\(\int \frac {(e x)^m (a+b x^2)^2}{\sqrt {c+d x}} \, dx\) [139]

Optimal result

Integrand size = 24, antiderivative size = 384 \[ \int \frac {(e x)^m \left (a+b x^2\right )^2}{\sqrt {c+d x}} \, dx=-\frac {8 b c (2+m) \left (2 b c^2 \left (12+7 m+m^2\right )+a d^2 \left (63+32 m+4 m^2\right )\right ) (e x)^{1+m} \sqrt {c+d x}}{d^4 e (3+2 m) (5+2 m) (7+2 m) (9+2 m)}+\frac {4 b \left (2 b c^2 \left (12+7 m+m^2\right )+a d^2 \left (63+32 m+4 m^2\right )\right ) (e x)^{2+m} \sqrt {c+d x}}{d^3 e^2 (5+2 m) (7+2 m) (9+2 m)}-\frac {4 b^2 c (4+m) (e x)^{3+m} \sqrt {c+d x}}{d^2 e^3 (7+2 m) (9+2 m)}+\frac {2 b^2 (e x)^{4+m} \sqrt {c+d x}}{d e^4 (9+2 m)}+\frac {2 \left (a^2 d^4 (5+2 m) \left (63+32 m+4 m^2\right )+\frac {4 b c^2 (1+m) (2+m) \left (2 b c^2 \left (12+7 m+m^2\right )+a d^2 \left (63+32 m+4 m^2\right )\right )}{\frac {3}{2}+m}\right ) \left (-\frac {d x}{c}\right )^{-m} (e x)^m \sqrt {c+d x} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m,\frac {3}{2},1+\frac {d x}{c}\right )}{d^5 (5+2 m) (7+2 m) (9+2 m)} \] Output:

-8*b*c*(2+m)*(2*b*c^2*(m^2+7*m+12)+a*d^2*(4*m^2+32*m+63))*(e*x)^(1+m)*(d*x 
+c)^(1/2)/d^4/e/(3+2*m)/(5+2*m)/(7+2*m)/(9+2*m)+4*b*(2*b*c^2*(m^2+7*m+12)+ 
a*d^2*(4*m^2+32*m+63))*(e*x)^(2+m)*(d*x+c)^(1/2)/d^3/e^2/(5+2*m)/(7+2*m)/( 
9+2*m)-4*b^2*c*(4+m)*(e*x)^(3+m)*(d*x+c)^(1/2)/d^2/e^3/(7+2*m)/(9+2*m)+2*b 
^2*(e*x)^(4+m)*(d*x+c)^(1/2)/d/e^4/(9+2*m)+2*(a^2*d^4*(5+2*m)*(4*m^2+32*m+ 
63)+4*b*c^2*(1+m)*(2+m)*(2*b*c^2*(m^2+7*m+12)+a*d^2*(4*m^2+32*m+63))/(3/2+ 
m))*(e*x)^m*(d*x+c)^(1/2)*hypergeom([1/2, -m],[3/2],1+d*x/c)/d^5/(5+2*m)/( 
7+2*m)/(9+2*m)/((-d*x/c)^m)
 

Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.53 \[ \int \frac {(e x)^m \left (a+b x^2\right )^2}{\sqrt {c+d x}} \, dx=\frac {2 \left (-\frac {d x}{c}\right )^{-m} (e x)^m \sqrt {c+d x} \left (315 \left (b c^2+a d^2\right )^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m,\frac {3}{2},1+\frac {d x}{c}\right )-420 b c \left (b c^2+a d^2\right ) (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-m,\frac {5}{2},1+\frac {d x}{c}\right )+126 b \left (3 b c^2+a d^2\right ) (c+d x)^2 \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-m,\frac {7}{2},1+\frac {d x}{c}\right )-180 b^2 c (c+d x)^3 \operatorname {Hypergeometric2F1}\left (\frac {7}{2},-m,\frac {9}{2},1+\frac {d x}{c}\right )+35 b^2 (c+d x)^4 \operatorname {Hypergeometric2F1}\left (\frac {9}{2},-m,\frac {11}{2},1+\frac {d x}{c}\right )\right )}{315 d^5} \] Input:

