3.858 \(\int \frac{e^{-2 \tanh ^{-1}(a+b x)}}{x^3} \, dx\)

Optimal. Leaf size=58 \[ \frac{2 b^2 \log (x)}{(a+1)^3}-\frac{2 b^2 \log (a+b x+1)}{(a+1)^3}+\frac{2 b}{(a+1)^2 x}-\frac{1-a}{2 (a+1) x^2} \]

[Out]

-(1 - a)/(2*(1 + a)*x^2) + (2*b)/((1 + a)^2*x) + (2*b^2*Log[x])/(1 + a)^3 - (2*b^2*Log[1 + a + b*x])/(1 + a)^3

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Rubi [A]  time = 0.0498517, antiderivative size = 58, 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, 77} \[ \frac{2 b^2 \log (x)}{(a+1)^3}-\frac{2 b^2 \log (a+b x+1)}{(a+1)^3}+\frac{2 b}{(a+1)^2 x}-\frac{1-a}{2 (a+1) x^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(2*ArcTanh[a + b*x])*x^3),x]

[Out]

-(1 - a)/(2*(1 + a)*x^2) + (2*b)/((1 + a)^2*x) + (2*b^2*Log[x])/(1 + a)^3 - (2*b^2*Log[1 + a + b*x])/(1 + a)^3

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 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]))))

Rubi steps

\begin{align*} \int \frac{e^{-2 \tanh ^{-1}(a+b x)}}{x^3} \, dx &=\int \frac{1-a-b x}{x^3 (1+a+b x)} \, dx\\ &=\int \left (\frac{1-a}{(1+a) x^3}-\frac{2 b}{(1+a)^2 x^2}+\frac{2 b^2}{(1+a)^3 x}-\frac{2 b^3}{(1+a)^3 (1+a+b x)}\right ) \, dx\\ &=-\frac{1-a}{2 (1+a) x^2}+\frac{2 b}{(1+a)^2 x}+\frac{2 b^2 \log (x)}{(1+a)^3}-\frac{2 b^2 \log (1+a+b x)}{(1+a)^3}\\ \end{align*}

Mathematica [A]  time = 0.0330436, size = 51, normalized size = 0.88 \[ \frac{(a+1) \left (a^2+4 b x-1\right )-4 b^2 x^2 \log (a+b x+1)+4 b^2 x^2 \log (x)}{2 (a+1)^3 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(E^(2*ArcTanh[a + b*x])*x^3),x]

[Out]

((1 + a)*(-1 + a^2 + 4*b*x) + 4*b^2*x^2*Log[x] - 4*b^2*x^2*Log[1 + a + b*x])/(2*(1 + a)^3*x^2)

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Maple [A]  time = 0.034, size = 63, normalized size = 1.1 \begin{align*} -2\,{\frac{{b}^{2}\ln \left ( bx+a+1 \right ) }{ \left ( 1+a \right ) ^{3}}}-{\frac{1}{ \left ( 2+2\,a \right ){x}^{2}}}+{\frac{a}{ \left ( 2+2\,a \right ){x}^{2}}}+2\,{\frac{b}{ \left ( 1+a \right ) ^{2}x}}+2\,{\frac{{b}^{2}\ln \left ( x \right ) }{ \left ( 1+a \right ) ^{3}}} \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^3,x)

[Out]

-2*b^2*ln(b*x+a+1)/(1+a)^3-1/2/(1+a)/x^2+1/2/(1+a)/x^2*a+2*b/(1+a)^2/x+2*b^2*ln(x)/(1+a)^3

