\(\int x^3 (a+b x)^n (c+d x)^{-n} \, dx\) [580]

Optimal result

Integrand size = 20, antiderivative size = 309 \[ \int x^3 (a+b x)^n (c+d x)^{-n} \, dx=\frac {\left (2 a b c d \left (6+n-n^2\right )+a^2 d^2 \left (18-7 n+n^2\right )+b^2 c^2 \left (6+5 n+n^2\right )\right ) (a+b x)^{1+n} (c+d x)^{1-n}}{24 b^3 d^3}-\frac {(a d (9-n)+b c (3+n)) (a+b x)^{2+n} (c+d x)^{1-n}}{12 b^3 d^2}+\frac {(a+b x)^{3+n} (c+d x)^{1-n}}{4 b^3 d}-\frac {\left (3 a b^2 c^2 d \left (2+n-2 n^2-n^3\right )+a^3 d^3 \left (6-11 n+6 n^2-n^3\right )+3 a^2 b c d^2 \left (2-n-2 n^2+n^3\right )+b^3 c^3 \left (6+11 n+6 n^2+n^3\right )\right ) (a+b x)^{1+n} (c+d x)^{1-n} \operatorname {Hypergeometric2F1}\left (1,2,2+n,-\frac {d (a+b x)}{b c-a d}\right )}{24 b^3 d^3 (b c-a d) (1+n)} \] Output:

1/24*(2*a*b*c*d*(-n^2+n+6)+a^2*d^2*(n^2-7*n+18)+b^2*c^2*(n^2+5*n+6))*(b*x+ 
a)^(1+n)*(d*x+c)^(1-n)/b^3/d^3-1/12*(a*d*(9-n)+b*c*(3+n))*(b*x+a)^(2+n)*(d 
*x+c)^(1-n)/b^3/d^2+1/4*(b*x+a)^(3+n)*(d*x+c)^(1-n)/b^3/d-1/24*(3*a*b^2*c^ 
2*d*(-n^3-2*n^2+n+2)+a^3*d^3*(-n^3+6*n^2-11*n+6)+3*a^2*b*c*d^2*(n^3-2*n^2- 
n+2)+b^3*c^3*(n^3+6*n^2+11*n+6))*(b*x+a)^(1+n)*(d*x+c)^(1-n)*hypergeom([1, 
 2],[2+n],-d*(b*x+a)/(-a*d+b*c))/b^3/d^3/(-a*d+b*c)/(1+n)
 

Mathematica [A] (warning: unable to verify)

Time = 0.28 (sec) , antiderivative size = 262, normalized size of antiderivative = 0.85 \[ \int x^3 (a+b x)^n (c+d x)^{-n} \, dx=\frac {(a+b x)^{1+n} (c+d x)^{-n} \left (b^3 d^2 (1+n) x^2 (c+d x)-(b c-a d)^2 (-a d (-3+n)+b c (3+n)) \left (\frac {b (c+d x)}{b c-a d}\right )^n \operatorname {Hypergeometric2F1}\left (-2+n,1+n,2+n,\frac {d (a+b x)}{-b c+a d}\right )+2 b c (b c-a d) (-a d (-2+n)+b c (3+n)) \left (\frac {b (c+d x)}{b c-a d}\right )^n \operatorname {Hypergeometric2F1}\left (-1+n,1+n,2+n,\frac {d (a+b x)}{-b c+a d}\right )-b^2 c^2 (-a d (-1+n)+b c (3+n)) \left (\frac {b (c+d x)}{b c-a d}\right )^n \operatorname {Hypergeometric2F1}\left (n,1+n,2+n,\frac {d (a+b x)}{-b c+a d}\right )\right )}{4 b^4 d^3 (1+n)} \] Input:

Integrate[(x^3*(a + b*x)^n)/(c + d*x)^n,x]
 

Output:

((a + b*x)^(1 + n)*(b^3*d^2*(1 + n)*x^2*(c + d*x) - (b*c - a*d)^2*(-(a*d*( 
-3 + n)) + b*c*(3 + n))*((b*(c + d*x))/(b*c - a*d))^n*Hypergeometric2F1[-2 
 + n, 1 + n, 2 + n, (d*(a + b*x))/(-(b*c) + a*d)] + 2*b*c*(b*c - a*d)*(-(a 
*d*(-2 + n)) + b*c*(3 + n))*((b*(c + d*x))/(b*c - a*d))^n*Hypergeometric2F 
1[-1 + n, 1 + n, 2 + n, (d*(a + b*x))/(-(b*c) + a*d)] - b^2*c^2*(-(a*d*(-1 
 + n)) + b*c*(3 + n))*((b*(c + d*x))/(b*c - a*d))^n*Hypergeometric2F1[n, 1 
 + n, 2 + n, (d*(a + b*x))/(-(b*c) + a*d)]))/(4*b^4*d^3*(1 + n)*(c + d*x)^ 
n)
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 306, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {111, 25, 164, 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.

