3.3.51 \(\int x^3 (d+e x)^2 (d^2-e^2 x^2)^p \, dx\) [251]

3.3.51.1 Optimal result

Integrand size = 25, antiderivative size = 149 \[ \int x^3 (d+e x)^2 \left (d^2-e^2 x^2\right )^p \, dx=-\frac {d^4 \left (d^2-e^2 x^2\right )^{1+p}}{e^4 (1+p)}+\frac {3 d^2 \left (d^2-e^2 x^2\right )^{2+p}}{2 e^4 (2+p)}-\frac {\left (d^2-e^2 x^2\right )^{3+p}}{2 e^4 (3+p)}+\frac {2}{5} d e x^5 \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},\frac {e^2 x^2}{d^2}\right ) \]

output

-d^4*(-e^2*x^2+d^2)^(p+1)/e^4/(p+1)+3/2*d^2*(-e^2*x^2+d^2)^(2+p)/e^4/(2+p) 
-1/2*(-e^2*x^2+d^2)^(3+p)/e^4/(3+p)+2/5*d*e*x^5*(-e^2*x^2+d^2)^p*hypergeom 
([5/2, -p],[7/2],e^2*x^2/d^2)/((1-e^2*x^2/d^2)^p)
 

3.3.51.2 Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.93 \[ \int x^3 (d+e x)^2 \left (d^2-e^2 x^2\right )^p \, dx=\frac {\left (d^2-e^2 x^2\right )^p \left (-\frac {5 \left (d^2-e^2 x^2\right ) \left (d^4 (5+p)+d^2 e^2 \left (5+6 p+p^2\right ) x^2+e^4 \left (2+3 p+p^2\right ) x^4\right )}{(1+p) (2+p) (3+p)}+4 d e^5 x^5 \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},\frac {e^2 x^2}{d^2}\right )\right )}{10 e^4} \]

input

Integrate[x^3*(d + e*x)^2*(d^2 - e^2*x^2)^p,x]
 

output

((d^2 - e^2*x^2)^p*((-5*(d^2 - e^2*x^2)*(d^4*(5 + p) + d^2*e^2*(5 + 6*p + 
p^2)*x^2 + e^4*(2 + 3*p + p^2)*x^4))/((1 + p)*(2 + p)*(3 + p)) + (4*d*e^5* 
x^5*Hypergeometric2F1[5/2, -p, 7/2, (e^2*x^2)/d^2])/(1 - (e^2*x^2)/d^2)^p) 
)/(10*e^4)
 

3.3.51.3 Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {543, 27, 279, 278, 354, 86, 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.

\(\displaystyle \int x^3 (d+e x)^2 \left (d^2-e^2 x^2\right )^p \, dx\)

\(\Big \downarrow \) 543

\(\displaystyle \int 2 d e x^4 \left (d^2-e^2 x^2\right )^pdx+\int x^3 \left (d^2-e^2 x^2\right )^p \left (d^2+e^2 x^2\right )dx\)

\(\Big \downarrow \) 27

\(\displaystyle 2 d e \int x^4 \left (d^2-e^2 x^2\right )^pdx+\int x^3 \left (d^2-e^2 x^2\right )^p \left (d^2+e^2 x^2\right )dx\)

\(\Big \downarrow \) 279

\(\displaystyle 2 d e \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \int x^4 \left (1-\frac {e^2 x^2}{d^2}\right )^pdx+\int x^3 \left (d^2-e^2 x^2\right )^p \left (d^2+e^2 x^2\right )dx\)

\(\Big \downarrow \) 278

\(\displaystyle \int x^3 \left (d^2-e^2 x^2\right )^p \left (d^2+e^2 x^2\right )dx+\frac {2}{5} d e x^5 \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},\frac {e^2 x^2}{d^2}\right )\)

\(\Big \downarrow \) 354

\(\displaystyle \frac {1}{2} \int x^2 \left (d^2-e^2 x^2\right )^p \left (d^2+e^2 x^2\right )dx^2+\frac {2}{5} d e x^5 \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},\frac {e^2 x^2}{d^2}\right )\)

