Integrand size = 22, antiderivative size = 274 \[ \int \left (c+e x^2\right ) \left (a+c x^2+b x^4\right )^p \, dx=c x \left (1+\frac {2 b x^2}{c-\sqrt {-4 a b+c^2}}\right )^{-p} \left (1+\frac {2 b x^2}{c+\sqrt {-4 a b+c^2}}\right )^{-p} \left (a+c x^2+b x^4\right )^p \operatorname {AppellF1}\left (\frac {1}{2},-p,-p,\frac {3}{2},-\frac {2 b x^2}{c-\sqrt {-4 a b+c^2}},-\frac {2 b x^2}{c+\sqrt {-4 a b+c^2}}\right )+\frac {1}{3} e x^3 \left (1+\frac {2 b x^2}{c-\sqrt {-4 a b+c^2}}\right )^{-p} \left (1+\frac {2 b x^2}{c+\sqrt {-4 a b+c^2}}\right )^{-p} \left (a+c x^2+b x^4\right )^p \operatorname {AppellF1}\left (\frac {3}{2},-p,-p,\frac {5}{2},-\frac {2 b x^2}{c-\sqrt {-4 a b+c^2}},-\frac {2 b x^2}{c+\sqrt {-4 a b+c^2}}\right ) \]
c*x*(b*x^4+c*x^2+a)^p*AppellF1(1/2,-p,-p,3/2,-2*b*x^2/(c-(-4*a*b+c^2)^(1/2 )),-2*b*x^2/(c+(-4*a*b+c^2)^(1/2)))/((1+2*b*x^2/(c-(-4*a*b+c^2)^(1/2)))^p) /((1+2*b*x^2/(c+(-4*a*b+c^2)^(1/2)))^p)+1/3*e*x^3*(b*x^4+c*x^2+a)^p*Appell F1(3/2,-p,-p,5/2,-2*b*x^2/(c-(-4*a*b+c^2)^(1/2)),-2*b*x^2/(c+(-4*a*b+c^2)^ (1/2)))/((1+2*b*x^2/(c-(-4*a*b+c^2)^(1/2)))^p)/((1+2*b*x^2/(c+(-4*a*b+c^2) ^(1/2)))^p)
Time = 0.48 (sec) , antiderivative size = 232, normalized size of antiderivative = 0.85 \[ \int \left (c+e x^2\right ) \left (a+c x^2+b x^4\right )^p \, dx=\frac {1}{3} x \left (\frac {c-\sqrt {-4 a b+c^2}+2 b x^2}{c-\sqrt {-4 a b+c^2}}\right )^{-p} \left (\frac {c+\sqrt {-4 a b+c^2}+2 b x^2}{c+\sqrt {-4 a b+c^2}}\right )^{-p} \left (a+c x^2+b x^4\right )^p \left (3 c \operatorname {AppellF1}\left (\frac {1}{2},-p,-p,\frac {3}{2},-\frac {2 b x^2}{c+\sqrt {-4 a b+c^2}},\frac {2 b x^2}{-c+\sqrt {-4 a b+c^2}}\right )+e x^2 \operatorname {AppellF1}\left (\frac {3}{2},-p,-p,\frac {5}{2},-\frac {2 b x^2}{c+\sqrt {-4 a b+c^2}},\frac {2 b x^2}{-c+\sqrt {-4 a b+c^2}}\right )\right ) \]
(x*(a + c*x^2 + b*x^4)^p*(3*c*AppellF1[1/2, -p, -p, 3/2, (-2*b*x^2)/(c + S qrt[-4*a*b + c^2]), (2*b*x^2)/(-c + Sqrt[-4*a*b + c^2])] + e*x^2*AppellF1[ 3/2, -p, -p, 5/2, (-2*b*x^2)/(c + Sqrt[-4*a*b + c^2]), (2*b*x^2)/(-c + Sqr t[-4*a*b + c^2])]))/(3*((c - Sqrt[-4*a*b + c^2] + 2*b*x^2)/(c - Sqrt[-4*a* b + c^2]))^p*((c + Sqrt[-4*a*b + c^2] + 2*b*x^2)/(c + Sqrt[-4*a*b + c^2])) ^p)
Time = 0.38 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1515, 2009}
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.
