\(\int \frac {(c+d x)^n (A+B x+C x^2+D x^3)}{(a+b x)^4} \, dx\) [181]

Optimal result

Integrand size = 30, antiderivative size = 332 \[ \int \frac {(c+d x)^n \left (A+B x+C x^2+D x^3\right )}{(a+b x)^4} \, dx=-\frac {\left (A b^3-a \left (b^2 B-a b C+a^2 D\right )\right ) (c+d x)^{1+n}}{3 b^3 (b c-a d) (a+b x)^3}+\frac {(a d D (1+3 n)-b (c D+C d n)) (c+d x)^{1+n}}{b^3 d^2 (1-n) n (a+b x)^2}+\frac {D (c+d x)^{1+n}}{b^3 d n (a+b x)}-\frac {\left (a b^2 d (1+n) \left (18 c^2 D+6 c C d n-B d^2 (1-n) n\right )-a^2 b d^2 (9 c D+C d n) \left (2+3 n+n^2\right )+a^3 d^3 D \left (6+11 n+6 n^2+n^3\right )-b^3 \left (6 c^3 D+6 c^2 C d n-3 B c d^2 (1-n) n+A d^3 n \left (2-3 n+n^2\right )\right )\right ) (c+d x)^{1+n} \operatorname {Hypergeometric2F1}\left (3,1+n,2+n,\frac {b (c+d x)}{b c-a d}\right )}{3 b^3 (b c-a d)^4 n \left (1-n^2\right )} \] Output:

-1/3*(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)^3+ 
(a*d*D*(1+3*n)-b*(C*d*n+D*c))*(d*x+c)^(1+n)/b^3/d^2/(1-n)/n/(b*x+a)^2+D*(d 
*x+c)^(1+n)/b^3/d/n/(b*x+a)-1/3*(a*b^2*d*(1+n)*(18*D*c^2+6*c*C*d*n-B*d^2*( 
1-n)*n)-a^2*b*d^2*(C*d*n+9*D*c)*(n^2+3*n+2)+a^3*d^3*D*(n^3+6*n^2+11*n+6)-b 
^3*(6*D*c^3+6*c^2*C*d*n-3*B*c*d^2*(1-n)*n+A*d^3*n*(n^2-3*n+2)))*(d*x+c)^(1 
+n)*hypergeom([3, 1+n],[2+n],b*(d*x+c)/(-a*d+b*c))/b^3/(-a*d+b*c)^4/n/(-n^ 
2+1)
 

Mathematica [A] (verified)

Time = 1.56 (sec) , antiderivative size = 220, normalized size of antiderivative = 0.66 \[ \int \frac {(c+d x)^n \left (A+B x+C x^2+D x^3\right )}{(a+b x)^4} \, dx=\frac {(c+d x)^{1+n} \left (-(b c-a d)^3 D \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {b (c+d x)}{b c-a d}\right )+d (b c-a d)^2 (b C-3 a D) \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,\frac {b (c+d x)}{b c-a d}\right )+d^2 (-b c+a d) \left (b^2 B-2 a b C+3 a^2 D\right ) \operatorname {Hypergeometric2F1}\left (3,1+n,2+n,\frac {b (c+d x)}{b c-a d}\right )+d^3 \left (A b^3-a \left (b^2 B-a b C+a^2 D\right )\right ) \operatorname {Hypergeometric2F1}\left (4,1+n,2+n,\frac {b (c+d x)}{b c-a d}\right )\right )}{b^3 (b c-a d)^4 (1+n)} \] Input:

Integrate[((c + d*x)^n*(A + B*x + C*x^2 + D*x^3))/(a + b*x)^4,x]
 

Output:

