\(\int \frac {e^{3 \coth ^{-1}(a x)}}{(c-a c x)^2} \, dx\) [183]

Optimal result

Integrand size = 18, antiderivative size = 33 \[ \int \frac {e^{3 \coth ^{-1}(a x)}}{(c-a c x)^2} \, dx=-\frac {a^4 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{5 c^2 \left (a-\frac {1}{x}\right )^5} \]

[Out]

-1/5*a^4*(1-1/a^2/x^2)^(5/2)/c^2/(a-1/x)^5

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6310, 6313, 665} \[ \int \frac {e^{3 \coth ^{-1}(a x)}}{(c-a c x)^2} \, dx=-\frac {a^4 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{5 c^2 \left (a-\frac {1}{x}\right )^5} \]

[In]

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

[Out]

-1/5*(a^4*(1 - 1/(a^2*x^2))^(5/2))/(c^2*(a - x^(-1))^5)

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^{3 \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 {\left (1-\frac {x^2}{a^2}\right )^{3/2}}{\left (1-\frac {x}{a}\right )^5} \, dx,x,\frac {1}{x}\right )}{a^2 c^2} \\ & = -\frac {a^4 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{5 c^2 \left (a-\frac {1}{x}\right )^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.09 \[ \int \frac {e^{3 \coth ^{-1}(a x)}}{(c-a c x)^2} \, dx=-\frac {\sqrt {1-\frac {1}{a^2 x^2}} x (1+a x)^2}{5 c^2 (-1+a x)^3} \]

[In]

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

[Out]

-1/5*(Sqrt[1 - 1/(a^2*x^2)]*x*(1 + a*x)^2)/(c^2*(-1 + a*x)^3)

Maple [A] (verified)

Time = 0.43 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.09

method	result	size
gosper	\(-\frac {a x +1}{5 \left (a x -1\right ) c^{2} \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} a}\)	\(36\)
default	\(-\frac {a x +1}{5 \left (a x -1\right ) c^{2} \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} a}\)	\(36\)
trager	\(-\frac {\left (a x +1\right ) \left (a^{2} x^{2}+2 a x +1\right ) \sqrt {-\frac {-a x +1}{a x +1}}}{5 a \,c^{2} \left (a x -1\right )^{3}}\)	\(51\)

[In]

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

[Out]

-1/5*(a*x+1)/(a*x-1)/c^2/((a*x-1)/(a*x+1))^(3/2)/a

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 77 vs. \(2 (29) = 58\).

Time = 0.26 (sec) , antiderivative size = 77, normalized size of antiderivative = 2.33 \[ \int \frac {e^{3 \coth ^{-1}(a x)}}{(c-a c x)^2} \, dx=-\frac {{\left (a^{3} x^{3} + 3 \, a^{2} x^{2} + 3 \, a x + 1\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{5 \, {\left (a^{4} c^{2} x^{3} - 3 \, a^{3} c^{2} x^{2} + 3 \, a^{2} c^{2} x - a c^{2}\right )}} \]

[In]

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

[Out]

-1/5*(a^3*x^3 + 3*a^2*x^2 + 3*a*x + 1)*sqrt((a*x - 1)/(a*x + 1))/(a^4*c^2*x^3 - 3*a^3*c^2*x^2 + 3*a^2*c^2*x -
a*c^2)

Sympy [F]

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

[In]

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

[Out]

Integral(1/(a**3*x**3*sqrt(a*x/(a*x + 1) - 1/(a*x + 1))/(a*x + 1) - 3*a**2*x**2*sqrt(a*x/(a*x + 1) - 1/(a*x +
1))/(a*x + 1) + 3*a*x*sqrt(a*x/(a*x + 1) - 1/(a*x + 1))/(a*x + 1) - sqrt(a*x/(a*x + 1) - 1/(a*x + 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 \[ \int \frac {e^{3 \coth ^{-1}(a x)}}{(c-a c x)^2} \, dx=-\frac {1}{5 \, a c^{2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{2}}} \]

[In]

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

[Out]

-1/5/(a*c^2*((a*x - 1)/(a*x + 1))^(5/2))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 69 vs. \(2 (29) = 58\).

Time = 0.33 (sec) , antiderivative size = 69, normalized size of antiderivative = 2.09 \[ \int \frac {e^{3 \coth ^{-1}(a x)}}{(c-a c x)^2} \, dx=-\frac {2 \, {\left (5 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )}^{4} x^{4} + 10 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )}^{2} x^{2} + 1\right )}}{5 \, {\left ({\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )} x - 1\right )}^{5} a c^{2}} \]

[In]

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

[Out]

-2/5*(5*(a + sqrt(a^2 - 1/x^2))^4*x^4 + 10*(a + sqrt(a^2 - 1/x^2))^2*x^2 + 1)/(((a + sqrt(a^2 - 1/x^2))*x - 1)
^5*a*c^2)

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.70 \[ \int \frac {e^{3 \coth ^{-1}(a x)}}{(c-a c x)^2} \, dx=-\frac {1}{5\,a\,c^2\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/2}} \]

[In]

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

[Out]

-1/(5*a*c^2*((a*x - 1)/(a*x + 1))^(5/2))