### 3.424 $$\int e^{-2 \coth ^{-1}(a x)} (c-\frac{c}{a x}) \, dx$$

Optimal. Leaf size=23 $\frac{c \log (x)}{a}-\frac{4 c \log (a x+1)}{a}+c x$

[Out]

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

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Rubi [A]  time = 0.0822374, antiderivative size = 23, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 20, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.2, Rules used = {6167, 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*ArcCoth[a*x]),x]

[Out]

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

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]

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 \coth ^{-1}(a x)} \left (c-\frac{c}{a x}\right ) \, dx &=-\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 = 0.0388513, size = 23, 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*ArcCoth[a*x]),x]

[Out]

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

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Maple [A]  time = 0.045, size = 24, 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)*(a*x-1),x)

[Out]

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

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Maxima [A]  time = 1.02825, size = 31, normalized size = 1.35 \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)/(a*x+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 = 1.85841, size = 55, normalized size = 2.39 \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)/(a*x+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 = 0.381285, size = 17, normalized size = 0.74 \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)/(a*x+1),x)

[Out]

c*x + c*(log(x) - 4*log(x + 1/a))/a

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Giac [A]  time = 1.13548, size = 34, normalized size = 1.48 \begin{align*} c x - \frac{4 \, c \log \left ({\left | a x + 1 \right |}\right )}{a} + \frac{c \log \left ({\left | x \right |}\right )}{a} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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