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

Optimal. Leaf size=28 \[ \frac{(1-a) \log (x)}{a+1}-\frac{2 \log (a+b x+1)}{a+1} \]

[Out]

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

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Rubi [A]  time = 0.0346215, antiderivative size = 28, 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} \[ \frac{(1-a) \log (x)}{a+1}-\frac{2 \log (a+b x+1)}{a+1} \]

Antiderivative was successfully verified.

[In]

Int[1/(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 = 0.0160087, size = 23, normalized size = 0.82 \[ \frac{-2 \log (a+b x+1)-a \log (x)+\log (x)}{a+1} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(E^(2*ArcTanh[a + b*x])*x),x]

[Out]

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

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Maple [A]  time = 0.033, size = 34, normalized size = 1.2 \begin{align*} -2\,{\frac{\ln \left ( bx+a+1 \right ) }{1+a}}+{\frac{\ln \left ( x \right ) }{1+a}}-{\frac{a\ln \left ( x \right ) }{1+a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.953797, size = 36, normalized size = 1.29 \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(1/(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 = 1.52606, size = 65, normalized size = 2.32 \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(1/(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 = 0.523708, size = 90, normalized size = 3.21 \begin{align*} - \frac{\left (a - 1\right ) \log{\left (x + \frac{- \frac{a^{2} \left (a - 1\right )}{a + 1} + a^{2} - \frac{2 a \left (a - 1\right )}{a + 1} - \frac{a - 1}{a + 1} - 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(1/(b*x+a+1)**2*(1-(b*x+a)**2)/x,x)

[Out]

-(a - 1)*log(x + (-a**2*(a - 1)/(a + 1) + a**2 - 2*a*(a - 1)/(a + 1) - (a - 1)/(a + 1) - 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 [B]  time = 1.1946, size = 89, normalized size = 3.18 \begin{align*} -b{\left (\frac{{\left (a - 1\right )} \log \left ({\left | -\frac{a}{b x + a + 1} - \frac{1}{b x + a + 1} + 1 \right |}\right )}{a b + b} - \frac{\log \left (\frac{{\left | b x + a + 1 \right |}}{{\left (b x + a + 1\right )}^{2}{\left | b \right |}}\right )}{b}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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