\(\int \frac {(A+B x) (a+b x+c x^2)^{3/2}}{x} \, dx\) [121]

Optimal result

Integrand size = 23, antiderivative size = 218 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^{3/2}}{x} \, dx=-\frac {\left (3 b^3 B-8 A b^2 c-12 a b B c-64 a A c^2+2 c \left (3 b^2 B-8 A b c-12 a B c\right ) x\right ) \sqrt {a+b x+c x^2}}{64 c^2}+\frac {(3 b B+8 A c+6 B c x) \left (a+b x+c x^2\right )^{3/2}}{24 c}-a^{3/2} A \text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )+\frac {\left (64 a A b c^2+\left (b^2-4 a c\right ) \left (3 b^2 B-8 A b c-12 a B c\right )\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{128 c^{5/2}} \] Output:

-1/64*(3*B*b^3-8*A*b^2*c-12*B*a*b*c-64*A*a*c^2+2*c*(-8*A*b*c-12*B*a*c+3*B* 
b^2)*x)*(c*x^2+b*x+a)^(1/2)/c^2+1/24*(6*B*c*x+8*A*c+3*B*b)*(c*x^2+b*x+a)^( 
3/2)/c-a^(3/2)*A*arctanh(1/2*(b*x+2*a)/a^(1/2)/(c*x^2+b*x+a)^(1/2))+1/128* 
(64*A*a*b*c^2+(-4*a*c+b^2)*(-8*A*b*c-12*B*a*c+3*B*b^2))*arctanh(1/2*(2*c*x 
+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/c^(5/2)
 

Mathematica [A] (verified)

Time = 1.25 (sec) , antiderivative size = 208, normalized size of antiderivative = 0.95 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^{3/2}}{x} \, dx=\frac {\sqrt {a+x (b+c x)} \left (-9 b^3 B+6 b^2 c (4 A+B x)+8 c^2 \left (32 a A+15 a B x+8 A c x^2+6 B c x^3\right )+4 b c (15 a B+2 c x (14 A+9 B x))\right )}{192 c^2}+2 a^{3/2} A \text {arctanh}\left (\frac {\sqrt {c} x-\sqrt {a+x (b+c x)}}{\sqrt {a}}\right )-\frac {\left (3 b^4 B-8 A b^3 c-24 a b^2 B c+96 a A b c^2+48 a^2 B c^2\right ) \log \left (c^2 \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )\right )}{128 c^{5/2}} \] Input:

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

Output:

(Sqrt[a + x*(b + c*x)]*(-9*b^3*B + 6*b^2*c*(4*A + B*x) + 8*c^2*(32*a*A + 1 
5*a*B*x + 8*A*c*x^2 + 6*B*c*x^3) + 4*b*c*(15*a*B + 2*c*x*(14*A + 9*B*x)))) 
/(192*c^2) + 2*a^(3/2)*A*ArcTanh[(Sqrt[c]*x - Sqrt[a + x*(b + c*x)])/Sqrt[ 
a]] - ((3*b^4*B - 8*A*b^3*c - 24*a*b^2*B*c + 96*a*A*b*c^2 + 48*a^2*B*c^2)* 
Log[c^2*(b + 2*c*x - 2*Sqrt[c]*Sqrt[a + x*(b + c*x)])])/(128*c^(5/2))
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.07, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {1231, 27, 1231, 27, 1269, 1092, 219, 1154, 219}

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.

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

\(\Big \downarrow \) 1231

\(\displaystyle \frac {\left (a+b x+c x^2\right )^{3/2} (8 A c+3 b B+6 B c x)}{24 c}-\frac {\int -\frac {\left (16 a A c+\left (8 A b c-3 B \left (b^2-4 a c\right )\right ) x\right ) \sqrt {c x^2+b x+a}}{2 x}dx}{8 c}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\left (16 a A c-\left (3 B b^2-8 A c b-12 a B c\right ) x\right ) \sqrt {c x^2+b x+a}}{x}dx}{16 c}+\frac {\left (a+b x+c x^2\right )^{3/2} (8 A c+3 b B+6 B c x)}{24 c}\)

