\(\int x (a+b x^2+c x^4)^p \, dx\) [1095]

Optimal result

Integrand size = 16, antiderivative size = 126 \[ \int x \left (a+b x^2+c x^4\right )^p \, dx=-\frac {2^p \left (-\frac {b-\sqrt {b^2-4 a c}+2 c x^2}{\sqrt {b^2-4 a c}}\right )^{-1-p} \left (a+b x^2+c x^4\right )^{1+p} \operatorname {Hypergeometric2F1}\left (-p,1+p,2+p,\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{2 \sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} (1+p)} \] Output:

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

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.07 \[ \int x \left (a+b x^2+c x^4\right )^p \, dx=\frac {2^{-2+p} \left (b-\sqrt {b^2-4 a c}+2 c x^2\right ) \left (\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{\sqrt {b^2-4 a c}}\right )^{-p} \left (a+b x^2+c x^4\right )^p \operatorname {Hypergeometric2F1}\left (-p,1+p,2+p,\frac {-b+\sqrt {b^2-4 a c}-2 c x^2}{2 \sqrt {b^2-4 a c}}\right )}{c (1+p)} \] Input:

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

Output:

(2^(-2 + p)*(b - Sqrt[b^2 - 4*a*c] + 2*c*x^2)*(a + b*x^2 + c*x^4)^p*Hyperg 
eometric2F1[-p, 1 + p, 2 + p, (-b + Sqrt[b^2 - 4*a*c] - 2*c*x^2)/(2*Sqrt[b 
^2 - 4*a*c])])/(c*(1 + p)*((b + Sqrt[b^2 - 4*a*c] + 2*c*x^2)/Sqrt[b^2 - 4* 
a*c])^p)
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1432, 1096}

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.

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

\(\Big \downarrow \) 1432

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

\(\Big \downarrow \) 1096

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 1096

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Rt[b^2 
 - 4*a*c, 2]}, Simp[(-(a + b*x + c*x^2)^(p + 1)/(q*(p + 1)*((q - b - 2*c*x) 
/(2*q))^(p + 1)))*Hypergeometric2F1[-p, p + 1, p + 2, (b + q + 2*c*x)/(2*q) 
], x]] /; FreeQ[{a, b, c, p}, x] &&  !IntegerQ[4*p] &&  !IntegerQ[3*p]
 

rule 1432

Int[(x_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[1/2 
 Subst[Int[(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int x \left (a+b x^2+c x^4\right )^p \, dx=\frac {2 \left (c \,x^{4}+b \,x^{2}+a \right )^{p} a +\left (c \,x^{4}+b \,x^{2}+a \right )^{p} b \,x^{2}-16 \left (\int \frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{p} x^{3}}{2 c p \,x^{4}+c \,x^{4}+2 b p \,x^{2}+b \,x^{2}+2 a p +a}d x \right ) a c \,p^{2}-8 \left (\int \frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{p} x^{3}}{2 c p \,x^{4}+c \,x^{4}+2 b p \,x^{2}+b \,x^{2}+2 a p +a}d x \right ) a c p +4 \left (\int \frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{p} x^{3}}{2 c p \,x^{4}+c \,x^{4}+2 b p \,x^{2}+b \,x^{2}+2 a p +a}d x \right ) b^{2} p^{2}+2 \left (\int \frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{p} x^{3}}{2 c p \,x^{4}+c \,x^{4}+2 b p \,x^{2}+b \,x^{2}+2 a p +a}d x \right ) b^{2} p}{2 b \left (2 p +1\right )} \] Input:

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

Output:

(2*(a + b*x**2 + c*x**4)**p*a + (a + b*x**2 + c*x**4)**p*b*x**2 - 16*int(( 
(a + b*x**2 + c*x**4)**p*x**3)/(2*a*p + a + 2*b*p*x**2 + b*x**2 + 2*c*p*x* 
*4 + c*x**4),x)*a*c*p**2 - 8*int(((a + b*x**2 + c*x**4)**p*x**3)/(2*a*p + 
a + 2*b*p*x**2 + b*x**2 + 2*c*p*x**4 + c*x**4),x)*a*c*p + 4*int(((a + b*x* 
*2 + c*x**4)**p*x**3)/(2*a*p + a + 2*b*p*x**2 + b*x**2 + 2*c*p*x**4 + c*x* 
*4),x)*b**2*p**2 + 2*int(((a + b*x**2 + c*x**4)**p*x**3)/(2*a*p + a + 2*b* 
p*x**2 + b*x**2 + 2*c*p*x**4 + c*x**4),x)*b**2*p)/(2*b*(2*p + 1))