\(\int (c+d x)^{5/2} (c^2-d^2 x^2)^p (A+B x+C x^2+D x^3) \, dx\) [241]

Optimal result

Integrand size = 39, antiderivative size = 372 \[ \int (c+d x)^{5/2} \left (c^2-d^2 x^2\right )^p \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {2 \left (4 c C d \left (13+17 p+4 p^2\right )-c^2 D \left (123+76 p+16 p^2\right )-B d^2 \left (143+96 p+16 p^2\right )\right ) (c+d x)^{5/2} \left (c^2-d^2 x^2\right )^{1+p}}{d^4 (9+4 p) (11+4 p) (13+4 p)}-\frac {2 (C d (13+4 p)-c D (17+8 p)) (c+d x)^{7/2} \left (c^2-d^2 x^2\right )^{1+p}}{d^4 (11+4 p) (13+4 p)}-\frac {2 D (c+d x)^{9/2} \left (c^2-d^2 x^2\right )^{1+p}}{d^4 (13+4 p)}-\frac {2^{\frac {5}{2}+p} c \left (15 c^3 D (29+8 p)+5 B c d^2 \left (143+96 p+16 p^2\right )+c^2 C d \left (559+276 p+32 p^2\right )+A d^3 \left (1287+1436 p+528 p^2+64 p^3\right )\right ) \sqrt {c+d x} \left (1+\frac {d x}{c}\right )^{-\frac {3}{2}-p} \left (c^2-d^2 x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (-\frac {5}{2}-p,1+p,2+p,\frac {c-d x}{2 c}\right )}{d^4 (1+p) (9+4 p) (11+4 p) (13+4 p)} \] Output:

2*(4*c*C*d*(4*p^2+17*p+13)-c^2*D*(16*p^2+76*p+123)-B*d^2*(16*p^2+96*p+143) 
)*(d*x+c)^(5/2)*(-d^2*x^2+c^2)^(p+1)/d^4/(9+4*p)/(11+4*p)/(13+4*p)-2*(C*d* 
(13+4*p)-c*D*(17+8*p))*(d*x+c)^(7/2)*(-d^2*x^2+c^2)^(p+1)/d^4/(11+4*p)/(13 
+4*p)-2*D*(d*x+c)^(9/2)*(-d^2*x^2+c^2)^(p+1)/d^4/(13+4*p)-2^(5/2+p)*c*(15* 
c^3*D*(29+8*p)+5*B*c*d^2*(16*p^2+96*p+143)+c^2*C*d*(32*p^2+276*p+559)+A*d^ 
3*(64*p^3+528*p^2+1436*p+1287))*(d*x+c)^(1/2)*(1+d*x/c)^(-3/2-p)*(-d^2*x^2 
+c^2)^(p+1)*hypergeom([p+1, -5/2-p],[2+p],1/2*(-d*x+c)/c)/d^4/(p+1)/(9+4*p 
)/(11+4*p)/(13+4*p)
 

Mathematica [C] (warning: unable to verify)

