\(\int \frac {(a+\frac {b}{c+d x})^{3/2}}{x^4} \, dx\) [27]

Optimal result

Integrand size = 19, antiderivative size = 234 \[ \int \frac {\left (a+\frac {b}{c+d x}\right )^{3/2}}{x^4} \, dx=-\frac {2 b d^3 \sqrt {a+\frac {b}{c+d x}}}{c^4}-\frac {\left (29 b^2+36 a b c+8 a^2 c^2\right ) d^2 (c+d x) \sqrt {a+\frac {b}{c+d x}}}{8 c^4 (b+a c) x}+\frac {(19 b+12 a c) d (c+d x)^2 \sqrt {a+\frac {b}{c+d x}}}{12 c^4 x^2}-\frac {(b+a c) (c+d x)^3 \sqrt {a+\frac {b}{c+d x}}}{3 c^4 x^3}+\frac {b \left (35 b^2+60 a b c+24 a^2 c^2\right ) d^3 \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+\frac {b}{c+d x}}}{\sqrt {b+a c}}\right )}{8 c^{9/2} (b+a c)^{3/2}} \] Output:

-2*b*d^3*(a+b/(d*x+c))^(1/2)/c^4-1/8*(8*a^2*c^2+36*a*b*c+29*b^2)*d^2*(d*x+ 
c)*(a+b/(d*x+c))^(1/2)/c^4/(a*c+b)/x+1/12*(12*a*c+19*b)*d*(d*x+c)^2*(a+b/( 
d*x+c))^(1/2)/c^4/x^2-1/3*(a*c+b)*(d*x+c)^3*(a+b/(d*x+c))^(1/2)/c^4/x^3+1/ 
8*b*(24*a^2*c^2+60*a*b*c+35*b^2)*d^3*arctanh(c^(1/2)*(a+b/(d*x+c))^(1/2)/( 
a*c+b)^(1/2))/c^(9/2)/(a*c+b)^(3/2)
 

Mathematica [A] (verified)

Time = 0.52 (sec) , antiderivative size = 212, normalized size of antiderivative = 0.91 \[ \int \frac {\left (a+\frac {b}{c+d x}\right )^{3/2}}{x^4} \, dx=-\frac {\sqrt {\frac {b+a c+a d x}{c+d x}} \left (8 a^2 c^2 \left (c^3+d^3 x^3\right )+2 a b c \left (8 c^3-7 c^2 d x+16 c d^2 x^2+55 d^3 x^3\right )+b^2 \left (8 c^3-14 c^2 d x+35 c d^2 x^2+105 d^3 x^3\right )\right )}{24 c^4 (b+a c) x^3}+\frac {b \left (35 b^2+60 a b c+24 a^2 c^2\right ) d^3 \arctan \left (\frac {\sqrt {c} \sqrt {\frac {b+a c+a d x}{c+d x}}}{\sqrt {-b-a c}}\right )}{8 c^{9/2} (-b-a c)^{3/2}} \] Input:

Integrate[(a + b/(c + d*x))^(3/2)/x^4,x]
 

Output:

-1/24*(Sqrt[(b + a*c + a*d*x)/(c + d*x)]*(8*a^2*c^2*(c^3 + d^3*x^3) + 2*a* 
b*c*(8*c^3 - 7*c^2*d*x + 16*c*d^2*x^2 + 55*d^3*x^3) + b^2*(8*c^3 - 14*c^2* 
d*x + 35*c*d^2*x^2 + 105*d^3*x^3)))/(c^4*(b + a*c)*x^3) + (b*(35*b^2 + 60* 
a*b*c + 24*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^(9/2)*(-b - a*c)^(3/2))
 

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.08, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.526, Rules used = {896, 941, 948, 100, 27, 87, 51, 60, 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.

