\(\int (a+b x+c x^2)^2 (d+f x^2)^q \, dx\) [9]

Optimal result

Integrand size = 22, antiderivative size = 246 \[ \int \left (a+b x+c x^2\right )^2 \left (d+f x^2\right )^q \, dx=-\frac {b (c d-a f) \left (d+f x^2\right )^{1+q}}{f^2 (1+q)}-\frac {\left (3 c^2 d-b^2 f (5+2 q)-2 a c f (5+2 q)\right ) x \left (d+f x^2\right )^{1+q}}{f^2 (3+2 q) (5+2 q)}+\frac {c^2 x^3 \left (d+f x^2\right )^{1+q}}{f (5+2 q)}+\frac {b c \left (d+f x^2\right )^{2+q}}{f^2 (2+q)}+\frac {\left (3 c^2 d^2-2 a c d f (5+2 q)-f (5+2 q) \left (b^2 d-a^2 f (3+2 q)\right )\right ) x \left (d+f x^2\right )^q \left (1+\frac {f x^2}{d}\right )^{-q} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-q,\frac {3}{2},-\frac {f x^2}{d}\right )}{f^2 (3+2 q) (5+2 q)} \] Output:

-b*(-a*f+c*d)*(f*x^2+d)^(1+q)/f^2/(1+q)-(3*c^2*d-b^2*f*(5+2*q)-2*a*c*f*(5+ 
2*q))*x*(f*x^2+d)^(1+q)/f^2/(3+2*q)/(5+2*q)+c^2*x^3*(f*x^2+d)^(1+q)/f/(5+2 
*q)+b*c*(f*x^2+d)^(2+q)/f^2/(2+q)+(3*c^2*d^2-2*a*c*d*f*(5+2*q)-f*(5+2*q)*( 
b^2*d-a^2*f*(3+2*q)))*x*(f*x^2+d)^q*hypergeom([1/2, -q],[3/2],-f*x^2/d)/f^ 
2/(3+2*q)/(5+2*q)/((1+f*x^2/d)^q)
 

Mathematica [A] (verified)

Time = 0.85 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.78 \[ \int \left (a+b x+c x^2\right )^2 \left (d+f x^2\right )^q \, dx=\frac {1}{15} \left (d+f x^2\right )^q \left (15 a^2 x \left (1+\frac {f x^2}{d}\right )^{-q} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-q,\frac {3}{2},-\frac {f x^2}{d}\right )+5 \left (b^2+2 a c\right ) x^3 \left (1+\frac {f x^2}{d}\right )^{-q} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-q,\frac {5}{2},-\frac {f x^2}{d}\right )+\frac {3 \left (\frac {5 b \left (d+f x^2\right ) \left (-c d+a f (2+q)+c f (1+q) x^2\right )}{f^2}+c^2 \left (2+3 q+q^2\right ) x^5 \left (1+\frac {f x^2}{d}\right )^{-q} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-q,\frac {7}{2},-\frac {f x^2}{d}\right )\right )}{(1+q) (2+q)}\right ) \] Input:

Integrate[(a + b*x + c*x^2)^2*(d + f*x^2)^q,x]
 

Output:

((d + f*x^2)^q*((15*a^2*x*Hypergeometric2F1[1/2, -q, 3/2, -((f*x^2)/d)])/( 
1 + (f*x^2)/d)^q + (5*(b^2 + 2*a*c)*x^3*Hypergeometric2F1[3/2, -q, 5/2, -( 
(f*x^2)/d)])/(1 + (f*x^2)/d)^q + (3*((5*b*(d + f*x^2)*(-(c*d) + a*f*(2 + q 
) + c*f*(1 + q)*x^2))/f^2 + (c^2*(2 + 3*q + q^2)*x^5*Hypergeometric2F1[5/2 
, -q, 7/2, -((f*x^2)/d)])/(1 + (f*x^2)/d)^q))/((1 + q)*(2 + q))))/15
 

Rubi [A] (verified)

Time = 0.80 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.10, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {2346, 2346, 27, 2346, 27, 455, 238, 237}

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.

