\(\int \frac {\sqrt {a+\frac {b}{c+d x}}}{x^4} \, dx\) [19]

Optimal result

Integrand size = 19, antiderivative size = 212 \[ \int \frac {\sqrt {a+\frac {b}{c+d x}}}{x^4} \, dx=-\frac {\left (11 b^2+20 a b c+8 a^2 c^2\right ) d^2 (c+d x) \sqrt {a+\frac {b}{c+d x}}}{8 c^3 (b+a c)^2 x}+\frac {(13 b+12 a c) d (c+d x)^2 \sqrt {a+\frac {b}{c+d x}}}{12 c^3 (b+a c) x^2}-\frac {(c+d x)^3 \sqrt {a+\frac {b}{c+d x}}}{3 c^3 x^3}+\frac {b \left (5 b^2+12 a b c+8 a^2 c^2\right ) d^3 \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+\frac {b}{c+d x}}}{\sqrt {b+a c}}\right )}{8 c^{7/2} (b+a c)^{5/2}} \] Output:

-1/8*(8*a^2*c^2+20*a*b*c+11*b^2)*d^2*(d*x+c)*(a+b/(d*x+c))^(1/2)/c^3/(a*c+ 
b)^2/x+1/12*(12*a*c+13*b)*d*(d*x+c)^2*(a+b/(d*x+c))^(1/2)/c^3/(a*c+b)/x^2- 
1/3*(d*x+c)^3*(a+b/(d*x+c))^(1/2)/c^3/x^3+1/8*b*(8*a^2*c^2+12*a*b*c+5*b^2) 
*d^3*arctanh(c^(1/2)*(a+b/(d*x+c))^(1/2)/(a*c+b)^(1/2))/c^(7/2)/(a*c+b)^(5 
/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.57 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.94 \[ \int \frac {\sqrt {a+\frac {b}{c+d x}}}{x^4} \, dx=-\frac {(c+d x) \sqrt {\frac {b+a c+a d x}{c+d x}} \left (8 a^2 c^2 \left (c^2-c d x+d^2 x^2\right )+2 a b c \left (8 c^2-9 c d x+13 d^2 x^2\right )+b^2 \left (8 c^2-10 c d x+15 d^2 x^2\right )\right )}{24 c^3 (b+a c)^2 x^3}-\frac {b \left (5 b^2+12 a b c+8 a^2 c^2\right ) d^3 \arctan \left (\frac {\sqrt {c} \sqrt {\frac {b+a c+a d x}{c+d x}}}{\sqrt {-b-a c}}\right )}{8 c^{7/2} (-b-a c)^{5/2}} \] Input:

Integrate[Sqrt[a + b/(c + d*x)]/x^4,x]
 

Output:

-1/24*((c + d*x)*Sqrt[(b + a*c + a*d*x)/(c + d*x)]*(8*a^2*c^2*(c^2 - c*d*x 
 + d^2*x^2) + 2*a*b*c*(8*c^2 - 9*c*d*x + 13*d^2*x^2) + b^2*(8*c^2 - 10*c*d 
*x + 15*d^2*x^2)))/(c^3*(b + a*c)^2*x^3) - (b*(5*b^2 + 12*a*b*c + 8*a^2*c^ 
2)*d^3*ArcTan[(Sqrt[c]*Sqrt[(b + a*c + a*d*x)/(c + d*x)])/Sqrt[-b - a*c]]) 
/(8*c^(7/2)*(-b - a*c)^(5/2))
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 224, 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, 941, 948, 100, 27, 87, 51, 73, 221}

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.

\(\displaystyle \int \frac {\sqrt {a+\frac {b}{c+d x}}}{x^4} \, dx\)

\(\Big \downarrow \) 896

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

\(\Big \downarrow \) 941

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

\(\Big \downarrow \) 948

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

\(\Big \downarrow \) 100

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 87

\(\displaystyle -d^3 \left (\frac {\left (a+\frac {b}{c+d x}\right )^{3/2}}{3 c^2 (a c+b) \left (1-\frac {c}{c+d x}\right )^3}-\frac {\frac {(4 a c+3 b) \left (a+\frac {b}{c+d x}\right )^{3/2}}{2 (a c+b) \left (1-\frac {c}{c+d x}\right )^2}-\frac {\left (8 a^2 c^2+12 a b c+5 b^2\right ) \int \frac {\sqrt {a+\frac {b}{c+d x}}}{\left (1-\frac {c}{c+d x}\right )^2}d\frac {1}{c+d x}}{4 (a c+b)}}{2 c^2 (a c+b)}\right )\)

