3.1.17 \(\int \frac {e^{2 \tanh ^{-1}(a x)}}{x^3} \, dx\) [17]

Optimal. Leaf size=33 \[ -\frac {1}{2 x^2}-\frac {2 a}{x}+2 a^2 \log (x)-2 a^2 \log (1-a x) \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6261, 78} \begin {gather*} 2 a^2 \log (x)-2 a^2 \log (1-a x)-\frac {2 a}{x}-\frac {1}{2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(2*ArcTanh[a*x])/x^3,x]

[Out]

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

Rule 78

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 6261

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(x_)^(m_.), x_Symbol] :> Int[x^m*((1 + a*x)^(n/2)/(1 - a*x)^(n/2)), x] /; Fre
eQ[{a, m, n}, x] &&  !IntegerQ[(n - 1)/2]

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 33, normalized size = 1.00 \begin {gather*} -\frac {1}{2 x^2}-\frac {2 a}{x}+2 a^2 \log (x)-2 a^2 \log (1-a x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 1.01, size = 31, normalized size = 0.94

method	result	size
norman	\(\frac {-\frac {1}{2}-2 a x}{x^{2}}+2 a^{2} \ln \left (x \right )-2 a^{2} \ln \left (a x -1\right )\)	\(30\)
default	\(-\frac {1}{2 x^{2}}-\frac {2 a}{x}+2 a^{2} \ln \left (x \right )-2 a^{2} \ln \left (a x -1\right )\)	\(31\)
risch	\(\frac {-\frac {1}{2}-2 a x}{x^{2}}+2 a^{2} \ln \left (-x \right )-2 a^{2} \ln \left (a x -1\right )\)	\(32\)
meijerg	\(\frac {a^{2} \left (-\ln \left (-a^{2} x^{2}+1\right )+2 \ln \left (x \right )+\ln \left (-a^{2}\right )\right )}{2}-\frac {a^{3} \left (-\frac {2}{x \sqrt {-a^{2}}}+\frac {2 a \arctanh \left (a x \right )}{\sqrt {-a^{2}}}\right )}{\sqrt {-a^{2}}}-\frac {a^{2} \left (\ln \left (-a^{2} x^{2}+1\right )-2 \ln \left (x \right )-\ln \left (-a^{2}\right )+\frac {1}{a^{2} x^{2}}\right )}{2}\)	\(106\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)^2/(-a^2*x^2+1)/x^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.26, size = 30, normalized size = 0.91 \begin {gather*} -2 \, a^{2} \log \left (a x - 1\right ) + 2 \, a^{2} \log \left (x\right ) - \frac {4 \, a x + 1}{2 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^2/(-a^2*x^2+1)/x^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.33, size = 35, normalized size = 1.06 \begin {gather*} -\frac {4 \, a^{2} x^{2} \log \left (a x - 1\right ) - 4 \, a^{2} x^{2} \log \left (x\right ) + 4 \, a x + 1}{2 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^2/(-a^2*x^2+1)/x^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.08, size = 27, normalized size = 0.82 \begin {gather*} - 2 a^{2} \left (- \log {\left (x \right )} + \log {\left (x - \frac {1}{a} \right )}\right ) - \frac {4 a x + 1}{2 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)**2/(-a**2*x**2+1)/x**3,x)

[Out]

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

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Giac [A]
time = 0.41, size = 32, normalized size = 0.97 \begin {gather*} -2 \, a^{2} \log \left ({\left | a x - 1 \right |}\right ) + 2 \, a^{2} \log \left ({\left | x \right |}\right ) - \frac {4 \, a x + 1}{2 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^2/(-a^2*x^2+1)/x^3,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.81, size = 24, normalized size = 0.73 \begin {gather*} 4\,a^2\,\mathrm {atanh}\left (2\,a\,x-1\right )-\frac {2\,a\,x+\frac {1}{2}}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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