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