3.132 \(\int (a^2+2 a b x^3+b^2 x^6)^p \, dx\)

Optimal. Leaf size=53 \[ \frac{x \left (a+b x^3\right ) \left (a^2+2 a b x^3+b^2 x^6\right )^p \, _2F_1\left (1,2 p+\frac{4}{3};\frac{4}{3};-\frac{b x^3}{a}\right )}{a} \]

[Out]

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

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Rubi [A]  time = 0.019247, antiderivative size = 55, normalized size of antiderivative = 1.04, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {1343, 246, 245} \[ x \left (\frac{b x^3}{a}+1\right )^{-2 p} \left (a^2+2 a b x^3+b^2 x^6\right )^p \, _2F_1\left (\frac{1}{3},-2 p;\frac{4}{3};-\frac{b x^3}{a}\right ) \]

Antiderivative was successfully verified.

[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 1343

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]

Rule 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^Fr
acPart[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 245

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])

Rubi steps

\begin{align*} \int \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx &=\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]  time = 0.166889, size = 204, normalized size = 3.85 \[ \frac{4^{-p} \left ((-1)^{2/3} \sqrt [3]{a}+\sqrt [3]{b} x\right ) \left (\frac{\sqrt [3]{a}+(-1)^{2/3} \sqrt [3]{b} x}{\left (1+\sqrt [3]{-1}\right ) \sqrt [3]{a}}\right )^{-2 p} \left (\frac{i \left (\frac{\sqrt [3]{b} x}{\sqrt [3]{a}}+1\right )}{\sqrt{3}+3 i}\right )^{-2 p} \left (\left (a+b x^3\right )^2\right )^p F_1\left (2 p+1;-2 p,-2 p;2 (p+1);-\frac{i \left (\sqrt [3]{b} x+(-1)^{2/3} \sqrt [3]{a}\right )}{\sqrt{3} \sqrt [3]{a}},\frac{-\frac{2 i \sqrt [3]{b} x}{\sqrt [3]{a}}+\sqrt{3}+i}{3 i+\sqrt{3}}\right )}{\sqrt [3]{b} (2 p+1)} \]

Warning: Unable to verify antiderivative.

[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), ((-I)*((-1)^(2/3)
*a^(1/3) + b^(1/3)*x))/(Sqrt[3]*a^(1/3)), (I + Sqrt[3] - ((2*I)*b^(1/3)*x)/a^(1/3))/(3*I + Sqrt[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))

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Maple [F]  time = 0.013, size = 0, normalized size = 0. \begin{align*} \int \left ({b}^{2}{x}^{6}+2\,ab{x}^{3}+{a}^{2} \right ) ^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a^{2} + 2 a b x^{3} + b^{2} x^{6}\right )^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)