((c + d*x)^(1 + n)*(-((b*c - a*d)^3*D*Hypergeometric2F1[1, 1 + n, 2 + n, ( 
b*(c + d*x))/(b*c - a*d)]) + d*(b*c - a*d)^2*(b*C - 3*a*D)*Hypergeometric2 
F1[2, 1 + n, 2 + n, (b*(c + d*x))/(b*c - a*d)] + d^2*(-(b*c) + a*d)*(b^2*B 
 - 2*a*b*C + 3*a^2*D)*Hypergeometric2F1[3, 1 + n, 2 + n, (b*(c + d*x))/(b* 
c - a*d)] + d^3*(A*b^3 - a*(b^2*B - a*b*C + a^2*D))*Hypergeometric2F1[4, 1 
 + n, 2 + n, (b*(c + d*x))/(b*c - a*d)]))/(b^3*(b*c - a*d)^4*(1 + n))
 

Rubi [A] (verified)

Time = 0.92 (sec) , antiderivative size = 506, normalized size of antiderivative = 1.52, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2124, 25, 1193, 27, 87, 78}

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.

\(\displaystyle \int \frac {(c+d x)^n \left (A+B x+C x^2+D x^3\right )}{(a+b x)^4} \, dx\)

\(\Big \downarrow \) 2124

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1193

\(\displaystyle \frac {-\frac {\int \frac {(c+d x)^n \left (d^2 D \left (n^2+6 n+5\right ) a^3-b d \left (3 c D (3 n+5)+C d \left (n^2+3 n+2\right )\right ) a^2+b^2 \left (12 D c^2+6 C d (n+1) c-B d^2 \left (1-n^2\right )\right ) a-b^3 \left (6 C c^2-3 B d (1-n) c+A d^2 \left (n^2-3 n+2\right )\right )-6 b (b c-a d)^2 D x\right )}{b^3 (a+b x)^2}dx}{2 (b c-a d)}-\frac {(c+d x)^{n+1} \left (a^3 (-d) D (n+7)+a^2 b (9 c D+C d (n+4))-a b^2 (B d (n+1)+6 c C)+b^3 (3 B c-A d (2-n))\right )}{2 b^3 (a+b x)^2 (b c-a d)}}{3 (b c-a d)}-\frac {(c+d x)^{n+1} \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right )}{3 b^3 (a+b x)^3 (b c-a d)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {\int \frac {(c+d x)^n \left (d^2 D \left (n^2+6 n+5\right ) a^3-b d \left (3 c D (3 n+5)+C d \left (n^2+3 n+2\right )\right ) a^2+b^2 \left (12 D c^2+6 C d (n+1) c-B d^2 \left (1-n^2\right )\right ) a-b^3 \left (6 C c^2-3 B d (1-n) c+A d^2 \left (n^2-3 n+2\right )\right )-6 b (b c-a d)^2 D x\right )}{(a+b x)^2}dx}{2 b^3 (b c-a d)}-\frac {(c+d x)^{n+1} \left (a^3 (-d) D (n+7)+a^2 b (9 c D+C d (n+4))-a b^2 (B d (n+1)+6 c C)+b^3 (3 B c-A d (2-n))\right )}{2 b^3 (a+b x)^2 (b c-a d)}}{3 (b c-a d)}-\frac {(c+d x)^{n+1} \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right )}{3 b^3 (a+b x)^3 (b c-a d)}\)

