3.1.14 \(\int e^{2 \tanh ^{-1}(a x)} \, dx\) [14]

Optimal. Leaf size=16 \[ -x-\frac {2 \log (1-a x)}{a} \]

[Out]

-x-2*ln(-a*x+1)/a

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Rubi [A]
time = 0.01, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6260, 45} \begin {gather*} -\frac {2 \log (1-a x)}{a}-x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(2*ArcTanh[a*x]),x]

[Out]

-x - (2*Log[1 - a*x])/a

Rule 45

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 6260

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

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 16, normalized size = 1.00 \begin {gather*} -x-\frac {2 \log (1-a x)}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(2*ArcTanh[a*x]),x]

[Out]

-x - (2*Log[1 - a*x])/a

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Maple [A]
time = 0.71, size = 16, normalized size = 1.00

method	result	size
default	\(-x -\frac {2 \ln \left (a x -1\right )}{a}\)	\(16\)
norman	\(-x -\frac {2 \ln \left (a x -1\right )}{a}\)	\(16\)
risch	\(-x -\frac {2 \ln \left (a x -1\right )}{a}\)	\(16\)
meijerg	\(-\frac {-\frac {2 x \left (-a^{2}\right )^{\frac {3}{2}}}{a^{2}}+\frac {2 \left (-a^{2}\right )^{\frac {3}{2}} \arctanh \left (a x \right )}{a^{3}}}{2 \sqrt {-a^{2}}}-\frac {\ln \left (-a^{2} x^{2}+1\right )}{a}+\frac {\arctanh \left (a x \right )}{a}\)	\(65\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)^2/(-a^2*x^2+1),x,method=_RETURNVERBOSE)

[Out]

-x-2/a*ln(a*x-1)

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Maxima [A]
time = 0.27, size = 15, normalized size = 0.94 \begin {gather*} -x - \frac {2 \, \log \left (a x - 1\right )}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x - 2*log(a*x - 1)/a

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Fricas [A]
time = 0.35, size = 17, normalized size = 1.06 \begin {gather*} -\frac {a x + 2 \, \log \left (a x - 1\right )}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a*x + 2*log(a*x - 1))/a

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Sympy [A]
time = 0.03, size = 12, normalized size = 0.75 \begin {gather*} - x - \frac {2 \log {\left (a x - 1 \right )}}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x - 2*log(a*x - 1)/a

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Giac [A]
time = 0.42, size = 16, normalized size = 1.00 \begin {gather*} -x - \frac {2 \, \log \left ({\left | a x - 1 \right |}\right )}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x - 2*log(abs(a*x - 1))/a

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Mupad [B]
time = 0.03, size = 15, normalized size = 0.94 \begin {gather*} -x-\frac {2\,\ln \left (a\,x-1\right )}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(a*x + 1)^2/(a^2*x^2 - 1),x)

[Out]

- x - (2*log(a*x - 1))/a

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