\(\int \frac {(c^2-d^2 x^2)^p (A+B x+C x^2+D x^3)}{c+d x} \, dx\) [238]

Optimal result

Integrand size = 37, antiderivative size = 246 \[ \int \frac {\left (c^2-d^2 x^2\right )^p \left (A+B x+C x^2+D x^3\right )}{c+d x} \, dx=-\frac {(C d-c D) \left (c^2-d^2 x^2\right )^{1+p}}{2 d^4 (1+p)}-\frac {D x \left (c^2-d^2 x^2\right )^{1+p}}{d^3 (3+2 p)}+\frac {\left (c^2 C d-B c d^2+A d^3-c^3 D\right ) \left (c^2-d^2 x^2\right )^{1+p}}{2 c d^4 p (c+d x)}-\frac {\left (3 c^3 D-c^2 C d (3+2 p)+B c d^2 (3+2 p)-A d^3 \left (3+8 p+4 p^2\right )\right ) x \left (c^2-d^2 x^2\right )^p \left (1-\frac {d^2 x^2}{c^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},\frac {d^2 x^2}{c^2}\right )}{2 c d^3 p (3+2 p)} \] Output:

-1/2*(C*d-D*c)*(-d^2*x^2+c^2)^(p+1)/d^4/(p+1)-D*x*(-d^2*x^2+c^2)^(p+1)/d^3 
/(3+2*p)+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/p/(d 
*x+c)-1/2*(3*D*c^3-c^2*C*d*(3+2*p)+B*c*d^2*(3+2*p)-A*d^3*(4*p^2+8*p+3))*x* 
(-d^2*x^2+c^2)^p*hypergeom([1/2, -p],[3/2],d^2*x^2/c^2)/c/d^3/p/(3+2*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 = 1.77 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.04 \[ \int \frac {\left (c^2-d^2 x^2\right )^p \left (A+B x+C x^2+D x^3\right )}{c+d x} \, dx=\frac {(c-d x)^p \left (6 B x^2 (c+d x)^p \left (1-\frac {d^2 x^2}{c^2}\right )^{-p} \operatorname {AppellF1}\left (2,-p,1-p,3,\frac {d x}{c},-\frac {d x}{c}\right )+4 C x^3 (c+d x)^p \left (1-\frac {d^2 x^2}{c^2}\right )^{-p} \operatorname {AppellF1}\left (3,-p,1-p,4,\frac {d x}{c},-\frac {d x}{c}\right )+3 D x^4 (c+d x)^p \left (1-\frac {d^2 x^2}{c^2}\right )^{-p} \operatorname {AppellF1}\left (4,-p,1-p,5,\frac {d x}{c},-\frac {d x}{c}\right )-\frac {3\ 2^{1+p} A (c-d x)^{1-p} \left (\frac {c+d x}{c}\right )^{-p} \left (c^2-d^2 x^2\right )^p \operatorname {Hypergeometric2F1}\left (1-p,1+p,2+p,\frac {c-d x}{2 c}\right )}{d+d p}\right )}{12 c} \] Input:

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 1.23 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.16, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.189, Rules used = {2170, 25, 2170, 27, 672, 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.

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

\(\Big \downarrow \) 2170

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2170

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 672

\(\displaystyle \frac {d^2 \left (-\frac {\left (-A d^3 \left (4 p^2+8 p+3\right )+B c d^2 (2 p+3)+3 c^3 D-c^2 C d (2 p+3)\right ) \int \frac {\left (c^2-d^2 x^2\right )^p}{c+d x}dx}{2 p+1}-\frac {\left (c^2-d^2 x^2\right )^{p+1} \left (B d^2 (2 p+3)+2 c^2 D (p+2)-c C d (2 p+3)\right )}{d (2 p+1) (c+d x)}\right )-\frac {d \left (c^2-d^2 x^2\right )^{p+1} (C d (2 p+3)-c D (4 p+5))}{2 (p+1)}}{d^5 (2 p+3)}-\frac {D (c+d x) \left (c^2-d^2 x^2\right )^{p+1}}{d^4 (2 p+3)}\)

\(\Big \downarrow \) 473

\(\displaystyle \frac {d^2 \left (-\frac {(c-d x)^{-p-1} \left (c^2-d^2 x^2\right )^{p+1} \left (\frac {d x}{c}+1\right )^{-p-1} \left (-A d^3 \left (4 p^2+8 p+3\right )+B c d^2 (2 p+3)+3 c^3 D-c^2 C d (2 p+3)\right ) \int (c-d x)^p \left (\frac {d x}{c}+1\right )^{p-1}dx}{c^2 (2 p+1)}-\frac {\left (c^2-d^2 x^2\right )^{p+1} \left (B d^2 (2 p+3)+2 c^2 D (p+2)-c C d (2 p+3)\right )}{d (2 p+1) (c+d x)}\right )-\frac {d \left (c^2-d^2 x^2\right )^{p+1} (C d (2 p+3)-c D (4 p+5))}{2 (p+1)}}{d^5 (2 p+3)}-\frac {D (c+d x) \left (c^2-d^2 x^2\right )^{p+1}}{d^4 (2 p+3)}\)

