3.9.30 \(\int e^{2 \tanh ^{-1}(a+b x)} \, dx\) [830]

Optimal. Leaf size=19 \[ -x-\frac {2 \log (1-a-b x)}{b} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.01, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6296, 45} \begin {gather*} -\frac {2 \log (-a-b x+1)}{b}-x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(2*ArcTanh[a + b*x]),x]

[Out]

-x - (2*Log[1 - a - b*x])/b

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 6296

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]
time = 0.01, size = 19, normalized size = 1.00 \begin {gather*} -x-\frac {2 \log (1-a-b x)}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(2*ArcTanh[a + b*x]),x]

[Out]

-x - (2*Log[1 - a - b*x])/b

________________________________________________________________________________________

Maple [A]
time = 0.06, size = 17, normalized size = 0.89

method	result	size
default	\(-x -\frac {2 \ln \left (b x +a -1\right )}{b}\)	\(17\)
norman	\(-x -\frac {2 \ln \left (b x +a -1\right )}{b}\)	\(17\)
risch	\(-x -\frac {2 \ln \left (b x +a -1\right )}{b}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.25, size = 16, normalized size = 0.84 \begin {gather*} -x - \frac {2 \, \log \left (b x + a - 1\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x - 2*log(b*x + a - 1)/b

________________________________________________________________________________________

Fricas [A]
time = 0.39, size = 18, normalized size = 0.95 \begin {gather*} -\frac {b x + 2 \, \log \left (b x + a - 1\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b*x + 2*log(b*x + a - 1))/b

________________________________________________________________________________________

Sympy [A]
time = 0.05, size = 14, normalized size = 0.74 \begin {gather*} - x - \frac {2 \log {\left (a + b x - 1 \right )}}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x - 2*log(a + b*x - 1)/b

________________________________________________________________________________________

Giac [A]
time = 0.41, size = 17, normalized size = 0.89 \begin {gather*} -x - \frac {2 \, \log \left ({\left | b x + a - 1 \right |}\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x - 2*log(abs(b*x + a - 1))/b

________________________________________________________________________________________

Mupad [B]
time = 0.04, size = 16, normalized size = 0.84 \begin {gather*} -x-\frac {2\,\ln \left (a+b\,x-1\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- x - (2*log(a + b*x - 1))/b

________________________________________________________________________________________