\(\Big \downarrow \) 51

\(\displaystyle -d^3 \left (\frac {\left (a+\frac {b}{c+d x}\right )^{3/2}}{3 c^2 (a c+b) \left (1-\frac {c}{c+d x}\right )^3}-\frac {\frac {(4 a c+3 b) \left (a+\frac {b}{c+d x}\right )^{3/2}}{2 (a c+b) \left (1-\frac {c}{c+d x}\right )^2}-\frac {\left (8 a^2 c^2+12 a b c+5 b^2\right ) \left (\frac {\sqrt {a+\frac {b}{c+d x}}}{c \left (1-\frac {c}{c+d x}\right )}-\frac {b \int \frac {1}{\sqrt {a+\frac {b}{c+d x}} \left (1-\frac {c}{c+d x}\right )}d\frac {1}{c+d x}}{2 c}\right )}{4 (a c+b)}}{2 c^2 (a c+b)}\right )\)

\(\Big \downarrow \) 73

\(\displaystyle -d^3 \left (\frac {\left (a+\frac {b}{c+d x}\right )^{3/2}}{3 c^2 (a c+b) \left (1-\frac {c}{c+d x}\right )^3}-\frac {\frac {(4 a c+3 b) \left (a+\frac {b}{c+d x}\right )^{3/2}}{2 (a c+b) \left (1-\frac {c}{c+d x}\right )^2}-\frac {\left (8 a^2 c^2+12 a b c+5 b^2\right ) \left (\frac {\sqrt {a+\frac {b}{c+d x}}}{c \left (1-\frac {c}{c+d x}\right )}-\frac {\int \frac {1}{\frac {a c}{b}-\frac {c}{b (c+d x)^2}+1}d\sqrt {a+\frac {b}{c+d x}}}{c}\right )}{4 (a c+b)}}{2 c^2 (a c+b)}\right )\)

\(\Big \downarrow \) 221

\(\displaystyle -d^3 \left (\frac {\left (a+\frac {b}{c+d x}\right )^{3/2}}{3 c^2 (a c+b) \left (1-\frac {c}{c+d x}\right )^3}-\frac {\frac {(4 a c+3 b) \left (a+\frac {b}{c+d x}\right )^{3/2}}{2 (a c+b) \left (1-\frac {c}{c+d x}\right )^2}-\frac {\left (8 a^2 c^2+12 a b c+5 b^2\right ) \left (\frac {\sqrt {a+\frac {b}{c+d x}}}{c \left (1-\frac {c}{c+d x}\right )}-\frac {b \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+\frac {b}{c+d x}}}{\sqrt {a c+b}}\right )}{c^{3/2} \sqrt {a c+b}}\right )}{4 (a c+b)}}{2 c^2 (a c+b)}\right )\)

Input:

Int[Sqrt[a + b/(c + d*x)]/x^4,x]
 

Output:

-(d^3*((a + b/(c + d*x))^(3/2)/(3*c^2*(b + a*c)*(1 - c/(c + d*x))^3) - ((( 
3*b + 4*a*c)*(a + b/(c + d*x))^(3/2))/(2*(b + a*c)*(1 - c/(c + d*x))^2) - 
((5*b^2 + 12*a*b*c + 8*a^2*c^2)*(Sqrt[a + b/(c + d*x)]/(c*(1 - c/(c + d*x) 
)) - (b*ArcTanh[(Sqrt[c]*Sqrt[a + b/(c + d*x)])/Sqrt[b + a*c]])/(c^(3/2)*S 
qrt[b + a*c])))/(4*(b + a*c)))/(2*c^2*(b + a*c))))
 

Defintions of rubi rules used

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 51

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

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 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

rule 100

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( 
p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d 
*e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1))   Int[(c + d*x)^ 
(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( 
p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x 
, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n 
 + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))
 

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 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 941

Int[((c_) + (d_.)*(x_)^(mn_.))^(q_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Sym 
bol] :> Int[(a + b*x^n)^p*((d + c*x^n)^q/x^(n*q)), x] /; FreeQ[{a, b, c, d, 
 n, p}, x] && EqQ[mn, -n] && IntegerQ[q] && (PosQ[n] ||  !IntegerQ[p])
 

rule 948

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. 
), x_Symbol] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^ 
p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ 
[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(2155\) vs. \(2(188)=376\).

