3.11.62 \(\int x^{-6-2 p} (a+b x^2)^p \, dx\) [1062]

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

[Out]

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

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Rubi [A]
time = 0.01, 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 = {372, 371} \begin {gather*} -\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} \end {gather*}

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 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 372

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])

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*}

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Mathematica [A]
time = 0.03, size = 66, normalized size = 1.25 \begin {gather*} -\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 {5}{2}-p,-p;-\frac {3}{2}-p;-\frac {b x^2}{a}\right )}{5+2 p} \end {gather*}

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

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: HeuristicGCDFailed} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: HeuristicGCDFailed >> no luck

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^p/x^(2*p + 6),x)

[Out]

int((a + b*x^2)^p/x^(2*p + 6), x)

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