\(\int \frac {1}{x^5 (a+\frac {b}{c+d x})^{3/2}} \, dx\) [52]

Optimal result

Integrand size = 19, antiderivative size = 329 \[ \int \frac {1}{x^5 \left (a+\frac {b}{c+d x}\right )^{3/2}} \, dx=-\frac {2 a^3 b d^4}{(b+a c)^5 \sqrt {a+\frac {b}{c+d x}}}-\frac {\left (3 b^3+24 a b^2 c+144 a^2 b c^2-64 a^3 c^3\right ) d^3 (c+d x) \sqrt {a+\frac {b}{c+d x}}}{64 c^2 (b+a c)^5 x}-\frac {\left (b^2+8 a b c+48 a^2 c^2\right ) d^2 (c+d x)^2 \sqrt {a+\frac {b}{c+d x}}}{32 c^2 (b+a c)^4 x^2}+\frac {(3 b+8 a c) d (c+d x)^3 \sqrt {a+\frac {b}{c+d x}}}{8 c^2 (b+a c)^3 x^3}-\frac {(c+d x)^4 \sqrt {a+\frac {b}{c+d x}}}{4 c^2 (b+a c)^2 x^4}-\frac {3 b \left (b^3+8 a b^2 c+48 a^2 b c^2-64 a^3 c^3\right ) d^4 \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+\frac {b}{c+d x}}}{\sqrt {b+a c}}\right )}{64 c^{5/2} (b+a c)^{11/2}} \] Output:

-2*a^3*b*d^4/(a*c+b)^5/(a+b/(d*x+c))^(1/2)-1/64*(-64*a^3*c^3+144*a^2*b*c^2 
+24*a*b^2*c+3*b^3)*d^3*(d*x+c)*(a+b/(d*x+c))^(1/2)/c^2/(a*c+b)^5/x-1/32*(4 
8*a^2*c^2+8*a*b*c+b^2)*d^2*(d*x+c)^2*(a+b/(d*x+c))^(1/2)/c^2/(a*c+b)^4/x^2 
+1/8*(8*a*c+3*b)*d*(d*x+c)^3*(a+b/(d*x+c))^(1/2)/c^2/(a*c+b)^3/x^3-1/4*(d* 
x+c)^4*(a+b/(d*x+c))^(1/2)/c^2/(a*c+b)^2/x^4-3/64*b*(-64*a^3*c^3+48*a^2*b* 
c^2+8*a*b^2*c+b^3)*d^4*arctanh(c^(1/2)*(a+b/(d*x+c))^(1/2)/(a*c+b)^(1/2))/ 
c^(5/2)/(a*c+b)^(11/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 1.28 (sec) , antiderivative size = 351, normalized size of antiderivative = 1.07 \[ \int \frac {1}{x^5 \left (a+\frac {b}{c+d x}\right )^{3/2}} \, dx=\frac {-\frac {\sqrt {c} (c+d x) \sqrt {\frac {b+a c+a d x}{c+d x}} \left (b^4 \left (16 c^3+24 c^2 d x+2 c d^2 x^2-3 d^3 x^3\right )+2 a^2 b^2 c \left (48 c^4+36 c^3 d x-39 c^2 d^2 x^2+34 c d^3 x^3-11 d^4 x^4\right )+a b^3 \left (64 c^4+72 c^3 d x-36 c^2 d^2 x^2-23 c d^3 x^3-3 d^4 x^4\right )+16 a^4 c^3 \left (c^4-d^4 x^4\right )+8 a^3 b c^2 \left (8 c^4+3 c^3 d x-5 c^2 d^2 x^2+11 c d^3 x^3+35 d^4 x^4\right )\right )}{(b+a c)^5 x^4 (b+a (c+d x))}-\frac {3 b \left (b^3+8 a b^2 c+48 a^2 b c^2-64 a^3 c^3\right ) d^4 \arctan \left (\frac {\sqrt {c} \sqrt {\frac {b+a c+a d x}{c+d x}}}{\sqrt {-b-a c}}\right )}{(-b-a c)^{11/2}}}{64 c^{5/2}} \] Input:

Integrate[1/(x^5*(a + b/(c + d*x))^(3/2)),x]
 

Output:

(-((Sqrt[c]*(c + d*x)*Sqrt[(b + a*c + a*d*x)/(c + d*x)]*(b^4*(16*c^3 + 24* 
c^2*d*x + 2*c*d^2*x^2 - 3*d^3*x^3) + 2*a^2*b^2*c*(48*c^4 + 36*c^3*d*x - 39 
*c^2*d^2*x^2 + 34*c*d^3*x^3 - 11*d^4*x^4) + a*b^3*(64*c^4 + 72*c^3*d*x - 3 
6*c^2*d^2*x^2 - 23*c*d^3*x^3 - 3*d^4*x^4) + 16*a^4*c^3*(c^4 - d^4*x^4) + 8 
*a^3*b*c^2*(8*c^4 + 3*c^3*d*x - 5*c^2*d^2*x^2 + 11*c*d^3*x^3 + 35*d^4*x^4) 
))/((b + a*c)^5*x^4*(b + a*(c + d*x)))) - (3*b*(b^3 + 8*a*b^2*c + 48*a^2*b 
*c^2 - 64*a^3*c^3)*d^4*ArcTan[(Sqrt[c]*Sqrt[(b + a*c + a*d*x)/(c + d*x)])/ 
Sqrt[-b - a*c]])/(-b - a*c)^(11/2))/(64*c^(5/2))
 

Rubi [A] (verified)

Time = 0.80 (sec) , antiderivative size = 332, normalized size of antiderivative = 1.01, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.632, Rules used = {896, 25, 941, 948, 25, 109, 27, 162, 52, 52, 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.

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

\(\Big \downarrow \) 896

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 941

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

\(\Big \downarrow \) 948

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 109

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 162

\(\displaystyle d^4 \left (\frac {\frac {\sqrt {a+\frac {b}{c+d x}} \left ((b-2 a c) (8 a c+b)-\frac {c \left (-32 a^2 c^2-8 a b c+3 b^2\right )}{c+d x}\right )}{8 c^2 (a c+b)^2 \left (1-\frac {c}{c+d x}\right )^4}-\frac {\left (-64 a^3 c^3+48 a^2 b c^2+8 a b^2 c+b^3\right ) \int \frac {1}{\sqrt {a+\frac {b}{c+d x}} \left (1-\frac {c}{c+d x}\right )^3}d\frac {1}{c+d x}}{16 c^2 (a c+b)^2}}{b (a c+b)}-\frac {2 a}{b (a c+b) (c+d x)^2 \left (1-\frac {c}{c+d x}\right )^4 \sqrt {a+\frac {b}{c+d x}}}\right )\)

\(\Big \downarrow \) 52

\(\displaystyle d^4 \left (\frac {\frac {\sqrt {a+\frac {b}{c+d x}} \left ((b-2 a c) (8 a c+b)-\frac {c \left (-32 a^2 c^2-8 a b c+3 b^2\right )}{c+d x}\right )}{8 c^2 (a c+b)^2 \left (1-\frac {c}{c+d x}\right )^4}-\frac {\left (-64 a^3 c^3+48 a^2 b c^2+8 a b^2 c+b^3\right ) \left (\frac {3 b \int \frac {1}{\sqrt {a+\frac {b}{c+d x}} \left (1-\frac {c}{c+d x}\right )^2}d\frac {1}{c+d x}}{4 (a c+b)}+\frac {\sqrt {a+\frac {b}{c+d x}}}{2 (a c+b) \left (1-\frac {c}{c+d x}\right )^2}\right )}{16 c^2 (a c+b)^2}}{b (a c+b)}-\frac {2 a}{b (a c+b) (c+d x)^2 \left (1-\frac {c}{c+d x}\right )^4 \sqrt {a+\frac {b}{c+d x}}}\right )\)

