\(\int \frac {e^{-2 \coth ^{-1}(a x)}}{c-\frac {c}{a^2 x^2}} \, dx\) [818]

Optimal result

Integrand size = 22, antiderivative size = 35 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{c-\frac {c}{a^2 x^2}} \, dx=\frac {x}{c}-\frac {1}{a c (1+a x)}-\frac {2 \log (1+a x)}{a c} \]

[Out]

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

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 35, 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, 6292, 6285, 45} \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{c-\frac {c}{a^2 x^2}} \, dx=-\frac {1}{a c (a x+1)}-\frac {2 \log (a x+1)}{a c}+\frac {x}{c} \]

[In]

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

[Out]

x/c - 1/(a*c*(1 + a*x)) - (2*Log[1 + a*x])/(a*c)

Rule 45

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 6285

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[x^m*(1 -
a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p
] || GtQ[c, 0])

Rule 6292

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_.), x_Symbol] :> Dist[d^p, Int[(u/x^(2*p))*(1
- a^2*x^2)^p*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[{a, c, d, n}, x] && EqQ[c + a^2*d, 0] && IntegerQ[p]

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.80 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{c-\frac {c}{a^2 x^2}} \, dx=\frac {x-\frac {1}{a+a^2 x}-\frac {2 \log (1+a x)}{a}}{c} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.03

method	result	size
risch	\(\frac {x}{c}-\frac {1}{a c \left (a x +1\right )}-\frac {2 \ln \left (a x +1\right )}{a c}\)	\(36\)
default	\(\frac {a^{2} \left (-\frac {2 \ln \left (a x +1\right )}{a^{3}}-\frac {1}{a^{3} \left (a x +1\right )}+\frac {x}{a^{2}}\right )}{c}\)	\(37\)
norman	\(\frac {\frac {a \,x^{2}}{c}+\frac {2 x}{c}}{a x +1}-\frac {2 \ln \left (a x +1\right )}{a c}\)	\(39\)
parallelrisch	\(\frac {a^{2} x^{2}-2 a \ln \left (a x +1\right ) x +2 a x -2 \ln \left (a x +1\right )}{c \left (a x +1\right ) a}\)	\(45\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.09 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{c-\frac {c}{a^2 x^2}} \, dx=\frac {a^{2} x^{2} + a x - 2 \, {\left (a x + 1\right )} \log \left (a x + 1\right ) - 1}{a^{2} c x + a c} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.03 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{c-\frac {c}{a^2 x^2}} \, dx=a^{2} \left (- \frac {1}{a^{4} c x + a^{3} c} + \frac {x}{a^{2} c} - \frac {2 \log {\left (a x + 1 \right )}}{a^{3} c}\right ) \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.97 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{c-\frac {c}{a^2 x^2}} \, dx=\frac {x}{c} - \frac {1}{a^{2} c x + a c} - \frac {2 \, \log \left (a x + 1\right )}{a c} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.03 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{c-\frac {c}{a^2 x^2}} \, dx=\frac {x}{c} - \frac {2 \, \log \left ({\left | a x + 1 \right |}\right )}{a c} - \frac {1}{{\left (a x + 1\right )} a c} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.94 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{c-\frac {c}{a^2 x^2}} \, dx=\frac {x}{c}-\frac {1}{a\,\left (c+a\,c\,x\right )}-\frac {2\,\ln \left (a\,x+1\right )}{a\,c} \]

[In]

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

[Out]

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