\(\int \frac {(e x)^m (a+b x^2)^2}{(c+d x)^{3/2}} \, dx\) [140]

Optimal result

Integrand size = 24, antiderivative size = 357 \[ \int \frac {(e x)^m \left (a+b x^2\right )^2}{(c+d x)^{3/2}} \, dx=\frac {2 \left (b c^2+a d^2\right )^2 (e x)^{1+m}}{c d^4 e \sqrt {c+d x}}+\frac {2 b \left (3 b c^2 \left (29+22 m+4 m^2\right )+2 a d^2 \left (35+24 m+4 m^2\right )\right ) (e x)^{1+m} \sqrt {c+d x}}{d^4 e (3+2 m) (5+2 m) (7+2 m)}-\frac {2 b^2 c (13+4 m) (e x)^{2+m} \sqrt {c+d x}}{d^3 e^2 (5+2 m) (7+2 m)}+\frac {2 b^2 (e x)^{3+m} \sqrt {c+d x}}{d^2 e^3 (7+2 m)}-\frac {2 \left (16 b^2 c^4 \left (24+50 m+35 m^2+10 m^3+m^4\right )+8 a b c^2 d^2 \left (70+153 m+115 m^2+36 m^3+4 m^4\right )+a^2 d^4 \left (105+352 m+344 m^2+128 m^3+16 m^4\right )\right ) \left (-\frac {d x}{c}\right )^{-m} (e x)^m \sqrt {c+d x} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m,\frac {3}{2},1+\frac {d x}{c}\right )}{c d^5 (3+2 m) (5+2 m) (7+2 m)} \] Output:

2*(a*d^2+b*c^2)^2*(e*x)^(1+m)/c/d^4/e/(d*x+c)^(1/2)+2*b*(3*b*c^2*(4*m^2+22 
*m+29)+2*a*d^2*(4*m^2+24*m+35))*(e*x)^(1+m)*(d*x+c)^(1/2)/d^4/e/(3+2*m)/(5 
+2*m)/(7+2*m)-2*b^2*c*(13+4*m)*(e*x)^(2+m)*(d*x+c)^(1/2)/d^3/e^2/(5+2*m)/( 
7+2*m)+2*b^2*(e*x)^(3+m)*(d*x+c)^(1/2)/d^2/e^3/(7+2*m)-2*(16*b^2*c^4*(m^4+ 
10*m^3+35*m^2+50*m+24)+8*a*b*c^2*d^2*(4*m^4+36*m^3+115*m^2+153*m+70)+a^2*d 
^4*(16*m^4+128*m^3+344*m^2+352*m+105))*(e*x)^m*(d*x+c)^(1/2)*hypergeom([1/ 
2, -m],[3/2],1+d*x/c)/c/d^5/(3+2*m)/(5+2*m)/(7+2*m)/((-d*x/c)^m)
 

Mathematica [A] (verified)

Time = 0.54 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.57 \[ \int \frac {(e x)^m \left (a+b x^2\right )^2}{(c+d x)^{3/2}} \, dx=\frac {2 \left (-\frac {d x}{c}\right )^{-m} (e x)^m \left (-105 \left (b c^2+a d^2\right )^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-m,\frac {1}{2},1+\frac {d x}{c}\right )-420 b c \left (b c^2+a d^2\right ) (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m,\frac {3}{2},1+\frac {d x}{c}\right )+70 b \left (3 b c^2+a d^2\right ) (c+d x)^2 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-m,\frac {5}{2},1+\frac {d x}{c}\right )-84 b^2 c (c+d x)^3 \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-m,\frac {7}{2},1+\frac {d x}{c}\right )+15 b^2 (c+d x)^4 \operatorname {Hypergeometric2F1}\left (\frac {7}{2},-m,\frac {9}{2},1+\frac {d x}{c}\right )\right )}{105 d^5 \sqrt {c+d x}} \] Input:

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

Output:

(2*(e*x)^m*(-105*(b*c^2 + a*d^2)^2*Hypergeometric2F1[-1/2, -m, 1/2, 1 + (d 
*x)/c] - 420*b*c*(b*c^2 + a*d^2)*(c + d*x)*Hypergeometric2F1[1/2, -m, 3/2, 
 1 + (d*x)/c] + 70*b*(3*b*c^2 + a*d^2)*(c + d*x)^2*Hypergeometric2F1[3/2, 
-m, 5/2, 1 + (d*x)/c] - 84*b^2*c*(c + d*x)^3*Hypergeometric2F1[5/2, -m, 7/ 
2, 1 + (d*x)/c] + 15*b^2*(c + d*x)^4*Hypergeometric2F1[7/2, -m, 9/2, 1 + ( 
d*x)/c]))/(105*d^5*(-((d*x)/c))^m*Sqrt[c + d*x])
 

Rubi [A] (verified)

Time = 0.87 (sec) , antiderivative size = 354, normalized size of antiderivative = 0.99, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {519, 27, 2125, 27, 1194, 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.

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

\(\Big \downarrow \) 519

\(\displaystyle \frac {2 (e x)^{m+1} \left (a d^2+b c^2\right )^2}{c d^4 e \sqrt {c+d x}}-\frac {2 \int \frac {(e x)^m \left (\frac {2 b^2 (m+1) c^4}{d^4}+\frac {b^2 x^2 c^2}{d^2}+\frac {4 a b (m+1) c^2}{d^2}-\frac {b^2 x^3 c}{d}-\frac {b \left (b c^2+2 a d^2\right ) x c}{d^3}+2 a^2 \left (m+\frac {1}{2}\right )\right )}{2 \sqrt {c+d x}}dx}{c}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 (e x)^{m+1} \left (a d^2+b c^2\right )^2}{c d^4 e \sqrt {c+d x}}-\frac {\int \frac {(e x)^m \left (\frac {2 b^2 (m+1) c^4}{d^4}+\frac {b^2 x^2 c^2}{d^2}+\frac {4 a b (m+1) c^2}{d^2}-\frac {b^2 x^3 c}{d}-\frac {b \left (b c^2+2 a d^2\right ) x c}{d^3}+a^2 (2 m+1)\right )}{\sqrt {c+d x}}dx}{c}\)

\(\Big \downarrow \) 2125

\(\displaystyle \frac {2 (e x)^{m+1} \left (a d^2+b c^2\right )^2}{c d^4 e \sqrt {c+d x}}-\frac {\frac {2 \int \frac {(e x)^m \left (\frac {b^2 c^2 (4 m+13) x^2 e^3}{d}+\frac {(2 m+7) \left (2 b^2 (m+1) c^4+4 a b d^2 (m+1) c^2+a^2 d^4 (2 m+1)\right ) e^3}{d^3}-\frac {b c \left (b c^2+2 a d^2\right ) (2 m+7) x e^3}{d^2}\right )}{2 \sqrt {c+d x}}dx}{d e^3 (2 m+7)}-\frac {2 b^2 c \sqrt {c+d x} (e x)^{m+3}}{d^2 e^3 (2 m+7)}}{c}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 (e x)^{m+1} \left (a d^2+b c^2\right )^2}{c d^4 e \sqrt {c+d x}}-\frac {\frac {\int \frac {(e x)^m \left (\frac {b^2 c^2 (4 m+13) x^2 e^3}{d}+\frac {(2 m+7) \left (2 b^2 (m+1) c^4+4 a b d^2 (m+1) c^2+a^2 d^4 (2 m+1)\right ) e^3}{d^3}-\frac {b c \left (b c^2+2 a d^2\right ) (2 m+7) x e^3}{d^2}\right )}{\sqrt {c+d x}}dx}{d e^3 (2 m+7)}-\frac {2 b^2 c \sqrt {c+d x} (e x)^{m+3}}{d^2 e^3 (2 m+7)}}{c}\)