\(\displaystyle \int \left (a+b x+c x^2\right )^2 \left (d+f x^2\right )^q \, dx\)

\(\Big \downarrow \) 2346

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

\(\Big \downarrow \) 2346

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2346

\(\displaystyle \frac {\frac {\frac {\int f \left ((q+2) \left (3 c^2 d^2-2 a c f (2 q+5) d-f (2 q+5) \left (b^2 d-a^2 f (2 q+3)\right )\right )-2 b f \left (4 q^2+16 q+15\right ) (c d-a f (q+2)) x\right ) \left (f x^2+d\right )^qdx}{f (2 q+3)}-\frac {(q+2) x \left (d+f x^2\right )^{q+1} \left (-2 a c f (2 q+5)+b^2 (-f) (2 q+5)+3 c^2 d\right )}{2 q+3}}{f (q+2)}+\frac {b c (2 q+5) x^2 \left (d+f x^2\right )^{q+1}}{q+2}}{f (2 q+5)}+\frac {c^2 x^3 \left (d+f x^2\right )^{q+1}}{f (2 q+5)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {\int \left ((q+2) \left (3 c^2 d^2-2 a c f (2 q+5) d-f (2 q+5) \left (b^2 d-a^2 f (2 q+3)\right )\right )-2 b f \left (4 q^2+16 q+15\right ) (c d-a f (q+2)) x\right ) \left (f x^2+d\right )^qdx}{2 q+3}-\frac {(q+2) x \left (d+f x^2\right )^{q+1} \left (-2 a c f (2 q+5)+b^2 (-f) (2 q+5)+3 c^2 d\right )}{2 q+3}}{f (q+2)}+\frac {b c (2 q+5) x^2 \left (d+f x^2\right )^{q+1}}{q+2}}{f (2 q+5)}+\frac {c^2 x^3 \left (d+f x^2\right )^{q+1}}{f (2 q+5)}\)

\(\Big \downarrow \) 455

\(\displaystyle \frac {\frac {\frac {(q+2) \left (-f (2 q+5) \left (b^2 d-a^2 f (2 q+3)\right )-2 a c d f (2 q+5)+3 c^2 d^2\right ) \int \left (f x^2+d\right )^qdx-\frac {b \left (4 q^2+16 q+15\right ) \left (d+f x^2\right )^{q+1} (c d-a f (q+2))}{q+1}}{2 q+3}-\frac {(q+2) x \left (d+f x^2\right )^{q+1} \left (-2 a c f (2 q+5)+b^2 (-f) (2 q+5)+3 c^2 d\right )}{2 q+3}}{f (q+2)}+\frac {b c (2 q+5) x^2 \left (d+f x^2\right )^{q+1}}{q+2}}{f (2 q+5)}+\frac {c^2 x^3 \left (d+f x^2\right )^{q+1}}{f (2 q+5)}\)

\(\Big \downarrow \) 238

\(\displaystyle \frac {\frac {\frac {(q+2) \left (d+f x^2\right )^q \left (\frac {f x^2}{d}+1\right )^{-q} \left (-f (2 q+5) \left (b^2 d-a^2 f (2 q+3)\right )-2 a c d f (2 q+5)+3 c^2 d^2\right ) \int \left (\frac {f x^2}{d}+1\right )^qdx-\frac {b \left (4 q^2+16 q+15\right ) \left (d+f x^2\right )^{q+1} (c d-a f (q+2))}{q+1}}{2 q+3}-\frac {(q+2) x \left (d+f x^2\right )^{q+1} \left (-2 a c f (2 q+5)+b^2 (-f) (2 q+5)+3 c^2 d\right )}{2 q+3}}{f (q+2)}+\frac {b c (2 q+5) x^2 \left (d+f x^2\right )^{q+1}}{q+2}}{f (2 q+5)}+\frac {c^2 x^3 \left (d+f x^2\right )^{q+1}}{f (2 q+5)}\)

\(\Big \downarrow \) 237

\(\displaystyle \frac {\frac {\frac {(q+2) x \left (d+f x^2\right )^q \left (\frac {f x^2}{d}+1\right )^{-q} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-q,\frac {3}{2},-\frac {f x^2}{d}\right ) \left (-f (2 q+5) \left (b^2 d-a^2 f (2 q+3)\right )-2 a c d f (2 q+5)+3 c^2 d^2\right )-\frac {b \left (4 q^2+16 q+15\right ) \left (d+f x^2\right )^{q+1} (c d-a f (q+2))}{q+1}}{2 q+3}-\frac {(q+2) x \left (d+f x^2\right )^{q+1} \left (-2 a c f (2 q+5)+b^2 (-f) (2 q+5)+3 c^2 d\right )}{2 q+3}}{f (q+2)}+\frac {b c (2 q+5) x^2 \left (d+f x^2\right )^{q+1}}{q+2}}{f (2 q+5)}+\frac {c^2 x^3 \left (d+f x^2\right )^{q+1}}{f (2 q+5)}\)

