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

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 28, normalized size of antiderivative = 1.00, 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 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]

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]

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 = 0.02, 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)

________________________________________________________________________________________

fricas [A]  time = 0.59, size = 23, normalized size = 0.82 \[ -\frac {{\left (a - 1\right )} \log \relax (x) + 2 \, \log \left (b x + a + 1\right )}{a + 1} \]

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)

________________________________________________________________________________________

giac [B]  time = 0.32, size = 66, normalized size = 2.36 \[ -b {\left (\frac {{\left (a - 1\right )} \log \left ({\left | -\frac {a}{b x + a + 1} - \frac {1}{b x + a + 1} + 1 \right |}\right )}{a b + b} - \frac {\log \left (\frac {{\left | b x + a + 1 \right |}}{{\left (b x + a + 1\right )}^{2} {\left | b \right |}}\right )}{b}\right )} \]

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)

________________________________________________________________________________________

maple [A]  time = 0.03, size = 34, normalized size = 1.21 \[ \frac {\ln \relax (x )}{1+a}-\frac {\ln \relax (x ) a}{1+a}-\frac {2 \ln \left (b x +a +1\right )}{1+a} \]

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]

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

________________________________________________________________________________________

maxima [A]  time = 0.31, size = 27, normalized size = 0.96 \[ -\frac {{\left (a - 1\right )} \log \relax (x)}{a + 1} - \frac {2 \, \log \left (b x + a + 1\right )}{a + 1} \]

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)

________________________________________________________________________________________

mupad [B]  time = 0.10, size = 28, normalized size = 1.00 \[ \frac {2\,\ln \relax (x)}{a+1}-\ln \relax (x)-\frac {2\,\ln \left (a+b\,x+1\right )}{a+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [B]  time = 0.46, size = 90, normalized size = 3.21 \[ - \frac {\left (a - 1\right ) \log {\left (x + \frac {- \frac {a^{2} \left (a - 1\right )}{a + 1} + a^{2} - \frac {2 a \left (a - 1\right )}{a + 1} - \frac {a - 1}{a + 1} - 1}{a b - 3 b} \right )}}{a + 1} - \frac {2 \log {\left (x + \frac {a^{2} - \frac {2 a^{2}}{a + 1} - \frac {4 a}{a + 1} - 1 - \frac {2}{a + 1}}{a b - 3 b} \right )}}{a + 1} \]

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)

________________________________________________________________________________________