\(\int \frac {(c^2-d^2 x^2)^p (A+B x+C x^2+D x^3)}{(c+d x)^3} \, dx\) [240]

Optimal result

Integrand size = 37, antiderivative size = 329 \[ \int \frac {\left (c^2-d^2 x^2\right )^p \left (A+B x+C x^2+D x^3\right )}{(c+d x)^3} \, dx=-\frac {\left (c^2 C d-B c d^2+A d^3-c^3 D\right ) \left (c^2-d^2 x^2\right )^{1+p}}{2 c d^4 (2-p) (c+d x)^3}-\frac {\left (3 B c d^2+c^3 D (11-4 p)+A d^3 (1-2 p)-c^2 C d (7-2 p)\right ) \left (c^2-d^2 x^2\right )^{1+p}}{4 c^2 d^4 (1-p) (2-p) (c+d x)^2}-\frac {D \left (c^2-d^2 x^2\right )^{1+p}}{d^4 (1+2 p) (c+d x)}-\frac {2^{-1+p} \left (3 B c d^2 p (1+2 p)+3 c^3 D (4+3 p)+A d^3 p \left (1-4 p^2\right )-c^2 C d \left (4+9 p+2 p^2\right )\right ) \left (\frac {c-d x}{c}\right )^{-p} \left (c^2-d^2 x^2\right )^p \operatorname {Hypergeometric2F1}\left (-p,p,1+p,\frac {c+d x}{2 c}\right )}{c^2 d^4 (1-p) (2-p) p (1+2 p)} \] Output:

-1/2*(A*d^3-B*c*d^2+C*c^2*d-D*c^3)*(-d^2*x^2+c^2)^(p+1)/c/d^4/(2-p)/(d*x+c 
)^3-1/4*(3*B*c*d^2+c^3*D*(11-4*p)+A*d^3*(1-2*p)-c^2*C*d*(7-2*p))*(-d^2*x^2 
+c^2)^(p+1)/c^2/d^4/(1-p)/(2-p)/(d*x+c)^2-D*(-d^2*x^2+c^2)^(p+1)/d^4/(1+2* 
p)/(d*x+c)-2^(-1+p)*(3*B*c*d^2*p*(1+2*p)+3*c^3*D*(4+3*p)+A*d^3*p*(-4*p^2+1 
)-c^2*C*d*(2*p^2+9*p+4))*(-d^2*x^2+c^2)^p*hypergeom([p, -p],[p+1],1/2*(d*x 
+c)/c)/c^2/d^4/(1-p)/(2-p)/p/(1+2*p)/(((-d*x+c)/c)^p)
 

Mathematica [A] (warning: unable to verify)

Time = 2.88 (sec) , antiderivative size = 265, normalized size of antiderivative = 0.81 \[ \int \frac {\left (c^2-d^2 x^2\right )^p \left (A+B x+C x^2+D x^3\right )}{(c+d x)^3} \, dx=\frac {2^{-3+p} \left (1+\frac {d x}{c}\right )^{-p} \left (c^2-d^2 x^2\right )^p \left (1-\frac {d^2 x^2}{c^2}\right )^{-p} \left (8 c^3 d D (1+p) x \left (\frac {1}{2}+\frac {d x}{2 c}\right )^p \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},\frac {d^2 x^2}{c^2}\right )+(c-d x) \left (1-\frac {d^2 x^2}{c^2}\right )^p \left (4 c^2 (-C d+3 c D) \operatorname {Hypergeometric2F1}\left (1-p,1+p,2+p,\frac {c-d x}{2 c}\right )-2 c \left (-2 c C d+B d^2+3 c^2 D\right ) \operatorname {Hypergeometric2F1}\left (2-p,1+p,2+p,\frac {c-d x}{2 c}\right )+\left (-c^2 C d+B c d^2-A d^3+c^3 D\right ) \operatorname {Hypergeometric2F1}\left (3-p,1+p,2+p,\frac {c-d x}{2 c}\right )\right )\right )}{c^3 d^4 (1+p)} \] Input:

Integrate[((c^2 - d^2*x^2)^p*(A + B*x + C*x^2 + D*x^3))/(c + d*x)^3,x]
 

Output:

(2^(-3 + p)*(c^2 - d^2*x^2)^p*(8*c^3*d*D*(1 + p)*x*(1/2 + (d*x)/(2*c))^p*H 
ypergeometric2F1[1/2, -p, 3/2, (d^2*x^2)/c^2] + (c - d*x)*(1 - (d^2*x^2)/c 
^2)^p*(4*c^2*(-(C*d) + 3*c*D)*Hypergeometric2F1[1 - p, 1 + p, 2 + p, (c - 
d*x)/(2*c)] - 2*c*(-2*c*C*d + B*d^2 + 3*c^2*D)*Hypergeometric2F1[2 - p, 1 
+ p, 2 + p, (c - d*x)/(2*c)] + (-(c^2*C*d) + B*c*d^2 - A*d^3 + c^3*D)*Hype 
rgeometric2F1[3 - p, 1 + p, 2 + p, (c - d*x)/(2*c)])))/(c^3*d^4*(1 + p)*(1 
 + (d*x)/c)^p*(1 - (d^2*x^2)/c^2)^p)
 

Rubi [A] (warning: unable to verify)

Time = 1.35 (sec) , antiderivative size = 314, normalized size of antiderivative = 0.95, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.162, Rules used = {2170, 2170, 27, 671, 473, 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.

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

\(\Big \downarrow \) 2170

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

\(\Big \downarrow \) 2170

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 671

\(\displaystyle -\frac {-\frac {d^2 \left (\frac {\left (A d^3 p \left (1-4 p^2\right )+3 B c d^2 p (2 p+1)+3 c^3 D (3 p+4)-c^2 C d \left (2 p^2+9 p+4\right )\right ) \int \frac {\left (c^2-d^2 x^2\right )^p}{(c+d x)^2}dx}{2 c (2-p)}-\frac {p (2 p+1) \left (c^2-d^2 x^2\right )^{p+1} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{2 c d (2-p) (c+d x)^3}\right )}{p}-\frac {d \left (c^2-d^2 x^2\right )^{p+1} (c D (4 p+3)-C (2 d p+d))}{2 p (c+d x)^2}}{d^5 (2 p+1)}-\frac {D \left (c^2-d^2 x^2\right )^{p+1}}{d^4 (2 p+1) (c+d x)}\)

\(\Big \downarrow \) 473

\(\displaystyle -\frac {-\frac {d^2 \left (\frac {(c-d x)^{-p-1} \left (\frac {d x}{c}+1\right )^{-p-1} \left (c^2-d^2 x^2\right )^{p+1} \left (A d^3 p \left (1-4 p^2\right )+3 B c d^2 p (2 p+1)+3 c^3 D (3 p+4)-c^2 C d \left (2 p^2+9 p+4\right )\right ) \int (c-d x)^p \left (\frac {d x}{c}+1\right )^{p-2}dx}{2 c^4 (2-p)}-\frac {p (2 p+1) \left (c^2-d^2 x^2\right )^{p+1} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{2 c d (2-p) (c+d x)^3}\right )}{p}-\frac {d \left (c^2-d^2 x^2\right )^{p+1} (c D (4 p+3)-C (2 d p+d))}{2 p (c+d x)^2}}{d^5 (2 p+1)}-\frac {D \left (c^2-d^2 x^2\right )^{p+1}}{d^4 (2 p+1) (c+d x)}\)

\(\Big \downarrow \) 79

\(\displaystyle -\frac {-\frac {d^2 \left (-\frac {p (2 p+1) \left (c^2-d^2 x^2\right )^{p+1} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{2 c d (2-p) (c+d x)^3}-\frac {2^{p-3} \left (c^2-d^2 x^2\right )^{p+1} \left (\frac {d x}{c}+1\right )^{-p-1} \operatorname {Hypergeometric2F1}\left (2-p,p+1,p+2,\frac {c-d x}{2 c}\right ) \left (A d^3 p \left (1-4 p^2\right )+3 B c d^2 p (2 p+1)+3 c^3 D (3 p+4)-c^2 C d \left (2 p^2+9 p+4\right )\right )}{c^4 d (2-p) (p+1)}\right )}{p}-\frac {d \left (c^2-d^2 x^2\right )^{p+1} (c D (4 p+3)-C (2 d p+d))}{2 p (c+d x)^2}}{d^5 (2 p+1)}-\frac {D \left (c^2-d^2 x^2\right )^{p+1}}{d^4 (2 p+1) (c+d x)}\)