Integrate[((e*x)^m*(a + b*x^2)^2)/Sqrt[c + d*x],x]
 

Output:

(2*(e*x)^m*Sqrt[c + d*x]*(315*(b*c^2 + a*d^2)^2*Hypergeometric2F1[1/2, -m, 
 3/2, 1 + (d*x)/c] - 420*b*c*(b*c^2 + a*d^2)*(c + d*x)*Hypergeometric2F1[3 
/2, -m, 5/2, 1 + (d*x)/c] + 126*b*(3*b*c^2 + a*d^2)*(c + d*x)^2*Hypergeome 
tric2F1[5/2, -m, 7/2, 1 + (d*x)/c] - 180*b^2*c*(c + d*x)^3*Hypergeometric2 
F1[7/2, -m, 9/2, 1 + (d*x)/c] + 35*b^2*(c + d*x)^4*Hypergeometric2F1[9/2, 
-m, 11/2, 1 + (d*x)/c]))/(315*d^5*(-((d*x)/c))^m)
 

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 361, normalized size of antiderivative = 0.94, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {521, 27, 2125, 27, 521, 27, 90, 77, 75}

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.

\(\displaystyle \int \frac {\left (a+b x^2\right )^2 (e x)^m}{\sqrt {c+d x}} \, dx\)

\(\Big \downarrow \) 521

\(\displaystyle \frac {2 \int \frac {(e x)^m \left (-2 b^2 c (m+4) x^3 e^4+2 a b d (2 m+9) x^2 e^4+a^2 d (2 m+9) e^4\right )}{2 \sqrt {c+d x}}dx}{d e^4 (2 m+9)}+\frac {2 b^2 \sqrt {c+d x} (e x)^{m+4}}{d e^4 (2 m+9)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {(e x)^m \left (-2 b^2 c (m+4) x^3 e^4+2 a b d (2 m+9) x^2 e^4+a^2 d (2 m+9) e^4\right )}{\sqrt {c+d x}}dx}{d e^4 (2 m+9)}+\frac {2 b^2 \sqrt {c+d x} (e x)^{m+4}}{d e^4 (2 m+9)}\)

\(\Big \downarrow \) 2125

\(\displaystyle \frac {\frac {2 \int \frac {e^7 (e x)^m \left (a^2 \left (4 m^2+32 m+63\right ) d^2+2 b \left (2 b \left (m^2+7 m+12\right ) c^2+a d^2 \left (4 m^2+32 m+63\right )\right ) x^2\right )}{2 \sqrt {c+d x}}dx}{d e^3 (2 m+7)}-\frac {4 b^2 c e (m+4) \sqrt {c+d x} (e x)^{m+3}}{d (2 m+7)}}{d e^4 (2 m+9)}+\frac {2 b^2 \sqrt {c+d x} (e x)^{m+4}}{d e^4 (2 m+9)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {e^4 \int \frac {(e x)^m \left (a^2 \left (4 m^2+32 m+63\right ) d^2+2 b \left (2 b \left (m^2+7 m+12\right ) c^2+a d^2 \left (4 m^2+32 m+63\right )\right ) x^2\right )}{\sqrt {c+d x}}dx}{d (2 m+7)}-\frac {4 b^2 c e (m+4) \sqrt {c+d x} (e x)^{m+3}}{d (2 m+7)}}{d e^4 (2 m+9)}+\frac {2 b^2 \sqrt {c+d x} (e x)^{m+4}}{d e^4 (2 m+9)}\)