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

Time = 4.38 (sec) , antiderivative size = 686, normalized size of antiderivative = 1.84 \[ \int (c+d x)^{5/2} \left (c^2-d^2 x^2\right )^p \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {(c-d x)^p \sqrt {c+d x} \left (30 c (B c+2 A d) x^2 (c+d x)^p \left (1-\frac {d^2 x^2}{c^2}\right )^{-p} \operatorname {AppellF1}\left (2,-p,-\frac {1}{2}-p,3,\frac {d x}{c},-\frac {d x}{c}\right )+20 \left (c^2 C+2 B c d+A d^2\right ) x^3 (c+d x)^p \left (1-\frac {d^2 x^2}{c^2}\right )^{-p} \operatorname {AppellF1}\left (3,-p,-\frac {1}{2}-p,4,\frac {d x}{c},-\frac {d x}{c}\right )+30 c C d x^4 (c+d x)^p \left (1-\frac {d^2 x^2}{c^2}\right )^{-p} \operatorname {AppellF1}\left (4,-p,-\frac {1}{2}-p,5,\frac {d x}{c},-\frac {d x}{c}\right )+15 B d^2 x^4 (c+d x)^p \left (1-\frac {d^2 x^2}{c^2}\right )^{-p} \operatorname {AppellF1}\left (4,-p,-\frac {1}{2}-p,5,\frac {d x}{c},-\frac {d x}{c}\right )+15 c^2 D x^4 (c+d x)^p \left (1-\frac {d^2 x^2}{c^2}\right )^{-p} \operatorname {AppellF1}\left (4,-p,-\frac {1}{2}-p,5,\frac {d x}{c},-\frac {d x}{c}\right )+12 C d^2 x^5 (c+d x)^p \left (1-\frac {d^2 x^2}{c^2}\right )^{-p} \operatorname {AppellF1}\left (5,-p,-\frac {1}{2}-p,6,\frac {d x}{c},-\frac {d x}{c}\right )+24 c d D x^5 (c+d x)^p \left (1-\frac {d^2 x^2}{c^2}\right )^{-p} \operatorname {AppellF1}\left (5,-p,-\frac {1}{2}-p,6,\frac {d x}{c},-\frac {d x}{c}\right )+10 d^2 D x^6 (c+d x)^p \left (1-\frac {d^2 x^2}{c^2}\right )^{-p} \operatorname {AppellF1}\left (6,-p,-\frac {1}{2}-p,7,\frac {d x}{c},-\frac {d x}{c}\right )-\frac {15\ 2^{\frac {5}{2}+p} A c^3 (c-d x)^{-p} \left (1+\frac {d x}{c}\right )^{-p} \left (c^2-d^2 x^2\right )^p \operatorname {Hypergeometric2F1}\left (-\frac {1}{2}-p,1+p,2+p,\frac {c-d x}{2 c}\right )}{d+d p}+\frac {15\ 2^{\frac {5}{2}+p} A c^2 x (c-d x)^{-p} \left (1+\frac {d x}{c}\right )^{-p} \left (c^2-d^2 x^2\right )^p \operatorname {Hypergeometric2F1}\left (-\frac {1}{2}-p,1+p,2+p,\frac {c-d x}{2 c}\right )}{1+p}\right )}{60 \sqrt {1+\frac {d x}{c}}} \] Input:

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

Output:

((c - d*x)^p*Sqrt[c + d*x]*((30*c*(B*c + 2*A*d)*x^2*(c + d*x)^p*AppellF1[2 
, -p, -1/2 - p, 3, (d*x)/c, -((d*x)/c)])/(1 - (d^2*x^2)/c^2)^p + (20*(c^2* 
C + 2*B*c*d + A*d^2)*x^3*(c + d*x)^p*AppellF1[3, -p, -1/2 - p, 4, (d*x)/c, 
 -((d*x)/c)])/(1 - (d^2*x^2)/c^2)^p + (30*c*C*d*x^4*(c + d*x)^p*AppellF1[4 
, -p, -1/2 - p, 5, (d*x)/c, -((d*x)/c)])/(1 - (d^2*x^2)/c^2)^p + (15*B*d^2 
*x^4*(c + d*x)^p*AppellF1[4, -p, -1/2 - p, 5, (d*x)/c, -((d*x)/c)])/(1 - ( 
d^2*x^2)/c^2)^p + (15*c^2*D*x^4*(c + d*x)^p*AppellF1[4, -p, -1/2 - p, 5, ( 
d*x)/c, -((d*x)/c)])/(1 - (d^2*x^2)/c^2)^p + (12*C*d^2*x^5*(c + d*x)^p*App 
ellF1[5, -p, -1/2 - p, 6, (d*x)/c, -((d*x)/c)])/(1 - (d^2*x^2)/c^2)^p + (2 
4*c*d*D*x^5*(c + d*x)^p*AppellF1[5, -p, -1/2 - p, 6, (d*x)/c, -((d*x)/c)]) 
/(1 - (d^2*x^2)/c^2)^p + (10*d^2*D*x^6*(c + d*x)^p*AppellF1[6, -p, -1/2 - 
p, 7, (d*x)/c, -((d*x)/c)])/(1 - (d^2*x^2)/c^2)^p - (15*2^(5/2 + p)*A*c^3* 
(c^2 - d^2*x^2)^p*Hypergeometric2F1[-1/2 - p, 1 + p, 2 + p, (c - d*x)/(2*c 
)])/((d + d*p)*(c - d*x)^p*(1 + (d*x)/c)^p) + (15*2^(5/2 + p)*A*c^2*x*(c^2 
 - d^2*x^2)^p*Hypergeometric2F1[-1/2 - p, 1 + p, 2 + p, (c - d*x)/(2*c)])/ 
((1 + p)*(c - d*x)^p*(1 + (d*x)/c)^p)))/(60*Sqrt[1 + (d*x)/c])
 

