\(\int x^2 (a+b x^2+c x^4)^p \, dx\) [1100]

Optimal result

Integrand size = 18, antiderivative size = 138 \[ \int x^2 \left (a+b x^2+c x^4\right )^p \, dx=\frac {1}{3} x^3 \left (1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+b x^2+c x^4\right )^p \operatorname {AppellF1}\left (\frac {3}{2},-p,-p,\frac {5}{2},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right ) \] Output:

1/3*x^3*(c*x^4+b*x^2+a)^p*AppellF1(3/2,-p,-p,5/2,-2*c*x^2/(b-(-4*a*c+b^2)^ 
(1/2)),-2*c*x^2/(b+(-4*a*c+b^2)^(1/2)))/((1+2*c*x^2/(b-(-4*a*c+b^2)^(1/2)) 
)^p)/((1+2*c*x^2/(b+(-4*a*c+b^2)^(1/2)))^p)
 

Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.20 \[ \int x^2 \left (a+b x^2+c x^4\right )^p \, dx=\frac {1}{3} x^3 \left (\frac {b-\sqrt {b^2-4 a c}+2 c x^2}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+b x^2+c x^4\right )^p \operatorname {AppellF1}\left (\frac {3}{2},-p,-p,\frac {5}{2},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right ) \] Input:

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

Output:

(x^3*(a + b*x^2 + c*x^4)^p*AppellF1[3/2, -p, -p, 5/2, (-2*c*x^2)/(b + Sqrt 
[b^2 - 4*a*c]), (2*c*x^2)/(-b + Sqrt[b^2 - 4*a*c])])/(3*((b - Sqrt[b^2 - 4 
*a*c] + 2*c*x^2)/(b - Sqrt[b^2 - 4*a*c]))^p*((b + Sqrt[b^2 - 4*a*c] + 2*c* 
x^2)/(b + Sqrt[b^2 - 4*a*c]))^p)
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {1461, 394}

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.

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

\(\Big \downarrow \) 1461

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

\(\Big \downarrow \) 394

\(\displaystyle \frac {1}{3} x^3 \left (\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1\right )^{-p} \left (\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1\right )^{-p} \left (a+b x^2+c x^4\right )^p \operatorname {AppellF1}\left (\frac {3}{2},-p,-p,\frac {5}{2},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )\)

Input:

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

Output:

(x^3*(a + b*x^2 + c*x^4)^p*AppellF1[3/2, -p, -p, 5/2, (-2*c*x^2)/(b - Sqrt 
[b^2 - 4*a*c]), (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])])/(3*(1 + (2*c*x^2)/(b 
- Sqrt[b^2 - 4*a*c]))^p*(1 + (2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]))^p)
 

Defintions of rubi rules used

rule 394

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/2 
, -p, -q, 1 + (m + 1)/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; FreeQ[{a, b, c, 
 d, e, m, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, 1] && (Int 
egerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 1461

Int[((d_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] 
:> Simp[a^IntPart[p]*((a + b*x^2 + c*x^4)^FracPart[p]/((1 + 2*c*(x^2/(b + R 
t[b^2 - 4*a*c, 2])))^FracPart[p]*(1 + 2*c*(x^2/(b - Rt[b^2 - 4*a*c, 2])))^F 
racPart[p]))   Int[(d*x)^m*(1 + 2*c*(x^2/(b + Sqrt[b^2 - 4*a*c])))^p*(1 + 2 
*c*(x^2/(b - Sqrt[b^2 - 4*a*c])))^p, x], x] /; FreeQ[{a, b, c, d, m, p}, x]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

Integral(x**2*(a + b*x**2 + c*x**4)**p, x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int x^2 \left (a+b x^2+c x^4\right )^p \, dx =\text {Too large to display} \] Input:

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

Output:

(2*(a + b*x**2 + c*x**4)**p*b*p*x + 4*(a + b*x**2 + c*x**4)**p*c*p*x**3 + 
(a + b*x**2 + c*x**4)**p*c*x**3 - 32*int((a + b*x**2 + c*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 + 16*c*p**2*x* 
*4 + 16*c*p*x**4 + 3*c*x**4),x)*a*b*p**3 - 32*int((a + b*x**2 + c*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 + 16* 
c*p**2*x**4 + 16*c*p*x**4 + 3*c*x**4),x)*a*b*p**2 - 6*int((a + b*x**2 + c* 
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 + 16*c*p**2*x**4 + 16*c*p*x**4 + 3*c*x**4),x)*a*b*p + 256*int(((a + b*x 
**2 + c*x**4)**p*x**2)/(16*a*p**2 + 16*a*p + 3*a + 16*b*p**2*x**2 + 16*b*p 
*x**2 + 3*b*x**2 + 16*c*p**2*x**4 + 16*c*p*x**4 + 3*c*x**4),x)*a*c*p**4 + 
320*int(((a + b*x**2 + c*x**4)**p*x**2)/(16*a*p**2 + 16*a*p + 3*a + 16*b*p 
**2*x**2 + 16*b*p*x**2 + 3*b*x**2 + 16*c*p**2*x**4 + 16*c*p*x**4 + 3*c*x** 
4),x)*a*c*p**3 + 112*int(((a + b*x**2 + c*x**4)**p*x**2)/(16*a*p**2 + 16*a 
*p + 3*a + 16*b*p**2*x**2 + 16*b*p*x**2 + 3*b*x**2 + 16*c*p**2*x**4 + 16*c 
*p*x**4 + 3*c*x**4),x)*a*c*p**2 + 12*int(((a + b*x**2 + c*x**4)**p*x**2)/( 
16*a*p**2 + 16*a*p + 3*a + 16*b*p**2*x**2 + 16*b*p*x**2 + 3*b*x**2 + 16*c* 
p**2*x**4 + 16*c*p*x**4 + 3*c*x**4),x)*a*c*p - 64*int(((a + b*x**2 + c*x** 
4)**p*x**2)/(16*a*p**2 + 16*a*p + 3*a + 16*b*p**2*x**2 + 16*b*p*x**2 + 3*b 
*x**2 + 16*c*p**2*x**4 + 16*c*p*x**4 + 3*c*x**4),x)*b**2*p**4 - 96*int(((a 
 + b*x**2 + c*x**4)**p*x**2)/(16*a*p**2 + 16*a*p + 3*a + 16*b*p**2*x**2...