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

\(\Big \downarrow \) 896

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

\(\Big \downarrow \) 941

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

\(\Big \downarrow \) 948

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

\(\Big \downarrow \) 100

\(\displaystyle -d^3 \left (\frac {\left (a+\frac {b}{c+d x}\right )^{5/2}}{3 c^2 (a c+b) \left (1-\frac {c}{c+d x}\right )^3}-\frac {\int \frac {\left (a+\frac {b}{c+d x}\right )^{3/2} \left (5 b+6 a c+\frac {6 c (b+a c)}{c+d x}\right )}{2 \left (1-\frac {c}{c+d x}\right )^3}d\frac {1}{c+d x}}{3 c^2 (a c+b)}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle -d^3 \left (\frac {\left (a+\frac {b}{c+d x}\right )^{5/2}}{3 c^2 (a c+b) \left (1-\frac {c}{c+d x}\right )^3}-\frac {\int \frac {\left (a+\frac {b}{c+d x}\right )^{3/2} \left (5 b+6 a c+\frac {6 c (b+a c)}{c+d x}\right )}{\left (1-\frac {c}{c+d x}\right )^3}d\frac {1}{c+d x}}{6 c^2 (a c+b)}\right )\)

\(\Big \downarrow \) 87

\(\displaystyle -d^3 \left (\frac {\left (a+\frac {b}{c+d x}\right )^{5/2}}{3 c^2 (a c+b) \left (1-\frac {c}{c+d x}\right )^3}-\frac {\frac {(12 a c+11 b) \left (a+\frac {b}{c+d x}\right )^{5/2}}{2 (a c+b) \left (1-\frac {c}{c+d x}\right )^2}-\frac {\left (24 a^2 c^2+60 a b c+35 b^2\right ) \int \frac {\left (a+\frac {b}{c+d x}\right )^{3/2}}{\left (1-\frac {c}{c+d x}\right )^2}d\frac {1}{c+d x}}{4 (a c+b)}}{6 c^2 (a c+b)}\right )\)

\(\Big \downarrow \) 51

\(\displaystyle -d^3 \left (\frac {\left (a+\frac {b}{c+d x}\right )^{5/2}}{3 c^2 (a c+b) \left (1-\frac {c}{c+d x}\right )^3}-\frac {\frac {(12 a c+11 b) \left (a+\frac {b}{c+d x}\right )^{5/2}}{2 (a c+b) \left (1-\frac {c}{c+d x}\right )^2}-\frac {\left (24 a^2 c^2+60 a b c+35 b^2\right ) \left (\frac {\left (a+\frac {b}{c+d x}\right )^{3/2}}{c \left (1-\frac {c}{c+d x}\right )}-\frac {3 b \int \frac {\sqrt {a+\frac {b}{c+d x}}}{1-\frac {c}{c+d x}}d\frac {1}{c+d x}}{2 c}\right )}{4 (a c+b)}}{6 c^2 (a c+b)}\right )\)

\(\Big \downarrow \) 60

\(\displaystyle -d^3 \left (\frac {\left (a+\frac {b}{c+d x}\right )^{5/2}}{3 c^2 (a c+b) \left (1-\frac {c}{c+d x}\right )^3}-\frac {\frac {(12 a c+11 b) \left (a+\frac {b}{c+d x}\right )^{5/2}}{2 (a c+b) \left (1-\frac {c}{c+d x}\right )^2}-\frac {\left (24 a^2 c^2+60 a b c+35 b^2\right ) \left (\frac {\left (a+\frac {b}{c+d x}\right )^{3/2}}{c \left (1-\frac {c}{c+d x}\right )}-\frac {3 b \left (\frac {(a c+b) \int \frac {1}{\sqrt {a+\frac {b}{c+d x}} \left (1-\frac {c}{c+d x}\right )}d\frac {1}{c+d x}}{c}-\frac {2 \sqrt {a+\frac {b}{c+d x}}}{c}\right )}{2 c}\right )}{4 (a c+b)}}{6 c^2 (a c+b)}\right )\)