\(\Big \downarrow \) 521

\(\displaystyle \frac {\frac {e^4 \left (\frac {2 \int \frac {e^2 (e x)^m \left (a^2 d^3 \left (8 m^3+84 m^2+286 m+315\right )-4 b c (m+2) \left (2 b \left (m^2+7 m+12\right ) c^2+a d^2 \left (4 m^2+32 m+63\right )\right ) x\right )}{2 \sqrt {c+d x}}dx}{d e^2 (2 m+5)}+\frac {4 b \sqrt {c+d x} (e x)^{m+2} \left (a d^2 \left (4 m^2+32 m+63\right )+2 b c^2 \left (m^2+7 m+12\right )\right )}{d e^2 (2 m+5)}\right )}{d (2 m+7)}-\frac {4 b^2 c e (m+4) \sqrt {c+d x} (e x)^{m+3}}{d (2 m+7)}}{d e^4 (2 m+9)}+\frac {2 b^2 \sqrt {c+d x} (e x)^{m+4}}{d e^4 (2 m+9)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {e^4 \left (\frac {\int \frac {(e x)^m \left (a^2 d^3 \left (8 m^3+84 m^2+286 m+315\right )-4 b c (m+2) \left (2 b \left (m^2+7 m+12\right ) c^2+a d^2 \left (4 m^2+32 m+63\right )\right ) x\right )}{\sqrt {c+d x}}dx}{d (2 m+5)}+\frac {4 b \sqrt {c+d x} (e x)^{m+2} \left (a d^2 \left (4 m^2+32 m+63\right )+2 b c^2 \left (m^2+7 m+12\right )\right )}{d e^2 (2 m+5)}\right )}{d (2 m+7)}-\frac {4 b^2 c e (m+4) \sqrt {c+d x} (e x)^{m+3}}{d (2 m+7)}}{d e^4 (2 m+9)}+\frac {2 b^2 \sqrt {c+d x} (e x)^{m+4}}{d e^4 (2 m+9)}\)

\(\Big \downarrow \) 90

\(\displaystyle \frac {\frac {e^4 \left (\frac {\frac {\left (a^2 d^4 \left (8 m^3+84 m^2+286 m+315\right )+\frac {4 b c^2 (m+1) (m+2) \left (a d^2 \left (4 m^2+32 m+63\right )+2 b c^2 \left (m^2+7 m+12\right )\right )}{m+\frac {3}{2}}\right ) \int \frac {(e x)^m}{\sqrt {c+d x}}dx}{d}-\frac {8 b c (m+2) \sqrt {c+d x} (e x)^{m+1} \left (a d^2 \left (4 m^2+32 m+63\right )+2 b c^2 \left (m^2+7 m+12\right )\right )}{d e (2 m+3)}}{d (2 m+5)}+\frac {4 b \sqrt {c+d x} (e x)^{m+2} \left (a d^2 \left (4 m^2+32 m+63\right )+2 b c^2 \left (m^2+7 m+12\right )\right )}{d e^2 (2 m+5)}\right )}{d (2 m+7)}-\frac {4 b^2 c e (m+4) \sqrt {c+d x} (e x)^{m+3}}{d (2 m+7)}}{d e^4 (2 m+9)}+\frac {2 b^2 \sqrt {c+d x} (e x)^{m+4}}{d e^4 (2 m+9)}\)

\(\Big \downarrow \) 77

\(\displaystyle \frac {\frac {e^4 \left (\frac {\frac {(e x)^m \left (-\frac {d x}{c}\right )^{-m} \left (a^2 d^4 \left (8 m^3+84 m^2+286 m+315\right )+\frac {4 b c^2 (m+1) (m+2) \left (a d^2 \left (4 m^2+32 m+63\right )+2 b c^2 \left (m^2+7 m+12\right )\right )}{m+\frac {3}{2}}\right ) \int \frac {\left (-\frac {d x}{c}\right )^m}{\sqrt {c+d x}}dx}{d}-\frac {8 b c (m+2) \sqrt {c+d x} (e x)^{m+1} \left (a d^2 \left (4 m^2+32 m+63\right )+2 b c^2 \left (m^2+7 m+12\right )\right )}{d e (2 m+3)}}{d (2 m+5)}+\frac {4 b \sqrt {c+d x} (e x)^{m+2} \left (a d^2 \left (4 m^2+32 m+63\right )+2 b c^2 \left (m^2+7 m+12\right )\right )}{d e^2 (2 m+5)}\right )}{d (2 m+7)}-\frac {4 b^2 c e (m+4) \sqrt {c+d x} (e x)^{m+3}}{d (2 m+7)}}{d e^4 (2 m+9)}+\frac {2 b^2 \sqrt {c+d x} (e x)^{m+4}}{d e^4 (2 m+9)}\)

