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

Optimal. Leaf size=25 \[ -\frac{c \log (x)}{a}+\frac{4 c \log (a x+1)}{a}-c x \]

[Out]

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

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Rubi [A]  time = 0.0656118, antiderivative size = 25, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {6131, 6129, 72} \[ -\frac{c \log (x)}{a}+\frac{4 c \log (a x+1)}{a}-c x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rubi steps

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

Mathematica [A]  time = 0.0364962, size = 25, normalized size = 1. \[ -\frac{c \log (x)}{a}+\frac{4 c \log (a x+1)}{a}-c x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.034, size = 26, normalized size = 1. \begin{align*} -cx-{\frac{c\ln \left ( x \right ) }{a}}+4\,{\frac{c\ln \left ( ax+1 \right ) }{a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.960578, size = 34, normalized size = 1.36 \begin{align*} -c x + \frac{4 \, c \log \left (a x + 1\right )}{a} - \frac{c \log \left (x\right )}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.29126, size = 57, normalized size = 2.28 \begin{align*} -\frac{a c x - 4 \, c \log \left (a x + 1\right ) + c \log \left (x\right )}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.426521, size = 19, normalized size = 0.76 \begin{align*} - c x - \frac{c \left (\log{\left (x \right )} - 4 \log{\left (x + \frac{1}{a} \right )}\right )}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c*x - c*(log(x) - 4*log(x + 1/a))/a

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Giac [B]  time = 1.18673, size = 76, normalized size = 3.04 \begin{align*} -\frac{{\left (a x + 1\right )} c}{a} - \frac{3 \, c \log \left (\frac{{\left | a x + 1 \right |}}{{\left (a x + 1\right )}^{2}{\left | a \right |}}\right )}{a} - \frac{c \log \left ({\left | -\frac{1}{a x + 1} + 1 \right |}\right )}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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