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]

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

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Rubi [A]  time = 0.05, antiderivative size = 58, 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} \[ \frac {2 b^2 \log (x)}{(a+1)^3}-\frac {2 b^2 \log (a+b x+1)}{(a+1)^3}+\frac {2 b}{(a+1)^2 x}-\frac {1-a}{2 (a+1) x^2} \]

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 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^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*}

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Mathematica [A]  time = 0.04, 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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fricas [A]  time = 0.60, size = 65, normalized size = 1.12 \[ -\frac {4 \, b^{2} x^{2} \log \left (b x + a + 1\right ) - 4 \, b^{2} x^{2} \log \relax (x) - a^{3} - 4 \, {\left (a + 1\right )} b x - a^{2} + a + 1}{2 \, {\left (a^{3} + 3 \, a^{2} + 3 \, a + 1\right )} x^{2}} \]

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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giac [B]  time = 0.18, size = 121, normalized size = 2.09 \[ \frac {2 \, b^{3} \log \left ({\left | -\frac {a}{b x + a + 1} - \frac {1}{b x + a + 1} + 1 \right |}\right )}{a^{3} b + 3 \, a^{2} b + 3 \, a b + b} - \frac {\frac {a b^{2} - 5 \, b^{2}}{a + 1} - \frac {2 \, {\left (a b^{3} - 3 \, b^{3}\right )}}{{\left (b x + a + 1\right )} b}}{2 \, {\left (a + 1\right )}^{2} {\left (\frac {a}{b x + a + 1} + \frac {1}{b x + a + 1} - 1\right )}^{2}} \]

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)

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maple [A]  time = 0.04, size = 63, normalized size = 1.09 \[ -\frac {1}{2 \left (1+a \right ) x^{2}}+\frac {a}{2 \left (1+a \right ) x^{2}}+\frac {2 b}{\left (1+a \right )^{2} x}+\frac {2 b^{2} \ln \relax (x )}{\left (1+a \right )^{3}}-\frac {2 b^{2} \ln \left (b x +a +1\right )}{\left (1+a \right )^{3}} \]

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]

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

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maxima [A]  time = 0.31, size = 74, normalized size = 1.28 \[ -\frac {2 \, b^{2} \log \left (b x + a + 1\right )}{a^{3} + 3 \, a^{2} + 3 \, a + 1} + \frac {2 \, b^{2} \log \relax (x)}{a^{3} + 3 \, a^{2} + 3 \, a + 1} + \frac {a^{2} + 4 \, b x - 1}{2 \, {\left (a^{2} + 2 \, a + 1\right )} x^{2}} \]

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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mupad [B]  time = 0.91, size = 66, normalized size = 1.14 \[ \frac {\frac {a-1}{2\,\left (a+1\right )}+\frac {2\,b\,x}{{\left (a+1\right )}^2}}{x^2}-\frac {4\,b^2\,\mathrm {atanh}\left (\frac {a^3+3\,a^2+3\,a+1}{{\left (a+1\right )}^3}+\frac {2\,b\,x}{a+1}\right )}{{\left (a+1\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 0.45, size = 209, normalized size = 3.60 \[ \frac {2 b^{2} \log {\left (x + \frac {- \frac {2 a^{4} b^{2}}{\left (a + 1\right )^{3}} - \frac {8 a^{3} b^{2}}{\left (a + 1\right )^{3}} - \frac {12 a^{2} b^{2}}{\left (a + 1\right )^{3}} + 2 a b^{2} - \frac {8 a b^{2}}{\left (a + 1\right )^{3}} + 2 b^{2} - \frac {2 b^{2}}{\left (a + 1\right )^{3}}}{4 b^{3}} \right )}}{\left (a + 1\right )^{3}} - \frac {2 b^{2} \log {\left (x + \frac {\frac {2 a^{4} b^{2}}{\left (a + 1\right )^{3}} + \frac {8 a^{3} b^{2}}{\left (a + 1\right )^{3}} + \frac {12 a^{2} b^{2}}{\left (a + 1\right )^{3}} + 2 a b^{2} + \frac {8 a b^{2}}{\left (a + 1\right )^{3}} + 2 b^{2} + \frac {2 b^{2}}{\left (a + 1\right )^{3}}}{4 b^{3}} \right )}}{\left (a + 1\right )^{3}} - \frac {- a^{2} - 4 b x + 1}{x^{2} \left (2 a^{2} + 4 a + 2\right )} \]

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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