\(\Big \downarrow \) 75

\(\displaystyle \frac {\frac {e^4 \left (\frac {\frac {2 \sqrt {c+d x} (e x)^m \left (-\frac {d x}{c}\right )^{-m} \left (a^2 d^4 \left (8 m^3+84 m^2+286 m+315\right )+\frac {4 b c^2 (m+1) (m+2) \left (a d^2 \left (4 m^2+32 m+63\right )+2 b c^2 \left (m^2+7 m+12\right )\right )}{m+\frac {3}{2}}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m,\frac {3}{2},\frac {d x}{c}+1\right )}{d^2}-\frac {8 b c (m+2) \sqrt {c+d x} (e x)^{m+1} \left (a d^2 \left (4 m^2+32 m+63\right )+2 b c^2 \left (m^2+7 m+12\right )\right )}{d e (2 m+3)}}{d (2 m+5)}+\frac {4 b \sqrt {c+d x} (e x)^{m+2} \left (a d^2 \left (4 m^2+32 m+63\right )+2 b c^2 \left (m^2+7 m+12\right )\right )}{d e^2 (2 m+5)}\right )}{d (2 m+7)}-\frac {4 b^2 c e (m+4) \sqrt {c+d x} (e x)^{m+3}}{d (2 m+7)}}{d e^4 (2 m+9)}+\frac {2 b^2 \sqrt {c+d x} (e x)^{m+4}}{d e^4 (2 m+9)}\)

Input:

Int[((e*x)^m*(a + b*x^2)^2)/Sqrt[c + d*x],x]
 

Output:

(2*b^2*(e*x)^(4 + m)*Sqrt[c + d*x])/(d*e^4*(9 + 2*m)) + ((-4*b^2*c*e*(4 + 
m)*(e*x)^(3 + m)*Sqrt[c + d*x])/(d*(7 + 2*m)) + (e^4*((4*b*(2*b*c^2*(12 + 
7*m + m^2) + a*d^2*(63 + 32*m + 4*m^2))*(e*x)^(2 + m)*Sqrt[c + d*x])/(d*e^ 
2*(5 + 2*m)) + ((-8*b*c*(2 + m)*(2*b*c^2*(12 + 7*m + m^2) + a*d^2*(63 + 32 
*m + 4*m^2))*(e*x)^(1 + m)*Sqrt[c + d*x])/(d*e*(3 + 2*m)) + (2*(a^2*d^4*(3 
15 + 286*m + 84*m^2 + 8*m^3) + (4*b*c^2*(1 + m)*(2 + m)*(2*b*c^2*(12 + 7*m 
 + m^2) + a*d^2*(63 + 32*m + 4*m^2)))/(3/2 + m))*(e*x)^m*Sqrt[c + d*x]*Hyp 
ergeometric2F1[1/2, -m, 3/2, 1 + (d*x)/c])/(d^2*(-((d*x)/c))^m))/(d*(5 + 2 
*m))))/(d*(7 + 2*m)))/(d*e^4*(9 + 2*m))
 

