\(\int \frac {x^{3/2} (A+B x)}{(b x+c x^2)^3} \, dx\) [101]

Optimal result

Integrand size = 22, antiderivative size = 106 \[ \int \frac {x^{3/2} (A+B x)}{\left (b x+c x^2\right )^3} \, dx=-\frac {2 A}{b^3 \sqrt {x}}+\frac {(b B-A c) \sqrt {x}}{2 b^2 (b+c x)^2}+\frac {(3 b B-7 A c) \sqrt {x}}{4 b^3 (b+c x)}+\frac {3 (b B-5 A c) \arctan \left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{4 b^{7/2} \sqrt {c}} \] Output:

-2*A/b^3/x^(1/2)+1/2*(-A*c+B*b)*x^(1/2)/b^2/(c*x+b)^2+1/4*(-7*A*c+3*B*b)*x 
^(1/2)/b^3/(c*x+b)+3/4*(-5*A*c+B*b)*arctan(c^(1/2)*x^(1/2)/b^(1/2))/b^(7/2 
)/c^(1/2)
 

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.91 \[ \int \frac {x^{3/2} (A+B x)}{\left (b x+c x^2\right )^3} \, dx=\frac {-8 A b^2+5 b^2 B x-25 A b c x+3 b B c x^2-15 A c^2 x^2}{4 b^3 \sqrt {x} (b+c x)^2}+\frac {3 (b B-5 A c) \arctan \left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{4 b^{7/2} \sqrt {c}} \] Input:

Integrate[(x^(3/2)*(A + B*x))/(b*x + c*x^2)^3,x]
 

Output:

(-8*A*b^2 + 5*b^2*B*x - 25*A*b*c*x + 3*b*B*c*x^2 - 15*A*c^2*x^2)/(4*b^3*Sq 
rt[x]*(b + c*x)^2) + (3*(b*B - 5*A*c)*ArcTan[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/( 
4*b^(7/2)*Sqrt[c])
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {9, 87, 52, 61, 73, 218}

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.

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

\(\Big \downarrow \) 9

\(\displaystyle \int \frac {A+B x}{x^{3/2} (b+c x)^3}dx\)

\(\Big \downarrow \) 87

\(\displaystyle -\frac {(b B-5 A c) \int \frac {1}{x^{3/2} (b+c x)^2}dx}{4 b c}-\frac {b B-A c}{2 b c \sqrt {x} (b+c x)^2}\)

\(\Big \downarrow \) 52

\(\displaystyle -\frac {(b B-5 A c) \left (\frac {3 \int \frac {1}{x^{3/2} (b+c x)}dx}{2 b}+\frac {1}{b \sqrt {x} (b+c x)}\right )}{4 b c}-\frac {b B-A c}{2 b c \sqrt {x} (b+c x)^2}\)

\(\Big \downarrow \) 61

\(\displaystyle -\frac {(b B-5 A c) \left (\frac {3 \left (-\frac {c \int \frac {1}{\sqrt {x} (b+c x)}dx}{b}-\frac {2}{b \sqrt {x}}\right )}{2 b}+\frac {1}{b \sqrt {x} (b+c x)}\right )}{4 b c}-\frac {b B-A c}{2 b c \sqrt {x} (b+c x)^2}\)

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 218

\(\displaystyle -\frac {(b B-5 A c) \left (\frac {3 \left (-\frac {2 \sqrt {c} \arctan \left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{b^{3/2}}-\frac {2}{b \sqrt {x}}\right )}{2 b}+\frac {1}{b \sqrt {x} (b+c x)}\right )}{4 b c}-\frac {b B-A c}{2 b c \sqrt {x} (b+c x)^2}\)

Input:

Int[(x^(3/2)*(A + B*x))/(b*x + c*x^2)^3,x]
 

Output:

-1/2*(b*B - A*c)/(b*c*Sqrt[x]*(b + c*x)^2) - ((b*B - 5*A*c)*(1/(b*Sqrt[x]* 
(b + c*x)) + (3*(-2/(b*Sqrt[x]) - (2*Sqrt[c]*ArcTan[(Sqrt[c]*Sqrt[x])/Sqrt 
[b]])/b^(3/2)))/(2*b)))/(4*b*c)
 