Time = 0.14 (sec) , antiderivative size = 2156, normalized size of antiderivative = 10.17

method	result	size
default	\(\text {Expression too large to display}\)	\(2156\)

Input:

int((a+b/(d*x+c))^(1/2)/x^4,x,method=_RETURNVERBOSE)
 

Output:

1/48*((a*d*x+a*c+b)/(d*x+c))^(1/2)*(d*x+c)*(-72*ln(1/2*(2*a*d^2*x+2*a*c*d+ 
2*(a*d^2*x^2+2*a*c*d*x+a*c^2+b*d*x+b*c)^(1/2)*(a*d^2)^(1/2)+b*d)/(a*d^2)^( 
1/2))*((a*c+b)*c)^(1/2)*a*b^3*c*d^4*x^3+51*(a*d^2)^(1/2)*ln((2*a*d*x*c+2*a 
*c^2+b*d*x+2*((a*c+b)*c)^(1/2)*(a*d^2*x^2+2*a*c*d*x+a*c^2+b*d*x+b*c)^(1/2) 
+2*b*c)/x)*a*b^3*c^2*d^3*x^3+72*((a*c+b)*c)^(1/2)*ln(1/2*(2*a*d^2*x+2*a*c* 
d+2*((a*d*x+a*c+b)*(d*x+c))^(1/2)*(a*d^2)^(1/2)+b*d)/(a*d^2)^(1/2))*a*b^3* 
c*d^4*x^3-96*(a*d^2*x^2+2*a*c*d*x+a*c^2+b*d*x+b*c)^(3/2)*(a*d^2)^(1/2)*((a 
*c+b)*c)^(1/2)*a^2*c^2*d^2*x^2+48*(a*d^2*x^2+2*a*c*d*x+a*c^2+b*d*x+b*c)^(3 
/2)*(a*d^2)^(1/2)*((a*c+b)*c)^(1/2)*a^2*c^3*d*x+36*(a*d^2*x^2+2*a*c*d*x+a* 
c^2+b*d*x+b*c)^(3/2)*(a*d^2)^(1/2)*((a*c+b)*c)^(1/2)*b^2*c*d*x-66*(a*d^2*x 
^2+2*a*c*d*x+a*c^2+b*d*x+b*c)^(3/2)*(a*d^2)^(1/2)*((a*c+b)*c)^(1/2)*b^2*d^ 
2*x^2-32*(a*d^2*x^2+2*a*c*d*x+a*c^2+b*d*x+b*c)^(3/2)*(a*d^2)^(1/2)*((a*c+b 
)*c)^(1/2)*a*b*c^3+96*(a*d^2*x^2+2*a*c*d*x+a*c^2+b*d*x+b*c)^(1/2)*(a*d^2)^ 
(1/2)*((a*c+b)*c)^(1/2)*a^3*c^2*d^4*x^4+15*(a*d^2)^(1/2)*ln((2*a*d*x*c+2*a 
*c^2+b*d*x+2*((a*c+b)*c)^(1/2)*(a*d^2*x^2+2*a*c*d*x+a*c^2+b*d*x+b*c)^(1/2) 
+2*b*c)/x)*b^4*c*d^3*x^3-72*ln(1/2*(2*a*d^2*x+2*a*c*d+2*(a*d^2*x^2+2*a*c*d 
*x+a*c^2+b*d*x+b*c)^(1/2)*(a*d^2)^(1/2)+b*d)/(a*d^2)^(1/2))*((a*c+b)*c)^(1 
/2)*a^2*b^2*c^2*d^4*x^3+60*(a*d^2)^(1/2)*ln((2*a*d*x*c+2*a*c^2+b*d*x+2*((a 
*c+b)*c)^(1/2)*(a*d^2*x^2+2*a*c*d*x+a*c^2+b*d*x+b*c)^(1/2)+2*b*c)/x)*a^2*b 
^2*c^3*d^3*x^3+72*((a*c+b)*c)^(1/2)*ln(1/2*(2*a*d^2*x+2*a*c*d+2*((a*d*x...
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 602, normalized size of antiderivative = 2.84 \[ \int \frac {\sqrt {a+\frac {b}{c+d x}}}{x^4} \, dx=\left [\frac {3 \, {\left (8 \, a^{2} b c^{2} + 12 \, a b^{2} c + 5 \, b^{3}\right )} \sqrt {a c^{2} + b c} d^{3} x^{3} \log \left (-\frac {2 \, a c^{2} + {\left (2 \, a c + b\right )} d x + 2 \, b c + 2 \, \sqrt {a c^{2} + b c} {\left (d x + c\right )} \sqrt {\frac {a d x + a c + b}{d x + c}}}{x}\right ) - 2 \, {\left (8 \, a^{3} c^{7} + 24 \, a^{2} b c^{6} + 24 \, a b^{2} c^{5} + 8 \, b^{3} c^{4} + {\left (8 \, a^{3} c^{4} + 34 \, a^{2} b c^{3} + 41 \, a b^{2} c^{2} + 15 \, b^{3} c\right )} d^{3} x^{3} + {\left (8 \, a^{2} b c^{4} + 13 \, a b^{2} c^{3} + 5 \, b^{3} c^{2}\right )} d^{2} x^{2} - 2 \, {\left (a^{2} b c^{5} + 2 \, a b^{2} c^{4} + b^{3} c^{3}\right )} d x\right )} \sqrt {\frac {a d x + a c + b}{d x + c}}}{48 \, {\left (a^{3} c^{7} + 3 \, a^{2} b c^{6} + 3 \, a b^{2} c^{5} + b^{3} c^{4}\right )} x^{3}}, -\frac {3 \, {\left (8 \, a^{2} b c^{2} + 12 \, a b^{2} c + 5 \, b^{3}\right )} \sqrt {-a c^{2} - b c} d^{3} x^{3} \arctan \left (\frac {\sqrt {-a c^{2} - b c} {\left (d x + c\right )} \sqrt {\frac {a d x + a c + b}{d x + c}}}{a c d x + a c^{2} + b c}\right ) + {\left (8 \, a^{3} c^{7} + 24 \, a^{2} b c^{6} + 24 \, a b^{2} c^{5} + 8 \, b^{3} c^{4} + {\left (8 \, a^{3} c^{4} + 34 \, a^{2} b c^{3} + 41 \, a b^{2} c^{2} + 15 \, b^{3} c\right )} d^{3} x^{3} + {\left (8 \, a^{2} b c^{4} + 13 \, a b^{2} c^{3} + 5 \, b^{3} c^{2}\right )} d^{2} x^{2} - 2 \, {\left (a^{2} b c^{5} + 2 \, a b^{2} c^{4} + b^{3} c^{3}\right )} d x\right )} \sqrt {\frac {a d x + a c + b}{d x + c}}}{24 \, {\left (a^{3} c^{7} + 3 \, a^{2} b c^{6} + 3 \, a b^{2} c^{5} + b^{3} c^{4}\right )} x^{3}}\right ] \] Input:

integrate((a+b/(d*x+c))^(1/2)/x^4,x, algorithm="fricas")
 

Output:

[1/48*(3*(8*a^2*b*c^2 + 12*a*b^2*c + 5*b^3)*sqrt(a*c^2 + b*c)*d^3*x^3*log( 
-(2*a*c^2 + (2*a*c + b)*d*x + 2*b*c + 2*sqrt(a*c^2 + b*c)*(d*x + c)*sqrt(( 
a*d*x + a*c + b)/(d*x + c)))/x) - 2*(8*a^3*c^7 + 24*a^2*b*c^6 + 24*a*b^2*c 
^5 + 8*b^3*c^4 + (8*a^3*c^4 + 34*a^2*b*c^3 + 41*a*b^2*c^2 + 15*b^3*c)*d^3* 
x^3 + (8*a^2*b*c^4 + 13*a*b^2*c^3 + 5*b^3*c^2)*d^2*x^2 - 2*(a^2*b*c^5 + 2* 
a*b^2*c^4 + b^3*c^3)*d*x)*sqrt((a*d*x + a*c + b)/(d*x + c)))/((a^3*c^7 + 3 
*a^2*b*c^6 + 3*a*b^2*c^5 + b^3*c^4)*x^3), -1/24*(3*(8*a^2*b*c^2 + 12*a*b^2 
*c + 5*b^3)*sqrt(-a*c^2 - b*c)*d^3*x^3*arctan(sqrt(-a*c^2 - b*c)*(d*x + c) 
*sqrt((a*d*x + a*c + b)/(d*x + c))/(a*c*d*x + a*c^2 + b*c)) + (8*a^3*c^7 + 
 24*a^2*b*c^6 + 24*a*b^2*c^5 + 8*b^3*c^4 + (8*a^3*c^4 + 34*a^2*b*c^3 + 41* 
a*b^2*c^2 + 15*b^3*c)*d^3*x^3 + (8*a^2*b*c^4 + 13*a*b^2*c^3 + 5*b^3*c^2)*d 
^2*x^2 - 2*(a^2*b*c^5 + 2*a*b^2*c^4 + b^3*c^3)*d*x)*sqrt((a*d*x + a*c + b) 
/(d*x + c)))/((a^3*c^7 + 3*a^2*b*c^6 + 3*a*b^2*c^5 + b^3*c^4)*x^3)]
 

