3.1062 \(\int x^{-6-2 p} (a+b x^2)^p \, dx\)

Optimal. Leaf size=53 \[ -\frac{x^{-2 p-5} \left (a+b x^2\right )^{p+1} \, _2F_1\left (-\frac{3}{2},1;\frac{1}{2} (-2 p-3);-\frac{b x^2}{a}\right )}{a (2 p+5)} \]

[Out]

-((x^(-5 - 2*p)*(a + b*x^2)^(1 + p)*Hypergeometric2F1[-3/2, 1, (-3 - 2*p)/2, -((b*x^2)/a)])/(a*(5 + 2*p)))

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Rubi [A]  time = 0.0203274, antiderivative size = 70, normalized size of antiderivative = 1.32, number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {365, 364} \[ -\frac{x^{-2 p-5} \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac{1}{2} (-2 p-5),-p;\frac{1}{2} (-2 p-3);-\frac{b x^2}{a}\right )}{2 p+5} \]

Antiderivative was successfully verified.

[In]

Int[x^(-6 - 2*p)*(a + b*x^2)^p,x]

[Out]

-((x^(-5 - 2*p)*(a + b*x^2)^p*Hypergeometric2F1[(-5 - 2*p)/2, -p, (-3 - 2*p)/2, -((b*x^2)/a)])/((5 + 2*p)*(1 +
(b*x^2)/a)^p))

Rule 365

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

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 x^{-6-2 p} \left (a+b x^2\right )^p \, dx &=\left (\left (a+b x^2\right )^p \left (1+\frac{b x^2}{a}\right )^{-p}\right ) \int x^{-6-2 p} \left (1+\frac{b x^2}{a}\right )^p \, dx\\ &=-\frac{x^{-5-2 p} \left (a+b x^2\right )^p \left (1+\frac{b x^2}{a}\right )^{-p} \, _2F_1\left (\frac{1}{2} (-5-2 p),-p;\frac{1}{2} (-3-2 p);-\frac{b x^2}{a}\right )}{5+2 p}\\ \end{align*}

Mathematica [A]  time = 0.0153312, size = 66, normalized size = 1.25 \[ -\frac{x^{-2 p-5} \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \, _2F_1\left (-p-\frac{5}{2},-p;-p-\frac{3}{2};-\frac{b x^2}{a}\right )}{2 p+5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((x^(-5 - 2*p)*(a + b*x^2)^p*Hypergeometric2F1[-5/2 - p, -p, -3/2 - p, -((b*x^2)/a)])/((5 + 2*p)*(1 + (b*x^2)
/a)^p))

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Maple [F]  time = 0.053, size = 0, normalized size = 0. \begin{align*} \int{x}^{-6-2\,p} \left ( b{x}^{2}+a \right ) ^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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