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

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

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 53, normalized size of antiderivative = 1.00, 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, 6131, 6129, 77} \[ \frac {5}{a c^2 (1-a x)}-\frac {1}{a c^2 (1-a x)^2}+\frac {4 \log (1-a x)}{a c^2}+\frac {x}{c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]))))

Rule 6129

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

Rule 6131

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.11, size = 51, normalized size = 0.96 \[ -\frac {5}{a c^2 (a x-1)}-\frac {1}{a c^2 (a x-1)^2}+\frac {4 \log (1-a x)}{a c^2}+\frac {x}{c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.56, size = 70, normalized size = 1.32 \[ \frac {a^{3} x^{3} - 2 \, a^{2} x^{2} - 4 \, a x + 4 \, {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (a x - 1\right ) + 4}{a^{3} c^{2} x^{2} - 2 \, a^{2} c^{2} x + a c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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giac [A]  time = 0.12, size = 42, normalized size = 0.79 \[ \frac {x}{c^{2}} + \frac {4 \, \log \left ({\left | a x - 1 \right |}\right )}{a c^{2}} - \frac {5 \, a x - 4}{{\left (a x - 1\right )}^{2} a c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 51, normalized size = 0.96 \[ \frac {x}{c^{2}}+\frac {4 \ln \left (a x -1\right )}{a \,c^{2}}-\frac {5}{a \,c^{2} \left (a x -1\right )}-\frac {1}{a \,c^{2} \left (a x -1\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.30, size = 55, normalized size = 1.04 \[ -\frac {5 \, a x - 4}{a^{3} c^{2} x^{2} - 2 \, a^{2} c^{2} x + a c^{2}} + \frac {x}{c^{2}} + \frac {4 \, \log \left (a x - 1\right )}{a c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.21, size = 54, normalized size = 1.02 \[ \frac {x}{c^2}-\frac {5\,x-\frac {4}{a}}{a^2\,c^2\,x^2-2\,a\,c^2\,x+c^2}+\frac {4\,\ln \left (a\,x-1\right )}{a\,c^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.23, size = 49, normalized size = 0.92 \[ \frac {- 5 a x + 4}{a^{3} c^{2} x^{2} - 2 a^{2} c^{2} x + a c^{2}} + \frac {x}{c^{2}} + \frac {4 \log {\left (a x - 1 \right )}}{a c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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