\(\int e^{-2 \coth ^{-1}(a x)} (c-a c x) \, dx\) [210]

Optimal result

Integrand size = 16, antiderivative size = 26 \[ \int e^{-2 \coth ^{-1}(a x)} (c-a c x) \, dx=3 c x-\frac {1}{2} a c x^2-\frac {4 c \log (1+a x)}{a} \]

[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 = 26, 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} \[ \int e^{-2 \coth ^{-1}(a x)} (c-a c x) \, dx=-\frac {1}{2} a c x^2-\frac {4 c \log (a x+1)}{a}+3 c x \]

[In]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int e^{-2 \coth ^{-1}(a x)} (c-a c x) \, dx=3 c x-\frac {1}{2} a c x^2-\frac {4 c \log (1+a x)}{a} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92

method	result	size
default	\(c \left (-\frac {a \,x^{2}}{2}+3 x -\frac {4 \ln \left (a x +1\right )}{a}\right )\)	\(24\)
norman	\(3 c x -\frac {a c \,x^{2}}{2}-\frac {4 c \ln \left (a x +1\right )}{a}\)	\(25\)
risch	\(3 c x -\frac {a c \,x^{2}}{2}-\frac {4 c \ln \left (a x +1\right )}{a}\)	\(25\)
parallelrisch	\(-\frac {a^{2} c \,x^{2}-6 a c x +8 c \ln \left (a x +1\right )}{2 a}\)	\(29\)
meijerg	\(-\frac {c \left (-\frac {a x \left (-3 a x +6\right )}{6}+\ln \left (a x +1\right )\right )}{a}+\frac {2 c \left (a x -\ln \left (a x +1\right )\right )}{a}-\frac {c \ln \left (a x +1\right )}{a}\)	\(55\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int e^{-2 \coth ^{-1}(a x)} (c-a c x) \, dx=-\frac {a^{2} c x^{2} - 6 \, a c x + 8 \, c \log \left (a x + 1\right )}{2 \, a} \]

[In]

integrate((-a*c*x+c)*(a*x-1)/(a*x+1),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 = 24, normalized size of antiderivative = 0.92 \[ \int e^{-2 \coth ^{-1}(a x)} (c-a c x) \, dx=- \frac {a c x^{2}}{2} + 3 c x - \frac {4 c \log {\left (a x + 1 \right )}}{a} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int e^{-2 \coth ^{-1}(a x)} (c-a c x) \, dx=-\frac {1}{2} \, a c x^{2} + 3 \, c x - \frac {4 \, c \log \left (a x + 1\right )}{a} \]

[In]

integrate((-a*c*x+c)*(a*x-1)/(a*x+1),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 = 0.26 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.35 \[ \int e^{-2 \coth ^{-1}(a x)} (c-a c x) \, dx=-\frac {4 \, c \log \left ({\left | a x + 1 \right |}\right )}{a} - \frac {a^{3} c x^{2} - 6 \, a^{2} c x}{2 \, a^{2}} \]

[In]

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

[Out]

-4*c*log(abs(a*x + 1))/a - 1/2*(a^3*c*x^2 - 6*a^2*c*x)/a^2

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int e^{-2 \coth ^{-1}(a x)} (c-a c x) \, dx=-\frac {c\,\left (8\,\ln \left (a\,x+1\right )-6\,a\,x+a^2\,x^2\right )}{2\,a} \]

[In]

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

[Out]

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