Input:

Int[((c^2 - d^2*x^2)^p*(A + B*x + C*x^2 + D*x^3))/(c + d*x)^3,x]
 

Output:

-((D*(c^2 - d^2*x^2)^(1 + p))/(d^4*(1 + 2*p)*(c + d*x))) - (-1/2*(d*(c*D*( 
3 + 4*p) - C*(d + 2*d*p))*(c^2 - d^2*x^2)^(1 + p))/(p*(c + d*x)^2) - (d^2* 
(-1/2*((c^2*C*d - B*c*d^2 + A*d^3 - c^3*D)*p*(1 + 2*p)*(c^2 - d^2*x^2)^(1 
+ p))/(c*d*(2 - p)*(c + d*x)^3) - (2^(-3 + p)*(3*B*c*d^2*p*(1 + 2*p) + 3*c 
^3*D*(4 + 3*p) + A*d^3*p*(1 - 4*p^2) - c^2*C*d*(4 + 9*p + 2*p^2))*(1 + (d* 
x)/c)^(-1 - p)*(c^2 - d^2*x^2)^(1 + p)*Hypergeometric2F1[2 - p, 1 + p, 2 + 
 p, (c - d*x)/(2*c)])/(c^4*d*(2 - p)*(1 + p))))/p)/(d^5*(1 + 2*p))
 

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 473

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
c^(n - 1)*((a + b*x^2)^(p + 1)/((1 + d*(x/c))^(p + 1)*(a/c + (b*x)/d)^(p + 
1)))   Int[(1 + d*(x/c))^(n + p)*(a/c + (b/d)*x)^p, x], x] /; FreeQ[{a, b, 
c, d, n}, x] && EqQ[b*c^2 + a*d^2, 0] && (IntegerQ[n] || GtQ[c, 0]) &&  !Gt 
Q[a, 0] &&  !(IntegerQ[n] && (IntegerQ[3*p] || IntegerQ[4*p]))
 

rule 671

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_ 
), x_Symbol] :> Simp[(d*g - e*f)*(d + e*x)^m*((a + c*x^2)^(p + 1)/(2*c*d*(m 
 + p + 1))), x] + Simp[(m*(g*c*d + c*e*f) + 2*e*c*f*(p + 1))/(e*(2*c*d)*(m 
+ p + 1))   Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, 
e, f, g, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && ((LtQ[m, -1] &&  !IGtQ[m + p 
 + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p 
 + 1, 0]
 

rule 2170

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : 
> With[{q = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[f*(d + e*x) 
^(m + q - 1)*((a + b*x^2)^(p + 1)/(b*e^(q - 1)*(m + q + 2*p + 1))), x] + Si 
mp[1/(b*e^q*(m + q + 2*p + 1))   Int[(d + e*x)^m*(a + b*x^2)^p*ExpandToSum[ 
b*e^q*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*(d + e*x)^q - 2*e*f*(m + 
 p + q)*(d + e*x)^(q - 2)*(a*e - b*d*x), x], x], x] /; NeQ[m + q + 2*p + 1, 
 0]] /; FreeQ[{a, b, d, e, m, p}, x] && PolyQ[Pq, x] && EqQ[b*d^2 + a*e^2, 
0] &&  !IGtQ[m, 0]
 

Maple [F]

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

Input:

int((-d^2*x^2+c^2)^p*(D*x^3+C*x^2+B*x+A)/(d*x+c)^3,x)
 

Output:

int((-d^2*x^2+c^2)^p*(D*x^3+C*x^2+B*x+A)/(d*x+c)^3,x)
 

Fricas [F]

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

integrate((-d^2*x^2+c^2)^p*(D*x^3+C*x^2+B*x+A)/(d*x+c)^3,x, algorithm="fri 
cas")
 

Output:

integral((D*x^3 + C*x^2 + B*x + A)*(-d^2*x^2 + c^2)^p/(d^3*x^3 + 3*c*d^2*x 
^2 + 3*c^2*d*x + c^3), x)
 

Sympy [F]

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

integrate((-d**2*x**2+c**2)**p*(D*x**3+C*x**2+B*x+A)/(d*x+c)**3,x)
 

Output:

Integral((-(-c + d*x)*(c + d*x))**p*(A + B*x + C*x**2 + D*x**3)/(c + d*x)* 
*3, x)
 

