3.1061 \(\int x^{-7-2 p} (a+b x^2)^p \, dx\)

Optimal. Leaf size=105 \[ -\frac{b^2 x^{-2 (p+1)} \left (a+b x^2\right )^{p+1}}{a^3 (p+1) (p+2) (p+3)}+\frac{b x^{-2 (p+2)} \left (a+b x^2\right )^{p+1}}{a^2 (p+2) (p+3)}-\frac{x^{-2 (p+3)} \left (a+b x^2\right )^{p+1}}{2 a (p+3)} \]

[Out]

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

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Rubi [A]  time = 0.0571453, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {271, 264} \[ -\frac{b^2 x^{-2 (p+1)} \left (a+b x^2\right )^{p+1}}{a^3 (p+1) (p+2) (p+3)}+\frac{b x^{-2 (p+2)} \left (a+b x^2\right )^{p+1}}{a^2 (p+2) (p+3)}-\frac{x^{-2 (p+3)} \left (a+b x^2\right )^{p+1}}{2 a (p+3)} \]

Antiderivative was successfully verified.

[In]

Int[x^(-7 - 2*p)*(a + b*x^2)^p,x]

[Out]

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

Rule 271

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]

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^{-7-2 p} \left (a+b x^2\right )^p \, dx &=-\frac{x^{-2 (3+p)} \left (a+b x^2\right )^{1+p}}{2 a (3+p)}-\frac{(2 b) \int x^{-5-2 p} \left (a+b x^2\right )^p \, dx}{a (3+p)}\\ &=\frac{b x^{-2 (2+p)} \left (a+b x^2\right )^{1+p}}{a^2 (2+p) (3+p)}-\frac{x^{-2 (3+p)} \left (a+b x^2\right )^{1+p}}{2 a (3+p)}+\frac{\left (2 b^2\right ) \int x^{-3-2 p} \left (a+b x^2\right )^p \, dx}{a^2 (2+p) (3+p)}\\ &=-\frac{b^2 x^{-2 (1+p)} \left (a+b x^2\right )^{1+p}}{a^3 (1+p) (2+p) (3+p)}+\frac{b x^{-2 (2+p)} \left (a+b x^2\right )^{1+p}}{a^2 (2+p) (3+p)}-\frac{x^{-2 (3+p)} \left (a+b x^2\right )^{1+p}}{2 a (3+p)}\\ \end{align*}

Mathematica [C]  time = 0.0139091, size = 62, normalized size = 0.59 \[ -\frac{x^{-2 (p+3)} \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \, _2F_1\left (-p-3,-p;-p-2;-\frac{b x^2}{a}\right )}{2 (p+3)} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(-7 - 2*p)*(a + b*x^2)^p,x]

[Out]

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

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Maple [A]  time = 0.004, size = 81, normalized size = 0.8 \begin{align*} -{\frac{ \left ( b{x}^{2}+a \right ) ^{1+p}{x}^{-6-2\,p} \left ( 2\,{b}^{2}{x}^{4}-2\,abp{x}^{2}+{a}^{2}{p}^{2}-2\,ab{x}^{2}+3\,{a}^{2}p+2\,{a}^{2} \right ) }{ \left ( 6+2\,p \right ) \left ( 2+p \right ) \left ( 1+p \right ){a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(-7-2*p)*(b*x^2+a)^p,x)

[Out]

-1/2*(b*x^2+a)^(1+p)*x^(-6-2*p)*(2*b^2*x^4-2*a*b*p*x^2+a^2*p^2-2*a*b*x^2+3*a^2*p+2*a^2)/(3+p)/(2+p)/(1+p)/a^3

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Maxima [A]  time = 1.65909, size = 113, normalized size = 1.08 \begin{align*} -\frac{{\left (2 \, b^{3} x^{6} - 2 \, a b^{2} p x^{4} +{\left (p^{2} + p\right )} a^{2} b x^{2} +{\left (p^{2} + 3 \, p + 2\right )} a^{3}\right )} e^{\left (p \log \left (b x^{2} + a\right ) - 2 \, p \log \left (x\right )\right )}}{2 \,{\left (p^{3} + 6 \, p^{2} + 11 \, p + 6\right )} a^{3} x^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-7-2*p)*(b*x^2+a)^p,x, algorithm="maxima")

[Out]

-1/2*(2*b^3*x^6 - 2*a*b^2*p*x^4 + (p^2 + p)*a^2*b*x^2 + (p^2 + 3*p + 2)*a^3)*e^(p*log(b*x^2 + a) - 2*p*log(x))
/((p^3 + 6*p^2 + 11*p + 6)*a^3*x^6)

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Fricas [A]  time = 1.61979, size = 219, normalized size = 2.09 \begin{align*} -\frac{{\left (2 \, b^{3} x^{7} - 2 \, a b^{2} p x^{5} +{\left (a^{2} b p^{2} + a^{2} b p\right )} x^{3} +{\left (a^{3} p^{2} + 3 \, a^{3} p + 2 \, a^{3}\right )} x\right )}{\left (b x^{2} + a\right )}^{p} x^{-2 \, p - 7}}{2 \,{\left (a^{3} p^{3} + 6 \, a^{3} p^{2} + 11 \, a^{3} p + 6 \, a^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-7-2*p)*(b*x^2+a)^p,x, algorithm="fricas")

[Out]

-1/2*(2*b^3*x^7 - 2*a*b^2*p*x^5 + (a^2*b*p^2 + a^2*b*p)*x^3 + (a^3*p^2 + 3*a^3*p + 2*a^3)*x)*(b*x^2 + a)^p*x^(
-2*p - 7)/(a^3*p^3 + 6*a^3*p^2 + 11*a^3*p + 6*a^3)

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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**(-7-2*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 - 7}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-7-2*p)*(b*x^2+a)^p,x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^p*x^(-2*p - 7), x)