\(\int e^{-2 \coth ^{-1}(a x)} (c-a c x)^3 \, dx\) [208]

Optimal result

Integrand size = 18, antiderivative size = 73 \[ \int e^{-2 \coth ^{-1}(a x)} (c-a c x)^3 \, dx=8 c^3 x-\frac {2 c^3 (1-a x)^2}{a}-\frac {2 c^3 (1-a x)^3}{3 a}-\frac {c^3 (1-a x)^4}{4 a}-\frac {16 c^3 \log (1+a x)}{a} \]

[Out]

8*c^3*x-2*c^3*(-a*x+1)^2/a-2/3*c^3*(-a*x+1)^3/a-1/4*c^3*(-a*x+1)^4/a-16*c^3*ln(a*x+1)/a

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6302, 6264, 45} \[ \int e^{-2 \coth ^{-1}(a x)} (c-a c x)^3 \, dx=-\frac {c^3 (1-a x)^4}{4 a}-\frac {2 c^3 (1-a x)^3}{3 a}-\frac {2 c^3 (1-a x)^2}{a}-\frac {16 c^3 \log (a x+1)}{a}+8 c^3 x \]

[In]

Int[(c - a*c*x)^3/E^(2*ArcCoth[a*x]),x]

[Out]

8*c^3*x - (2*c^3*(1 - a*x)^2)/a - (2*c^3*(1 - a*x)^3)/(3*a) - (c^3*(1 - a*x)^4)/(4*a) - (16*c^3*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)^3 \, dx \\ & = -\left (c^3 \int \frac {(1-a x)^4}{1+a x} \, dx\right ) \\ & = -\left (c^3 \int \left (-8-4 (1-a x)-2 (1-a x)^2-(1-a x)^3+\frac {16}{1+a x}\right ) \, dx\right ) \\ & = 8 c^3 x-\frac {2 c^3 (1-a x)^2}{a}-\frac {2 c^3 (1-a x)^3}{3 a}-\frac {c^3 (1-a x)^4}{4 a}-\frac {16 c^3 \log (1+a x)}{a} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.66 \[ \int e^{-2 \coth ^{-1}(a x)} (c-a c x)^3 \, dx=-\frac {c^3 \left (35-180 a x+66 a^2 x^2-20 a^3 x^3+3 a^4 x^4+192 \log (1+a x)\right )}{12 a} \]

[In]

Integrate[(c - a*c*x)^3/E^(2*ArcCoth[a*x]),x]

[Out]

-1/12*(c^3*(35 - 180*a*x + 66*a^2*x^2 - 20*a^3*x^3 + 3*a^4*x^4 + 192*Log[1 + a*x]))/a

Maple [A] (verified)

Time = 0.52 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.58

method	result	size
default	\(c^{3} \left (-\frac {a^{3} x^{4}}{4}+\frac {5 a^{2} x^{3}}{3}-\frac {11 a \,x^{2}}{2}+15 x -\frac {16 \ln \left (a x +1\right )}{a}\right )\)	\(42\)
norman	\(15 c^{3} x -\frac {11 a \,c^{3} x^{2}}{2}+\frac {5 a^{2} c^{3} x^{3}}{3}-\frac {a^{3} c^{3} x^{4}}{4}-\frac {16 c^{3} \ln \left (a x +1\right )}{a}\)	\(53\)
risch	\(15 c^{3} x -\frac {11 a \,c^{3} x^{2}}{2}+\frac {5 a^{2} c^{3} x^{3}}{3}-\frac {a^{3} c^{3} x^{4}}{4}-\frac {16 c^{3} \ln \left (a x +1\right )}{a}\)	\(53\)
parallelrisch	\(-\frac {3 a^{4} c^{3} x^{4}-20 a^{3} c^{3} x^{3}+66 a^{2} c^{3} x^{2}-180 a \,c^{3} x +192 c^{3} \ln \left (a x +1\right )}{12 a}\)	\(58\)
meijerg	\(-\frac {c^{3} \left (-\frac {a x \left (-15 a^{3} x^{3}+20 a^{2} x^{2}-30 a x +60\right )}{60}+\ln \left (a x +1\right )\right )}{a}+\frac {4 c^{3} \left (\frac {a x \left (4 a^{2} x^{2}-6 a x +12\right )}{12}-\ln \left (a x +1\right )\right )}{a}-\frac {6 c^{3} \left (-\frac {a x \left (-3 a x +6\right )}{6}+\ln \left (a x +1\right )\right )}{a}+\frac {4 c^{3} \left (a x -\ln \left (a x +1\right )\right )}{a}-\frac {c^{3} \ln \left (a x +1\right )}{a}\)	\(137\)

