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

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*(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)

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Rubi [A]  time = 0.0796972, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 18, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.222, Rules used = {6167, 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*ArcCoth[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 6167

Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Dist[(-1)^(n/2), Int[u*E^(n*ArcTanh[a*x]), x], x] /; Free
Q[a, x] && IntegerQ[n/2]

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])

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 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx &=-\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*}

Mathematica [A]  time = 0.0312029, 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*ArcCoth[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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Maple [A]  time = 0.046, size = 75, normalized size = 1.1 \begin{align*} -{\frac{\ln \left ( ax+1 \right ) }{16\,{c}^{5}a}}+{\frac{1}{6\,{c}^{5}a \left ( ax-1 \right ) ^{3}}}-{\frac{1}{8\,{c}^{5}a \left ( ax-1 \right ) ^{2}}}+{\frac{1}{8\,{c}^{5}a \left ( ax-1 \right ) }}+{\frac{\ln \left ( ax-1 \right ) }{16\,{c}^{5}a}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16/c^5/a*ln(a*x+1)+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)

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Maxima [A]  time = 1.03516, size = 113, normalized size = 1.64 \begin{align*} \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}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x-1)/(a*x+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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Fricas [A]  time = 1.55361, size = 251, normalized size = 3.64 \begin{align*} \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 )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x-1)/(a*x+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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Sympy [A]  time = 1.11456, size = 76, normalized size = 1.1 \begin{align*} \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}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x-1)/(a*x+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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Giac [A]  time = 1.13845, size = 120, normalized size = 1.74 \begin{align*} -\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}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x-1)/(a*x+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)