\(\Big \downarrow \) 1231

\(\displaystyle \frac {-\frac {\int -\frac {128 a^2 A c^2+\left (64 a A b c^2+\left (b^2-4 a c\right ) \left (3 B b^2-8 A c b-12 a B c\right )\right ) x}{2 x \sqrt {c x^2+b x+a}}dx}{4 c}-\frac {\sqrt {a+b x+c x^2} \left (2 c x \left (-12 a B c-8 A b c+3 b^2 B\right )-64 a A c^2-12 a b B c-8 A b^2 c+3 b^3 B\right )}{4 c}}{16 c}+\frac {\left (a+b x+c x^2\right )^{3/2} (8 A c+3 b B+6 B c x)}{24 c}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {128 a^2 A c^2+\left (64 a A b c^2+\left (b^2-4 a c\right ) \left (3 B b^2-8 A c b-12 a B c\right )\right ) x}{x \sqrt {c x^2+b x+a}}dx}{8 c}-\frac {\sqrt {a+b x+c x^2} \left (2 c x \left (-12 a B c-8 A b c+3 b^2 B\right )-64 a A c^2-12 a b B c-8 A b^2 c+3 b^3 B\right )}{4 c}}{16 c}+\frac {\left (a+b x+c x^2\right )^{3/2} (8 A c+3 b B+6 B c x)}{24 c}\)

\(\Big \downarrow \) 1269

\(\displaystyle \frac {\frac {128 a^2 A c^2 \int \frac {1}{x \sqrt {c x^2+b x+a}}dx+\left (\left (b^2-4 a c\right ) \left (-12 a B c-8 A b c+3 b^2 B\right )+64 a A b c^2\right ) \int \frac {1}{\sqrt {c x^2+b x+a}}dx}{8 c}-\frac {\sqrt {a+b x+c x^2} \left (2 c x \left (-12 a B c-8 A b c+3 b^2 B\right )-64 a A c^2-12 a b B c-8 A b^2 c+3 b^3 B\right )}{4 c}}{16 c}+\frac {\left (a+b x+c x^2\right )^{3/2} (8 A c+3 b B+6 B c x)}{24 c}\)

\(\Big \downarrow \) 1092

\(\displaystyle \frac {\frac {128 a^2 A c^2 \int \frac {1}{x \sqrt {c x^2+b x+a}}dx+2 \left (\left (b^2-4 a c\right ) \left (-12 a B c-8 A b c+3 b^2 B\right )+64 a A b c^2\right ) \int \frac {1}{4 c-\frac {(b+2 c x)^2}{c x^2+b x+a}}d\frac {b+2 c x}{\sqrt {c x^2+b x+a}}}{8 c}-\frac {\sqrt {a+b x+c x^2} \left (2 c x \left (-12 a B c-8 A b c+3 b^2 B\right )-64 a A c^2-12 a b B c-8 A b^2 c+3 b^3 B\right )}{4 c}}{16 c}+\frac {\left (a+b x+c x^2\right )^{3/2} (8 A c+3 b B+6 B c x)}{24 c}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {128 a^2 A c^2 \int \frac {1}{x \sqrt {c x^2+b x+a}}dx+\frac {\left (\left (b^2-4 a c\right ) \left (-12 a B c-8 A b c+3 b^2 B\right )+64 a A b c^2\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c}}}{8 c}-\frac {\sqrt {a+b x+c x^2} \left (2 c x \left (-12 a B c-8 A b c+3 b^2 B\right )-64 a A c^2-12 a b B c-8 A b^2 c+3 b^3 B\right )}{4 c}}{16 c}+\frac {\left (a+b x+c x^2\right )^{3/2} (8 A c+3 b B+6 B c x)}{24 c}\)