\(\displaystyle \int \left (c+e x^2\right ) \left (a+b x^4+c x^2\right )^p \, dx\) |
\(\Big \downarrow \) 1515 |
\(\displaystyle \int \left (e x^2 \left (a+b x^4+c x^2\right )^p+c \left (a+b x^4+c x^2\right )^p\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} e x^3 \left (\frac {2 b x^2}{c-\sqrt {c^2-4 a b}}+1\right )^{-p} \left (a+b x^4+c x^2\right )^p \left (\frac {2 b x^2}{\sqrt {c^2-4 a b}+c}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{2},-p,-p,\frac {5}{2},-\frac {2 b x^2}{c-\sqrt {c^2-4 a b}},-\frac {2 b x^2}{c+\sqrt {c^2-4 a b}}\right )+c x \left (\frac {2 b x^2}{c-\sqrt {c^2-4 a b}}+1\right )^{-p} \left (a+b x^4+c x^2\right )^p \left (\frac {2 b x^2}{\sqrt {c^2-4 a b}+c}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,-p,\frac {3}{2},-\frac {2 b x^2}{c-\sqrt {c^2-4 a b}},-\frac {2 b x^2}{c+\sqrt {c^2-4 a b}}\right )\) |
(c*x*(a + c*x^2 + b*x^4)^p*AppellF1[1/2, -p, -p, 3/2, (-2*b*x^2)/(c - Sqrt [-4*a*b + c^2]), (-2*b*x^2)/(c + Sqrt[-4*a*b + c^2])])/((1 + (2*b*x^2)/(c - Sqrt[-4*a*b + c^2]))^p*(1 + (2*b*x^2)/(c + Sqrt[-4*a*b + c^2]))^p) + (e* x^3*(a + c*x^2 + b*x^4)^p*AppellF1[3/2, -p, -p, 5/2, (-2*b*x^2)/(c - Sqrt[ -4*a*b + c^2]), (-2*b*x^2)/(c + Sqrt[-4*a*b + c^2])])/(3*(1 + (2*b*x^2)/(c - Sqrt[-4*a*b + c^2]))^p*(1 + (2*b*x^2)/(c + Sqrt[-4*a*b + c^2]))^p)
3.5.2.3.1 Defintions of rubi rules used
Int[((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symb ol] :> Int[ExpandIntegrand[(d + e*x^2)*(a + b*x^2 + c*x^4)^p, x], x] /; Fre eQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
\[\int \left (e \,x^{2}+c \right ) \left (b \,x^{4}+c \,x^{2}+a \right )^{p}d x\]
\[ \int \left (c+e x^2\right ) \left (a+c x^2+b x^4\right )^p \, dx=\int { {\left (e x^{2} + c\right )} {\left (b x^{4} + c x^{2} + a\right )}^{p} \,d x } \]
\[ \int \left (c+e x^2\right ) \left (a+c x^2+b x^4\right )^p \, dx=\int \left (c + e x^{2}\right ) \left (a + b x^{4} + c x^{2}\right )^{p}\, dx \]
\[ \int \left (c+e x^2\right ) \left (a+c x^2+b x^4\right )^p \, dx=\int { {\left (e x^{2} + c\right )} {\left (b x^{4} + c x^{2} + a\right )}^{p} \,d x } \]
\[ \int \left (c+e x^2\right ) \left (a+c x^2+b x^4\right )^p \, dx=\int { {\left (e x^{2} + c\right )} {\left (b x^{4} + c x^{2} + a\right )}^{p} \,d x } \]
Timed out. \[ \int \left (c+e x^2\right ) \left (a+c x^2+b x^4\right )^p \, dx=\int \left (e\,x^2+c\right )\,{\left (b\,x^4+c\,x^2+a\right )}^p \,d x \]