\(\int \frac {(e x)^m (a+b x^2)^2}{(c+d x)^{5/2}} \, dx\) [141]

Optimal result

Integrand size = 24, antiderivative size = 337 \[ \int \frac {(e x)^m \left (a+b x^2\right )^2}{(c+d x)^{5/2}} \, dx=\frac {2 \left (b c^2+a d^2\right )^2 (e x)^{1+m}}{3 c d^4 e (c+d x)^{3/2}}+\frac {2 \left (b c^2+a d^2\right ) \left (a d^2 (1-2 m)-b c^2 (11+2 m)\right ) (e x)^{1+m}}{3 c^2 d^4 e \sqrt {c+d x}}-\frac {2 b^2 c (17+8 m) (e x)^{1+m} \sqrt {c+d x}}{d^4 e (3+2 m) (5+2 m)}+\frac {2 b^2 (e x)^{1+m} (c+d x)^{3/2}}{d^4 e (5+2 m)}-\frac {2 \left (a^2 d^4 \left (15+16 m-56 m^2-64 m^3-16 m^4\right )-16 b^2 c^4 \left (24+50 m+35 m^2+10 m^3+m^4\right )-8 a b c^2 d^2 \left (30+77 m+71 m^2+28 m^3+4 m^4\right )\right ) \left (-\frac {d x}{c}\right )^{-m} (e x)^m \sqrt {c+d x} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m,\frac {3}{2},1+\frac {d x}{c}\right )}{3 c^2 d^5 (3+2 m) (5+2 m)} \] Output:

2/3*(a*d^2+b*c^2)^2*(e*x)^(1+m)/c/d^4/e/(d*x+c)^(3/2)+2/3*(a*d^2+b*c^2)*(a 
*d^2*(1-2*m)-b*c^2*(11+2*m))*(e*x)^(1+m)/c^2/d^4/e/(d*x+c)^(1/2)-2*b^2*c*( 
17+8*m)*(e*x)^(1+m)*(d*x+c)^(1/2)/d^4/e/(3+2*m)/(5+2*m)+2*b^2*(e*x)^(1+m)* 
(d*x+c)^(3/2)/d^4/e/(5+2*m)-2/3*(a^2*d^4*(-16*m^4-64*m^3-56*m^2+16*m+15)-1 
6*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+28*m^3+71*m^2+7 
7*m+30))*(e*x)^m*(d*x+c)^(1/2)*hypergeom([1/2, -m],[3/2],1+d*x/c)/c^2/d^5/ 
(3+2*m)/(5+2*m)/((-d*x/c)^m)
 

Mathematica [A] (verified)

Time = 0.90 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.61 \[ \int \frac {(e x)^m \left (a+b x^2\right )^2}{(c+d x)^{5/2}} \, dx=\frac {2 \left (-\frac {d x}{c}\right )^{-m} (e x)^m \left (-5 \left (b c^2+a d^2\right )^2 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-m,-\frac {1}{2},1+\frac {d x}{c}\right )+60 b c \left (b c^2+a d^2\right ) (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-m,\frac {1}{2},1+\frac {d x}{c}\right )+30 b \left (3 b c^2+a d^2\right ) (c+d x)^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m,\frac {3}{2},1+\frac {d x}{c}\right )-20 b^2 c (c+d x)^3 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-m,\frac {5}{2},1+\frac {d x}{c}\right )+3 b^2 (c+d x)^4 \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-m,\frac {7}{2},1+\frac {d x}{c}\right )\right )}{15 d^5 (c+d x)^{3/2}} \] Input:

Integrate[((e*x)^m*(a + b*x^2)^2)/(c + d*x)^(5/2),x]
 

Output:

(2*(e*x)^m*(-5*(b*c^2 + a*d^2)^2*Hypergeometric2F1[-3/2, -m, -1/2, 1 + (d* 
x)/c] + 60*b*c*(b*c^2 + a*d^2)*(c + d*x)*Hypergeometric2F1[-1/2, -m, 1/2, 
1 + (d*x)/c] + 30*b*(3*b*c^2 + a*d^2)*(c + d*x)^2*Hypergeometric2F1[1/2, - 
m, 3/2, 1 + (d*x)/c] - 20*b^2*c*(c + d*x)^3*Hypergeometric2F1[3/2, -m, 5/2 
, 1 + (d*x)/c] + 3*b^2*(c + d*x)^4*Hypergeometric2F1[5/2, -m, 7/2, 1 + (d* 
x)/c]))/(15*d^5*(-((d*x)/c))^m*(c + d*x)^(3/2))
 

Rubi [A] (verified)

Time = 0.85 (sec) , antiderivative size = 337, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {519, 27, 2124, 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.

\(\displaystyle \int \frac {\left (a+b x^2\right )^2 (e x)^m}{(c+d x)^{5/2}} \, dx\)

\(\Big \downarrow \) 519

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2124

\(\displaystyle \frac {\frac {2 \int -\frac {(e x)^m \left (\frac {6 b^2 e x c^3}{d^3}-\frac {3 b^2 e x^2 c^2}{d^2}+e \left (-\frac {4 b^2 \left (m^2+6 m+5\right ) c^4}{d^4}-\frac {8 a b \left (m^2+3 m+2\right ) c^2}{d^2}+a^2 \left (1-4 m^2\right )\right )\right )}{2 \sqrt {c+d x}}dx}{c e}+\frac {2 (e x)^{m+1} \left (a d^2+b c^2\right ) \left (a d^2 (1-2 m)-b c^2 (2 m+11)\right )}{c d^4 e \sqrt {c+d x}}}{3 c}+\frac {2 (e x)^{m+1} \left (a d^2+b c^2\right )^2}{3 c d^4 e (c+d x)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {2 (e x)^{m+1} \left (a d^2+b c^2\right ) \left (a d^2 (1-2 m)-b c^2 (2 m+11)\right )}{c d^4 e \sqrt {c+d x}}-\frac {\int \frac {(e x)^m \left (\frac {6 b^2 e x c^3}{d^3}-\frac {3 b^2 e x^2 c^2}{d^2}+e \left (-\frac {4 b^2 \left (m^2+6 m+5\right ) c^4}{d^4}-\frac {8 a b \left (m^2+3 m+2\right ) c^2}{d^2}+a^2 \left (1-4 m^2\right )\right )\right )}{\sqrt {c+d x}}dx}{c e}}{3 c}+\frac {2 (e x)^{m+1} \left (a d^2+b c^2\right )^2}{3 c d^4 e (c+d x)^{3/2}}\)

\(\Big \downarrow \) 1194

\(\displaystyle \frac {\frac {2 (e x)^{m+1} \left (a d^2+b c^2\right ) \left (a d^2 (1-2 m)-b c^2 (2 m+11)\right )}{c d^4 e \sqrt {c+d x}}-\frac {\frac {2 \int \frac {e^3 (e x)^m \left (6 b^2 d (3 m+7) x c^3+(2 m+5) \left (-4 b^2 \left (m^2+6 m+5\right ) c^4-8 a b d^2 \left (m^2+3 m+2\right ) c^2+a^2 d^4 \left (1-4 m^2\right )\right )\right )}{2 d^3 \sqrt {c+d x}}dx}{d e^2 (2 m+5)}-\frac {6 b^2 c^2 \sqrt {c+d x} (e x)^{m+2}}{d^3 e (2 m+5)}}{c e}}{3 c}+\frac {2 (e x)^{m+1} \left (a d^2+b c^2\right )^2}{3 c d^4 e (c+d x)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {2 (e x)^{m+1} \left (a d^2+b c^2\right ) \left (a d^2 (1-2 m)-b c^2 (2 m+11)\right )}{c d^4 e \sqrt {c+d x}}-\frac {\frac {e \int \frac {(e x)^m \left (6 b^2 d (3 m+7) x c^3+(2 m+5) \left (-4 b^2 \left (m^2+6 m+5\right ) c^4-8 a b d^2 \left (m^2+3 m+2\right ) c^2+a^2 d^4 \left (1-4 m^2\right )\right )\right )}{\sqrt {c+d x}}dx}{d^4 (2 m+5)}-\frac {6 b^2 c^2 \sqrt {c+d x} (e x)^{m+2}}{d^3 e (2 m+5)}}{c e}}{3 c}+\frac {2 (e x)^{m+1} \left (a d^2+b c^2\right )^2}{3 c d^4 e (c+d x)^{3/2}}\)