\(\Big \downarrow \) 1194

\(\displaystyle \frac {2 (e x)^{m+1} \left (a d^2+b c^2\right )^2}{c d^4 e \sqrt {c+d x}}-\frac {\frac {\frac {2 \int \frac {e^5 (e x)^m \left (\left (4 m^2+24 m+35\right ) \left (2 b^2 (m+1) c^4+4 a b d^2 (m+1) c^2+a^2 d^4 (2 m+1)\right )-b c d \left (3 b \left (4 m^2+22 m+29\right ) c^2+2 a d^2 \left (4 m^2+24 m+35\right )\right ) x\right )}{2 d^2 \sqrt {c+d x}}dx}{d e^2 (2 m+5)}+\frac {2 b^2 c^2 e (4 m+13) \sqrt {c+d x} (e x)^{m+2}}{d^2 (2 m+5)}}{d e^3 (2 m+7)}-\frac {2 b^2 c \sqrt {c+d x} (e x)^{m+3}}{d^2 e^3 (2 m+7)}}{c}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 (e x)^{m+1} \left (a d^2+b c^2\right )^2}{c d^4 e \sqrt {c+d x}}-\frac {\frac {\frac {e^3 \int \frac {(e x)^m \left (\left (4 m^2+24 m+35\right ) \left (2 b^2 (m+1) c^4+4 a b d^2 (m+1) c^2+a^2 d^4 (2 m+1)\right )-b c d \left (3 b \left (4 m^2+22 m+29\right ) c^2+2 a d^2 \left (4 m^2+24 m+35\right )\right ) x\right )}{\sqrt {c+d x}}dx}{d^3 (2 m+5)}+\frac {2 b^2 c^2 e (4 m+13) \sqrt {c+d x} (e x)^{m+2}}{d^2 (2 m+5)}}{d e^3 (2 m+7)}-\frac {2 b^2 c \sqrt {c+d x} (e x)^{m+3}}{d^2 e^3 (2 m+7)}}{c}\)

\(\Big \downarrow \) 90

\(\displaystyle \frac {2 (e x)^{m+1} \left (a d^2+b c^2\right )^2}{c d^4 e \sqrt {c+d x}}-\frac {\frac {\frac {e^3 \left (\frac {\left (a^2 d^4 \left (16 m^4+128 m^3+344 m^2+352 m+105\right )+8 a b c^2 d^2 \left (4 m^4+36 m^3+115 m^2+153 m+70\right )+16 b^2 c^4 \left (m^4+10 m^3+35 m^2+50 m+24\right )\right ) \int \frac {(e x)^m}{\sqrt {c+d x}}dx}{2 m+3}-\frac {2 b c \sqrt {c+d x} (e x)^{m+1} \left (2 a d^2 \left (4 m^2+24 m+35\right )+3 b c^2 \left (4 m^2+22 m+29\right )\right )}{e (2 m+3)}\right )}{d^3 (2 m+5)}+\frac {2 b^2 c^2 e (4 m+13) \sqrt {c+d x} (e x)^{m+2}}{d^2 (2 m+5)}}{d e^3 (2 m+7)}-\frac {2 b^2 c \sqrt {c+d x} (e x)^{m+3}}{d^2 e^3 (2 m+7)}}{c}\)

