3.831 \(\int \frac{e^{2 \tanh ^{-1}(a+b x)}}{x} \, dx\)

Optimal. Leaf size=33 \[ \frac{(a+1) \log (x)}{1-a}-\frac{2 \log (-a-b x+1)}{1-a} \]

[Out]

((1 + a)*Log[x])/(1 - a) - (2*Log[1 - a - b*x])/(1 - a)

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Rubi [A]  time = 0.0377747, antiderivative size = 33, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {6163, 72} \[ \frac{(a+1) \log (x)}{1-a}-\frac{2 \log (-a-b x+1)}{1-a} \]

Antiderivative was successfully verified.

[In]

Int[E^(2*ArcTanh[a + b*x])/x,x]

[Out]

((1 + a)*Log[x])/(1 - a) - (2*Log[1 - a - b*x])/(1 - a)

Rule 6163

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

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rubi steps

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

Mathematica [A]  time = 0.0173799, size = 26, normalized size = 0.79 \[ \frac{2 \log (-a-b x+1)-(a+1) \log (x)}{a-1} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(2*ArcTanh[a + b*x])/x,x]

[Out]

(-((1 + a)*Log[x]) + 2*Log[1 - a - b*x])/(-1 + a)

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Maple [A]  time = 0.033, size = 35, normalized size = 1.1 \begin{align*} -{\frac{\ln \left ( x \right ) }{a-1}}-{\frac{a\ln \left ( x \right ) }{a-1}}+2\,{\frac{\ln \left ( bx+a-1 \right ) }{a-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.953696, size = 36, normalized size = 1.09 \begin{align*} -\frac{{\left (a + 1\right )} \log \left (x\right )}{a - 1} + \frac{2 \, \log \left (b x + a - 1\right )}{a - 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.88123, size = 65, normalized size = 1.97 \begin{align*} -\frac{{\left (a + 1\right )} \log \left (x\right ) - 2 \, \log \left (b x + a - 1\right )}{a - 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 0.532849, size = 88, normalized size = 2.67 \begin{align*} - \frac{\left (a + 1\right ) \log{\left (x + \frac{a^{2} - \frac{a^{2} \left (a + 1\right )}{a - 1} + \frac{2 a \left (a + 1\right )}{a - 1} - 1 - \frac{a + 1}{a - 1}}{a b + 3 b} \right )}}{a - 1} + \frac{2 \log{\left (x + \frac{a^{2} + \frac{2 a^{2}}{a - 1} - \frac{4 a}{a - 1} - 1 + \frac{2}{a - 1}}{a b + 3 b} \right )}}{a - 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.20083, size = 46, normalized size = 1.39 \begin{align*} \frac{2 \, b \log \left ({\left | b x + a - 1 \right |}\right )}{a b - b} - \frac{{\left (a + 1\right )} \log \left ({\left | x \right |}\right )}{a - 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b*log(abs(b*x + a - 1))/(a*b - b) - (a + 1)*log(abs(x))/(a - 1)