3.3.12 \(\int \frac {e^{-2 \tanh ^{-1}(a x)}}{(c-a c x)^2} \, dx\) [212]

Optimal. Leaf size=11 \[ \frac {\tanh ^{-1}(a x)}{a c^2} \]

[Out]

arctanh(a*x)/a/c^2

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Rubi [A]
time = 0.02, antiderivative size = 11, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6264, 35, 212} \begin {gather*} \frac {\tanh ^{-1}(a x)}{a c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(E^(2*ArcTanh[a*x])*(c - a*c*x)^2),x]

[Out]

ArcTanh[a*x]/(a*c^2)

Rule 35

Int[1/(((a_) + (b_.)*(x_))*((c_) + (d_.)*(x_))), x_Symbol] :> Int[1/(a*c + b*d*x^2), x] /; FreeQ[{a, b, c, d},
x] && EqQ[b*c + a*d, 0]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

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])

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 11, normalized size = 1.00 \begin {gather*} \frac {\tanh ^{-1}(a x)}{a c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(E^(2*ArcTanh[a*x])*(c - a*c*x)^2),x]

[Out]

ArcTanh[a*x]/(a*c^2)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(27\) vs. \(2(11)=22\).
time = 0.72, size = 28, normalized size = 2.55

method	result	size
default	\(\frac {\frac {\ln \left (a x +1\right )}{2 a}-\frac {\ln \left (a x -1\right )}{2 a}}{c^{2}}\)	\(28\)
norman	\(-\frac {\ln \left (a x -1\right )}{2 a \,c^{2}}+\frac {\ln \left (a x +1\right )}{2 a \,c^{2}}\)	\(30\)
risch	\(-\frac {\ln \left (a x -1\right )}{2 a \,c^{2}}+\frac {\ln \left (-a x -1\right )}{2 a \,c^{2}}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 29 vs. \(2 (11) = 22\).
time = 0.26, size = 29, normalized size = 2.64 \begin {gather*} \frac {\log \left (a x + 1\right )}{2 \, a c^{2}} - \frac {\log \left (a x - 1\right )}{2 \, a c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 23 vs. \(2 (11) = 22\).
time = 0.38, size = 23, normalized size = 2.09 \begin {gather*} \frac {\log \left (a x + 1\right ) - \log \left (a x - 1\right )}{2 \, a c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 22 vs. \(2 (8) = 16\).
time = 0.07, size = 22, normalized size = 2.00 \begin {gather*} - \frac {\frac {\log {\left (x - \frac {1}{a} \right )}}{2} - \frac {\log {\left (x + \frac {1}{a} \right )}}{2}}{a c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 25 vs. \(2 (11) = 22\).
time = 0.44, size = 25, normalized size = 2.27 \begin {gather*} \frac {\log \left ({\left | -\frac {2 \, c}{a c x - c} - 1 \right |}\right )}{2 \, a c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.07, size = 11, normalized size = 1.00 \begin {gather*} \frac {\mathrm {atanh}\left (a\,x\right )}{a\,c^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

atanh(a*x)/(a*c^2)

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