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]

-(b^2*x^6+2*a*b*x^3+a^2)^p*hypergeom([-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 = 0.02, antiderivative size = 58, normalized size of antiderivative = 1.00, 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 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])

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]

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*}

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Mathematica [A]  time = 0.01, 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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fricas [F]  time = 0.81, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p}}{x^{2}}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p}}{x^{2}}\,{d x} \]

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)

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int \frac {\left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )^{p}}{x^{2}}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p}}{x^{2}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^p}{x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\left (a + b x^{3}\right )^{2}\right )^{p}}{x^{2}}\, dx \]

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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