3.215 \(\int \frac {e^{-2 \tanh ^{-1}(a x)}}{(c-a c x)^5} \, dx\)

Optimal. Leaf size=69 \[ \frac {1}{8 a c^5 (1-a x)}+\frac {1}{8 a c^5 (1-a x)^2}+\frac {1}{6 a c^5 (1-a x)^3}+\frac {\tanh ^{-1}(a x)}{8 a c^5} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 69, 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} \[ \frac {1}{8 a c^5 (1-a x)}+\frac {1}{8 a c^5 (1-a x)^2}+\frac {1}{6 a c^5 (1-a x)^3}+\frac {\tanh ^{-1}(a x)}{8 a c^5} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(2*ArcTanh[a*x])*(c - a*c*x)^5),x]

[Out]

1/(6*a*c^5*(1 - a*x)^3) + 1/(8*a*c^5*(1 - a*x)^2) + 1/(8*a*c^5*(1 - a*x)) + ArcTanh[a*x]/(8*a*c^5)

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)^5} \, dx &=\frac {\int \frac {1}{(1-a x)^4 (1+a x)} \, dx}{c^5}\\ &=\frac {\int \left (\frac {1}{2 (-1+a x)^4}-\frac {1}{4 (-1+a x)^3}+\frac {1}{8 (-1+a x)^2}-\frac {1}{8 \left (-1+a^2 x^2\right )}\right ) \, dx}{c^5}\\ &=\frac {1}{6 a c^5 (1-a x)^3}+\frac {1}{8 a c^5 (1-a x)^2}+\frac {1}{8 a c^5 (1-a x)}-\frac {\int \frac {1}{-1+a^2 x^2} \, dx}{8 c^5}\\ &=\frac {1}{6 a c^5 (1-a x)^3}+\frac {1}{8 a c^5 (1-a x)^2}+\frac {1}{8 a c^5 (1-a x)}+\frac {\tanh ^{-1}(a x)}{8 a c^5}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 44, normalized size = 0.64 \[ \frac {-3 a^2 x^2+9 a x+3 (a x-1)^3 \tanh ^{-1}(a x)-10}{24 a c^5 (a x-1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(E^(2*ArcTanh[a*x])*(c - a*c*x)^5),x]

[Out]

(-10 + 9*a*x - 3*a^2*x^2 + 3*(-1 + a*x)^3*ArcTanh[a*x])/(24*a*c^5*(-1 + a*x)^3)

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fricas [A]  time = 0.51, size = 113, normalized size = 1.64 \[ -\frac {6 \, a^{2} x^{2} - 18 \, a x - 3 \, {\left (a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1\right )} \log \left (a x + 1\right ) + 3 \, {\left (a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1\right )} \log \left (a x - 1\right ) + 20}{48 \, {\left (a^{4} c^{5} x^{3} - 3 \, a^{3} c^{5} x^{2} + 3 \, a^{2} c^{5} x - a c^{5}\right )}} \]

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)^5,x, algorithm="fricas")

[Out]

-1/48*(6*a^2*x^2 - 18*a*x - 3*(a^3*x^3 - 3*a^2*x^2 + 3*a*x - 1)*log(a*x + 1) + 3*(a^3*x^3 - 3*a^2*x^2 + 3*a*x
- 1)*log(a*x - 1) + 20)/(a^4*c^5*x^3 - 3*a^3*c^5*x^2 + 3*a^2*c^5*x - a*c^5)

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giac [A]  time = 0.20, size = 89, normalized size = 1.29 \[ \frac {\log \left ({\left | -\frac {2 \, c}{a c x - c} - 1 \right |}\right )}{16 \, a c^{5}} - \frac {\frac {3 \, a^{2} c^{2}}{a c x - c} - \frac {3 \, a^{2} c^{3}}{{\left (a c x - c\right )}^{2}} + \frac {4 \, a^{2} c^{4}}{{\left (a c x - c\right )}^{3}}}{24 \, a^{3} c^{6}} \]

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)^5,x, algorithm="giac")

[Out]

1/16*log(abs(-2*c/(a*c*x - c) - 1))/(a*c^5) - 1/24*(3*a^2*c^2/(a*c*x - c) - 3*a^2*c^3/(a*c*x - c)^2 + 4*a^2*c^
4/(a*c*x - c)^3)/(a^3*c^6)

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maple [A]  time = 0.04, size = 75, normalized size = 1.09 \[ -\frac {1}{6 c^{5} a \left (a x -1\right )^{3}}+\frac {1}{8 c^{5} a \left (a x -1\right )^{2}}-\frac {1}{8 c^{5} a \left (a x -1\right )}-\frac {\ln \left (a x -1\right )}{16 c^{5} a}+\frac {\ln \left (a x +1\right )}{16 c^{5} a} \]

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)^5,x)

[Out]

-1/6/c^5/a/(a*x-1)^3+1/8/c^5/a/(a*x-1)^2-1/8/c^5/a/(a*x-1)-1/16/c^5/a*ln(a*x-1)+1/16/c^5/a*ln(a*x+1)

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maxima [A]  time = 0.35, size = 84, normalized size = 1.22 \[ -\frac {3 \, a^{2} x^{2} - 9 \, a x + 10}{24 \, {\left (a^{4} c^{5} x^{3} - 3 \, a^{3} c^{5} x^{2} + 3 \, a^{2} c^{5} x - a c^{5}\right )}} + \frac {\log \left (a x + 1\right )}{16 \, a c^{5}} - \frac {\log \left (a x - 1\right )}{16 \, a c^{5}} \]

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)^5,x, algorithm="maxima")

[Out]

-1/24*(3*a^2*x^2 - 9*a*x + 10)/(a^4*c^5*x^3 - 3*a^3*c^5*x^2 + 3*a^2*c^5*x - a*c^5) + 1/16*log(a*x + 1)/(a*c^5)
- 1/16*log(a*x - 1)/(a*c^5)

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mupad [B]  time = 0.83, size = 64, normalized size = 0.93 \[ \frac {\frac {a\,x^2}{8}-\frac {3\,x}{8}+\frac {5}{12\,a}}{-a^3\,c^5\,x^3+3\,a^2\,c^5\,x^2-3\,a\,c^5\,x+c^5}+\frac {\mathrm {atanh}\left (a\,x\right )}{8\,a\,c^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(a^2*x^2 - 1)/((c - a*c*x)^5*(a*x + 1)^2),x)

[Out]

((a*x^2)/8 - (3*x)/8 + 5/(12*a))/(c^5 + 3*a^2*c^5*x^2 - a^3*c^5*x^3 - 3*a*c^5*x) + atanh(a*x)/(8*a*c^5)

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sympy [A]  time = 0.37, size = 76, normalized size = 1.10 \[ \frac {- 3 a^{2} x^{2} + 9 a x - 10}{24 a^{4} c^{5} x^{3} - 72 a^{3} c^{5} x^{2} + 72 a^{2} c^{5} x - 24 a c^{5}} + \frac {- \frac {\log {\left (x - \frac {1}{a} \right )}}{16} + \frac {\log {\left (x + \frac {1}{a} \right )}}{16}}{a c^{5}} \]

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)**5,x)

[Out]

(-3*a**2*x**2 + 9*a*x - 10)/(24*a**4*c**5*x**3 - 72*a**3*c**5*x**2 + 72*a**2*c**5*x - 24*a*c**5) + (-log(x - 1
/a)/16 + log(x + 1/a)/16)/(a*c**5)

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