\(\int \frac {(c+d x)^n (A+B x+C x^2+D x^3)}{(a+b x)^{3/2}} \, dx\) [185]

Optimal result

Integrand size = 32, antiderivative size = 413 \[ \int \frac {(c+d x)^n \left (A+B x+C x^2+D x^3\right )}{(a+b x)^{3/2}} \, dx=-\frac {2 \left (A b^3-a \left (b^2 B-a b C+a^2 D\right )\right ) (c+d x)^{1+n}}{b^3 (b c-a d) \sqrt {a+b x}}-\frac {2 (3 b c D+6 a d D (2+n)-b C d (5+2 n)) \sqrt {a+b x} (c+d x)^{1+n}}{b^3 d^2 (3+2 n) (5+2 n)}+\frac {2 D (a+b x)^{3/2} (c+d x)^{1+n}}{b^3 d (5+2 n)}+\frac {2 \left (2 (b c-a d) \left (\frac {b c}{2}+a d (1+n)\right ) (3 b c D+6 a d D (2+n)-b C d (5+2 n))-2 d \left (\frac {3}{2}+n\right ) \left (4 a^3 d^2 D \left (2+3 n+n^2\right )-b^3 d (5+2 n) (B c+A d (1+2 n))-2 a^2 b d \left (3 c D+C d \left (5+7 n+2 n^2\right )\right )+a b^2 \left (3 c^2 D+c C d (5+2 n)+2 B d^2 \left (5+7 n+2 n^2\right )\right )\right )\right ) \sqrt {a+b x} (c+d x)^n \left (\frac {b (c+d x)}{b c-a d}\right )^{-n} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-n,\frac {3}{2},-\frac {d (a+b x)}{b c-a d}\right )}{b^4 d^2 (b c-a d) (3+2 n) (5+2 n)} \] Output:

-2*(A*b^3-a*(B*b^2-C*a*b+D*a^2))*(d*x+c)^(1+n)/b^3/(-a*d+b*c)/(b*x+a)^(1/2 
)-2*(3*D*b*c+6*a*d*D*(2+n)-b*C*d*(5+2*n))*(b*x+a)^(1/2)*(d*x+c)^(1+n)/b^3/ 
d^2/(3+2*n)/(5+2*n)+2*D*(b*x+a)^(3/2)*(d*x+c)^(1+n)/b^3/d/(5+2*n)+2*(2*(-a 
*d+b*c)*(1/2*b*c+a*d*(1+n))*(3*D*b*c+6*a*d*D*(2+n)-b*C*d*(5+2*n))-2*d*(3/2 
+n)*(4*a^3*d^2*D*(n^2+3*n+2)-b^3*d*(5+2*n)*(B*c+A*d*(1+2*n))-2*a^2*b*d*(3* 
D*c+C*d*(2*n^2+7*n+5))+a*b^2*(3*D*c^2+c*C*d*(5+2*n)+2*B*d^2*(2*n^2+7*n+5)) 
))*(b*x+a)^(1/2)*(d*x+c)^n*hypergeom([1/2, -n],[3/2],-d*(b*x+a)/(-a*d+b*c) 
)/b^4/d^2/(-a*d+b*c)/(3+2*n)/(5+2*n)/((b*(d*x+c)/(-a*d+b*c))^n)
 

Mathematica [A] (verified)

Time = 1.42 (sec) , antiderivative size = 232, normalized size of antiderivative = 0.56 \[ \int \frac {(c+d x)^n \left (A+B x+C x^2+D x^3\right )}{(a+b x)^{3/2}} \, dx=\frac {2 (c+d x)^n \left (\frac {b (c+d x)}{b c-a d}\right )^{-n} \left (-15 \left (A b^3-a \left (b^2 B-a b C+a^2 D\right )\right ) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-n,\frac {1}{2},\frac {d (a+b x)}{-b c+a d}\right )+(a+b x) \left (15 \left (b^2 B-2 a b C+3 a^2 D\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-n,\frac {3}{2},\frac {d (a+b x)}{-b c+a d}\right )-(a+b x) \left (-5 (b C-3 a D) \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-n,\frac {5}{2},\frac {d (a+b x)}{-b c+a d}\right )-3 D (a+b x) \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-n,\frac {7}{2},\frac {d (a+b x)}{-b c+a d}\right )\right )\right )\right )}{15 b^4 \sqrt {a+b x}} \] Input:

Integrate[((c + d*x)^n*(A + B*x + C*x^2 + D*x^3))/(a + b*x)^(3/2),x]
 

Output:

(2*(c + d*x)^n*(-15*(A*b^3 - a*(b^2*B - a*b*C + a^2*D))*Hypergeometric2F1[ 
-1/2, -n, 1/2, (d*(a + b*x))/(-(b*c) + a*d)] + (a + b*x)*(15*(b^2*B - 2*a* 
b*C + 3*a^2*D)*Hypergeometric2F1[1/2, -n, 3/2, (d*(a + b*x))/(-(b*c) + a*d 
)] - (a + b*x)*(-5*(b*C - 3*a*D)*Hypergeometric2F1[3/2, -n, 5/2, (d*(a + b 
*x))/(-(b*c) + a*d)] - 3*D*(a + b*x)*Hypergeometric2F1[5/2, -n, 7/2, (d*(a 
 + b*x))/(-(b*c) + a*d)]))))/(15*b^4*Sqrt[a + b*x]*((b*(c + d*x))/(b*c - a 
*d))^n)
 

Rubi [A] (verified)

Time = 0.90 (sec) , antiderivative size = 421, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.219, Rules used = {2124, 27, 1194, 27, 90, 80, 79}

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.

\(\displaystyle \int \frac {(c+d x)^n \left (A+B x+C x^2+D x^3\right )}{(a+b x)^{3/2}} \, dx\)

\(\Big \downarrow \) 2124

\(\displaystyle -\frac {2 \int \frac {(c+d x)^n \left (-\left (\left (c-\frac {a d}{b}\right ) D x^2\right )-\frac {(b c-a d) (b C-a D) x}{b^2}+\frac {2 d D (n+1) a^3-b (c D+2 C d (n+1)) a^2+b^2 (c C+2 B d (n+1)) a-b^3 (B c+A d (2 n+1))}{b^3}\right )}{2 \sqrt {a+b x}}dx}{b c-a d}-\frac {2 (c+d x)^{n+1} \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{\sqrt {a+b x} (b c-a d)}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\int \frac {(c+d x)^n \left (-\left (\left (c-\frac {a d}{b}\right ) D x^2\right )-\frac {(b c-a d) (b C-a D) x}{b^2}+\frac {2 d D (n+1) a^3-b (c D+2 C d (n+1)) a^2+b^2 (c C+2 B d (n+1)) a-b^3 (B c+A d (2 n+1))}{b^3}\right )}{\sqrt {a+b x}}dx}{b c-a d}-\frac {2 (c+d x)^{n+1} \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{\sqrt {a+b x} (b c-a d)}\)

\(\Big \downarrow \) 1194

\(\displaystyle -\frac {\frac {2 \int \frac {(c+d x)^n \left (\frac {4 d^2 D \left (n^2+3 n+2\right ) a^3}{b}-2 d \left (3 c D+C d \left (2 n^2+7 n+5\right )\right ) a^2+b \left (3 D c^2+C d (2 n+5) c+2 B d^2 \left (2 n^2+7 n+5\right )\right ) a-b^2 d (2 n+5) (B c+A d (2 n+1))+(b c-a d) (3 b c D+6 a d (n+2) D-b C d (2 n+5)) x\right )}{2 \sqrt {a+b x}}dx}{b^2 d (2 n+5)}-\frac {2 D (a+b x)^{3/2} (b c-a d) (c+d x)^{n+1}}{b^3 d (2 n+5)}}{b c-a d}-\frac {2 (c+d x)^{n+1} \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{\sqrt {a+b x} (b c-a d)}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {\int \frac {(c+d x)^n \left (\frac {4 d^2 D \left (n^2+3 n+2\right ) a^3}{b}-2 d \left (3 c D+C d \left (2 n^2+7 n+5\right )\right ) a^2+b \left (3 D c^2+C d (2 n+5) c+2 B d^2 \left (2 n^2+7 n+5\right )\right ) a-b^2 d (2 n+5) (B c+A d (2 n+1))+(b c-a d) (3 b c D+6 a d (n+2) D-b C d (2 n+5)) x\right )}{\sqrt {a+b x}}dx}{b^2 d (2 n+5)}-\frac {2 D (a+b x)^{3/2} (b c-a d) (c+d x)^{n+1}}{b^3 d (2 n+5)}}{b c-a d}-\frac {2 (c+d x)^{n+1} \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{\sqrt {a+b x} (b c-a d)}\)