\(\Big \downarrow \) 90

\(\displaystyle \frac {\frac {2 (e x)^{m+1} \left (a d^2+b c^2\right ) \left (a d^2 (1-2 m)-b c^2 (2 m+11)\right )}{c d^4 e \sqrt {c+d x}}-\frac {\frac {e \left (\frac {12 b^2 c^3 (3 m+7) \sqrt {c+d x} (e x)^{m+1}}{e (2 m+3)}-\left (\frac {6 b^2 c^4 (m+1) (3 m+7)}{m+\frac {3}{2}}-(2 m+5) \left (a^2 d^4 \left (1-4 m^2\right )-8 a b c^2 d^2 \left (m^2+3 m+2\right )-4 b^2 c^4 \left (m^2+6 m+5\right )\right )\right ) \int \frac {(e x)^m}{\sqrt {c+d x}}dx\right )}{d^4 (2 m+5)}-\frac {6 b^2 c^2 \sqrt {c+d x} (e x)^{m+2}}{d^3 e (2 m+5)}}{c e}}{3 c}+\frac {2 (e x)^{m+1} \left (a d^2+b c^2\right )^2}{3 c d^4 e (c+d x)^{3/2}}\)

\(\Big \downarrow \) 77

\(\displaystyle \frac {\frac {2 (e x)^{m+1} \left (a d^2+b c^2\right ) \left (a d^2 (1-2 m)-b c^2 (2 m+11)\right )}{c d^4 e \sqrt {c+d x}}-\frac {\frac {e \left (\frac {12 b^2 c^3 (3 m+7) \sqrt {c+d x} (e x)^{m+1}}{e (2 m+3)}-(e x)^m \left (-\frac {d x}{c}\right )^{-m} \left (\frac {6 b^2 c^4 (m+1) (3 m+7)}{m+\frac {3}{2}}-(2 m+5) \left (a^2 d^4 \left (1-4 m^2\right )-8 a b c^2 d^2 \left (m^2+3 m+2\right )-4 b^2 c^4 \left (m^2+6 m+5\right )\right )\right ) \int \frac {\left (-\frac {d x}{c}\right )^m}{\sqrt {c+d x}}dx\right )}{d^4 (2 m+5)}-\frac {6 b^2 c^2 \sqrt {c+d x} (e x)^{m+2}}{d^3 e (2 m+5)}}{c e}}{3 c}+\frac {2 (e x)^{m+1} \left (a d^2+b c^2\right )^2}{3 c d^4 e (c+d x)^{3/2}}\)