Defintions of rubi rules used

rule 9

Int[(u_.)*(Px_)^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> With[{r = Expon[Px, 
x, Min]}, Simp[1/e^(p*r)   Int[u*(e*x)^(m + p*r)*ExpandToSum[Px/x^r, x]^p, 
x], x] /; IGtQ[r, 0]] /; FreeQ[{e, m}, x] && PolyQ[Px, x] && IntegerQ[p] && 
  !MonomialQ[Px, 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 61

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] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0 
] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[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 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

Maple [A] (verified)

Time = 0.84 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.79

method	result	size
derivativedivides	\(-\frac {2 A}{b^{3} \sqrt {x}}-\frac {2 \left (\frac {\left (\frac {7}{8} A \,c^{2}-\frac {3}{8} B b c \right ) x^{\frac {3}{2}}+\frac {b \left (9 A c -5 B b \right ) \sqrt {x}}{8}}{\left (c x +b \right )^{2}}+\frac {3 \left (5 A c -B b \right ) \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{8 \sqrt {b c}}\right )}{b^{3}}\)	\(84\)
default	\(-\frac {2 A}{b^{3} \sqrt {x}}-\frac {2 \left (\frac {\left (\frac {7}{8} A \,c^{2}-\frac {3}{8} B b c \right ) x^{\frac {3}{2}}+\frac {b \left (9 A c -5 B b \right ) \sqrt {x}}{8}}{\left (c x +b \right )^{2}}+\frac {3 \left (5 A c -B b \right ) \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{8 \sqrt {b c}}\right )}{b^{3}}\)	\(84\)
risch	\(-\frac {2 A}{b^{3} \sqrt {x}}-\frac {\frac {2 \left (\frac {7}{8} A \,c^{2}-\frac {3}{8} B b c \right ) x^{\frac {3}{2}}+\frac {b \left (9 A c -5 B b \right ) \sqrt {x}}{4}}{\left (c x +b \right )^{2}}+\frac {3 \left (5 A c -B b \right ) \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \sqrt {b c}}}{b^{3}}\)	\(85\)

Input:

int(x^(3/2)*(B*x+A)/(c*x^2+b*x)^3,x,method=_RETURNVERBOSE)
 

Output:

-2*A/b^3/x^(1/2)-2/b^3*(((7/8*A*c^2-3/8*B*b*c)*x^(3/2)+1/8*b*(9*A*c-5*B*b) 
*x^(1/2))/(c*x+b)^2+3/8*(5*A*c-B*b)/(b*c)^(1/2)*arctan(c*x^(1/2)/(b*c)^(1/ 
2)))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 331, normalized size of antiderivative = 3.12 \[ \int \frac {x^{3/2} (A+B x)}{\left (b x+c x^2\right )^3} \, dx=\left [\frac {3 \, {\left ({\left (B b c^{2} - 5 \, A c^{3}\right )} x^{3} + 2 \, {\left (B b^{2} c - 5 \, A b c^{2}\right )} x^{2} + {\left (B b^{3} - 5 \, A b^{2} c\right )} x\right )} \sqrt {-b c} \log \left (\frac {c x - b + 2 \, \sqrt {-b c} \sqrt {x}}{c x + b}\right ) - 2 \, {\left (8 \, A b^{3} c - 3 \, {\left (B b^{2} c^{2} - 5 \, A b c^{3}\right )} x^{2} - 5 \, {\left (B b^{3} c - 5 \, A b^{2} c^{2}\right )} x\right )} \sqrt {x}}{8 \, {\left (b^{4} c^{3} x^{3} + 2 \, b^{5} c^{2} x^{2} + b^{6} c x\right )}}, -\frac {3 \, {\left ({\left (B b c^{2} - 5 \, A c^{3}\right )} x^{3} + 2 \, {\left (B b^{2} c - 5 \, A b c^{2}\right )} x^{2} + {\left (B b^{3} - 5 \, A b^{2} c\right )} x\right )} \sqrt {b c} \arctan \left (\frac {\sqrt {b c}}{c \sqrt {x}}\right ) + {\left (8 \, A b^{3} c - 3 \, {\left (B b^{2} c^{2} - 5 \, A b c^{3}\right )} x^{2} - 5 \, {\left (B b^{3} c - 5 \, A b^{2} c^{2}\right )} x\right )} \sqrt {x}}{4 \, {\left (b^{4} c^{3} x^{3} + 2 \, b^{5} c^{2} x^{2} + b^{6} c x\right )}}\right ] \] Input:

integrate(x^(3/2)*(B*x+A)/(c*x^2+b*x)^3,x, algorithm="fricas")
 