\(\Big \downarrow \) 87

\(\displaystyle \frac {-\frac {\frac {\left (a^3 d^3 D \left (n^3+6 n^2+11 n+6\right )-a^2 b d^2 \left (n^2+3 n+2\right ) (9 c D+C d n)+a b^2 d (n+1) \left (-B d^2 (1-n) n+18 c^2 D+6 c C d n\right )-\left (b^3 \left (A d^3 n \left (n^2-3 n+2\right )-3 B c d^2 (1-n) n+6 c^3 D+6 c^2 C d n\right )\right )\right ) \int \frac {(c+d x)^n}{a+b x}dx}{b c-a d}-\frac {(c+d x)^{n+1} \left (a^3 d^2 D \left (n^2+6 n+11\right )-a^2 b d \left (9 c D (n+3)+C d \left (n^2+3 n+2\right )\right )+a b^2 \left (-B d^2 \left (1-n^2\right )+18 c^2 D+6 c C d (n+1)\right )-b^3 \left (A d^2 \left (n^2-3 n+2\right )-3 B c d (1-n)+6 c^2 C\right )\right )}{(a+b x) (b c-a d)}}{2 b^3 (b c-a d)}-\frac {(c+d x)^{n+1} \left (a^3 (-d) D (n+7)+a^2 b (9 c D+C d (n+4))-a b^2 (B d (n+1)+6 c C)+b^3 (3 B c-A d (2-n))\right )}{2 b^3 (a+b x)^2 (b c-a d)}}{3 (b c-a d)}-\frac {(c+d x)^{n+1} \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right )}{3 b^3 (a+b x)^3 (b c-a d)}\)

\(\Big \downarrow \) 78

\(\displaystyle \frac {-\frac {-\frac {(c+d x)^{n+1} \left (a^3 d^2 D \left (n^2+6 n+11\right )-a^2 b d \left (9 c D (n+3)+C d \left (n^2+3 n+2\right )\right )+a b^2 \left (-B d^2 \left (1-n^2\right )+18 c^2 D+6 c C d (n+1)\right )-b^3 \left (A d^2 \left (n^2-3 n+2\right )-3 B c d (1-n)+6 c^2 C\right )\right )}{(a+b x) (b c-a d)}-\frac {(c+d x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {b (c+d x)}{b c-a d}\right ) \left (a^3 d^3 D \left (n^3+6 n^2+11 n+6\right )-a^2 b d^2 \left (n^2+3 n+2\right ) (9 c D+C d n)+a b^2 d (n+1) \left (-B d^2 (1-n) n+18 c^2 D+6 c C d n\right )-\left (b^3 \left (A d^3 n \left (n^2-3 n+2\right )-3 B c d^2 (1-n) n+6 c^3 D+6 c^2 C d n\right )\right )\right )}{(n+1) (b c-a d)^2}}{2 b^3 (b c-a d)}-\frac {(c+d x)^{n+1} \left (a^3 (-d) D (n+7)+a^2 b (9 c D+C d (n+4))-a b^2 (B d (n+1)+6 c C)+b^3 (3 B c-A d (2-n))\right )}{2 b^3 (a+b x)^2 (b c-a d)}}{3 (b c-a d)}-\frac {(c+d x)^{n+1} \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right )}{3 b^3 (a+b x)^3 (b c-a d)}\)

Input:

Int[((c + d*x)^n*(A + B*x + C*x^2 + D*x^3))/(a + b*x)^4,x]
 

Output:

-1/3*((A*b^3 - a*(b^2*B - a*b*C + a^2*D))*(c + d*x)^(1 + n))/(b^3*(b*c - a 
*d)*(a + b*x)^3) + (-1/2*((b^3*(3*B*c - A*d*(2 - n)) - a^3*d*D*(7 + n) - a 
*b^2*(6*c*C + B*d*(1 + n)) + a^2*b*(9*c*D + C*d*(4 + n)))*(c + d*x)^(1 + n 
))/(b^3*(b*c - a*d)*(a + b*x)^2) - (-(((a^3*d^2*D*(11 + 6*n + n^2) + a*b^2 
*(18*c^2*D + 6*c*C*d*(1 + n) - B*d^2*(1 - n^2)) - b^3*(6*c^2*C - 3*B*c*d*( 
1 - n) + A*d^2*(2 - 3*n + n^2)) - a^2*b*d*(9*c*D*(3 + n) + C*d*(2 + 3*n + 
n^2)))*(c + d*x)^(1 + n))/((b*c - a*d)*(a + b*x))) - ((a*b^2*d*(1 + n)*(18 
*c^2*D + 6*c*C*d*n - B*d^2*(1 - n)*n) - a^2*b*d^2*(9*c*D + C*d*n)*(2 + 3*n 
 + n^2) + a^3*d^3*D*(6 + 11*n + 6*n^2 + n^3) - b^3*(6*c^3*D + 6*c^2*C*d*n 
- 3*B*c*d^2*(1 - n)*n + A*d^3*n*(2 - 3*n + n^2)))*(c + d*x)^(1 + n)*Hyperg 
eometric2F1[1, 1 + n, 2 + n, (b*(c + d*x))/(b*c - a*d)])/((b*c - a*d)^2*(1 
 + n)))/(2*b^3*(b*c - a*d)))/(3*(b*c - a*d))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 78