\(\Big \downarrow \) 75

\(\displaystyle \frac {\frac {2 (e x)^{m+1} \left (a d^2+b c^2\right ) \left (a d^2 (1-2 m)-b c^2 (2 m+11)\right )}{c d^4 e \sqrt {c+d x}}-\frac {\frac {e \left (\frac {12 b^2 c^3 (3 m+7) \sqrt {c+d x} (e x)^{m+1}}{e (2 m+3)}-\frac {2 \sqrt {c+d x} (e x)^m \left (-\frac {d x}{c}\right )^{-m} \left (\frac {6 b^2 c^4 (m+1) (3 m+7)}{m+\frac {3}{2}}-(2 m+5) \left (a^2 d^4 \left (1-4 m^2\right )-8 a b c^2 d^2 \left (m^2+3 m+2\right )-4 b^2 c^4 \left (m^2+6 m+5\right )\right )\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m,\frac {3}{2},\frac {d x}{c}+1\right )}{d}\right )}{d^4 (2 m+5)}-\frac {6 b^2 c^2 \sqrt {c+d x} (e x)^{m+2}}{d^3 e (2 m+5)}}{c e}}{3 c}+\frac {2 (e x)^{m+1} \left (a d^2+b c^2\right )^2}{3 c d^4 e (c+d x)^{3/2}}\)

Input:

Int[((e*x)^m*(a + b*x^2)^2)/(c + d*x)^(5/2),x]
 

Output:

(2*(b*c^2 + a*d^2)^2*(e*x)^(1 + m))/(3*c*d^4*e*(c + d*x)^(3/2)) + ((2*(b*c 
^2 + a*d^2)*(a*d^2*(1 - 2*m) - b*c^2*(11 + 2*m))*(e*x)^(1 + m))/(c*d^4*e*S 
qrt[c + d*x]) - ((-6*b^2*c^2*(e*x)^(2 + m)*Sqrt[c + d*x])/(d^3*e*(5 + 2*m) 
) + (e*((12*b^2*c^3*(7 + 3*m)*(e*x)^(1 + m)*Sqrt[c + d*x])/(e*(3 + 2*m)) - 
 (2*((6*b^2*c^4*(1 + m)*(7 + 3*m))/(3/2 + m) - (5 + 2*m)*(a^2*d^4*(1 - 4*m 
^2) - 8*a*b*c^2*d^2*(2 + 3*m + m^2) - 4*b^2*c^4*(5 + 6*m + m^2)))*(e*x)^m* 
Sqrt[c + d*x]*Hypergeometric2F1[1/2, -m, 3/2, 1 + (d*x)/c])/(d*(-((d*x)/c) 
)^m)))/(d^4*(5 + 2*m)))/(c*e))/(3*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 2124

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] : 
> With[{Qx = PolynomialQuotient[Px, a + b*x, x], R = PolynomialRemainder[Px 
, a + b*x, x]}, Simp[R*(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((m + 1)*(b*c - 
 a*d))), x] + Simp[1/((m + 1)*(b*c - a*d))   Int[(a + b*x)^(m + 1)*(c + d*x 
)^n*ExpandToSum[(m + 1)*(b*c - a*d)*Qx - d*R*(m + n + 2), x], x], x]] /; Fr 
eeQ[{a, b, c, d, n}, x] && PolyQ[Px, x] && LtQ[m, -1] && (IntegerQ[m] ||  ! 
ILtQ[n, -1])
 

Maple [F]

\[\int \frac {\left (e x \right )^{m} \left (b \,x^{2}+a \right )^{2}}{\left (d x +c \right )^{\frac {5}{2}}}d x\]

Input:

int((e*x)^m*(b*x^2+a)^2/(d*x+c)^(5/2),x)
 

Output:

int((e*x)^m*(b*x^2+a)^2/(d*x+c)^(5/2),x)
 

Fricas [F]

\[ \int \frac {(e x)^m \left (a+b x^2\right )^2}{(c+d x)^{5/2}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{2} \left (e x\right )^{m}}{{\left (d x + c\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((e*x)^m*(b*x^2+a)^2/(d*x+c)^(5/2),x, algorithm="fricas")
 

Output:

integral((b^2*x^4 + 2*a*b*x^2 + a^2)*sqrt(d*x + c)*(e*x)^m/(d^3*x^3 + 3*c* 
d^2*x^2 + 3*c^2*d*x + c^3), x)
 