[In]

int((-a*c*x+c)^3*(a*x-1)/(a*x+1),x,method=_RETURNVERBOSE)

[Out]

c^3*(-1/4*a^3*x^4+5/3*a^2*x^3-11/2*a*x^2+15*x-16*ln(a*x+1)/a)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.78 \[ \int e^{-2 \coth ^{-1}(a x)} (c-a c x)^3 \, dx=-\frac {3 \, a^{4} c^{3} x^{4} - 20 \, a^{3} c^{3} x^{3} + 66 \, a^{2} c^{3} x^{2} - 180 \, a c^{3} x + 192 \, c^{3} \log \left (a x + 1\right )}{12 \, a} \]

[In]

integrate((-a*c*x+c)^3*(a*x-1)/(a*x+1),x, algorithm="fricas")

[Out]

-1/12*(3*a^4*c^3*x^4 - 20*a^3*c^3*x^3 + 66*a^2*c^3*x^2 - 180*a*c^3*x + 192*c^3*log(a*x + 1))/a

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.77 \[ \int e^{-2 \coth ^{-1}(a x)} (c-a c x)^3 \, dx=- \frac {a^{3} c^{3} x^{4}}{4} + \frac {5 a^{2} c^{3} x^{3}}{3} - \frac {11 a c^{3} x^{2}}{2} + 15 c^{3} x - \frac {16 c^{3} \log {\left (a x + 1 \right )}}{a} \]

[In]

integrate((-a*c*x+c)**3*(a*x-1)/(a*x+1),x)

[Out]

-a**3*c**3*x**4/4 + 5*a**2*c**3*x**3/3 - 11*a*c**3*x**2/2 + 15*c**3*x - 16*c**3*log(a*x + 1)/a

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.71 \[ \int e^{-2 \coth ^{-1}(a x)} (c-a c x)^3 \, dx=-\frac {1}{4} \, a^{3} c^{3} x^{4} + \frac {5}{3} \, a^{2} c^{3} x^{3} - \frac {11}{2} \, a c^{3} x^{2} + 15 \, c^{3} x - \frac {16 \, c^{3} \log \left (a x + 1\right )}{a} \]

[In]

integrate((-a*c*x+c)^3*(a*x-1)/(a*x+1),x, algorithm="maxima")

[Out]

-1/4*a^3*c^3*x^4 + 5/3*a^2*c^3*x^3 - 11/2*a*c^3*x^2 + 15*c^3*x - 16*c^3*log(a*x + 1)/a

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.88 \[ \int e^{-2 \coth ^{-1}(a x)} (c-a c x)^3 \, dx=-\frac {16 \, c^{3} \log \left ({\left | a x + 1 \right |}\right )}{a} - \frac {3 \, a^{7} c^{3} x^{4} - 20 \, a^{6} c^{3} x^{3} + 66 \, a^{5} c^{3} x^{2} - 180 \, a^{4} c^{3} x}{12 \, a^{4}} \]

[In]

integrate((-a*c*x+c)^3*(a*x-1)/(a*x+1),x, algorithm="giac")

[Out]

-16*c^3*log(abs(a*x + 1))/a - 1/12*(3*a^7*c^3*x^4 - 20*a^6*c^3*x^3 + 66*a^5*c^3*x^2 - 180*a^4*c^3*x)/a^4

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.71 \[ \int e^{-2 \coth ^{-1}(a x)} (c-a c x)^3 \, dx=15\,c^3\,x-\frac {11\,a\,c^3\,x^2}{2}+\frac {5\,a^2\,c^3\,x^3}{3}-\frac {a^3\,c^3\,x^4}{4}-\frac {16\,c^3\,\ln \left (a\,x+1\right )}{a} \]

[In]

int(((c - a*c*x)^3*(a*x - 1))/(a*x + 1),x)

[Out]

15*c^3*x - (11*a*c^3*x^2)/2 + (5*a^2*c^3*x^3)/3 - (a^3*c^3*x^4)/4 - (16*c^3*log(a*x + 1))/a