3.1060 \(\int \frac{e^{2 \tanh ^{-1}(a x)} x^2}{(c-a^2 c x^2)^2} \, dx\)

Optimal. Leaf size=51 \[ -\frac{3}{4 a^3 c^2 (1-a x)}+\frac{1}{4 a^3 c^2 (1-a x)^2}+\frac{\tanh ^{-1}(a x)}{4 a^3 c^2} \]

[Out]

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

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Rubi [A]  time = 0.105425, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {6150, 88, 207} \[ -\frac{3}{4 a^3 c^2 (1-a x)}+\frac{1}{4 a^3 c^2 (1-a x)^2}+\frac{\tanh ^{-1}(a x)}{4 a^3 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.026789, size = 35, normalized size = 0.69 \[ \frac{3 a x+(a x-1)^2 \tanh ^{-1}(a x)-2}{4 a^3 c^2 (a x-1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.032, size = 60, normalized size = 1.2 \begin{align*}{\frac{\ln \left ( ax+1 \right ) }{8\,{c}^{2}{a}^{3}}}+{\frac{1}{4\,{c}^{2}{a}^{3} \left ( ax-1 \right ) ^{2}}}+{\frac{3}{4\,{c}^{2}{a}^{3} \left ( ax-1 \right ) }}-{\frac{\ln \left ( ax-1 \right ) }{8\,{c}^{2}{a}^{3}}} \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)^2,x)

[Out]

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

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Maxima [A]  time = 0.947161, size = 89, normalized size = 1.75 \begin{align*} \frac{3 \, a x - 2}{4 \,{\left (a^{5} c^{2} x^{2} - 2 \, a^{4} c^{2} x + a^{3} c^{2}\right )}} + \frac{\log \left (a x + 1\right )}{8 \, a^{3} c^{2}} - \frac{\log \left (a x - 1\right )}{8 \, a^{3} c^{2}} \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)^2,x, algorithm="maxima")

[Out]

1/4*(3*a*x - 2)/(a^5*c^2*x^2 - 2*a^4*c^2*x + a^3*c^2) + 1/8*log(a*x + 1)/(a^3*c^2) - 1/8*log(a*x - 1)/(a^3*c^2
)

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Fricas [A]  time = 2.32425, size = 174, normalized size = 3.41 \begin{align*} \frac{6 \, a x +{\left (a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (a x + 1\right ) -{\left (a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (a x - 1\right ) - 4}{8 \,{\left (a^{5} c^{2} x^{2} - 2 \, a^{4} c^{2} x + a^{3} c^{2}\right )}} \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)^2,x, algorithm="fricas")

[Out]

1/8*(6*a*x + (a^2*x^2 - 2*a*x + 1)*log(a*x + 1) - (a^2*x^2 - 2*a*x + 1)*log(a*x - 1) - 4)/(a^5*c^2*x^2 - 2*a^4
*c^2*x + a^3*c^2)

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Sympy [A]  time = 0.468091, size = 60, normalized size = 1.18 \begin{align*} \frac{3 a x - 2}{4 a^{5} c^{2} x^{2} - 8 a^{4} c^{2} x + 4 a^{3} c^{2}} - \frac{\frac{\log{\left (x - \frac{1}{a} \right )}}{8} - \frac{\log{\left (x + \frac{1}{a} \right )}}{8}}{a^{3} c^{2}} \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)**2,x)

[Out]

(3*a*x - 2)/(4*a**5*c**2*x**2 - 8*a**4*c**2*x + 4*a**3*c**2) - (log(x - 1/a)/8 - log(x + 1/a)/8)/(a**3*c**2)

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Giac [A]  time = 1.1554, size = 70, normalized size = 1.37 \begin{align*} \frac{\log \left ({\left | a x + 1 \right |}\right )}{8 \, a^{3} c^{2}} - \frac{\log \left ({\left | a x - 1 \right |}\right )}{8 \, a^{3} c^{2}} + \frac{3 \, a x - 2}{4 \,{\left (a x - 1\right )}^{2} a^{3} c^{2}} \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)^2,x, algorithm="giac")

[Out]

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