3.1052 \(\int \frac {e^{2 \tanh ^{-1}(a x)} x}{c-a^2 c x^2} \, dx\)

Optimal. Leaf size=30 \[ \frac {1}{a^2 c (1-a x)}+\frac {\log (1-a x)}{a^2 c} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {6150, 43} \[ \frac {1}{a^2 c (1-a x)}+\frac {\log (1-a x)}{a^2 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.02, size = 23, normalized size = 0.77 \[ \frac {\frac {1}{1-a x}+\log (1-a x)}{a^2 c} \]

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[1 - a*x])/(a^2*c)

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fricas [A]  time = 0.64, size = 30, normalized size = 1.00 \[ \frac {{\left (a x - 1\right )} \log \left (a x - 1\right ) - 1}{a^{3} c x - a^{2} c} \]

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) - 1)/(a^3*c*x - a^2*c)

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giac [A]  time = 0.23, size = 30, normalized size = 1.00 \[ \frac {\log \left ({\left | a x - 1 \right |}\right )}{a^{2} c} - \frac {1}{{\left (a x - 1\right )} a^{2} c} \]

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))/(a^2*c) - 1/((a*x - 1)*a^2*c)

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maple [A]  time = 0.03, size = 30, normalized size = 1.00 \[ -\frac {1}{c \,a^{2} \left (a x -1\right )}+\frac {\ln \left (a x -1\right )}{c \,a^{2}} \]

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),x)

[Out]

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

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maxima [A]  time = 0.32, size = 31, normalized size = 1.03 \[ -\frac {1}{a^{3} c x - a^{2} c} + \frac {\log \left (a x - 1\right )}{a^{2} c} \]

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]

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

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mupad [B]  time = 0.04, size = 27, normalized size = 0.90 \[ \frac {1}{a^2\,\left (c-a\,c\,x\right )}+\frac {\ln \left (a\,x-1\right )}{a^2\,c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.13, size = 24, normalized size = 0.80 \[ - \frac {1}{a^{3} c x - a^{2} c} + \frac {\log {\left (a x - 1 \right )}}{a^{2} c} \]

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

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