\(\int \frac {1}{x^3 (a+b \sqrt {c+d x})} \, dx\) [117]

Optimal result

Integrand size = 19, antiderivative size = 204 \[ \int \frac {1}{x^3 \left (a+b \sqrt {c+d x}\right )} \, dx=-\frac {a-b \sqrt {c+d x}}{2 \left (a^2-b^2 c\right ) x^2}-\frac {b d \left (4 a b c-\left (a^2+3 b^2 c\right ) \sqrt {c+d x}\right )}{4 c \left (a^2-b^2 c\right )^2 x}-\frac {b \left (a^4-6 a^2 b^2 c-3 b^4 c^2\right ) d^2 \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{4 c^{3/2} \left (a^2-b^2 c\right )^3}+\frac {a b^4 d^2 \log (x)}{\left (a^2-b^2 c\right )^3}-\frac {2 a b^4 d^2 \log \left (a+b \sqrt {c+d x}\right )}{\left (a^2-b^2 c\right )^3} \] Output:

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

Mathematica [A] (verified)

Time = 0.55 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.97 \[ \int \frac {1}{x^3 \left (a+b \sqrt {c+d x}\right )} \, dx=\frac {b \left (a^4-6 a^2 b^2 c-3 b^4 c^2\right ) d^2 x^2 \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )+\sqrt {c} \left (\left (a^2-b^2 c\right ) \left (2 a^3 c-2 a b^2 c (c-2 d x)+b^3 c (2 c-3 d x) \sqrt {c+d x}-a^2 b \sqrt {c+d x} (2 c+d x)\right )-4 a b^4 c d^2 x^2 \log (-d x)+8 a b^4 c d^2 x^2 \log \left (a+b \sqrt {c+d x}\right )\right )}{4 c^{3/2} \left (-a^2+b^2 c\right )^3 x^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.20, 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, 25, 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.

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

\(\Big \downarrow \) 896

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1732

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

\(\Big \downarrow \) 593

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 686

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 657

\(\displaystyle -2 d^2 \left (\frac {b \left (\frac {\int \left (\frac {8 a b^4 c}{\left (a^2-b^2 c\right ) \left (a+b \sqrt {c+d x}\right )}-\frac {a^4-6 b^2 c a^2+8 b^3 c \sqrt {c+d x} a-3 b^4 c^2}{\left (a^2-b^2 c\right ) d x}\right )d\sqrt {c+d x}}{2 c \left (a^2-b^2 c\right )}+\frac {4 a b c-\left (a^2+3 b^2 c\right ) \sqrt {c+d x}}{2 c d x \left (a^2-b^2 c\right )}\right )}{4 \left (a^2-b^2 c\right )}+\frac {a-b \sqrt {c+d x}}{4 d^2 x^2 \left (a^2-b^2 c\right )}\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle -2 d^2 \left (\frac {a-b \sqrt {c+d x}}{4 d^2 x^2 \left (a^2-b^2 c\right )}+\frac {b \left (\frac {4 a b c-\left (a^2+3 b^2 c\right ) \sqrt {c+d x}}{2 c d x \left (a^2-b^2 c\right )}+\frac {-\frac {4 a b^3 c \log (-d x)}{a^2-b^2 c}+\frac {8 a b^3 c \log \left (a+b \sqrt {c+d x}\right )}{a^2-b^2 c}+\frac {\left (a^4-6 a^2 b^2 c-3 b^4 c^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{\sqrt {c} \left (a^2-b^2 c\right )}}{2 c \left (a^2-b^2 c\right )}\right )}{4 \left (a^2-b^2 c\right )}\right )\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 = 0.09 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.11

