\(\int e^{4 \coth ^{-1}(a x)} (c-\frac {c}{a^2 x^2}) \, dx\) [801]

Optimal result

Integrand size = 20, antiderivative size = 33 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx=\frac {c}{a^2 x}+c x-\frac {4 c \log (x)}{a}+\frac {8 c \log (1-a x)}{a} \]

[Out]

c/a^2/x+c*x-4*c*ln(x)/a+8*c*ln(-a*x+1)/a

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 33, 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, 90} \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx=\frac {c}{a^2 x}-\frac {4 c \log (x)}{a}+\frac {8 c \log (1-a x)}{a}+c x \]

[In]

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

[Out]

c/(a^2*x) + c*x - (4*c*Log[x])/a + (8*c*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 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^{4 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx \\ & = -\frac {c \int \frac {e^{4 \text {arctanh}(a x)} \left (1-a^2 x^2\right )}{x^2} \, dx}{a^2} \\ & = -\frac {c \int \frac {(1+a x)^3}{x^2 (1-a x)} \, dx}{a^2} \\ & = -\frac {c \int \left (-a^2+\frac {1}{x^2}+\frac {4 a}{x}-\frac {8 a^2}{-1+a x}\right ) \, dx}{a^2} \\ & = \frac {c}{a^2 x}+c x-\frac {4 c \log (x)}{a}+\frac {8 c \log (1-a x)}{a} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx=\frac {c}{a^2 x}+c x-\frac {4 c \log (x)}{a}+\frac {8 c \log (1-a x)}{a} \]

[In]

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

[Out]

c/(a^2*x) + c*x - (4*c*Log[x])/a + (8*c*Log[1 - a*x])/a

Maple [A] (verified)

Time = 0.64 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88

method	result	size
default	\(\frac {c \left (a^{2} x +\frac {1}{x}-4 a \ln \left (x \right )+8 a \ln \left (a x -1\right )\right )}{a^{2}}\)	\(29\)
risch	\(\frac {c}{a^{2} x}+c x -\frac {4 c \ln \left (x \right )}{a}+\frac {8 c \ln \left (-a x +1\right )}{a}\)	\(34\)
parallelrisch	\(-\frac {-a^{2} c \,x^{2}+4 c \ln \left (x \right ) a x -8 c \ln \left (a x -1\right ) a x -c}{a^{2} x}\)	\(40\)
norman	\(\frac {a^{2} c \,x^{3}-\frac {c}{a}}{x \left (a x -1\right ) a}-\frac {4 c \ln \left (x \right )}{a}+\frac {8 c \ln \left (a x -1\right )}{a}\)	\(51\)
meijerg	\(-\frac {c \left (-\frac {a x \left (-3 a x +6\right )}{3 \left (-a x +1\right )}-2 \ln \left (-a x +1\right )\right )}{a}+\frac {2 c \left (\frac {a x}{-a x +1}+\ln \left (-a x +1\right )\right )}{a}-\frac {2 c \left (\frac {2 a x}{-2 a x +2}-\ln \left (-a x +1\right )+1+\ln \left (x \right )+\ln \left (-a \right )\right )}{a}+\frac {c \left (-\frac {3 a x}{-3 a x +3}+2 \ln \left (-a x +1\right )-1-2 \ln \left (x \right )-2 \ln \left (-a \right )+\frac {1}{a x}\right )}{a}\)	\(141\)

[In]

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

[Out]

c/a^2*(a^2*x+1/x-4*a*ln(x)+8*a*ln(a*x-1))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.06 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx=\frac {a^{2} c x^{2} + 8 \, a c x \log \left (a x - 1\right ) - 4 \, a c x \log \left (x\right ) + c}{a^{2} x} \]

[In]

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

[Out]

(a^2*c*x^2 + 8*a*c*x*log(a*x - 1) - 4*a*c*x*log(x) + c)/(a^2*x)

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.79 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx=c x + \frac {4 c \left (- \log {\left (x \right )} + 2 \log {\left (x - \frac {1}{a} \right )}\right )}{a} + \frac {c}{a^{2} x} \]

[In]

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

[Out]

c*x + 4*c*(-log(x) + 2*log(x - 1/a))/a + c/(a**2*x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx=c x + \frac {8 \, c \log \left (a x - 1\right )}{a} - \frac {4 \, c \log \left (x\right )}{a} + \frac {c}{a^{2} x} \]

[In]

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

[Out]

c*x + 8*c*log(a*x - 1)/a - 4*c*log(x)/a + c/(a^2*x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.00 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx=-\frac {4 \, c \log \left (\frac {{\left | a x - 1 \right |}}{{\left (a x - 1\right )}^{2} {\left | a \right |}}\right )}{a} - \frac {4 \, c \log \left ({\left | -\frac {1}{a x - 1} - 1 \right |}\right )}{a} + \frac {{\left (a x - 1\right )} c}{a {\left (\frac {1}{a x - 1} + 1\right )}} \]

[In]

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

[Out]

-4*c*log(abs(a*x - 1)/((a*x - 1)^2*abs(a)))/a - 4*c*log(abs(-1/(a*x - 1) - 1))/a + (a*x - 1)*c/(a*(1/(a*x - 1)
+ 1))

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx=c\,x+\frac {c}{a^2\,x}-\frac {4\,c\,\ln \left (x\right )}{a}+\frac {8\,c\,\ln \left (a\,x-1\right )}{a} \]

[In]

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

[Out]

c*x + c/(a^2*x) - (4*c*log(x))/a + (8*c*log(a*x - 1))/a