Input:

Int[(a + b*x + c*x^2)^2*(d + f*x^2)^q,x]
 

Output:

(c^2*x^3*(d + f*x^2)^(1 + q))/(f*(5 + 2*q)) + ((b*c*(5 + 2*q)*x^2*(d + f*x 
^2)^(1 + q))/(2 + q) + (-(((2 + q)*(3*c^2*d - b^2*f*(5 + 2*q) - 2*a*c*f*(5 
 + 2*q))*x*(d + f*x^2)^(1 + q))/(3 + 2*q)) + (-((b*(15 + 16*q + 4*q^2)*(c* 
d - a*f*(2 + q))*(d + f*x^2)^(1 + q))/(1 + q)) + ((2 + q)*(3*c^2*d^2 - 2*a 
*c*d*f*(5 + 2*q) - f*(5 + 2*q)*(b^2*d - a^2*f*(3 + 2*q)))*x*(d + f*x^2)^q* 
Hypergeometric2F1[1/2, -q, 3/2, -((f*x^2)/d)])/(1 + (f*x^2)/d)^q)/(3 + 2*q 
))/(f*(2 + q)))/(f*(5 + 2*q))
 

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 237

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[- 
p, 1/2, 1/2 + 1, (-b)*(x^2/a)], x] /; FreeQ[{a, b, p}, x] &&  !IntegerQ[2*p 
] && GtQ[a, 0]
 

rule 238

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^2) 
^FracPart[p]/(1 + b*(x^2/a))^FracPart[p])   Int[(1 + b*(x^2/a))^p, x], x] / 
; FreeQ[{a, b, p}, x] &&  !IntegerQ[2*p] &&  !GtQ[a, 0]
 

rule 455

Int[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*(( 
a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Simp[c   Int[(a + b*x^2)^p, x], x] 
/; FreeQ[{a, b, c, d, p}, x] &&  !LeQ[p, -1]
 

rule 2346

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], 
e = Coeff[Pq, x, Expon[Pq, x]]}, Simp[e*x^(q - 1)*((a + b*x^2)^(p + 1)/(b*( 
q + 2*p + 1))), x] + Simp[1/(b*(q + 2*p + 1))   Int[(a + b*x^2)^p*ExpandToS 
um[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], 
x], x]] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] &&  !LeQ[p, -1]
 

Maple [F]

\[\int \left (c \,x^{2}+b x +a \right )^{2} \left (f \,x^{2}+d \right )^{q}d x\]

Input:

int((c*x^2+b*x+a)^2*(f*x^2+d)^q,x)
 

Output:

int((c*x^2+b*x+a)^2*(f*x^2+d)^q,x)
 

Fricas [F]

\[ \int \left (a+b x+c x^2\right )^2 \left (d+f x^2\right )^q \, dx=\int { {\left (c x^{2} + b x + a\right )}^{2} {\left (f x^{2} + d\right )}^{q} \,d x } \] Input:

integrate((c*x^2+b*x+a)^2*(f*x^2+d)^q,x, algorithm="fricas")
 

Output:

integral((c^2*x^4 + 2*b*c*x^3 + 2*a*b*x + (b^2 + 2*a*c)*x^2 + a^2)*(f*x^2 
+ d)^q, x)
 

Sympy [A] (verification not implemented)