\(\Big \downarrow \) 77

\(\displaystyle \frac {2 (e x)^{m+1} \left (a d^2+b c^2\right )^2}{c d^4 e \sqrt {c+d x}}-\frac {\frac {\frac {e^3 \left (\frac {(e x)^m \left (-\frac {d x}{c}\right )^{-m} \left (a^2 d^4 \left (16 m^4+128 m^3+344 m^2+352 m+105\right )+8 a b c^2 d^2 \left (4 m^4+36 m^3+115 m^2+153 m+70\right )+16 b^2 c^4 \left (m^4+10 m^3+35 m^2+50 m+24\right )\right ) \int \frac {\left (-\frac {d x}{c}\right )^m}{\sqrt {c+d x}}dx}{2 m+3}-\frac {2 b c \sqrt {c+d x} (e x)^{m+1} \left (2 a d^2 \left (4 m^2+24 m+35\right )+3 b c^2 \left (4 m^2+22 m+29\right )\right )}{e (2 m+3)}\right )}{d^3 (2 m+5)}+\frac {2 b^2 c^2 e (4 m+13) \sqrt {c+d x} (e x)^{m+2}}{d^2 (2 m+5)}}{d e^3 (2 m+7)}-\frac {2 b^2 c \sqrt {c+d x} (e x)^{m+3}}{d^2 e^3 (2 m+7)}}{c}\)

\(\Big \downarrow \) 75

\(\displaystyle \frac {2 (e x)^{m+1} \left (a d^2+b c^2\right )^2}{c d^4 e \sqrt {c+d x}}-\frac {\frac {\frac {e^3 \left (\frac {2 \sqrt {c+d x} (e x)^m \left (-\frac {d x}{c}\right )^{-m} \left (a^2 d^4 \left (16 m^4+128 m^3+344 m^2+352 m+105\right )+8 a b c^2 d^2 \left (4 m^4+36 m^3+115 m^2+153 m+70\right )+16 b^2 c^4 \left (m^4+10 m^3+35 m^2+50 m+24\right )\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m,\frac {3}{2},\frac {d x}{c}+1\right )}{d (2 m+3)}-\frac {2 b c \sqrt {c+d x} (e x)^{m+1} \left (2 a d^2 \left (4 m^2+24 m+35\right )+3 b c^2 \left (4 m^2+22 m+29\right )\right )}{e (2 m+3)}\right )}{d^3 (2 m+5)}+\frac {2 b^2 c^2 e (4 m+13) \sqrt {c+d x} (e x)^{m+2}}{d^2 (2 m+5)}}{d e^3 (2 m+7)}-\frac {2 b^2 c \sqrt {c+d x} (e x)^{m+3}}{d^2 e^3 (2 m+7)}}{c}\)

Input:

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

Output:

(2*(b*c^2 + a*d^2)^2*(e*x)^(1 + m))/(c*d^4*e*Sqrt[c + d*x]) - ((-2*b^2*c*( 
e*x)^(3 + m)*Sqrt[c + d*x])/(d^2*e^3*(7 + 2*m)) + ((2*b^2*c^2*e*(13 + 4*m) 
*(e*x)^(2 + m)*Sqrt[c + d*x])/(d^2*(5 + 2*m)) + (e^3*((-2*b*c*(3*b*c^2*(29 
 + 22*m + 4*m^2) + 2*a*d^2*(35 + 24*m + 4*m^2))*(e*x)^(1 + m)*Sqrt[c + d*x 
])/(e*(3 + 2*m)) + (2*(16*b^2*c^4*(24 + 50*m + 35*m^2 + 10*m^3 + m^4) + 8* 
a*b*c^2*d^2*(70 + 153*m + 115*m^2 + 36*m^3 + 4*m^4) + a^2*d^4*(105 + 352*m 
 + 344*m^2 + 128*m^3 + 16*m^4))*(e*x)^m*Sqrt[c + d*x]*Hypergeometric2F1[1/ 
2, -m, 3/2, 1 + (d*x)/c])/(d*(3 + 2*m)*(-((d*x)/c))^m)))/(d^3*(5 + 2*m)))/ 
(d*e^3*(7 + 2*m)))/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 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 519

Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), 
 x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x^2)^p, c + d*x, x], R = 
 PolynomialRemainder[(a + b*x^2)^p, c + d*x, x]}, Simp[(-R)*(e*x)^(m + 1)*( 
(c + d*x)^(n + 1)/(c*e*(n + 1))), x] + Simp[1/(c*(n + 1))   Int[(e*x)^m*(c 
+ d*x)^(n + 1)*ExpandToSum[c*(n + 1)*Qx + R*(m + n + 2), x], x], x]] /; Fre 
eQ[{a, b, c, d, e, m}, x] && IGtQ[p, 0] && LtQ[n, -1] &&  !IntegerQ[m]
 

rule 1194

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

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]

\[\int \frac {\left (e x \right )^{m} \left (b \,x^{2}+a \right )^{2}}{\left (d x +c \right )^{\frac {3}{2}}}d x\]

Input:

int((e*x)^m*(b*x^2+a)^2/(d*x+c)^(3/2),x)
 

Output:

int((e*x)^m*(b*x^2+a)^2/(d*x+c)^(3/2),x)
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

integrate((e*x)**m*(b*x**2+a)**2/(d*x+c)**(3/2),x)
 

Output:

Integral((e*x)**m*(a + b*x**2)**2/(c + d*x)**(3/2), x)
 

Maxima [F]

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

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

Output:

integrate((b*x^2 + a)^2*(e*x)^m/(d*x + c)^(3/2), x)
 

Giac [F]

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

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

Output:

integrate((b*x^2 + a)^2*(e*x)^m/(d*x + c)^(3/2), x)
 

Mupad [F(-1)]

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

int(((e*x)^m*(a + b*x^2)^2)/(c + d*x)^(3/2),x)
 

Output:

int(((e*x)^m*(a + b*x^2)^2)/(c + d*x)^(3/2), x)
 

Reduce [F]

\[ \int \frac {(e x)^m \left (a+b x^2\right )^2}{(c+d x)^{3/2}} \, dx=\text {too large to display} \] Input:

int((e*x)^m*(b*x^2+a)^2/(d*x+c)^(3/2),x)
 

Output:

(2*e**m*(16*x**m*sqrt(c + d*x)*a**2*d**4*m**4 + 128*x**m*sqrt(c + d*x)*a** 
2*d**4*m**3 + 344*x**m*sqrt(c + d*x)*a**2*d**4*m**2 + 352*x**m*sqrt(c + d* 
x)*a**2*d**4*m + 105*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 + 288*x**m*sqrt(c + d*x)*a*b*c**2*d**2*m**3 + 920*x**m* 
sqrt(c + d*x)*a*b*c**2*d**2*m**2 + 1224*x**m*sqrt(c + d*x)*a*b*c**2*d**2*m 
 + 560*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 - 240*x**m*sqrt(c + d*x)*a*b*c*d**3*m**3*x - 536*x**m*sqrt(c + d*x 
)*a*b*c*d**3*m**2*x - 228*x**m*sqrt(c + d*x)*a*b*c*d**3*m*x + 280*x**m*sqr 
t(c + d*x)*a*b*c*d**3*x + 32*x**m*sqrt(c + d*x)*a*b*d**4*m**4*x**2 + 192*x 
**m*sqrt(c + d*x)*a*b*d**4*m**3*x**2 + 272*x**m*sqrt(c + d*x)*a*b*d**4*m** 
2*x**2 - 48*x**m*sqrt(c + d*x)*a*b*d**4*m*x**2 - 70*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*sqrt(c + d 
*x)*b**2*c**3*d*m**4*x - 136*x**m*sqrt(c + d*x)*b**2*c**3*d*m**3*x - 344*x 
**m*sqrt(c + d*x)*b**2*c**3*d*m**2*x - 176*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 + 112*x**m*sqrt(c + d*x)*b**2*c**2*d**2*m**3*x**2 + 188* 
x**m*sqrt(c + d*x)*b**2*c**2*d**2*m**2*x**2 - 28*x**m*sqrt(c + d*x)*b**2*c 
**2*d**2*m*x**2 - 48*x**m*sqrt(c + d*x)*b**2*c**2*d**2*x**2 - 16*x**m*s...