\(\Big \downarrow \) 86

\(\displaystyle \frac {1}{2} \int \left (\frac {2 d^4 \left (d^2-e^2 x^2\right )^p}{e^2}-\frac {3 d^2 \left (d^2-e^2 x^2\right )^{p+1}}{e^2}+\frac {\left (d^2-e^2 x^2\right )^{p+2}}{e^2}\right )dx^2+\frac {2}{5} d e x^5 \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},\frac {e^2 x^2}{d^2}\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2}{5} d e x^5 \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},\frac {e^2 x^2}{d^2}\right )+\frac {1}{2} \left (\frac {3 d^2 \left (d^2-e^2 x^2\right )^{p+2}}{e^4 (p+2)}-\frac {\left (d^2-e^2 x^2\right )^{p+3}}{e^4 (p+3)}-\frac {2 d^4 \left (d^2-e^2 x^2\right )^{p+1}}{e^4 (p+1)}\right )\)

input

Int[x^3*(d + e*x)^2*(d^2 - e^2*x^2)^p,x]
 

output

((-2*d^4*(d^2 - e^2*x^2)^(1 + p))/(e^4*(1 + p)) + (3*d^2*(d^2 - e^2*x^2)^( 
2 + p))/(e^4*(2 + p)) - (d^2 - e^2*x^2)^(3 + p)/(e^4*(3 + p)))/2 + (2*d*e* 
x^5*(d^2 - e^2*x^2)^p*Hypergeometric2F1[5/2, -p, 7/2, (e^2*x^2)/d^2])/(5*( 
1 - (e^2*x^2)/d^2)^p)
 

3.3.51.3.1 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 86

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ 
.), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; 
 FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 
] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p 
+ 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
 

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 354

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S 
ymbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x 
, x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ 
[(m - 1)/2]
 

rule 543

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> Module[{k}, Int[x^m*Sum[Binomial[n, 2*k]*c^(n - 2*k)*d^(2*k)*x^(2*k), 
 {k, 0, n/2}]*(a + b*x^2)^p, x] + Int[x^(m + 1)*Sum[Binomial[n, 2*k + 1]*c^ 
(n - 2*k - 1)*d^(2*k + 1)*x^(2*k), {k, 0, (n - 1)/2}]*(a + b*x^2)^p, x]] /; 
 FreeQ[{a, b, c, d, p}, x] && IGtQ[n, 1] && IntegerQ[m] &&  !IntegerQ[2*p] 
&&  !(EqQ[m, 1] && EqQ[b*c^2 + a*d^2, 0])
 

rule 2009

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

3.3.51.4 Maple [F]

\[\int x^{3} \left (e x +d \right )^{2} \left (-e^{2} x^{2}+d^{2}\right )^{p}d x\]

input

int(x^3*(e*x+d)^2*(-e^2*x^2+d^2)^p,x)
 

output

int(x^3*(e*x+d)^2*(-e^2*x^2+d^2)^p,x)
 

3.3.51.5 Fricas [F]

\[ \int x^3 (d+e x)^2 \left (d^2-e^2 x^2\right )^p \, dx=\int { {\left (e x + d\right )}^{2} {\left (-e^{2} x^{2} + d^{2}\right )}^{p} x^{3} \,d x } \]

input

integrate(x^3*(e*x+d)^2*(-e^2*x^2+d^2)^p,x, algorithm="fricas")
 

output

integral((e^2*x^5 + 2*d*e*x^4 + d^2*x^3)*(-e^2*x^2 + d^2)^p, x)
 

3.3.51.6 Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 348 vs. \(2 (124) = 248\).

Time = 2.13 (sec) , antiderivative size = 1328, normalized size of antiderivative = 8.91 \[ \int x^3 (d+e x)^2 \left (d^2-e^2 x^2\right )^p \, dx=\text {Too large to display} \]

input

integrate(x**3*(e*x+d)**2*(-e**2*x**2+d**2)**p,x)
 

