\(\int (c+d x)^2 (c^2-d^2 x^2)^p (A+B x+C x^2+D x^3) \, dx\) [235]

Optimal result

Integrand size = 37, antiderivative size = 355 \[ \int (c+d x)^2 \left (c^2-d^2 x^2\right )^p \left (A+B x+C x^2+D x^3\right ) \, dx=-\frac {c \left (c^2 C d+B c d^2+A d^3+c^3 D\right ) \left (c^2-d^2 x^2\right )^{1+p}}{d^4 (1+p)}-\frac {\left (6 c^3 D+2 c^2 C d (4+p)+2 B c d^2 (5+2 p)+A d^3 (5+2 p)\right ) x \left (c^2-d^2 x^2\right )^{1+p}}{d^3 (3+2 p) (5+2 p)}-\frac {(C d+2 c D) x^3 \left (c^2-d^2 x^2\right )^{1+p}}{d (5+2 p)}+\frac {\left (2 c C d+B d^2+3 c^2 D\right ) \left (c^2-d^2 x^2\right )^{2+p}}{2 d^4 (2+p)}-\frac {D \left (c^2-d^2 x^2\right )^{3+p}}{2 d^4 (3+p)}+\frac {2 c^2 \left (3 c^3 D+c^2 C d (4+p)+B c d^2 (5+2 p)+A d^3 \left (10+9 p+2 p^2\right )\right ) x \left (c^2-d^2 x^2\right )^p \left (1-\frac {d^2 x^2}{c^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},\frac {d^2 x^2}{c^2}\right )}{d^3 (3+2 p) (5+2 p)} \] Output:

-c*(A*d^3+B*c*d^2+C*c^2*d+D*c^3)*(-d^2*x^2+c^2)^(p+1)/d^4/(p+1)-(6*D*c^3+2 
*c^2*C*d*(4+p)+2*B*c*d^2*(5+2*p)+A*d^3*(5+2*p))*x*(-d^2*x^2+c^2)^(p+1)/d^3 
/(3+2*p)/(5+2*p)-(C*d+2*D*c)*x^3*(-d^2*x^2+c^2)^(p+1)/d/(5+2*p)+1/2*(B*d^2 
+2*C*c*d+3*D*c^2)*(-d^2*x^2+c^2)^(2+p)/d^4/(2+p)-1/2*D*(-d^2*x^2+c^2)^(3+p 
)/d^4/(3+p)+2*c^2*(3*D*c^3+c^2*C*d*(4+p)+B*c*d^2*(5+2*p)+A*d^3*(2*p^2+9*p+ 
10))*x*(-d^2*x^2+c^2)^p*hypergeom([1/2, -p],[3/2],d^2*x^2/c^2)/d^3/(3+2*p) 
/(5+2*p)/((1-d^2*x^2/c^2)^p)
 

Mathematica [A] (verified)

Time = 1.77 (sec) , antiderivative size = 338, normalized size of antiderivative = 0.95 \[ \int (c+d x)^2 \left (c^2-d^2 x^2\right )^p \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {1}{30} \left (c^2-d^2 x^2\right )^p \left (-\frac {15 c (B c+2 A d) \left (c^2-d^2 x^2\right )}{d^2 (1+p)}-\frac {15 c^4 D \left (c^2-d^2 x^2\right )}{d^4 (1+p)}+\frac {15 D \left (-c^2+d^2 x^2\right )^3}{d^4 (3+p)}+\frac {30 D \left (c^3-c d^2 x^2\right )^2}{d^4 (2+p)}-\frac {15 \left (2 c C d+B d^2+c^2 D\right ) \left (c^2-d^2 x^2\right ) \left (c^2+d^2 (1+p) x^2\right )}{d^4 (1+p) (2+p)}+30 A c^2 x \left (1-\frac {d^2 x^2}{c^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},\frac {d^2 x^2}{c^2}\right )+10 \left (c^2 C+2 B c d+A d^2\right ) x^3 \left (1-\frac {d^2 x^2}{c^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},\frac {d^2 x^2}{c^2}\right )+6 d (C d+2 c D) x^5 \left (1-\frac {d^2 x^2}{c^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},\frac {d^2 x^2}{c^2}\right )\right ) \] Input:

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

Output:

((c^2 - d^2*x^2)^p*((-15*c*(B*c + 2*A*d)*(c^2 - d^2*x^2))/(d^2*(1 + p)) - 
(15*c^4*D*(c^2 - d^2*x^2))/(d^4*(1 + p)) + (15*D*(-c^2 + d^2*x^2)^3)/(d^4* 
(3 + p)) + (30*D*(c^3 - c*d^2*x^2)^2)/(d^4*(2 + p)) - (15*(2*c*C*d + B*d^2 
 + c^2*D)*(c^2 - d^2*x^2)*(c^2 + d^2*(1 + p)*x^2))/(d^4*(1 + p)*(2 + p)) + 
 (30*A*c^2*x*Hypergeometric2F1[1/2, -p, 3/2, (d^2*x^2)/c^2])/(1 - (d^2*x^2 
)/c^2)^p + (10*(c^2*C + 2*B*c*d + A*d^2)*x^3*Hypergeometric2F1[3/2, -p, 5/ 
2, (d^2*x^2)/c^2])/(1 - (d^2*x^2)/c^2)^p + (6*d*(C*d + 2*c*D)*x^5*Hypergeo 
metric2F1[5/2, -p, 7/2, (d^2*x^2)/c^2])/(1 - (d^2*x^2)/c^2)^p))/30
 

Rubi [A] (verified)

Time = 2.81 (sec) , antiderivative size = 439, normalized size of antiderivative = 1.24, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.324, Rules used = {2346, 27, 2346, 25, 2346, 27, 2346, 25, 27, 455, 238, 237}

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.

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

\(\Big \downarrow \) 2346

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2346

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2346

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2346

\(\displaystyle \frac {\frac {\frac {-\frac {\int -c d^5 \left (2 c \left (p^2+5 p+6\right ) \left (3 D c^3+C d (p+4) c^2+B d^2 (2 p+5) c+A d^3 \left (2 p^2+9 p+10\right )\right )+d \left (4 p^2+16 p+15\right ) \left (D (p+5) c^3+2 C d (p+3) c^2+B d^2 (p+3)^2 c+2 A d^3 \left (p^2+5 p+6\right )\right ) x\right ) \left (c^2-d^2 x^2\right )^pdx}{d^2 (2 p+3)}-\frac {d^3 \left (p^2+5 p+6\right ) x \left (c^2-d^2 x^2\right )^{p+1} \left (A d^3 (2 p+5)+2 B c d^2 (2 p+5)+6 c^3 D+2 c^2 C d (p+4)\right )}{2 p+3}}{d^2 (p+2)}-\frac {d^2 (2 p+5) x^2 \left (c^2-d^2 x^2\right )^{p+1} \left (B d^2 (p+3)+c^2 D (p+5)+2 c C d (p+3)\right )}{2 (p+2)}}{d^2 (2 p+5)}-\frac {d (p+3) x^3 (2 c D+C d) \left (c^2-d^2 x^2\right )^{p+1}}{2 p+5}}{d^2 (p+3)}-\frac {D x^4 \left (c^2-d^2 x^2\right )^{p+1}}{2 (p+3)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\frac {\frac {\int c d^5 \left (2 c \left (p^2+5 p+6\right ) \left (3 D c^3+C d (p+4) c^2+B d^2 (2 p+5) c+A d^3 \left (2 p^2+9 p+10\right )\right )+d \left (4 p^2+16 p+15\right ) \left (D (p+5) c^3+2 C d (p+3) c^2+B d^2 (p+3)^2 c+2 A d^3 \left (p^2+5 p+6\right )\right ) x\right ) \left (c^2-d^2 x^2\right )^pdx}{d^2 (2 p+3)}-\frac {d^3 \left (p^2+5 p+6\right ) x \left (c^2-d^2 x^2\right )^{p+1} \left (A d^3 (2 p+5)+2 B c d^2 (2 p+5)+6 c^3 D+2 c^2 C d (p+4)\right )}{2 p+3}}{d^2 (p+2)}-\frac {d^2 (2 p+5) x^2 \left (c^2-d^2 x^2\right )^{p+1} \left (B d^2 (p+3)+c^2 D (p+5)+2 c C d (p+3)\right )}{2 (p+2)}}{d^2 (2 p+5)}-\frac {d (p+3) x^3 (2 c D+C d) \left (c^2-d^2 x^2\right )^{p+1}}{2 p+5}}{d^2 (p+3)}-\frac {D x^4 \left (c^2-d^2 x^2\right )^{p+1}}{2 (p+3)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {\frac {c d^3 \int \left (2 c \left (p^2+5 p+6\right ) \left (3 D c^3+C d (p+4) c^2+B d^2 (2 p+5) c+A d^3 \left (2 p^2+9 p+10\right )\right )+d \left (4 p^2+16 p+15\right ) \left (D (p+5) c^3+2 C d (p+3) c^2+B d^2 (p+3)^2 c+2 A d^3 \left (p^2+5 p+6\right )\right ) x\right ) \left (c^2-d^2 x^2\right )^pdx}{2 p+3}-\frac {d^3 \left (p^2+5 p+6\right ) x \left (c^2-d^2 x^2\right )^{p+1} \left (A d^3 (2 p+5)+2 B c d^2 (2 p+5)+6 c^3 D+2 c^2 C d (p+4)\right )}{2 p+3}}{d^2 (p+2)}-\frac {d^2 (2 p+5) x^2 \left (c^2-d^2 x^2\right )^{p+1} \left (B d^2 (p+3)+c^2 D (p+5)+2 c C d (p+3)\right )}{2 (p+2)}}{d^2 (2 p+5)}-\frac {d (p+3) x^3 (2 c D+C d) \left (c^2-d^2 x^2\right )^{p+1}}{2 p+5}}{d^2 (p+3)}-\frac {D x^4 \left (c^2-d^2 x^2\right )^{p+1}}{2 (p+3)}\)