method	result	size
derivativedivides	\(2 d^{2} \left (-\frac {a \,b^{4} \ln \left (a +b \sqrt {d x +c}\right )}{\left (-b^{2} c +a^{2}\right )^{3}}-\frac {\frac {-\frac {b \left (-3 b^{4} c^{2}+2 a^{2} b^{2} c +a^{4}\right ) \left (d x +c \right )^{\frac {3}{2}}}{8 c}+\left (-\frac {1}{2} a \,b^{4} c +\frac {1}{2} a^{3} b^{2}\right ) \left (d x +c \right )+\left (\frac {3}{4} a^{2} b^{3} c -\frac {1}{8} a^{4} b -\frac {5}{8} b^{5} c^{2}\right ) \sqrt {d x +c}+\frac {3 a \,b^{4} c^{2}}{4}-a^{3} b^{2} c +\frac {a^{5}}{4}}{d^{2} x^{2}}+\frac {b \left (-4 a \,b^{3} c \ln \left (-d x \right )+\frac {\left (-3 b^{4} c^{2}-6 a^{2} b^{2} c +a^{4}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{\sqrt {c}}\right )}{8 c}}{\left (-b^{2} c +a^{2}\right )^{3}}\right )\)	\(227\)
default	\(2 d^{2} \left (-\frac {a \,b^{4} \ln \left (a +b \sqrt {d x +c}\right )}{\left (-b^{2} c +a^{2}\right )^{3}}-\frac {\frac {-\frac {b \left (-3 b^{4} c^{2}+2 a^{2} b^{2} c +a^{4}\right ) \left (d x +c \right )^{\frac {3}{2}}}{8 c}+\left (-\frac {1}{2} a \,b^{4} c +\frac {1}{2} a^{3} b^{2}\right ) \left (d x +c \right )+\left (\frac {3}{4} a^{2} b^{3} c -\frac {1}{8} a^{4} b -\frac {5}{8} b^{5} c^{2}\right ) \sqrt {d x +c}+\frac {3 a \,b^{4} c^{2}}{4}-a^{3} b^{2} c +\frac {a^{5}}{4}}{d^{2} x^{2}}+\frac {b \left (-4 a \,b^{3} c \ln \left (-d x \right )+\frac {\left (-3 b^{4} c^{2}-6 a^{2} b^{2} c +a^{4}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{\sqrt {c}}\right )}{8 c}}{\left (-b^{2} c +a^{2}\right )^{3}}\right )\)	\(227\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.46 (sec) , antiderivative size = 531, normalized size of antiderivative = 2.60 \[ \int \frac {1}{x^3 \left (a+b \sqrt {c+d x}\right )} \, dx=\left [\frac {16 \, a b^{4} c^{2} d^{2} x^{2} \log \left (\sqrt {d x + c} b + a\right ) - 8 \, a b^{4} c^{2} d^{2} x^{2} \log \left (x\right ) + 4 \, a b^{4} c^{4} - 8 \, a^{3} b^{2} c^{3} + 4 \, a^{5} c^{2} + {\left (3 \, b^{5} c^{2} + 6 \, a^{2} b^{3} c - a^{4} b\right )} \sqrt {c} d^{2} x^{2} \log \left (\frac {d x - 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) - 8 \, {\left (a b^{4} c^{3} - a^{3} b^{2} c^{2}\right )} d x - 2 \, {\left (2 \, b^{5} c^{4} - 4 \, a^{2} b^{3} c^{3} + 2 \, a^{4} b c^{2} - {\left (3 \, b^{5} c^{3} - 2 \, a^{2} b^{3} c^{2} - a^{4} b c\right )} d x\right )} \sqrt {d x + c}}{8 \, {\left (b^{6} c^{5} - 3 \, a^{2} b^{4} c^{4} + 3 \, a^{4} b^{2} c^{3} - a^{6} c^{2}\right )} x^{2}}, \frac {8 \, a b^{4} c^{2} d^{2} x^{2} \log \left (\sqrt {d x + c} b + a\right ) - 4 \, a b^{4} c^{2} d^{2} x^{2} \log \left (x\right ) + 2 \, a b^{4} c^{4} - 4 \, a^{3} b^{2} c^{3} + 2 \, a^{5} c^{2} + {\left (3 \, b^{5} c^{2} + 6 \, a^{2} b^{3} c - a^{4} b\right )} \sqrt {-c} d^{2} x^{2} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x + c}}\right ) - 4 \, {\left (a b^{4} c^{3} - a^{3} b^{2} c^{2}\right )} d x - {\left (2 \, b^{5} c^{4} - 4 \, a^{2} b^{3} c^{3} + 2 \, a^{4} b c^{2} - {\left (3 \, b^{5} c^{3} - 2 \, a^{2} b^{3} c^{2} - a^{4} b c\right )} d x\right )} \sqrt {d x + c}}{4 \, {\left (b^{6} c^{5} - 3 \, a^{2} b^{4} c^{4} + 3 \, a^{4} b^{2} c^{3} - a^{6} c^{2}\right )} x^{2}}\right ] \] Input:

