \] Input:

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

Output:

1/4*d^2*(4*(b^6*c + 5*a^2*b^4)*log(d*x)/(b^8*c^4 - 4*a^2*b^6*c^3 + 6*a^4*b 
^4*c^2 - 4*a^6*b^2*c + a^8) - 8*(b^6*c + 5*a^2*b^4)*log(sqrt(d*x + c)*b + 
a)/(b^8*c^4 - 4*a^2*b^6*c^3 + 6*a^4*b^4*c^2 - 4*a^6*b^2*c + a^8) - (15*a*b 
^5*c^2 + 10*a^3*b^3*c - a^5*b)*log((sqrt(d*x + c) - sqrt(c))/(sqrt(d*x + c 
) + sqrt(c)))/((b^8*c^5 - 4*a^2*b^6*c^4 + 6*a^4*b^4*c^3 - 4*a^6*b^2*c^2 + 
a^8*c)*sqrt(c)) - 2*(7*a*b^4*c^3 + 6*a^3*b^2*c^2 - a^5*c + (11*a*b^4*c + a 
^3*b^2)*(d*x + c)^2 - (2*b^5*c^2 - a^2*b^3*c - a^4*b)*(d*x + c)^(3/2) - (1 
9*a*b^4*c^2 + 5*a^3*b^2*c)*(d*x + c) + 3*(b^5*c^3 - a^2*b^3*c^2)*sqrt(d*x 
+ c))/(a*b^6*c^6 - 3*a^3*b^4*c^5 + 3*a^5*b^2*c^4 - a^7*c^3 + (b^7*c^4 - 3* 
a^2*b^5*c^3 + 3*a^4*b^3*c^2 - a^6*b*c)*(d*x + c)^(5/2) + (a*b^6*c^4 - 3*a^ 
3*b^4*c^3 + 3*a^5*b^2*c^2 - a^7*c)*(d*x + c)^2 - 2*(b^7*c^5 - 3*a^2*b^5*c^ 
4 + 3*a^4*b^3*c^3 - a^6*b*c^2)*(d*x + c)^(3/2) - 2*(a*b^6*c^5 - 3*a^3*b^4* 
c^4 + 3*a^5*b^2*c^3 - a^7*c^2)*(d*x + c) + (b^7*c^6 - 3*a^2*b^5*c^5 + 3*a^ 
4*b^3*c^4 - a^6*b*c^3)*sqrt(d*x + c)))
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 521, normalized size of antiderivative = 1.74 \[ \int \frac {1}{x^3 \left (a+b \sqrt {c+d x}\right )^2} \, dx=\frac {{\left (b^{6} c d^{2} + 5 \, a^{2} b^{4} d^{2}\right )} \log \left (-d x\right )}{b^{8} c^{4} - 4 \, a^{2} b^{6} c^{3} + 6 \, a^{4} b^{4} c^{2} - 4 \, a^{6} b^{2} c + a^{8}} - \frac {2 \, {\left (b^{7} c d^{2} + 5 \, a^{2} b^{5} d^{2}\right )} \log \left ({\left | \sqrt {d x + c} b + a \right |}\right )}{b^{9} c^{4} - 4 \, a^{2} b^{7} c^{3} + 6 \, a^{4} b^{5} c^{2} - 4 \, a^{6} b^{3} c + a^{8} b} - \frac {{\left (15 \, a b^{5} c^{2} d^{2} + 10 \, a^{3} b^{3} c d^{2} - a^{5} b d^{2}\right )} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{2 \, {\left (b^{8} c^{5} - 4 \, a^{2} b^{6} c^{4} + 6 \, a^{4} b^{4} c^{3} - 4 \, a^{6} b^{2} c^{2} + a^{8} c\right )} \sqrt {-c}} - \frac {7 \, a b^{6} c^{4} d^{2} - a^{3} b^{4} c^{3} d^{2} - 7 \, a^{5} b^{2} c^{2} d^{2} + a^{7} c d^{2} + {\left (11 \, a b^{6} c^{2} d^{2} - 10 \, a^{3} b^{4} c d^{2} - a^{5} b^{2} d^{2}\right )} {\left (d x + c\right )}^{2} - {\left (2 \, b^{7} c^{3} d^{2} - 3 \, a^{2} b^{5} c^{2} d^{2} + a^{6} b d^{2}\right )} {\left (d x + c\right )}^{\frac {3}{2}} - {\left (19 \, a b^{6} c^{3} d^{2} - 14 \, a^{3} b^{4} c^{2} d^{2} - 5 \, a^{5} b^{2} c d^{2}\right )} {\left (d x + c\right )} + 3 \, {\left (b^{7} c^{4} d^{2} - 2 \, a^{2} b^{5} c^{3} d^{2} + a^{4} b^{3} c^{2} d^{2}\right )} \sqrt {d x + c}}{2 \, {\left (b^{2} c - a^{2}\right )}^{4} {\left (\sqrt {d x + c} b + a\right )} c d^{2} x^{2}} \] Input:

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

Output:

(b^6*c*d^2 + 5*a^2*b^4*d^2)*log(-d*x)/(b^8*c^4 - 4*a^2*b^6*c^3 + 6*a^4*b^4 
*c^2 - 4*a^6*b^2*c + a^8) - 2*(b^7*c*d^2 + 5*a^2*b^5*d^2)*log(abs(sqrt(d*x 
 + c)*b + a))/(b^9*c^4 - 4*a^2*b^7*c^3 + 6*a^4*b^5*c^2 - 4*a^6*b^3*c + a^8 
*b) - 1/2*(15*a*b^5*c^2*d^2 + 10*a^3*b^3*c*d^2 - a^5*b*d^2)*arctan(sqrt(d* 
x + c)/sqrt(-c))/((b^8*c^5 - 4*a^2*b^6*c^4 + 6*a^4*b^4*c^3 - 4*a^6*b^2*c^2 
 + a^8*c)*sqrt(-c)) - 1/2*(7*a*b^6*c^4*d^2 - a^3*b^4*c^3*d^2 - 7*a^5*b^2*c 
^2*d^2 + a^7*c*d^2 + (11*a*b^6*c^2*d^2 - 10*a^3*b^4*c*d^2 - a^5*b^2*d^2)*( 
d*x + c)^2 - (2*b^7*c^3*d^2 - 3*a^2*b^5*c^2*d^2 + a^6*b*d^2)*(d*x + c)^(3/ 
2) - (19*a*b^6*c^3*d^2 - 14*a^3*b^4*c^2*d^2 - 5*a^5*b^2*c*d^2)*(d*x + c) + 
 3*(b^7*c^4*d^2 - 2*a^2*b^5*c^3*d^2 + a^4*b^3*c^2*d^2)*sqrt(d*x + c))/((b^ 
2*c - a^2)^4*(sqrt(d*x + c)*b + a)*c*d^2*x^2)
 

Mupad [B] (verification not implemented)

Time = 11.82 (sec) , antiderivative size = 1441, normalized size of antiderivative = 4.82 \[ \int \frac {1}{x^3 \left (a+b \sqrt {c+d x}\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