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

rule 87

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

rule 1193

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

\[\int \frac {\left (x d +c \right )^{n} \left (D x^{3}+C \,x^{2}+B x +A \right )}{\left (b x +a \right )^{4}}d x\]

Input:

int((d*x+c)^n*(D*x^3+C*x^2+B*x+A)/(b*x+a)^4,x)
 

Output:

int((d*x+c)^n*(D*x^3+C*x^2+B*x+A)/(b*x+a)^4,x)
 

Fricas [F]

\[ \int \frac {(c+d x)^n \left (A+B x+C x^2+D x^3\right )}{(a+b x)^4} \, dx=\int { \frac {{\left (D x^{3} + C x^{2} + B x + A\right )} {\left (d x + c\right )}^{n}}{{\left (b x + a\right )}^{4}} \,d x } \] Input:

integrate((d*x+c)^n*(D*x^3+C*x^2+B*x+A)/(b*x+a)^4,x, algorithm="fricas")
 

Output:

integral((D*x^3 + C*x^2 + B*x + A)*(d*x + c)^n/(b^4*x^4 + 4*a*b^3*x^3 + 6* 
a^2*b^2*x^2 + 4*a^3*b*x + a^4), x)
 

Sympy [F]

\[ \int \frac {(c+d x)^n \left (A+B x+C x^2+D x^3\right )}{(a+b x)^4} \, dx=\int \frac {\left (c + d x\right )^{n} \left (A + B x + C x^{2} + D x^{3}\right )}{\left (a + b x\right )^{4}}\, dx \] Input:

integrate((d*x+c)**n*(D*x**3+C*x**2+B*x+A)/(b*x+a)**4,x)
 

Output:

Integral((c + d*x)**n*(A + B*x + C*x**2 + D*x**3)/(a + b*x)**4, x)
 

Maxima [F]

\[ \int \frac {(c+d x)^n \left (A+B x+C x^2+D x^3\right )}{(a+b x)^4} \, dx=\int { \frac {{\left (D x^{3} + C x^{2} + B x + A\right )} {\left (d x + c\right )}^{n}}{{\left (b x + a\right )}^{4}} \,d x } \] Input:

integrate((d*x+c)^n*(D*x^3+C*x^2+B*x+A)/(b*x+a)^4,x, algorithm="maxima")
 

Output:

integrate((D*x^3 + C*x^2 + B*x + A)*(d*x + c)^n/(b*x + a)^4, x)
 

Giac [F]

\[ \int \frac {(c+d x)^n \left (A+B x+C x^2+D x^3\right )}{(a+b x)^4} \, dx=\int { \frac {{\left (D x^{3} + C x^{2} + B x + A\right )} {\left (d x + c\right )}^{n}}{{\left (b x + a\right )}^{4}} \,d x } \] Input:

integrate((d*x+c)^n*(D*x^3+C*x^2+B*x+A)/(b*x+a)^4,x, algorithm="giac")
 