Maxima [F]

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

integrate((-d^2*x^2+c^2)^p*(D*x^3+C*x^2+B*x+A)/(d*x+c)^3,x, algorithm="max 
ima")
 

Output:

integrate((D*x^3 + C*x^2 + B*x + A)*(-d^2*x^2 + c^2)^p/(d*x + c)^3, x)
 

Giac [F]

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

integrate((-d^2*x^2+c^2)^p*(D*x^3+C*x^2+B*x+A)/(d*x+c)^3,x, algorithm="gia 
c")
 

Output:

integrate((D*x^3 + C*x^2 + B*x + A)*(-d^2*x^2 + c^2)^p/(d*x + c)^3, x)
 

Mupad [F(-1)]

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

int(((c^2 - d^2*x^2)^p*(A + B*x + C*x^2 + x^3*D))/(c + d*x)^3,x)
 

Output:

int(((c^2 - d^2*x^2)^p*(A + B*x + C*x^2 + x^3*D))/(c + d*x)^3, x)
 

Reduce [F]

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

int((-d^2*x^2+c^2)^p*(D*x^3+C*x^2+B*x+A)/(d*x+c)^3,x)
 

Output:

( - 4*(c**2 - d**2*x**2)**p*a*d**2*p**3 + (c**2 - d**2*x**2)**p*a*d**2*p + 
 2*(c**2 - d**2*x**2)**p*b*c*d*p**2 + (c**2 - d**2*x**2)**p*b*c*d*p + 4*(c 
**2 - d**2*x**2)**p*b*d**2*p**2*x + 2*(c**2 - d**2*x**2)**p*b*d**2*p*x - 2 
*(c**2 - d**2*x**2)**p*c**3*p**2 + 2*(c**2 - d**2*x**2)**p*c**3*p + 2*(c** 
2 - d**2*x**2)**p*c**3 - 4*(c**2 - d**2*x**2)**p*c**2*d*p**2*x + 4*(c**2 - 
 d**2*x**2)**p*c**2*d*p*x + 4*(c**2 - d**2*x**2)**p*c**2*d*x - 4*(c**2 - d 
**2*x**2)**p*c*d**2*p*x**2 + 2*(c**2 - d**2*x**2)**p*c*d**2*x**2 + 4*(c**2 
 - d**2*x**2)**p*d**3*p**2*x**3 - 2*(c**2 - d**2*x**2)**p*d**3*p*x**3 - 32 
*int(((c**2 - d**2*x**2)**p*x)/(4*c**4*p**2 - c**4 + 8*c**3*d*p**2*x - 2*c 
**3*d*x - 8*c*d**3*p**2*x**3 + 2*c*d**3*x**3 - 4*d**4*p**2*x**4 + d**4*x** 
4),x)*a*c**2*d**4*p**6 + 16*int(((c**2 - d**2*x**2)**p*x)/(4*c**4*p**2 - c 
**4 + 8*c**3*d*p**2*x - 2*c**3*d*x - 8*c*d**3*p**2*x**3 + 2*c*d**3*x**3 - 
4*d**4*p**2*x**4 + d**4*x**4),x)*a*c**2*d**4*p**4 - 2*int(((c**2 - d**2*x* 
*2)**p*x)/(4*c**4*p**2 - c**4 + 8*c**3*d*p**2*x - 2*c**3*d*x - 8*c*d**3*p* 
*2*x**3 + 2*c*d**3*x**3 - 4*d**4*p**2*x**4 + d**4*x**4),x)*a*c**2*d**4*p** 
2 - 64*int(((c**2 - d**2*x**2)**p*x)/(4*c**4*p**2 - c**4 + 8*c**3*d*p**2*x 
 - 2*c**3*d*x - 8*c*d**3*p**2*x**3 + 2*c*d**3*x**3 - 4*d**4*p**2*x**4 + d* 
*4*x**4),x)*a*c*d**5*p**6*x + 32*int(((c**2 - d**2*x**2)**p*x)/(4*c**4*p** 
2 - c**4 + 8*c**3*d*p**2*x - 2*c**3*d*x - 8*c*d**3*p**2*x**3 + 2*c*d**3*x* 
*3 - 4*d**4*p**2*x**4 + d**4*x**4),x)*a*c*d**5*p**4*x - 4*int(((c**2 - ...