### 3.210 $$\int e^{-2 \coth ^{-1}(a x)} (c-a c x) \, dx$$

Optimal. Leaf size=26 $-\frac{1}{2} a c x^2-\frac{4 c \log (a x+1)}{a}+3 c x$

[Out]

3*c*x - (a*c*x^2)/2 - (4*c*Log[1 + a*x])/a

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Rubi [A]  time = 0.0350002, antiderivative size = 26, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 16, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.188, Rules used = {6167, 6129, 43} $-\frac{1}{2} a c x^2-\frac{4 c \log (a x+1)}{a}+3 c x$

Antiderivative was successfully veriﬁed.

[In]

Int[(c - a*c*x)/E^(2*ArcCoth[a*x]),x]

[Out]

3*c*x - (a*c*x^2)/2 - (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 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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int e^{-2 \coth ^{-1}(a x)} (c-a c x) \, dx &=-\int e^{-2 \tanh ^{-1}(a x)} (c-a c x) \, dx\\ &=-\left (c \int \frac{(1-a x)^2}{1+a x} \, dx\right )\\ &=-\left (c \int \left (-3+a x+\frac{4}{1+a x}\right ) \, dx\right )\\ &=3 c x-\frac{1}{2} a c x^2-\frac{4 c \log (1+a x)}{a}\\ \end{align*}

Mathematica [A]  time = 0.0092851, size = 26, normalized size = 1. $-\frac{1}{2} a c x^2-\frac{4 c \log (a x+1)}{a}+3 c x$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

3*c*x - (a*c*x^2)/2 - (4*c*Log[1 + a*x])/a

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Maple [A]  time = 0.043, size = 25, normalized size = 1. \begin{align*} 3\,cx-{\frac{ac{x}^{2}}{2}}-4\,{\frac{c\ln \left ( ax+1 \right ) }{a}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*c*x-1/2*a*c*x^2-4*c*ln(a*x+1)/a

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Maxima [A]  time = 1.02372, size = 32, normalized size = 1.23 \begin{align*} -\frac{1}{2} \, a c x^{2} + 3 \, c x - \frac{4 \, c \log \left (a x + 1\right )}{a} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a*c*x^2 + 3*c*x - 4*c*log(a*x + 1)/a

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Fricas [A]  time = 1.46784, size = 66, normalized size = 2.54 \begin{align*} -\frac{a^{2} c x^{2} - 6 \, a c x + 8 \, c \log \left (a x + 1\right )}{2 \, a} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(a^2*c*x^2 - 6*a*c*x + 8*c*log(a*x + 1))/a

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Sympy [A]  time = 1.09001, size = 24, normalized size = 0.92 \begin{align*} - \frac{a c x^{2}}{2} + 3 c x - \frac{4 c \log{\left (a x + 1 \right )}}{a} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*c*x**2/2 + 3*c*x - 4*c*log(a*x + 1)/a

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Giac [A]  time = 1.12306, size = 47, normalized size = 1.81 \begin{align*} -\frac{4 \, c \log \left ({\left | a x + 1 \right |}\right )}{a} - \frac{a^{3} c x^{2} - 6 \, a^{2} c x}{2 \, a^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*c*log(abs(a*x + 1))/a - 1/2*(a^3*c*x^2 - 6*a^2*c*x)/a^2