\(\int x^2 (c+d x) (c^2-d^2 x^2)^p \, dx\) [1197]

Optimal result

Integrand size = 23, antiderivative size = 119 \[ \int x^2 (c+d x) \left (c^2-d^2 x^2\right )^p \, dx=-\frac {c^2 \left (c^2-d^2 x^2\right )^{1+p}}{2 d^3 (1+p)}+\frac {\left (c^2-d^2 x^2\right )^{2+p}}{2 d^3 (2+p)}+\frac {1}{3} c x^3 \left (c^2-d^2 x^2\right )^p \left (1-\frac {d^2 x^2}{c^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},\frac {d^2 x^2}{c^2}\right ) \] Output:

-1/2*c^2*(-d^2*x^2+c^2)^(p+1)/d^3/(p+1)+1/2*(-d^2*x^2+c^2)^(2+p)/d^3/(2+p) 
+1/3*c*x^3*(-d^2*x^2+c^2)^p*hypergeom([3/2, -p],[5/2],d^2*x^2/c^2)/((1-d^2 
*x^2/c^2)^p)
 

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.87 \[ \int x^2 (c+d x) \left (c^2-d^2 x^2\right )^p \, dx=\frac {1}{6} \left (c^2-d^2 x^2\right )^p \left (-\frac {3 \left (c^2-d^2 x^2\right ) \left (c^2+d^2 (1+p) x^2\right )}{d^3 (1+p) (2+p)}+2 c x^3 \left (1-\frac {d^2 x^2}{c^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},\frac {d^2 x^2}{c^2}\right )\right ) \] Input:

Integrate[x^2*(c + d*x)*(c^2 - d^2*x^2)^p,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.01, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {542, 243, 53, 279, 278, 2009}

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.

\(\displaystyle \int x^2 (c+d x) \left (c^2-d^2 x^2\right )^p \, dx\)

\(\Big \downarrow \) 542

\(\displaystyle c \int x^2 \left (c^2-d^2 x^2\right )^pdx+d \int x^3 \left (c^2-d^2 x^2\right )^pdx\)

\(\Big \downarrow \) 243

\(\displaystyle c \int x^2 \left (c^2-d^2 x^2\right )^pdx+\frac {1}{2} d \int x^2 \left (c^2-d^2 x^2\right )^pdx^2\)

\(\Big \downarrow \) 53

\(\displaystyle c \int x^2 \left (c^2-d^2 x^2\right )^pdx+\frac {1}{2} d \int \left (\frac {c^2 \left (c^2-d^2 x^2\right )^p}{d^2}-\frac {\left (c^2-d^2 x^2\right )^{p+1}}{d^2}\right )dx^2\)

\(\Big \downarrow \) 279

\(\displaystyle c \left (c^2-d^2 x^2\right )^p \left (1-\frac {d^2 x^2}{c^2}\right )^{-p} \int x^2 \left (1-\frac {d^2 x^2}{c^2}\right )^pdx+\frac {1}{2} d \int \left (\frac {c^2 \left (c^2-d^2 x^2\right )^p}{d^2}-\frac {\left (c^2-d^2 x^2\right )^{p+1}}{d^2}\right )dx^2\)

\(\Big \downarrow \) 278

\(\displaystyle \frac {1}{2} d \int \left (\frac {c^2 \left (c^2-d^2 x^2\right )^p}{d^2}-\frac {\left (c^2-d^2 x^2\right )^{p+1}}{d^2}\right )dx^2+\frac {1}{3} c x^3 \left (c^2-d^2 x^2\right )^p \left (1-\frac {d^2 x^2}{c^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},\frac {d^2 x^2}{c^2}\right )\)

\(\Big \downarrow \) 2009

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

Input:

Int[x^2*(c + d*x)*(c^2 - d^2*x^2)^p,x]
 

Output:

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

Defintions of rubi rules used

rule 53

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int 
[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, 
x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) 
|| LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
 

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 278

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

rule 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntP 
art[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p])   Int[(c*x)^m* 
(1 + b*(x^2/a))^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && 
!(ILtQ[p, 0] || GtQ[a, 0])
 

rule 542

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

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [F]

\[\int x^{2} \left (d x +c \right ) \left (-d^{2} x^{2}+c^{2}\right )^{p}d x\]

Input:

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

Output:

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

Fricas [F]

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

integrate(x^2*(d*x+c)*(-d^2*x^2+c^2)^p,x, algorithm="fricas")
 

Output:

integral((d*x^3 + c*x^2)*(-d^2*x^2 + c^2)^p, x)
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 347 vs. \(2 (95) = 190\).

