\(\int e^{-2 \coth ^{-1}(a x)} (c-\frac {c}{a x})^3 \, dx\) [422]

Optimal result

Integrand size = 22, antiderivative size = 54 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^3 \, dx=-\frac {c^3}{2 a^3 x^2}+\frac {5 c^3}{a^2 x}+c^3 x+\frac {11 c^3 \log (x)}{a}-\frac {16 c^3 \log (1+a x)}{a} \]

[Out]

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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {6302, 6266, 6264, 90} \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^3 \, dx=-\frac {c^3}{2 a^3 x^2}+\frac {5 c^3}{a^2 x}+\frac {11 c^3 \log (x)}{a}-\frac {16 c^3 \log (a x+1)}{a}+c^3 x \]

[In]

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

[Out]

-1/2*c^3/(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 90

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]))

Rule 6264

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 6266

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 6302

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]

Rubi steps \begin{align*} \text {integral}& = -\int e^{-2 \text {arctanh}(a x)} \left (c-\frac {c}{a x}\right )^3 \, dx \\ & = \frac {c^3 \int \frac {e^{-2 \text {arctanh}(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] (verified)

Time = 0.07 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.85 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^3 \, dx=-\frac {c^3 \left (\frac {1}{2 x^2}-\frac {5 a}{x}-a^3 x-11 a^2 \log (x)+16 a^2 \log (1+a x)\right )}{a^3} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.62 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.80

method	result	size
default	\(\frac {c^{3} \left (-16 a^{2} \ln \left (a x +1\right )+a^{3} x -\frac {1}{2 x^{2}}+\frac {5 a}{x}+11 a^{2} \ln \left (x \right )\right )}{a^{3}}\)	\(43\)
risch	\(c^{3} x +\frac {5 a \,c^{3} x -\frac {1}{2} c^{3}}{a^{3} x^{2}}+\frac {11 c^{3} \ln \left (-x \right )}{a}-\frac {16 c^{3} \ln \left (a x +1\right )}{a}\)	\(53\)
norman	\(\frac {a^{2} c^{3} x^{3}-\frac {c^{3}}{2 a}+5 c^{3} x}{a^{2} x^{2}}+\frac {11 c^{3} \ln \left (x \right )}{a}-\frac {16 c^{3} \ln \left (a x +1\right )}{a}\)	\(58\)
parallelrisch	\(\frac {2 a^{3} c^{3} x^{3}+22 c^{3} \ln \left (x \right ) a^{2} x^{2}-32 c^{3} \ln \left (a x +1\right ) a^{2} x^{2}+10 a \,c^{3} x -c^{3}}{2 a^{3} x^{2}}\)	\(63\)
meijerg	\(\frac {c^{3} \left (a x -\ln \left (a x +1\right )\right )}{a}-\frac {4 c^{3} \ln \left (a x +1\right )}{a}+\frac {6 c^{3} \left (-\ln \left (a x +1\right )+\ln \left (x \right )+\ln \left (a \right )\right )}{a}-\frac {4 c^{3} \left (\ln \left (a x +1\right )-\ln \left (x \right )-\ln \left (a \right )-\frac {1}{a x}\right )}{a}+\frac {c^{3} \left (-\ln \left (a x +1\right )+\ln \left (x \right )+\ln \left (a \right )-\frac {1}{2 a^{2} x^{2}}+\frac {1}{a x}\right )}{a}\)	\(122\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.15 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^3 \, dx=\frac {2 \, a^{3} c^{3} x^{3} - 32 \, a^{2} c^{3} x^{2} \log \left (a x + 1\right ) + 22 \, a^{2} c^{3} x^{2} \log \left (x\right ) + 10 \, a c^{3} x - c^{3}}{2 \, a^{3} x^{2}} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.78 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^3 \, dx=c^{3} x + \frac {c^{3} \cdot \left (11 \log {\left (x \right )} - 16 \log {\left (x + \frac {1}{a} \right )}\right )}{a} + \frac {10 a c^{3} x - c^{3}}{2 a^{3} x^{2}} \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.94 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^3 \, dx=c^{3} x - \frac {16 \, c^{3} \log \left (a x + 1\right )}{a} + \frac {11 \, c^{3} \log \left (x\right )}{a} + \frac {10 \, a c^{3} x - c^{3}}{2 \, a^{3} x^{2}} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.98 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^3 \, dx=c^{3} x - \frac {16 \, c^{3} \log \left ({\left | a x + 1 \right |}\right )}{a} + \frac {11 \, c^{3} \log \left ({\left | x \right |}\right )}{a} + \frac {10 \, a c^{3} x - c^{3}}{2 \, a^{3} x^{2}} \]

[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)

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.94 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^3 \, dx=c^3\,x-\frac {\frac {c^3}{2}-5\,a\,c^3\,x}{a^3\,x^2}+\frac {11\,c^3\,\ln \left (x\right )}{a}-\frac {16\,c^3\,\ln \left (a\,x+1\right )}{a} \]

[In]

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

[Out]

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