\(\int (e x)^m \sqrt {c+d x} (a+b x^2)^2 \, dx\) [138]

Optimal result

Integrand size = 24, antiderivative size = 386 \[ \int (e x)^m \sqrt {c+d x} \left (a+b x^2\right )^2 \, dx=-\frac {8 b c (2+m) \left (2 b c^2 \left (12+7 m+m^2\right )+a d^2 \left (99+40 m+4 m^2\right )\right ) (e x)^{1+m} (c+d x)^{3/2}}{d^4 e (5+2 m) (7+2 m) (9+2 m) (11+2 m)}+\frac {4 b \left (2 b c^2 \left (12+7 m+m^2\right )+a d^2 \left (99+40 m+4 m^2\right )\right ) (e x)^{2+m} (c+d x)^{3/2}}{d^3 e^2 (7+2 m) (9+2 m) (11+2 m)}-\frac {4 b^2 c (4+m) (e x)^{3+m} (c+d x)^{3/2}}{d^2 e^3 (9+2 m) (11+2 m)}+\frac {2 b^2 (e x)^{4+m} (c+d x)^{3/2}}{d e^4 (11+2 m)}+\frac {2 \left (a^2 d^4 (7+2 m) \left (99+40 m+4 m^2\right )+\frac {4 b c^2 (1+m) (2+m) \left (2 b c^2 \left (12+7 m+m^2\right )+a d^2 \left (99+40 m+4 m^2\right )\right )}{\frac {5}{2}+m}\right ) \left (-\frac {d x}{c}\right )^{-m} (e x)^m (c+d x)^{3/2} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-m,\frac {5}{2},1+\frac {d x}{c}\right )}{3 d^5 (7+2 m) (9+2 m) (11+2 m)} \] Output:

-8*b*c*(2+m)*(2*b*c^2*(m^2+7*m+12)+a*d^2*(4*m^2+40*m+99))*(e*x)^(1+m)*(d*x 
+c)^(3/2)/d^4/e/(5+2*m)/(7+2*m)/(9+2*m)/(11+2*m)+4*b*(2*b*c^2*(m^2+7*m+12) 
+a*d^2*(4*m^2+40*m+99))*(e*x)^(2+m)*(d*x+c)^(3/2)/d^3/e^2/(7+2*m)/(9+2*m)/ 
(11+2*m)-4*b^2*c*(4+m)*(e*x)^(3+m)*(d*x+c)^(3/2)/d^2/e^3/(9+2*m)/(11+2*m)+ 
2*b^2*(e*x)^(4+m)*(d*x+c)^(3/2)/d/e^4/(11+2*m)+2/3*(a^2*d^4*(7+2*m)*(4*m^2 
+40*m+99)+4*b*c^2*(1+m)*(2+m)*(2*b*c^2*(m^2+7*m+12)+a*d^2*(4*m^2+40*m+99)) 
/(5/2+m))*(e*x)^m*(d*x+c)^(3/2)*hypergeom([3/2, -m],[5/2],1+d*x/c)/d^5/(7+ 
2*m)/(9+2*m)/(11+2*m)/((-d*x/c)^m)
 

Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.53 \[ \int (e x)^m \sqrt {c+d x} \left (a+b x^2\right )^2 \, dx=\frac {2 \left (-\frac {d x}{c}\right )^{-m} (e x)^m (c+d x)^{3/2} \left (1155 \left (b c^2+a d^2\right )^2 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-m,\frac {5}{2},1+\frac {d x}{c}\right )-2772 b c \left (b c^2+a d^2\right ) (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-m,\frac {7}{2},1+\frac {d x}{c}\right )+990 b \left (3 b c^2+a d^2\right ) (c+d x)^2 \operatorname {Hypergeometric2F1}\left (\frac {7}{2},-m,\frac {9}{2},1+\frac {d x}{c}\right )-1540 b^2 c (c+d x)^3 \operatorname {Hypergeometric2F1}\left (\frac {9}{2},-m,\frac {11}{2},1+\frac {d x}{c}\right )+315 b^2 (c+d x)^4 \operatorname {Hypergeometric2F1}\left (\frac {11}{2},-m,\frac {13}{2},1+\frac {d x}{c}\right )\right )}{3465 d^5} \] Input:

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

Output:

(2*(e*x)^m*(c + d*x)^(3/2)*(1155*(b*c^2 + a*d^2)^2*Hypergeometric2F1[3/2, 
-m, 5/2, 1 + (d*x)/c] - 2772*b*c*(b*c^2 + a*d^2)*(c + d*x)*Hypergeometric2 
F1[5/2, -m, 7/2, 1 + (d*x)/c] + 990*b*(3*b*c^2 + a*d^2)*(c + d*x)^2*Hyperg 
eometric2F1[7/2, -m, 9/2, 1 + (d*x)/c] - 1540*b^2*c*(c + d*x)^3*Hypergeome 
tric2F1[9/2, -m, 11/2, 1 + (d*x)/c] + 315*b^2*(c + d*x)^4*Hypergeometric2F 
1[11/2, -m, 13/2, 1 + (d*x)/c]))/(3465*d^5*(-((d*x)/c))^m)
 

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 363, 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.

