\(\int (a^2+2 a b x^3+b^2 x^6)^p \, dx\) [132]

Optimal result

Integrand size = 20, antiderivative size = 53 \[ \int \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx=\frac {x \left (a+b x^3\right ) \left (a^2+2 a b x^3+b^2 x^6\right )^p \operatorname {Hypergeometric2F1}\left (1,\frac {4}{3}+2 p,\frac {4}{3},-\frac {b x^3}{a}\right )}{a} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.04, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {1357, 252, 251} \[ \int \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx=x \left (\frac {b x^3}{a}+1\right )^{-2 p} \left (a^2+2 a b x^3+b^2 x^6\right )^p \operatorname {Hypergeometric2F1}\left (\frac {1}{3},-2 p,\frac {4}{3},-\frac {b x^3}{a}\right ) \]

[In]

Int[(a^2 + 2*a*b*x^3 + b^2*x^6)^p,x]

[Out]

(x*(a^2 + 2*a*b*x^3 + b^2*x^6)^p*Hypergeometric2F1[1/3, -2*p, 4/3, -((b*x^3)/a)])/(1 + (b*x^3)/a)^(2*p)

Rule 251

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-p, 1/n, 1/n + 1, (-b)*(x^n/a)],
x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p
] || GtQ[a, 0])

Rule 252

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^Fra
cPart[p]), Int[(1 + b*(x^n/a))^p, x], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILt
Q[Simplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 1357

Int[((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_), x_Symbol] :> Dist[(a + b*x^n + c*x^(2*n))^p/(b + 2*c*x
^n)^(2*p), Int[(b + 2*c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0]

Rubi steps \begin{align*} \text {integral}& = \left (\left (2 a b+2 b^2 x^3\right )^{-2 p} \left (a^2+2 a b x^3+b^2 x^6\right )^p\right ) \int \left (2 a b+2 b^2 x^3\right )^{2 p} \, dx \\ & = \left (\left (1+\frac {b x^3}{a}\right )^{-2 p} \left (a^2+2 a b x^3+b^2 x^6\right )^p\right ) \int \left (1+\frac {b x^3}{a}\right )^{2 p} \, dx \\ & = x \left (1+\frac {b x^3}{a}\right )^{-2 p} \left (a^2+2 a b x^3+b^2 x^6\right )^p \, _2F_1\left (\frac {1}{3},-2 p;\frac {4}{3};-\frac {b x^3}{a}\right ) \\ \end{align*}

Mathematica [C] (warning: unable to verify)

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

Time = 0.16 (sec) , antiderivative size = 211, normalized size of antiderivative = 3.98 \[ \int \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx=\frac {4^{-p} \left ((-1)^{2/3} \sqrt [3]{a}+\sqrt [3]{b} x\right ) \left (\frac {\sqrt [3]{a}+(-1)^{2/3} \sqrt [3]{b} x}{\left (1+\sqrt [3]{-1}\right ) \sqrt [3]{a}}\right )^{-2 p} \left (\frac {i \left (1+\frac {\sqrt [3]{b} x}{\sqrt [3]{a}}\right )}{3 i+\sqrt {3}}\right )^{-2 p} \left (\left (a+b x^3\right )^2\right )^p \operatorname {AppellF1}\left (1+2 p,-2 p,-2 p,2 (1+p),-\frac {(-1)^{2/3} \left ((-1)^{2/3} \sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (1+\sqrt [3]{-1}\right ) \sqrt [3]{a}},\frac {i+\sqrt {3}-\frac {2 i \sqrt [3]{b} x}{\sqrt [3]{a}}}{3 i+\sqrt {3}}\right )}{\sqrt [3]{b} (1+2 p)} \]

[In]

Integrate[(a^2 + 2*a*b*x^3 + b^2*x^6)^p,x]

[Out]

(((-1)^(2/3)*a^(1/3) + b^(1/3)*x)*((a + b*x^3)^2)^p*AppellF1[1 + 2*p, -2*p, -2*p, 2*(1 + p), -(((-1)^(2/3)*((-
1)^(2/3)*a^(1/3) + b^(1/3)*x))/((1 + (-1)^(1/3))*a^(1/3))), (I + Sqrt[3] - ((2*I)*b^(1/3)*x)/a^(1/3))/(3*I + S
qrt[3])])/(4^p*b^(1/3)*(1 + 2*p)*((a^(1/3) + (-1)^(2/3)*b^(1/3)*x)/((1 + (-1)^(1/3))*a^(1/3)))^(2*p)*((I*(1 +
(b^(1/3)*x)/a^(1/3)))/(3*I + Sqrt[3]))^(2*p))

Maple [F]

\[\int \left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )^{p}d x\]

[In]

int((b^2*x^6+2*a*b*x^3+a^2)^p,x)

[Out]

int((b^2*x^6+2*a*b*x^3+a^2)^p,x)

Fricas [F]

\[ \int \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx=\int { {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} \,d x } \]

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^p,x, algorithm="fricas")

[Out]

integral((b^2*x^6 + 2*a*b*x^3 + a^2)^p, x)

Sympy [F]

\[ \int \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx=\int \left (a^{2} + 2 a b x^{3} + b^{2} x^{6}\right )^{p}\, dx \]

[In]

integrate((b**2*x**6+2*a*b*x**3+a**2)**p,x)

[Out]

Integral((a**2 + 2*a*b*x**3 + b**2*x**6)**p, x)

Maxima [F]

\[ \int \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx=\int { {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} \,d x } \]

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^p,x, algorithm="maxima")

[Out]

integrate((b^2*x^6 + 2*a*b*x^3 + a^2)^p, x)

Giac [F]

\[ \int \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx=\int { {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} \,d x } \]

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^p,x, algorithm="giac")

[Out]

integrate((b^2*x^6 + 2*a*b*x^3 + a^2)^p, x)

Mupad [F(-1)]

Timed out. \[ \int \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx=\int {\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^p \,d x \]

[In]

int((a^2 + b^2*x^6 + 2*a*b*x^3)^p,x)

[Out]

int((a^2 + b^2*x^6 + 2*a*b*x^3)^p, x)