3.1063 \(\int \frac {e^{2 \tanh ^{-1}(a x)}}{x (c-a^2 c x^2)^2} \, dx\)

Optimal. Leaf size=64 \[ \frac {3}{4 c^2 (1-a x)}+\frac {1}{4 c^2 (1-a x)^2}-\frac {7 \log (1-a x)}{8 c^2}-\frac {\log (a x+1)}{8 c^2}+\frac {\log (x)}{c^2} \]

[Out]

1/4/c^2/(-a*x+1)^2+3/4/c^2/(-a*x+1)+ln(x)/c^2-7/8*ln(-a*x+1)/c^2-1/8*ln(a*x+1)/c^2

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Rubi [A]  time = 0.10, antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {6150, 72} \[ \frac {3}{4 c^2 (1-a x)}+\frac {1}{4 c^2 (1-a x)^2}-\frac {7 \log (1-a x)}{8 c^2}-\frac {\log (a x+1)}{8 c^2}+\frac {\log (x)}{c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

1/(4*c^2*(1 - a*x)^2) + 3/(4*c^2*(1 - a*x)) + Log[x]/c^2 - (7*Log[1 - a*x])/(8*c^2) - Log[1 + a*x]/(8*c^2)

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rule 6150

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[x^m*(1 -
a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p
] || GtQ[c, 0])

Rubi steps

\begin {align*} \int \frac {e^{2 \tanh ^{-1}(a x)}}{x \left (c-a^2 c x^2\right )^2} \, dx &=\frac {\int \frac {1}{x (1-a x)^3 (1+a x)} \, dx}{c^2}\\ &=\frac {\int \left (\frac {1}{x}-\frac {a}{2 (-1+a x)^3}+\frac {3 a}{4 (-1+a x)^2}-\frac {7 a}{8 (-1+a x)}-\frac {a}{8 (1+a x)}\right ) \, dx}{c^2}\\ &=\frac {1}{4 c^2 (1-a x)^2}+\frac {3}{4 c^2 (1-a x)}+\frac {\log (x)}{c^2}-\frac {7 \log (1-a x)}{8 c^2}-\frac {\log (1+a x)}{8 c^2}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 48, normalized size = 0.75 \[ \frac {\frac {6}{1-a x}+\frac {2}{(a x-1)^2}-7 \log (1-a x)-\log (a x+1)+8 \log (x)}{8 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6/(1 - a*x) + 2/(-1 + a*x)^2 + 8*Log[x] - 7*Log[1 - a*x] - Log[1 + a*x])/(8*c^2)

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fricas [A]  time = 0.56, size = 89, normalized size = 1.39 \[ -\frac {6 \, a x + {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (a x + 1\right ) + 7 \, {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (a x - 1\right ) - 8 \, {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \log \relax (x) - 8}{8 \, {\left (a^{2} c^{2} x^{2} - 2 \, a c^{2} x + c^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(6*a*x + (a^2*x^2 - 2*a*x + 1)*log(a*x + 1) + 7*(a^2*x^2 - 2*a*x + 1)*log(a*x - 1) - 8*(a^2*x^2 - 2*a*x +
1)*log(x) - 8)/(a^2*c^2*x^2 - 2*a*c^2*x + c^2)

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giac [A]  time = 0.18, size = 50, normalized size = 0.78 \[ -\frac {\log \left ({\left | a x + 1 \right |}\right )}{8 \, c^{2}} - \frac {7 \, \log \left ({\left | a x - 1 \right |}\right )}{8 \, c^{2}} + \frac {\log \left ({\left | x \right |}\right )}{c^{2}} - \frac {3 \, a x - 4}{4 \, {\left (a x - 1\right )}^{2} c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*log(abs(a*x + 1))/c^2 - 7/8*log(abs(a*x - 1))/c^2 + log(abs(x))/c^2 - 1/4*(3*a*x - 4)/((a*x - 1)^2*c^2)

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maple [A]  time = 0.04, size = 54, normalized size = 0.84 \[ \frac {\ln \relax (x )}{c^{2}}+\frac {1}{4 c^{2} \left (a x -1\right )^{2}}-\frac {3}{4 c^{2} \left (a x -1\right )}-\frac {7 \ln \left (a x -1\right )}{8 c^{2}}-\frac {\ln \left (a x +1\right )}{8 c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

ln(x)/c^2+1/4/c^2/(a*x-1)^2-3/4/c^2/(a*x-1)-7/8/c^2*ln(a*x-1)-1/8*ln(a*x+1)/c^2

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maxima [A]  time = 0.35, size = 60, normalized size = 0.94 \[ -\frac {3 \, a x - 4}{4 \, {\left (a^{2} c^{2} x^{2} - 2 \, a c^{2} x + c^{2}\right )}} - \frac {\log \left (a x + 1\right )}{8 \, c^{2}} - \frac {7 \, \log \left (a x - 1\right )}{8 \, c^{2}} + \frac {\log \relax (x)}{c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(3*a*x - 4)/(a^2*c^2*x^2 - 2*a*c^2*x + c^2) - 1/8*log(a*x + 1)/c^2 - 7/8*log(a*x - 1)/c^2 + log(x)/c^2

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mupad [B]  time = 0.96, size = 60, normalized size = 0.94 \[ \frac {\ln \relax (x)}{c^2}-\frac {7\,\ln \left (a\,x-1\right )}{8\,c^2}-\frac {\ln \left (a\,x+1\right )}{8\,c^2}-\frac {\frac {3\,a\,x}{4}-1}{a^2\,c^2\,x^2-2\,a\,c^2\,x+c^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x)/c^2 - (7*log(a*x - 1))/(8*c^2) - log(a*x + 1)/(8*c^2) - ((3*a*x)/4 - 1)/(c^2 + a^2*c^2*x^2 - 2*a*c^2*x)

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sympy [A]  time = 0.48, size = 58, normalized size = 0.91 \[ - \frac {3 a x - 4}{4 a^{2} c^{2} x^{2} - 8 a c^{2} x + 4 c^{2}} - \frac {- \log {\relax (x )} + \frac {7 \log {\left (x - \frac {1}{a} \right )}}{8} + \frac {\log {\left (x + \frac {1}{a} \right )}}{8}}{c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(3*a*x - 4)/(4*a**2*c**2*x**2 - 8*a*c**2*x + 4*c**2) - (-log(x) + 7*log(x - 1/a)/8 + log(x + 1/a)/8)/c**2

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