### 3.802 $$\int \frac{e^{4 \coth ^{-1}(a x)}}{c-\frac{c}{a^2 x^2}} \, dx$$

Optimal. Leaf size=53 $\frac{5}{a c (1-a x)}-\frac{1}{a c (1-a x)^2}+\frac{4 \log (1-a x)}{a c}+\frac{x}{c}$

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.168444, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 22, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.182, Rules used = {6167, 6157, 6150, 77} $\frac{5}{a c (1-a x)}-\frac{1}{a c (1-a x)^2}+\frac{4 \log (1-a x)}{a c}+\frac{x}{c}$

Antiderivative was successfully veriﬁed.

[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 6167

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]

Rule 6157

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

Rule 6150

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 77

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]))))

Rubi steps

\begin{align*} \int \frac{e^{4 \coth ^{-1}(a x)}}{c-\frac{c}{a^2 x^2}} \, dx &=\int \frac{e^{4 \tanh ^{-1}(a x)}}{c-\frac{c}{a^2 x^2}} \, dx\\ &=-\frac{a^2 \int \frac{e^{4 \tanh ^{-1}(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]  time = 0.0337655, size = 53, normalized size = 1. $\frac{5}{a c (1-a x)}-\frac{1}{a c (1-a x)^2}+\frac{4 \log (1-a x)}{a c}+\frac{x}{c}$

Antiderivative was successfully veriﬁed.

[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]  time = 0.043, size = 51, normalized size = 1. \begin{align*}{\frac{x}{c}}-{\frac{1}{ac \left ( ax-1 \right ) ^{2}}}+4\,{\frac{\ln \left ( ax-1 \right ) }{ac}}-5\,{\frac{1}{ac \left ( ax-1 \right ) }} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.04098, size = 66, normalized size = 1.25 \begin{align*} -\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} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

Fricas [A]  time = 1.58226, size = 140, normalized size = 2.64 \begin{align*} \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} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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]  time = 0.424671, size = 41, normalized size = 0.77 \begin{align*} - \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} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

Giac [A]  time = 1.13249, size = 100, normalized size = 1.89 \begin{align*} \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}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)