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

Optimal. Leaf size=48 \[ -\frac {4}{a c (1-a x)}+\frac {2}{a c (1-a x)^2}-\frac {\log (1-a x)}{a c} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6129, 43} \[ -\frac {4}{a c (1-a x)}+\frac {2}{a c (1-a x)^2}-\frac {\log (1-a x)}{a c} \]

Antiderivative was successfully verified.

[In]

Int[E^(4*ArcTanh[a*x])/(c - a*c*x),x]

[Out]

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

Rule 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 6129

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

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 36, normalized size = 0.75 \[ \frac {4 a x+(a x-1)^2 (-\log (1-a x))-2}{a c (a x-1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.55, size = 49, normalized size = 1.02 \[ \frac {4 \, a x - {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (a x - 1\right ) - 2}{a^{3} c x^{2} - 2 \, a^{2} c x + a c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.18, size = 37, normalized size = 0.77 \[ -\frac {\log \left ({\left | a x - 1 \right |}\right )}{a c} + \frac {2 \, {\left (2 \, a x - 1\right )}}{{\left (a x - 1\right )}^{2} a c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 46, normalized size = 0.96 \[ \frac {2}{c a \left (a x -1\right )^{2}}-\frac {\ln \left (a x -1\right )}{c a}+\frac {4}{c a \left (a x -1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.32, size = 44, normalized size = 0.92 \[ \frac {2 \, {\left (2 \, a x - 1\right )}}{a^{3} c x^{2} - 2 \, a^{2} c x + a c} - \frac {\log \left (a x - 1\right )}{a c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.07, size = 42, normalized size = 0.88 \[ \frac {4\,x-\frac {2}{a}}{c\,a^2\,x^2-2\,c\,a\,x+c}-\frac {\ln \left (a\,x-1\right )}{a\,c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.22, size = 37, normalized size = 0.77 \[ - \frac {- 4 a x + 2}{a^{3} c x^{2} - 2 a^{2} c x + a c} - \frac {\log {\left (a x - 1 \right )}}{a c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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