Integrand size = 14, antiderivative size = 133 \[ \int \left (a+b x^2+c x^4\right )^p \, dx=x \left (1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+b x^2+c x^4\right )^p \operatorname {AppellF1}\left (\frac {1}{2},-p,-p,\frac {3}{2},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right ) \] Output:
x*(c*x^4+b*x^2+a)^p*AppellF1(1/2,-p,-p,3/2,-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)
Time = 0.15 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.21 \[ \int \left (a+b x^2+c x^4\right )^p \, dx=x \left (\frac {b-\sqrt {b^2-4 a c}+2 c x^2}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+b x^2+c x^4\right )^p \operatorname {AppellF1}\left (\frac {1}{2},-p,-p,\frac {3}{2},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right ) \] Input:
Integrate[(a + b*x^2 + c*x^4)^p,x]
Output:
(x*(a + b*x^2 + c*x^4)^p*AppellF1[1/2, -p, -p, 3/2, (-2*c*x^2)/(b + Sqrt[b ^2 - 4*a*c]), (2*c*x^2)/(-b + Sqrt[b^2 - 4*a*c])])/(((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)
Time = 0.40 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1418, 333}
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.
\(\displaystyle \int \left (a+b x^2+c x^4\right )^p \, dx\) |
\(\Big \downarrow \) 1418 |
\(\displaystyle \left (\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1\right )^{-p} \left (\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1\right )^{-p} \left (a+b x^2+c x^4\right )^p \int \left (\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1\right )^p \left (\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}+1\right )^pdx\) |
\(\Big \downarrow \) 333 |
\(\displaystyle x \left (\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1\right )^{-p} \left (\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1\right )^{-p} \left (a+b x^2+c x^4\right )^p \operatorname {AppellF1}\left (\frac {1}{2},-p,-p,\frac {3}{2},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )\) |
Input:
Int[(a + b*x^2 + c*x^4)^p,x]
Output:
(x*(a + b*x^2 + c*x^4)^p*AppellF1[1/2, -p, -p, 3/2, (-2*c*x^2)/(b - Sqrt[b ^2 - 4*a*c]), (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])])/((1 + (2*c*x^2)/(b - Sq rt[b^2 - 4*a*c]))^p*(1 + (2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]))^p)
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[a^p*c^q*x*AppellF1[1/2, -p, -q, 3/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; F reeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> With[{q = Rt[b^ 2 - 4*a*c, 2]}, Simp[a^IntPart[p]*((a + b*x^2 + c*x^4)^FracPart[p]/((1 + 2* c*(x^2/(b + q)))^FracPart[p]*(1 + 2*c*(x^2/(b - q)))^FracPart[p])) Int[(1 + 2*c*(x^2/(b + q)))^p*(1 + 2*c*(x^2/(b - q)))^p, x], x]] /; FreeQ[{a, b, c, p}, x] && NeQ[b^2 - 4*a*c, 0]
\[\int \left (c \,x^{4}+b \,x^{2}+a \right )^{p}d x\]
Input:
int((c*x^4+b*x^2+a)^p,x)
Output:
int((c*x^4+b*x^2+a)^p,x)
\[ \int \left (a+b x^2+c x^4\right )^p \, dx=\int { {\left (c x^{4} + b x^{2} + a\right )}^{p} \,d x } \] Input:
integrate((c*x^4+b*x^2+a)^p,x, algorithm="fricas")
Output:
integral((c*x^4 + b*x^2 + a)^p, x)
\[ \int \left (a+b x^2+c x^4\right )^p \, dx=\int \left (a + b x^{2} + c x^{4}\right )^{p}\, dx \] Input:
integrate((c*x**4+b*x**2+a)**p,x)
Output:
Integral((a + b*x**2 + c*x**4)**p, x)
\[ \int \left (a+b x^2+c x^4\right )^p \, dx=\int { {\left (c x^{4} + b x^{2} + a\right )}^{p} \,d x } \] Input:
integrate((c*x^4+b*x^2+a)^p,x, algorithm="maxima")
Output:
integrate((c*x^4 + b*x^2 + a)^p, x)
\[ \int \left (a+b x^2+c x^4\right )^p \, dx=\int { {\left (c x^{4} + b x^{2} + a\right )}^{p} \,d x } \] Input:
integrate((c*x^4+b*x^2+a)^p,x, algorithm="giac")
Output:
integrate((c*x^4 + b*x^2 + a)^p, x)
Timed out. \[ \int \left (a+b x^2+c x^4\right )^p \, dx=\int {\left (c\,x^4+b\,x^2+a\right )}^p \,d x \] Input:
int((a + b*x^2 + c*x^4)^p,x)
Output:
int((a + b*x^2 + c*x^4)^p, x)
\[ \int \left (a+b x^2+c x^4\right )^p \, dx=\frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{p} x +16 \left (\int \frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{p}}{4 c p \,x^{4}+c \,x^{4}+4 b p \,x^{2}+b \,x^{2}+4 a p +a}d x \right ) a \,p^{2}+4 \left (\int \frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{p}}{4 c p \,x^{4}+c \,x^{4}+4 b p \,x^{2}+b \,x^{2}+4 a p +a}d x \right ) a p +8 \left (\int \frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{p} x^{2}}{4 c p \,x^{4}+c \,x^{4}+4 b p \,x^{2}+b \,x^{2}+4 a p +a}d x \right ) b \,p^{2}+2 \left (\int \frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{p} x^{2}}{4 c p \,x^{4}+c \,x^{4}+4 b p \,x^{2}+b \,x^{2}+4 a p +a}d x \right ) b p}{4 p +1} \] Input:
int((c*x^4+b*x^2+a)^p,x)
Output:
((a + b*x**2 + c*x**4)**p*x + 16*int((a + b*x**2 + c*x**4)**p/(4*a*p + a + 4*b*p*x**2 + b*x**2 + 4*c*p*x**4 + c*x**4),x)*a*p**2 + 4*int((a + b*x**2 + c*x**4)**p/(4*a*p + a + 4*b*p*x**2 + b*x**2 + 4*c*p*x**4 + c*x**4),x)*a* p + 8*int(((a + b*x**2 + c*x**4)**p*x**2)/(4*a*p + a + 4*b*p*x**2 + b*x**2 + 4*c*p*x**4 + c*x**4),x)*b*p**2 + 2*int(((a + b*x**2 + c*x**4)**p*x**2)/ (4*a*p + a + 4*b*p*x**2 + b*x**2 + 4*c*p*x**4 + c*x**4),x)*b*p)/(4*p + 1)