3.28 \(\int e^{4 \tanh ^{-1}(a x)} x \, dx\)

Optimal. Leaf size=39 \[ \frac{4}{a^2 (1-a x)}+\frac{8 \log (1-a x)}{a^2}+\frac{4 x}{a}+\frac{x^2}{2} \]

[Out]

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

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Rubi [A]  time = 0.0285469, antiderivative size = 39, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {6126, 77} \[ \frac{4}{a^2 (1-a x)}+\frac{8 \log (1-a x)}{a^2}+\frac{4 x}{a}+\frac{x^2}{2} \]

Antiderivative was successfully verified.

[In]

Int[E^(4*ArcTanh[a*x])*x,x]

[Out]

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

Rule 6126

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(x_)^(m_.), x_Symbol] :> Int[(x^m*(1 + a*x)^(n/2))/(1 - a*x)^(n/2), x] /; Fre
eQ[{a, m, n}, x] &&  !IntegerQ[(n - 1)/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]))))

Rubi steps

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

Mathematica [A]  time = 0.0291175, size = 39, normalized size = 1. \[ \frac{4}{a^2 (1-a x)}+\frac{8 \log (1-a x)}{a^2}+\frac{4 x}{a}+\frac{x^2}{2} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(4*ArcTanh[a*x])*x,x]

[Out]

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

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Maple [A]  time = 0.036, size = 36, normalized size = 0.9 \begin{align*}{\frac{{x}^{2}}{2}}+4\,{\frac{x}{a}}-4\,{\frac{1}{{a}^{2} \left ( ax-1 \right ) }}+8\,{\frac{\ln \left ( ax-1 \right ) }{{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)^4/(-a^2*x^2+1)^2*x,x)

[Out]

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

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Maxima [A]  time = 0.967439, size = 55, normalized size = 1.41 \begin{align*} \frac{a x^{2} + 8 \, x}{2 \, a} - \frac{4}{a^{3} x - a^{2}} + \frac{8 \, \log \left (a x - 1\right )}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^4/(-a^2*x^2+1)^2*x,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.83649, size = 109, normalized size = 2.79 \begin{align*} \frac{a^{3} x^{3} + 7 \, a^{2} x^{2} - 8 \, a x + 16 \,{\left (a x - 1\right )} \log \left (a x - 1\right ) - 8}{2 \,{\left (a^{3} x - a^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^4/(-a^2*x^2+1)^2*x,x, algorithm="fricas")

[Out]

1/2*(a^3*x^3 + 7*a^2*x^2 - 8*a*x + 16*(a*x - 1)*log(a*x - 1) - 8)/(a^3*x - a^2)

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Sympy [A]  time = 0.389765, size = 31, normalized size = 0.79 \begin{align*} \frac{x^{2}}{2} - \frac{4}{a^{3} x - a^{2}} + \frac{4 x}{a} + \frac{8 \log{\left (a x - 1 \right )}}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)**4/(-a**2*x**2+1)**2*x,x)

[Out]

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

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Giac [A]  time = 1.17703, size = 59, normalized size = 1.51 \begin{align*} \frac{8 \, \log \left ({\left | a x - 1 \right |}\right )}{a^{2}} + \frac{a^{4} x^{2} + 8 \, a^{3} x}{2 \, a^{4}} - \frac{4}{{\left (a x - 1\right )} a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^4/(-a^2*x^2+1)^2*x,x, algorithm="giac")

[Out]

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