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

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

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6129, 44, 207} \[ \frac {1}{4 a c^4 (1-a x)}+\frac {1}{4 a c^4 (1-a x)^2}+\frac {\tanh ^{-1}(a x)}{4 a c^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

Rule 6129

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

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 35, normalized size = 0.69 \[ \frac {-a x+(a x-1)^2 \tanh ^{-1}(a x)+2}{4 a c^4 (a x-1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.52, size = 76, normalized size = 1.49 \[ -\frac {2 \, a x - {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (a x + 1\right ) + {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (a x - 1\right ) - 4}{8 \, {\left (a^{3} c^{4} x^{2} - 2 \, a^{2} c^{4} x + a c^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.58, size = 58, normalized size = 1.14 \[ -\frac {\log \left ({\left | -\frac {2}{a x + 1} + 1 \right |}\right )}{8 \, a c^{4}} - \frac {\frac {3}{a} - \frac {8}{{\left (a x + 1\right )} a}}{16 \, c^{4} {\left (\frac {2}{a x + 1} - 1\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*log(abs(-2/(a*x + 1) + 1))/(a*c^4) - 1/16*(3/a - 8/((a*x + 1)*a))/(c^4*(2/(a*x + 1) - 1)^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.33, size = 63, normalized size = 1.24 \[ -\frac {a x - 2}{4 \, {\left (a^{3} c^{4} x^{2} - 2 \, a^{2} c^{4} x + a c^{4}\right )}} + \frac {\log \left (a x + 1\right )}{8 \, a c^{4}} - \frac {\log \left (a x - 1\right )}{8 \, a c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.82, size = 47, normalized size = 0.92 \[ \frac {\mathrm {atanh}\left (a\,x\right )}{4\,a\,c^4}-\frac {\frac {x}{4}-\frac {1}{2\,a}}{a^2\,c^4\,x^2-2\,a\,c^4\,x+c^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.28, size = 56, normalized size = 1.10 \[ - \frac {a x - 2}{4 a^{3} c^{4} x^{2} - 8 a^{2} c^{4} x + 4 a c^{4}} - \frac {\frac {\log {\left (x - \frac {1}{a} \right )}}{8} - \frac {\log {\left (x + \frac {1}{a} \right )}}{8}}{a c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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