\(\int \frac {e^{2 \coth ^{-1}(a x)}}{(c-a^2 c x^2)^3} \, dx\) [571]

Optimal result

Integrand size = 22, antiderivative size = 86 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=-\frac {1}{12 a c^3 (1-a x)^3}-\frac {1}{8 a c^3 (1-a x)^2}-\frac {3}{16 a c^3 (1-a x)}+\frac {1}{16 a c^3 (1+a x)}-\frac {\text {arctanh}(a x)}{4 a c^3} \]

[Out]

-1/12/a/c^3/(-a*x+1)^3-1/8/a/c^3/(-a*x+1)^2-3/16/a/c^3/(-a*x+1)+1/16/a/c^3/(a*x+1)-1/4*arctanh(a*x)/a/c^3

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {6302, 6275, 46, 213} \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=-\frac {\text {arctanh}(a x)}{4 a c^3}-\frac {3}{16 a c^3 (1-a x)}+\frac {1}{16 a c^3 (a x+1)}-\frac {1}{8 a c^3 (1-a x)^2}-\frac {1}{12 a c^3 (1-a x)^3} \]

[In]

Int[E^(2*ArcCoth[a*x])/(c - a^2*c*x^2)^3,x]

[Out]

-1/12*1/(a*c^3*(1 - a*x)^3) - 1/(8*a*c^3*(1 - a*x)^2) - 3/(16*a*c^3*(1 - a*x)) + 1/(16*a*c^3*(1 + a*x)) - ArcT
anh[a*x]/(4*a*c^3)

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 6275

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 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)}}{\left (c-a^2 c x^2\right )^3} \, dx \\ & = -\frac {\int \frac {1}{(1-a x)^4 (1+a x)^2} \, dx}{c^3} \\ & = -\frac {\int \left (\frac {1}{4 (-1+a x)^4}-\frac {1}{4 (-1+a x)^3}+\frac {3}{16 (-1+a x)^2}+\frac {1}{16 (1+a x)^2}-\frac {1}{4 \left (-1+a^2 x^2\right )}\right ) \, dx}{c^3} \\ & = -\frac {1}{12 a c^3 (1-a x)^3}-\frac {1}{8 a c^3 (1-a x)^2}-\frac {3}{16 a c^3 (1-a x)}+\frac {1}{16 a c^3 (1+a x)}+\frac {\int \frac {1}{-1+a^2 x^2} \, dx}{4 c^3} \\ & = -\frac {1}{12 a c^3 (1-a x)^3}-\frac {1}{8 a c^3 (1-a x)^2}-\frac {3}{16 a c^3 (1-a x)}+\frac {1}{16 a c^3 (1+a x)}-\frac {\text {arctanh}(a x)}{4 a c^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.73 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {4+a x-6 a^2 x^2+3 a^3 x^3-3 (-1+a x)^3 (1+a x) \text {arctanh}(a x)}{12 a c^3 (-1+a x)^3 (1+a x)} \]

[In]

Integrate[E^(2*ArcCoth[a*x])/(c - a^2*c*x^2)^3,x]

[Out]

(4 + a*x - 6*a^2*x^2 + 3*a^3*x^3 - 3*(-1 + a*x)^3*(1 + a*x)*ArcTanh[a*x])/(12*a*c^3*(-1 + a*x)^3*(1 + a*x))

Maple [A] (verified)

Time = 0.61 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.88

method	result	size
default	\(\frac {\frac {1}{16 a \left (a x +1\right )}-\frac {\ln \left (a x +1\right )}{8 a}+\frac {1}{12 a \left (a x -1\right )^{3}}-\frac {1}{8 \left (a x -1\right )^{2} a}+\frac {3}{16 a \left (a x -1\right )}+\frac {\ln \left (a x -1\right )}{8 a}}{c^{3}}\)	\(76\)
risch	\(\frac {\frac {a^{2} x^{3}}{4}-\frac {a \,x^{2}}{2}+\frac {x}{12}+\frac {1}{3 a}}{\left (a x -1\right )^{2} \left (a^{2} x^{2}-1\right ) c^{3}}+\frac {\ln \left (-a x +1\right )}{8 a \,c^{3}}-\frac {\ln \left (a x +1\right )}{8 a \,c^{3}}\)	\(76\)
norman	\(\frac {\frac {3 x}{4 c}+\frac {a \,x^{2}}{4 c}-\frac {11 a^{2} x^{3}}{12 c}-\frac {a^{3} x^{4}}{12 c}+\frac {a^{4} x^{5}}{3 c}}{\left (a x +1\right )^{2} \left (a x -1\right )^{3} c^{2}}+\frac {\ln \left (a x -1\right )}{8 a \,c^{3}}-\frac {\ln \left (a x +1\right )}{8 a \,c^{3}}\)	\(97\)
parallelrisch	\(\frac {3 \ln \left (a x -1\right ) x^{4} a^{4}-3 \ln \left (a x +1\right ) x^{4} a^{4}+8 a^{4} x^{4}-6 a^{3} \ln \left (a x -1\right ) x^{3}+6 a^{3} \ln \left (a x +1\right ) x^{3}-10 a^{3} x^{3}-12 a^{2} x^{2}+6 a \ln \left (a x -1\right ) x -6 a \ln \left (a x +1\right ) x +18 a x -3 \ln \left (a x -1\right )+3 \ln \left (a x +1\right )}{24 c^{3} \left (a x -1\right )^{2} \left (a^{2} x^{2}-1\right ) a}\)	\(148\)

