3.32 \(\int \frac {e^{4 \tanh ^{-1}(a x)}}{x^3} \, dx\)

Optimal. Leaf size=46 \[ \frac {4 a^2}{1-a x}+8 a^2 \log (x)-8 a^2 \log (1-a x)-\frac {4 a}{x}-\frac {1}{2 x^2} \]

[Out]

-1/2/x^2-4*a/x+4*a^2/(-a*x+1)+8*a^2*ln(x)-8*a^2*ln(-a*x+1)

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Rubi [A]  time = 0.04, antiderivative size = 46, 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 = {6126, 88} \[ \frac {4 a^2}{1-a x}+8 a^2 \log (x)-8 a^2 \log (1-a x)-\frac {4 a}{x}-\frac {1}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[E^(4*ArcTanh[a*x])/x^3,x]

[Out]

-1/(2*x^2) - (4*a)/x + (4*a^2)/(1 - a*x) + 8*a^2*Log[x] - 8*a^2*Log[1 - a*x]

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 6126

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^{4 \tanh ^{-1}(a x)}}{x^3} \, dx &=\int \frac {(1+a x)^2}{x^3 (1-a x)^2} \, dx\\ &=\int \left (\frac {1}{x^3}+\frac {4 a}{x^2}+\frac {8 a^2}{x}+\frac {4 a^3}{(-1+a x)^2}-\frac {8 a^3}{-1+a x}\right ) \, dx\\ &=-\frac {1}{2 x^2}-\frac {4 a}{x}+\frac {4 a^2}{1-a x}+8 a^2 \log (x)-8 a^2 \log (1-a x)\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 46, normalized size = 1.00 \[ \frac {4 a^2}{1-a x}+8 a^2 \log (x)-8 a^2 \log (1-a x)-\frac {4 a}{x}-\frac {1}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(4*ArcTanh[a*x])/x^3,x]

[Out]

-1/2*1/x^2 - (4*a)/x + (4*a^2)/(1 - a*x) + 8*a^2*Log[x] - 8*a^2*Log[1 - a*x]

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fricas [A]  time = 0.55, size = 73, normalized size = 1.59 \[ -\frac {16 \, a^{2} x^{2} - 7 \, a x + 16 \, {\left (a^{3} x^{3} - a^{2} x^{2}\right )} \log \left (a x - 1\right ) - 16 \, {\left (a^{3} x^{3} - a^{2} x^{2}\right )} \log \relax (x) - 1}{2 \, {\left (a x^{3} - x^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^4/(-a^2*x^2+1)^2/x^3,x, algorithm="fricas")

[Out]

-1/2*(16*a^2*x^2 - 7*a*x + 16*(a^3*x^3 - a^2*x^2)*log(a*x - 1) - 16*(a^3*x^3 - a^2*x^2)*log(x) - 1)/(a*x^3 - x
^2)

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giac [A]  time = 0.39, size = 47, normalized size = 1.02 \[ -8 \, a^{2} \log \left ({\left | a x - 1 \right |}\right ) + 8 \, a^{2} \log \left ({\left | x \right |}\right ) - \frac {16 \, a^{2} x^{2} - 7 \, a x - 1}{2 \, {\left (a x - 1\right )} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^4/(-a^2*x^2+1)^2/x^3,x, algorithm="giac")

[Out]

-8*a^2*log(abs(a*x - 1)) + 8*a^2*log(abs(x)) - 1/2*(16*a^2*x^2 - 7*a*x - 1)/((a*x - 1)*x^2)

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maple [A]  time = 0.04, size = 43, normalized size = 0.93 \[ -\frac {1}{2 x^{2}}-\frac {4 a}{x}+8 a^{2} \ln \relax (x )-\frac {4 a^{2}}{a x -1}-8 a^{2} \ln \left (a x -1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)^4/(-a^2*x^2+1)^2/x^3,x)

[Out]

-1/2/x^2-4*a/x+8*a^2*ln(x)-4*a^2/(a*x-1)-8*a^2*ln(a*x-1)

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maxima [A]  time = 0.30, size = 48, normalized size = 1.04 \[ -8 \, a^{2} \log \left (a x - 1\right ) + 8 \, a^{2} \log \relax (x) - \frac {16 \, a^{2} x^{2} - 7 \, a x - 1}{2 \, {\left (a x^{3} - x^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^4/(-a^2*x^2+1)^2/x^3,x, algorithm="maxima")

[Out]

-8*a^2*log(a*x - 1) + 8*a^2*log(x) - 1/2*(16*a^2*x^2 - 7*a*x - 1)/(a*x^3 - x^2)

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mupad [B]  time = 0.06, size = 41, normalized size = 0.89 \[ 16\,a^2\,\mathrm {atanh}\left (2\,a\,x-1\right )+\frac {-8\,a^2\,x^2+\frac {7\,a\,x}{2}+\frac {1}{2}}{a\,x^3-x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x + 1)^4/(x^3*(a^2*x^2 - 1)^2),x)

[Out]

16*a^2*atanh(2*a*x - 1) + ((7*a*x)/2 - 8*a^2*x^2 + 1/2)/(a*x^3 - x^2)

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sympy [A]  time = 0.25, size = 41, normalized size = 0.89 \[ 8 a^{2} \left (\log {\relax (x )} - \log {\left (x - \frac {1}{a} \right )}\right ) + \frac {- 16 a^{2} x^{2} + 7 a x + 1}{2 a x^{3} - 2 x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)**4/(-a**2*x**2+1)**2/x**3,x)

[Out]

8*a**2*(log(x) - log(x - 1/a)) + (-16*a**2*x**2 + 7*a*x + 1)/(2*a*x**3 - 2*x**2)

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