Rubi [A] (verified)

Time = 1.55 (sec) , antiderivative size = 359, normalized size of antiderivative = 0.97, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.205, Rules used = {2170, 27, 2170, 27, 672, 474, 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.

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

\(\Big \downarrow \) 2170

\(\displaystyle -\frac {2 \int -\frac {1}{2} (c+d x)^{5/2} \left (c^2-d^2 x^2\right )^p \left ((C d (4 p+13)-c D (8 p+17)) x^2 d^4+\left (D (5-4 p) c^2+B d^2 (4 p+13)\right ) x d^3+\left (9 D c^3+A d^3 (4 p+13)\right ) d^2\right )dx}{d^5 (4 p+13)}-\frac {2 D (c+d x)^{9/2} \left (c^2-d^2 x^2\right )^{p+1}}{d^4 (4 p+13)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int (c+d x)^{5/2} \left (c^2-d^2 x^2\right )^p \left ((C d (4 p+13)-c D (8 p+17)) x^2 d^4+\left (D (5-4 p) c^2+B d^2 (4 p+13)\right ) x d^3+\left (9 D c^3+A d^3 (4 p+13)\right ) d^2\right )dx}{d^5 (4 p+13)}-\frac {2 D (c+d x)^{9/2} \left (c^2-d^2 x^2\right )^{p+1}}{d^4 (4 p+13)}\)

\(\Big \downarrow \) 2170

\(\displaystyle \frac {-\frac {2 \int \frac {1}{2} d^6 (c+d x)^{5/2} \left (20 D (p+1) c^3-7 C d (4 p+13) c^2-A d^3 \left (16 p^2+96 p+143\right )+d \left (-D \left (16 p^2+76 p+123\right ) c^2+4 C d \left (4 p^2+17 p+13\right ) c-B d^2 \left (16 p^2+96 p+143\right )\right ) x\right ) \left (c^2-d^2 x^2\right )^pdx}{d^4 (4 p+11)}-\frac {2 d (c+d x)^{7/2} \left (c^2-d^2 x^2\right )^{p+1} (C d (4 p+13)-c D (8 p+17))}{4 p+11}}{d^5 (4 p+13)}-\frac {2 D (c+d x)^{9/2} \left (c^2-d^2 x^2\right )^{p+1}}{d^4 (4 p+13)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {d^2 \int (c+d x)^{5/2} \left (20 D (p+1) c^3-7 C d (4 p+13) c^2-A d^3 \left (16 p^2+96 p+143\right )+d \left (-D \left (16 p^2+76 p+123\right ) c^2+4 C d \left (4 p^2+17 p+13\right ) c-B d^2 \left (16 p^2+96 p+143\right )\right ) x\right ) \left (c^2-d^2 x^2\right )^pdx}{4 p+11}-\frac {2 d (c+d x)^{7/2} \left (c^2-d^2 x^2\right )^{p+1} (C d (4 p+13)-c D (8 p+17))}{4 p+11}}{d^5 (4 p+13)}-\frac {2 D (c+d x)^{9/2} \left (c^2-d^2 x^2\right )^{p+1}}{d^4 (4 p+13)}\)

\(\Big \downarrow \) 672

\(\displaystyle \frac {-\frac {d^2 \left (-\frac {\left (A d^3 \left (64 p^3+528 p^2+1436 p+1287\right )+5 B c d^2 \left (16 p^2+96 p+143\right )+15 c^3 D (8 p+29)+c^2 C d \left (32 p^2+276 p+559\right )\right ) \int (c+d x)^{5/2} \left (c^2-d^2 x^2\right )^pdx}{4 p+9}-\frac {2 (c+d x)^{5/2} \left (c^2-d^2 x^2\right )^{p+1} \left (-B d^2 \left (16 p^2+96 p+143\right )+c^2 (-D) \left (16 p^2+76 p+123\right )+4 c C d \left (4 p^2+17 p+13\right )\right )}{d (4 p+9)}\right )}{4 p+11}-\frac {2 d (c+d x)^{7/2} \left (c^2-d^2 x^2\right )^{p+1} (C d (4 p+13)-c D (8 p+17))}{4 p+11}}{d^5 (4 p+13)}-\frac {2 D (c+d x)^{9/2} \left (c^2-d^2 x^2\right )^{p+1}}{d^4 (4 p+13)}\)

