3.5.9 \(\int \frac {e^{4 \coth ^{-1}(a x)}}{c-\frac {c}{a x}} \, dx\) [409]

Optimal. Leaf size=53 \[ \frac {x}{c}-\frac {2}{a c (1-a x)^2}+\frac {8}{a c (1-a x)}+\frac {5 \log (1-a x)}{a c} \]

[Out]

x/c-2/a/c/(-a*x+1)^2+8/a/c/(-a*x+1)+5*ln(-a*x+1)/a/c

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Rubi [A]
time = 0.09, 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 = {6302, 6266, 6264, 78} \begin {gather*} \frac {8}{a c (1-a x)}-\frac {2}{a c (1-a x)^2}+\frac {5 \log (1-a x)}{a c}+\frac {x}{c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x/c - 2/(a*c*(1 - a*x)^2) + 8/(a*c*(1 - a*x)) + (5*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 6264

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 6266

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 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*} \int \frac {e^{4 \coth ^{-1}(a x)}}{c-\frac {c}{a x}} \, dx &=\int \frac {e^{4 \tanh ^{-1}(a x)}}{c-\frac {c}{a x}} \, dx\\ &=-\frac {a \int \frac {e^{4 \tanh ^{-1}(a x)} x}{1-a x} \, dx}{c}\\ &=-\frac {a \int \frac {x (1+a x)^2}{(1-a x)^3} \, dx}{c}\\ &=-\frac {a \int \left (-\frac {1}{a}-\frac {4}{a (-1+a x)^3}-\frac {8}{a (-1+a x)^2}-\frac {5}{a (-1+a x)}\right ) \, dx}{c}\\ &=\frac {x}{c}-\frac {2}{a c (1-a x)^2}+\frac {8}{a c (1-a x)}+\frac {5 \log (1-a x)}{a c}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 51, normalized size = 0.96 \begin {gather*} -\frac {a \left (-\frac {x}{a}+\frac {2}{a^2 (1-a x)^2}-\frac {8}{a^2 (1-a x)}-\frac {5 \log (1-a x)}{a^2}\right )}{c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*(-(x/a) + 2/(a^2*(1 - a*x)^2) - 8/(a^2*(1 - a*x)) - (5*Log[1 - a*x])/a^2))/c)

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Maple [A]
time = 0.13, size = 47, normalized size = 0.89

method	result	size
risch	\(\frac {x}{c}+\frac {-8 c x +\frac {6 c}{a}}{c^{2} \left (a x -1\right )^{2}}+\frac {5 \ln \left (a x -1\right )}{a c}\)	\(43\)
default	\(\frac {a \left (\frac {x}{a}-\frac {8}{a^{2} \left (a x -1\right )}-\frac {2}{a^{2} \left (a x -1\right )^{2}}+\frac {5 \ln \left (a x -1\right )}{a^{2}}\right )}{c}\)	\(47\)
norman	\(\frac {\frac {a^{2} x^{3}}{c}-\frac {8 a \,x^{2}}{c}+\frac {5 x}{c}}{\left (a x -1\right )^{2}}+\frac {5 \ln \left (a x -1\right )}{a c}\)	\(50\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a/c*(x/a-8/a^2/(a*x-1)-2/a^2/(a*x-1)^2+5/a^2*ln(a*x-1))

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Maxima [A]
time = 0.27, size = 49, normalized size = 0.92 \begin {gather*} -\frac {2 \, {\left (4 \, a x - 3\right )}}{a^{3} c x^{2} - 2 \, a^{2} c x + a c} + \frac {x}{c} + \frac {5 \, \log \left (a x - 1\right )}{a c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.33, size = 64, normalized size = 1.21 \begin {gather*} \frac {a^{3} x^{3} - 2 \, a^{2} x^{2} - 7 \, a x + 5 \, {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (a x - 1\right ) + 6}{a^{3} c x^{2} - 2 \, a^{2} c x + a c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^3*x^3 - 2*a^2*x^2 - 7*a*x + 5*(a^2*x^2 - 2*a*x + 1)*log(a*x - 1) + 6)/(a^3*c*x^2 - 2*a^2*c*x + a*c)

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Sympy [A]
time = 0.13, size = 41, normalized size = 0.77 \begin {gather*} \frac {- 8 a x + 6}{a^{3} c x^{2} - 2 a^{2} c x + a c} + \frac {x}{c} + \frac {5 \log {\left (a x - 1 \right )}}{a c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-8*a*x + 6)/(a**3*c*x**2 - 2*a**2*c*x + a*c) + x/c + 5*log(a*x - 1)/(a*c)

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Giac [A]
time = 0.39, size = 74, normalized size = 1.40 \begin {gather*} \frac {a x - 1}{a c} - \frac {5 \, \log \left (\frac {{\left | a x - 1 \right |}}{{\left (a x - 1\right )}^{2} {\left | a \right |}}\right )}{a c} - \frac {2 \, {\left (\frac {4 \, a^{3} c}{a x - 1} + \frac {a^{3} c}{{\left (a x - 1\right )}^{2}}\right )}}{a^{4} c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.06, size = 48, normalized size = 0.91 \begin {gather*} \frac {x}{c}-\frac {8\,x-\frac {6}{a}}{c\,a^2\,x^2-2\,c\,a\,x+c}+\frac {5\,\ln \left (a\,x-1\right )}{a\,c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/c - (8*x - 6/a)/(c + a^2*c*x^2 - 2*a*c*x) + (5*log(a*x - 1))/(a*c)

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