Time = 12.57 (sec) , antiderivative size = 500, normalized size of antiderivative = 2.03 \[ \int \left (a+b x+c x^2\right )^2 \left (d+f x^2\right )^q \, dx=a^{2} d^{q} x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - q \\ \frac {3}{2} \end {matrix}\middle | {\frac {f x^{2} e^{i \pi }}{d}} \right )} + 2 a b \left (\begin {cases} \frac {d^{q} x^{2}}{2} & \text {for}\: f = 0 \\\frac {\begin {cases} \frac {\left (d + f x^{2}\right )^{q + 1}}{q + 1} & \text {for}\: q \neq -1 \\\log {\left (d + f x^{2} \right )} & \text {otherwise} \end {cases}}{2 f} & \text {otherwise} \end {cases}\right ) + \frac {2 a c d^{q} x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - q \\ \frac {5}{2} \end {matrix}\middle | {\frac {f x^{2} e^{i \pi }}{d}} \right )}}{3} + \frac {b^{2} d^{q} x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - q \\ \frac {5}{2} \end {matrix}\middle | {\frac {f x^{2} e^{i \pi }}{d}} \right )}}{3} + 2 b c \left (\begin {cases} \frac {d^{q} x^{4}}{4} & \text {for}\: f = 0 \\\frac {d \log {\left (x - \sqrt {- \frac {d}{f}} \right )}}{2 d f^{2} + 2 f^{3} x^{2}} + \frac {d \log {\left (x + \sqrt {- \frac {d}{f}} \right )}}{2 d f^{2} + 2 f^{3} x^{2}} + \frac {d}{2 d f^{2} + 2 f^{3} x^{2}} + \frac {f x^{2} \log {\left (x - \sqrt {- \frac {d}{f}} \right )}}{2 d f^{2} + 2 f^{3} x^{2}} + \frac {f x^{2} \log {\left (x + \sqrt {- \frac {d}{f}} \right )}}{2 d f^{2} + 2 f^{3} x^{2}} & \text {for}\: q = -2 \\- \frac {d \log {\left (x - \sqrt {- \frac {d}{f}} \right )}}{2 f^{2}} - \frac {d \log {\left (x + \sqrt {- \frac {d}{f}} \right )}}{2 f^{2}} + \frac {x^{2}}{2 f} & \text {for}\: q = -1 \\- \frac {d^{2} \left (d + f x^{2}\right )^{q}}{2 f^{2} q^{2} + 6 f^{2} q + 4 f^{2}} + \frac {d f q x^{2} \left (d + f x^{2}\right )^{q}}{2 f^{2} q^{2} + 6 f^{2} q + 4 f^{2}} + \frac {f^{2} q x^{4} \left (d + f x^{2}\right )^{q}}{2 f^{2} q^{2} + 6 f^{2} q + 4 f^{2}} + \frac {f^{2} x^{4} \left (d + f x^{2}\right )^{q}}{2 f^{2} q^{2} + 6 f^{2} q + 4 f^{2}} & \text {otherwise} \end {cases}\right ) + \frac {c^{2} d^{q} x^{5} {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{2}, - q \\ \frac {7}{2} \end {matrix}\middle | {\frac {f x^{2} e^{i \pi }}{d}} \right )}}{5} \] Input:

integrate((c*x**2+b*x+a)**2*(f*x**2+d)**q,x)
 

Output:

a**2*d**q*x*hyper((1/2, -q), (3/2,), f*x**2*exp_polar(I*pi)/d) + 2*a*b*Pie 
cewise((d**q*x**2/2, Eq(f, 0)), (Piecewise(((d + f*x**2)**(q + 1)/(q + 1), 
 Ne(q, -1)), (log(d + f*x**2), True))/(2*f), True)) + 2*a*c*d**q*x**3*hype 
r((3/2, -q), (5/2,), f*x**2*exp_polar(I*pi)/d)/3 + b**2*d**q*x**3*hyper((3 
/2, -q), (5/2,), f*x**2*exp_polar(I*pi)/d)/3 + 2*b*c*Piecewise((d**q*x**4/ 
4, Eq(f, 0)), (d*log(x - sqrt(-d/f))/(2*d*f**2 + 2*f**3*x**2) + d*log(x + 
sqrt(-d/f))/(2*d*f**2 + 2*f**3*x**2) + d/(2*d*f**2 + 2*f**3*x**2) + f*x**2 
*log(x - sqrt(-d/f))/(2*d*f**2 + 2*f**3*x**2) + f*x**2*log(x + sqrt(-d/f)) 
/(2*d*f**2 + 2*f**3*x**2), Eq(q, -2)), (-d*log(x - sqrt(-d/f))/(2*f**2) - 
d*log(x + sqrt(-d/f))/(2*f**2) + x**2/(2*f), Eq(q, -1)), (-d**2*(d + f*x** 
2)**q/(2*f**2*q**2 + 6*f**2*q + 4*f**2) + d*f*q*x**2*(d + f*x**2)**q/(2*f* 
*2*q**2 + 6*f**2*q + 4*f**2) + f**2*q*x**4*(d + f*x**2)**q/(2*f**2*q**2 + 
6*f**2*q + 4*f**2) + f**2*x**4*(d + f*x**2)**q/(2*f**2*q**2 + 6*f**2*q + 4 
*f**2), True)) + c**2*d**q*x**5*hyper((5/2, -q), (7/2,), f*x**2*exp_polar( 
I*pi)/d)/5
 