Sympy [F]

\[ \int \frac {\sqrt {a+\frac {b}{c+d x}}}{x^4} \, dx=\int \frac {\sqrt {\frac {a c + a d x + b}{c + d x}}}{x^{4}}\, dx \] Input:

integrate((a+b/(d*x+c))**(1/2)/x**4,x)
 

Output:

Integral(sqrt((a*c + a*d*x + b)/(c + d*x))/x**4, x)
 

Maxima [F]

\[ \int \frac {\sqrt {a+\frac {b}{c+d x}}}{x^4} \, dx=\int { \frac {\sqrt {a + \frac {b}{d x + c}}}{x^{4}} \,d x } \] Input:

integrate((a+b/(d*x+c))^(1/2)/x^4,x, algorithm="maxima")
 

Output:

integrate(sqrt(a + b/(d*x + c))/x^4, x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1447 vs. \(2 (188) = 376\).

Time = 0.18 (sec) , antiderivative size = 1447, normalized size of antiderivative = 6.83 \[ \int \frac {\sqrt {a+\frac {b}{c+d x}}}{x^4} \, dx=\text {Too large to display} \] Input:

integrate((a+b/(d*x+c))^(1/2)/x^4,x, algorithm="giac")
 

Output:

-1/8*(8*a^2*b*c^2*d^3*sgn(d*x + c) + 12*a*b^2*c*d^3*sgn(d*x + c) + 5*b^3*d 
^3*sgn(d*x + c))*arctan(-(sqrt(a*d^2)*x - sqrt(a*d^2*x^2 + 2*a*c*d*x + a*c 
^2 + b*d*x + b*c))/sqrt(-a*c^2 - b*c))/((a^2*c^5 + 2*a*b*c^4 + b^2*c^3)*sq 
rt(-a*c^2 - b*c)) - 1/24*(64*a^(11/2)*c^8*d^2*abs(d)*sgn(d*x + c) + 192*(s 
qrt(a*d^2)*x - sqrt(a*d^2*x^2 + 2*a*c*d*x + a*c^2 + b*d*x + b*c))*a^5*c^7* 
d^3*sgn(d*x + c) + 192*(sqrt(a*d^2)*x - sqrt(a*d^2*x^2 + 2*a*c*d*x + a*c^2 
 + b*d*x + b*c))^2*a^(9/2)*c^6*d^2*abs(d)*sgn(d*x + c) + 304*a^(9/2)*b*c^7 
*d^2*abs(d)*sgn(d*x + c) + 64*(sqrt(a*d^2)*x - sqrt(a*d^2*x^2 + 2*a*c*d*x 
+ a*c^2 + b*d*x + b*c))^3*a^4*c^5*d^3*sgn(d*x + c) + 744*(sqrt(a*d^2)*x - 
sqrt(a*d^2*x^2 + 2*a*c*d*x + a*c^2 + b*d*x + b*c))*a^4*b*c^6*d^3*sgn(d*x + 
 c) + 528*(sqrt(a*d^2)*x - sqrt(a*d^2*x^2 + 2*a*c*d*x + a*c^2 + b*d*x + b* 
c))^2*a^(7/2)*b*c^5*d^2*abs(d)*sgn(d*x + c) + 576*a^(7/2)*b^2*c^6*d^2*abs( 
d)*sgn(d*x + c) + 64*(sqrt(a*d^2)*x - sqrt(a*d^2*x^2 + 2*a*c*d*x + a*c^2 + 
 b*d*x + b*c))^3*a^3*b*c^4*d^3*sgn(d*x + c) + 1116*(sqrt(a*d^2)*x - sqrt(a 
*d^2*x^2 + 2*a*c*d*x + a*c^2 + b*d*x + b*c))*a^3*b^2*c^5*d^3*sgn(d*x + c) 
+ 480*(sqrt(a*d^2)*x - sqrt(a*d^2*x^2 + 2*a*c*d*x + a*c^2 + b*d*x + b*c))^ 
2*a^(5/2)*b^2*c^4*d^2*abs(d)*sgn(d*x + c) + 544*a^(5/2)*b^3*c^5*d^2*abs(d) 
*sgn(d*x + c) + 24*(sqrt(a*d^2)*x - sqrt(a*d^2*x^2 + 2*a*c*d*x + a*c^2 + b 
*d*x + b*c))^5*a^2*b*c^2*d^3*sgn(d*x + c) - 96*(sqrt(a*d^2)*x - sqrt(a*d^2 
*x^2 + 2*a*c*d*x + a*c^2 + b*d*x + b*c))^3*a^2*b^2*c^3*d^3*sgn(d*x + c)...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+\frac {b}{c+d x}}}{x^4} \, dx=\int \frac {\sqrt {a+\frac {b}{c+d\,x}}}{x^4} \,d x \] Input:

