3.11.63 \(\int x^{-5-2 p} (a+b x^2)^p \, dx\) [1063]

Optimal. Leaf size=67 \[ \frac {b x^{-2 (1+p)} \left (a+b x^2\right )^{1+p}}{2 a^2 (1+p) (2+p)}-\frac {x^{-2 (2+p)} \left (a+b x^2\right )^{1+p}}{2 a (2+p)} \]

[Out]

1/2*b*(b*x^2+a)^(1+p)/a^2/(1+p)/(2+p)/(x^(2+2*p))-1/2*(b*x^2+a)^(1+p)/a/(2+p)/(x^(4+2*p))

________________________________________________________________________________________

Rubi [A]
time = 0.01, antiderivative size = 67, normalized size of antiderivative = 1.00, 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 = {277, 270} \begin {gather*} \frac {b x^{-2 (p+1)} \left (a+b x^2\right )^{p+1}}{2 a^2 (p+1) (p+2)}-\frac {x^{-2 (p+2)} \left (a+b x^2\right )^{p+1}}{2 a (p+2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(-5 - 2*p)*(a + b*x^2)^p,x]

[Out]

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

Rule 277

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.03, size = 62, normalized size = 0.93 \begin {gather*} -\frac {x^{-2 (2+p)} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \, _2F_1\left (-2-p,-p;-1-p;-\frac {b x^2}{a}\right )}{2 (2+p)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(-5 - 2*p)*(a + b*x^2)^p,x]

[Out]

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

________________________________________________________________________________________

Maple [A]
time = 0.07, size = 45, normalized size = 0.67

method	result	size
gosper	\(-\frac {\left (b \,x^{2}+a \right )^{1+p} x^{-4-2 p} \left (-b \,x^{2}+a p +a \right )}{2 \left (2+p \right ) \left (1+p \right ) a^{2}}\)	\(45\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(-5-2*p)*(b*x^2+a)^p,x,method=_RETURNVERBOSE)

[Out]

-1/2*(b*x^2+a)^(1+p)*x^(-4-2*p)*(-b*x^2+a*p+a)/(2+p)/(1+p)/a^2

________________________________________________________________________________________

Maxima [A]
time = 0.29, size = 59, normalized size = 0.88 \begin {gather*} \frac {{\left (b^{2} x^{4} - a b p x^{2} - a^{2} {\left (p + 1\right )}\right )} e^{\left (p \log \left (b x^{2} + a\right ) - 2 \, p \log \left (x\right )\right )}}{2 \, {\left (p^{2} + 3 \, p + 2\right )} a^{2} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-5-2*p)*(b*x^2+a)^p,x, algorithm="maxima")

[Out]

1/2*(b^2*x^4 - a*b*p*x^2 - a^2*(p + 1))*e^(p*log(b*x^2 + a) - 2*p*log(x))/((p^2 + 3*p + 2)*a^2*x^4)

________________________________________________________________________________________

Fricas [A]
time = 0.88, size = 67, normalized size = 1.00 \begin {gather*} \frac {{\left (b^{2} x^{5} - a b p x^{3} - {\left (a^{2} p + a^{2}\right )} x\right )} {\left (b x^{2} + a\right )}^{p} x^{-2 \, p - 5}}{2 \, {\left (a^{2} p^{2} + 3 \, a^{2} p + 2 \, a^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-5-2*p)*(b*x^2+a)^p,x, algorithm="fricas")

[Out]

1/2*(b^2*x^5 - a*b*p*x^3 - (a^2*p + a^2)*x)*(b*x^2 + a)^p*x^(-2*p - 5)/(a^2*p^2 + 3*a^2*p + 2*a^2)

________________________________________________________________________________________

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**(-5-2*p)*(b*x**2+a)**p,x)

[Out]

Timed out

________________________________________________________________________________________

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^(-5-2*p)*(b*x^2+a)^p,x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^p*x^(-2*p - 5), x)

________________________________________________________________________________________

Mupad [B]
time = 5.02, size = 96, normalized size = 1.43 \begin {gather*} -{\left (b\,x^2+a\right )}^p\,\left (\frac {x\,\left (p+1\right )}{2\,x^{2\,p+5}\,\left (p^2+3\,p+2\right )}-\frac {b^2\,x^5}{2\,a^2\,x^{2\,p+5}\,\left (p^2+3\,p+2\right )}+\frac {b\,p\,x^3}{2\,a\,x^{2\,p+5}\,\left (p^2+3\,p+2\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^p/x^(2*p + 5),x)

[Out]

-(a + b*x^2)^p*((x*(p + 1))/(2*x^(2*p + 5)*(3*p + p^2 + 2)) - (b^2*x^5)/(2*a^2*x^(2*p + 5)*(3*p + p^2 + 2)) +
(b*p*x^3)/(2*a*x^(2*p + 5)*(3*p + p^2 + 2)))

________________________________________________________________________________________