3.1022 \(\int \frac{(-a+b x^n)^p (a+b x^n)^p}{x^2} \, dx\)

Optimal. Leaf size=76 \[ -\frac{\left (b x^n-a\right )^p \left (a+b x^n\right )^p \left (1-\frac{b^2 x^{2 n}}{a^2}\right )^{-p} \, _2F_1\left (-\frac{1}{2 n},-p;1-\frac{1}{2 n};\frac{b^2 x^{2 n}}{a^2}\right )}{x} \]

[Out]

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

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Rubi [A]  time = 0.0520478, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {366, 365, 364} \[ -\frac{\left (b x^n-a\right )^p \left (a+b x^n\right )^p \left (1-\frac{b^2 x^{2 n}}{a^2}\right )^{-p} \, _2F_1\left (-\frac{1}{2 n},-p;1-\frac{1}{2 n};\frac{b^2 x^{2 n}}{a^2}\right )}{x} \]

Antiderivative was successfully verified.

[In]

Int[((-a + b*x^n)^p*(a + b*x^n)^p)/x^2,x]

[Out]

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

Rule 366

Int[((c_.)*(x_))^(m_.)*((a1_) + (b1_.)*(x_)^(n_))^(p_)*((a2_) + (b2_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[((a1
+ b1*x^n)^FracPart[p]*(a2 + b2*x^n)^FracPart[p])/(a1*a2 + b1*b2*x^(2*n))^FracPart[p], Int[(c*x)^m*(a1*a2 + b1
*b2*x^(2*n))^p, x], x] /; FreeQ[{a1, b1, a2, b2, c, m, n, p}, x] && EqQ[a2*b1 + a1*b2, 0] &&  !IntegerQ[p]

Rule 365

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])

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int \frac{\left (-a+b x^n\right )^p \left (a+b x^n\right )^p}{x^2} \, dx &=\left (\left (-a+b x^n\right )^p \left (a+b x^n\right )^p \left (-a^2+b^2 x^{2 n}\right )^{-p}\right ) \int \frac{\left (-a^2+b^2 x^{2 n}\right )^p}{x^2} \, dx\\ &=\left (\left (-a+b x^n\right )^p \left (a+b x^n\right )^p \left (1-\frac{b^2 x^{2 n}}{a^2}\right )^{-p}\right ) \int \frac{\left (1-\frac{b^2 x^{2 n}}{a^2}\right )^p}{x^2} \, dx\\ &=-\frac{\left (-a+b x^n\right )^p \left (a+b x^n\right )^p \left (1-\frac{b^2 x^{2 n}}{a^2}\right )^{-p} \, _2F_1\left (-\frac{1}{2 n},-p;1-\frac{1}{2 n};\frac{b^2 x^{2 n}}{a^2}\right )}{x}\\ \end{align*}

Mathematica [A]  time = 0.015318, size = 78, normalized size = 1.03 \[ -\frac{\left (b x^n-a\right )^p \left (a+b x^n\right )^p \left (1-\frac{b^2 x^{2 n}}{a^2}\right )^{-p} \text{HypergeometricPFQ}\left (\left \{-\frac{1}{2 n},-p\right \},\left \{1-\frac{1}{2 n}\right \},\frac{b^2 x^{2 n}}{a^2}\right )}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((-a + b*x^n)^p*(a + b*x^n)^p*HypergeometricPFQ[{-1/(2*n), -p}, {1 - 1/(2*n)}, (b^2*x^(2*n))/a^2])/(x*(1 - (
b^2*x^(2*n))/a^2)^p))

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Maple [F]  time = 0.144, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( -a+b{x}^{n} \right ) ^{p} \left ( a+b{x}^{n} \right ) ^{p}}{{x}^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{n} + a\right )}^{p}{\left (b x^{n} - a\right )}^{p}}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b x^{n} + a\right )}^{p}{\left (b x^{n} - a\right )}^{p}}{x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x^n + a)^p*(b*x^n - a)^p/x^2, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (- a + b x^{n}\right )^{p} \left (a + b x^{n}\right )^{p}}{x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((-a + b*x**n)**p*(a + b*x**n)**p/x**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{n} + a\right )}^{p}{\left (b x^{n} - a\right )}^{p}}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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