\(\Big \downarrow \) 474

\(\displaystyle \frac {-\frac {d^2 \left (-\frac {c^2 \sqrt {c+d x} \left (A d^3 \left (64 p^3+528 p^2+1436 p+1287\right )+5 B c d^2 \left (16 p^2+96 p+143\right )+15 c^3 D (8 p+29)+c^2 C d \left (32 p^2+276 p+559\right )\right ) \int \left (\frac {d x}{c}+1\right )^{5/2} \left (c^2-d^2 x^2\right )^pdx}{(4 p+9) \sqrt {\frac {d x}{c}+1}}-\frac {2 (c+d x)^{5/2} \left (c^2-d^2 x^2\right )^{p+1} \left (-B d^2 \left (16 p^2+96 p+143\right )+c^2 (-D) \left (16 p^2+76 p+123\right )+4 c C d \left (4 p^2+17 p+13\right )\right )}{d (4 p+9)}\right )}{4 p+11}-\frac {2 d (c+d x)^{7/2} \left (c^2-d^2 x^2\right )^{p+1} (C d (4 p+13)-c D (8 p+17))}{4 p+11}}{d^5 (4 p+13)}-\frac {2 D (c+d x)^{9/2} \left (c^2-d^2 x^2\right )^{p+1}}{d^4 (4 p+13)}\)

\(\Big \downarrow \) 473

\(\displaystyle \frac {-\frac {d^2 \left (-\frac {c^2 \sqrt {c+d x} \left (c^2-c d x\right )^{-p-1} \left (c^2-d^2 x^2\right )^{p+1} \left (\frac {d x}{c}+1\right )^{-p-\frac {3}{2}} \left (A d^3 \left (64 p^3+528 p^2+1436 p+1287\right )+5 B c d^2 \left (16 p^2+96 p+143\right )+15 c^3 D (8 p+29)+c^2 C d \left (32 p^2+276 p+559\right )\right ) \int \left (\frac {d x}{c}+1\right )^{p+\frac {5}{2}} \left (c^2-c d x\right )^pdx}{4 p+9}-\frac {2 (c+d x)^{5/2} \left (c^2-d^2 x^2\right )^{p+1} \left (-B d^2 \left (16 p^2+96 p+143\right )+c^2 (-D) \left (16 p^2+76 p+123\right )+4 c C d \left (4 p^2+17 p+13\right )\right )}{d (4 p+9)}\right )}{4 p+11}-\frac {2 d (c+d x)^{7/2} \left (c^2-d^2 x^2\right )^{p+1} (C d (4 p+13)-c D (8 p+17))}{4 p+11}}{d^5 (4 p+13)}-\frac {2 D (c+d x)^{9/2} \left (c^2-d^2 x^2\right )^{p+1}}{d^4 (4 p+13)}\)

\(\Big \downarrow \) 79

\(\displaystyle \frac {-\frac {d^2 \left (\frac {c 2^{p+\frac {5}{2}} \sqrt {c+d x} \left (\frac {d x}{c}+1\right )^{-p-\frac {3}{2}} \left (c^2-d^2 x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (-p-\frac {5}{2},p+1,p+2,\frac {c-d x}{2 c}\right ) \left (A d^3 \left (64 p^3+528 p^2+1436 p+1287\right )+5 B c d^2 \left (16 p^2+96 p+143\right )+15 c^3 D (8 p+29)+c^2 C d \left (32 p^2+276 p+559\right )\right )}{d (p+1) (4 p+9)}-\frac {2 (c+d x)^{5/2} \left (c^2-d^2 x^2\right )^{p+1} \left (-B d^2 \left (16 p^2+96 p+143\right )+c^2 (-D) \left (16 p^2+76 p+123\right )+4 c C d \left (4 p^2+17 p+13\right )\right )}{d (4 p+9)}\right )}{4 p+11}-\frac {2 d (c+d x)^{7/2} \left (c^2-d^2 x^2\right )^{p+1} (C d (4 p+13)-c D (8 p+17))}{4 p+11}}{d^5 (4 p+13)}-\frac {2 D (c+d x)^{9/2} \left (c^2-d^2 x^2\right )^{p+1}}{d^4 (4 p+13)}\)

