3.944 \(\int \frac {e^{\tanh ^{-1}(a x)}}{(1-a^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=56 \[ \frac {1}{4 a (1-a x)}-\frac {1}{8 a (a x+1)}+\frac {1}{8 a (1-a x)^2}+\frac {3 \tanh ^{-1}(a x)}{8 a} \]

[Out]

1/8/a/(-a*x+1)^2+1/4/a/(-a*x+1)-1/8/a/(a*x+1)+3/8*arctanh(a*x)/a

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Rubi [A]  time = 0.06, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {6140, 44, 207} \[ \frac {1}{4 a (1-a x)}-\frac {1}{8 a (a x+1)}+\frac {1}{8 a (1-a x)^2}+\frac {3 \tanh ^{-1}(a x)}{8 a} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[a*x]/(1 - a^2*x^2)^(5/2),x]

[Out]

1/(8*a*(1 - a*x)^2) + 1/(4*a*(1 - a*x)) - 1/(8*a*(1 + a*x)) + (3*ArcTanh[a*x])/(8*a)

Rule 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 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])

Rubi steps

\begin {align*} \int \frac {e^{\tanh ^{-1}(a x)}}{\left (1-a^2 x^2\right )^{5/2}} \, dx &=\int \frac {1}{(1-a x)^3 (1+a x)^2} \, dx\\ &=\int \left (-\frac {1}{4 (-1+a x)^3}+\frac {1}{4 (-1+a x)^2}+\frac {1}{8 (1+a x)^2}-\frac {3}{8 \left (-1+a^2 x^2\right )}\right ) \, dx\\ &=\frac {1}{8 a (1-a x)^2}+\frac {1}{4 a (1-a x)}-\frac {1}{8 a (1+a x)}-\frac {3}{8} \int \frac {1}{-1+a^2 x^2} \, dx\\ &=\frac {1}{8 a (1-a x)^2}+\frac {1}{4 a (1-a x)}-\frac {1}{8 a (1+a x)}+\frac {3 \tanh ^{-1}(a x)}{8 a}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 53, normalized size = 0.95 \[ \frac {-3 a^2 x^2+3 a x+3 (a x-1)^2 (a x+1) \tanh ^{-1}(a x)+2}{8 a (a x-1)^2 (a x+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcTanh[a*x]/(1 - a^2*x^2)^(5/2),x]

[Out]

(2 + 3*a*x - 3*a^2*x^2 + 3*(-1 + a*x)^2*(1 + a*x)*ArcTanh[a*x])/(8*a*(-1 + a*x)^2*(1 + a*x))

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fricas [B]  time = 0.60, size = 99, normalized size = 1.77 \[ -\frac {6 \, a^{2} x^{2} - 6 \, a x - 3 \, {\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \log \left (a x + 1\right ) + 3 \, {\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \log \left (a x - 1\right ) - 4}{16 \, {\left (a^{4} x^{3} - a^{3} x^{2} - a^{2} x + a\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(6*a^2*x^2 - 6*a*x - 3*(a^3*x^3 - a^2*x^2 - a*x + 1)*log(a*x + 1) + 3*(a^3*x^3 - a^2*x^2 - a*x + 1)*log(
a*x - 1) - 4)/(a^4*x^3 - a^3*x^2 - a^2*x + a)

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giac [A]  time = 0.33, size = 58, normalized size = 1.04 \[ \frac {3 \, \log \left ({\left | a x + 1 \right |}\right )}{16 \, a} - \frac {3 \, \log \left ({\left | a x - 1 \right |}\right )}{16 \, a} - \frac {3 \, a^{2} x^{2} - 3 \, a x - 2}{8 \, {\left (a x + 1\right )} {\left (a x - 1\right )}^{2} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/16*log(abs(a*x + 1))/a - 3/16*log(abs(a*x - 1))/a - 1/8*(3*a^2*x^2 - 3*a*x - 2)/((a*x + 1)*(a*x - 1)^2*a)

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maple [A]  time = 0.03, size = 60, normalized size = 1.07 \[ \frac {1}{8 a \left (a x -1\right )^{2}}-\frac {1}{4 a \left (a x -1\right )}-\frac {3 \ln \left (a x -1\right )}{16 a}-\frac {1}{8 a \left (a x +1\right )}+\frac {3 \ln \left (a x +1\right )}{16 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/a/(a*x-1)^2-1/4/a/(a*x-1)-3/16/a*ln(a*x-1)-1/8/a/(a*x+1)+3/16*ln(a*x+1)/a

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maxima [A]  time = 0.31, size = 64, normalized size = 1.14 \[ -\frac {3 \, a^{2} x^{2} - 3 \, a x - 2}{8 \, {\left (a^{4} x^{3} - a^{3} x^{2} - a^{2} x + a\right )}} + \frac {3 \, \log \left (a x + 1\right )}{16 \, a} - \frac {3 \, \log \left (a x - 1\right )}{16 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(3*a^2*x^2 - 3*a*x - 2)/(a^4*x^3 - a^3*x^2 - a^2*x + a) + 3/16*log(a*x + 1)/a - 3/16*log(a*x - 1)/a

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mupad [B]  time = 0.06, size = 49, normalized size = 0.88 \[ \frac {3\,\mathrm {atanh}\left (a\,x\right )}{8\,a}-\frac {\frac {3\,x}{8}-\frac {3\,a\,x^2}{8}+\frac {1}{4\,a}}{-a^3\,x^3+a^2\,x^2+a\,x-1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(a*x + 1)/(a^2*x^2 - 1)^3,x)

[Out]

(3*atanh(a*x))/(8*a) - ((3*x)/8 - (3*a*x^2)/8 + 1/(4*a))/(a*x + a^2*x^2 - a^3*x^3 - 1)

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sympy [A]  time = 0.30, size = 65, normalized size = 1.16 \[ - \frac {3 a^{2} x^{2} - 3 a x - 2}{8 a^{4} x^{3} - 8 a^{3} x^{2} - 8 a^{2} x + 8 a} - \frac {\frac {3 \log {\left (x - \frac {1}{a} \right )}}{16} - \frac {3 \log {\left (x + \frac {1}{a} \right )}}{16}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(3*a**2*x**2 - 3*a*x - 2)/(8*a**4*x**3 - 8*a**3*x**2 - 8*a**2*x + 8*a) - (3*log(x - 1/a)/16 - 3*log(x + 1/a)/
16)/a

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