\(\int e^{4 \coth ^{-1}(a x)} (c-a c x) \, dx\) [192]

Optimal result

Integrand size = 16, antiderivative size = 27 \[ \int e^{4 \coth ^{-1}(a x)} (c-a c x) \, dx=-3 c x-\frac {1}{2} a c x^2-\frac {4 c \log (1-a x)}{a} \]

[Out]

-3*c*x-1/2*a*c*x^2-4*c*ln(-a*x+1)/a

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {6302, 6264, 45} \[ \int e^{4 \coth ^{-1}(a x)} (c-a c x) \, dx=-\frac {1}{2} a c x^2-\frac {4 c \log (1-a x)}{a}-3 c x \]

[In]

Int[E^(4*ArcCoth[a*x])*(c - a*c*x),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int e^{4 \coth ^{-1}(a x)} (c-a c x) \, dx=c \left (-3 x-\frac {a x^2}{2}-\frac {4 \log (1-a x)}{a}\right ) \]

[In]

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

[Out]

c*(-3*x - (a*x^2)/2 - (4*Log[1 - a*x])/a)

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89

method	result	size
default	\(c \left (-\frac {a \,x^{2}}{2}-3 x -\frac {4 \ln \left (a x -1\right )}{a}\right )\)	\(24\)
risch	\(-\frac {a c \,x^{2}}{2}-3 c x -\frac {4 c \ln \left (a x -1\right )}{a}\)	\(25\)
parallelrisch	\(-\frac {a^{2} c \,x^{2}+6 a c x +8 c \ln \left (a x -1\right )}{2 a}\)	\(29\)
norman	\(\frac {3 c x -\frac {5}{2} a c \,x^{2}-\frac {1}{2} a^{2} c \,x^{3}}{a x -1}-\frac {4 c \ln \left (a x -1\right )}{a}\)	\(43\)
meijerg	\(-\frac {c \left (\frac {a x \left (-2 a^{2} x^{2}-6 a x +12\right )}{-4 a x +4}+3 \ln \left (-a x +1\right )\right )}{a}+\frac {c \left (-\frac {a x \left (-3 a x +6\right )}{3 \left (-a x +1\right )}-2 \ln \left (-a x +1\right )\right )}{a}+\frac {c \left (\frac {a x}{-a x +1}+\ln \left (-a x +1\right )\right )}{a}+\frac {c x}{-a x +1}\)	\(112\)

[In]

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

[Out]

c*(-1/2*a*x^2-3*x-4/a*ln(a*x-1))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int e^{4 \coth ^{-1}(a x)} (c-a c x) \, dx=-\frac {a^{2} c x^{2} + 6 \, a c x + 8 \, c \log \left (a x - 1\right )}{2 \, a} \]

[In]

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

[Out]

-1/2*(a^2*c*x^2 + 6*a*c*x + 8*c*log(a*x - 1))/a

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int e^{4 \coth ^{-1}(a x)} (c-a c x) \, dx=- \frac {a c x^{2}}{2} - 3 c x - \frac {4 c \log {\left (a x - 1 \right )}}{a} \]

[In]

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

[Out]

-a*c*x**2/2 - 3*c*x - 4*c*log(a*x - 1)/a

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int e^{4 \coth ^{-1}(a x)} (c-a c x) \, dx=-\frac {1}{2} \, a c x^{2} - 3 \, c x - \frac {4 \, c \log \left (a x - 1\right )}{a} \]

[In]

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

[Out]

-1/2*a*c*x^2 - 3*c*x - 4*c*log(a*x - 1)/a

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.85 \[ \int e^{4 \coth ^{-1}(a x)} (c-a c x) \, dx=-\frac {{\left (a x - 1\right )}^{2} {\left (c + \frac {8 \, c}{a x - 1}\right )}}{2 \, a} + \frac {4 \, c \log \left (\frac {{\left | a x - 1 \right |}}{{\left (a x - 1\right )}^{2} {\left | a \right |}}\right )}{a} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 4.14 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int e^{4 \coth ^{-1}(a x)} (c-a c x) \, dx=-\frac {c\,\left (8\,\ln \left (a\,x-1\right )+6\,a\,x+a^2\,x^2\right )}{2\,a} \]

[In]

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

[Out]

-(c*(8*log(a*x - 1) + 6*a*x + a^2*x^2))/(2*a)