3.1.31 \(\int \frac {(c+d x)^n (A+B x+C x^2+D x^3)}{(a+b x)^3} \, dx\) [31]

3.1.31.1 Optimal result

Integrand size = 30, antiderivative size = 329 \[ \int \frac {(c+d x)^n \left (A+B x+C x^2+D x^3\right )}{(a+b x)^3} \, dx=\frac {D (c+d x)^{1+n}}{b^3 d (1+n)}-\frac {\left (A b^3-a \left (b^2 B-a b C+a^2 D\right )\right ) (c+d x)^{1+n}}{2 b^3 (b c-a d) (a+b x)^2}-\frac {\left (b^3 (2 B c-A d (1-n))-a^3 d D (5+n)-a b^2 (4 c C+B d (1+n))+a^2 b (6 c D+C d (3+n))\right ) (c+d x)^{1+n}}{2 b^3 (b c-a d)^2 (a+b x)}-\frac {\left (b^3 \left (2 c^2 C+2 B c d n-A d^2 (1-n) n\right )-a^3 d^2 D \left (6+5 n+n^2\right )+a^2 b d (2+n) (6 c D+C d (1+n))-a b^2 \left (6 c^2 D+4 c C d (1+n)+B d^2 n (1+n)\right )\right ) (c+d x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {b (c+d x)}{b c-a d}\right )}{2 b^3 (b c-a d)^3 (1+n)} \]

output

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

3.1.31.2 Mathematica [A] (verified)

Time = 1.31 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.57 \[ \int \frac {(c+d x)^n \left (A+B x+C x^2+D x^3\right )}{(a+b x)^3} \, dx=\frac {(c+d x)^{1+n} \left (\frac {D}{d}-\frac {(b C-3 a D) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {b (c+d x)}{b c-a d}\right )}{b c-a d}+\frac {d \left (b^2 B-2 a b C+3 a^2 D\right ) \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,\frac {b (c+d x)}{b c-a d}\right )}{(b c-a d)^2}-\frac {d^2 \left (A b^3-a \left (b^2 B-a b C+a^2 D\right )\right ) \operatorname {Hypergeometric2F1}\left (3,1+n,2+n,\frac {b (c+d x)}{b c-a d}\right )}{(b c-a d)^3}\right )}{b^3 (1+n)} \]

input

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

output

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

3.1.31.3 Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 360, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {2124, 25, 1193, 25, 27, 90, 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.

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

\(\Big \downarrow \) 2124

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1193

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 90

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

\(\Big \downarrow \) 78

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

input

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

output

-1/2*((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)^2) + (-(((b^3*(2*B*c - A*d*(1 - n)) - a^3*d*D*(5 + n) - a*b^ 
2*(4*c*C + B*d*(1 + n)) + a^2*b*(6*c*D + C*d*(3 + n)))*(c + d*x)^(1 + n))/ 
(b^3*(b*c - a*d)*(a + b*x))) + ((2*(b*c - a*d)^2*D*(c + d*x)^(1 + n))/(d*( 
1 + n)) - ((b^3*(2*c^2*C + 2*B*c*d*n - A*d^2*(1 - n)*n) - a^3*d^2*D*(6 + 5 
*n + n^2) + a^2*b*d*(2 + n)*(6*c*D + C*d*(1 + n)) - a*b^2*(6*c^2*D + 4*c*C 
*d*(1 + n) + B*d^2*n*(1 + n)))*(c + d*x)^(1 + n)*Hypergeometric2F1[1, 1 + 
n, 2 + n, (b*(c + d*x))/(b*c - a*d)])/((b*c - a*d)*(1 + n)))/(b^3*(b*c - a 
*d)))/(2*(b*c - a*d))
 

3.1.31.3.1 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 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 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])
 

3.1.31.4 Maple [F]

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

input

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

output

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

3.1.31.5 Fricas [F]

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

input

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

output

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

3.1.31.6 Sympy [F]

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

input

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

output

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

3.1.31.7 Maxima [F]

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

input

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

output

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

3.1.31.8 Giac [F]

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

input

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

output

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

3.1.31.9 Mupad [F(-1)]

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

input

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

output

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