Output:

[1/8*(3*((B*b*c^2 - 5*A*c^3)*x^3 + 2*(B*b^2*c - 5*A*b*c^2)*x^2 + (B*b^3 - 
5*A*b^2*c)*x)*sqrt(-b*c)*log((c*x - b + 2*sqrt(-b*c)*sqrt(x))/(c*x + b)) - 
 2*(8*A*b^3*c - 3*(B*b^2*c^2 - 5*A*b*c^3)*x^2 - 5*(B*b^3*c - 5*A*b^2*c^2)* 
x)*sqrt(x))/(b^4*c^3*x^3 + 2*b^5*c^2*x^2 + b^6*c*x), -1/4*(3*((B*b*c^2 - 5 
*A*c^3)*x^3 + 2*(B*b^2*c - 5*A*b*c^2)*x^2 + (B*b^3 - 5*A*b^2*c)*x)*sqrt(b* 
c)*arctan(sqrt(b*c)/(c*sqrt(x))) + (8*A*b^3*c - 3*(B*b^2*c^2 - 5*A*b*c^3)* 
x^2 - 5*(B*b^3*c - 5*A*b^2*c^2)*x)*sqrt(x))/(b^4*c^3*x^3 + 2*b^5*c^2*x^2 + 
 b^6*c*x)]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1598 vs. \(2 (104) = 208\).

Time = 69.84 (sec) , antiderivative size = 1598, normalized size of antiderivative = 15.08 \[ \int \frac {x^{3/2} (A+B x)}{\left (b x+c x^2\right )^3} \, dx=\text {Too large to display} \] Input:

integrate(x**(3/2)*(B*x+A)/(c*x**2+b*x)**3,x)
 

Output:

Piecewise((zoo*(-2*A/(7*x**(7/2)) - 2*B/(5*x**(5/2))), Eq(b, 0) & Eq(c, 0) 
), ((-2*A/sqrt(x) + 2*B*sqrt(x))/b**3, Eq(c, 0)), ((-2*A/(7*x**(7/2)) - 2* 
B/(5*x**(5/2)))/c**3, Eq(b, 0)), (-15*A*b**2*c*sqrt(x)*log(sqrt(x) - sqrt( 
-b/c))/(8*b**5*c*sqrt(x)*sqrt(-b/c) + 16*b**4*c**2*x**(3/2)*sqrt(-b/c) + 8 
*b**3*c**3*x**(5/2)*sqrt(-b/c)) + 15*A*b**2*c*sqrt(x)*log(sqrt(x) + sqrt(- 
b/c))/(8*b**5*c*sqrt(x)*sqrt(-b/c) + 16*b**4*c**2*x**(3/2)*sqrt(-b/c) + 8* 
b**3*c**3*x**(5/2)*sqrt(-b/c)) - 16*A*b**2*c*sqrt(-b/c)/(8*b**5*c*sqrt(x)* 
sqrt(-b/c) + 16*b**4*c**2*x**(3/2)*sqrt(-b/c) + 8*b**3*c**3*x**(5/2)*sqrt( 
-b/c)) - 30*A*b*c**2*x**(3/2)*log(sqrt(x) - sqrt(-b/c))/(8*b**5*c*sqrt(x)* 
sqrt(-b/c) + 16*b**4*c**2*x**(3/2)*sqrt(-b/c) + 8*b**3*c**3*x**(5/2)*sqrt( 
-b/c)) + 30*A*b*c**2*x**(3/2)*log(sqrt(x) + sqrt(-b/c))/(8*b**5*c*sqrt(x)* 
sqrt(-b/c) + 16*b**4*c**2*x**(3/2)*sqrt(-b/c) + 8*b**3*c**3*x**(5/2)*sqrt( 
-b/c)) - 50*A*b*c**2*x*sqrt(-b/c)/(8*b**5*c*sqrt(x)*sqrt(-b/c) + 16*b**4*c 
**2*x**(3/2)*sqrt(-b/c) + 8*b**3*c**3*x**(5/2)*sqrt(-b/c)) - 15*A*c**3*x** 
(5/2)*log(sqrt(x) - sqrt(-b/c))/(8*b**5*c*sqrt(x)*sqrt(-b/c) + 16*b**4*c** 
2*x**(3/2)*sqrt(-b/c) + 8*b**3*c**3*x**(5/2)*sqrt(-b/c)) + 15*A*c**3*x**(5 
/2)*log(sqrt(x) + sqrt(-b/c))/(8*b**5*c*sqrt(x)*sqrt(-b/c) + 16*b**4*c**2* 
x**(3/2)*sqrt(-b/c) + 8*b**3*c**3*x**(5/2)*sqrt(-b/c)) - 30*A*c**3*x**2*sq 
rt(-b/c)/(8*b**5*c*sqrt(x)*sqrt(-b/c) + 16*b**4*c**2*x**(3/2)*sqrt(-b/c) + 
 8*b**3*c**3*x**(5/2)*sqrt(-b/c)) + 3*B*b**3*sqrt(x)*log(sqrt(x) - sqrt...
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.92 \[ \int \frac {x^{3/2} (A+B x)}{\left (b x+c x^2\right )^3} \, dx=-\frac {8 \, A b^{2} - 3 \, {\left (B b c - 5 \, A c^{2}\right )} x^{2} - 5 \, {\left (B b^{2} - 5 \, A b c\right )} x}{4 \, {\left (b^{3} c^{2} x^{\frac {5}{2}} + 2 \, b^{4} c x^{\frac {3}{2}} + b^{5} \sqrt {x}\right )}} + \frac {3 \, {\left (B b - 5 \, A c\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \, \sqrt {b c} b^{3}} \] Input:

integrate(x^(3/2)*(B*x+A)/(c*x^2+b*x)^3,x, algorithm="maxima")
                                                                                    
                                                                                    
 

Output:

-1/4*(8*A*b^2 - 3*(B*b*c - 5*A*c^2)*x^2 - 5*(B*b^2 - 5*A*b*c)*x)/(b^3*c^2* 
x^(5/2) + 2*b^4*c*x^(3/2) + b^5*sqrt(x)) + 3/4*(B*b - 5*A*c)*arctan(c*sqrt 
(x)/sqrt(b*c))/(sqrt(b*c)*b^3)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.81 \[ \int \frac {x^{3/2} (A+B x)}{\left (b x+c x^2\right )^3} \, dx=\frac {3 \, {\left (B b - 5 \, A c\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \, \sqrt {b c} b^{3}} - \frac {2 \, A}{b^{3} \sqrt {x}} + \frac {3 \, B b c x^{\frac {3}{2}} - 7 \, A c^{2} x^{\frac {3}{2}} + 5 \, B b^{2} \sqrt {x} - 9 \, A b c \sqrt {x}}{4 \, {\left (c x + b\right )}^{2} b^{3}} \] Input:

integrate(x^(3/2)*(B*x+A)/(c*x^2+b*x)^3,x, algorithm="giac")
 

Output:

3/4*(B*b - 5*A*c)*arctan(c*sqrt(x)/sqrt(b*c))/(sqrt(b*c)*b^3) - 2*A/(b^3*s 
qrt(x)) + 1/4*(3*B*b*c*x^(3/2) - 7*A*c^2*x^(3/2) + 5*B*b^2*sqrt(x) - 9*A*b 
*c*sqrt(x))/((c*x + b)^2*b^3)
 

Mupad [B] (verification not implemented)