\(\Big \downarrow \) 73

\(\displaystyle -d^3 \left (\frac {\left (a+\frac {b}{c+d x}\right )^{5/2}}{3 c^2 (a c+b) \left (1-\frac {c}{c+d x}\right )^3}-\frac {\frac {(12 a c+11 b) \left (a+\frac {b}{c+d x}\right )^{5/2}}{2 (a c+b) \left (1-\frac {c}{c+d x}\right )^2}-\frac {\left (24 a^2 c^2+60 a b c+35 b^2\right ) \left (\frac {\left (a+\frac {b}{c+d x}\right )^{3/2}}{c \left (1-\frac {c}{c+d x}\right )}-\frac {3 b \left (\frac {2 (a c+b) \int \frac {1}{\frac {a c}{b}-\frac {c}{b (c+d x)^2}+1}d\sqrt {a+\frac {b}{c+d x}}}{b c}-\frac {2 \sqrt {a+\frac {b}{c+d x}}}{c}\right )}{2 c}\right )}{4 (a c+b)}}{6 c^2 (a c+b)}\right )\)

\(\Big \downarrow \) 221

\(\displaystyle -d^3 \left (\frac {\left (a+\frac {b}{c+d x}\right )^{5/2}}{3 c^2 (a c+b) \left (1-\frac {c}{c+d x}\right )^3}-\frac {\frac {(12 a c+11 b) \left (a+\frac {b}{c+d x}\right )^{5/2}}{2 (a c+b) \left (1-\frac {c}{c+d x}\right )^2}-\frac {\left (24 a^2 c^2+60 a b c+35 b^2\right ) \left (\frac {\left (a+\frac {b}{c+d x}\right )^{3/2}}{c \left (1-\frac {c}{c+d x}\right )}-\frac {3 b \left (\frac {2 \sqrt {a c+b} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+\frac {b}{c+d x}}}{\sqrt {a c+b}}\right )}{c^{3/2}}-\frac {2 \sqrt {a+\frac {b}{c+d x}}}{c}\right )}{2 c}\right )}{4 (a c+b)}}{6 c^2 (a c+b)}\right )\)

Input:

Int[(a + b/(c + d*x))^(3/2)/x^4,x]
 

Output:

-(d^3*((a + b/(c + d*x))^(5/2)/(3*c^2*(b + a*c)*(1 - c/(c + d*x))^3) - ((( 
11*b + 12*a*c)*(a + b/(c + d*x))^(5/2))/(2*(b + a*c)*(1 - c/(c + d*x))^2) 
- ((35*b^2 + 60*a*b*c + 24*a^2*c^2)*((a + b/(c + d*x))^(3/2)/(c*(1 - c/(c 
+ d*x))) - (3*b*((-2*Sqrt[a + b/(c + d*x)])/c + (2*Sqrt[b + a*c]*ArcTanh[( 
Sqrt[c]*Sqrt[a + b/(c + d*x)])/Sqrt[b + a*c]])/c^(3/2)))/(2*c)))/(4*(b + a 
*c)))/(6*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 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 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. \(5517\) vs. \(2(208)=416\).

Time = 0.16 (sec) , antiderivative size = 5518, normalized size of antiderivative = 23.58

method	result	size
default	\(\text {Expression too large to display}\)	\(5518\)

Input:

int((a+b/(d*x+c))^(3/2)/x^4,x,method=_RETURNVERBOSE)
                                                                                    
                                                                                    
 

Output:

result too large to display
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 580, normalized size of antiderivative = 2.48 \[ \int \frac {\left (a+\frac {b}{c+d x}\right )^{3/2}}{x^4} \, dx=\left [\frac {3 \, {\left (24 \, a^{2} b c^{2} + 60 \, a b^{2} c + 35 \, b^{3}\right )} \sqrt {a c^{2} + b c} d^{3} x^{3} \log \left (-\frac {2 \, a c^{2} + {\left (2 \, a c + b\right )} d x + 2 \, b c + 2 \, \sqrt {a c^{2} + b c} {\left (d x + c\right )} \sqrt {\frac {a d x + a c + b}{d x + c}}}{x}\right ) - 2 \, {\left (8 \, a^{3} c^{7} + 24 \, a^{2} b c^{6} + 24 \, a b^{2} c^{5} + 8 \, b^{3} c^{4} + {\left (8 \, a^{3} c^{4} + 118 \, a^{2} b c^{3} + 215 \, a b^{2} c^{2} + 105 \, b^{3} c\right )} d^{3} x^{3} + {\left (32 \, a^{2} b c^{4} + 67 \, a b^{2} c^{3} + 35 \, b^{3} c^{2}\right )} d^{2} x^{2} - 14 \, {\left (a^{2} b c^{5} + 2 \, a b^{2} c^{4} + b^{3} c^{3}\right )} d x\right )} \sqrt {\frac {a d x + a c + b}{d x + c}}}{48 \, {\left (a^{2} c^{7} + 2 \, a b c^{6} + b^{2} c^{5}\right )} x^{3}}, -\frac {3 \, {\left (24 \, a^{2} b c^{2} + 60 \, a b^{2} c + 35 \, b^{3}\right )} \sqrt {-a c^{2} - b c} d^{3} x^{3} \arctan \left (\frac {\sqrt {-a c^{2} - b c} {\left (d x + c\right )} \sqrt {\frac {a d x + a c + b}{d x + c}}}{a c d x + a c^{2} + b c}\right ) + {\left (8 \, a^{3} c^{7} + 24 \, a^{2} b c^{6} + 24 \, a b^{2} c^{5} + 8 \, b^{3} c^{4} + {\left (8 \, a^{3} c^{4} + 118 \, a^{2} b c^{3} + 215 \, a b^{2} c^{2} + 105 \, b^{3} c\right )} d^{3} x^{3} + {\left (32 \, a^{2} b c^{4} + 67 \, a b^{2} c^{3} + 35 \, b^{3} c^{2}\right )} d^{2} x^{2} - 14 \, {\left (a^{2} b c^{5} + 2 \, a b^{2} c^{4} + b^{3} c^{3}\right )} d x\right )} \sqrt {\frac {a d x + a c + b}{d x + c}}}{24 \, {\left (a^{2} c^{7} + 2 \, a b c^{6} + b^{2} c^{5}\right )} x^{3}}\right ] \] Input:

integrate((a+b/(d*x+c))^(3/2)/x^4,x, algorithm="fricas")
 

Output:

[1/48*(3*(24*a^2*b*c^2 + 60*a*b^2*c + 35*b^3)*sqrt(a*c^2 + b*c)*d^3*x^3*lo 
g(-(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 + 118*a^2*b*c^3 + 215*a*b^2*c^2 + 105*b^3*c) 
*d^3*x^3 + (32*a^2*b*c^4 + 67*a*b^2*c^3 + 35*b^3*c^2)*d^2*x^2 - 14*(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^2 
*c^7 + 2*a*b*c^6 + b^2*c^5)*x^3), -1/24*(3*(24*a^2*b*c^2 + 60*a*b^2*c + 35 
*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 + 118*a^2*b*c^3 + 215*a*b^ 
2*c^2 + 105*b^3*c)*d^3*x^3 + (32*a^2*b*c^4 + 67*a*b^2*c^3 + 35*b^3*c^2)*d^ 
2*x^2 - 14*(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^2*c^7 + 2*a*b*c^6 + b^2*c^5)*x^3)]
 

Sympy [F]

\[ \int \frac {\left (a+\frac {b}{c+d x}\right )^{3/2}}{x^4} \, dx=\int \frac {\left (\frac {a c + a d x + b}{c + d x}\right )^{\frac {3}{2}}}{x^{4}}\, dx \] Input:

integrate((a+b/(d*x+c))**(3/2)/x**4,x)
 

Output:

Integral(((a*c + a*d*x + b)/(c + d*x))**(3/2)/x**4, x)
 

