### 3.192 $$\int e^{4 \coth ^{-1}(a x)} (c-a c x) \, dx$$

Optimal. Leaf size=27 $-\frac{1}{2} a c x^2-\frac{4 c \log (1-a x)}{a}-3 c x$

[Out]

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

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Rubi [A]  time = 0.0366372, antiderivative size = 27, 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} $-\frac{1}{2} a c x^2-\frac{4 c \log (1-a x)}{a}-3 c x$

Antiderivative was successfully veriﬁed.

[In]

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

Mathematica [A]  time = 0.0122167, size = 26, normalized size = 0.96 $c \left (-\frac{a x^2}{2}-\frac{4 \log (1-a x)}{a}-3 x\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.043, size = 25, normalized size = 0.9 \begin{align*} -{\frac{ac{x}^{2}}{2}}-3\,cx-4\,{\frac{c\ln \left ( ax-1 \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*(-a*c*x+c),x)

[Out]

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

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Maxima [A]  time = 1.02561, size = 32, normalized size = 1.19 \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(1/(a*x-1)^2*(a*x+1)^2*(-a*c*x+c),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 = 1.50781, size = 66, normalized size = 2.44 \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(1/(a*x-1)^2*(a*x+1)^2*(-a*c*x+c),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 = 0.470965, size = 26, normalized size = 0.96 \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(1/(a*x-1)**2*(a*x+1)**2*(-a*c*x+c),x)

[Out]

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

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Giac [A]  time = 1.17871, size = 68, normalized size = 2.52 \begin{align*} -\frac{{\left (a x - 1\right )}^{2}{\left (c + \frac{8 \, c}{a x - 1}\right )}}{2 \, a} + \frac{4 \, c \log \left (\frac{{\left | a x - 1 \right |}}{{\left (a x - 1\right )}^{2}{\left | 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*(-a*c*x+c),x, algorithm="giac")

[Out]

-1/2*(a*x - 1)^2*(c + 8*c/(a*x - 1))/a + 4*c*log(abs(a*x - 1)/((a*x - 1)^2*abs(a)))/a