### 3.784 $$\int e^{2 \coth ^{-1}(a x)} (c-\frac{c}{a^2 x^2}) \, dx$$

Optimal. Leaf size=21 $-\frac{c}{a^2 x}+\frac{2 c \log (x)}{a}+c x$

[Out]

-(c/(a^2*x)) + c*x + (2*c*Log[x])/a

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Rubi [A]  time = 0.0854281, antiderivative size = 21, 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, 6157, 6150, 43} $-\frac{c}{a^2 x}+\frac{2 c \log (x)}{a}+c x$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-(c/(a^2*x)) + c*x + (2*c*Log[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 6157

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_.), x_Symbol] :> Dist[d^p, Int[(u*(1 - a^2*x^
2)^p*E^(n*ArcTanh[a*x]))/x^(2*p), x], x] /; FreeQ[{a, c, d, n}, x] && EqQ[c + a^2*d, 0] && IntegerQ[p]

Rule 6150

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[x^m*(1 -
a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[a^2*c + d, 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)} \left (c-\frac{c}{a^2 x^2}\right ) \, dx &=-\int e^{2 \tanh ^{-1}(a x)} \left (c-\frac{c}{a^2 x^2}\right ) \, dx\\ &=\frac{c \int \frac{e^{2 \tanh ^{-1}(a x)} \left (1-a^2 x^2\right )}{x^2} \, dx}{a^2}\\ &=\frac{c \int \frac{(1+a x)^2}{x^2} \, dx}{a^2}\\ &=\frac{c \int \left (a^2+\frac{1}{x^2}+\frac{2 a}{x}\right ) \, dx}{a^2}\\ &=-\frac{c}{a^2 x}+c x+\frac{2 c \log (x)}{a}\\ \end{align*}

Mathematica [A]  time = 0.0136718, size = 21, normalized size = 1. $-\frac{c}{a^2 x}+\frac{2 c \log (x)}{a}+c x$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-(c/(a^2*x)) + c*x + (2*c*Log[x])/a

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Maple [A]  time = 0.042, size = 22, normalized size = 1.1 \begin{align*} -{\frac{c}{{a}^{2}x}}+cx+2\,{\frac{c\ln \left ( x \right ) }{a}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.04415, size = 28, normalized size = 1.33 \begin{align*} c x + \frac{2 \, c \log \left (x\right )}{a} - \frac{c}{a^{2} x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c*x + 2*c*log(x)/a - c/(a^2*x)

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Fricas [A]  time = 1.44905, size = 57, normalized size = 2.71 \begin{align*} \frac{a^{2} c x^{2} + 2 \, a c x \log \left (x\right ) - c}{a^{2} x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.264319, size = 20, normalized size = 0.95 \begin{align*} \frac{a^{2} c x + 2 a c \log{\left (x \right )} - \frac{c}{x}}{a^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a**2*c*x + 2*a*c*log(x) - c/x)/a**2

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Giac [A]  time = 1.13647, size = 30, normalized size = 1.43 \begin{align*} c x + \frac{2 \, c \log \left ({\left | x \right |}\right )}{a} - \frac{c}{a^{2} x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c*x + 2*c*log(abs(x))/a - c/(a^2*x)