\(\displaystyle \int x^3 (a+b x)^n (c+d x)^{-n} \, dx\)

\(\Big \downarrow \) 111

\(\displaystyle \frac {\int -x (a+b x)^n (c+d x)^{-n} (2 a c+(a d (3-n)+b c (n+3)) x)dx}{4 b d}+\frac {x^2 (a+b x)^{n+1} (c+d x)^{1-n}}{4 b d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {x^2 (a+b x)^{n+1} (c+d x)^{1-n}}{4 b d}-\frac {\int x (a+b x)^n (c+d x)^{-n} (2 a c+(a d (3-n)+b c (n+3)) x)dx}{4 b d}\)

\(\Big \downarrow \) 164

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

\(\Big \downarrow \) 80

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

\(\Big \downarrow \) 79

\(\displaystyle \frac {x^2 (a+b x)^{n+1} (c+d x)^{1-n}}{4 b d}-\frac {\frac {(a+b x)^{n+1} (c+d x)^{-n} \left (a^3 d^3 \left (-n^3+6 n^2-11 n+6\right )+3 a^2 b c d^2 \left (n^3-2 n^2-n+2\right )+3 a b^2 c^2 d \left (-n^3-2 n^2+n+2\right )+b^3 c^3 \left (n^3+6 n^2+11 n+6\right )\right ) \left (\frac {b (c+d x)}{b c-a d}\right )^n \operatorname {Hypergeometric2F1}\left (n,n+1,n+2,-\frac {d (a+b x)}{b c-a d}\right )}{6 b^3 d^2 (n+1)}-\frac {(a+b x)^{n+1} (c+d x)^{1-n} \left (a^2 d^2 \left (n^2-5 n+6\right )+2 a b c d \left (3-n^2\right )-2 b d x (a d (3-n)+b c (n+3))+b^2 c^2 \left (n^2+5 n+6\right )\right )}{6 b^2 d^2}}{4 b d}\)

Input:

Int[(x^3*(a + b*x)^n)/(c + d*x)^n,x]
 

Output:

(x^2*(a + b*x)^(1 + n)*(c + d*x)^(1 - n))/(4*b*d) - (-1/6*((a + b*x)^(1 + 
n)*(c + d*x)^(1 - n)*(2*a*b*c*d*(3 - n^2) + a^2*d^2*(6 - 5*n + n^2) + b^2* 
c^2*(6 + 5*n + n^2) - 2*b*d*(a*d*(3 - n) + b*c*(3 + n))*x))/(b^2*d^2) + (( 
3*a*b^2*c^2*d*(2 + n - 2*n^2 - n^3) + a^3*d^3*(6 - 11*n + 6*n^2 - n^3) + 3 
*a^2*b*c*d^2*(2 - n - 2*n^2 + n^3) + b^3*c^3*(6 + 11*n + 6*n^2 + n^3))*(a 
+ b*x)^(1 + n)*((b*(c + d*x))/(b*c - a*d))^n*Hypergeometric2F1[n, 1 + n, 2 
 + n, -((d*(a + b*x))/(b*c - a*d))])/(6*b^3*d^2*(1 + n)*(c + d*x)^n))/(4*b 
*d)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 111

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[b*(a + b*x)^(m - 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 
)/(d*f*(m + n + p + 1))), x] + Simp[1/(d*f*(m + n + p + 1))   Int[(a + b*x) 
^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m 
 - 1) + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m 
 + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] & 
& GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegerQ[m]
 

rule 164

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_ 
))*((g_.) + (h_.)*(x_)), x_] :> Simp[(-(a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - 
 b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x))*(a + b*x)^(m + 1)*(( 
c + d*x)^(n + 1)/(b^2*d^2*(m + n + 2)*(m + n + 3))), x] + Simp[(a^2*d^2*f*h 
*(n + 1)*(n + 2) + a*b*d*(n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 
3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*(m + n + 3) + 
d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3))   Int[( 
a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] 
&& NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]
 

Maple [F]

\[\int x^{3} \left (b x +a \right )^{n} \left (x d +c \right )^{-n}d x\]

Input:

int(x^3*(b*x+a)^n/((d*x+c)^n),x)
 

Output:

int(x^3*(b*x+a)^n/((d*x+c)^n),x)
 

Fricas [F]

\[ \int x^3 (a+b x)^n (c+d x)^{-n} \, dx=\int { \frac {{\left (b x + a\right )}^{n} x^{3}}{{\left (d x + c\right )}^{n}} \,d x } \] Input:

integrate(x^3*(b*x+a)^n/((d*x+c)^n),x, algorithm="fricas")
 

Output:

integral((b*x + a)^n*x^3/(d*x + c)^n, x)
 

Sympy [F(-2)]

Exception generated. \[ \int x^3 (a+b x)^n (c+d x)^{-n} \, dx=\text {Exception raised: HeuristicGCDFailed} \] Input:

integrate(x**3*(b*x+a)**n/((d*x+c)**n),x)
 

Output:

Exception raised: HeuristicGCDFailed >> no luck
 

Maxima [F]

\[ \int x^3 (a+b x)^n (c+d x)^{-n} \, dx=\int { \frac {{\left (b x + a\right )}^{n} x^{3}}{{\left (d x + c\right )}^{n}} \,d x } \] Input:

integrate(x^3*(b*x+a)^n/((d*x+c)^n),x, algorithm="maxima")
 

Output:

integrate((b*x + a)^n*x^3/(d*x + c)^n, x)
 

Giac [F]

\[ \int x^3 (a+b x)^n (c+d x)^{-n} \, dx=\int { \frac {{\left (b x + a\right )}^{n} x^{3}}{{\left (d x + c\right )}^{n}} \,d x } \] Input:

integrate(x^3*(b*x+a)^n/((d*x+c)^n),x, algorithm="giac")
 

Output:

integrate((b*x + a)^n*x^3/(d*x + c)^n, x)
 

Mupad [F(-1)]

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

int((x^3*(a + b*x)^n)/(c + d*x)^n,x)
 

Output:

int((x^3*(a + b*x)^n)/(c + d*x)^n, x)
 

Reduce [F]

\[ \int x^3 (a+b x)^n (c+d x)^{-n} \, dx=\int \frac {\left (b x +a \right )^{n} x^{3}}{\left (d x +c \right )^{n}}d x \] Input:

int(x^3*(b*x+a)^n/((d*x+c)^n),x)
 

Output:

int(((a + b*x)**n*x**3)/(c + d*x)**n,x)