\(\int \frac {(c^2-d^2 x^2)^p (A+B x+C x^2+D x^3)}{(c+d x)^2} \, dx\) [239]

Optimal result

Integrand size = 37, antiderivative size = 286 \[ \int \frac {\left (c^2-d^2 x^2\right )^p \left (A+B x+C x^2+D x^3\right )}{(c+d x)^2} \, dx=-\frac {D \left (c^2-d^2 x^2\right )^{1+p}}{2 d^4 (1+p)}-\frac {\left (c^2 C d-B c d^2+A d^3-c^3 D\right ) \left (c^2-d^2 x^2\right )^{1+p}}{2 c d^4 (1-p) (c+d x)^2}+\frac {\left (B c d^2+c^3 D (3-2 p)-c^2 C d (2-p)-A d^3 p\right ) \left (c^2-d^2 x^2\right )^{1+p}}{2 c^2 d^4 (1-p) p (c+d x)}+\frac {\left (3 c^3 D-c^2 C d (2+p)+B c d^2 (1+2 p)-A d^3 p (1+2 p)\right ) x \left (c^2-d^2 x^2\right )^p \left (1-\frac {d^2 x^2}{c^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},\frac {d^2 x^2}{c^2}\right )}{2 c^2 d^3 (1-p) p} \] Output:

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

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.

Time = 2.38 (sec) , antiderivative size = 255, normalized size of antiderivative = 0.89 \[ \int \frac {\left (c^2-d^2 x^2\right )^p \left (A+B x+C x^2+D x^3\right )}{(c+d x)^2} \, dx=\frac {(c-d x)^p \left (6 B x^2 (c+d x)^p \left (1-\frac {d^2 x^2}{c^2}\right )^{-p} \operatorname {AppellF1}\left (2,-p,2-p,3,\frac {d x}{c},-\frac {d x}{c}\right )+4 C x^3 (c+d x)^p \left (1-\frac {d^2 x^2}{c^2}\right )^{-p} \operatorname {AppellF1}\left (3,-p,2-p,4,\frac {d x}{c},-\frac {d x}{c}\right )+3 D x^4 (c+d x)^p \left (1-\frac {d^2 x^2}{c^2}\right )^{-p} \operatorname {AppellF1}\left (4,-p,2-p,5,\frac {d x}{c},-\frac {d x}{c}\right )-\frac {3\ 2^p A (c-d x)^{1-p} \left (\frac {c+d x}{c}\right )^{-p} \left (c^2-d^2 x^2\right )^p \operatorname {Hypergeometric2F1}\left (2-p,1+p,2+p,\frac {c-d x}{2 c}\right )}{d+d p}\right )}{12 c^2} \] Input:

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

Output:

((c - d*x)^p*((6*B*x^2*(c + d*x)^p*AppellF1[2, -p, 2 - p, 3, (d*x)/c, -((d 
*x)/c)])/(1 - (d^2*x^2)/c^2)^p + (4*C*x^3*(c + d*x)^p*AppellF1[3, -p, 2 - 
p, 4, (d*x)/c, -((d*x)/c)])/(1 - (d^2*x^2)/c^2)^p + (3*D*x^4*(c + d*x)^p*A 
ppellF1[4, -p, 2 - p, 5, (d*x)/c, -((d*x)/c)])/(1 - (d^2*x^2)/c^2)^p - (3* 
2^p*A*(c - d*x)^(1 - p)*(c^2 - d^2*x^2)^p*Hypergeometric2F1[2 - p, 1 + p, 
2 + p, (c - d*x)/(2*c)])/((d + d*p)*((c + d*x)/c)^p)))/(12*c^2)
 

Rubi [A] (warning: unable to verify)

Time = 1.21 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.01, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.189, Rules used = {2170, 27, 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.

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

\(\Big \downarrow \) 2170

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2170

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 671

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

\(\Big \downarrow \) 473

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

\(\Big \downarrow \) 79

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

Input:

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

Output:

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

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Giac [F(-2)]

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

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro 
unding error%%%{-1,[0,1,1,3,0,0,0,0]%%%}+%%%{1,[0,1,0,2,1,1,0,0]%%%}+%%%{- 
1,[0,1,0,
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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