\(\int e^{2 \coth ^{-1}(a x)} (c-\frac {c}{a x})^3 \, dx\) [389]

Optimal result

Integrand size = 22, antiderivative size = 39 \[ \int e^{2 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^3 \, dx=-\frac {c^3}{2 a^3 x^2}+\frac {c^3}{a^2 x}+c^3 x-\frac {c^3 \log (x)}{a} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 39, 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, 76} \[ \int e^{2 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^3 \, dx=-\frac {c^3}{2 a^3 x^2}+\frac {c^3}{a^2 x}-\frac {c^3 \log (x)}{a}+c^3 x \]

[In]

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

[Out]

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

Rule 76

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] &&  !(ILtQ[n
+ p + 2, 0] && GtQ[n + 2*p, 0])

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.97 \[ \int e^{2 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^3 \, dx=-\frac {c^3 \left (3 a^2+\frac {1}{x^2}-\frac {2 a}{x}-2 a^3 x+2 a^2 \log (x)\right )}{2 a^3} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.59 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.79

method	result	size
default	\(\frac {c^{3} \left (a^{3} x -a^{2} \ln \left (x \right )-\frac {1}{2 x^{2}}+\frac {a}{x}\right )}{a^{3}}\)	\(31\)
risch	\(c^{3} x +\frac {a \,c^{3} x -\frac {1}{2} c^{3}}{a^{3} x^{2}}-\frac {c^{3} \ln \left (x \right )}{a}\)	\(36\)
norman	\(\frac {c^{3} x +a^{2} c^{3} x^{3}-\frac {c^{3}}{2 a}}{a^{2} x^{2}}-\frac {c^{3} \ln \left (x \right )}{a}\)	\(43\)
parallelrisch	\(-\frac {-2 a^{3} c^{3} x^{3}+2 c^{3} \ln \left (x \right ) a^{2} x^{2}-2 a \,c^{3} x +c^{3}}{2 a^{3} x^{2}}\)	\(44\)
meijerg	\(-\frac {c^{3} \left (-a x -\ln \left (-a x +1\right )\right )}{a}-\frac {2 c^{3} \ln \left (-a x +1\right )}{a}+\frac {2 c^{3} \left (\ln \left (-a x +1\right )-\ln \left (x \right )-\ln \left (-a \right )+\frac {1}{a x}\right )}{a}+\frac {c^{3} \left (-\ln \left (-a x +1\right )+\ln \left (x \right )+\ln \left (-a \right )-\frac {1}{2 a^{2} x^{2}}-\frac {1}{a x}\right )}{a}\)	\(111\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.15 \[ \int e^{2 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^3 \, dx=\frac {2 \, a^{3} c^{3} x^{3} - 2 \, a^{2} c^{3} x^{2} \log \left (x\right ) + 2 \, a c^{3} x - c^{3}}{2 \, a^{3} x^{2}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.95 \[ \int e^{2 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^3 \, dx=\frac {a^{3} c^{3} x - a^{2} c^{3} \log {\left (x \right )} + \frac {2 a c^{3} x - c^{3}}{2 x^{2}}}{a^{3}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.95 \[ \int e^{2 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^3 \, dx=c^{3} x - \frac {c^{3} \log \left (x\right )}{a} + \frac {2 \, a c^{3} x - c^{3}}{2 \, a^{3} x^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.97 \[ \int e^{2 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^3 \, dx=c^{3} x - \frac {c^{3} \log \left ({\left | x \right |}\right )}{a} + \frac {2 \, a c^{3} x - c^{3}}{2 \, a^{3} x^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 4.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.90 \[ \int e^{2 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^3 \, dx=\frac {c^3\,\left (2\,a\,x+2\,a^3\,x^3-2\,a^2\,x^2\,\ln \left (x\right )-1\right )}{2\,a^3\,x^2} \]

[In]

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

[Out]

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