[In]

int(1/(a*x-1)*(a*x+1)/(-a^2*c*x^2+c)^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.41 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {6 \, a^{3} x^{3} - 12 \, a^{2} x^{2} + 2 \, a x - 3 \, {\left (a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, a x - 1\right )} \log \left (a x + 1\right ) + 3 \, {\left (a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, a x - 1\right )} \log \left (a x - 1\right ) + 8}{24 \, {\left (a^{5} c^{3} x^{4} - 2 \, a^{4} c^{3} x^{3} + 2 \, a^{2} c^{3} x - a c^{3}\right )}} \]

[In]

integrate(1/(a*x-1)*(a*x+1)/(-a^2*c*x^2+c)^3,x, algorithm="fricas")

[Out]

1/24*(6*a^3*x^3 - 12*a^2*x^2 + 2*a*x - 3*(a^4*x^4 - 2*a^3*x^3 + 2*a*x - 1)*log(a*x + 1) + 3*(a^4*x^4 - 2*a^3*x
^3 + 2*a*x - 1)*log(a*x - 1) + 8)/(a^5*c^3*x^4 - 2*a^4*c^3*x^3 + 2*a^2*c^3*x - a*c^3)

Sympy [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.99 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=- \frac {- 3 a^{3} x^{3} + 6 a^{2} x^{2} - a x - 4}{12 a^{5} c^{3} x^{4} - 24 a^{4} c^{3} x^{3} + 24 a^{2} c^{3} x - 12 a c^{3}} - \frac {- \frac {\log {\left (x - \frac {1}{a} \right )}}{8} + \frac {\log {\left (x + \frac {1}{a} \right )}}{8}}{a c^{3}} \]

[In]

integrate(1/(a*x-1)*(a*x+1)/(-a**2*c*x**2+c)**3,x)

[Out]

-(-3*a**3*x**3 + 6*a**2*x**2 - a*x - 4)/(12*a**5*c**3*x**4 - 24*a**4*c**3*x**3 + 24*a**2*c**3*x - 12*a*c**3) -
(-log(x - 1/a)/8 + log(x + 1/a)/8)/(a*c**3)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.06 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {3 \, a^{3} x^{3} - 6 \, a^{2} x^{2} + a x + 4}{12 \, {\left (a^{5} c^{3} x^{4} - 2 \, a^{4} c^{3} x^{3} + 2 \, a^{2} c^{3} x - a c^{3}\right )}} - \frac {\log \left (a x + 1\right )}{8 \, a c^{3}} + \frac {\log \left (a x - 1\right )}{8 \, a c^{3}} \]

[In]

integrate(1/(a*x-1)*(a*x+1)/(-a^2*c*x^2+c)^3,x, algorithm="maxima")

[Out]

1/12*(3*a^3*x^3 - 6*a^2*x^2 + a*x + 4)/(a^5*c^3*x^4 - 2*a^4*c^3*x^3 + 2*a^2*c^3*x - a*c^3) - 1/8*log(a*x + 1)/
(a*c^3) + 1/8*log(a*x - 1)/(a*c^3)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.86 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=-\frac {\log \left ({\left | a x + 1 \right |}\right )}{8 \, a c^{3}} + \frac {\log \left ({\left | a x - 1 \right |}\right )}{8 \, a c^{3}} + \frac {3 \, a^{3} x^{3} - 6 \, a^{2} x^{2} + a x + 4}{12 \, {\left (a x + 1\right )} {\left (a x - 1\right )}^{3} a c^{3}} \]

[In]

integrate(1/(a*x-1)*(a*x+1)/(-a^2*c*x^2+c)^3,x, algorithm="giac")

[Out]

-1/8*log(abs(a*x + 1))/(a*c^3) + 1/8*log(abs(a*x - 1))/(a*c^3) + 1/12*(3*a^3*x^3 - 6*a^2*x^2 + a*x + 4)/((a*x
+ 1)*(a*x - 1)^3*a*c^3)

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.85 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=-\frac {\frac {x}{12}-\frac {a\,x^2}{2}+\frac {1}{3\,a}+\frac {a^2\,x^3}{4}}{-a^4\,c^3\,x^4+2\,a^3\,c^3\,x^3-2\,a\,c^3\,x+c^3}-\frac {\mathrm {atanh}\left (a\,x\right )}{4\,a\,c^3} \]

[In]

int((a*x + 1)/((c - a^2*c*x^2)^3*(a*x - 1)),x)

[Out]

- (x/12 - (a*x^2)/2 + 1/(3*a) + (a^2*x^3)/4)/(c^3 + 2*a^3*c^3*x^3 - a^4*c^3*x^4 - 2*a*c^3*x) - atanh(a*x)/(4*a
*c^3)