\(\int x^5 (a+b x^3+c x^6)^p \, dx\) [258]

Optimal result

Integrand size = 18, antiderivative size = 161 \[ \int x^5 \left (a+b x^3+c x^6\right )^p \, dx=\frac {\left (a+b x^3+c x^6\right )^{1+p}}{6 c (1+p)}+\frac {2^p b \left (-\frac {b-\sqrt {b^2-4 a c}+2 c x^3}{\sqrt {b^2-4 a c}}\right )^{-1-p} \left (a+b x^3+c x^6\right )^{1+p} \operatorname {Hypergeometric2F1}\left (-p,1+p,2+p,\frac {b+\sqrt {b^2-4 a c}+2 c x^3}{2 \sqrt {b^2-4 a c}}\right )}{3 c \sqrt {b^2-4 a c} (1+p)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1371, 654, 638} \[ \int x^5 \left (a+b x^3+c x^6\right )^p \, dx=\frac {b 2^p \left (a+b x^3+c x^6\right )^{p+1} \left (-\frac {-\sqrt {b^2-4 a c}+b+2 c x^3}{\sqrt {b^2-4 a c}}\right )^{-p-1} \operatorname {Hypergeometric2F1}\left (-p,p+1,p+2,\frac {2 c x^3+b+\sqrt {b^2-4 a c}}{2 \sqrt {b^2-4 a c}}\right )}{3 c (p+1) \sqrt {b^2-4 a c}}+\frac {\left (a+b x^3+c x^6\right )^{p+1}}{6 c (p+1)} \]

[In]

Int[x^5*(a + b*x^3 + c*x^6)^p,x]

[Out]

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

Rule 638

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] && NeQ[b^2 - 4*a*c, 0] &&  !IntegerQ[4*p]

Rule 654

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 1371

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]]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \text {Subst}\left (\int x \left (a+b x+c x^2\right )^p \, dx,x,x^3\right ) \\ & = \frac {\left (a+b x^3+c x^6\right )^{1+p}}{6 c (1+p)}-\frac {b \text {Subst}\left (\int \left (a+b x+c x^2\right )^p \, dx,x,x^3\right )}{6 c} \\ & = \frac {\left (a+b x^3+c x^6\right )^{1+p}}{6 c (1+p)}+\frac {2^p b \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 \sqrt {b^2-4 a c} (1+p)} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.

Time = 0.27 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.01 \[ \int x^5 \left (a+b x^3+c x^6\right )^p \, dx=\frac {1}{6} x^6 \left (\frac {b-\sqrt {b^2-4 a c}+2 c x^3}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x^3}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+b x^3+c x^6\right )^p \operatorname {AppellF1}\left (2,-p,-p,3,-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}},\frac {2 c x^3}{-b+\sqrt {b^2-4 a c}}\right ) \]

[In]

Integrate[x^5*(a + b*x^3 + c*x^6)^p,x]

[Out]

(x^6*(a + b*x^3 + c*x^6)^p*AppellF1[2, -p, -p, 3, (-2*c*x^3)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^3)/(-b + Sqrt[b^2
- 4*a*c])])/(6*((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]

\[\int x^{5} \left (c \,x^{6}+b \,x^{3}+a \right )^{p}d x\]

[In]

int(x^5*(c*x^6+b*x^3+a)^p,x)

[Out]

int(x^5*(c*x^6+b*x^3+a)^p,x)

Fricas [F]

\[ \int x^5 \left (a+b x^3+c x^6\right )^p \, dx=\int { {\left (c x^{6} + b x^{3} + a\right )}^{p} x^{5} \,d x } \]

[In]

integrate(x^5*(c*x^6+b*x^3+a)^p,x, algorithm="fricas")

[Out]

integral((c*x^6 + b*x^3 + a)^p*x^5, x)

Sympy [F(-1)]

Timed out. \[ \int x^5 \left (a+b x^3+c x^6\right )^p \, dx=\text {Timed out} \]

[In]

integrate(x**5*(c*x**6+b*x**3+a)**p,x)

[Out]

Timed out

Maxima [F]

\[ \int x^5 \left (a+b x^3+c x^6\right )^p \, dx=\int { {\left (c x^{6} + b x^{3} + a\right )}^{p} x^{5} \,d x } \]

[In]

integrate(x^5*(c*x^6+b*x^3+a)^p,x, algorithm="maxima")

[Out]

integrate((c*x^6 + b*x^3 + a)^p*x^5, x)

Giac [F]

\[ \int x^5 \left (a+b x^3+c x^6\right )^p \, dx=\int { {\left (c x^{6} + b x^{3} + a\right )}^{p} x^{5} \,d x } \]

[In]

integrate(x^5*(c*x^6+b*x^3+a)^p,x, algorithm="giac")

[Out]

integrate((c*x^6 + b*x^3 + a)^p*x^5, x)

Mupad [F(-1)]

Timed out. \[ \int x^5 \left (a+b x^3+c x^6\right )^p \, dx=\int x^5\,{\left (c\,x^6+b\,x^3+a\right )}^p \,d x \]

[In]

int(x^5*(a + b*x^3 + c*x^6)^p,x)

[Out]

int(x^5*(a + b*x^3 + c*x^6)^p, x)