\(\Big \downarrow \) 455

\(\displaystyle \frac {\frac {\frac {\frac {c d^3 \left (2 c \left (p^2+5 p+6\right ) \left (A d^3 \left (2 p^2+9 p+10\right )+B c d^2 (2 p+5)+3 c^3 D+c^2 C d (p+4)\right ) \int \left (c^2-d^2 x^2\right )^pdx-\frac {\left (4 p^2+16 p+15\right ) \left (c^2-d^2 x^2\right )^{p+1} \left (2 A d^3 \left (p^2+5 p+6\right )+B c d^2 (p+3)^2+c^3 D (p+5)+2 c^2 C d (p+3)\right )}{2 d (p+1)}\right )}{2 p+3}-\frac {d^3 \left (p^2+5 p+6\right ) x \left (c^2-d^2 x^2\right )^{p+1} \left (A d^3 (2 p+5)+2 B c d^2 (2 p+5)+6 c^3 D+2 c^2 C d (p+4)\right )}{2 p+3}}{d^2 (p+2)}-\frac {d^2 (2 p+5) x^2 \left (c^2-d^2 x^2\right )^{p+1} \left (B d^2 (p+3)+c^2 D (p+5)+2 c C d (p+3)\right )}{2 (p+2)}}{d^2 (2 p+5)}-\frac {d (p+3) x^3 (2 c D+C d) \left (c^2-d^2 x^2\right )^{p+1}}{2 p+5}}{d^2 (p+3)}-\frac {D x^4 \left (c^2-d^2 x^2\right )^{p+1}}{2 (p+3)}\)