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

\(\Big \downarrow \) 521

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2125

\(\displaystyle \frac {\frac {2 \int \frac {1}{2} e^7 (e x)^m \sqrt {c+d x} \left (a^2 \left (4 m^2+40 m+99\right ) d^2+2 b \left (2 b \left (m^2+7 m+12\right ) c^2+a d^2 \left (4 m^2+40 m+99\right )\right ) x^2\right )dx}{d e^3 (2 m+9)}-\frac {4 b^2 c e (m+4) (c+d x)^{3/2} (e x)^{m+3}}{d (2 m+9)}}{d e^4 (2 m+11)}+\frac {2 b^2 (c+d x)^{3/2} (e x)^{m+4}}{d e^4 (2 m+11)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {e^4 \int (e x)^m \sqrt {c+d x} \left (a^2 \left (4 m^2+40 m+99\right ) d^2+2 b \left (2 b \left (m^2+7 m+12\right ) c^2+a d^2 \left (4 m^2+40 m+99\right )\right ) x^2\right )dx}{d (2 m+9)}-\frac {4 b^2 c e (m+4) (c+d x)^{3/2} (e x)^{m+3}}{d (2 m+9)}}{d e^4 (2 m+11)}+\frac {2 b^2 (c+d x)^{3/2} (e x)^{m+4}}{d e^4 (2 m+11)}\)

\(\Big \downarrow \) 521

\(\displaystyle \frac {\frac {e^4 \left (\frac {2 \int \frac {1}{2} e^2 (e x)^m \sqrt {c+d x} \left (a^2 d^3 \left (8 m^3+108 m^2+478 m+693\right )-4 b c (m+2) \left (2 b \left (m^2+7 m+12\right ) c^2+a d^2 \left (4 m^2+40 m+99\right )\right ) x\right )dx}{d e^2 (2 m+7)}+\frac {4 b (c+d x)^{3/2} (e x)^{m+2} \left (a d^2 \left (4 m^2+40 m+99\right )+2 b c^2 \left (m^2+7 m+12\right )\right )}{d e^2 (2 m+7)}\right )}{d (2 m+9)}-\frac {4 b^2 c e (m+4) (c+d x)^{3/2} (e x)^{m+3}}{d (2 m+9)}}{d e^4 (2 m+11)}+\frac {2 b^2 (c+d x)^{3/2} (e x)^{m+4}}{d e^4 (2 m+11)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {e^4 \left (\frac {\int (e x)^m \sqrt {c+d x} \left (a^2 d^3 \left (8 m^3+108 m^2+478 m+693\right )-4 b c (m+2) \left (2 b \left (m^2+7 m+12\right ) c^2+a d^2 \left (4 m^2+40 m+99\right )\right ) x\right )dx}{d (2 m+7)}+\frac {4 b (c+d x)^{3/2} (e x)^{m+2} \left (a d^2 \left (4 m^2+40 m+99\right )+2 b c^2 \left (m^2+7 m+12\right )\right )}{d e^2 (2 m+7)}\right )}{d (2 m+9)}-\frac {4 b^2 c e (m+4) (c+d x)^{3/2} (e x)^{m+3}}{d (2 m+9)}}{d e^4 (2 m+11)}+\frac {2 b^2 (c+d x)^{3/2} (e x)^{m+4}}{d e^4 (2 m+11)}\)

\(\Big \downarrow \) 90

\(\displaystyle \frac {\frac {e^4 \left (\frac {\frac {\left (a^2 d^4 \left (8 m^3+108 m^2+478 m+693\right )+\frac {4 b c^2 (m+1) (m+2) \left (a d^2 \left (4 m^2+40 m+99\right )+2 b c^2 \left (m^2+7 m+12\right )\right )}{m+\frac {5}{2}}\right ) \int (e x)^m \sqrt {c+d x}dx}{d}-\frac {8 b c (m+2) (c+d x)^{3/2} (e x)^{m+1} \left (a d^2 \left (4 m^2+40 m+99\right )+2 b c^2 \left (m^2+7 m+12\right )\right )}{d e (2 m+5)}}{d (2 m+7)}+\frac {4 b (c+d x)^{3/2} (e x)^{m+2} \left (a d^2 \left (4 m^2+40 m+99\right )+2 b c^2 \left (m^2+7 m+12\right )\right )}{d e^2 (2 m+7)}\right )}{d (2 m+9)}-\frac {4 b^2 c e (m+4) (c+d x)^{3/2} (e x)^{m+3}}{d (2 m+9)}}{d e^4 (2 m+11)}+\frac {2 b^2 (c+d x)^{3/2} (e x)^{m+4}}{d e^4 (2 m+11)}\)