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 75

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x 
)^(n + 1)/(d*(n + 1)*(-d/(b*c))^m))*Hypergeometric2F1[-m, n + 1, n + 2, 1 + 
 d*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (IntegerQ[m] 
 || GtQ[-d/(b*c), 0])
 

rule 77

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((-b)*(c/ 
d))^IntPart[m]*((b*x)^FracPart[m]/((-d)*(x/c))^FracPart[m])   Int[((-d)*(x/ 
c))^m*(c + d*x)^n, x], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && 
 !IntegerQ[n] &&  !GtQ[c, 0] &&  !GtQ[-d/(b*c), 0]
 

rule 90

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), 
 x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p 
+ 2))   Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, 
p}, x] && NeQ[n + p + 2, 0]
 

rule 521

Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), 
 x_Symbol] :> Simp[b^p*(e*x)^(m + 2*p)*((c + d*x)^(n + 1)/(d*e^(2*p)*(m + n 
 + 2*p + 1))), x] + Simp[1/(d*e^(2*p)*(m + n + 2*p + 1))   Int[(e*x)^m*(c + 
 d*x)^n*ExpandToSum[d*(m + n + 2*p + 1)*(e^(2*p)*(a + b*x^2)^p - b^p*(e*x)^ 
(2*p)) - b^p*(e*c)*(m + 2*p)*(e*x)^(2*p - 1), x], x], x] /; FreeQ[{a, b, c, 
 d, e}, x] && IGtQ[p, 0] && NeQ[m + n + 2*p + 1, 0] &&  !IntegerQ[m] &&  !I 
ntegerQ[n]
 

rule 2125

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] 
:> With[{q = Expon[Px, x], k = Coeff[Px, x, Expon[Px, x]]}, Simp[k*(a + b*x 
)^(m + q)*((c + d*x)^(n + 1)/(d*b^q*(m + n + q + 1))), x] + Simp[1/(d*b^q*( 
m + n + q + 1))   Int[(a + b*x)^m*(c + d*x)^n*ExpandToSum[d*b^q*(m + n + q 
+ 1)*Px - d*k*(m + n + q + 1)*(a + b*x)^q - k*(b*c - a*d)*(m + q)*(a + b*x) 
^(q - 1), x], x], x] /; NeQ[m + n + q + 1, 0]] /; FreeQ[{a, b, c, d, m, n}, 
 x] && PolyQ[Px, x]
 

Maple [F]

\[\int \frac {\left (e x \right )^{m} \left (b \,x^{2}+a \right )^{2}}{\sqrt {d x +c}}d x\]

Input:

int((e*x)^m*(b*x^2+a)^2/(d*x+c)^(1/2),x)
 

Output:

int((e*x)^m*(b*x^2+a)^2/(d*x+c)^(1/2),x)
 

Fricas [F]

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

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

Output:

integral((b^2*x^4 + 2*a*b*x^2 + a^2)*(e*x)^m/sqrt(d*x + c), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 2.58 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.35 \[ \int \frac {(e x)^m \left (a+b x^2\right )^2}{\sqrt {c+d x}} \, dx=\frac {a^{2} e^{m} x^{m + 1} \Gamma \left (m + 1\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, m + 1 \\ m + 2 \end {matrix}\middle | {\frac {d x e^{i \pi }}{c}} \right )}}{\sqrt {c} \Gamma \left (m + 2\right )} + \frac {2 a b e^{m} x^{m + 3} \Gamma \left (m + 3\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, m + 3 \\ m + 4 \end {matrix}\middle | {\frac {d x e^{i \pi }}{c}} \right )}}{\sqrt {c} \Gamma \left (m + 4\right )} + \frac {b^{2} e^{m} x^{m + 5} \Gamma \left (m + 5\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, m + 5 \\ m + 6 \end {matrix}\middle | {\frac {d x e^{i \pi }}{c}} \right )}}{\sqrt {c} \Gamma \left (m + 6\right )} \] Input:

integrate((e*x)**m*(b*x**2+a)**2/(d*x+c)**(1/2),x)
 

Output:

a**2*e**m*x**(m + 1)*gamma(m + 1)*hyper((1/2, m + 1), (m + 2,), d*x*exp_po 
lar(I*pi)/c)/(sqrt(c)*gamma(m + 2)) + 2*a*b*e**m*x**(m + 3)*gamma(m + 3)*h 
yper((1/2, m + 3), (m + 4,), d*x*exp_polar(I*pi)/c)/(sqrt(c)*gamma(m + 4)) 
 + b**2*e**m*x**(m + 5)*gamma(m + 5)*hyper((1/2, m + 5), (m + 6,), d*x*exp 
_polar(I*pi)/c)/(sqrt(c)*gamma(m + 6))
 

