### 3.216 $$\int e^{-3 \coth ^{-1}(a x)} (c-a c x)^p \, dx$$

Optimal. Leaf size=94 $\frac{x \left (1-\frac{1}{a x}\right )^{3/2} \left (\frac{a-\frac{1}{x}}{a+\frac{1}{x}}\right )^{-p-\frac{3}{2}} (c-a c x)^p \text{Hypergeometric2F1}\left (-p-\frac{3}{2},-p-1,-p,\frac{2}{x \left (a+\frac{1}{x}\right )}\right )}{(p+1) \sqrt{\frac{1}{a x}+1}}$

[Out]

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

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Rubi [A]  time = 0.125535, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 18, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.167, Rules used = {6176, 6181, 132} $\frac{x \left (1-\frac{1}{a x}\right )^{3/2} \left (\frac{a-\frac{1}{x}}{a+\frac{1}{x}}\right )^{-p-\frac{3}{2}} (c-a c x)^p \, _2F_1\left (-p-\frac{3}{2},-p-1;-p;\frac{2}{\left (a+\frac{1}{x}\right ) x}\right )}{(p+1) \sqrt{\frac{1}{a x}+1}}$

Antiderivative was successfully veriﬁed.

[In]

Int[(c - a*c*x)^p/E^(3*ArcCoth[a*x]),x]

[Out]

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

Rule 6176

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_))^(p_), x_Symbol] :> Dist[(c + d*x)^p/(x^p*(1 + c/(d
*x))^p), Int[u*x^p*(1 + c/(d*x))^p*E^(n*ArcCoth[a*x]), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[a^2*c^2 - d^
2, 0] &&  !IntegerQ[n/2] &&  !IntegerQ[p]

Rule 6181

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_))^(p_.)*(x_)^(m_), x_Symbol] :> -Dist[c^p*x^m*(1/x)^m, Sub
st[Int[((1 + (d*x)/c)^p*(1 + x/a)^(n/2))/(x^(m + 2)*(1 - x/a)^(n/2)), x], x, 1/x], x] /; FreeQ[{a, c, d, m, n,
p}, x] && EqQ[c^2 - a^2*d^2, 0] &&  !IntegerQ[n/2] && (IntegerQ[p] || GtQ[c, 0]) &&  !IntegerQ[m]

Rule 132

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x
)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1)*Hypergeometric2F1[m + 1, -n, m + 2, -(((d*e - c*f)*(a + b*x))/((b*c -
a*d)*(e + f*x)))])/(((b*e - a*f)*(m + 1))*(((b*e - a*f)*(c + d*x))/((b*c - a*d)*(e + f*x)))^n), x] /; FreeQ[{a
, b, c, d, e, f, m, n, p}, x] && EqQ[m + n + p + 2, 0] &&  !IntegerQ[n]

Rubi steps

\begin{align*} \int e^{-3 \coth ^{-1}(a x)} (c-a c x)^p \, dx &=\left (\left (1-\frac{1}{a x}\right )^{-p} x^{-p} (c-a c x)^p\right ) \int e^{-3 \coth ^{-1}(a x)} \left (1-\frac{1}{a x}\right )^p x^p \, dx\\ &=-\left (\left (\left (1-\frac{1}{a x}\right )^{-p} \left (\frac{1}{x}\right )^p (c-a c x)^p\right ) \operatorname{Subst}\left (\int \frac{x^{-2-p} \left (1-\frac{x}{a}\right )^{\frac{3}{2}+p}}{\left (1+\frac{x}{a}\right )^{3/2}} \, dx,x,\frac{1}{x}\right )\right )\\ &=\frac{\left (\frac{a-\frac{1}{x}}{a+\frac{1}{x}}\right )^{-\frac{3}{2}-p} \left (1-\frac{1}{a x}\right )^{3/2} x (c-a c x)^p \, _2F_1\left (-\frac{3}{2}-p,-1-p;-p;\frac{2}{\left (a+\frac{1}{x}\right ) x}\right )}{(1+p) \sqrt{1+\frac{1}{a x}}}\\ \end{align*}

Mathematica [A]  time = 0.0613283, size = 96, normalized size = 1.02 $\frac{\sqrt{1-\frac{1}{a x}} (a x+1) \left (\frac{a x-1}{a x+1}\right )^{-p-\frac{1}{2}} (c-a c x)^p \text{Hypergeometric2F1}\left (-p-\frac{3}{2},-p-1,-p,\frac{2}{a x+1}\right )}{a (p+1) \sqrt{\frac{1}{a x}+1}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(c - a*c*x)^p/E^(3*ArcCoth[a*x]),x]

[Out]

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

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Maple [F]  time = 0.365, size = 0, normalized size = 0. \begin{align*} \int \left ( -acx+c \right ) ^{p} \left ({\frac{ax-1}{ax+1}} \right ) ^{{\frac{3}{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((-a*c*x+c)^p*((a*x-1)/(a*x+1))^(3/2),x)

[Out]

int((-a*c*x+c)^p*((a*x-1)/(a*x+1))^(3/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (-a c x + c\right )}^{p} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{3}{2}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a*c*x+c)^p*((a*x-1)/(a*x+1))^(3/2),x, algorithm="maxima")

[Out]

integrate((-a*c*x + c)^p*((a*x - 1)/(a*x + 1))^(3/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (a x - 1\right )}{\left (-a c x + c\right )}^{p} \sqrt{\frac{a x - 1}{a x + 1}}}{a x + 1}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a*c*x+c)^p*((a*x-1)/(a*x+1))^(3/2),x, algorithm="fricas")

[Out]

integral((a*x - 1)*(-a*c*x + c)^p*sqrt((a*x - 1)/(a*x + 1))/(a*x + 1), x)

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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((-a*c*x+c)**p*((a*x-1)/(a*x+1))**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (-a c x + c\right )}^{p} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{3}{2}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a*c*x+c)^p*((a*x-1)/(a*x+1))^(3/2),x, algorithm="giac")

[Out]

integrate((-a*c*x + c)^p*((a*x - 1)/(a*x + 1))^(3/2), x)