\(\Big \downarrow \) 52

\(\displaystyle d^4 \left (\frac {\frac {\sqrt {a+\frac {b}{c+d x}} \left ((b-2 a c) (8 a c+b)-\frac {c \left (-32 a^2 c^2-8 a b c+3 b^2\right )}{c+d x}\right )}{8 c^2 (a c+b)^2 \left (1-\frac {c}{c+d x}\right )^4}-\frac {\left (-64 a^3 c^3+48 a^2 b c^2+8 a b^2 c+b^3\right ) \left (\frac {3 b \left (\frac {b \int \frac {1}{\sqrt {a+\frac {b}{c+d x}} \left (1-\frac {c}{c+d x}\right )}d\frac {1}{c+d x}}{2 (a c+b)}+\frac {\sqrt {a+\frac {b}{c+d x}}}{(a c+b) \left (1-\frac {c}{c+d x}\right )}\right )}{4 (a c+b)}+\frac {\sqrt {a+\frac {b}{c+d x}}}{2 (a c+b) \left (1-\frac {c}{c+d x}\right )^2}\right )}{16 c^2 (a c+b)^2}}{b (a c+b)}-\frac {2 a}{b (a c+b) (c+d x)^2 \left (1-\frac {c}{c+d x}\right )^4 \sqrt {a+\frac {b}{c+d x}}}\right )\)

\(\Big \downarrow \) 73

\(\displaystyle d^4 \left (\frac {\frac {\sqrt {a+\frac {b}{c+d x}} \left ((b-2 a c) (8 a c+b)-\frac {c \left (-32 a^2 c^2-8 a b c+3 b^2\right )}{c+d x}\right )}{8 c^2 (a c+b)^2 \left (1-\frac {c}{c+d x}\right )^4}-\frac {\left (-64 a^3 c^3+48 a^2 b c^2+8 a b^2 c+b^3\right ) \left (\frac {3 b \left (\frac {\int \frac {1}{\frac {a c}{b}-\frac {c}{b (c+d x)^2}+1}d\sqrt {a+\frac {b}{c+d x}}}{a c+b}+\frac {\sqrt {a+\frac {b}{c+d x}}}{(a c+b) \left (1-\frac {c}{c+d x}\right )}\right )}{4 (a c+b)}+\frac {\sqrt {a+\frac {b}{c+d x}}}{2 (a c+b) \left (1-\frac {c}{c+d x}\right )^2}\right )}{16 c^2 (a c+b)^2}}{b (a c+b)}-\frac {2 a}{b (a c+b) (c+d x)^2 \left (1-\frac {c}{c+d x}\right )^4 \sqrt {a+\frac {b}{c+d x}}}\right )\)

\(\Big \downarrow \) 221

\(\displaystyle d^4 \left (\frac {\frac {\sqrt {a+\frac {b}{c+d x}} \left ((b-2 a c) (8 a c+b)-\frac {c \left (-32 a^2 c^2-8 a b c+3 b^2\right )}{c+d x}\right )}{8 c^2 (a c+b)^2 \left (1-\frac {c}{c+d x}\right )^4}-\frac {\left (-64 a^3 c^3+48 a^2 b c^2+8 a b^2 c+b^3\right ) \left (\frac {3 b \left (\frac {b \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+\frac {b}{c+d x}}}{\sqrt {a c+b}}\right )}{\sqrt {c} (a c+b)^{3/2}}+\frac {\sqrt {a+\frac {b}{c+d x}}}{(a c+b) \left (1-\frac {c}{c+d x}\right )}\right )}{4 (a c+b)}+\frac {\sqrt {a+\frac {b}{c+d x}}}{2 (a c+b) \left (1-\frac {c}{c+d x}\right )^2}\right )}{16 c^2 (a c+b)^2}}{b (a c+b)}-\frac {2 a}{b (a c+b) (c+d x)^2 \left (1-\frac {c}{c+d x}\right )^4 \sqrt {a+\frac {b}{c+d x}}}\right )\)

Input:

Int[1/(x^5*(a + b/(c + d*x))^(3/2)),x]
 

