### 3.422 $$\int e^{-2 \coth ^{-1}(a x)} (c-\frac{c}{a x})^3 \, dx$$

Optimal. Leaf size=54 $-\frac{c^3}{2 a^3 x^2}+\frac{5 c^3}{a^2 x}+\frac{11 c^3 \log (x)}{a}-\frac{16 c^3 \log (a x+1)}{a}+c^3 x$

[Out]

-c^3/(2*a^3*x^2) + (5*c^3)/(a^2*x) + c^3*x + (11*c^3*Log[x])/a - (16*c^3*Log[1 + a*x])/a

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Rubi [A]  time = 0.141078, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 22, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.182, Rules used = {6167, 6131, 6129, 88} $-\frac{c^3}{2 a^3 x^2}+\frac{5 c^3}{a^2 x}+\frac{11 c^3 \log (x)}{a}-\frac{16 c^3 \log (a x+1)}{a}+c^3 x$

Antiderivative was successfully veriﬁed.

[In]

Int[(c - c/(a*x))^3/E^(2*ArcCoth[a*x]),x]

[Out]

-c^3/(2*a^3*x^2) + (5*c^3)/(a^2*x) + c^3*x + (11*c^3*Log[x])/a - (16*c^3*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 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

\begin{align*} \int e^{-2 \coth ^{-1}(a x)} \left (c-\frac{c}{a x}\right )^3 \, dx &=-\int e^{-2 \tanh ^{-1}(a x)} \left (c-\frac{c}{a x}\right )^3 \, dx\\ &=\frac{c^3 \int \frac{e^{-2 \tanh ^{-1}(a x)} (1-a x)^3}{x^3} \, dx}{a^3}\\ &=\frac{c^3 \int \frac{(1-a x)^4}{x^3 (1+a x)} \, dx}{a^3}\\ &=\frac{c^3 \int \left (a^3+\frac{1}{x^3}-\frac{5 a}{x^2}+\frac{11 a^2}{x}-\frac{16 a^3}{1+a x}\right ) \, dx}{a^3}\\ &=-\frac{c^3}{2 a^3 x^2}+\frac{5 c^3}{a^2 x}+c^3 x+\frac{11 c^3 \log (x)}{a}-\frac{16 c^3 \log (1+a x)}{a}\\ \end{align*}

Mathematica [A]  time = 0.128238, size = 56, normalized size = 1.04 $-\frac{c^3}{2 a^3 x^2}+\frac{5 c^3}{a^2 x}+\frac{11 c^3 \log (a x)}{a}-\frac{16 c^3 \log (a x+1)}{a}+c^3 x$

Warning: Unable to verify antiderivative.

[In]

Integrate[(c - c/(a*x))^3/E^(2*ArcCoth[a*x]),x]

[Out]

-c^3/(2*a^3*x^2) + (5*c^3)/(a^2*x) + c^3*x + (11*c^3*Log[a*x])/a - (16*c^3*Log[1 + a*x])/a

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Maple [A]  time = 0.046, size = 53, normalized size = 1. \begin{align*} -{\frac{{c}^{3}}{2\,{x}^{2}{a}^{3}}}+5\,{\frac{{c}^{3}}{{a}^{2}x}}+{c}^{3}x+11\,{\frac{{c}^{3}\ln \left ( x \right ) }{a}}-16\,{\frac{{c}^{3}\ln \left ( ax+1 \right ) }{a}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((c-c/a/x)^3/(a*x+1)*(a*x-1),x)

[Out]

-1/2*c^3/x^2/a^3+5*c^3/a^2/x+c^3*x+11*c^3*ln(x)/a-16*c^3*ln(a*x+1)/a

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Maxima [A]  time = 1.09595, size = 69, normalized size = 1.28 \begin{align*} c^{3} x - \frac{16 \, c^{3} \log \left (a x + 1\right )}{a} + \frac{11 \, c^{3} \log \left (x\right )}{a} + \frac{10 \, a c^{3} x - c^{3}}{2 \, a^{3} x^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a/x)^3*(a*x-1)/(a*x+1),x, algorithm="maxima")

[Out]

c^3*x - 16*c^3*log(a*x + 1)/a + 11*c^3*log(x)/a + 1/2*(10*a*c^3*x - c^3)/(a^3*x^2)

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Fricas [A]  time = 1.84189, size = 140, normalized size = 2.59 \begin{align*} \frac{2 \, a^{3} c^{3} x^{3} - 32 \, a^{2} c^{3} x^{2} \log \left (a x + 1\right ) + 22 \, a^{2} c^{3} x^{2} \log \left (x\right ) + 10 \, a c^{3} x - c^{3}}{2 \, a^{3} x^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a/x)^3*(a*x-1)/(a*x+1),x, algorithm="fricas")

[Out]

1/2*(2*a^3*c^3*x^3 - 32*a^2*c^3*x^2*log(a*x + 1) + 22*a^2*c^3*x^2*log(x) + 10*a*c^3*x - c^3)/(a^3*x^2)

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Sympy [A]  time = 0.56804, size = 42, normalized size = 0.78 \begin{align*} c^{3} x + \frac{c^{3} \left (11 \log{\left (x \right )} - 16 \log{\left (x + \frac{1}{a} \right )}\right )}{a} + \frac{10 a c^{3} x - c^{3}}{2 a^{3} x^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a/x)**3*(a*x-1)/(a*x+1),x)

[Out]

c**3*x + c**3*(11*log(x) - 16*log(x + 1/a))/a + (10*a*c**3*x - c**3)/(2*a**3*x**2)

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Giac [A]  time = 1.17912, size = 72, normalized size = 1.33 \begin{align*} c^{3} x - \frac{16 \, c^{3} \log \left ({\left | a x + 1 \right |}\right )}{a} + \frac{11 \, c^{3} \log \left ({\left | x \right |}\right )}{a} + \frac{10 \, a c^{3} x - c^{3}}{2 \, a^{3} x^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a/x)^3*(a*x-1)/(a*x+1),x, algorithm="giac")

[Out]

c^3*x - 16*c^3*log(abs(a*x + 1))/a + 11*c^3*log(abs(x))/a + 1/2*(10*a*c^3*x - c^3)/(a^3*x^2)