\(\Big \downarrow \) 90

\(\displaystyle -\frac {\frac {\left (\frac {4 a^3 d^2 D \left (n^2+3 n+2\right )}{b}-2 a^2 d \left (3 c D+C d \left (2 n^2+7 n+5\right )\right )+a b \left (2 B d^2 \left (2 n^2+7 n+5\right )+3 c^2 D+c C d (2 n+5)\right )-\frac {(b c-a d) (2 a d (n+1)+b c) (6 a d D (n+2)+3 b c D-b C d (2 n+5))}{b d (2 n+3)}-b^2 d (2 n+5) (A d (2 n+1)+B c)\right ) \int \frac {(c+d x)^n}{\sqrt {a+b x}}dx+\frac {2 \sqrt {a+b x} (b c-a d) (c+d x)^{n+1} (6 a d D (n+2)+3 b c D-b C d (2 n+5))}{b d (2 n+3)}}{b^2 d (2 n+5)}-\frac {2 D (a+b x)^{3/2} (b c-a d) (c+d x)^{n+1}}{b^3 d (2 n+5)}}{b c-a d}-\frac {2 (c+d x)^{n+1} \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{\sqrt {a+b x} (b c-a d)}\)

\(\Big \downarrow \) 80

\(\displaystyle -\frac {\frac {(c+d x)^n \left (\frac {b (c+d x)}{b c-a d}\right )^{-n} \left (\frac {4 a^3 d^2 D \left (n^2+3 n+2\right )}{b}-2 a^2 d \left (3 c D+C d \left (2 n^2+7 n+5\right )\right )+a b \left (2 B d^2 \left (2 n^2+7 n+5\right )+3 c^2 D+c C d (2 n+5)\right )-\frac {(b c-a d) (2 a d (n+1)+b c) (6 a d D (n+2)+3 b c D-b C d (2 n+5))}{b d (2 n+3)}-b^2 d (2 n+5) (A d (2 n+1)+B c)\right ) \int \frac {\left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^n}{\sqrt {a+b x}}dx+\frac {2 \sqrt {a+b x} (b c-a d) (c+d x)^{n+1} (6 a d D (n+2)+3 b c D-b C d (2 n+5))}{b d (2 n+3)}}{b^2 d (2 n+5)}-\frac {2 D (a+b x)^{3/2} (b c-a d) (c+d x)^{n+1}}{b^3 d (2 n+5)}}{b c-a d}-\frac {2 (c+d x)^{n+1} \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{\sqrt {a+b x} (b c-a d)}\)

\(\Big \downarrow \) 79

\(\displaystyle -\frac {2 (c+d x)^{n+1} \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{\sqrt {a+b x} (b c-a d)}-\frac {\frac {\frac {2 \sqrt {a+b x} (c+d x)^n \left (\frac {b (c+d x)}{b c-a d}\right )^{-n} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-n,\frac {3}{2},-\frac {d (a+b x)}{b c-a d}\right ) \left (\frac {4 a^3 d^2 D \left (n^2+3 n+2\right )}{b}-2 a^2 d \left (3 c D+C d \left (2 n^2+7 n+5\right )\right )+a b \left (2 B d^2 \left (2 n^2+7 n+5\right )+3 c^2 D+c C d (2 n+5)\right )-\frac {(b c-a d) (2 a d (n+1)+b c) (6 a d D (n+2)+3 b c D-b C d (2 n+5))}{b d (2 n+3)}-b^2 d (2 n+5) (A d (2 n+1)+B c)\right )}{b}+\frac {2 \sqrt {a+b x} (b c-a d) (c+d x)^{n+1} (6 a d D (n+2)+3 b c D-b C d (2 n+5))}{b d (2 n+3)}}{b^2 d (2 n+5)}-\frac {2 D (a+b x)^{3/2} (b c-a d) (c+d x)^{n+1}}{b^3 d (2 n+5)}}{b c-a d}\)

Input:

Int[((c + d*x)^n*(A + B*x + C*x^2 + D*x^3))/(a + b*x)^(3/2),x]
 

Output:

(-2*(A - (a*(b^2*B - a*b*C + a^2*D))/b^3)*(c + d*x)^(1 + n))/((b*c - a*d)* 
Sqrt[a + b*x]) - ((-2*(b*c - a*d)*D*(a + b*x)^(3/2)*(c + d*x)^(1 + n))/(b^ 
3*d*(5 + 2*n)) + ((2*(b*c - a*d)*(3*b*c*D + 6*a*d*D*(2 + n) - b*C*d*(5 + 2 
*n))*Sqrt[a + b*x]*(c + d*x)^(1 + n))/(b*d*(3 + 2*n)) + (2*((4*a^3*d^2*D*( 
2 + 3*n + n^2))/b - b^2*d*(5 + 2*n)*(B*c + A*d*(1 + 2*n)) - ((b*c - a*d)*( 
b*c + 2*a*d*(1 + n))*(3*b*c*D + 6*a*d*D*(2 + n) - b*C*d*(5 + 2*n)))/(b*d*( 
3 + 2*n)) - 2*a^2*d*(3*c*D + C*d*(5 + 7*n + 2*n^2)) + a*b*(3*c^2*D + c*C*d 
*(5 + 2*n) + 2*B*d^2*(5 + 7*n + 2*n^2)))*Sqrt[a + b*x]*(c + d*x)^n*Hyperge 
ometric2F1[1/2, -n, 3/2, -((d*(a + b*x))/(b*c - a*d))])/(b*((b*(c + d*x))/ 
(b*c - a*d))^n))/(b^2*d*(5 + 2*n)))/(b*c - a*d)
 

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 79

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

rule 80

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c 
 + d*x)^FracPart[n]/((b/(b*c - a*d))^IntPart[n]*(b*((c + d*x)/(b*c - a*d))) 
^FracPart[n])   Int[(a + b*x)^m*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c - a*d) 
), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] &&  !IntegerQ[m] &&  !Integ 
erQ[n] && (RationalQ[m] ||  !SimplerQ[n + 1, m + 1])
 

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 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]

\[\int \frac {\left (x d +c \right )^{n} \left (D x^{3}+C \,x^{2}+B x +A \right )}{\left (b x +a \right )^{\frac {3}{2}}}d x\]

Input:

int((d*x+c)^n*(D*x^3+C*x^2+B*x+A)/(b*x+a)^(3/2),x)
 

Output:

int((d*x+c)^n*(D*x^3+C*x^2+B*x+A)/(b*x+a)^(3/2),x)
 

Fricas [F]

\[ \int \frac {(c+d x)^n \left (A+B x+C x^2+D x^3\right )}{(a+b x)^{3/2}} \, dx=\int { \frac {{\left (D x^{3} + C x^{2} + B x + A\right )} {\left (d x + c\right )}^{n}}{{\left (b x + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((d*x+c)^n*(D*x^3+C*x^2+B*x+A)/(b*x+a)^(3/2),x, algorithm="fricas 
")
 

Output:

integral((D*x^3 + C*x^2 + B*x + A)*sqrt(b*x + a)*(d*x + c)^n/(b^2*x^2 + 2* 
a*b*x + a^2), x)
 

Sympy [F(-2)]

Exception generated. \[ \int \frac {(c+d x)^n \left (A+B x+C x^2+D x^3\right )}{(a+b x)^{3/2}} \, dx=\text {Exception raised: HeuristicGCDFailed} \] Input:

integrate((d*x+c)**n*(D*x**3+C*x**2+B*x+A)/(b*x+a)**(3/2),x)
 

Output:

Exception raised: HeuristicGCDFailed >> no luck
 

Maxima [F]

\[ \int \frac {(c+d x)^n \left (A+B x+C x^2+D x^3\right )}{(a+b x)^{3/2}} \, dx=\int { \frac {{\left (D x^{3} + C x^{2} + B x + A\right )} {\left (d x + c\right )}^{n}}{{\left (b x + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((d*x+c)^n*(D*x^3+C*x^2+B*x+A)/(b*x+a)^(3/2),x, algorithm="maxima 
")
 

Output:

integrate((D*x^3 + C*x^2 + B*x + A)*(d*x + c)^n/(b*x + a)^(3/2), x)
 

