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

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Rubi [A]  time = 0.11, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, 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 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])

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^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*}

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Mathematica [A]  time = 0.03, 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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fricas [A]  time = 0.58, size = 78, normalized size = 1.53 \[ \frac {6 \, a x + {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (a x + 1\right ) - {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (a x - 1\right ) - 4}{8 \, {\left (a^{5} c^{2} x^{2} - 2 \, a^{4} c^{2} x + a^{3} c^{2}\right )}} \]

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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giac [A]  time = 0.30, size = 52, normalized size = 1.02 \[ \frac {\log \left ({\left | a x + 1 \right |}\right )}{8 \, a^{3} c^{2}} - \frac {\log \left ({\left | a x - 1 \right |}\right )}{8 \, a^{3} c^{2}} + \frac {3 \, a x - 2}{4 \, {\left (a x - 1\right )}^{2} a^{3} c^{2}} \]

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)

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maple [A]  time = 0.03, size = 60, normalized size = 1.18 \[ \frac {1}{4 c^{2} a^{3} \left (a x -1\right )^{2}}+\frac {3}{4 c^{2} a^{3} \left (a x -1\right )}-\frac {\ln \left (a x -1\right )}{8 c^{2} a^{3}}+\frac {\ln \left (a x +1\right )}{8 c^{2} a^{3}} \]

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]

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

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maxima [A]  time = 0.32, size = 66, normalized size = 1.29 \[ \frac {3 \, a x - 2}{4 \, {\left (a^{5} c^{2} x^{2} - 2 \, a^{4} c^{2} x + a^{3} c^{2}\right )}} + \frac {\log \left (a x + 1\right )}{8 \, a^{3} c^{2}} - \frac {\log \left (a x - 1\right )}{8 \, a^{3} c^{2}} \]

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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mupad [B]  time = 0.07, size = 49, normalized size = 0.96 \[ \frac {\frac {3\,x}{4\,a^2}-\frac {1}{2\,a^3}}{a^2\,c^2\,x^2-2\,a\,c^2\,x+c^2}+\frac {\mathrm {atanh}\left (a\,x\right )}{4\,a^3\,c^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.30, size = 61, normalized size = 1.20 \[ - \frac {- 3 a x + 2}{4 a^{5} c^{2} x^{2} - 8 a^{4} c^{2} x + 4 a^{3} c^{2}} - \frac {\frac {\log {\left (x - \frac {1}{a} \right )}}{8} - \frac {\log {\left (x + \frac {1}{a} \right )}}{8}}{a^{3} c^{2}} \]

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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