3.134 \(\int \frac{(a^2+2 a b x^3+b^2 x^6)^p}{x^2} \, dx\)

Optimal. Leaf size=58 \[ -\frac{\left (\frac{b x^3}{a}+1\right )^{-2 p} \left (a^2+2 a b x^3+b^2 x^6\right )^p \, _2F_1\left (-\frac{1}{3},-2 p;\frac{2}{3};-\frac{b x^3}{a}\right )}{x} \]

[Out]

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

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Rubi [A]  time = 0.0183031, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {1356, 364} \[ -\frac{\left (\frac{b x^3}{a}+1\right )^{-2 p} \left (a^2+2 a b x^3+b^2 x^6\right )^p \, _2F_1\left (-\frac{1}{3},-2 p;\frac{2}{3};-\frac{b x^3}{a}\right )}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1356

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a
+ b*x^n + c*x^(2*n))^FracPart[p])/(1 + (2*c*x^n)/b)^(2*FracPart[p]), Int[(d*x)^m*(1 + (2*c*x^n)/b)^(2*p), x],
x] /; FreeQ[{a, b, c, d, m, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[2*p]

Rule 364

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

Rubi steps

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

Mathematica [A]  time = 0.0064322, size = 49, normalized size = 0.84 \[ -\frac{\left (\left (a+b x^3\right )^2\right )^p \left (\frac{b x^3}{a}+1\right )^{-2 p} \, _2F_1\left (-\frac{1}{3},-2 p;\frac{2}{3};-\frac{b x^3}{a}\right )}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.033, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ({b}^{2}{x}^{6}+2\,ab{x}^{3}+{a}^{2} \right ) ^{p}}{{x}^{2}}}\, 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^2,x)

[Out]

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

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

[Out]

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

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

[Out]

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

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

[Out]

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

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

[Out]

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