Integrand size = 18, antiderivative size = 224 \[ \int x^8 \left (a+b x^3+c x^6\right )^p \, dx=-\frac {b (2+p) \left (a+b x^3+c x^6\right )^{1+p}}{6 c^2 (1+p) (3+2 p)}+\frac {x^3 \left (a+b x^3+c x^6\right )^{1+p}}{3 c (3+2 p)}+\frac {2^p \left (2 a c-b^2 (2+p)\right ) \left (-\frac {b-\sqrt {b^2-4 a c}+2 c x^3}{\sqrt {b^2-4 a c}}\right )^{-1-p} \left (a+b x^3+c x^6\right )^{1+p} \operatorname {Hypergeometric2F1}\left (-p,1+p,2+p,\frac {b+\sqrt {b^2-4 a c}+2 c x^3}{2 \sqrt {b^2-4 a c}}\right )}{3 c^2 \sqrt {b^2-4 a c} (1+p) (3+2 p)} \]
-1/6*b*(2+p)*(c*x^6+b*x^3+a)^(p+1)/c^2/(2*p^2+5*p+3)+1/3*x^3*(c*x^6+b*x^3+ a)^(p+1)/c/(3+2*p)+1/3*2^p*(2*a*c-b^2*(2+p))*(c*x^6+b*x^3+a)^(p+1)*hyperge om([-p, p+1],[2+p],1/2*(b+2*c*x^3+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))* ((-b-2*c*x^3+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(-1-p)/c^2/(p+1)/(3+2 *p)/(-4*a*c+b^2)^(1/2)
Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
Time = 0.33 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.72 \[ \int x^8 \left (a+b x^3+c x^6\right )^p \, dx=\frac {1}{9} x^9 \left (\frac {b-\sqrt {b^2-4 a c}+2 c x^3}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x^3}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+b x^3+c x^6\right )^p \operatorname {AppellF1}\left (3,-p,-p,4,-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}},\frac {2 c x^3}{-b+\sqrt {b^2-4 a c}}\right ) \]
(x^9*(a + b*x^3 + c*x^6)^p*AppellF1[3, -p, -p, 4, (-2*c*x^3)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^3)/(-b + Sqrt[b^2 - 4*a*c])])/(9*((b - Sqrt[b^2 - 4*a*c ] + 2*c*x^3)/(b - Sqrt[b^2 - 4*a*c]))^p*((b + Sqrt[b^2 - 4*a*c] + 2*c*x^3) /(b + Sqrt[b^2 - 4*a*c]))^p)
Time = 0.34 (sec) , antiderivative size = 222, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {1693, 1166, 25, 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.
\(\displaystyle \int x^8 \left (a+b x^3+c x^6\right )^p \, dx\) |
\(\Big \downarrow \) 1693 |
\(\displaystyle \frac {1}{3} \int x^6 \left (c x^6+b x^3+a\right )^pdx^3\) |
\(\Big \downarrow \) 1166 |
\(\displaystyle \frac {1}{3} \left (\frac {\int -\left (\left (b (p+2) x^3+a\right ) \left (c x^6+b x^3+a\right )^p\right )dx^3}{c (2 p+3)}+\frac {x^3 \left (a+b x^3+c x^6\right )^{p+1}}{c (2 p+3)}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{3} \left (\frac {x^3 \left (a+b x^3+c x^6\right )^{p+1}}{c (2 p+3)}-\frac {\int \left (b (p+2) x^3+a\right ) \left (c x^6+b x^3+a\right )^pdx^3}{c (2 p+3)}\right )\) |
\(\Big \downarrow \) 1160 |
\(\displaystyle \frac {1}{3} \left (\frac {x^3 \left (a+b x^3+c x^6\right )^{p+1}}{c (2 p+3)}-\frac {\frac {\left (2 a c-b^2 (p+2)\right ) \int \left (c x^6+b x^3+a\right )^pdx^3}{2 c}+\frac {b (p+2) \left (a+b x^3+c x^6\right )^{p+1}}{2 c (p+1)}}{c (2 p+3)}\right )\) |
\(\Big \downarrow \) 1096 |
\(\displaystyle \frac {1}{3} \left (\frac {x^3 \left (a+b x^3+c x^6\right )^{p+1}}{c (2 p+3)}-\frac {\frac {b (p+2) \left (a+b x^3+c x^6\right )^{p+1}}{2 c (p+1)}-\frac {2^p \left (2 a c-b^2 (p+2)\right ) \left (-\frac {-\sqrt {b^2-4 a c}+b+2 c x^3}{\sqrt {b^2-4 a c}}\right )^{-p-1} \left (a+b x^3+c x^6\right )^{p+1} \operatorname {Hypergeometric2F1}\left (-p,p+1,p+2,\frac {2 c x^3+b+\sqrt {b^2-4 a c}}{2 \sqrt {b^2-4 a c}}\right )}{c (p+1) \sqrt {b^2-4 a c}}}{c (2 p+3)}\right )\) |
((x^3*(a + b*x^3 + c*x^6)^(1 + p))/(c*(3 + 2*p)) - ((b*(2 + p)*(a + b*x^3 + c*x^6)^(1 + p))/(2*c*(1 + p)) - (2^p*(2*a*c - b^2*(2 + p))*(-((b - Sqrt[ b^2 - 4*a*c] + 2*c*x^3)/Sqrt[b^2 - 4*a*c]))^(-1 - p)*(a + b*x^3 + c*x^6)^( 1 + p)*Hypergeometric2F1[-p, 1 + p, 2 + p, (b + Sqrt[b^2 - 4*a*c] + 2*c*x^ 3)/(2*Sqrt[b^2 - 4*a*c])])/(c*Sqrt[b^2 - 4*a*c]*(1 + p)))/(c*(3 + 2*p)))/3
3.3.57.3.1 Defintions of rubi rules used
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[((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]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[e*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 1))), x] + Simp[1/(c*(m + 2*p + 1)) Int[(d + e*x)^(m - 2)*Simp[c*d^2*(m + 2*p + 1) - e*(a*e*(m - 1) + b*d*(p + 1)) + e*(2*c*d - b*e)*(m + p)*x, x]* (a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && If[Ration alQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadrat icQ[a, b, c, d, e, m, p, x]
Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol ] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[n2, 2*n] && IntegerQ [Simplify[(m + 1)/n]]
\[\int x^{8} \left (c \,x^{6}+b \,x^{3}+a \right )^{p}d x\]
\[ \int x^8 \left (a+b x^3+c x^6\right )^p \, dx=\int { {\left (c x^{6} + b x^{3} + a\right )}^{p} x^{8} \,d x } \]
Timed out. \[ \int x^8 \left (a+b x^3+c x^6\right )^p \, dx=\text {Timed out} \]
\[ \int x^8 \left (a+b x^3+c x^6\right )^p \, dx=\int { {\left (c x^{6} + b x^{3} + a\right )}^{p} x^{8} \,d x } \]
\[ \int x^8 \left (a+b x^3+c x^6\right )^p \, dx=\int { {\left (c x^{6} + b x^{3} + a\right )}^{p} x^{8} \,d x } \]
Timed out. \[ \int x^8 \left (a+b x^3+c x^6\right )^p \, dx=\int x^8\,{\left (c\,x^6+b\,x^3+a\right )}^p \,d x \]