\(\int x^2 (a^2+2 a b x^2+b^2 x^4)^p \, dx\) [692]

Optimal result

Integrand size = 24, antiderivative size = 60 \[ \int x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\frac {1}{3} x^3 \left (1+\frac {b x^2}{a}\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-2 p,\frac {5}{2},-\frac {b x^2}{a}\right ) \] Output:

1/3*x^3*(b^2*x^4+2*a*b*x^2+a^2)^p*hypergeom([3/2, -2*p],[5/2],-b*x^2/a)/(( 
1+b*x^2/a)^(2*p))
 

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.85 \[ \int x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\frac {1}{3} x^3 \left (\left (a+b x^2\right )^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-2 p} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-2 p,\frac {5}{2},-\frac {b x^2}{a}\right ) \] Input:

Integrate[x^2*(a^2 + 2*a*b*x^2 + b^2*x^4)^p,x]
 

Output:

(x^3*((a + b*x^2)^2)^p*Hypergeometric2F1[3/2, -2*p, 5/2, -((b*x^2)/a)])/(3 
*(1 + (b*x^2)/a)^(2*p))
 

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {1385, 278}

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.

\(\displaystyle \int x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx\)

\(\Big \downarrow \) 1385

\(\displaystyle \left (\frac {b x^2}{a}+1\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p \int x^2 \left (\frac {b x^2}{a}+1\right )^{2 p}dx\)

\(\Big \downarrow \) 278

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

Input:

Int[x^2*(a^2 + 2*a*b*x^2 + b^2*x^4)^p,x]
 

Output:

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

Defintions of rubi rules used

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 1385

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> S 
imp[a^IntPart[p]*((a + b*x^n + c*x^(2*n))^FracPart[p]/(1 + 2*c*(x^n/b))^(2* 
FracPart[p]))   Int[u*(1 + 2*c*(x^n/b))^(2*p), x], x] /; FreeQ[{a, b, c, n, 
 p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[2*p] && NeQ[u, 
 x^(n - 1)] && NeQ[u, x^(2*n - 1)]
 

Maple [F]

\[\int x^{2} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p}d x\]

Input:

int(x^2*(b^2*x^4+2*a*b*x^2+a^2)^p,x)
 

Output:

int(x^2*(b^2*x^4+2*a*b*x^2+a^2)^p,x)
 

Fricas [F]

\[ \int x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\int { {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} x^{2} \,d x } \] Input:

integrate(x^2*(b^2*x^4+2*a*b*x^2+a^2)^p,x, algorithm="fricas")
 

Output:

integral((b^2*x^4 + 2*a*b*x^2 + a^2)^p*x^2, x)
 

Sympy [F]

\[ \int x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\int x^{2} \left (\left (a + b x^{2}\right )^{2}\right )^{p}\, dx \] Input:

integrate(x**2*(b**2*x**4+2*a*b*x**2+a**2)**p,x)
 

Output:

Integral(x**2*((a + b*x**2)**2)**p, x)
 

Maxima [F]

\[ \int x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\int { {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} x^{2} \,d x } \] Input:

integrate(x^2*(b^2*x^4+2*a*b*x^2+a^2)^p,x, algorithm="maxima")
 

Output:

integrate((b^2*x^4 + 2*a*b*x^2 + a^2)^p*x^2, x)
 

Giac [F]

\[ \int x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\int { {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} x^{2} \,d x } \] Input:

integrate(x^2*(b^2*x^4+2*a*b*x^2+a^2)^p,x, algorithm="giac")
 

Output:

integrate((b^2*x^4 + 2*a*b*x^2 + a^2)^p*x^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\int x^2\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^p \,d x \] Input:

int(x^2*(a^2 + b^2*x^4 + 2*a*b*x^2)^p,x)
 

Output:

int(x^2*(a^2 + b^2*x^4 + 2*a*b*x^2)^p, x)
 

Reduce [F]

\[ \int x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\frac {4 \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p} a p x +4 \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p} b p \,x^{3}+\left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p} b \,x^{3}-64 \left (\int \frac {\left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p}}{16 b \,p^{2} x^{2}+16 b p \,x^{2}+16 a \,p^{2}+3 b \,x^{2}+16 a p +3 a}d x \right ) a^{2} p^{3}-64 \left (\int \frac {\left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p}}{16 b \,p^{2} x^{2}+16 b p \,x^{2}+16 a \,p^{2}+3 b \,x^{2}+16 a p +3 a}d x \right ) a^{2} p^{2}-12 \left (\int \frac {\left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p}}{16 b \,p^{2} x^{2}+16 b p \,x^{2}+16 a \,p^{2}+3 b \,x^{2}+16 a p +3 a}d x \right ) a^{2} p}{b \left (16 p^{2}+16 p +3\right )} \] Input:

int(x^2*(b^2*x^4+2*a*b*x^2+a^2)^p,x)
 

Output:

(4*(a**2 + 2*a*b*x**2 + b**2*x**4)**p*a*p*x + 4*(a**2 + 2*a*b*x**2 + b**2* 
x**4)**p*b*p*x**3 + (a**2 + 2*a*b*x**2 + b**2*x**4)**p*b*x**3 - 64*int((a* 
*2 + 2*a*b*x**2 + b**2*x**4)**p/(16*a*p**2 + 16*a*p + 3*a + 16*b*p**2*x**2 
 + 16*b*p*x**2 + 3*b*x**2),x)*a**2*p**3 - 64*int((a**2 + 2*a*b*x**2 + b**2 
*x**4)**p/(16*a*p**2 + 16*a*p + 3*a + 16*b*p**2*x**2 + 16*b*p*x**2 + 3*b*x 
**2),x)*a**2*p**2 - 12*int((a**2 + 2*a*b*x**2 + b**2*x**4)**p/(16*a*p**2 + 
 16*a*p + 3*a + 16*b*p**2*x**2 + 16*b*p*x**2 + 3*b*x**2),x)*a**2*p)/(b*(16 
*p**2 + 16*p + 3))