Time = 1.68 (sec) , antiderivative size = 382, normalized size of antiderivative = 3.21 \[ \int x^2 (c+d x) \left (c^2-d^2 x^2\right )^p \, dx=\frac {c c^{2 p} x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - p \\ \frac {5}{2} \end {matrix}\middle | {\frac {d^{2} x^{2} e^{2 i \pi }}{c^{2}}} \right )}}{3} + d \left (\begin {cases} \frac {x^{4} \left (c^{2}\right )^{p}}{4} & \text {for}\: d = 0 \\- \frac {c^{2} \log {\left (- \frac {c}{d} + x \right )}}{- 2 c^{2} d^{4} + 2 d^{6} x^{2}} - \frac {c^{2} \log {\left (\frac {c}{d} + x \right )}}{- 2 c^{2} d^{4} + 2 d^{6} x^{2}} - \frac {c^{2}}{- 2 c^{2} d^{4} + 2 d^{6} x^{2}} + \frac {d^{2} x^{2} \log {\left (- \frac {c}{d} + x \right )}}{- 2 c^{2} d^{4} + 2 d^{6} x^{2}} + \frac {d^{2} x^{2} \log {\left (\frac {c}{d} + x \right )}}{- 2 c^{2} d^{4} + 2 d^{6} x^{2}} & \text {for}\: p = -2 \\- \frac {c^{2} \log {\left (- \frac {c}{d} + x \right )}}{2 d^{4}} - \frac {c^{2} \log {\left (\frac {c}{d} + x \right )}}{2 d^{4}} - \frac {x^{2}}{2 d^{2}} & \text {for}\: p = -1 \\- \frac {c^{4} \left (c^{2} - d^{2} x^{2}\right )^{p}}{2 d^{4} p^{2} + 6 d^{4} p + 4 d^{4}} - \frac {c^{2} d^{2} p x^{2} \left (c^{2} - d^{2} x^{2}\right )^{p}}{2 d^{4} p^{2} + 6 d^{4} p + 4 d^{4}} + \frac {d^{4} p x^{4} \left (c^{2} - d^{2} x^{2}\right )^{p}}{2 d^{4} p^{2} + 6 d^{4} p + 4 d^{4}} + \frac {d^{4} x^{4} \left (c^{2} - d^{2} x^{2}\right )^{p}}{2 d^{4} p^{2} + 6 d^{4} p + 4 d^{4}} & \text {otherwise} \end {cases}\right ) \] Input:

integrate(x**2*(d*x+c)*(-d**2*x**2+c**2)**p,x)
 

Output:

c*c**(2*p)*x**3*hyper((3/2, -p), (5/2,), d**2*x**2*exp_polar(2*I*pi)/c**2) 
/3 + 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**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))
 

Maxima [F]

