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

Optimal. Leaf size=54 \[ \frac {4 a^3}{1-a x}+12 a^3 \log (x)-12 a^3 \log (1-a x)-\frac {8 a^2}{x}-\frac {2 a}{x^2}-\frac {1}{3 x^3} \]

[Out]

-1/3/x^3-2*a/x^2-8*a^2/x+4*a^3/(-a*x+1)+12*a^3*ln(x)-12*a^3*ln(-a*x+1)

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Rubi [A]  time = 0.04, antiderivative size = 54, 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} \[ \frac {4 a^3}{1-a x}-\frac {8 a^2}{x}+12 a^3 \log (x)-12 a^3 \log (1-a x)-\frac {2 a}{x^2}-\frac {1}{3 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(3*x^3) - (2*a)/x^2 - (8*a^2)/x + (4*a^3)/(1 - a*x) + 12*a^3*Log[x] - 12*a^3*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^4} \, dx &=\int \frac {(1+a x)^2}{x^4 (1-a x)^2} \, dx\\ &=\int \left (\frac {1}{x^4}+\frac {4 a}{x^3}+\frac {8 a^2}{x^2}+\frac {12 a^3}{x}+\frac {4 a^4}{(-1+a x)^2}-\frac {12 a^4}{-1+a x}\right ) \, dx\\ &=-\frac {1}{3 x^3}-\frac {2 a}{x^2}-\frac {8 a^2}{x}+\frac {4 a^3}{1-a x}+12 a^3 \log (x)-12 a^3 \log (1-a x)\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 54, normalized size = 1.00 \[ \frac {4 a^3}{1-a x}+12 a^3 \log (x)-12 a^3 \log (1-a x)-\frac {8 a^2}{x}-\frac {2 a}{x^2}-\frac {1}{3 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*1/x^3 - (2*a)/x^2 - (8*a^2)/x + (4*a^3)/(1 - a*x) + 12*a^3*Log[x] - 12*a^3*Log[1 - a*x]

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fricas [A]  time = 0.45, size = 81, normalized size = 1.50 \[ -\frac {36 \, a^{3} x^{3} - 18 \, a^{2} x^{2} - 5 \, a x + 36 \, {\left (a^{4} x^{4} - a^{3} x^{3}\right )} \log \left (a x - 1\right ) - 36 \, {\left (a^{4} x^{4} - a^{3} x^{3}\right )} \log \relax (x) - 1}{3 \, {\left (a x^{4} - x^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(36*a^3*x^3 - 18*a^2*x^2 - 5*a*x + 36*(a^4*x^4 - a^3*x^3)*log(a*x - 1) - 36*(a^4*x^4 - a^3*x^3)*log(x) -
1)/(a*x^4 - x^3)

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giac [A]  time = 0.18, size = 55, normalized size = 1.02 \[ -12 \, a^{3} \log \left ({\left | a x - 1 \right |}\right ) + 12 \, a^{3} \log \left ({\left | x \right |}\right ) - \frac {36 \, a^{3} x^{3} - 18 \, a^{2} x^{2} - 5 \, a x - 1}{3 \, {\left (a x - 1\right )} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-12*a^3*log(abs(a*x - 1)) + 12*a^3*log(abs(x)) - 1/3*(36*a^3*x^3 - 18*a^2*x^2 - 5*a*x - 1)/((a*x - 1)*x^3)

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maple [A]  time = 0.03, size = 51, normalized size = 0.94 \[ -\frac {1}{3 x^{3}}-\frac {2 a}{x^{2}}-\frac {8 a^{2}}{x}+12 a^{3} \ln \relax (x )-\frac {4 a^{3}}{a x -1}-12 a^{3} \ln \left (a x -1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3/x^3-2*a/x^2-8*a^2/x+12*a^3*ln(x)-4*a^3/(a*x-1)-12*a^3*ln(a*x-1)

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maxima [A]  time = 0.31, size = 56, normalized size = 1.04 \[ -12 \, a^{3} \log \left (a x - 1\right ) + 12 \, a^{3} \log \relax (x) - \frac {36 \, a^{3} x^{3} - 18 \, a^{2} x^{2} - 5 \, a x - 1}{3 \, {\left (a x^{4} - x^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-12*a^3*log(a*x - 1) + 12*a^3*log(x) - 1/3*(36*a^3*x^3 - 18*a^2*x^2 - 5*a*x - 1)/(a*x^4 - x^3)

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mupad [B]  time = 0.82, size = 49, normalized size = 0.91 \[ 24\,a^3\,\mathrm {atanh}\left (2\,a\,x-1\right )+\frac {-12\,a^3\,x^3+6\,a^2\,x^2+\frac {5\,a\,x}{3}+\frac {1}{3}}{a\,x^4-x^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

24*a^3*atanh(2*a*x - 1) + ((5*a*x)/3 + 6*a^2*x^2 - 12*a^3*x^3 + 1/3)/(a*x^4 - x^3)

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sympy [A]  time = 0.27, size = 49, normalized size = 0.91 \[ 12 a^{3} \left (\log {\relax (x )} - \log {\left (x - \frac {1}{a} \right )}\right ) + \frac {- 36 a^{3} x^{3} + 18 a^{2} x^{2} + 5 a x + 1}{3 a x^{4} - 3 x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

12*a**3*(log(x) - log(x - 1/a)) + (-36*a**3*x**3 + 18*a**2*x**2 + 5*a*x + 1)/(3*a*x**4 - 3*x**3)

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