\(\Big \downarrow \) 238

\(\displaystyle \frac {\frac {\frac {\frac {c d^3 \left (2 c \left (p^2+5 p+6\right ) \left (c^2-d^2 x^2\right )^p \left (1-\frac {d^2 x^2}{c^2}\right )^{-p} \left (A d^3 \left (2 p^2+9 p+10\right )+B c d^2 (2 p+5)+3 c^3 D+c^2 C d (p+4)\right ) \int \left (1-\frac {d^2 x^2}{c^2}\right )^pdx-\frac {\left (4 p^2+16 p+15\right ) \left (c^2-d^2 x^2\right )^{p+1} \left (2 A d^3 \left (p^2+5 p+6\right )+B c d^2 (p+3)^2+c^3 D (p+5)+2 c^2 C d (p+3)\right )}{2 d (p+1)}\right )}{2 p+3}-\frac {d^3 \left (p^2+5 p+6\right ) x \left (c^2-d^2 x^2\right )^{p+1} \left (A d^3 (2 p+5)+2 B c d^2 (2 p+5)+6 c^3 D+2 c^2 C d (p+4)\right )}{2 p+3}}{d^2 (p+2)}-\frac {d^2 (2 p+5) x^2 \left (c^2-d^2 x^2\right )^{p+1} \left (B d^2 (p+3)+c^2 D (p+5)+2 c C d (p+3)\right )}{2 (p+2)}}{d^2 (2 p+5)}-\frac {d (p+3) x^3 (2 c D+C d) \left (c^2-d^2 x^2\right )^{p+1}}{2 p+5}}{d^2 (p+3)}-\frac {D x^4 \left (c^2-d^2 x^2\right )^{p+1}}{2 (p+3)}\)

\(\Big \downarrow \) 237

\(\displaystyle \frac {\frac {\frac {\frac {c d^3 \left (2 c \left (p^2+5 p+6\right ) x \left (c^2-d^2 x^2\right )^p \left (1-\frac {d^2 x^2}{c^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},\frac {d^2 x^2}{c^2}\right ) \left (A d^3 \left (2 p^2+9 p+10\right )+B c d^2 (2 p+5)+3 c^3 D+c^2 C d (p+4)\right )-\frac {\left (4 p^2+16 p+15\right ) \left (c^2-d^2 x^2\right )^{p+1} \left (2 A d^3 \left (p^2+5 p+6\right )+B c d^2 (p+3)^2+c^3 D (p+5)+2 c^2 C d (p+3)\right )}{2 d (p+1)}\right )}{2 p+3}-\frac {d^3 \left (p^2+5 p+6\right ) x \left (c^2-d^2 x^2\right )^{p+1} \left (A d^3 (2 p+5)+2 B c d^2 (2 p+5)+6 c^3 D+2 c^2 C d (p+4)\right )}{2 p+3}}{d^2 (p+2)}-\frac {d^2 (2 p+5) x^2 \left (c^2-d^2 x^2\right )^{p+1} \left (B d^2 (p+3)+c^2 D (p+5)+2 c C d (p+3)\right )}{2 (p+2)}}{d^2 (2 p+5)}-\frac {d (p+3) x^3 (2 c D+C d) \left (c^2-d^2 x^2\right )^{p+1}}{2 p+5}}{d^2 (p+3)}-\frac {D x^4 \left (c^2-d^2 x^2\right )^{p+1}}{2 (p+3)}\)

Input:

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

Output:

-1/2*(D*x^4*(c^2 - d^2*x^2)^(1 + p))/(3 + p) + (-((d*(C*d + 2*c*D)*(3 + p) 
*x^3*(c^2 - d^2*x^2)^(1 + p))/(5 + 2*p)) + (-1/2*(d^2*(5 + 2*p)*(2*c*C*d*( 
3 + p) + B*d^2*(3 + p) + c^2*D*(5 + p))*x^2*(c^2 - d^2*x^2)^(1 + p))/(2 + 
p) + (-((d^3*(6 + 5*p + p^2)*(6*c^3*D + 2*c^2*C*d*(4 + p) + 2*B*c*d^2*(5 + 
 2*p) + A*d^3*(5 + 2*p))*x*(c^2 - d^2*x^2)^(1 + p))/(3 + 2*p)) + (c*d^3*(- 
1/2*((15 + 16*p + 4*p^2)*(2*c^2*C*d*(3 + p) + B*c*d^2*(3 + p)^2 + c^3*D*(5 
 + p) + 2*A*d^3*(6 + 5*p + p^2))*(c^2 - d^2*x^2)^(1 + p))/(d*(1 + p)) + (2 
*c*(6 + 5*p + p^2)*(3*c^3*D + c^2*C*d*(4 + p) + B*c*d^2*(5 + 2*p) + A*d^3* 
(10 + 9*p + 2*p^2))*x*(c^2 - d^2*x^2)^p*Hypergeometric2F1[1/2, -p, 3/2, (d 
^2*x^2)/c^2])/(1 - (d^2*x^2)/c^2)^p))/(3 + 2*p))/(d^2*(2 + p)))/(d^2*(5 + 
2*p)))/(d^2*(3 + 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 237