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

integrate(x^2*(d*x+c)*(-d^2*x^2+c^2)^p,x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate(x^2*(d*x+c)*(-d^2*x^2+c^2)^p,x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int x^2 (c+d x) \left (c^2-d^2 x^2\right )^p \, dx=\frac {-4 \left (-d^{2} x^{2}+c^{2}\right )^{p} c^{4} p^{2}-8 \left (-d^{2} x^{2}+c^{2}\right )^{p} c^{4} p -3 \left (-d^{2} x^{2}+c^{2}\right )^{p} c^{4}-4 \left (-d^{2} x^{2}+c^{2}\right )^{p} c^{3} d \,p^{3} x -12 \left (-d^{2} x^{2}+c^{2}\right )^{p} c^{3} d \,p^{2} x -8 \left (-d^{2} x^{2}+c^{2}\right )^{p} c^{3} d p x -4 \left (-d^{2} x^{2}+c^{2}\right )^{p} c^{2} d^{2} p^{3} x^{2}-8 \left (-d^{2} x^{2}+c^{2}\right )^{p} c^{2} d^{2} p^{2} x^{2}-3 \left (-d^{2} x^{2}+c^{2}\right )^{p} c^{2} d^{2} p \,x^{2}+4 \left (-d^{2} x^{2}+c^{2}\right )^{p} c \,d^{3} p^{3} x^{3}+14 \left (-d^{2} x^{2}+c^{2}\right )^{p} c \,d^{3} p^{2} x^{3}+14 \left (-d^{2} x^{2}+c^{2}\right )^{p} c \,d^{3} p \,x^{3}+4 \left (-d^{2} x^{2}+c^{2}\right )^{p} c \,d^{3} x^{3}+4 \left (-d^{2} x^{2}+c^{2}\right )^{p} d^{4} p^{3} x^{4}+12 \left (-d^{2} x^{2}+c^{2}\right )^{p} d^{4} p^{2} x^{4}+11 \left (-d^{2} x^{2}+c^{2}\right )^{p} d^{4} p \,x^{4}+3 \left (-d^{2} x^{2}+c^{2}\right )^{p} d^{4} x^{4}+16 \left (\int \frac {\left (-d^{2} x^{2}+c^{2}\right )^{p}}{-4 d^{2} p^{2} x^{2}-8 d^{2} p \,x^{2}+4 c^{2} p^{2}-3 d^{2} x^{2}+8 c^{2} p +3 c^{2}}d x \right ) c^{5} d \,p^{5}+80 \left (\int \frac {\left (-d^{2} x^{2}+c^{2}\right )^{p}}{-4 d^{2} p^{2} x^{2}-8 d^{2} p \,x^{2}+4 c^{2} p^{2}-3 d^{2} x^{2}+8 c^{2} p +3 c^{2}}d x \right ) c^{5} d \,p^{4}+140 \left (\int \frac {\left (-d^{2} x^{2}+c^{2}\right )^{p}}{-4 d^{2} p^{2} x^{2}-8 d^{2} p \,x^{2}+4 c^{2} p^{2}-3 d^{2} x^{2}+8 c^{2} p +3 c^{2}}d x \right ) c^{5} d \,p^{3}+100 \left (\int \frac {\left (-d^{2} x^{2}+c^{2}\right )^{p}}{-4 d^{2} p^{2} x^{2}-8 d^{2} p \,x^{2}+4 c^{2} p^{2}-3 d^{2} x^{2}+8 c^{2} p +3 c^{2}}d x \right ) c^{5} d \,p^{2}+24 \left (\int \frac {\left (-d^{2} x^{2}+c^{2}\right )^{p}}{-4 d^{2} p^{2} x^{2}-8 d^{2} p \,x^{2}+4 c^{2} p^{2}-3 d^{2} x^{2}+8 c^{2} p +3 c^{2}}d x \right ) c^{5} d p}{2 d^{3} \left (4 p^{4}+20 p^{3}+35 p^{2}+25 p +6\right )} \] Input:

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

Output:

( - 4*(c**2 - d**2*x**2)**p*c**4*p**2 - 8*(c**2 - d**2*x**2)**p*c**4*p - 3 
*(c**2 - d**2*x**2)**p*c**4 - 4*(c**2 - d**2*x**2)**p*c**3*d*p**3*x - 12*( 
c**2 - d**2*x**2)**p*c**3*d*p**2*x - 8*(c**2 - d**2*x**2)**p*c**3*d*p*x - 
4*(c**2 - d**2*x**2)**p*c**2*d**2*p**3*x**2 - 8*(c**2 - d**2*x**2)**p*c**2 
*d**2*p**2*x**2 - 3*(c**2 - d**2*x**2)**p*c**2*d**2*p*x**2 + 4*(c**2 - d** 
2*x**2)**p*c*d**3*p**3*x**3 + 14*(c**2 - d**2*x**2)**p*c*d**3*p**2*x**3 + 
14*(c**2 - d**2*x**2)**p*c*d**3*p*x**3 + 4*(c**2 - d**2*x**2)**p*c*d**3*x* 
*3 + 4*(c**2 - d**2*x**2)**p*d**4*p**3*x**4 + 12*(c**2 - d**2*x**2)**p*d** 
4*p**2*x**4 + 11*(c**2 - d**2*x**2)**p*d**4*p*x**4 + 3*(c**2 - d**2*x**2)* 
*p*d**4*x**4 + 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)*c**5*d*p**5 + 80*i 
nt((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)*c**5*d*p**4 + 140*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)*c**5*d*p**3 + 100*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) 
*c**5*d*p**2 + 24*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)*c**5*d*p)/(2*d**3* 
(4*p**4 + 20*p**3 + 35*p**2 + 25*p + 6))