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

Optimal. Leaf size=48 \[ \frac {2 b \log (x)}{(1-a)^2}-\frac {2 b \log (-a-b x+1)}{(1-a)^2}-\frac {a+1}{(1-a) x} \]

[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 = 48, 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} \[ \frac {2 b \log (x)}{(1-a)^2}-\frac {2 b \log (-a-b x+1)}{(1-a)^2}-\frac {a+1}{(1-a) x} \]

Antiderivative was successfully verified.

[In]

Int[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 = 34, normalized size = 0.71 \[ \frac {a^2-2 b x \log (-a-b x+1)+2 b x \log (x)-1}{(a-1)^2 x} \]

Antiderivative was successfully verified.

[In]

Integrate[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 = 1.18, size = 39, normalized size = 0.81 \[ -\frac {2 \, b x \log \left (b x + a - 1\right ) - 2 \, b x \log \relax (x) - a^{2} + 1}{{\left (a^{2} - 2 \, a + 1\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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 [A]  time = 0.14, size = 57, normalized size = 1.19 \[ -\frac {2 \, b^{2} \log \left ({\left | b x + a - 1 \right |}\right )}{a^{2} b - 2 \, a b + b} + \frac {2 \, b \log \left ({\left | x \right |}\right )}{a^{2} - 2 \, a + 1} + \frac {a^{2} - 1}{{\left (a - 1\right )}^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 46, normalized size = 0.96 \[ \frac {1}{\left (a -1\right ) x}+\frac {a}{\left (a -1\right ) x}+\frac {2 b \ln \relax (x )}{\left (a -1\right )^{2}}-\frac {2 b \ln \left (b x +a -1\right )}{\left (a -1\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.31, size = 48, normalized size = 1.00 \[ -\frac {2 \, b \log \left (b x + a - 1\right )}{a^{2} - 2 \, a + 1} + \frac {2 \, b \log \relax (x)}{a^{2} - 2 \, a + 1} + \frac {a + 1}{{\left (a - 1\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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 = 0.10, size = 47, normalized size = 0.98 \[ \frac {a+1}{x\,\left (a-1\right )}-\frac {4\,b\,\mathrm {atanh}\left (\frac {2\,b\,x+\frac {a^2-2\,a+1}{a-1}}{a-1}\right )}{{\left (a-1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 0.35, size = 144, normalized size = 3.00 \[ \frac {2 b \log {\left (x + \frac {- \frac {2 a^{3} b}{\left (a - 1\right )^{2}} + \frac {6 a^{2} b}{\left (a - 1\right )^{2}} + 2 a b - \frac {6 a b}{\left (a - 1\right )^{2}} - 2 b + \frac {2 b}{\left (a - 1\right )^{2}}}{4 b^{2}} \right )}}{\left (a - 1\right )^{2}} - \frac {2 b \log {\left (x + \frac {\frac {2 a^{3} b}{\left (a - 1\right )^{2}} - \frac {6 a^{2} b}{\left (a - 1\right )^{2}} + 2 a b + \frac {6 a b}{\left (a - 1\right )^{2}} - 2 b - \frac {2 b}{\left (a - 1\right )^{2}}}{4 b^{2}} \right )}}{\left (a - 1\right )^{2}} - \frac {- a - 1}{x \left (a - 1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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 - (-a - 1)/(x*(a - 1))

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