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

Optimal. Leaf size=41 \[ -\frac {2 b \log (x)}{(a+1)^2}+\frac {2 b \log (a+b x+1)}{(a+1)^2}-\frac {1-a}{(a+1) x} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 41, 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, 77} \[ -\frac {2 b \log (x)}{(a+1)^2}+\frac {2 b \log (a+b x+1)}{(a+1)^2}-\frac {1-a}{(a+1) x} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(2*ArcTanh[a + b*x])*x^2),x]

[Out]

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

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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

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Mathematica [A]  time = 0.02, size = 31, normalized size = 0.76 \[ \frac {a^2+2 b x \log (a+b x+1)-2 b x \log (x)-1}{(a+1)^2 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 2.38, size = 36, normalized size = 0.88 \[ \frac {2 \, b x \log \left (b x + a + 1\right ) - 2 \, b x \log \relax (x) + a^{2} - 1}{{\left (a^{2} + 2 \, a + 1\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.19, size = 80, normalized size = 1.95 \[ -\frac {2 \, b^{2} \log \left ({\left | -\frac {a}{b x + a + 1} - \frac {1}{b x + a + 1} + 1 \right |}\right )}{a^{2} b + 2 \, a b + b} - \frac {a b - b}{{\left (a + 1\right )}^{2} {\left (\frac {a}{b x + a + 1} + \frac {1}{b x + a + 1} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 47, normalized size = 1.15 \[ -\frac {1}{\left (1+a \right ) x}+\frac {a}{\left (1+a \right ) x}-\frac {2 b \ln \relax (x )}{\left (1+a \right )^{2}}+\frac {2 b \ln \left (b x +a +1\right )}{\left (1+a \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.31, size = 48, normalized size = 1.17 \[ \frac {2 \, b \log \left (b x + a + 1\right )}{a^{2} + 2 \, a + 1} - \frac {2 \, b \log \relax (x)}{a^{2} + 2 \, a + 1} + \frac {a - 1}{{\left (a + 1\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.92, size = 47, normalized size = 1.15 \[ \frac {a-1}{x\,\left (a+1\right )}+\frac {4\,b\,\mathrm {atanh}\left (\frac {2\,b\,x+\frac {a^2+2\,a+1}{a+1}}{a+1}\right )}{{\left (a+1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a - 1)/(x*(a + 1)) + (4*b*atanh((2*b*x + (2*a + a^2 + 1)/(a + 1))/(a + 1)))/(a + 1)^2

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sympy [B]  time = 0.35, size = 143, normalized size = 3.49 \[ - \frac {2 b \log {\left (x + \frac {- \frac {2 a^{3} b}{\left (a + 1\right )^{2}} - \frac {6 a^{2} b}{\left (a + 1\right )^{2}} + 2 a b - \frac {6 a b}{\left (a + 1\right )^{2}} + 2 b - \frac {2 b}{\left (a + 1\right )^{2}}}{4 b^{2}} \right )}}{\left (a + 1\right )^{2}} + \frac {2 b \log {\left (x + \frac {\frac {2 a^{3} b}{\left (a + 1\right )^{2}} + \frac {6 a^{2} b}{\left (a + 1\right )^{2}} + 2 a b + \frac {6 a b}{\left (a + 1\right )^{2}} + 2 b + \frac {2 b}{\left (a + 1\right )^{2}}}{4 b^{2}} \right )}}{\left (a + 1\right )^{2}} - \frac {1 - a}{x \left (a + 1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*b*log(x + (-2*a**3*b/(a + 1)**2 - 6*a**2*b/(a + 1)**2 + 2*a*b - 6*a*b/(a + 1)**2 + 2*b - 2*b/(a + 1)**2)/(4
*b**2))/(a + 1)**2 + 2*b*log(x + (2*a**3*b/(a + 1)**2 + 6*a**2*b/(a + 1)**2 + 2*a*b + 6*a*b/(a + 1)**2 + 2*b +
2*b/(a + 1)**2)/(4*b**2))/(a + 1)**2 - (1 - a)/(x*(a + 1))

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