\(\int \frac {e^{\coth ^{-1}(a x)}}{(c-a c x)^2} \, dx\) [163]

Optimal result

Integrand size = 16, antiderivative size = 33 \[ \int \frac {e^{\coth ^{-1}(a x)}}{(c-a c x)^2} \, dx=-\frac {a^2 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{3 c^2 \left (a-\frac {1}{x}\right )^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {6310, 6313, 665} \[ \int \frac {e^{\coth ^{-1}(a x)}}{(c-a c x)^2} \, dx=-\frac {a^2 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{3 c^2 \left (a-\frac {1}{x}\right )^3} \]

[In]

Int[E^ArcCoth[a*x]/(c - a*c*x)^2,x]

[Out]

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

Rule 665

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^m*((a + c*x^2)^(p + 1)/
(2*c*d*(p + 1))), x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + 2*p
+ 2, 0]

Rule 6310

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

Rule 6313

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[-c^n, Subst[Int[(c +
d*x)^(p - n)*((1 - x^2/a^2)^(n/2)/x^(m + 2)), x], x, 1/x], x] /; FreeQ[{a, c, d, p}, x] && EqQ[c + a*d, 0] &&
IntegerQ[(n - 1)/2] && IntegerQ[m] && (IntegerQ[p] || EqQ[p, n/2] || EqQ[p, n/2 + 1] || LtQ[-5, m, -1]) && In
tegerQ[2*p]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {e^{\coth ^{-1}(a x)}}{\left (1-\frac {1}{a x}\right )^2 x^2} \, dx}{a^2 c^2} \\ & = -\frac {\text {Subst}\left (\int \frac {\sqrt {1-\frac {x^2}{a^2}}}{\left (1-\frac {x}{a}\right )^3} \, dx,x,\frac {1}{x}\right )}{a^2 c^2} \\ & = -\frac {a^2 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{3 c^2 \left (a-\frac {1}{x}\right )^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.03 \[ \int \frac {e^{\coth ^{-1}(a x)}}{(c-a c x)^2} \, dx=-\frac {\sqrt {1-\frac {1}{a^2 x^2}} x (1+a x)}{3 c^2 (-1+a x)^2} \]

[In]

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

[Out]

-1/3*(Sqrt[1 - 1/(a^2*x^2)]*x*(1 + a*x))/(c^2*(-1 + a*x)^2)

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.09

method	result	size
gosper	\(-\frac {a x +1}{3 \left (a x -1\right ) c^{2} \sqrt {\frac {a x -1}{a x +1}}\, a}\)	\(36\)
default	\(-\frac {a x +1}{3 \left (a x -1\right ) c^{2} \sqrt {\frac {a x -1}{a x +1}}\, a}\)	\(36\)
trager	\(-\frac {\left (a x +1\right )^{2} \sqrt {-\frac {-a x +1}{a x +1}}}{3 a \,c^{2} \left (a x -1\right )^{2}}\)	\(40\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.73 \[ \int \frac {e^{\coth ^{-1}(a x)}}{(c-a c x)^2} \, dx=-\frac {{\left (a^{2} x^{2} + 2 \, a x + 1\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{3 \, {\left (a^{3} c^{2} x^{2} - 2 \, a^{2} c^{2} x + a c^{2}\right )}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.70 \[ \int \frac {e^{\coth ^{-1}(a x)}}{(c-a c x)^2} \, dx=-\frac {1}{3 \, a c^{2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.48 \[ \int \frac {e^{\coth ^{-1}(a x)}}{(c-a c x)^2} \, dx=-\frac {2 \, {\left (3 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )}^{2} x^{2} + 1\right )}}{3 \, {\left ({\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )} x - 1\right )}^{3} a c^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 4.30 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.70 \[ \int \frac {e^{\coth ^{-1}(a x)}}{(c-a c x)^2} \, dx=-\frac {1}{3\,a\,c^2\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}} \]

[In]

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

[Out]

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