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[- 
p, 1/2, 1/2 + 1, (-b)*(x^2/a)], x] /; FreeQ[{a, b, p}, x] &&  !IntegerQ[2*p 
] && GtQ[a, 0]
 

rule 238

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^2) 
^FracPart[p]/(1 + b*(x^2/a))^FracPart[p])   Int[(1 + b*(x^2/a))^p, x], x] / 
; FreeQ[{a, b, p}, x] &&  !IntegerQ[2*p] &&  !GtQ[a, 0]
 

rule 455

Int[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*(( 
a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Simp[c   Int[(a + b*x^2)^p, x], x] 
/; FreeQ[{a, b, c, d, p}, x] &&  !LeQ[p, -1]
 

rule 2346

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], 
e = Coeff[Pq, x, Expon[Pq, x]]}, Simp[e*x^(q - 1)*((a + b*x^2)^(p + 1)/(b*( 
q + 2*p + 1))), x] + Simp[1/(b*(q + 2*p + 1))   Int[(a + b*x^2)^p*ExpandToS 
um[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], 
x], x]] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] &&  !LeQ[p, -1]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((D*d^2*x^5 + (2*D*c*d + C*d^2)*x^4 + (D*c^2 + 2*C*c*d + B*d^2)*x^ 
3 + A*c^2 + (C*c^2 + 2*B*c*d + A*d^2)*x^2 + (B*c^2 + 2*A*c*d)*x)*(-d^2*x^2 
 + c^2)^p, x)
 

Sympy [A] (verification not implemented)

Time = 5.51 (sec) , antiderivative size = 2343, normalized size of antiderivative = 6.60 \[ \int (c+d x)^2 \left (c^2-d^2 x^2\right )^p \left (A+B x+C x^2+D x^3\right ) \, dx=\text {Too large to display} \] Input:

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

Output:

A*c**2*c**(2*p)*x*hyper((1/2, -p), (3/2,), d**2*x**2*exp_polar(2*I*pi)/c** 
2) + 2*A*c*d*Piecewise((x**2*(c**2)**p/2, Eq(d**2, 0)), (-Piecewise(((c**2 
 - d**2*x**2)**(p + 1)/(p + 1), Ne(p, -1)), (log(c**2 - d**2*x**2), True)) 
/(2*d**2), True)) + A*c**(2*p)*d**2*x**3*hyper((3/2, -p), (5/2,), d**2*x** 
2*exp_polar(2*I*pi)/c**2)/3 + B*c**2*Piecewise((x**2*(c**2)**p/2, Eq(d**2, 
 0)), (-Piecewise(((c**2 - d**2*x**2)**(p + 1)/(p + 1), Ne(p, -1)), (log(c 
**2 - d**2*x**2), True))/(2*d**2), True)) + 2*B*c*c**(2*p)*d*x**3*hyper((3 
/2, -p), (5/2,), d**2*x**2*exp_polar(2*I*pi)/c**2)/3 + B*d**2*Piecewise((x 
**4*(c**2)**p/4, Eq(d, 0)), (-c**2*log(-c/d + x)/(-2*c**2*d**4 + 2*d**6*x* 
*2) - c**2*log(c/d + x)/(-2*c**2*d**4 + 2*d**6*x**2) - c**2/(-2*c**2*d**4 
+ 2*d**6*x**2) + d**2*x**2*log(-c/d + x)/(-2*c**2*d**4 + 2*d**6*x**2) + d* 
*2*x**2*log(c/d + x)/(-2*c**2*d**4 + 2*d**6*x**2), Eq(p, -2)), (-c**2*log( 
-c/d + x)/(2*d**4) - c**2*log(c/d + x)/(2*d**4) - x**2/(2*d**2), Eq(p, -1) 
), (-c**4*(c**2 - d**2*x**2)**p/(2*d**4*p**2 + 6*d**4*p + 4*d**4) - c**2*d 
**2*p*x**2*(c**2 - d**2*x**2)**p/(2*d**4*p**2 + 6*d**4*p + 4*d**4) + d**4* 
p*x**4*(c**2 - d**2*x**2)**p/(2*d**4*p**2 + 6*d**4*p + 4*d**4) + d**4*x**4 
*(c**2 - d**2*x**2)**p/(2*d**4*p**2 + 6*d**4*p + 4*d**4), True)) + C*c**2* 
c**(2*p)*x**3*hyper((3/2, -p), (5/2,), d**2*x**2*exp_polar(2*I*pi)/c**2)/3 
 + 2*C*c*d*Piecewise((x**4*(c**2)**p/4, Eq(d, 0)), (-c**2*log(-c/d + x)/(- 
2*c**2*d**4 + 2*d**6*x**2) - c**2*log(c/d + x)/(-2*c**2*d**4 + 2*d**6*x...
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

