\(\int e^{2 \coth ^{-1}(a x)} (c-\frac {c}{a^2 x^2}) \, dx\) [784]

Optimal result

Integrand size = 20, antiderivative size = 21 \[ \int e^{2 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx=-\frac {c}{a^2 x}+c x+\frac {2 c \log (x)}{a} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6302, 6292, 6285, 45} \[ \int e^{2 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx=-\frac {c}{a^2 x}+\frac {2 c \log (x)}{a}+c x \]

[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 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 6285

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 6292

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_.), x_Symbol] :> Dist[d^p, Int[(u/x^(2*p))*(1
- a^2*x^2)^p*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[{a, c, d, n}, x] && EqQ[c + a^2*d, 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^2 x^2}\right ) \, dx \\ & = \frac {c \int \frac {e^{2 \text {arctanh}(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] (verified)

Time = 0.02 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int e^{2 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx=-\frac {c}{a^2 x}+c x+\frac {2 c \log (x)}{a} \]

[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

Maple [A] (verified)

Time = 0.57 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05

method	result	size
default	\(\frac {c \left (a^{2} x +2 a \ln \left (x \right )-\frac {1}{x}\right )}{a^{2}}\)	\(22\)
risch	\(-\frac {c}{a^{2} x}+c x +\frac {2 c \ln \left (x \right )}{a}\)	\(22\)
parallelrisch	\(\frac {a^{2} c \,x^{2}+2 c \ln \left (x \right ) a x -c}{a^{2} x}\)	\(27\)
norman	\(\frac {a c \,x^{2}-\frac {c}{a}}{a x}+\frac {2 c \ln \left (x \right )}{a}\)	\(30\)
meijerg	\(-\frac {c \left (-a x -\ln \left (-a x +1\right )\right )}{a}+\frac {c \left (-\ln \left (-a x +1\right )+\ln \left (x \right )+\ln \left (-a \right )\right )}{a}+\frac {c \ln \left (-a x +1\right )}{a}-\frac {c \left (\ln \left (-a x +1\right )-\ln \left (x \right )-\ln \left (-a \right )+\frac {1}{a x}\right )}{a}\)	\(86\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24 \[ \int e^{2 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx=\frac {a^{2} c x^{2} + 2 \, a c x \log \left (x\right ) - c}{a^{2} x} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int e^{2 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx=\frac {a^{2} c x + 2 a c \log {\left (x \right )} - \frac {c}{x}}{a^{2}} \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int e^{2 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx=c x + \frac {2 \, c \log \left (x\right )}{a} - \frac {c}{a^{2} x} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05 \[ \int e^{2 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx=c x + \frac {2 \, c \log \left ({\left | x \right |}\right )}{a} - \frac {c}{a^{2} x} \]

[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)

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int e^{2 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx=\frac {c\,\left (a^2\,x^2+2\,a\,x\,\ln \left (x\right )-1\right )}{a^2\,x} \]

[In]

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

[Out]

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