\(\Big \downarrow \) 77

\(\displaystyle \frac {\frac {e^4 \left (\frac {\frac {(e x)^m \left (-\frac {d x}{c}\right )^{-m} \left (a^2 d^4 \left (8 m^3+108 m^2+478 m+693\right )+\frac {4 b c^2 (m+1) (m+2) \left (a d^2 \left (4 m^2+40 m+99\right )+2 b c^2 \left (m^2+7 m+12\right )\right )}{m+\frac {5}{2}}\right ) \int \left (-\frac {d x}{c}\right )^m \sqrt {c+d x}dx}{d}-\frac {8 b c (m+2) (c+d x)^{3/2} (e x)^{m+1} \left (a d^2 \left (4 m^2+40 m+99\right )+2 b c^2 \left (m^2+7 m+12\right )\right )}{d e (2 m+5)}}{d (2 m+7)}+\frac {4 b (c+d x)^{3/2} (e x)^{m+2} \left (a d^2 \left (4 m^2+40 m+99\right )+2 b c^2 \left (m^2+7 m+12\right )\right )}{d e^2 (2 m+7)}\right )}{d (2 m+9)}-\frac {4 b^2 c e (m+4) (c+d x)^{3/2} (e x)^{m+3}}{d (2 m+9)}}{d e^4 (2 m+11)}+\frac {2 b^2 (c+d x)^{3/2} (e x)^{m+4}}{d e^4 (2 m+11)}\)

\(\Big \downarrow \) 75

\(\displaystyle \frac {\frac {e^4 \left (\frac {\frac {2 (c+d x)^{3/2} (e x)^m \left (-\frac {d x}{c}\right )^{-m} \left (a^2 d^4 \left (8 m^3+108 m^2+478 m+693\right )+\frac {4 b c^2 (m+1) (m+2) \left (a d^2 \left (4 m^2+40 m+99\right )+2 b c^2 \left (m^2+7 m+12\right )\right )}{m+\frac {5}{2}}\right ) \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-m,\frac {5}{2},\frac {d x}{c}+1\right )}{3 d^2}-\frac {8 b c (m+2) (c+d x)^{3/2} (e x)^{m+1} \left (a d^2 \left (4 m^2+40 m+99\right )+2 b c^2 \left (m^2+7 m+12\right )\right )}{d e (2 m+5)}}{d (2 m+7)}+\frac {4 b (c+d x)^{3/2} (e x)^{m+2} \left (a d^2 \left (4 m^2+40 m+99\right )+2 b c^2 \left (m^2+7 m+12\right )\right )}{d e^2 (2 m+7)}\right )}{d (2 m+9)}-\frac {4 b^2 c e (m+4) (c+d x)^{3/2} (e x)^{m+3}}{d (2 m+9)}}{d e^4 (2 m+11)}+\frac {2 b^2 (c+d x)^{3/2} (e x)^{m+4}}{d e^4 (2 m+11)}\)

Input:

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

Output:

(2*b^2*(e*x)^(4 + m)*(c + d*x)^(3/2))/(d*e^4*(11 + 2*m)) + ((-4*b^2*c*e*(4 
 + m)*(e*x)^(3 + m)*(c + d*x)^(3/2))/(d*(9 + 2*m)) + (e^4*((4*b*(2*b*c^2*( 
12 + 7*m + m^2) + a*d^2*(99 + 40*m + 4*m^2))*(e*x)^(2 + m)*(c + d*x)^(3/2) 
)/(d*e^2*(7 + 2*m)) + ((-8*b*c*(2 + m)*(2*b*c^2*(12 + 7*m + m^2) + a*d^2*( 
99 + 40*m + 4*m^2))*(e*x)^(1 + m)*(c + d*x)^(3/2))/(d*e*(5 + 2*m)) + (2*(a 
^2*d^4*(693 + 478*m + 108*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*(99 + 40*m + 4*m^2)))/(5/2 + m))*(e*x)^m*(c + d* 
x)^(3/2)*Hypergeometric2F1[3/2, -m, 5/2, 1 + (d*x)/c])/(3*d^2*(-((d*x)/c)) 
^m))/(d*(7 + 2*m))))/(d*(9 + 2*m)))/(d*e^4*(11 + 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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 3.07 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.36 \[ \int (e x)^m \sqrt {c+d x} \left (a+b x^2\right )^2 \, dx=\frac {a^{2} \sqrt {c} e^{m} x^{m + 1} \Gamma \left (m + 1\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, m + 1 \\ m + 2 \end {matrix}\middle | {\frac {d x e^{i \pi }}{c}} \right )}}{\Gamma \left (m + 2\right )} + \frac {2 a b \sqrt {c} e^{m} x^{m + 3} \Gamma \left (m + 3\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, m + 3 \\ m + 4 \end {matrix}\middle | {\frac {d x e^{i \pi }}{c}} \right )}}{\Gamma \left (m + 4\right )} + \frac {b^{2} \sqrt {c} e^{m} x^{m + 5} \Gamma \left (m + 5\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, m + 5 \\ m + 6 \end {matrix}\middle | {\frac {d x e^{i \pi }}{c}} \right )}}{\Gamma \left (m + 6\right )} \] Input:

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

Output:

a**2*sqrt(c)*e**m*x**(m + 1)*gamma(m + 1)*hyper((-1/2, m + 1), (m + 2,), d 
*x*exp_polar(I*pi)/c)/gamma(m + 2) + 2*a*b*sqrt(c)*e**m*x**(m + 3)*gamma(m 
 + 3)*hyper((-1/2, m + 3), (m + 4,), d*x*exp_polar(I*pi)/c)/gamma(m + 4) + 
 b**2*sqrt(c)*e**m*x**(m + 5)*gamma(m + 5)*hyper((-1/2, m + 5), (m + 6,), 
d*x*exp_polar(I*pi)/c)/gamma(m + 6)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

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

Output:

(2*e**m*(16*x**m*sqrt(c + d*x)*a**2*c*d**4*m**4 + 256*x**m*sqrt(c + d*x)*a 
**2*c*d**4*m**3 + 1496*x**m*sqrt(c + d*x)*a**2*c*d**4*m**2 + 3776*x**m*sqr 
t(c + d*x)*a**2*c*d**4*m + 3465*x**m*sqrt(c + d*x)*a**2*c*d**4 + 32*x**m*s 
qrt(c + d*x)*a**2*d**5*m**5*x + 528*x**m*sqrt(c + d*x)*a**2*d**5*m**4*x + 
3248*x**m*sqrt(c + d*x)*a**2*d**5*m**3*x + 9048*x**m*sqrt(c + d*x)*a**2*d* 
*5*m**2*x + 10706*x**m*sqrt(c + d*x)*a**2*d**5*m*x + 3465*x**m*sqrt(c + d* 
x)*a**2*d**5*x + 32*x**m*sqrt(c + d*x)*a*b*c**3*d**2*m**4 + 416*x**m*sqrt( 
c + d*x)*a*b*c**3*d**2*m**3 + 1816*x**m*sqrt(c + d*x)*a*b*c**3*d**2*m**2 + 
 3016*x**m*sqrt(c + d*x)*a*b*c**3*d**2*m + 1584*x**m*sqrt(c + d*x)*a*b*c** 
3*d**2 - 32*x**m*sqrt(c + d*x)*a*b*c**2*d**3*m**4*x - 400*x**m*sqrt(c + d* 
x)*a*b*c**2*d**3*m**3*x - 1624*x**m*sqrt(c + d*x)*a*b*c**2*d**3*m**2*x - 2 
300*x**m*sqrt(c + d*x)*a*b*c**2*d**3*m*x - 792*x**m*sqrt(c + d*x)*a*b*c**2 
*d**3*x + 32*x**m*sqrt(c + d*x)*a*b*c*d**4*m**4*x**2 + 384*x**m*sqrt(c + d 
*x)*a*b*c*d**4*m**3*x**2 + 1456*x**m*sqrt(c + d*x)*a*b*c*d**4*m**2*x**2 + 
1824*x**m*sqrt(c + d*x)*a*b*c*d**4*m*x**2 + 594*x**m*sqrt(c + d*x)*a*b*c*d 
**4*x**2 + 64*x**m*sqrt(c + d*x)*a*b*d**5*m**5*x**3 + 928*x**m*sqrt(c + d* 
x)*a*b*d**5*m**4*x**3 + 4832*x**m*sqrt(c + d*x)*a*b*d**5*m**3*x**3 + 10928 
*x**m*sqrt(c + d*x)*a*b*d**5*m**2*x**3 + 10308*x**m*sqrt(c + d*x)*a*b*d**5 
*m*x**3 + 2970*x**m*sqrt(c + d*x)*a*b*d**5*x**3 + 16*x**m*sqrt(c + d*x)*b* 
*2*c**5*m**4 + 160*x**m*sqrt(c + d*x)*b**2*c**5*m**3 + 560*x**m*sqrt(c ...