Input:

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

Output:

(-2*D*(c + d*x)^(9/2)*(c^2 - d^2*x^2)^(1 + p))/(d^4*(13 + 4*p)) + ((-2*d*( 
C*d*(13 + 4*p) - c*D*(17 + 8*p))*(c + d*x)^(7/2)*(c^2 - d^2*x^2)^(1 + p))/ 
(11 + 4*p) - (d^2*((-2*(4*c*C*d*(13 + 17*p + 4*p^2) - c^2*D*(123 + 76*p + 
16*p^2) - B*d^2*(143 + 96*p + 16*p^2))*(c + d*x)^(5/2)*(c^2 - d^2*x^2)^(1 
+ p))/(d*(9 + 4*p)) + (2^(5/2 + p)*c*(15*c^3*D*(29 + 8*p) + 5*B*c*d^2*(143 
 + 96*p + 16*p^2) + c^2*C*d*(559 + 276*p + 32*p^2) + A*d^3*(1287 + 1436*p 
+ 528*p^2 + 64*p^3))*Sqrt[c + d*x]*(1 + (d*x)/c)^(-3/2 - p)*(c^2 - d^2*x^2 
)^(1 + p)*Hypergeometric2F1[-5/2 - p, 1 + p, 2 + p, (c - d*x)/(2*c)])/(d*( 
1 + p)*(9 + 4*p))))/(11 + 4*p))/(d^5*(13 + 4*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 474

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

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]

\[\int \left (d x +c \right )^{\frac {5}{2}} \left (-d^{2} x^{2}+c^{2}\right )^{p} \left (D x^{3}+C \,x^{2}+B x +A \right )d x\]

Input:

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

Output:

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

Fricas [F]

\[ \int (c+d x)^{5/2} \left (c^2-d^2 x^2\right )^p \left (A+B x+C x^2+D x^3\right ) \, dx=\int { {\left (D x^{3} + C x^{2} + B x + A\right )} {\left (d x + c\right )}^{\frac {5}{2}} {\left (-d^{2} x^{2} + c^{2}\right )}^{p} \,d x } \] Input:

integrate((d*x+c)^(5/2)*(-d^2*x^2+c^2)^p*(D*x^3+C*x^2+B*x+A),x, algorithm= 
"fricas")
 

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)*sqrt(d*x 
+ c)*(-d^2*x^2 + c^2)^p, x)
 

Sympy [F]

\[ \int (c+d x)^{5/2} \left (c^2-d^2 x^2\right )^p \left (A+B x+C x^2+D x^3\right ) \, dx=\int \left (- \left (- c + d x\right ) \left (c + d x\right )\right )^{p} \left (c + d x\right )^{\frac {5}{2}} \left (A + B x + C x^{2} + D x^{3}\right )\, dx \] Input:

integrate((d*x+c)**(5/2)*(-d**2*x**2+c**2)**p*(D*x**3+C*x**2+B*x+A),x)
 

Output:

Integral((-(-c + d*x)*(c + d*x))**p*(c + d*x)**(5/2)*(A + B*x + C*x**2 + D 
*x**3), x)
 

Maxima [F]

\[ \int (c+d x)^{5/2} \left (c^2-d^2 x^2\right )^p \left (A+B x+C x^2+D x^3\right ) \, dx=\int { {\left (D x^{3} + C x^{2} + B x + A\right )} {\left (d x + c\right )}^{\frac {5}{2}} {\left (-d^{2} x^{2} + c^{2}\right )}^{p} \,d x } \] Input:

integrate((d*x+c)^(5/2)*(-d^2*x^2+c^2)^p*(D*x^3+C*x^2+B*x+A),x, algorithm= 
"maxima")
 

