3.1065 \(\int x^{-3-2 p} (a+b x^2)^p \, dx\)

Optimal. Leaf size=30 \[ -\frac{x^{-2 (p+1)} \left (a+b x^2\right )^{p+1}}{2 a (p+1)} \]

[Out]

-(a + b*x^2)^(1 + p)/(2*a*(1 + p)*x^(2*(1 + p)))

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Rubi [A]  time = 0.0060034, antiderivative size = 30, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.059, Rules used = {264} \[ -\frac{x^{-2 (p+1)} \left (a+b x^2\right )^{p+1}}{2 a (p+1)} \]

Antiderivative was successfully verified.

[In]

Int[x^(-3 - 2*p)*(a + b*x^2)^p,x]

[Out]

-(a + b*x^2)^(1 + p)/(2*a*(1 + p)*x^(2*(1 + p)))

Rule 264

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

Rubi steps

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

Mathematica [A]  time = 0.0116172, size = 29, normalized size = 0.97 \[ \frac{x^{-2 p-2} \left (a+b x^2\right )^{p+1}}{a (-2 p-2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(-2 - 2*p)*(a + b*x^2)^(1 + p))/(a*(-2 - 2*p))

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Maple [A]  time = 0.002, size = 29, normalized size = 1. \begin{align*} -{\frac{{x}^{-2-2\,p} \left ( b{x}^{2}+a \right ) ^{1+p}}{2\,a \left ( 1+p \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*x^(-2-2*p)*(b*x^2+a)^(1+p)/a/(1+p)

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Maxima [A]  time = 1.83406, size = 50, normalized size = 1.67 \begin{align*} -\frac{{\left (b x^{2} + a\right )} e^{\left (p \log \left (b x^{2} + a\right ) - 2 \, p \log \left (x\right )\right )}}{2 \, a{\left (p + 1\right )} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-3-2*p)*(b*x^2+a)^p,x, algorithm="maxima")

[Out]

-1/2*(b*x^2 + a)*e^(p*log(b*x^2 + a) - 2*p*log(x))/(a*(p + 1)*x^2)

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Fricas [A]  time = 1.61178, size = 77, normalized size = 2.57 \begin{align*} -\frac{{\left (b x^{3} + a x\right )}{\left (b x^{2} + a\right )}^{p} x^{-2 \, p - 3}}{2 \,{\left (a p + a\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-3-2*p)*(b*x^2+a)^p,x, algorithm="fricas")

[Out]

-1/2*(b*x^3 + a*x)*(b*x^2 + a)^p*x^(-2*p - 3)/(a*p + a)

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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**(-3-2*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 - 3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-3-2*p)*(b*x^2+a)^p,x, algorithm="giac")

[Out]

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