Time = 5.39 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.09 \[ \int \frac {x^{3/2} (A+B x)}{\left (b x+c x^2\right )^3} \, dx=-\frac {\frac {2\,A}{b}+\frac {5\,x\,\left (5\,A\,c-B\,b\right )}{4\,b^2}+\frac {3\,c\,x^2\,\left (5\,A\,c-B\,b\right )}{4\,b^3}}{b^2\,\sqrt {x}+c^2\,x^{5/2}+2\,b\,c\,x^{3/2}}-\frac {3\,\mathrm {atan}\left (\frac {3\,\sqrt {c}\,\sqrt {x}\,\left (5\,A\,c-B\,b\right )}{\sqrt {b}\,\left (15\,A\,c-3\,B\,b\right )}\right )\,\left (5\,A\,c-B\,b\right )}{4\,b^{7/2}\,\sqrt {c}} \] Input:

int((x^(3/2)*(A + B*x))/(b*x + c*x^2)^3,x)
 

Output:

- ((2*A)/b + (5*x*(5*A*c - B*b))/(4*b^2) + (3*c*x^2*(5*A*c - B*b))/(4*b^3) 
)/(b^2*x^(1/2) + c^2*x^(5/2) + 2*b*c*x^(3/2)) - (3*atan((3*c^(1/2)*x^(1/2) 
*(5*A*c - B*b))/(b^(1/2)*(15*A*c - 3*B*b)))*(5*A*c - B*b))/(4*b^(7/2)*c^(1 
/2))
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 237, normalized size of antiderivative = 2.24 \[ \int \frac {x^{3/2} (A+B x)}{\left (b x+c x^2\right )^3} \, dx=\frac {-15 \sqrt {x}\, \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {\sqrt {x}\, c}{\sqrt {c}\, \sqrt {b}}\right ) a \,b^{2} c -30 \sqrt {x}\, \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {\sqrt {x}\, c}{\sqrt {c}\, \sqrt {b}}\right ) a b \,c^{2} x -15 \sqrt {x}\, \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {\sqrt {x}\, c}{\sqrt {c}\, \sqrt {b}}\right ) a \,c^{3} x^{2}+3 \sqrt {x}\, \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {\sqrt {x}\, c}{\sqrt {c}\, \sqrt {b}}\right ) b^{4}+6 \sqrt {x}\, \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {\sqrt {x}\, c}{\sqrt {c}\, \sqrt {b}}\right ) b^{3} c x +3 \sqrt {x}\, \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {\sqrt {x}\, c}{\sqrt {c}\, \sqrt {b}}\right ) b^{2} c^{2} x^{2}-8 a \,b^{3} c -25 a \,b^{2} c^{2} x -15 a b \,c^{3} x^{2}+5 b^{4} c x +3 b^{3} c^{2} x^{2}}{4 \sqrt {x}\, b^{4} c \left (c^{2} x^{2}+2 b c x +b^{2}\right )} \] Input:

int(x^(3/2)*(B*x+A)/(c*x^2+b*x)^3,x)
 

Output:

( - 15*sqrt(x)*sqrt(c)*sqrt(b)*atan((sqrt(x)*c)/(sqrt(c)*sqrt(b)))*a*b**2* 
c - 30*sqrt(x)*sqrt(c)*sqrt(b)*atan((sqrt(x)*c)/(sqrt(c)*sqrt(b)))*a*b*c** 
2*x - 15*sqrt(x)*sqrt(c)*sqrt(b)*atan((sqrt(x)*c)/(sqrt(c)*sqrt(b)))*a*c** 
3*x**2 + 3*sqrt(x)*sqrt(c)*sqrt(b)*atan((sqrt(x)*c)/(sqrt(c)*sqrt(b)))*b** 
4 + 6*sqrt(x)*sqrt(c)*sqrt(b)*atan((sqrt(x)*c)/(sqrt(c)*sqrt(b)))*b**3*c*x 
 + 3*sqrt(x)*sqrt(c)*sqrt(b)*atan((sqrt(x)*c)/(sqrt(c)*sqrt(b)))*b**2*c**2 
*x**2 - 8*a*b**3*c - 25*a*b**2*c**2*x - 15*a*b*c**3*x**2 + 5*b**4*c*x + 3* 
b**3*c**2*x**2)/(4*sqrt(x)*b**4*c*(b**2 + 2*b*c*x + c**2*x**2))