### 3.408 $$\int e^{4 \coth ^{-1}(a x)} (c-\frac{c}{a x}) \, dx$$

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

[Out]

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

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Rubi [A]  time = 0.0816088, antiderivative size = 25, 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} $-\frac{c \log (x)}{a}+\frac{4 c \log (1-a x)}{a}+c x$

Antiderivative was successfully veriﬁed.

[In]

Int[E^(4*ArcCoth[a*x])*(c - c/(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^{4 \coth ^{-1}(a x)} \left (c-\frac{c}{a x}\right ) \, dx &=\int e^{4 \tanh ^{-1}(a x)} \left (c-\frac{c}{a x}\right ) \, dx\\ &=-\frac{c \int \frac{e^{4 \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 = 0.0385445, size = 25, normalized size = 1. $-\frac{c \log (x)}{a}+\frac{4 c \log (1-a x)}{a}+c x$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.043, size = 25, normalized size = 1. \begin{align*} cx+4\,{\frac{c\ln \left ( ax-1 \right ) }{a}}-{\frac{c\ln \left ( x \right ) }{a}} \end{align*}

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)

[Out]

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

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Maxima [A]  time = 1.12228, size = 32, normalized size = 1.28 \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(1/(a*x-1)^2*(a*x+1)^2*(c-c/a/x),x, algorithm="maxima")

[Out]

c*x + 4*c*log(a*x - 1)/a - c*log(x)/a

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Fricas [A]  time = 1.54833, size = 55, normalized size = 2.2 \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(1/(a*x-1)^2*(a*x+1)^2*(c-c/a/x),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.389204, size = 17, normalized size = 0.68 \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(1/(a*x-1)**2*(a*x+1)**2*(c-c/a/x),x)

[Out]

c*x + c*(-log(x) + 4*log(x - 1/a))/a

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Giac [B]  time = 1.15558, size = 74, normalized size = 2.96 \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(1/(a*x-1)^2*(a*x+1)^2*(c-c/a/x),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