Maxima [F]

\[ \int \left (a+b x+c x^2\right )^2 \left (d+f x^2\right )^q \, dx=\int { {\left (c x^{2} + b x + a\right )}^{2} {\left (f x^{2} + d\right )}^{q} \,d x } \] Input:

integrate((c*x^2+b*x+a)^2*(f*x^2+d)^q,x, algorithm="maxima")
 

Output:

integrate((c*x^2 + b*x + a)^2*(f*x^2 + d)^q, x)
 

Giac [F]

\[ \int \left (a+b x+c x^2\right )^2 \left (d+f x^2\right )^q \, dx=\int { {\left (c x^{2} + b x + a\right )}^{2} {\left (f x^{2} + d\right )}^{q} \,d x } \] Input:

integrate((c*x^2+b*x+a)^2*(f*x^2+d)^q,x, algorithm="giac")
 

Output:

integrate((c*x^2 + b*x + a)^2*(f*x^2 + d)^q, x)
 

Mupad [F(-1)]

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

int((d + f*x^2)^q*(a + b*x + c*x^2)^2,x)
 

Output:

int((d + f*x^2)^q*(a + b*x + c*x^2)^2, x)
 

Reduce [F]

\[ \int \left (a+b x+c x^2\right )^2 \left (d+f x^2\right )^q \, dx=\text {too large to display} \] Input:

int((c*x^2+b*x+a)^2*(f*x^2+d)^q,x)
 

Output:

(4*(d + f*x**2)**q*a**2*f**2*q**4*x + 28*(d + f*x**2)**q*a**2*f**2*q**3*x 
+ 71*(d + f*x**2)**q*a**2*f**2*q**2*x + 77*(d + f*x**2)**q*a**2*f**2*q*x + 
 30*(d + f*x**2)**q*a**2*f**2*x + 8*(d + f*x**2)**q*a*b*d*f*q**4 + 52*(d + 
 f*x**2)**q*a*b*d*f*q**3 + 118*(d + f*x**2)**q*a*b*d*f*q**2 + 107*(d + f*x 
**2)**q*a*b*d*f*q + 30*(d + f*x**2)**q*a*b*d*f + 8*(d + f*x**2)**q*a*b*f** 
2*q**4*x**2 + 52*(d + f*x**2)**q*a*b*f**2*q**3*x**2 + 118*(d + f*x**2)**q* 
a*b*f**2*q**2*x**2 + 107*(d + f*x**2)**q*a*b*f**2*q*x**2 + 30*(d + f*x**2) 
**q*a*b*f**2*x**2 + 8*(d + f*x**2)**q*a*c*d*f*q**4*x + 44*(d + f*x**2)**q* 
a*c*d*f*q**3*x + 76*(d + f*x**2)**q*a*c*d*f*q**2*x + 40*(d + f*x**2)**q*a* 
c*d*f*q*x + 8*(d + f*x**2)**q*a*c*f**2*q**4*x**3 + 48*(d + f*x**2)**q*a*c* 
f**2*q**3*x**3 + 98*(d + f*x**2)**q*a*c*f**2*q**2*x**3 + 78*(d + f*x**2)** 
q*a*c*f**2*q*x**3 + 20*(d + f*x**2)**q*a*c*f**2*x**3 + 4*(d + f*x**2)**q*b 
**2*d*f*q**4*x + 22*(d + f*x**2)**q*b**2*d*f*q**3*x + 38*(d + f*x**2)**q*b 
**2*d*f*q**2*x + 20*(d + f*x**2)**q*b**2*d*f*q*x + 4*(d + f*x**2)**q*b**2* 
f**2*q**4*x**3 + 24*(d + f*x**2)**q*b**2*f**2*q**3*x**3 + 49*(d + f*x**2)* 
*q*b**2*f**2*q**2*x**3 + 39*(d + f*x**2)**q*b**2*f**2*q*x**3 + 10*(d + f*x 
**2)**q*b**2*f**2*x**3 - 8*(d + f*x**2)**q*b*c*d**2*q**3 - 36*(d + f*x**2) 
**q*b*c*d**2*q**2 - 46*(d + f*x**2)**q*b*c*d**2*q - 15*(d + f*x**2)**q*b*c 
*d**2 + 8*(d + f*x**2)**q*b*c*d*f*q**4*x**2 + 36*(d + f*x**2)**q*b*c*d*f*q 
**3*x**2 + 46*(d + f*x**2)**q*b*c*d*f*q**2*x**2 + 15*(d + f*x**2)**q*b*...