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

Output:

[1/8*(16*a*b^4*c^2*d^2*x^2*log(sqrt(d*x + c)*b + a) - 8*a*b^4*c^2*d^2*x^2* 
log(x) + 4*a*b^4*c^4 - 8*a^3*b^2*c^3 + 4*a^5*c^2 + (3*b^5*c^2 + 6*a^2*b^3* 
c - a^4*b)*sqrt(c)*d^2*x^2*log((d*x - 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x) - 
8*(a*b^4*c^3 - a^3*b^2*c^2)*d*x - 2*(2*b^5*c^4 - 4*a^2*b^3*c^3 + 2*a^4*b*c 
^2 - (3*b^5*c^3 - 2*a^2*b^3*c^2 - a^4*b*c)*d*x)*sqrt(d*x + c))/((b^6*c^5 - 
 3*a^2*b^4*c^4 + 3*a^4*b^2*c^3 - a^6*c^2)*x^2), 1/4*(8*a*b^4*c^2*d^2*x^2*l 
og(sqrt(d*x + c)*b + a) - 4*a*b^4*c^2*d^2*x^2*log(x) + 2*a*b^4*c^4 - 4*a^3 
*b^2*c^3 + 2*a^5*c^2 + (3*b^5*c^2 + 6*a^2*b^3*c - a^4*b)*sqrt(-c)*d^2*x^2* 
arctan(sqrt(-c)/sqrt(d*x + c)) - 4*(a*b^4*c^3 - a^3*b^2*c^2)*d*x - (2*b^5* 
c^4 - 4*a^2*b^3*c^3 + 2*a^4*b*c^2 - (3*b^5*c^3 - 2*a^2*b^3*c^2 - a^4*b*c)* 
d*x)*sqrt(d*x + c))/((b^6*c^5 - 3*a^2*b^4*c^4 + 3*a^4*b^2*c^3 - a^6*c^2)*x 
^2)]
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 367, normalized size of antiderivative = 1.80 \[ \int \frac {1}{x^3 \left (a+b \sqrt {c+d x}\right )} \, dx=-\frac {1}{8} \, {\left (\frac {8 \, a b^{4} \log \left (d x\right )}{b^{6} c^{3} - 3 \, a^{2} b^{4} c^{2} + 3 \, a^{4} b^{2} c - a^{6}} - \frac {16 \, a b^{4} \log \left (\sqrt {d x + c} b + a\right )}{b^{6} c^{3} - 3 \, a^{2} b^{4} c^{2} + 3 \, a^{4} b^{2} c - a^{6}} - \frac {{\left (3 \, b^{5} c^{2} + 6 \, a^{2} b^{3} c - a^{4} b\right )} \log \left (\frac {\sqrt {d x + c} - \sqrt {c}}{\sqrt {d x + c} + \sqrt {c}}\right )}{{\left (b^{6} c^{4} - 3 \, a^{2} b^{4} c^{3} + 3 \, a^{4} b^{2} c^{2} - a^{6} c\right )} \sqrt {c}} + \frac {2 \, {\left (4 \, {\left (d x + c\right )} a b^{2} c - 6 \, a b^{2} c^{2} + 2 \, a^{3} c - {\left (3 \, b^{3} c + a^{2} b\right )} {\left (d x + c\right )}^{\frac {3}{2}} + {\left (5 \, b^{3} c^{2} - a^{2} b c\right )} \sqrt {d x + c}\right )}}{b^{4} c^{5} - 2 \, a^{2} b^{2} c^{4} + a^{4} c^{3} + {\left (b^{4} c^{3} - 2 \, a^{2} b^{2} c^{2} + a^{4} c\right )} {\left (d x + c\right )}^{2} - 2 \, {\left (b^{4} c^{4} - 2 \, a^{2} b^{2} c^{3} + a^{4} c^{2}\right )} {\left (d x + c\right )}}\right )} d^{2} \] Input:

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

Output:

-1/8*(8*a*b^4*log(d*x)/(b^6*c^3 - 3*a^2*b^4*c^2 + 3*a^4*b^2*c - a^6) - 16* 
a*b^4*log(sqrt(d*x + c)*b + a)/(b^6*c^3 - 3*a^2*b^4*c^2 + 3*a^4*b^2*c - a^ 
6) - (3*b^5*c^2 + 6*a^2*b^3*c - a^4*b)*log((sqrt(d*x + c) - sqrt(c))/(sqrt 
(d*x + c) + sqrt(c)))/((b^6*c^4 - 3*a^2*b^4*c^3 + 3*a^4*b^2*c^2 - a^6*c)*s 
qrt(c)) + 2*(4*(d*x + c)*a*b^2*c - 6*a*b^2*c^2 + 2*a^3*c - (3*b^3*c + a^2* 
b)*(d*x + c)^(3/2) + (5*b^3*c^2 - a^2*b*c)*sqrt(d*x + c))/(b^4*c^5 - 2*a^2 
*b^2*c^4 + a^4*c^3 + (b^4*c^3 - 2*a^2*b^2*c^2 + a^4*c)*(d*x + c)^2 - 2*(b^ 
4*c^4 - 2*a^2*b^2*c^3 + a^4*c^2)*(d*x + c)))*d^2
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 375, normalized size of antiderivative = 1.84 \[ \int \frac {1}{x^3 \left (a+b \sqrt {c+d x}\right )} \, dx=\frac {2 \, a b^{5} d^{2} \log \left ({\left | \sqrt {d x + c} b + a \right |}\right )}{b^{7} c^{3} - 3 \, a^{2} b^{5} c^{2} + 3 \, a^{4} b^{3} c - a^{6} b} - \frac {a b^{4} d^{2} \log \left (d x\right )}{b^{6} c^{3} - 3 \, a^{2} b^{4} c^{2} + 3 \, a^{4} b^{2} c - a^{6}} + \frac {{\left (3 \, b^{5} c^{2} d^{2} + 6 \, a^{2} b^{3} c d^{2} - a^{4} b d^{2}\right )} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{4 \, {\left (b^{6} c^{4} - 3 \, a^{2} b^{4} c^{3} + 3 \, a^{4} b^{2} c^{2} - a^{6} c\right )} \sqrt {-c}} + \frac {6 \, a b^{4} c^{3} d^{2} - 8 \, a^{3} b^{2} c^{2} d^{2} + 2 \, a^{5} c d^{2} + {\left (3 \, b^{5} c^{2} d^{2} - 2 \, a^{2} b^{3} c d^{2} - a^{4} b d^{2}\right )} {\left (d x + c\right )}^{\frac {3}{2}} - 4 \, {\left (a b^{4} c^{2} d^{2} - a^{3} b^{2} c d^{2}\right )} {\left (d x + c\right )} - {\left (5 \, b^{5} c^{3} d^{2} - 6 \, a^{2} b^{3} c^{2} d^{2} + a^{4} b c d^{2}\right )} \sqrt {d x + c}}{4 \, {\left (b^{2} c - a^{2}\right )}^{3} c d^{2} x^{2}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 11.25 (sec) , antiderivative size = 1094, normalized size of antiderivative = 5.36 \[ \int \frac {1}{x^3 \left (a+b \sqrt {c+d x}\right )} \, dx =\text {Too large to display} \] Input:

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

Output:

(log((b^5*d^4*(3*b^2*c + a^2)^2*(c + d*x)^(1/2))/(16*c^2*(b^2*c - a^2)^4) 
- (a*b^4*d^4*(15*b^4*c^2 - a^4 + 2*a^2*b^2*c))/(16*c^2*(b^2*c - a^2)^4) - 
(b*d^2*(c^3)^(1/2)*((b^2*d^2*(3*b^2*c - a^2))/(4*c*(b^2*c - a^2)) + (b^2*d 
^2*(c^3)^(1/2)*(a^2*(c + d*x)^(1/2) + 4*a*b*c + 3*b^2*c*(c + d*x)^(1/2))*( 
3*b^4*c^2 - a^4 + 6*a^2*b^2*c + 8*a*b^3*(c^3)^(1/2)))/(4*c^3*(b^2*c - a^2) 
^3) - (a*b^3*d^2*(9*b^2*c - a^2)*(c + d*x)^(1/2))/(2*c*(b^2*c - a^2)^2))*( 
3*b^4*c^2 - a^4 + 6*a^2*b^2*c + 8*a*b^3*(c^3)^(1/2)))/(8*c^3*(b^2*c - a^2) 
^3))*(8*a*b^4*c^3*d^2 - a^4*b*d^2*(c^3)^(1/2) + 3*b^5*c^2*d^2*(c^3)^(1/2) 
+ 6*a^2*b^3*c*d^2*(c^3)^(1/2)))/(8*(a^6*c^3 - b^6*c^6 - 3*a^4*b^2*c^4 + 3* 
a^2*b^4*c^5)) - ((a^3*d^2 - 3*a*b^2*c*d^2)/(2*(a^4 + b^4*c^2 - 2*a^2*b^2*c 
)) - ((a^2*b*d^2 + 3*b^3*c*d^2)*(c + d*x)^(3/2))/(4*c*(a^4 + b^4*c^2 - 2*a 
^2*b^2*c)) + (b*d^2*(5*b^2*c - a^2)*(c + d*x)^(1/2))/(4*(a^4 + b^4*c^2 - 2 
*a^2*b^2*c)) + (a*b^2*d^2*(c + d*x))/(a^4 + b^4*c^2 - 2*a^2*b^2*c))/((c + 
d*x)^2 - 2*c*(c + d*x) + c^2) + (log((b^5*d^4*(3*b^2*c + a^2)^2*(c + d*x)^ 
(1/2))/(16*c^2*(b^2*c - a^2)^4) - (a*b^4*d^4*(15*b^4*c^2 - a^4 + 2*a^2*b^2 
*c))/(16*c^2*(b^2*c - a^2)^4) - (b*d^2*(c^3)^(1/2)*((b^2*d^2*(3*b^2*c - a^ 
2))/(4*c*(b^2*c - a^2)) + (b^2*d^2*(c^3)^(1/2)*(a^2*(c + d*x)^(1/2) + 4*a* 
b*c + 3*b^2*c*(c + d*x)^(1/2))*(a^4 - 3*b^4*c^2 - 6*a^2*b^2*c + 8*a*b^3*(c 
^3)^(1/2)))/(4*c^3*(b^2*c - a^2)^3) - (a*b^3*d^2*(9*b^2*c - a^2)*(c + d*x) 
^(1/2))/(2*c*(b^2*c - a^2)^2))*(a^4 - 3*b^4*c^2 - 6*a^2*b^2*c + 8*a*b^3...
 

