\(\int e^{4 \coth ^{-1}(a x)} (c-\frac {c}{a x})^2 \, dx\) [407]

Optimal result

Integrand size = 22, antiderivative size = 27 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^2 \, dx=-\frac {c^2}{a^2 x}+c^2 x+\frac {2 c^2 \log (x)}{a} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 27, 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, 45} \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^2 \, dx=-\frac {c^2}{a^2 x}+\frac {2 c^2 \log (x)}{a}+c^2 x \]

[In]

Int[E^(4*ArcCoth[a*x])*(c - c/(a*x))^2,x]

[Out]

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

Rule 45

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])

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^{4 \text {arctanh}(a x)} \left (c-\frac {c}{a x}\right )^2 \, dx \\ & = \frac {c^2 \int \frac {e^{4 \text {arctanh}(a x)} (1-a x)^2}{x^2} \, dx}{a^2} \\ & = \frac {c^2 \int \frac {(1+a x)^2}{x^2} \, dx}{a^2} \\ & = \frac {c^2 \int \left (a^2+\frac {1}{x^2}+\frac {2 a}{x}\right ) \, dx}{a^2} \\ & = -\frac {c^2}{a^2 x}+c^2 x+\frac {2 c^2 \log (x)}{a} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^2 \, dx=\frac {c^2 \left (-\frac {1}{x}+a^2 x+2 a \log (x)\right )}{a^2} \]

[In]

Integrate[E^(4*ArcCoth[a*x])*(c - c/(a*x))^2,x]

[Out]

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

Maple [A] (verified)

Time = 0.61 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89

method	result	size
default	\(\frac {c^{2} \left (a^{2} x +2 a \ln \left (x \right )-\frac {1}{x}\right )}{a^{2}}\)	\(24\)
risch	\(-\frac {c^{2}}{a^{2} x}+c^{2} x +\frac {2 c^{2} \ln \left (x \right )}{a}\)	\(28\)
parallelrisch	\(\frac {a^{2} c^{2} x^{2}+2 c^{2} \ln \left (x \right ) a x -c^{2}}{a^{2} x}\)	\(33\)
norman	\(\frac {\frac {c^{2}}{a}-2 a \,c^{2} x^{2}+a^{2} c^{2} x^{3}}{\left (a x -1\right ) a x}+\frac {2 c^{2} \ln \left (x \right )}{a}\)	\(53\)
meijerg	\(-\frac {c^{2} \left (-\frac {a x \left (-3 a x +6\right )}{3 \left (-a x +1\right )}-2 \ln \left (-a x +1\right )\right )}{a}-\frac {2 c^{2} x}{-a x +1}-\frac {c^{2} \left (-\frac {3 a x}{-3 a x +3}+2 \ln \left (-a x +1\right )-1-2 \ln \left (x \right )-2 \ln \left (-a \right )+\frac {1}{a x}\right )}{a}\)	\(100\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^2 \, dx=\frac {a^{2} c^{2} x^{2} + 2 \, a c^{2} x \log \left (x\right ) - c^{2}}{a^{2} x} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^2 \, dx=\frac {a^{2} c^{2} x + 2 a c^{2} \log {\left (x \right )} - \frac {c^{2}}{x}}{a^{2}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^2 \, dx=c^{2} x + \frac {2 \, c^{2} \log \left (x\right )}{a} - \frac {c^{2}}{a^{2} x} \]

[In]

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

[Out]

c^2*x + 2*c^2*log(x)/a - c^2/(a^2*x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 94 vs. \(2 (27) = 54\).

Time = 0.27 (sec) , antiderivative size = 94, normalized size of antiderivative = 3.48 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^2 \, dx=-\frac {2 \, c^{2} \log \left (\frac {{\left | a x - 1 \right |}}{{\left (a x - 1\right )}^{2} {\left | a \right |}}\right )}{a} + \frac {2 \, c^{2} \log \left ({\left | -\frac {1}{a x - 1} - 1 \right |}\right )}{a} + \frac {c^{2} + \frac {2 \, c^{2}}{a x - 1}}{a^{2} {\left (\frac {1}{{\left (a x - 1\right )} a} + \frac {1}{{\left (a x - 1\right )}^{2} a}\right )}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 3.86 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^2 \, dx=\frac {c^2\,\left (a^2\,x^2+2\,a\,x\,\ln \left (x\right )-1\right )}{a^2\,x} \]

[In]

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

[Out]

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