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Maxima [A]  time = 0.95116, size = 100, normalized size = 1.72 \begin{align*} -\frac{2 \, b^{2} \log \left (b x + a + 1\right )}{a^{3} + 3 \, a^{2} + 3 \, a + 1} + \frac{2 \, b^{2} \log \left (x\right )}{a^{3} + 3 \, a^{2} + 3 \, a + 1} + \frac{a^{2} + 4 \, b x - 1}{2 \,{\left (a^{2} + 2 \, a + 1\right )} x^{2}} \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^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.5681, size = 162, normalized size = 2.79 \begin{align*} -\frac{4 \, b^{2} x^{2} \log \left (b x + a + 1\right ) - 4 \, b^{2} x^{2} \log \left (x\right ) - a^{3} - 4 \,{\left (a + 1\right )} b x - a^{2} + a + 1}{2 \,{\left (a^{3} + 3 \, a^{2} + 3 \, a + 1\right )} x^{2}} \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^3,x, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 0.651894, size = 209, normalized size = 3.6 \begin{align*} \frac{2 b^{2} \log{\left (x + \frac{- \frac{2 a^{4} b^{2}}{\left (a + 1\right )^{3}} - \frac{8 a^{3} b^{2}}{\left (a + 1\right )^{3}} - \frac{12 a^{2} b^{2}}{\left (a + 1\right )^{3}} + 2 a b^{2} - \frac{8 a b^{2}}{\left (a + 1\right )^{3}} + 2 b^{2} - \frac{2 b^{2}}{\left (a + 1\right )^{3}}}{4 b^{3}} \right )}}{\left (a + 1\right )^{3}} - \frac{2 b^{2} \log{\left (x + \frac{\frac{2 a^{4} b^{2}}{\left (a + 1\right )^{3}} + \frac{8 a^{3} b^{2}}{\left (a + 1\right )^{3}} + \frac{12 a^{2} b^{2}}{\left (a + 1\right )^{3}} + 2 a b^{2} + \frac{8 a b^{2}}{\left (a + 1\right )^{3}} + 2 b^{2} + \frac{2 b^{2}}{\left (a + 1\right )^{3}}}{4 b^{3}} \right )}}{\left (a + 1\right )^{3}} + \frac{a^{2} + 4 b x - 1}{x^{2} \left (2 a^{2} + 4 a + 2\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**3,x)

[Out]

2*b**2*log(x + (-2*a**4*b**2/(a + 1)**3 - 8*a**3*b**2/(a + 1)**3 - 12*a**2*b**2/(a + 1)**3 + 2*a*b**2 - 8*a*b*
*2/(a + 1)**3 + 2*b**2 - 2*b**2/(a + 1)**3)/(4*b**3))/(a + 1)**3 - 2*b**2*log(x + (2*a**4*b**2/(a + 1)**3 + 8*
a**3*b**2/(a + 1)**3 + 12*a**2*b**2/(a + 1)**3 + 2*a*b**2 + 8*a*b**2/(a + 1)**3 + 2*b**2 + 2*b**2/(a + 1)**3)/
(4*b**3))/(a + 1)**3 + (a**2 + 4*b*x - 1)/(x**2*(2*a**2 + 4*a + 2))

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Giac [B]  time = 1.19617, size = 163, normalized size = 2.81 \begin{align*} \frac{2 \, b^{3} \log \left ({\left | -\frac{a}{b x + a + 1} - \frac{1}{b x + a + 1} + 1 \right |}\right )}{a^{3} b + 3 \, a^{2} b + 3 \, a b + b} - \frac{\frac{a b^{2} - 5 \, b^{2}}{a + 1} - \frac{2 \,{\left (a b^{3} - 3 \, b^{3}\right )}}{{\left (b x + a + 1\right )} b}}{2 \,{\left (a + 1\right )}^{2}{\left (\frac{a}{b x + a + 1} + \frac{1}{b x + a + 1} - 1\right )}^{2}} \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^3,x, algorithm="giac")

[Out]

2*b^3*log(abs(-a/(b*x + a + 1) - 1/(b*x + a + 1) + 1))/(a^3*b + 3*a^2*b + 3*a*b + b) - 1/2*((a*b^2 - 5*b^2)/(a
+ 1) - 2*(a*b^3 - 3*b^3)/((b*x + a + 1)*b))/((a + 1)^2*(a/(b*x + a + 1) + 1/(b*x + a + 1) - 1)^2)