3.1322 \(\int \frac {e^{n \tanh ^{-1}(a x)}}{(c-a^2 c x^2)^2} \, dx\)

Optimal. Leaf size=72 \[ \frac {2 e^{n \tanh ^{-1}(a x)}}{a c^2 n \left (4-n^2\right )}-\frac {(n-2 a x) e^{n \tanh ^{-1}(a x)}}{a c^2 \left (4-n^2\right ) \left (1-a^2 x^2\right )} \]

[Out]

2*exp(n*arctanh(a*x))/a/c^2/n/(-n^2+4)-exp(n*arctanh(a*x))*(-2*a*x+n)/a/c^2/(-n^2+4)/(-a^2*x^2+1)

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Rubi [A]  time = 0.07, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {6136, 6137} \[ \frac {2 e^{n \tanh ^{-1}(a x)}}{a c^2 n \left (4-n^2\right )}-\frac {(n-2 a x) e^{n \tanh ^{-1}(a x)}}{a c^2 \left (4-n^2\right ) \left (1-a^2 x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*E^(n*ArcTanh[a*x]))/(a*c^2*n*(4 - n^2)) - (E^(n*ArcTanh[a*x])*(n - 2*a*x))/(a*c^2*(4 - n^2)*(1 - a^2*x^2))

Rule 6136

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((n + 2*a*(p + 1)*x)*(c + d*x^2
)^(p + 1)*E^(n*ArcTanh[a*x]))/(a*c*(n^2 - 4*(p + 1)^2)), x] - Dist[(2*(p + 1)*(2*p + 3))/(c*(n^2 - 4*(p + 1)^2
)), Int[(c + d*x^2)^(p + 1)*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[{a, c, d, n}, x] && EqQ[a^2*c + d, 0] && LtQ[p
, -1] &&  !IntegerQ[n] && NeQ[n^2 - 4*(p + 1)^2, 0] && IntegerQ[2*p]

Rule 6137

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))/((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[E^(n*ArcTanh[a*x])/(a*c*n), x] /; F
reeQ[{a, c, d, n}, x] && EqQ[a^2*c + d, 0] &&  !IntegerQ[n/2]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 68, normalized size = 0.94 \[ -\frac {(1-a x)^{-\frac {n}{2}-1} (a x+1)^{\frac {n}{2}-1} \left (-2 a^2 x^2+2 a n x-n^2+2\right )}{a c^2 (n-2) n (n+2)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [A]  time = 0.46, size = 79, normalized size = 1.10 \[ \frac {{\left (2 \, a^{2} x^{2} - 2 \, a n x + n^{2} - 2\right )} \left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{a c^{2} n^{3} - 4 \, a c^{2} n - {\left (a^{3} c^{2} n^{3} - 4 \, a^{3} c^{2} n\right )} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a^2*x^2 - 2*a*n*x + n^2 - 2)*((a*x + 1)/(a*x - 1))^(1/2*n)/(a*c^2*n^3 - 4*a*c^2*n - (a^3*c^2*n^3 - 4*a^3*c^
2*n)*x^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 55, normalized size = 0.76 \[ -\frac {{\mathrm e}^{n \arctanh \left (a x \right )} \left (2 a^{2} x^{2}-2 n a x +n^{2}-2\right )}{\left (a^{2} x^{2}-1\right ) c^{2} a n \left (n^{2}-4\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-exp(n*arctanh(a*x))*(2*a^2*x^2-2*a*n*x+n^2-2)/(a^2*x^2-1)/c^2/a/n/(n^2-4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.21, size = 93, normalized size = 1.29 \[ \frac {{\left (a\,x+1\right )}^{n/2}\,\left (\frac {2\,x^2}{a\,c^2\,n\,\left (n^2-4\right )}-\frac {2\,x}{a^2\,c^2\,\left (n^2-4\right )}+\frac {n^2-2}{a^3\,c^2\,n\,\left (n^2-4\right )}\right )}{\left (\frac {1}{a^2}-x^2\right )\,{\left (1-a\,x\right )}^{n/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*atanh(a*x))/(c - a^2*c*x^2)^2,x)

[Out]

