3.763 \(\int e^{2 p \coth ^{-1}(a x)} (c-a^2 c x^2)^p \, dx\)

Optimal. Leaf size=51 \[ \frac {x \left (1-\frac {1}{a^2 x^2}\right )^{-p} \left (\frac {1}{a x}+1\right )^{2 p+1} \left (c-a^2 c x^2\right )^p}{2 p+1} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {6192, 6196, 37} \[ \frac {x \left (1-\frac {1}{a^2 x^2}\right )^{-p} \left (\frac {1}{a x}+1\right )^{2 p+1} \left (c-a^2 c x^2\right )^p}{2 p+1} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 37

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

Rule 6192

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

Rule 6196

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

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 36, normalized size = 0.71 \[ \frac {(a x+1) \left (c-a^2 c x^2\right )^p e^{2 p \coth ^{-1}(a x)}}{2 a p+a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(E^(2*p*ArcCoth[a*x])*(1 + a*x)*(c - a^2*c*x^2)^p)/(a + 2*a*p)

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fricas [A]  time = 0.75, size = 42, normalized size = 0.82 \[ \frac {{\left (a x - 1\right )} {\left (-a^{2} c x^{2} + c\right )}^{p} \left (\frac {a x - 1}{a x + 1}\right )^{p}}{2 \, a p + a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (-a^{2} c x^{2} + c\right )}^{p} \left (\frac {a x - 1}{a x + 1}\right )^{p}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 38, normalized size = 0.75 \[ \frac {\left (a x +1\right ) {\mathrm e}^{2 p \,\mathrm {arccoth}\left (a x \right )} \left (-a^{2} c \,x^{2}+c \right )^{p}}{a \left (1+2 p \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(2*p*arccoth(a*x))*(-a^2*c*x^2+c)^p,x)

[Out]

(a*x+1)/a/(1+2*p)*exp(2*p*arccoth(a*x))*(-a^2*c*x^2+c)^p

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maxima [A]  time = 0.33, size = 36, normalized size = 0.71 \[ \frac {{\left (a \left (-c\right )^{p} x - \left (-c\right )^{p}\right )} {\left (a x - 1\right )}^{2 \, p}}{a {\left (2 \, p + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.34, size = 59, normalized size = 1.16 \[ \frac {{\left (c-a^2\,c\,x^2\right )}^p\,\left (a\,x+1\right )\,{\left (\frac {a\,x+1}{a\,x}\right )}^p}{a\,\left (2\,p+1\right )\,{\left (\frac {a\,x-1}{a\,x}\right )}^p} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(2*p*acoth(a*x))*(c - a^2*c*x^2)^p,x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \begin {cases} - \frac {i x}{\sqrt {c}} & \text {for}\: a = 0 \wedge p = - \frac {1}{2} \\c^{p} x e^{i \pi p} & \text {for}\: a = 0 \\\int \frac {e^{- \operatorname {acoth}{\left (a x \right )}}}{\sqrt {- c \left (a x - 1\right ) \left (a x + 1\right )}}\, dx & \text {for}\: p = - \frac {1}{2} \\\frac {a x \left (- a^{2} c x^{2} + c\right )^{p} e^{2 p \operatorname {acoth}{\left (a x \right )}}}{2 a p + a} + \frac {\left (- a^{2} c x^{2} + c\right )^{p} e^{2 p \operatorname {acoth}{\left (a x \right )}}}{2 a p + a} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*p*acoth(a*x))*(-a**2*c*x**2+c)**p,x)

[Out]

Piecewise((-I*x/sqrt(c), Eq(a, 0) & Eq(p, -1/2)), (c**p*x*exp(I*pi*p), Eq(a, 0)), (Integral(exp(-acoth(a*x))/s
qrt(-c*(a*x - 1)*(a*x + 1)), x), Eq(p, -1/2)), (a*x*(-a**2*c*x**2 + c)**p*exp(2*p*acoth(a*x))/(2*a*p + a) + (-
a**2*c*x**2 + c)**p*exp(2*p*acoth(a*x))/(2*a*p + a), True))

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