\(\Big \downarrow \) 79

\(\displaystyle \frac {d^2 \left (\frac {2^{p-1} \left (\frac {d x}{c}+1\right )^{-p-1} \left (c^2-d^2 x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1-p,p+1,p+2,\frac {c-d x}{2 c}\right ) \left (-A d^3 \left (4 p^2+8 p+3\right )+B c d^2 (2 p+3)+3 c^3 D-c^2 C d (2 p+3)\right )}{c^2 d (p+1) (2 p+1)}-\frac {\left (c^2-d^2 x^2\right )^{p+1} \left (B d^2 (2 p+3)+2 c^2 D (p+2)-c C d (2 p+3)\right )}{d (2 p+1) (c+d x)}\right )-\frac {d \left (c^2-d^2 x^2\right )^{p+1} (C d (2 p+3)-c D (4 p+5))}{2 (p+1)}}{d^5 (2 p+3)}-\frac {D (c+d x) \left (c^2-d^2 x^2\right )^{p+1}}{d^4 (2 p+3)}\)

Input:

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

Output:

-((D*(c + d*x)*(c^2 - d^2*x^2)^(1 + p))/(d^4*(3 + 2*p))) + (-1/2*(d*(C*d*( 
3 + 2*p) - c*D*(5 + 4*p))*(c^2 - d^2*x^2)^(1 + p))/(1 + p) + d^2*(-(((2*c^ 
2*D*(2 + p) - c*C*d*(3 + 2*p) + B*d^2*(3 + 2*p))*(c^2 - d^2*x^2)^(1 + p))/ 
(d*(1 + 2*p)*(c + d*x))) + (2^(-1 + p)*(3*c^3*D - c^2*C*d*(3 + 2*p) + B*c* 
d^2*(3 + 2*p) - A*d^3*(3 + 8*p + 4*p^2))*(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^ 
2*d*(1 + p)*(1 + 2*p))))/(d^5*(3 + 2*p))
 

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 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 672

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_ 
), x_Symbol] :> Simp[g*(d + e*x)^m*((a + c*x^2)^(p + 1)/(c*(m + 2*p + 2))), 
 x] + Simp[(m*(d*g + e*f) + 2*e*f*(p + 1))/(e*(m + 2*p + 2))   Int[(d + e*x 
)^m*(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] && NeQ[m + 2*p + 2, 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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(4*(c**2 - d**2*x**2)**p*a*d*p**2 + 8*(c**2 - d**2*x**2)**p*a*d*p + 3*(c** 
2 - d**2*x**2)**p*a*d - 4*(c**2 - d**2*x**2)**p*b*c*p**2 - 8*(c**2 - d**2* 
x**2)**p*b*c*p - 3*(c**2 - d**2*x**2)**p*b*c + 4*(c**2 - d**2*x**2)**p*b*d 
*p**2*x + 6*(c**2 - d**2*x**2)**p*b*d*p*x - 4*(c**2 - d**2*x**2)**p*c**2*p 
**2*x + 4*(c**2 - d**2*x**2)**p*d**2*p**2*x**3 + 2*(c**2 - d**2*x**2)**p*d 
**2*p*x**3 + 32*int((c**2 - d**2*x**2)**p/(4*c**2*p**2 + 8*c**2*p + 3*c**2 
 - 4*d**2*p**2*x**2 - 8*d**2*p*x**2 - 3*d**2*x**2),x)*a*c*d**2*p**5 + 128* 
int((c**2 - d**2*x**2)**p/(4*c**2*p**2 + 8*c**2*p + 3*c**2 - 4*d**2*p**2*x 
**2 - 8*d**2*p*x**2 - 3*d**2*x**2),x)*a*c*d**2*p**4 + 176*int((c**2 - d**2 
*x**2)**p/(4*c**2*p**2 + 8*c**2*p + 3*c**2 - 4*d**2*p**2*x**2 - 8*d**2*p*x 
**2 - 3*d**2*x**2),x)*a*c*d**2*p**3 + 96*int((c**2 - d**2*x**2)**p/(4*c**2 
*p**2 + 8*c**2*p + 3*c**2 - 4*d**2*p**2*x**2 - 8*d**2*p*x**2 - 3*d**2*x**2 
),x)*a*c*d**2*p**2 + 18*int((c**2 - d**2*x**2)**p/(4*c**2*p**2 + 8*c**2*p 
+ 3*c**2 - 4*d**2*p**2*x**2 - 8*d**2*p*x**2 - 3*d**2*x**2),x)*a*c*d**2*p - 
 16*int((c**2 - d**2*x**2)**p/(4*c**2*p**2 + 8*c**2*p + 3*c**2 - 4*d**2*p* 
*2*x**2 - 8*d**2*p*x**2 - 3*d**2*x**2),x)*b*c**2*d*p**4 - 56*int((c**2 - d 
**2*x**2)**p/(4*c**2*p**2 + 8*c**2*p + 3*c**2 - 4*d**2*p**2*x**2 - 8*d**2* 
p*x**2 - 3*d**2*x**2),x)*b*c**2*d*p**3 - 60*int((c**2 - d**2*x**2)**p/(4*c 
**2*p**2 + 8*c**2*p + 3*c**2 - 4*d**2*p**2*x**2 - 8*d**2*p*x**2 - 3*d**2*x 
**2),x)*b*c**2*d*p**2 - 18*int((c**2 - d**2*x**2)**p/(4*c**2*p**2 + 8*c...