((a*x + 1)^(n/2)*((2*x^2)/(a*c^2*n*(n^2 - 4)) - (2*x)/(a^2*c^2*(n^2 - 4)) + (n^2 - 2)/(a^3*c^2*n*(n^2 - 4))))/
((1/a^2 - x^2)*(1 - a*x)^(n/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \begin {cases} \tilde {\infty } x e^{- \infty n} & \text {for}\: a = - \frac {1}{x} \\\tilde {\infty } x e^{\infty n} & \text {for}\: a = \frac {1}{x} \\\tilde {\infty } \int e^{n \operatorname {atanh}{\left (a x \right )}}\, dx & \text {for}\: c = 0 \\- \frac {a^{2} x^{2} \operatorname {atanh}{\left (a x \right )}}{4 a^{3} c^{2} x^{2} e^{2 \operatorname {atanh}{\left (a x \right )}} - 4 a c^{2} e^{2 \operatorname {atanh}{\left (a x \right )}}} - \frac {2 a x \operatorname {atanh}{\left (a x \right )}}{4 a^{3} c^{2} x^{2} e^{2 \operatorname {atanh}{\left (a x \right )}} - 4 a c^{2} e^{2 \operatorname {atanh}{\left (a x \right )}}} + \frac {a x}{4 a^{3} c^{2} x^{2} e^{2 \operatorname {atanh}{\left (a x \right )}} - 4 a c^{2} e^{2 \operatorname {atanh}{\left (a x \right )}}} - \frac {\operatorname {atanh}{\left (a x \right )}}{4 a^{3} c^{2} x^{2} e^{2 \operatorname {atanh}{\left (a x \right )}} - 4 a c^{2} e^{2 \operatorname {atanh}{\left (a x \right )}}} + \frac {2}{4 a^{3} c^{2} x^{2} e^{2 \operatorname {atanh}{\left (a x \right )}} - 4 a c^{2} e^{2 \operatorname {atanh}{\left (a x \right )}}} & \text {for}\: n = -2 \\- \frac {a^{2} x^{2} \log {\left (x - \frac {1}{a} \right )}}{4 a^{3} c^{2} x^{2} - 4 a c^{2}} + \frac {a^{2} x^{2} \log {\left (x + \frac {1}{a} \right )}}{4 a^{3} c^{2} x^{2} - 4 a c^{2}} - \frac {2 a x}{4 a^{3} c^{2} x^{2} - 4 a c^{2}} + \frac {\log {\left (x - \frac {1}{a} \right )}}{4 a^{3} c^{2} x^{2} - 4 a c^{2}} - \frac {\log {\left (x + \frac {1}{a} \right )}}{4 a^{3} c^{2} x^{2} - 4 a c^{2}} & \text {for}\: n = 0 \\\frac {\int \frac {e^{2 \operatorname {atanh}{\left (a x \right )}}}{a^{4} x^{4} - 2 a^{2} x^{2} + 1}\, dx}{c^{2}} & \text {for}\: n = 2 \\- \frac {2 a^{2} x^{2} e^{n \operatorname {atanh}{\left (a x \right )}}}{a^{3} c^{2} n^{3} x^{2} - 4 a^{3} c^{2} n x^{2} - a c^{2} n^{3} + 4 a c^{2} n} + \frac {2 a n x e^{n \operatorname {atanh}{\left (a x \right )}}}{a^{3} c^{2} n^{3} x^{2} - 4 a^{3} c^{2} n x^{2} - a c^{2} n^{3} + 4 a c^{2} n} - \frac {n^{2} e^{n \operatorname {atanh}{\left (a x \right )}}}{a^{3} c^{2} n^{3} x^{2} - 4 a^{3} c^{2} n x^{2} - a c^{2} n^{3} + 4 a c^{2} n} + \frac {2 e^{n \operatorname {atanh}{\left (a x \right )}}}{a^{3} c^{2} n^{3} x^{2} - 4 a^{3} c^{2} n x^{2} - a c^{2} n^{3} + 4 a c^{2} n} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*x*exp(-oo*n), Eq(a, -1/x)), (zoo*x*exp(oo*n), Eq(a, 1/x)), (zoo*Integral(exp(n*atanh(a*x)), x),
Eq(c, 0)), (-a**2*x**2*atanh(a*x)/(4*a**3*c**2*x**2*exp(2*atanh(a*x)) - 4*a*c**2*exp(2*atanh(a*x))) - 2*a*x*a
tanh(a*x)/(4*a**3*c**2*x**2*exp(2*atanh(a*x)) - 4*a*c**2*exp(2*atanh(a*x))) + a*x/(4*a**3*c**2*x**2*exp(2*atan
h(a*x)) - 4*a*c**2*exp(2*atanh(a*x))) - atanh(a*x)/(4*a**3*c**2*x**2*exp(2*atanh(a*x)) - 4*a*c**2*exp(2*atanh(
a*x))) + 2/(4*a**3*c**2*x**2*exp(2*atanh(a*x)) - 4*a*c**2*exp(2*atanh(a*x))), Eq(n, -2)), (-a**2*x**2*log(x -
1/a)/(4*a**3*c**2*x**2 - 4*a*c**2) + a**2*x**2*log(x + 1/a)/(4*a**3*c**2*x**2 - 4*a*c**2) - 2*a*x/(4*a**3*c**2
*x**2 - 4*a*c**2) + log(x - 1/a)/(4*a**3*c**2*x**2 - 4*a*c**2) - log(x + 1/a)/(4*a**3*c**2*x**2 - 4*a*c**2), E
q(n, 0)), (Integral(exp(2*atanh(a*x))/(a**4*x**4 - 2*a**2*x**2 + 1), x)/c**2, Eq(n, 2)), (-2*a**2*x**2*exp(n*a
tanh(a*x))/(a**3*c**2*n**3*x**2 - 4*a**3*c**2*n*x**2 - a*c**2*n**3 + 4*a*c**2*n) + 2*a*n*x*exp(n*atanh(a*x))/(
a**3*c**2*n**3*x**2 - 4*a**3*c**2*n*x**2 - a*c**2*n**3 + 4*a*c**2*n) - n**2*exp(n*atanh(a*x))/(a**3*c**2*n**3*
x**2 - 4*a**3*c**2*n*x**2 - a*c**2*n**3 + 4*a*c**2*n) + 2*exp(n*atanh(a*x))/(a**3*c**2*n**3*x**2 - 4*a**3*c**2
*n*x**2 - a*c**2*n**3 + 4*a*c**2*n), True))

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