3.2928 \(\int \frac{(a+b (c x)^n)^p}{x^2} \, dx\)

Optimal. Leaf size=62 \[ -\frac{\left (a+b (c x)^n\right )^p \left (\frac{b (c x)^n}{a}+1\right )^{-p} \, _2F_1\left (-\frac{1}{n},-p;-\frac{1-n}{n};-\frac{b (c x)^n}{a}\right )}{x} \]

[Out]

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

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Rubi [A]  time = 0.0350772, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {367, 12, 365, 364} \[ -\frac{\left (a+b (c x)^n\right )^p \left (\frac{b (c x)^n}{a}+1\right )^{-p} \, _2F_1\left (-\frac{1}{n},-p;-\frac{1-n}{n};-\frac{b (c x)^n}{a}\right )}{x} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*(c*x)^n)^p/x^2,x]

[Out]

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

Rule 367

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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

Mathematica [A]  time = 0.0394991, size = 59, normalized size = 0.95 \[ -\frac{\left (a+b (c x)^n\right )^p \left (\frac{b (c x)^n}{a}+1\right )^{-p} \, _2F_1\left (-\frac{1}{n},-p;1-\frac{1}{n};-\frac{b (c x)^n}{a}\right )}{x} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*(c*x)^n)^p/x^2,x]

[Out]

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

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Maple [F]  time = 0.063, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+b \left ( cx \right ) ^{n} \right ) ^{p}}{{x}^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(((c*x)^n*b + 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 (\left (c x\right )^{n} b + a\right )}^{p}}{x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(((c*x)^n*b + 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 \left (c x\right )^{n}\right )^{p}}{x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*(c*x)**n)**p/x**2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(((c*x)^n*b + a)^p/x^2, x)