3.1364 \(\int e^{2 (1+p) \tanh ^{-1}(a x)} (c-a^2 c x^2)^{-p} \, dx\)

Optimal. Leaf size=95 \[ \frac {\left (1-a^2 x^2\right )^p (1-a x)^{1-2 p} \left (c-a^2 c x^2\right )^{-p}}{a (1-2 p)}+\frac {\left (1-a^2 x^2\right )^p (1-a x)^{-2 p} \left (c-a^2 c x^2\right )^{-p}}{a p} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6143, 6140, 43} \[ \frac {\left (1-a^2 x^2\right )^p (1-a x)^{1-2 p} \left (c-a^2 c x^2\right )^{-p}}{a (1-2 p)}+\frac {\left (1-a^2 x^2\right )^p (1-a x)^{-2 p} \left (c-a^2 c x^2\right )^{-p}}{a p} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 6140

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

Rule 6143

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

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 58, normalized size = 0.61 \[ \frac {(1-a x)^{-2 p} (a p x+p-1) \left (1-a^2 x^2\right )^p \left (c-a^2 c x^2\right )^{-p}}{a p (2 p-1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.48, size = 81, normalized size = 0.85 \[ -\frac {{\left (a^{2} p x^{2} - a x - p + 1\right )} \left (\frac {a x + 1}{a x - 1}\right )^{p + 1}}{{\left (2 \, a p^{2} - a p + {\left (2 \, a^{2} p^{2} - a^{2} p\right )} x\right )} {\left (-a^{2} c x^{2} + c\right )}^{p}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 60, normalized size = 0.63 \[ -\frac {\left (a x -1\right ) \left (a p x +p -1\right ) {\mathrm e}^{2 \left (1+p \right ) \arctanh \left (a x \right )} \left (-a^{2} c \,x^{2}+c \right )^{-p}}{a p \left (2 p -1\right ) \left (a x +1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.36, size = 41, normalized size = 0.43 \[ -\frac {a p x + p - 1}{{\left (2 \, p^{2} - p\right )} {\left (a x - 1\right )}^{2 \, p} a \left (-c\right )^{p}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.01, size = 86, normalized size = 0.91 \[ -\frac {p\,{\left (a\,x+1\right )}^p-{\left (a\,x+1\right )}^p+a\,p\,x\,{\left (a\,x+1\right )}^p}{a\,p\,{\left (c-a^2\,c\,x^2\right )}^p\,{\left (1-a\,x\right )}^p-2\,a\,p^2\,{\left (c-a^2\,c\,x^2\right )}^p\,{\left (1-a\,x\right )}^p} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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