Reduce [B] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 458, normalized size of antiderivative = 2.25 \[ \int \frac {1}{x^3 \left (a+b \sqrt {c+d x}\right )} \, dx=\frac {4 \sqrt {d x +c}\, a^{4} b \,c^{2}+2 \sqrt {d x +c}\, a^{4} b c d x -8 \sqrt {d x +c}\, a^{2} b^{3} c^{3}+4 \sqrt {d x +c}\, a^{2} b^{3} c^{2} d x +4 \sqrt {d x +c}\, b^{5} c^{4}-6 \sqrt {d x +c}\, b^{5} c^{3} d x +\sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) a^{4} b \,d^{2} x^{2}-6 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) a^{2} b^{3} c \,d^{2} x^{2}-3 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) b^{5} c^{2} d^{2} x^{2}-\sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) a^{4} b \,d^{2} x^{2}+6 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) a^{2} b^{3} c \,d^{2} x^{2}+3 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) b^{5} c^{2} d^{2} x^{2}+8 \,\mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) a \,b^{4} c^{2} d^{2} x^{2}+8 \,\mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) a \,b^{4} c^{2} d^{2} x^{2}-16 \,\mathrm {log}\left (\sqrt {d x +c}\, b +a \right ) a \,b^{4} c^{2} d^{2} x^{2}-4 a^{5} c^{2}+8 a^{3} b^{2} c^{3}-8 a^{3} b^{2} c^{2} d x -4 a^{3} b^{2} c \,d^{2} x^{2}-4 a \,b^{4} c^{4}+8 a \,b^{4} c^{3} d x +4 a \,b^{4} c^{2} d^{2} x^{2}}{8 c^{2} x^{2} \left (-b^{6} c^{3}+3 a^{2} b^{4} c^{2}-3 a^{4} b^{2} c +a^{6}\right )} \] Input:

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

Output:

(4*sqrt(c + d*x)*a**4*b*c**2 + 2*sqrt(c + d*x)*a**4*b*c*d*x - 8*sqrt(c + d 
*x)*a**2*b**3*c**3 + 4*sqrt(c + d*x)*a**2*b**3*c**2*d*x + 4*sqrt(c + d*x)* 
b**5*c**4 - 6*sqrt(c + d*x)*b**5*c**3*d*x + sqrt(c)*log(sqrt(c + d*x) - sq 
rt(c))*a**4*b*d**2*x**2 - 6*sqrt(c)*log(sqrt(c + d*x) - sqrt(c))*a**2*b**3 
*c*d**2*x**2 - 3*sqrt(c)*log(sqrt(c + d*x) - sqrt(c))*b**5*c**2*d**2*x**2 
- sqrt(c)*log(sqrt(c + d*x) + sqrt(c))*a**4*b*d**2*x**2 + 6*sqrt(c)*log(sq 
rt(c + d*x) + sqrt(c))*a**2*b**3*c*d**2*x**2 + 3*sqrt(c)*log(sqrt(c + d*x) 
 + sqrt(c))*b**5*c**2*d**2*x**2 + 8*log(sqrt(c + d*x) - sqrt(c))*a*b**4*c* 
*2*d**2*x**2 + 8*log(sqrt(c + d*x) + sqrt(c))*a*b**4*c**2*d**2*x**2 - 16*l 
og(sqrt(c + d*x)*b + a)*a*b**4*c**2*d**2*x**2 - 4*a**5*c**2 + 8*a**3*b**2* 
c**3 - 8*a**3*b**2*c**2*d*x - 4*a**3*b**2*c*d**2*x**2 - 4*a*b**4*c**4 + 8* 
a*b**4*c**3*d*x + 4*a*b**4*c**2*d**2*x**2)/(8*c**2*x**2*(a**6 - 3*a**4*b** 
2*c + 3*a**2*b**4*c**2 - b**6*c**3))