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

Optimal. Leaf size=41 \[ \frac {1}{8 a^2 (a x+1)}+\frac {1}{8 a^2 (1-a x)^2}-\frac {\tanh ^{-1}(a x)}{8 a^2} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {6150, 77, 207} \[ \frac {1}{8 a^2 (a x+1)}+\frac {1}{8 a^2 (1-a x)^2}-\frac {\tanh ^{-1}(a x)}{8 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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 6150

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[x^m*(1 -
a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, m, 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)} x}{\left (1-a^2 x^2\right )^{5/2}} \, dx &=\int \frac {x}{(1-a x)^3 (1+a x)^2} \, dx\\ &=\int \left (-\frac {1}{4 a (-1+a x)^3}-\frac {1}{8 a (1+a x)^2}+\frac {1}{8 a \left (-1+a^2 x^2\right )}\right ) \, dx\\ &=\frac {1}{8 a^2 (1-a x)^2}+\frac {1}{8 a^2 (1+a x)}+\frac {\int \frac {1}{-1+a^2 x^2} \, dx}{8 a}\\ &=\frac {1}{8 a^2 (1-a x)^2}+\frac {1}{8 a^2 (1+a x)}-\frac {\tanh ^{-1}(a x)}{8 a^2}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 28, normalized size = 0.68 \[ \frac {\frac {1}{a x+1}+\frac {1}{(a x-1)^2}-\tanh ^{-1}(a x)}{8 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.53, size = 100, normalized size = 2.44 \[ \frac {2 \, a^{2} x^{2} - 2 \, a x - {\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \log \left (a x + 1\right ) + {\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \log \left (a x - 1\right ) + 4}{16 \, {\left (a^{5} x^{3} - a^{4} x^{2} - a^{3} x + a^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(2*a^2*x^2 - 2*a*x - (a^3*x^3 - a^2*x^2 - a*x + 1)*log(a*x + 1) + (a^3*x^3 - a^2*x^2 - a*x + 1)*log(a*x -
1) + 4)/(a^5*x^3 - a^4*x^2 - a^3*x + a^2)

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giac [A]  time = 0.16, size = 57, normalized size = 1.39 \[ -\frac {\log \left ({\left | a x + 1 \right |}\right )}{16 \, a^{2}} + \frac {\log \left ({\left | a x - 1 \right |}\right )}{16 \, a^{2}} + \frac {a^{2} x^{2} - a x + 2}{8 \, {\left (a x + 1\right )} {\left (a x - 1\right )}^{2} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 48, normalized size = 1.17 \[ \frac {1}{8 a^{2} \left (a x -1\right )^{2}}+\frac {\ln \left (a x -1\right )}{16 a^{2}}+\frac {1}{8 a^{2} \left (a x +1\right )}-\frac {\ln \left (a x +1\right )}{16 a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.32, size = 65, normalized size = 1.59 \[ \frac {a^{2} x^{2} - a x + 2}{8 \, {\left (a^{5} x^{3} - a^{4} x^{2} - a^{3} x + a^{2}\right )}} - \frac {\log \left (a x + 1\right )}{16 \, a^{2}} + \frac {\log \left (a x - 1\right )}{16 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(a^2*x^2 - a*x + 2)/(a^5*x^3 - a^4*x^2 - a^3*x + a^2) - 1/16*log(a*x + 1)/a^2 + 1/16*log(a*x - 1)/a^2

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mupad [B]  time = 0.93, size = 51, normalized size = 1.24 \[ -\frac {\frac {1}{4\,a^2}-\frac {x}{8\,a}+\frac {x^2}{8}}{-a^3\,x^3+a^2\,x^2+a\,x-1}-\frac {\mathrm {atanh}\left (a\,x\right )}{8\,a^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x*(a*x + 1))/(a^2*x^2 - 1)^3,x)

[Out]

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

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sympy [A]  time = 0.29, size = 61, normalized size = 1.49 \[ - \frac {- a^{2} x^{2} + a x - 2}{8 a^{5} x^{3} - 8 a^{4} x^{2} - 8 a^{3} x + 8 a^{2}} - \frac {- \frac {\log {\left (x - \frac {1}{a} \right )}}{16} + \frac {\log {\left (x + \frac {1}{a} \right )}}{16}}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-a**2*x**2 + a*x - 2)/(8*a**5*x**3 - 8*a**4*x**2 - 8*a**3*x + 8*a**2) - (-log(x - 1/a)/16 + log(x + 1/a)/16)
/a**2

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