### 3.765 $$\int e^{4 \coth ^{-1}(a x)} (c-a^2 c x^2)^p \, dx$$

Optimal. Leaf size=63 $\frac{c 2^{p+2} (a x+1)^{1-p} \left (c-a^2 c x^2\right )^{p-1} \text{Hypergeometric2F1}\left (-p-2,p-1,p,\frac{1}{2} (1-a x)\right )}{a (1-p)}$

[Out]

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

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Rubi [A]  time = 0.10418, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.182, Rules used = {6167, 6141, 678, 69} $\frac{c 2^{p+2} (a x+1)^{1-p} \left (c-a^2 c x^2\right )^{p-1} \, _2F_1\left (-p-2,p-1;p;\frac{1}{2} (1-a x)\right )}{a (1-p)}$

Antiderivative was successfully veriﬁed.

[In]

Int[E^(4*ArcCoth[a*x])*(c - a^2*c*x^2)^p,x]

[Out]

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

Rule 6167

Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Dist[(-1)^(n/2), Int[u*E^(n*ArcTanh[a*x]), x], x] /; Free
Q[a, x] && IntegerQ[n/2]

Rule 6141

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

Rule 678

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(d^(m - 1)*(a + c*x^2)^(p + 1))/((1
+ (e*x)/d)^(p + 1)*(a/d + (c*x)/e)^(p + 1)), Int[(1 + (e*x)/d)^(m + p)*(a/d + (c*x)/e)^p, x], x] /; FreeQ[{a,
c, d, e, m}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && (IntegerQ[m] || GtQ[d, 0]) &&  !(IGtQ[m, 0] && (
IntegerQ[3*p] || IntegerQ[4*p]))

Rule 69

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

Rubi steps

\begin{align*} \int e^{4 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^p \, dx &=\int e^{4 \tanh ^{-1}(a x)} \left (c-a^2 c x^2\right )^p \, dx\\ &=c^2 \int (1+a x)^4 \left (c-a^2 c x^2\right )^{-2+p} \, dx\\ &=\left (c^2 (1+a x)^{1-p} (c-a c x)^{1-p} \left (c-a^2 c x^2\right )^{-1+p}\right ) \int (1+a x)^{2+p} (c-a c x)^{-2+p} \, dx\\ &=\frac{2^{2+p} c (1+a x)^{1-p} \left (c-a^2 c x^2\right )^{-1+p} \, _2F_1\left (-2-p,-1+p;p;\frac{1}{2} (1-a x)\right )}{a (1-p)}\\ \end{align*}

Mathematica [A]  time = 0.0226024, size = 72, normalized size = 1.14 $-\frac{2^{p+2} (1-a x)^{p-1} \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \text{Hypergeometric2F1}\left (-p-2,p-1,p,\frac{1}{2} (1-a x)\right )}{a (p-1)}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[E^(4*ArcCoth[a*x])*(c - a^2*c*x^2)^p,x]

[Out]

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

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Maple [F]  time = 0.677, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( ax+1 \right ) ^{2} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{p}}{ \left ( ax-1 \right ) ^{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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