( - 16*(c**2 - d**2*x**2)**p*a*c**3*d**2*p**5 - 152*(c**2 - d**2*x**2)**p* 
a*c**3*d**2*p**4 - 548*(c**2 - d**2*x**2)**p*a*c**3*d**2*p**3 - 922*(c**2 
- d**2*x**2)**p*a*c**3*d**2*p**2 - 702*(c**2 - d**2*x**2)**p*a*c**3*d**2*p 
 - 180*(c**2 - d**2*x**2)**p*a*c**3*d**2 + 12*(c**2 - d**2*x**2)**p*a*c**2 
*d**3*p**4*x + 102*(c**2 - d**2*x**2)**p*a*c**2*d**3*p**3*x + 312*(c**2 - 
d**2*x**2)**p*a*c**2*d**3*p**2*x + 402*(c**2 - d**2*x**2)**p*a*c**2*d**3*p 
*x + 180*(c**2 - d**2*x**2)**p*a*c**2*d**3*x + 16*(c**2 - d**2*x**2)**p*a* 
c*d**4*p**5*x**2 + 152*(c**2 - d**2*x**2)**p*a*c*d**4*p**4*x**2 + 548*(c** 
2 - d**2*x**2)**p*a*c*d**4*p**3*x**2 + 922*(c**2 - d**2*x**2)**p*a*c*d**4* 
p**2*x**2 + 702*(c**2 - d**2*x**2)**p*a*c*d**4*p*x**2 + 180*(c**2 - d**2*x 
**2)**p*a*c*d**4*x**2 + 8*(c**2 - d**2*x**2)**p*a*d**5*p**5*x**3 + 72*(c** 
2 - d**2*x**2)**p*a*d**5*p**4*x**3 + 242*(c**2 - d**2*x**2)**p*a*d**5*p**3 
*x**3 + 372*(c**2 - d**2*x**2)**p*a*d**5*p**2*x**3 + 254*(c**2 - d**2*x**2 
)**p*a*d**5*p*x**3 + 60*(c**2 - d**2*x**2)**p*a*d**5*x**3 - 8*(c**2 - d**2 
*x**2)**p*b*c**4*d*p**5 - 84*(c**2 - d**2*x**2)**p*b*c**4*d*p**4 - 334*(c* 
*2 - d**2*x**2)**p*b*c**4*d*p**3 - 615*(c**2 - d**2*x**2)**p*b*c**4*d*p**2 
 - 504*(c**2 - d**2*x**2)**p*b*c**4*d*p - 135*(c**2 - d**2*x**2)**p*b*c**4 
*d - 16*(c**2 - d**2*x**2)**p*b*c**3*d**2*p**5*x - 136*(c**2 - d**2*x**2)* 
*p*b*c**3*d**2*p**4*x - 416*(c**2 - d**2*x**2)**p*b*c**3*d**2*p**3*x - 536 
*(c**2 - d**2*x**2)**p*b*c**3*d**2*p**2*x - 240*(c**2 - d**2*x**2)**p*b...