\(\int \frac {e^{2 \coth ^{-1}(a x)}}{(c-a^2 c x^2)^2} \, dx\) [570]

Optimal result

Integrand size = 22, antiderivative size = 51 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^2} \, dx=-\frac {1}{4 a c^2 (1-a x)^2}-\frac {1}{4 a c^2 (1-a x)}-\frac {\text {arctanh}(a x)}{4 a c^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 51, 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} \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^2} \, dx=-\frac {\text {arctanh}(a x)}{4 a c^2}-\frac {1}{4 a c^2 (1-a x)}-\frac {1}{4 a c^2 (1-a x)^2} \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.69 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^2} \, dx=\frac {-2+a x-(-1+a x)^2 \text {arctanh}(a x)}{4 a c^2 (-1+a x)^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.57 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00

method	result	size
risch	\(\frac {\frac {x}{4}-\frac {1}{2 a}}{\left (a x -1\right )^{2} c^{2}}+\frac {\ln \left (-a x +1\right )}{8 a \,c^{2}}-\frac {\ln \left (a x +1\right )}{8 a \,c^{2}}\)	\(51\)
default	\(\frac {-\frac {\ln \left (a x +1\right )}{8 a}-\frac {1}{4 \left (a x -1\right )^{2} a}+\frac {1}{4 a \left (a x -1\right )}+\frac {\ln \left (a x -1\right )}{8 a}}{c^{2}}\)	\(52\)
norman	\(\frac {-\frac {3}{4 a c}+\frac {a \,x^{2}}{2 c}-\frac {a^{2} x^{3}}{4 c}}{c \left (a x +1\right ) \left (a x -1\right )^{2}}+\frac {\ln \left (a x -1\right )}{8 a \,c^{2}}-\frac {\ln \left (a x +1\right )}{8 a \,c^{2}}\)	\(77\)
parallelrisch	\(\frac {a^{2} \ln \left (a x -1\right ) x^{2}-a^{2} \ln \left (a x +1\right ) x^{2}+4 a^{2} x^{2}-2 a \ln \left (a x -1\right ) x +2 a \ln \left (a x +1\right ) x -6 a x +\ln \left (a x -1\right )-\ln \left (a x +1\right )}{8 c^{2} \left (a x -1\right )^{2} a}\)	\(90\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.49 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^2} \, dx=\frac {2 \, a x - {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (a x + 1\right ) + {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (a x - 1\right ) - 4}{8 \, {\left (a^{3} c^{2} x^{2} - 2 \, a^{2} c^{2} x + a c^{2}\right )}} \]

[In]

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

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.06 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^2} \, dx=\frac {a x - 2}{4 a^{3} c^{2} x^{2} - 8 a^{2} c^{2} x + 4 a c^{2}} + \frac {\frac {\log {\left (x - \frac {1}{a} \right )}}{8} - \frac {\log {\left (x + \frac {1}{a} \right )}}{8}}{a c^{2}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.24 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^2} \, dx=\frac {a x - 2}{4 \, {\left (a^{3} c^{2} x^{2} - 2 \, a^{2} c^{2} x + a c^{2}\right )}} - \frac {\log \left (a x + 1\right )}{8 \, a c^{2}} + \frac {\log \left (a x - 1\right )}{8 \, a c^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^2} \, dx=-\frac {\log \left ({\left | a x + 1 \right |}\right )}{8 \, a c^{2}} + \frac {\log \left ({\left | a x - 1 \right |}\right )}{8 \, a c^{2}} + \frac {a x - 2}{4 \, {\left (a x - 1\right )}^{2} a c^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 3.89 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.90 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^2} \, dx=\frac {\frac {x}{4}-\frac {1}{2\,a}}{a^2\,c^2\,x^2-2\,a\,c^2\,x+c^2}-\frac {\mathrm {atanh}\left (a\,x\right )}{4\,a\,c^2} \]

[In]

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

[Out]

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