\(\Big \downarrow \) 1154

\(\displaystyle \frac {\frac {\frac {\left (\left (b^2-4 a c\right ) \left (-12 a B c-8 A b c+3 b^2 B\right )+64 a A b c^2\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c}}-256 a^2 A c^2 \int \frac {1}{4 a-\frac {(2 a+b x)^2}{c x^2+b x+a}}d\frac {2 a+b x}{\sqrt {c x^2+b x+a}}}{8 c}-\frac {\sqrt {a+b x+c x^2} \left (2 c x \left (-12 a B c-8 A b c+3 b^2 B\right )-64 a A c^2-12 a b B c-8 A b^2 c+3 b^3 B\right )}{4 c}}{16 c}+\frac {\left (a+b x+c x^2\right )^{3/2} (8 A c+3 b B+6 B c x)}{24 c}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\frac {\left (\left (b^2-4 a c\right ) \left (-12 a B c-8 A b c+3 b^2 B\right )+64 a A b c^2\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c}}-128 a^{3/2} A c^2 \text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{8 c}-\frac {\sqrt {a+b x+c x^2} \left (2 c x \left (-12 a B c-8 A b c+3 b^2 B\right )-64 a A c^2-12 a b B c-8 A b^2 c+3 b^3 B\right )}{4 c}}{16 c}+\frac {\left (a+b x+c x^2\right )^{3/2} (8 A c+3 b B+6 B c x)}{24 c}\)

Input:

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

Output:

((3*b*B + 8*A*c + 6*B*c*x)*(a + b*x + c*x^2)^(3/2))/(24*c) + (-1/4*((3*b^3 
*B - 8*A*b^2*c - 12*a*b*B*c - 64*a*A*c^2 + 2*c*(3*b^2*B - 8*A*b*c - 12*a*B 
*c)*x)*Sqrt[a + b*x + c*x^2])/c + (-128*a^(3/2)*A*c^2*ArcTanh[(2*a + b*x)/ 
(2*Sqrt[a]*Sqrt[a + b*x + c*x^2])] + ((64*a*A*b*c^2 + (b^2 - 4*a*c)*(3*b^2 
*B - 8*A*b*c - 12*a*B*c))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c* 
x^2])])/Sqrt[c])/(8*c))/(16*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 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 1092

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2   Subst[I 
nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a 
, b, c}, x]
 

rule 1154

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym 
bol] :> Simp[-2   Subst[Int[1/(4*c*d^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, ( 
2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c 
, d, e}, x]
 

rule 1231

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x)^(m + 1)*(c*e*f*(m + 2*p + 2) 
 - g*(c*d + 2*c*d*p - b*e*p) + g*c*e*(m + 2*p + 1)*x)*((a + b*x + c*x^2)^p/ 
(c*e^2*(m + 2*p + 1)*(m + 2*p + 2))), x] - Simp[p/(c*e^2*(m + 2*p + 1)*(m + 
 2*p + 2))   Int[(d + e*x)^m*(a + b*x + c*x^2)^(p - 1)*Simp[c*e*f*(b*d - 2* 
a*e)*(m + 2*p + 2) + g*(a*e*(b*e - 2*c*d*m + b*e*m) + b*d*(b*e*p - c*d - 2* 
c*d*p)) + (c*e*f*(2*c*d - b*e)*(m + 2*p + 2) + g*(b^2*e^2*(p + m + 1) - 2*c 
^2*d^2*(1 + 2*p) - c*e*(b*d*(m - 2*p) + 2*a*e*(m + 2*p + 1))))*x, x], x], x 
] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && GtQ[p, 0] && (IntegerQ[p] ||  !R 
ationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])) &&  !ILtQ[m + 2*p, 0] && (Integer 
Q[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
 

rule 1269

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g/e   Int[(d + e*x)^(m + 1)*(a + b*x + 
 c*x^2)^p, x], x] + Simp[(e*f - d*g)/e   Int[(d + e*x)^m*(a + b*x + c*x^2)^ 
p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] &&  !IGtQ[m, 0]
 

Maple [A] (verified)

Time = 1.20 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.24

method	result	size
default	\(B \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{8 c}+\frac {3 \left (4 a c -b^{2}\right ) \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )+A \left (\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{3}+\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{2}+a \left (\sqrt {c \,x^{2}+b x +a}+\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 \sqrt {c}}-\sqrt {a}\, \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right )\right )\right )\)	\(271\)

Input:

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

Output:

B*(1/8*(2*c*x+b)*(c*x^2+b*x+a)^(3/2)/c+3/16*(4*a*c-b^2)/c*(1/4*(2*c*x+b)*( 
c*x^2+b*x+a)^(1/2)/c+1/8*(4*a*c-b^2)/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2 
+b*x+a)^(1/2))))+A*(1/3*(c*x^2+b*x+a)^(3/2)+1/2*b*(1/4*(2*c*x+b)*(c*x^2+b* 
x+a)^(1/2)/c+1/8*(4*a*c-b^2)/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^ 
(1/2)))+a*((c*x^2+b*x+a)^(1/2)+1/2*b*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^ 
(1/2))/c^(1/2)-a^(1/2)*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x)))
 

Fricas [A] (verification not implemented)

Time = 2.63 (sec) , antiderivative size = 1023, normalized size of antiderivative = 4.69 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^{3/2}}{x} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/768*(384*A*a^(3/2)*c^3*log(-(8*a*b*x + (b^2 + 4*a*c)*x^2 - 4*sqrt(c*x^2 
 + b*x + a)*(b*x + 2*a)*sqrt(a) + 8*a^2)/x^2) + 3*(3*B*b^4 + 48*(B*a^2 + 2 
*A*a*b)*c^2 - 8*(3*B*a*b^2 + A*b^3)*c)*sqrt(c)*log(-8*c^2*x^2 - 8*b*c*x - 
b^2 - 4*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) + 4*(48*B*c^4*x 
^3 - 9*B*b^3*c + 256*A*a*c^3 + 12*(5*B*a*b + 2*A*b^2)*c^2 + 8*(9*B*b*c^3 + 
 8*A*c^4)*x^2 + 2*(3*B*b^2*c^2 + 4*(15*B*a + 14*A*b)*c^3)*x)*sqrt(c*x^2 + 
b*x + a))/c^3, 1/384*(192*A*a^(3/2)*c^3*log(-(8*a*b*x + (b^2 + 4*a*c)*x^2 
- 4*sqrt(c*x^2 + b*x + a)*(b*x + 2*a)*sqrt(a) + 8*a^2)/x^2) - 3*(3*B*b^4 + 
 48*(B*a^2 + 2*A*a*b)*c^2 - 8*(3*B*a*b^2 + A*b^3)*c)*sqrt(-c)*arctan(1/2*s 
qrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c)) + 2*(48 
*B*c^4*x^3 - 9*B*b^3*c + 256*A*a*c^3 + 12*(5*B*a*b + 2*A*b^2)*c^2 + 8*(9*B 
*b*c^3 + 8*A*c^4)*x^2 + 2*(3*B*b^2*c^2 + 4*(15*B*a + 14*A*b)*c^3)*x)*sqrt( 
c*x^2 + b*x + a))/c^3, 1/768*(768*A*sqrt(-a)*a*c^3*arctan(1/2*sqrt(c*x^2 + 
 b*x + a)*(b*x + 2*a)*sqrt(-a)/(a*c*x^2 + a*b*x + a^2)) + 3*(3*B*b^4 + 48* 
(B*a^2 + 2*A*a*b)*c^2 - 8*(3*B*a*b^2 + A*b^3)*c)*sqrt(c)*log(-8*c^2*x^2 - 
8*b*c*x - b^2 - 4*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) + 4*( 
48*B*c^4*x^3 - 9*B*b^3*c + 256*A*a*c^3 + 12*(5*B*a*b + 2*A*b^2)*c^2 + 8*(9 
*B*b*c^3 + 8*A*c^4)*x^2 + 2*(3*B*b^2*c^2 + 4*(15*B*a + 14*A*b)*c^3)*x)*sqr 
t(c*x^2 + b*x + a))/c^3, 1/384*(384*A*sqrt(-a)*a*c^3*arctan(1/2*sqrt(c*x^2 
 + b*x + a)*(b*x + 2*a)*sqrt(-a)/(a*c*x^2 + a*b*x + a^2)) - 3*(3*B*b^4 ...
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^{3/2}}{x} \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` for 
 more deta
 

Giac [F(-2)]

Exception generated. \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^{3/2}}{x} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Degree mismatch inside factorisatio 
n over extensionNot implemented, e.g. for multivariate mod/approx polynomi 
alsError:
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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