Maxima [F]

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

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

Output:

integrate((b*x^2 + a)^2*(e*x)^m/sqrt(d*x + c), x)
 

Giac [F]

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

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

Output:

integrate((b*x^2 + a)^2*(e*x)^m/sqrt(d*x + c), x)
 

Mupad [F(-1)]

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

int(((e*x)^m*(a + b*x^2)^2)/(c + d*x)^(1/2),x)
 

Output:

int(((e*x)^m*(a + b*x^2)^2)/(c + d*x)^(1/2), x)
 

Reduce [F]

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

int((e*x)^m*(b*x^2+a)^2/(d*x+c)^(1/2),x)
 

Output:

(2*e**m*(16*x**m*sqrt(c + d*x)*a**2*d**4*m**4 + 192*x**m*sqrt(c + d*x)*a** 
2*d**4*m**3 + 824*x**m*sqrt(c + d*x)*a**2*d**4*m**2 + 1488*x**m*sqrt(c + d 
*x)*a**2*d**4*m + 945*x**m*sqrt(c + d*x)*a**2*d**4 + 32*x**m*sqrt(c + d*x) 
*a*b*c**2*d**2*m**4 + 352*x**m*sqrt(c + d*x)*a*b*c**2*d**2*m**3 + 1336*x** 
m*sqrt(c + d*x)*a*b*c**2*d**2*m**2 + 2024*x**m*sqrt(c + d*x)*a*b*c**2*d**2 
*m + 1008*x**m*sqrt(c + d*x)*a*b*c**2*d**2 - 32*x**m*sqrt(c + d*x)*a*b*c*d 
**3*m**4*x - 336*x**m*sqrt(c + d*x)*a*b*c*d**3*m**3*x - 1176*x**m*sqrt(c + 
 d*x)*a*b*c*d**3*m**2*x - 1516*x**m*sqrt(c + d*x)*a*b*c*d**3*m*x - 504*x** 
m*sqrt(c + d*x)*a*b*c*d**3*x + 32*x**m*sqrt(c + d*x)*a*b*d**4*m**4*x**2 + 
320*x**m*sqrt(c + d*x)*a*b*d**4*m**3*x**2 + 1040*x**m*sqrt(c + d*x)*a*b*d* 
*4*m**2*x**2 + 1200*x**m*sqrt(c + d*x)*a*b*d**4*m*x**2 + 378*x**m*sqrt(c + 
 d*x)*a*b*d**4*x**2 + 16*x**m*sqrt(c + d*x)*b**2*c**4*m**4 + 160*x**m*sqrt 
(c + d*x)*b**2*c**4*m**3 + 560*x**m*sqrt(c + d*x)*b**2*c**4*m**2 + 800*x** 
m*sqrt(c + d*x)*b**2*c**4*m + 384*x**m*sqrt(c + d*x)*b**2*c**4 - 16*x**m*s 
qrt(c + d*x)*b**2*c**3*d*m**4*x - 152*x**m*sqrt(c + d*x)*b**2*c**3*d*m**3* 
x - 488*x**m*sqrt(c + d*x)*b**2*c**3*d*m**2*x - 592*x**m*sqrt(c + d*x)*b** 
2*c**3*d*m*x - 192*x**m*sqrt(c + d*x)*b**2*c**3*d*x + 16*x**m*sqrt(c + d*x 
)*b**2*c**2*d**2*m**4*x**2 + 144*x**m*sqrt(c + d*x)*b**2*c**2*d**2*m**3*x* 
*2 + 428*x**m*sqrt(c + d*x)*b**2*c**2*d**2*m**2*x**2 + 468*x**m*sqrt(c + d 
*x)*b**2*c**2*d**2*m*x**2 + 144*x**m*sqrt(c + d*x)*b**2*c**2*d**2*x**2 ...