int((a + b/(c + d*x))^(1/2)/x^4,x)
 

Output:

int((a + b/(c + d*x))^(1/2)/x^4, x)
 

Reduce [B] (verification not implemented)

Time = 0.57 (sec) , antiderivative size = 1294, normalized size of antiderivative = 6.10 \[ \int \frac {\sqrt {a+\frac {b}{c+d x}}}{x^4} \, dx =\text {Too large to display} \] Input:

int((a+b/(d*x+c))^(1/2)/x^4,x)
 

Output:

( - 32*sqrt(c + d*x)*sqrt(a*c + a*d*x + b)*a**4*c**7 + 32*sqrt(c + d*x)*sq 
rt(a*c + a*d*x + b)*a**4*c**6*d*x - 32*sqrt(c + d*x)*sqrt(a*c + a*d*x + b) 
*a**4*c**5*d**2*x**2 - 112*sqrt(c + d*x)*sqrt(a*c + a*d*x + b)*a**3*b*c**6 
 + 120*sqrt(c + d*x)*sqrt(a*c + a*d*x + b)*a**3*b*c**5*d*x - 152*sqrt(c + 
d*x)*sqrt(a*c + a*d*x + b)*a**3*b*c**4*d**2*x**2 - 144*sqrt(c + d*x)*sqrt( 
a*c + a*d*x + b)*a**2*b**2*c**5 + 164*sqrt(c + d*x)*sqrt(a*c + a*d*x + b)* 
a**2*b**2*c**4*d*x - 232*sqrt(c + d*x)*sqrt(a*c + a*d*x + b)*a**2*b**2*c** 
3*d**2*x**2 - 80*sqrt(c + d*x)*sqrt(a*c + a*d*x + b)*a*b**3*c**4 + 96*sqrt 
(c + d*x)*sqrt(a*c + a*d*x + b)*a*b**3*c**3*d*x - 142*sqrt(c + d*x)*sqrt(a 
*c + a*d*x + b)*a*b**3*c**2*d**2*x**2 - 16*sqrt(c + d*x)*sqrt(a*c + a*d*x 
+ b)*b**4*c**3 + 20*sqrt(c + d*x)*sqrt(a*c + a*d*x + b)*b**4*c**2*d*x - 30 
*sqrt(c + d*x)*sqrt(a*c + a*d*x + b)*b**4*c*d**2*x**2 - 48*sqrt(c)*sqrt(a* 
c + b)*log(sqrt(a*c + a*d*x + b) - sqrt(2*sqrt(c)*sqrt(a)*sqrt(a*c + b) + 
2*a*c + b) + sqrt(a)*sqrt(c + d*x))*a**3*b*c**3*d**3*x**3 - 96*sqrt(c)*sqr 
t(a*c + b)*log(sqrt(a*c + a*d*x + b) - sqrt(2*sqrt(c)*sqrt(a)*sqrt(a*c + b 
) + 2*a*c + b) + sqrt(a)*sqrt(c + d*x))*a**2*b**2*c**2*d**3*x**3 - 66*sqrt 
(c)*sqrt(a*c + b)*log(sqrt(a*c + a*d*x + b) - sqrt(2*sqrt(c)*sqrt(a)*sqrt( 
a*c + b) + 2*a*c + b) + sqrt(a)*sqrt(c + d*x))*a*b**3*c*d**3*x**3 - 15*sqr 
t(c)*sqrt(a*c + b)*log(sqrt(a*c + a*d*x + b) - sqrt(2*sqrt(c)*sqrt(a)*sqrt 
(a*c + b) + 2*a*c + b) + sqrt(a)*sqrt(c + d*x))*b**4*d**3*x**3 - 48*sqr...