3.1021 \(\int \frac{(-a+b x^n)^p (a+b x^n)^p}{x} \, dx\)

Optimal. Leaf size=72 \[ -\frac{\left (a^2-b^2 x^{2 n}\right ) \left (b x^n-a\right )^p \left (a+b x^n\right )^p \, _2F_1\left (1,p+1;p+2;1-\frac{b^2 x^{2 n}}{a^2}\right )}{2 a^2 n (p+1)} \]

[Out]

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

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Rubi [A]  time = 0.0671256, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {267, 126, 266, 65} \[ -\frac{\left (a^2-b^2 x^{2 n}\right ) \left (b x^n-a\right )^p \left (a+b x^n\right )^p \, _2F_1\left (1,p+1;p+2;1-\frac{b^2 x^{2 n}}{a^2}\right )}{2 a^2 n (p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 267

Int[(x_)^(m_.)*((a1_) + (b1_.)*(x_)^(n_))^(p_)*((a2_) + (b2_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a1 + b1*x)^p*(a2 + b2*x)^p, x], x, x^n], x] /; FreeQ[{a1, b1, a2, b2, m, n, p}
, x] && EqQ[a2*b1 + a1*b2, 0] && IntegerQ[Simplify[(m + 1)/(2*n)]]

Rule 126

Int[((f_.)*(x_))^(p_.)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Dist[((a + b*x)^Fra
cPart[m]*(c + d*x)^FracPart[m])/(a*c + b*d*x^2)^FracPart[m], Int[(a*c + b*d*x^2)^m*(f*x)^p, x], x] /; FreeQ[{a
, b, c, d, f, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[m - n, 0]

Rule 266

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

Rule 65

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)*Hypergeometric2F1[-m, n +
1, n + 2, 1 + (d*x)/c])/(d*(n + 1)*(-(d/(b*c)))^m), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Inte
gerQ[m] || GtQ[-(d/(b*c)), 0])

Rubi steps

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

Mathematica [A]  time = 0.0241013, size = 73, normalized size = 1.01 \[ \frac{\left (b^2 x^{2 n}-a^2\right ) \left (b x^n-a\right )^p \left (a+b x^n\right )^p \, _2F_1\left (1,p+1;p+2;1-\frac{b^2 x^{2 n}}{a^2}\right )}{2 a^2 n (p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.238, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( -a+b{x}^{n} \right ) ^{p} \left ( a+b{x}^{n} \right ) ^{p}}{x}}\, 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,x)

[Out]

int((-a+b*x^n)^p*(a+b*x^n)^p/x,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}\,{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,x, algorithm="maxima")

[Out]

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

[Out]

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

[Out]

Integral((-a + b*x**n)**p*(a + b*x**n)**p/x, 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}\,{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,x, algorithm="giac")

[Out]

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