3.11.65 \(\int x^{-3-2 p} (a+b x^2)^p \, dx\) [1065]

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

[Out]

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

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Rubi [A]
time = 0.00, antiderivative size = 30, normalized size of antiderivative = 1.00, 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 = {270} \begin {gather*} -\frac {x^{-2 (p+1)} \left (a+b x^2\right )^{p+1}}{2 a (p+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(-3 - 2*p)*(a + b*x^2)^p,x]

[Out]

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

Rule 270

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*}

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Mathematica [A]
time = 0.03, size = 29, normalized size = 0.97 \begin {gather*} \frac {x^{-2-2 p} \left (a+b x^2\right )^{1+p}}{a (-2-2 p)} \end {gather*}

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 = 0.07, size = 29, normalized size = 0.97

method	result	size
gosper	\(-\frac {x^{-2-2 p} \left (b \,x^{2}+a \right )^{1+p}}{2 a \left (1+p \right )}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.32, size = 37, normalized size = 1.23 \begin {gather*} -\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 {gather*}

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 = 1.06, size = 34, normalized size = 1.13 \begin {gather*} -\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 {gather*}

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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, 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^(-3-2*p)*(b*x^2+a)^p,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 5.05, size = 52, normalized size = 1.73 \begin {gather*} -{\left (b\,x^2+a\right )}^p\,\left (\frac {x}{2\,x^{2\,p+3}\,\left (p+1\right )}+\frac {b\,x^3}{2\,a\,x^{2\,p+3}\,\left (p+1\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^p/x^(2*p + 3),x)

[Out]

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

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