3.1235 \(\int \frac{e^{-2 \tanh ^{-1}(a x)}}{(c-a^2 c x^2)^4} \, dx\)

Optimal. Leaf size=119 \[ \frac{5}{64 a c^4 (1-a x)}-\frac{5}{32 a c^4 (a x+1)}+\frac{1}{64 a c^4 (1-a x)^2}-\frac{3}{32 a c^4 (a x+1)^2}-\frac{1}{16 a c^4 (a x+1)^3}-\frac{1}{32 a c^4 (a x+1)^4}+\frac{15 \tanh ^{-1}(a x)}{64 a c^4} \]

[Out]

1/(64*a*c^4*(1 - a*x)^2) + 5/(64*a*c^4*(1 - a*x)) - 1/(32*a*c^4*(1 + a*x)^4) - 1/(16*a*c^4*(1 + a*x)^3) - 3/(3
2*a*c^4*(1 + a*x)^2) - 5/(32*a*c^4*(1 + a*x)) + (15*ArcTanh[a*x])/(64*a*c^4)

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Rubi [A]  time = 0.0868269, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {6140, 44, 207} \[ \frac{5}{64 a c^4 (1-a x)}-\frac{5}{32 a c^4 (a x+1)}+\frac{1}{64 a c^4 (1-a x)^2}-\frac{3}{32 a c^4 (a x+1)^2}-\frac{1}{16 a c^4 (a x+1)^3}-\frac{1}{32 a c^4 (a x+1)^4}+\frac{15 \tanh ^{-1}(a x)}{64 a c^4} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(2*ArcTanh[a*x])*(c - a^2*c*x^2)^4),x]

[Out]

1/(64*a*c^4*(1 - a*x)^2) + 5/(64*a*c^4*(1 - a*x)) - 1/(32*a*c^4*(1 + a*x)^4) - 1/(16*a*c^4*(1 + a*x)^3) - 3/(3
2*a*c^4*(1 + a*x)^2) - 5/(32*a*c^4*(1 + a*x)) + (15*ArcTanh[a*x])/(64*a*c^4)

Rule 6140

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

Rule 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

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)}}{\left (c-a^2 c x^2\right )^4} \, dx &=\frac{\int \frac{1}{(1-a x)^3 (1+a x)^5} \, dx}{c^4}\\ &=\frac{\int \left (-\frac{1}{32 (-1+a x)^3}+\frac{5}{64 (-1+a x)^2}+\frac{1}{8 (1+a x)^5}+\frac{3}{16 (1+a x)^4}+\frac{3}{16 (1+a x)^3}+\frac{5}{32 (1+a x)^2}-\frac{15}{64 \left (-1+a^2 x^2\right )}\right ) \, dx}{c^4}\\ &=\frac{1}{64 a c^4 (1-a x)^2}+\frac{5}{64 a c^4 (1-a x)}-\frac{1}{32 a c^4 (1+a x)^4}-\frac{1}{16 a c^4 (1+a x)^3}-\frac{3}{32 a c^4 (1+a x)^2}-\frac{5}{32 a c^4 (1+a x)}-\frac{15 \int \frac{1}{-1+a^2 x^2} \, dx}{64 c^4}\\ &=\frac{1}{64 a c^4 (1-a x)^2}+\frac{5}{64 a c^4 (1-a x)}-\frac{1}{32 a c^4 (1+a x)^4}-\frac{1}{16 a c^4 (1+a x)^3}-\frac{3}{32 a c^4 (1+a x)^2}-\frac{5}{32 a c^4 (1+a x)}+\frac{15 \tanh ^{-1}(a x)}{64 a c^4}\\ \end{align*}

Mathematica [A]  time = 0.061964, size = 80, normalized size = 0.67 \[ \frac{-15 a^5 x^5-30 a^4 x^4+10 a^3 x^3+50 a^2 x^2+17 a x+15 (a x-1)^2 (a x+1)^4 \tanh ^{-1}(a x)-16}{64 a (a x-1)^2 (a c x+c)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(E^(2*ArcTanh[a*x])*(c - a^2*c*x^2)^4),x]

[Out]

(-16 + 17*a*x + 50*a^2*x^2 + 10*a^3*x^3 - 30*a^4*x^4 - 15*a^5*x^5 + 15*(-1 + a*x)^2*(1 + a*x)^4*ArcTanh[a*x])/
(64*a*(-1 + a*x)^2*(c + a*c*x)^4)

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Maple [A]  time = 0.036, size = 120, normalized size = 1. \begin{align*} -{\frac{1}{32\,a{c}^{4} \left ( ax+1 \right ) ^{4}}}-{\frac{1}{16\,a{c}^{4} \left ( ax+1 \right ) ^{3}}}-{\frac{3}{32\,a{c}^{4} \left ( ax+1 \right ) ^{2}}}-{\frac{5}{32\,a{c}^{4} \left ( ax+1 \right ) }}+{\frac{15\,\ln \left ( ax+1 \right ) }{128\,a{c}^{4}}}+{\frac{1}{64\,a{c}^{4} \left ( ax-1 \right ) ^{2}}}-{\frac{5}{64\,a{c}^{4} \left ( ax-1 \right ) }}-{\frac{15\,\ln \left ( ax-1 \right ) }{128\,a{c}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32/a/c^4/(a*x+1)^4-1/16/a/c^4/(a*x+1)^3-3/32/a/c^4/(a*x+1)^2-5/32/a/c^4/(a*x+1)+15/128*ln(a*x+1)/a/c^4+1/64
/c^4/a/(a*x-1)^2-5/64/c^4/a/(a*x-1)-15/128/c^4/a*ln(a*x-1)

