\(\int \frac {e^{4 \coth ^{-1}(a x)}}{c-\frac {c}{a^2 x^2}} \, dx\) [802]

Optimal result

Integrand size = 22, antiderivative size = 53 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{c-\frac {c}{a^2 x^2}} \, dx=\frac {x}{c}-\frac {1}{a c (1-a x)^2}+\frac {5}{a c (1-a x)}+\frac {4 \log (1-a x)}{a c} \]

[Out]

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

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 53, 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, 78} \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{c-\frac {c}{a^2 x^2}} \, dx=\frac {5}{a c (1-a x)}-\frac {1}{a c (1-a x)^2}+\frac {4 \log (1-a x)}{a c}+\frac {x}{c} \]

[In]

Int[E^(4*ArcCoth[a*x])/(c - c/(a^2*x^2)),x]

[Out]

x/c - 1/(a*c*(1 - a*x)^2) + 5/(a*c*(1 - a*x)) + (4*Log[1 - a*x])/(a*c)

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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^{4 \text {arctanh}(a x)}}{c-\frac {c}{a^2 x^2}} \, dx \\ & = -\frac {a^2 \int \frac {e^{4 \text {arctanh}(a x)} x^2}{1-a^2 x^2} \, dx}{c} \\ & = -\frac {a^2 \int \frac {x^2 (1+a x)}{(1-a x)^3} \, dx}{c} \\ & = -\frac {a^2 \int \left (-\frac {1}{a^2}-\frac {2}{a^2 (-1+a x)^3}-\frac {5}{a^2 (-1+a x)^2}-\frac {4}{a^2 (-1+a x)}\right ) \, dx}{c} \\ & = \frac {x}{c}-\frac {1}{a c (1-a x)^2}+\frac {5}{a c (1-a x)}+\frac {4 \log (1-a x)}{a c} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{c-\frac {c}{a^2 x^2}} \, dx=\frac {x}{c}-\frac {1}{a c (1-a x)^2}+\frac {5}{a c (1-a x)}+\frac {4 \log (1-a x)}{a c} \]

[In]

Integrate[E^(4*ArcCoth[a*x])/(c - c/(a^2*x^2)),x]

[Out]

x/c - 1/(a*c*(1 - a*x)^2) + 5/(a*c*(1 - a*x)) + (4*Log[1 - a*x])/(a*c)

Maple [A] (verified)

Time = 0.63 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.81

method	result	size
risch	\(\frac {x}{c}+\frac {-5 c x +\frac {4 c}{a}}{c^{2} \left (a x -1\right )^{2}}+\frac {4 \ln \left (a x -1\right )}{a c}\)	\(43\)
default	\(\frac {a^{2} \left (\frac {x}{a^{2}}-\frac {1}{a^{3} \left (a x -1\right )^{2}}-\frac {5}{a^{3} \left (a x -1\right )}+\frac {4 \ln \left (a x -1\right )}{a^{3}}\right )}{c}\)	\(49\)
norman	\(\frac {\frac {a^{2} x^{3}}{c}-\frac {6 a \,x^{2}}{c}+\frac {4 x}{c}}{\left (a x -1\right )^{2}}+\frac {4 \ln \left (a x -1\right )}{a c}\)	\(50\)
parallelrisch	\(\frac {a^{3} x^{3}+4 a^{2} \ln \left (a x -1\right ) x^{2}-6 a^{2} x^{2}-8 a \ln \left (a x -1\right ) x +4 a x +4 \ln \left (a x -1\right )}{\left (a x -1\right )^{2} c a}\)	\(67\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.21 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{c-\frac {c}{a^2 x^2}} \, dx=\frac {a^{3} x^{3} - 2 \, a^{2} x^{2} - 4 \, a x + 4 \, {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (a x - 1\right ) + 4}{a^{3} c x^{2} - 2 \, a^{2} c x + a c} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.77 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{c-\frac {c}{a^2 x^2}} \, dx=\frac {- 5 a x + 4}{a^{3} c x^{2} - 2 a^{2} c x + a c} + \frac {x}{c} + \frac {4 \log {\left (a x - 1 \right )}}{a c} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.92 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{c-\frac {c}{a^2 x^2}} \, dx=-\frac {5 \, a x - 4}{a^{3} c x^{2} - 2 \, a^{2} c x + a c} + \frac {x}{c} + \frac {4 \, \log \left (a x - 1\right )}{a c} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.40 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{c-\frac {c}{a^2 x^2}} \, dx=\frac {a x - 1}{a c} - \frac {4 \, \log \left (\frac {{\left | a x - 1 \right |}}{{\left (a x - 1\right )}^{2} {\left | a \right |}}\right )}{a c} - \frac {\frac {5 \, a^{3} c}{a x - 1} + \frac {a^{3} c}{{\left (a x - 1\right )}^{2}}}{a^{4} c^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 3.88 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.91 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{c-\frac {c}{a^2 x^2}} \, dx=\frac {x}{c}-\frac {5\,x-\frac {4}{a}}{c\,a^2\,x^2-2\,c\,a\,x+c}+\frac {4\,\ln \left (a\,x-1\right )}{a\,c} \]

[In]

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

[Out]

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