output

d**2*Piecewise((x**4*(d**2)**p/4, Eq(e, 0)), (-d**2*log(-d/e + x)/(-2*d**2 
*e**4 + 2*e**6*x**2) - d**2*log(d/e + x)/(-2*d**2*e**4 + 2*e**6*x**2) - d* 
*2/(-2*d**2*e**4 + 2*e**6*x**2) + e**2*x**2*log(-d/e + x)/(-2*d**2*e**4 + 
2*e**6*x**2) + e**2*x**2*log(d/e + x)/(-2*d**2*e**4 + 2*e**6*x**2), Eq(p, 
-2)), (-d**2*log(-d/e + x)/(2*e**4) - d**2*log(d/e + x)/(2*e**4) - x**2/(2 
*e**2), Eq(p, -1)), (-d**4*(d**2 - e**2*x**2)**p/(2*e**4*p**2 + 6*e**4*p + 
 4*e**4) - d**2*e**2*p*x**2*(d**2 - e**2*x**2)**p/(2*e**4*p**2 + 6*e**4*p 
+ 4*e**4) + e**4*p*x**4*(d**2 - e**2*x**2)**p/(2*e**4*p**2 + 6*e**4*p + 4* 
e**4) + e**4*x**4*(d**2 - e**2*x**2)**p/(2*e**4*p**2 + 6*e**4*p + 4*e**4), 
 True)) + 2*d*d**(2*p)*e*x**5*hyper((5/2, -p), (7/2,), e**2*x**2*exp_polar 
(2*I*pi)/d**2)/5 + e**2*Piecewise((x**6*(d**2)**p/6, Eq(e, 0)), (-2*d**4*l 
og(-d/e + x)/(4*d**4*e**6 - 8*d**2*e**8*x**2 + 4*e**10*x**4) - 2*d**4*log( 
d/e + x)/(4*d**4*e**6 - 8*d**2*e**8*x**2 + 4*e**10*x**4) - 3*d**4/(4*d**4* 
e**6 - 8*d**2*e**8*x**2 + 4*e**10*x**4) + 4*d**2*e**2*x**2*log(-d/e + x)/( 
4*d**4*e**6 - 8*d**2*e**8*x**2 + 4*e**10*x**4) + 4*d**2*e**2*x**2*log(d/e 
+ x)/(4*d**4*e**6 - 8*d**2*e**8*x**2 + 4*e**10*x**4) + 4*d**2*e**2*x**2/(4 
*d**4*e**6 - 8*d**2*e**8*x**2 + 4*e**10*x**4) - 2*e**4*x**4*log(-d/e + x)/ 
(4*d**4*e**6 - 8*d**2*e**8*x**2 + 4*e**10*x**4) - 2*e**4*x**4*log(d/e + x) 
/(4*d**4*e**6 - 8*d**2*e**8*x**2 + 4*e**10*x**4), Eq(p, -3)), (-2*d**4*log 
(-d/e + x)/(-2*d**2*e**6 + 2*e**8*x**2) - 2*d**4*log(d/e + x)/(-2*d**2*...
 

3.3.51.7 Maxima [F]

\[ \int x^3 (d+e x)^2 \left (d^2-e^2 x^2\right )^p \, dx=\int { {\left (e x + d\right )}^{2} {\left (-e^{2} x^{2} + d^{2}\right )}^{p} x^{3} \,d x } \]

input

integrate(x^3*(e*x+d)^2*(-e^2*x^2+d^2)^p,x, algorithm="maxima")
 

output

1/2*(e^4*(p + 1)*x^4 - d^2*e^2*p*x^2 - d^4)*(-e^2*x^2 + d^2)^p*d^2/((p^2 + 
 3*p + 2)*e^4) + integrate((e^2*x^5 + 2*d*e*x^4)*e^(p*log(e*x + d) + p*log 
(-e*x + d)), x)
 

3.3.51.8 Giac [F]

\[ \int x^3 (d+e x)^2 \left (d^2-e^2 x^2\right )^p \, dx=\int { {\left (e x + d\right )}^{2} {\left (-e^{2} x^{2} + d^{2}\right )}^{p} x^{3} \,d x } \]

input

integrate(x^3*(e*x+d)^2*(-e^2*x^2+d^2)^p,x, algorithm="giac")
 

output

integrate((e*x + d)^2*(-e^2*x^2 + d^2)^p*x^3, x)
 

3.3.51.9 Mupad [F(-1)]

Timed out. \[ \int x^3 (d+e x)^2 \left (d^2-e^2 x^2\right )^p \, dx=\int x^3\,{\left (d^2-e^2\,x^2\right )}^p\,{\left (d+e\,x\right )}^2 \,d x \]

input

int(x^3*(d^2 - e^2*x^2)^p*(d + e*x)^2,x)
 

output

int(x^3*(d^2 - e^2*x^2)^p*(d + e*x)^2, x)