Output:

integrate((D*x^3 + C*x^2 + B*x + A)*(d*x + c)^n/(b*x + a)^4, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x)^n \left (A+B x+C x^2+D x^3\right )}{(a+b x)^4} \, dx=\int \frac {{\left (c+d\,x\right )}^n\,\left (A+B\,x+C\,x^2+x^3\,D\right )}{{\left (a+b\,x\right )}^4} \,d x \] Input:

int(((c + d*x)^n*(A + B*x + C*x^2 + x^3*D))/(a + b*x)^4,x)
 

Output:

int(((c + d*x)^n*(A + B*x + C*x^2 + x^3*D))/(a + b*x)^4, x)
 

Reduce [F]

\[ \int \frac {(c+d x)^n \left (A+B x+C x^2+D x^3\right )}{(a+b x)^4} \, dx=\text {too large to display} \] Input:

int((d*x+c)^n*(D*x^3+C*x^2+B*x+A)/(b*x+a)^4,x)
                                                                                    
                                                                                    
 

Output:

( - (c + d*x)**n*a**3*c*d**2*n**2 - 5*(c + d*x)**n*a**3*c*d**2*n - 6*(c + 
d*x)**n*a**3*c*d**2 + (c + d*x)**n*a**3*d**3*n**3*x + 5*(c + d*x)**n*a**3* 
d**3*n**2*x + 6*(c + d*x)**n*a**3*d**3*n*x + 2*(c + d*x)**n*a**2*b*c**2*d* 
n**2 + 8*(c + d*x)**n*a**2*b*c**2*d*n - 2*(c + d*x)**n*a**2*b*c*d**2*n**3* 
x - 11*(c + d*x)**n*a**2*b*c*d**2*n**2*x - 15*(c + d*x)**n*a**2*b*c*d**2*n 
*x - 18*(c + d*x)**n*a**2*b*c*d**2*x - (c + d*x)**n*a**2*b*d**3*n**3*x**2 
- (c + d*x)**n*a**2*b*d**3*n**2*x**2 + 6*(c + d*x)**n*a**2*b*d**3*n*x**2 + 
 (c + d*x)**n*a*b**3*c*d*n**3 - 4*(c + d*x)**n*a*b**3*c*d*n**2 + 3*(c + d* 
x)**n*a*b**3*c*d*n + (c + d*x)**n*a*b**3*d**2*n**3*x - (c + d*x)**n*a*b**3 
*d**2*n**2*x - (c + d*x)**n*a*b**2*c**3*n**2 - (c + d*x)**n*a*b**2*c**3*n 
+ (c + d*x)**n*a*b**2*c**2*d*n**3*x + 7*(c + d*x)**n*a*b**2*c**2*d*n**2*x 
+ 24*(c + d*x)**n*a*b**2*c**2*d*n*x + 2*(c + d*x)**n*a*b**2*c*d**2*n**3*x* 
*2 - (c + d*x)**n*a*b**2*c*d**2*n**2*x**2 + 3*(c + d*x)**n*a*b**2*c*d**2*n 
*x**2 - 18*(c + d*x)**n*a*b**2*c*d**2*x**2 + (c + d*x)**n*a*b**2*d**3*n**3 
*x**3 - 3*(c + d*x)**n*a*b**2*d**3*n**2*x**3 + 2*(c + d*x)**n*a*b**2*d**3* 
n*x**3 - 3*(c + d*x)**n*b**4*c*d*n**2*x + 3*(c + d*x)**n*b**4*c*d*n*x - 3* 
(c + d*x)**n*b**3*c**3*n**2*x - 3*(c + d*x)**n*b**3*c**3*n*x - 6*(c + d*x) 
**n*b**3*c**2*d*n**2*x**2 + 12*(c + d*x)**n*b**3*c**2*d*n*x**2 - 3*(c + d* 
x)**n*b**3*c*d**2*n**2*x**3 + 9*(c + d*x)**n*b**3*c*d**2*n*x**3 - 6*(c + d 
*x)**n*b**3*c*d**2*x**3 - int(((c + d*x)**n*x)/(a**5*c*d*n**3 - 3*a**5*...