3.11.54 \(\int \frac {e^{2 \tanh ^{-1}(a x)}}{x (c-a^2 c x^2)} \, dx\) [1054]

Optimal. Leaf size=31 \[ \frac {1}{c (1-a x)}+\frac {\log (x)}{c}-\frac {\log (1-a x)}{c} \]

[Out]

1/c/(-a*x+1)+ln(x)/c-ln(-a*x+1)/c

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Rubi [A]
time = 0.06, antiderivative size = 31, 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 = {6285, 46} \begin {gather*} \frac {1}{c (1-a x)}-\frac {\log (1-a x)}{c}+\frac {\log (x)}{c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 46

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x
)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 6285

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

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Mathematica [A]
time = 0.02, size = 24, normalized size = 0.77 \begin {gather*} \frac {\frac {1}{1-a x}+\log (x)-\log (1-a x)}{c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.07, size = 25, normalized size = 0.81

method	result	size
default	\(\frac {\ln \left (x \right )-\frac {1}{a x -1}-\ln \left (a x -1\right )}{c}\)	\(25\)
risch	\(-\frac {1}{c \left (a x -1\right )}-\frac {\ln \left (a x -1\right )}{c}+\frac {\ln \left (-x \right )}{c}\)	\(33\)
norman	\(\frac {-\frac {a x}{c}-\frac {a^{2} x^{2}}{c}}{a^{2} x^{2}-1}+\frac {\ln \left (x \right )}{c}-\frac {\ln \left (a x -1\right )}{c}\)	\(50\)

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),x,method=_RETURNVERBOSE)

[Out]

1/c*(ln(x)-1/(a*x-1)-ln(a*x-1))

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Maxima [A]
time = 0.26, size = 30, normalized size = 0.97 \begin {gather*} -\frac {\log \left (a x - 1\right )}{c} + \frac {\log \left (x\right )}{c} - \frac {1}{a c x - c} \end {gather*}

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),x, algorithm="maxima")

[Out]

-log(a*x - 1)/c + log(x)/c - 1/(a*c*x - c)

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Fricas [A]
time = 0.33, size = 35, normalized size = 1.13 \begin {gather*} -\frac {{\left (a x - 1\right )} \log \left (a x - 1\right ) - {\left (a x - 1\right )} \log \left (x\right ) + 1}{a c x - c} \end {gather*}

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),x, algorithm="fricas")

[Out]

-((a*x - 1)*log(a*x - 1) - (a*x - 1)*log(x) + 1)/(a*c*x - c)

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Sympy [A]
time = 0.09, size = 19, normalized size = 0.61 \begin {gather*} - \frac {1}{a c x - c} + \frac {\log {\left (x \right )} - \log {\left (x - \frac {1}{a} \right )}}{c} \end {gather*}

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),x)

[Out]

-1/(a*c*x - c) + (log(x) - log(x - 1/a))/c

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Giac [A]
time = 0.42, size = 32, normalized size = 1.03 \begin {gather*} -\frac {\log \left ({\left | a x - 1 \right |}\right )}{c} + \frac {\log \left ({\left | x \right |}\right )}{c} - \frac {1}{{\left (a x - 1\right )} c} \end {gather*}

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),x, algorithm="giac")

[Out]

-log(abs(a*x - 1))/c + log(abs(x))/c - 1/((a*x - 1)*c)

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Mupad [B]
time = 0.91, size = 22, normalized size = 0.71 \begin {gather*} \frac {2\,\mathrm {atanh}\left (2\,a\,x-1\right )}{c}+\frac {1}{c-a\,c\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*atanh(2*a*x - 1))/c + 1/(c - a*c*x)

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