3.5.62 \(\int \frac {e^{2 \tanh ^{-1}(a x)}}{c-\frac {c}{a x}} \, dx\) [462]

Optimal. Leaf size=38 \[ -\frac {x}{c}-\frac {2}{a c (1-a x)}-\frac {3 \log (1-a x)}{a c} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {6266, 6264, 78} \begin {gather*} -\frac {2}{a c (1-a x)}-\frac {3 \log (1-a x)}{a c}-\frac {x}{c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(x/c) - 2/(a*c*(1 - a*x)) - (3*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]

Rubi steps

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

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Mathematica [A]
time = 0.02, size = 31, normalized size = 0.82 \begin {gather*} -\frac {a x+\frac {2}{1-a x}+3 \log (1-a x)}{a c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*x + 2/(1 - a*x) + 3*Log[1 - a*x])/(a*c))

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Maple [A]
time = 0.74, size = 36, normalized size = 0.95

method	result	size
default	\(\frac {a \left (-\frac {x}{a}+\frac {2}{\left (a x -1\right ) a^{2}}-\frac {3 \ln \left (a x -1\right )}{a^{2}}\right )}{c}\)	\(36\)
risch	\(-\frac {x}{c}+\frac {2}{a c \left (a x -1\right )}-\frac {3 \ln \left (a x -1\right )}{a c}\)	\(37\)
norman	\(\frac {-\frac {a \,x^{2}}{c}+\frac {3 x}{c}}{a x -1}-\frac {3 \ln \left (a x -1\right )}{a c}\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 36, normalized size = 0.95 \begin {gather*} -\frac {x}{c} + \frac {2}{a^{2} c x - a c} - \frac {3 \, \log \left (a x - 1\right )}{a c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.34, size = 41, normalized size = 1.08 \begin {gather*} -\frac {a^{2} x^{2} - a x + 3 \, {\left (a x - 1\right )} \log \left (a x - 1\right ) - 2}{a^{2} c x - a c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.08, size = 26, normalized size = 0.68 \begin {gather*} \frac {2}{a^{2} c x - a c} - \frac {x}{c} - \frac {3 \log {\left (a x - 1 \right )}}{a c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.41, size = 37, normalized size = 0.97 \begin {gather*} -\frac {x}{c} - \frac {3 \, \log \left ({\left | a x - 1 \right |}\right )}{a c} + \frac {2}{{\left (a x - 1\right )} a c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.06, size = 35, normalized size = 0.92 \begin {gather*} -\frac {x}{c}-\frac {2}{a\,\left (c-a\,c\,x\right )}-\frac {3\,\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^2*x^2 - 1)),x)

[Out]

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

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