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Maxima [A]  time = 1.00513, size = 189, normalized size = 1.59 \begin{align*} -\frac{15 \, a^{5} x^{5} + 30 \, a^{4} x^{4} - 10 \, a^{3} x^{3} - 50 \, a^{2} x^{2} - 17 \, a x + 16}{64 \,{\left (a^{7} c^{4} x^{6} + 2 \, a^{6} c^{4} x^{5} - a^{5} c^{4} x^{4} - 4 \, a^{4} c^{4} x^{3} - a^{3} c^{4} x^{2} + 2 \, a^{2} c^{4} x + a c^{4}\right )}} + \frac{15 \, \log \left (a x + 1\right )}{128 \, a c^{4}} - \frac{15 \, \log \left (a x - 1\right )}{128 \, a c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/64*(15*a^5*x^5 + 30*a^4*x^4 - 10*a^3*x^3 - 50*a^2*x^2 - 17*a*x + 16)/(a^7*c^4*x^6 + 2*a^6*c^4*x^5 - a^5*c^4
*x^4 - 4*a^4*c^4*x^3 - a^3*c^4*x^2 + 2*a^2*c^4*x + a*c^4) + 15/128*log(a*x + 1)/(a*c^4) - 15/128*log(a*x - 1)/
(a*c^4)

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Fricas [B]  time = 2.35021, size = 458, normalized size = 3.85 \begin{align*} -\frac{30 \, a^{5} x^{5} + 60 \, a^{4} x^{4} - 20 \, a^{3} x^{3} - 100 \, a^{2} x^{2} - 34 \, a x - 15 \,{\left (a^{6} x^{6} + 2 \, a^{5} x^{5} - a^{4} x^{4} - 4 \, a^{3} x^{3} - a^{2} x^{2} + 2 \, a x + 1\right )} \log \left (a x + 1\right ) + 15 \,{\left (a^{6} x^{6} + 2 \, a^{5} x^{5} - a^{4} x^{4} - 4 \, a^{3} x^{3} - a^{2} x^{2} + 2 \, a x + 1\right )} \log \left (a x - 1\right ) + 32}{128 \,{\left (a^{7} c^{4} x^{6} + 2 \, a^{6} c^{4} x^{5} - a^{5} c^{4} x^{4} - 4 \, a^{4} c^{4} x^{3} - a^{3} c^{4} x^{2} + 2 \, a^{2} c^{4} x + a c^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/128*(30*a^5*x^5 + 60*a^4*x^4 - 20*a^3*x^3 - 100*a^2*x^2 - 34*a*x - 15*(a^6*x^6 + 2*a^5*x^5 - a^4*x^4 - 4*a^
3*x^3 - a^2*x^2 + 2*a*x + 1)*log(a*x + 1) + 15*(a^6*x^6 + 2*a^5*x^5 - a^4*x^4 - 4*a^3*x^3 - a^2*x^2 + 2*a*x +
1)*log(a*x - 1) + 32)/(a^7*c^4*x^6 + 2*a^6*c^4*x^5 - a^5*c^4*x^4 - 4*a^4*c^4*x^3 - a^3*c^4*x^2 + 2*a^2*c^4*x +
a*c^4)

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Sympy [A]  time = 1.04576, size = 143, normalized size = 1.2 \begin{align*} - \frac{15 a^{5} x^{5} + 30 a^{4} x^{4} - 10 a^{3} x^{3} - 50 a^{2} x^{2} - 17 a x + 16}{64 a^{7} c^{4} x^{6} + 128 a^{6} c^{4} x^{5} - 64 a^{5} c^{4} x^{4} - 256 a^{4} c^{4} x^{3} - 64 a^{3} c^{4} x^{2} + 128 a^{2} c^{4} x + 64 a c^{4}} - \frac{\frac{15 \log{\left (x - \frac{1}{a} \right )}}{128} - \frac{15 \log{\left (x + \frac{1}{a} \right )}}{128}}{a c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(15*a**5*x**5 + 30*a**4*x**4 - 10*a**3*x**3 - 50*a**2*x**2 - 17*a*x + 16)/(64*a**7*c**4*x**6 + 128*a**6*c**4*
x**5 - 64*a**5*c**4*x**4 - 256*a**4*c**4*x**3 - 64*a**3*c**4*x**2 + 128*a**2*c**4*x + 64*a*c**4) - (15*log(x -
1/a)/128 - 15*log(x + 1/a)/128)/(a*c**4)

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Giac [A]  time = 1.16423, size = 165, normalized size = 1.39 \begin{align*} -\frac{15 \, \log \left ({\left | -\frac{2}{a x + 1} + 1 \right |}\right )}{128 \, a c^{4}} + \frac{\frac{24}{a x + 1} - 11}{256 \, a c^{4}{\left (\frac{2}{a x + 1} - 1\right )}^{2}} - \frac{\frac{5 \, a^{11} c^{12}}{a x + 1} + \frac{3 \, a^{11} c^{12}}{{\left (a x + 1\right )}^{2}} + \frac{2 \, a^{11} c^{12}}{{\left (a x + 1\right )}^{3}} + \frac{a^{11} c^{12}}{{\left (a x + 1\right )}^{4}}}{32 \, a^{12} c^{16}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15/128*log(abs(-2/(a*x + 1) + 1))/(a*c^4) + 1/256*(24/(a*x + 1) - 11)/(a*c^4*(2/(a*x + 1) - 1)^2) - 1/32*(5*a
^11*c^12/(a*x + 1) + 3*a^11*c^12/(a*x + 1)^2 + 2*a^11*c^12/(a*x + 1)^3 + a^11*c^12/(a*x + 1)^4)/(a^12*c^16)