\(\int \frac {e^{-2 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx\) [215]

Optimal result

Integrand size = 18, antiderivative size = 69 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=-\frac {1}{6 a c^5 (1-a x)^3}-\frac {1}{8 a c^5 (1-a x)^2}-\frac {1}{8 a c^5 (1-a x)}-\frac {\text {arctanh}(a x)}{8 a c^5} \]

[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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6302, 6264, 46, 213} \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=-\frac {\text {arctanh}(a x)}{8 a c^5}-\frac {1}{8 a c^5 (1-a x)}-\frac {1}{8 a c^5 (1-a x)^2}-\frac {1}{6 a c^5 (1-a x)^3} \]

[In]

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

[Out]

-1/6*1/(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 46

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] && Lt
Q[m + n + 2, 0])

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 6264

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 6302

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]

Rubi steps \begin{align*} \text {integral}& = -\int \frac {e^{-2 \text {arctanh}(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 {\text {arctanh}(a x)}{8 a c^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.64 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=\frac {10-9 a x+3 a^2 x^2-3 (-1+a x)^3 \text {arctanh}(a x)}{24 a c^5 (-1+a x)^3} \]

[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)

Maple [A] (verified)

Time = 0.49 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.83

method	result	size
risch	\(\frac {\frac {a \,x^{2}}{8}-\frac {3 x}{8}+\frac {5}{12 a}}{\left (a x -1\right )^{3} c^{5}}-\frac {\ln \left (a x +1\right )}{16 c^{5} a}+\frac {\ln \left (-a x +1\right )}{16 c^{5} a}\)	\(57\)
default	\(\frac {-\frac {\ln \left (a x +1\right )}{16 a}+\frac {1}{6 a \left (a x -1\right )^{3}}-\frac {1}{8 \left (a x -1\right )^{2} a}+\frac {1}{8 a \left (a x -1\right )}+\frac {\ln \left (a x -1\right )}{16 a}}{c^{5}}\)	\(64\)
norman	\(\frac {-\frac {7 x}{8 c}+\frac {2 a \,x^{2}}{c}-\frac {37 a^{2} x^{3}}{24 c}+\frac {5 a^{3} x^{4}}{12 c}}{c^{4} \left (a x -1\right )^{4}}+\frac {\ln \left (a x -1\right )}{16 a \,c^{5}}-\frac {\ln \left (a x +1\right )}{16 c^{5} a}\)	\(79\)
parallelrisch	\(\frac {3 a^{3} \ln \left (a x -1\right ) x^{3}-3 a^{3} \ln \left (a x +1\right ) x^{3}+20 a^{3} x^{3}-9 a^{2} \ln \left (a x -1\right ) x^{2}+9 a^{2} \ln \left (a x +1\right ) x^{2}-54 a^{2} x^{2}+9 a \ln \left (a x -1\right ) x -9 a \ln \left (a x +1\right ) x +42 a x -3 \ln \left (a x -1\right )+3 \ln \left (a x +1\right )}{48 c^{5} \left (a x -1\right )^{3} a}\)	\(129\)

[In]

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

[Out]

(1/8*a*x^2-3/8*x+5/12/a)/(a*x-1)^3/c^5-1/16/c^5/a*ln(a*x+1)+1/16/c^5/a*ln(-a*x+1)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.64 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=\frac {6 \, a^{2} x^{2} - 18 \, a x - 3 \, {\left (a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1\right )} \log \left (a x + 1\right ) + 3 \, {\left (a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1\right )} \log \left (a x - 1\right ) + 20}{48 \, {\left (a^{4} c^{5} x^{3} - 3 \, a^{3} c^{5} x^{2} + 3 \, a^{2} c^{5} x - a c^{5}\right )}} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.13 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=- \frac {- 3 a^{2} x^{2} + 9 a x - 10}{24 a^{4} c^{5} x^{3} - 72 a^{3} c^{5} x^{2} + 72 a^{2} c^{5} x - 24 a c^{5}} - \frac {- \frac {\log {\left (x - \frac {1}{a} \right )}}{16} + \frac {\log {\left (x + \frac {1}{a} \right )}}{16}}{a c^{5}} \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.22 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=\frac {3 \, a^{2} x^{2} - 9 \, a x + 10}{24 \, {\left (a^{4} c^{5} x^{3} - 3 \, a^{3} c^{5} x^{2} + 3 \, a^{2} c^{5} x - a c^{5}\right )}} - \frac {\log \left (a x + 1\right )}{16 \, a c^{5}} + \frac {\log \left (a x - 1\right )}{16 \, a c^{5}} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.29 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=-\frac {\log \left ({\left | -\frac {2 \, c}{a c x - c} - 1 \right |}\right )}{16 \, a c^{5}} + \frac {\frac {3 \, a^{2} c^{2}}{a c x - c} - \frac {3 \, a^{2} c^{3}}{{\left (a c x - c\right )}^{2}} + \frac {4 \, a^{2} c^{4}}{{\left (a c x - c\right )}^{3}}}{24 \, a^{3} c^{6}} \]

[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)

Mupad [B] (verification not implemented)

Time = 4.43 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.94 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=-\frac {\frac {a\,x^2}{8}-\frac {3\,x}{8}+\frac {5}{12\,a}}{-a^3\,c^5\,x^3+3\,a^2\,c^5\,x^2-3\,a\,c^5\,x+c^5}-\frac {\mathrm {atanh}\left (a\,x\right )}{8\,a\,c^5} \]

[In]

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