(((5*a^3*b^2*d^2 + 19*a*b^4*c*d^2)*(c + d*x))/(2*(b^2*c - a^2)*(a^4 + b^4* 
c^2 - 2*a^2*b^2*c)) + ((a^3*b^2*d^2 + 11*a*b^4*c*d^2)*(c + d*x)^2)/(2*c*(a 
^6 - b^6*c^3 - 3*a^4*b^2*c + 3*a^2*b^4*c^2)) - (a*(7*b^4*c^2*d^2 - a^4*d^2 
 + 6*a^2*b^2*c*d^2))/(2*(b^2*c - a^2)*(a^4 + b^4*c^2 - 2*a^2*b^2*c)) + (b* 
(a^2*d^2 + 2*b^2*c*d^2)*(c + d*x)^(3/2))/(2*c*(a^4 + b^4*c^2 - 2*a^2*b^2*c 
)) - (3*b^3*c*d^2*(c + d*x)^(1/2))/(2*(a^4 + b^4*c^2 - 2*a^2*b^2*c)))/(a*( 
c + d*x)^2 + b*(c + d*x)^(5/2) + a*c^2 - 2*a*c*(c + d*x) - 2*b*c*(c + d*x) 
^(3/2) + b*c^2*(c + d*x)^(1/2)) + log(a + b*(c + d*x)^(1/2))*((10*b^4*d^2) 
/(b^2*c - a^2)^3 - (12*b^6*c*d^2)/(b^2*c - a^2)^4) + (log((a*b^4*d^4*(a^6 
- 44*b^6*c^3 + 2*a^4*b^2*c - 103*a^2*b^4*c^2))/(4*c^2*(b^2*c - a^2)^6) - ( 
b*d^2*((b^2*d^2*(a^2*(c + d*x)^(1/2) + 4*a*b*c + 3*b^2*c*(c + d*x)^(1/2))* 
(a^5*(c^3)^(1/2) + 4*b^5*c^4 + 20*a^2*b^3*c^3 - 10*a^3*b^2*c*(c^3)^(1/2) - 
 15*a*b^4*c^2*(c^3)^(1/2)))/(2*c^3*(b^2*c - a^2)^4) - (b^3*d^2*(c + d*x)^( 
1/2)*(6*b^4*c^2 - a^4 + 19*a^2*b^2*c))/(c*(b^2*c - a^2)^3) + (a*b^2*d^2*(7 
*b^2*c - a^2))/(2*c*(b^2*c - a^2)^2))*(a^5*(c^3)^(1/2) + 4*b^5*c^4 + 20*a^ 
2*b^3*c^3 - 10*a^3*b^2*c*(c^3)^(1/2) - 15*a*b^4*c^2*(c^3)^(1/2)))/(4*c^3*( 
b^2*c - a^2)^4) + (a^2*b^5*d^4*(11*b^2*c + a^2)^2*(c + d*x)^(1/2))/(4*c^2* 
(b^2*c - a^2)^6))*(4*b^6*c^4*d^2 + 20*a^2*b^4*c^3*d^2 + a^5*b*d^2*(c^3)^(1 
/2) - 10*a^3*b^3*c*d^2*(c^3)^(1/2) - 15*a*b^5*c^2*d^2*(c^3)^(1/2)))/(4*(a^ 
8*c^3 + b^8*c^7 - 4*a^6*b^2*c^4 + 6*a^4*b^4*c^5 - 4*a^2*b^6*c^6)) + (lo...
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 1103, normalized size of antiderivative = 3.69 \[ \int \frac {1}{x^3 \left (a+b \sqrt {c+d x}\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(sqrt(c)*sqrt(c + d*x)*log(sqrt(c + d*x) - sqrt(c))*a**5*b**2*d**2*x**2 - 
10*sqrt(c)*sqrt(c + d*x)*log(sqrt(c + d*x) - sqrt(c))*a**3*b**4*c*d**2*x** 
2 - 15*sqrt(c)*sqrt(c + d*x)*log(sqrt(c + d*x) - sqrt(c))*a*b**6*c**2*d**2 
*x**2 - sqrt(c)*sqrt(c + d*x)*log(sqrt(c + d*x) + sqrt(c))*a**5*b**2*d**2* 
x**2 + 10*sqrt(c)*sqrt(c + d*x)*log(sqrt(c + d*x) + sqrt(c))*a**3*b**4*c*d 
**2*x**2 + 15*sqrt(c)*sqrt(c + d*x)*log(sqrt(c + d*x) + sqrt(c))*a*b**6*c* 
*2*d**2*x**2 + 20*sqrt(c + d*x)*log(sqrt(c + d*x) - sqrt(c))*a**2*b**5*c** 
2*d**2*x**2 + 4*sqrt(c + d*x)*log(sqrt(c + d*x) - sqrt(c))*b**7*c**3*d**2* 
x**2 + 20*sqrt(c + d*x)*log(sqrt(c + d*x) + sqrt(c))*a**2*b**5*c**2*d**2*x 
**2 + 4*sqrt(c + d*x)*log(sqrt(c + d*x) + sqrt(c))*b**7*c**3*d**2*x**2 - 4 
0*sqrt(c + d*x)*log(sqrt(c + d*x)*b + a)*a**2*b**5*c**2*d**2*x**2 - 8*sqrt 
(c + d*x)*log(sqrt(c + d*x)*b + a)*b**7*c**3*d**2*x**2 + 2*sqrt(c + d*x)*a 
**6*b*c**2 + 2*sqrt(c + d*x)*a**6*b*c*d*x - 6*sqrt(c + d*x)*a**4*b**3*c**3 
 - 2*sqrt(c + d*x)*a**4*b**3*c*d**2*x**2 + 6*sqrt(c + d*x)*a**2*b**5*c**4 
- 6*sqrt(c + d*x)*a**2*b**5*c**3*d*x - 20*sqrt(c + d*x)*a**2*b**5*c**2*d** 
2*x**2 - 2*sqrt(c + d*x)*b**7*c**5 + 4*sqrt(c + d*x)*b**7*c**4*d*x + 22*sq 
rt(c + d*x)*b**7*c**3*d**2*x**2 + sqrt(c)*log(sqrt(c + d*x) - sqrt(c))*a** 
6*b*d**2*x**2 - 10*sqrt(c)*log(sqrt(c + d*x) - sqrt(c))*a**4*b**3*c*d**2*x 
**2 - 15*sqrt(c)*log(sqrt(c + d*x) - sqrt(c))*a**2*b**5*c**2*d**2*x**2 - s 
qrt(c)*log(sqrt(c + d*x) + sqrt(c))*a**6*b*d**2*x**2 + 10*sqrt(c)*log(s...