Giac [F]

\[ \int \frac {(c+d x)^n \left (A+B x+C x^2+D x^3\right )}{(a+b x)^{3/2}} \, dx=\int { \frac {{\left (D x^{3} + C x^{2} + B x + A\right )} {\left (d x + c\right )}^{n}}{{\left (b x + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((d*x+c)^n*(D*x^3+C*x^2+B*x+A)/(b*x+a)^(3/2),x, algorithm="giac")
 

Output:

integrate((D*x^3 + C*x^2 + B*x + A)*(d*x + c)^n/(b*x + a)^(3/2), x)
 

Mupad [F(-1)]

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

int(((c + d*x)^n*(A + B*x + C*x^2 + x^3*D))/(a + b*x)^(3/2),x)
 

Output:

int(((c + d*x)^n*(A + B*x + C*x^2 + x^3*D))/(a + b*x)^(3/2), x)
 

Reduce [F]

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

int((d*x+c)^n*(D*x^3+C*x^2+B*x+A)/(b*x+a)^(3/2),x)
                                                                                    
                                                                                    
 

Output:

(2*( - 8*(c + d*x)**n*sqrt(a + b*x)*a**3*c*d**2*n**2 - 40*(c + d*x)**n*sqr 
t(a + b*x)*a**3*c*d**2*n - 48*(c + d*x)**n*sqrt(a + b*x)*a**3*c*d**2 + 8*( 
c + d*x)**n*sqrt(a + b*x)*a**3*d**3*n**3*x + 40*(c + d*x)**n*sqrt(a + b*x) 
*a**3*d**3*n**2*x + 48*(c + d*x)**n*sqrt(a + b*x)*a**3*d**3*n*x + 16*(c + 
d*x)**n*sqrt(a + b*x)*a**2*b*c**2*d*n**2 + 64*(c + d*x)**n*sqrt(a + b*x)*a 
**2*b*c**2*d*n + 40*(c + d*x)**n*sqrt(a + b*x)*a**2*b*c**2*d - 16*(c + d*x 
)**n*sqrt(a + b*x)*a**2*b*c*d**2*n**3*x - 68*(c + d*x)**n*sqrt(a + b*x)*a* 
*2*b*c*d**2*n**2*x - 60*(c + d*x)**n*sqrt(a + b*x)*a**2*b*c*d**2*n*x - 24* 
(c + d*x)**n*sqrt(a + b*x)*a**2*b*c*d**2*x - 8*(c + d*x)**n*sqrt(a + b*x)* 
a**2*b*d**3*n**3*x**2 - 28*(c + d*x)**n*sqrt(a + b*x)*a**2*b*d**3*n**2*x** 
2 - 12*(c + d*x)**n*sqrt(a + b*x)*a**2*b*d**3*n*x**2 + 8*(c + d*x)**n*sqrt 
(a + b*x)*a*b**3*c*d*n**3 + 28*(c + d*x)**n*sqrt(a + b*x)*a*b**3*c*d*n**2 
+ 14*(c + d*x)**n*sqrt(a + b*x)*a*b**3*c*d*n - 15*(c + d*x)**n*sqrt(a + b* 
x)*a*b**3*c*d + 8*(c + d*x)**n*sqrt(a + b*x)*a*b**3*d**2*n**3*x + 32*(c + 
d*x)**n*sqrt(a + b*x)*a*b**3*d**2*n**2*x + 30*(c + d*x)**n*sqrt(a + b*x)*a 
*b**3*d**2*n*x - 8*(c + d*x)**n*sqrt(a + b*x)*a*b**2*c**3*n**2 - 8*(c + d* 
x)**n*sqrt(a + b*x)*a*b**2*c**3*n + 8*(c + d*x)**n*sqrt(a + b*x)*a*b**2*c* 
*2*d*n**3*x + 16*(c + d*x)**n*sqrt(a + b*x)*a*b**2*c**2*d*n**2*x + 32*(c + 
 d*x)**n*sqrt(a + b*x)*a*b**2*c**2*d*n*x + 20*(c + d*x)**n*sqrt(a + b*x)*a 
*b**2*c**2*d*x + 16*(c + d*x)**n*sqrt(a + b*x)*a*b**2*c*d**2*n**3*x**2 ...