Sympy [F]

\[ \int \frac {(e x)^m \left (a+b x^2\right )^2}{(c+d x)^{5/2}} \, dx=\int \frac {\left (e x\right )^{m} \left (a + b x^{2}\right )^{2}}{\left (c + d x\right )^{\frac {5}{2}}}\, dx \] Input:

integrate((e*x)**m*(b*x**2+a)**2/(d*x+c)**(5/2),x)
 

Output:

Integral((e*x)**m*(a + b*x**2)**2/(c + d*x)**(5/2), x)
 

Maxima [F]

\[ \int \frac {(e x)^m \left (a+b x^2\right )^2}{(c+d x)^{5/2}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{2} \left (e x\right )^{m}}{{\left (d x + c\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((e*x)^m*(b*x^2+a)^2/(d*x+c)^(5/2),x, algorithm="maxima")
 

Output:

integrate((b*x^2 + a)^2*(e*x)^m/(d*x + c)^(5/2), x)
 

Giac [F]

\[ \int \frac {(e x)^m \left (a+b x^2\right )^2}{(c+d x)^{5/2}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{2} \left (e x\right )^{m}}{{\left (d x + c\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((e*x)^m*(b*x^2+a)^2/(d*x+c)^(5/2),x, algorithm="giac")
 

Output:

integrate((b*x^2 + a)^2*(e*x)^m/(d*x + c)^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e x)^m \left (a+b x^2\right )^2}{(c+d x)^{5/2}} \, dx=\int \frac {{\left (e\,x\right )}^m\,{\left (b\,x^2+a\right )}^2}{{\left (c+d\,x\right )}^{5/2}} \,d x \] Input:

int(((e*x)^m*(a + b*x^2)^2)/(c + d*x)^(5/2),x)
 

Output:

int(((e*x)^m*(a + b*x^2)^2)/(c + d*x)^(5/2), x)
 

Reduce [F]

\[ \int \frac {(e x)^m \left (a+b x^2\right )^2}{(c+d x)^{5/2}} \, dx=\text {too large to display} \] Input:

int((e*x)^m*(b*x^2+a)^2/(d*x+c)^(5/2),x)
 

Output:

(2*e**m*(16*x**m*sqrt(c + d*x)*a**2*d**4*m**4 + 64*x**m*sqrt(c + d*x)*a**2 
*d**4*m**3 + 56*x**m*sqrt(c + d*x)*a**2*d**4*m**2 - 16*x**m*sqrt(c + d*x)* 
a**2*d**4*m - 15*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 + 224*x**m*sqrt(c + d*x)*a*b*c**2*d**2*m**3 + 568*x**m*sqrt 
(c + d*x)*a*b*c**2*d**2*m**2 + 616*x**m*sqrt(c + d*x)*a*b*c**2*d**2*m + 24 
0*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 - 144*x**m*sqrt(c + d*x)*a*b*c*d**3*m**3*x - 88*x**m*sqrt(c + d*x)*a*b* 
c*d**3*m**2*x + 324*x**m*sqrt(c + d*x)*a*b*c*d**3*m*x + 360*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 + 64*x**m*sqr 
t(c + d*x)*a*b*d**4*m**3*x**2 - 112*x**m*sqrt(c + d*x)*a*b*d**4*m**2*x**2 
- 144*x**m*sqrt(c + d*x)*a*b*d**4*m*x**2 + 90*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 - 120*x**m*sqrt(c + d*x)*b**2*c**3*d*m**3*x - 200*x**m*sq 
rt(c + d*x)*b**2*c**3*d*m**2*x + 240*x**m*sqrt(c + d*x)*b**2*c**3*d*m*x + 
576*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 + 80*x**m*sqrt(c + d*x)*b**2*c**2*d**2*m**3*x**2 - 20*x**m*sqr 
t(c + d*x)*b**2*c**2*d**2*m**2*x**2 - 300*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 - 16*x**m*sqrt(c ...