Output:

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

Giac [F]

\[ \int (c+d x)^{5/2} \left (c^2-d^2 x^2\right )^p \left (A+B x+C x^2+D x^3\right ) \, dx=\int { {\left (D x^{3} + C x^{2} + B x + A\right )} {\left (d x + c\right )}^{\frac {5}{2}} {\left (-d^{2} x^{2} + c^{2}\right )}^{p} \,d x } \] Input:

integrate((d*x+c)^(5/2)*(-d^2*x^2+c^2)^p*(D*x^3+C*x^2+B*x+A),x, algorithm= 
"giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int (c+d x)^{5/2} \left (c^2-d^2 x^2\right )^p \left (A+B x+C x^2+D x^3\right ) \, dx=\int {\left (c^2-d^2\,x^2\right )}^p\,{\left (c+d\,x\right )}^{5/2}\,\left (A+B\,x+C\,x^2+x^3\,D\right ) \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int (c+d x)^{5/2} \left (c^2-d^2 x^2\right )^p \left (A+B x+C x^2+D x^3\right ) \, dx=\text {too large to display} \] Input:

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

Output:

(2*(4096*sqrt(c + d*x)*(c**2 - d**2*x**2)**p*a*c**3*d**2*p**6 + 49152*sqrt 
(c + d*x)*(c**2 - d**2*x**2)**p*a*c**3*d**2*p**5 + 232704*sqrt(c + d*x)*(c 
**2 - d**2*x**2)**p*a*c**3*d**2*p**4 + 544128*sqrt(c + d*x)*(c**2 - d**2*x 
**2)**p*a*c**3*d**2*p**3 + 632720*sqrt(c + d*x)*(c**2 - d**2*x**2)**p*a*c* 
*3*d**2*p**2 + 304680*sqrt(c + d*x)*(c**2 - d**2*x**2)**p*a*c**3*d**2*p + 
19305*sqrt(c + d*x)*(c**2 - d**2*x**2)**p*a*c**3*d**2 + 2048*sqrt(c + d*x) 
*(c**2 - d**2*x**2)**p*a*c**2*d**3*p**4*x + 19776*sqrt(c + d*x)*(c**2 - d* 
*2*x**2)**p*a*c**2*d**3*p**3*x + 69712*sqrt(c + d*x)*(c**2 - d**2*x**2)**p 
*a*c**2*d**3*p**2*x + 105804*sqrt(c + d*x)*(c**2 - d**2*x**2)**p*a*c**2*d* 
*3*p*x + 57915*sqrt(c + d*x)*(c**2 - d**2*x**2)**p*a*c**2*d**3*x + 2048*sq 
rt(c + d*x)*(c**2 - d**2*x**2)**p*a*c*d**4*p**5*x**2 + 22272*sqrt(c + d*x) 
*(c**2 - d**2*x**2)**p*a*c*d**4*p**4*x**2 + 93184*sqrt(c + d*x)*(c**2 - d* 
*2*x**2)**p*a*c*d**4*p**3*x**2 + 185568*sqrt(c + d*x)*(c**2 - d**2*x**2)** 
p*a*c*d**4*p**2*x**2 + 172728*sqrt(c + d*x)*(c**2 - d**2*x**2)**p*a*c*d**4 
*p*x**2 + 57915*sqrt(c + d*x)*(c**2 - d**2*x**2)**p*a*c*d**4*x**2 + 1024*s 
qrt(c + d*x)*(c**2 - d**2*x**2)**p*a*d**5*p**5*x**3 + 10496*sqrt(c + d*x)* 
(c**2 - d**2*x**2)**p*a*d**5*p**4*x**3 + 40832*sqrt(c + d*x)*(c**2 - d**2* 
x**2)**p*a*d**5*p**3*x**3 + 74464*sqrt(c + d*x)*(c**2 - d**2*x**2)**p*a*d* 
*5*p**2*x**3 + 62724*sqrt(c + d*x)*(c**2 - d**2*x**2)**p*a*d**5*p*x**3 + 1 
9305*sqrt(c + d*x)*(c**2 - d**2*x**2)**p*a*d**5*x**3 + 4096*sqrt(c + d*...