\(\int x^3 (a+b x^2+c x^4)^p \, dx\) [1094]

Optimal result

Integrand size = 18, antiderivative size = 160 \[ \int x^3 \left (a+b x^2+c x^4\right )^p \, dx=\frac {\left (a+b x^2+c x^4\right )^{1+p}}{4 c (1+p)}+\frac {2^{-1+p} b \left (-\frac {b-\sqrt {b^2-4 a c}+2 c x^2}{\sqrt {b^2-4 a c}}\right )^{-1-p} \left (a+b x^2+c x^4\right )^{1+p} \operatorname {Hypergeometric2F1}\left (-p,1+p,2+p,\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{2 \sqrt {b^2-4 a c}}\right )}{c \sqrt {b^2-4 a c} (1+p)} \] Output:

1/4*(c*x^4+b*x^2+a)^(p+1)/c/(p+1)+2^(-1+p)*b*(-(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))/c/(-4*a*c+b^2) 
^(1/2)/(p+1)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.

Time = 0.27 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.01 \[ \int x^3 \left (a+b x^2+c x^4\right )^p \, dx=\frac {1}{4} x^4 \left (\frac {b-\sqrt {b^2-4 a c}+2 c x^2}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+b x^2+c x^4\right )^p \operatorname {AppellF1}\left (2,-p,-p,3,-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right ) \] Input:

Integrate[x^3*(a + b*x^2 + c*x^4)^p,x]
 

Output:

(x^4*(a + b*x^2 + c*x^4)^p*AppellF1[2, -p, -p, 3, (-2*c*x^2)/(b + Sqrt[b^2 
 - 4*a*c]), (2*c*x^2)/(-b + Sqrt[b^2 - 4*a*c])])/(4*((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 = 0.42 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.01, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1434, 1160, 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.

\(\displaystyle \int x^3 \left (a+b x^2+c x^4\right )^p \, dx\)

\(\Big \downarrow \) 1434

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

\(\Big \downarrow \) 1160

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

\(\Big \downarrow \) 1096

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

Input:

Int[x^3*(a + b*x^2 + c*x^4)^p,x]
 

Output:

((a + b*x^2 + c*x^4)^(1 + p)/(2*c*(1 + p)) + (2^p*b*(-((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)*Hy 
pergeometric2F1[-p, 1 + p, 2 + p, (b + Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(2*Sqr 
t[b^2 - 4*a*c])])/(c*Sqrt[b^2 - 4*a*c]*(1 + p)))/2
 

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 1160

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

rule 1434

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

Maple [F]

\[\int x^{3} \left (c \,x^{4}+b \,x^{2}+a \right )^{p}d x\]

Input:

int(x^3*(c*x^4+b*x^2+a)^p,x)
 

Output:

int(x^3*(c*x^4+b*x^2+a)^p,x)
 

Fricas [F]

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

integrate(x^3*(c*x^4+b*x^2+a)^p,x, algorithm="fricas")
 

Output:

integral((c*x^4 + b*x^2 + a)^p*x^3, x)
 

Sympy [F]

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

integrate(x**3*(c*x**4+b*x**2+a)**p,x)
 

Output:

Integral(x**3*(a + b*x**2 + c*x**4)**p, x)
 

Maxima [F]

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

integrate(x^3*(c*x^4+b*x^2+a)^p,x, algorithm="maxima")
 

Output:

integrate((c*x^4 + b*x^2 + a)^p*x^3, x)
 

Giac [F]

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

integrate(x^3*(c*x^4+b*x^2+a)^p,x, algorithm="giac")
 

Output:

integrate((c*x^4 + b*x^2 + a)^p*x^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int x^3 \left (a+b x^2+c x^4\right )^p \, dx=\int x^3\,{\left (c\,x^4+b\,x^2+a\right )}^p \,d x \] Input:

int(x^3*(a + b*x^2 + c*x^4)^p,x)
 

Output:

int(x^3*(a + b*x^2 + c*x^4)^p, x)
 

Reduce [F]

\[ \int x^3 \left (a+b x^2+c x^4\right )^p \, dx=\frac {-\left (c \,x^{4}+b \,x^{2}+a \right )^{p} a +\left (c \,x^{4}+b \,x^{2}+a \right )^{p} b p \,x^{2}+2 \left (c \,x^{4}+b \,x^{2}+a \right )^{p} c p \,x^{4}+\left (c \,x^{4}+b \,x^{2}+a \right )^{p} c \,x^{4}+16 \left (\int \frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{p} x^{3}}{2 c p \,x^{4}+c \,x^{4}+2 b p \,x^{2}+b \,x^{2}+2 a p +a}d x \right ) a c \,p^{3}+24 \left (\int \frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{p} x^{3}}{2 c p \,x^{4}+c \,x^{4}+2 b p \,x^{2}+b \,x^{2}+2 a p +a}d x \right ) a c \,p^{2}+8 \left (\int \frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{p} x^{3}}{2 c p \,x^{4}+c \,x^{4}+2 b p \,x^{2}+b \,x^{2}+2 a p +a}d x \right ) a c p -4 \left (\int \frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{p} x^{3}}{2 c p \,x^{4}+c \,x^{4}+2 b p \,x^{2}+b \,x^{2}+2 a p +a}d x \right ) b^{2} p^{3}-6 \left (\int \frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{p} x^{3}}{2 c p \,x^{4}+c \,x^{4}+2 b p \,x^{2}+b \,x^{2}+2 a p +a}d x \right ) b^{2} p^{2}-2 \left (\int \frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{p} x^{3}}{2 c p \,x^{4}+c \,x^{4}+2 b p \,x^{2}+b \,x^{2}+2 a p +a}d x \right ) b^{2} p}{4 c \left (2 p^{2}+3 p +1\right )} \] Input:

int(x^3*(c*x^4+b*x^2+a)^p,x)
 

Output:

( - (a + b*x**2 + c*x**4)**p*a + (a + b*x**2 + c*x**4)**p*b*p*x**2 + 2*(a 
+ b*x**2 + c*x**4)**p*c*p*x**4 + (a + b*x**2 + c*x**4)**p*c*x**4 + 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**3 + 24*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**3 - 6*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)/(4*c*(2*p**2 + 3*p + 1))