3.257 \(\int x^8 (a+b x^3+c x^6)^p \, dx\)

Optimal. Leaf size=224 \[ \frac{2^p \left (2 a c-b^2 (p+2)\right ) \left (a+b x^3+c x^6\right )^{p+1} \left (-\frac{-\sqrt{b^2-4 a c}+b+2 c x^3}{\sqrt{b^2-4 a c}}\right )^{-p-1} \, _2F_1\left (-p,p+1;p+2;\frac{2 c x^3+b+\sqrt{b^2-4 a c}}{2 \sqrt{b^2-4 a c}}\right )}{3 c^2 (p+1) (2 p+3) \sqrt{b^2-4 a c}}-\frac{b (p+2) \left (a+b x^3+c x^6\right )^{p+1}}{6 c^2 (p+1) (2 p+3)}+\frac{x^3 \left (a+b x^3+c x^6\right )^{p+1}}{3 c (2 p+3)} \]

[Out]

-(b*(2 + p)*(a + b*x^3 + c*x^6)^(1 + p))/(6*c^2*(1 + p)*(3 + 2*p)) + (x^3*(a + b*x^3 + c*x^6)^(1 + p))/(3*c*(3
+ 2*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])])/(3*c^2*Sqrt[b^2 - 4*a*c]*(1 + p)*(3 + 2*p))

________________________________________________________________________________________

Rubi [A]  time = 0.243492, antiderivative size = 224, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {1357, 742, 640, 624} \[ \frac{2^p \left (2 a c-b^2 (p+2)\right ) \left (a+b x^3+c x^6\right )^{p+1} \left (-\frac{-\sqrt{b^2-4 a c}+b+2 c x^3}{\sqrt{b^2-4 a c}}\right )^{-p-1} \, _2F_1\left (-p,p+1;p+2;\frac{2 c x^3+b+\sqrt{b^2-4 a c}}{2 \sqrt{b^2-4 a c}}\right )}{3 c^2 (p+1) (2 p+3) \sqrt{b^2-4 a c}}-\frac{b (p+2) \left (a+b x^3+c x^6\right )^{p+1}}{6 c^2 (p+1) (2 p+3)}+\frac{x^3 \left (a+b x^3+c x^6\right )^{p+1}}{3 c (2 p+3)} \]

Antiderivative was successfully verified.

[In]

Int[x^8*(a + b*x^3 + c*x^6)^p,x]

[Out]

-(b*(2 + p)*(a + b*x^3 + c*x^6)^(1 + p))/(6*c^2*(1 + p)*(3 + 2*p)) + (x^3*(a + b*x^3 + c*x^6)^(1 + p))/(3*c*(3
+ 2*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])])/(3*c^2*Sqrt[b^2 - 4*a*c]*(1 + p)*(3 + 2*p))

Rule 1357

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(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] && NeQ[
b^2 - 4*a*c, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 742

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 1)), x] + Dist[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] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0]
&& If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, b, c, d, e, m,
p, x]

Rule 640

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] + Dist[(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[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 624

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)*Hypergeometric2F1[-p, p + 1, p + 2, (b + q + 2*c*x)/(2*q)])/(q*(p + 1)*((q - b - 2*c*x)/(2*q))^(p
+ 1)), x]] /; FreeQ[{a, b, c, p}, x] && NeQ[b^2 - 4*a*c, 0] &&  !IntegerQ[4*p]

Rubi steps

\begin{align*} \int x^8 \left (a+b x^3+c x^6\right )^p \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int x^2 \left (a+b x+c x^2\right )^p \, dx,x,x^3\right )\\ &=\frac{x^3 \left (a+b x^3+c x^6\right )^{1+p}}{3 c (3+2 p)}+\frac{\operatorname{Subst}\left (\int (-a-b (2+p) x) \left (a+b x+c x^2\right )^p \, dx,x,x^3\right )}{3 c (3+2 p)}\\ &=-\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{\left (2 a c-b^2 (2+p)\right ) \operatorname{Subst}\left (\int \left (a+b x+c x^2\right )^p \, dx,x,x^3\right )}{6 c^2 (3+2 p)}\\ &=-\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} \, _2F_1\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)}\\ \end{align*}

Mathematica [C]  time = 0.203059, size = 162, normalized size = 0.72 \[ \frac{1}{9} x^9 \left (\frac{-\sqrt{b^2-4 a c}+b+2 c x^3}{b-\sqrt{b^2-4 a c}}\right )^{-p} \left (\frac{\sqrt{b^2-4 a c}+b+2 c x^3}{\sqrt{b^2-4 a c}+b}\right )^{-p} \left (a+b x^3+c x^6\right )^p F_1\left (3;-p,-p;4;-\frac{2 c x^3}{b+\sqrt{b^2-4 a c}},\frac{2 c x^3}{\sqrt{b^2-4 a c}-b}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^8*(a + b*x^3 + c*x^6)^p,x]

[Out]

(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)

________________________________________________________________________________________

Maple [F]  time = 0.02, size = 0, normalized size = 0. \begin{align*} \int{x}^{8} \left ( c{x}^{6}+b{x}^{3}+a \right ) ^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8*(c*x^6+b*x^3+a)^p,x)

[Out]

int(x^8*(c*x^6+b*x^3+a)^p,x)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (c x^{6} + b x^{3} + a\right )}^{p} x^{8}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(c*x^6+b*x^3+a)^p,x, algorithm="maxima")

[Out]

integrate((c*x^6 + b*x^3 + a)^p*x^8, x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (c x^{6} + b x^{3} + a\right )}^{p} x^{8}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(c*x^6+b*x^3+a)^p,x, algorithm="fricas")

[Out]

integral((c*x^6 + b*x^3 + a)^p*x^8, x)

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**8*(c*x**6+b*x**3+a)**p,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (c x^{6} + b x^{3} + a\right )}^{p} x^{8}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(c*x^6+b*x^3+a)^p,x, algorithm="giac")

[Out]

integrate((c*x^6 + b*x^3 + a)^p*x^8, x)