Maxima [F]

\[ \int \frac {\left (a+\frac {b}{c+d x}\right )^{3/2}}{x^4} \, dx=\int { \frac {{\left (a + \frac {b}{d x + c}\right )}^{\frac {3}{2}}}{x^{4}} \,d x } \] Input:

integrate((a+b/(d*x+c))^(3/2)/x^4,x, algorithm="maxima")
 

Output:

integrate((a + b/(d*x + c))^(3/2)/x^4, x)
 

Giac [F]

\[ \int \frac {\left (a+\frac {b}{c+d x}\right )^{3/2}}{x^4} \, dx=\int { \frac {{\left (a + \frac {b}{d x + c}\right )}^{\frac {3}{2}}}{x^{4}} \,d x } \] Input:

integrate((a+b/(d*x+c))^(3/2)/x^4,x, algorithm="giac")
 

Output:

undef
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+\frac {b}{c+d x}\right )^{3/2}}{x^4} \, dx=\int \frac {{\left (a+\frac {b}{c+d\,x}\right )}^{3/2}}{x^4} \,d x \] Input:

int((a + b/(c + d*x))^(3/2)/x^4,x)
 

Output:

int((a + b/(c + d*x))^(3/2)/x^4, x)
 

Reduce [B] (verification not implemented)

Time = 19.73 (sec) , antiderivative size = 2266, normalized size of antiderivative = 9.68 \[ \int \frac {\left (a+\frac {b}{c+d x}\right )^{3/2}}{x^4} \, dx =\text {Too large to display} \] Input:

int((a+b/(d*x+c))^(3/2)/x^4,x)
 

Output:

( - 192*sqrt(c + d*x)*sqrt(a*c + a*d*x + b)*a**4*c**8 - 192*sqrt(c + d*x)* 
sqrt(a*c + a*d*x + b)*a**4*c**5*d**3*x**3 - 688*sqrt(c + d*x)*sqrt(a*c + a 
*d*x + b)*a**3*b*c**7 + 336*sqrt(c + d*x)*sqrt(a*c + a*d*x + b)*a**3*b*c** 
6*d*x - 768*sqrt(c + d*x)*sqrt(a*c + a*d*x + b)*a**3*b*c**5*d**2*x**2 - 29 
44*sqrt(c + d*x)*sqrt(a*c + a*d*x + b)*a**3*b*c**4*d**3*x**3 - 912*sqrt(c 
+ d*x)*sqrt(a*c + a*d*x + b)*a**2*b**2*c**6 + 868*sqrt(c + d*x)*sqrt(a*c + 
 a*d*x + b)*a**2*b**2*c**5*d*x - 2056*sqrt(c + d*x)*sqrt(a*c + a*d*x + b)* 
a**2*b**2*c**4*d**2*x**2 - 6812*sqrt(c + d*x)*sqrt(a*c + a*d*x + b)*a**2*b 
**2*c**3*d**3*x**3 - 528*sqrt(c + d*x)*sqrt(a*c + a*d*x + b)*a*b**3*c**5 + 
 728*sqrt(c + d*x)*sqrt(a*c + a*d*x + b)*a*b**3*c**4*d*x - 1778*sqrt(c + d 
*x)*sqrt(a*c + a*d*x + b)*a*b**3*c**3*d**2*x**2 - 5530*sqrt(c + d*x)*sqrt( 
a*c + a*d*x + b)*a*b**3*c**2*d**3*x**3 - 112*sqrt(c + d*x)*sqrt(a*c + a*d* 
x + b)*b**4*c**4 + 196*sqrt(c + d*x)*sqrt(a*c + a*d*x + b)*b**4*c**3*d*x - 
 490*sqrt(c + d*x)*sqrt(a*c + a*d*x + b)*b**4*c**2*d**2*x**2 - 1470*sqrt(c 
 + d*x)*sqrt(a*c + a*d*x + b)*b**4*c*d**3*x**3 - 864*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**4*d**3*x**3 - 864*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**4*x**4 - 2664*sqrt(c)*sq 
rt(a*c + b)*log(sqrt(a*c + a*d*x + b) - sqrt(2*sqrt(c)*sqrt(a)*sqrt(a*c...