3.1051 \(\int \frac{e^{2 \tanh ^{-1}(a x)} x^2}{c-a^2 c x^2} \, dx\)

Optimal. Leaf size=39 \[ \frac{x}{a^2 c}+\frac{1}{a^3 c (1-a x)}+\frac{2 \log (1-a x)}{a^3 c} \]

[Out]

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

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Rubi [A]  time = 0.0955794, antiderivative size = 39, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.08, Rules used = {6150, 43} \[ \frac{x}{a^2 c}+\frac{1}{a^3 c (1-a x)}+\frac{2 \log (1-a x)}{a^3 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

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])

Rubi steps

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

Mathematica [A]  time = 0.024749, size = 28, normalized size = 0.72 \[ \frac{a x+\frac{1}{1-a x}+2 \log (1-a x)}{a^3 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.035, size = 39, normalized size = 1. \begin{align*}{\frac{x}{{a}^{2}c}}-{\frac{1}{{a}^{3}c \left ( ax-1 \right ) }}+2\,{\frac{\ln \left ( ax-1 \right ) }{{a}^{3}c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.953028, size = 54, normalized size = 1.38 \begin{align*} -\frac{1}{a^{4} c x - a^{3} c} + \frac{x}{a^{2} c} + \frac{2 \, \log \left (a x - 1\right )}{a^{3} c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.02149, size = 89, normalized size = 2.28 \begin{align*} \frac{a^{2} x^{2} - a x + 2 \,{\left (a x - 1\right )} \log \left (a x - 1\right ) - 1}{a^{4} c x - a^{3} c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.330464, size = 32, normalized size = 0.82 \begin{align*} - \frac{1}{a^{4} c x - a^{3} c} + \frac{x}{a^{2} c} + \frac{2 \log{\left (a x - 1 \right )}}{a^{3} c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.13432, size = 53, normalized size = 1.36 \begin{align*} \frac{x}{a^{2} c} + \frac{2 \, \log \left ({\left | a x - 1 \right |}\right )}{a^{3} c} - \frac{1}{{\left (a x - 1\right )} a^{3} c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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