Output:

d^4*((-2*a)/(b*(b + a*c)*(c + d*x)^2*Sqrt[a + b/(c + d*x)]*(1 - c/(c + d*x 
))^4) + ((Sqrt[a + b/(c + d*x)]*((b - 2*a*c)*(b + 8*a*c) - (c*(3*b^2 - 8*a 
*b*c - 32*a^2*c^2))/(c + d*x)))/(8*c^2*(b + a*c)^2*(1 - c/(c + d*x))^4) - 
((b^3 + 8*a*b^2*c + 48*a^2*b*c^2 - 64*a^3*c^3)*(Sqrt[a + b/(c + d*x)]/(2*( 
b + a*c)*(1 - c/(c + d*x))^2) + (3*b*(Sqrt[a + b/(c + d*x)]/((b + a*c)*(1 
- c/(c + d*x))) + (b*ArcTanh[(Sqrt[c]*Sqrt[a + b/(c + d*x)])/Sqrt[b + a*c] 
])/(Sqrt[c]*(b + a*c)^(3/2))))/(4*(b + a*c))))/(16*c^2*(b + a*c)^2))/(b*(b 
 + a*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 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[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 109

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

rule 162

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_) 
)*((g_.) + (h_.)*(x_)), x_] :> Simp[((b^3*c*e*g*(m + 2) - a^3*d*f*h*(n + 2) 
 - a^2*b*(c*f*h*m - d*(f*g + e*h)*(m + n + 3)) - a*b^2*(c*(f*g + e*h) + d*e 
*g*(2*m + n + 4)) + b*(a^2*d*f*h*(m - n) - a*b*(2*c*f*h*(m + 1) - d*(f*g + 
e*h)*(n + 1)) + b^2*(c*(f*g + e*h)*(m + 1) - d*e*g*(m + n + 2)))*x)/(b^2*(b 
*c - a*d)^2*(m + 1)*(m + 2)))*(a + b*x)^(m + 1)*(c + d*x)^(n + 1), x] + Sim 
p[(f*(h/b^2) - (d*(m + n + 3)*(a^2*d*f*h*(m - n) - a*b*(2*c*f*h*(m + 1) - d 
*(f*g + e*h)*(n + 1)) + b^2*(c*(f*g + e*h)*(m + 1) - d*e*g*(m + n + 2))))/( 
b^2*(b*c - a*d)^2*(m + 1)*(m + 2)))   Int[(a + b*x)^(m + 2)*(c + d*x)^n, x] 
, x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && (LtQ[m, -2] || (EqQ[m + 
 n + 3, 0] &&  !LtQ[n, -2]))
 

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. \(6949\) vs. \(2(299)=598\).

Time = 0.25 (sec) , antiderivative size = 6950, normalized size of antiderivative = 21.12

method	result	size
default	\(\text {Expression too large to display}\)	\(6950\)

Input:

int(1/x^5/(a+b/(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
                                                                                    
                                                                                    
 

Output:

result too large to display
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 695 vs. \(2 (301) = 602\).

Time = 0.15 (sec) , antiderivative size = 1398, normalized size of antiderivative = 4.25 \[ \int \frac {1}{x^5 \left (a+\frac {b}{c+d x}\right )^{3/2}} \, dx=\text {Too large to display} \] Input:

integrate(1/x^5/(a+b/(d*x+c))^(3/2),x, algorithm="fricas")
 

Output:

[1/128*(3*((64*a^4*b*c^3 - 48*a^3*b^2*c^2 - 8*a^2*b^3*c - a*b^4)*d^5*x^5 + 
 (64*a^4*b*c^4 + 16*a^3*b^2*c^3 - 56*a^2*b^3*c^2 - 9*a*b^4*c - b^5)*d^4*x^ 
4)*sqrt(a*c^2 + b*c)*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*(16*a^5*c^10 
+ 80*a^4*b*c^9 + 160*a^3*b^2*c^8 + 160*a^2*b^3*c^7 + 80*a*b^4*c^6 - (16*a^ 
5*c^5 - 264*a^4*b*c^4 - 258*a^3*b^2*c^3 + 25*a^2*b^3*c^2 + 3*a*b^4*c)*d^5* 
x^5 + 16*b^5*c^5 - (16*a^5*c^6 - 352*a^4*b*c^5 - 414*a^3*b^2*c^4 - 20*a^2* 
b^3*c^3 + 29*a*b^4*c^2 + 3*b^5*c)*d^4*x^4 + (48*a^4*b*c^6 + 38*a^3*b^2*c^5 
 - 69*a^2*b^3*c^4 - 60*a*b^4*c^3 - b^5*c^2)*d^3*x^3 - 2*(8*a^4*b*c^7 + 11* 
a^3*b^2*c^6 - 15*a^2*b^3*c^5 - 31*a*b^4*c^4 - 13*b^5*c^3)*d^2*x^2 + 8*(2*a 
^5*c^9 + 13*a^4*b*c^8 + 32*a^3*b^2*c^7 + 38*a^2*b^3*c^6 + 22*a*b^4*c^5 + 5 
*b^5*c^4)*d*x)*sqrt((a*d*x + a*c + b)/(d*x + c)))/((a^7*c^9 + 6*a^6*b*c^8 
+ 15*a^5*b^2*c^7 + 20*a^4*b^3*c^6 + 15*a^3*b^4*c^5 + 6*a^2*b^5*c^4 + a*b^6 
*c^3)*d*x^5 + (a^7*c^10 + 7*a^6*b*c^9 + 21*a^5*b^2*c^8 + 35*a^4*b^3*c^7 + 
35*a^3*b^4*c^6 + 21*a^2*b^5*c^5 + 7*a*b^6*c^4 + b^7*c^3)*x^4), -1/64*(3*(( 
64*a^4*b*c^3 - 48*a^3*b^2*c^2 - 8*a^2*b^3*c - a*b^4)*d^5*x^5 + (64*a^4*b*c 
^4 + 16*a^3*b^2*c^3 - 56*a^2*b^3*c^2 - 9*a*b^4*c - b^5)*d^4*x^4)*sqrt(-a*c 
^2 - b*c)*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)) + (16*a^5*c^10 + 80*a^4*b*c^9 + 160*a^3*b^2 
*c^8 + 160*a^2*b^3*c^7 + 80*a*b^4*c^6 - (16*a^5*c^5 - 264*a^4*b*c^4 - 2...
 

Sympy [F]

\[ \int \frac {1}{x^5 \left (a+\frac {b}{c+d x}\right )^{3/2}} \, dx=\int \frac {1}{x^{5} \left (\frac {a c + a d x + b}{c + d x}\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(1/x**5/(a+b/(d*x+c))**(3/2),x)
 

Output:

Integral(1/(x**5*((a*c + a*d*x + b)/(c + d*x))**(3/2)), x)
 

Maxima [F]

\[ \int \frac {1}{x^5 \left (a+\frac {b}{c+d x}\right )^{3/2}} \, dx=\int { \frac {1}{{\left (a + \frac {b}{d x + c}\right )}^{\frac {3}{2}} x^{5}} \,d x } \] Input:

integrate(1/x^5/(a+b/(d*x+c))^(3/2),x, algorithm="maxima")
 

Output:

integrate(1/((a + b/(d*x + c))^(3/2)*x^5), x)
 

Giac [F(-1)]

Timed out. \[ \int \frac {1}{x^5 \left (a+\frac {b}{c+d x}\right )^{3/2}} \, dx=\text {Timed out} \] Input:

integrate(1/x^5/(a+b/(d*x+c))^(3/2),x, algorithm="giac")
 

Output:

Timed out
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^5 \left (a+\frac {b}{c+d x}\right )^{3/2}} \, dx=\int \frac {1}{x^5\,{\left (a+\frac {b}{c+d\,x}\right )}^{3/2}} \,d x \] Input:

int(1/(x^5*(a + b/(c + d*x))^(3/2)),x)
 

Output:

int(1/(x^5*(a + b/(c + d*x))^(3/2)), x)
 

Reduce [F]

\[ \int \frac {1}{x^5 \left (a+\frac {b}{c+d x}\right )^{3/2}} \, dx=\int \frac {1}{x^{5} \left (a +\frac {b}{d x +c}\right )^{\frac {3}{2}}}d x \] Input:

int(1/x^5/(a+b/(d*x+c))^(3/2),x)
 

Output:

int(1/x^5/(a+b/(d*x+c))^(3/2),x)