\(\int e^{4 \coth ^{-1}(a x)} (c-a c x)^3 \, dx\) [190]

Optimal result

Integrand size = 18, antiderivative size = 35 \[ \int e^{4 \coth ^{-1}(a x)} (c-a c x)^3 \, dx=\frac {2 c^3 (1+a x)^3}{3 a}-\frac {c^3 (1+a x)^4}{4 a} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 35, 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} \[ \int e^{4 \coth ^{-1}(a x)} (c-a c x)^3 \, dx=\frac {2 c^3 (a x+1)^3}{3 a}-\frac {c^3 (a x+1)^4}{4 a} \]

[In]

Int[E^(4*ArcCoth[a*x])*(c - a*c*x)^3,x]

[Out]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.86 \[ \int e^{4 \coth ^{-1}(a x)} (c-a c x)^3 \, dx=-\frac {1}{12} c^3 x \left (-12-6 a x+4 a^2 x^2+3 a^3 x^3\right ) \]

[In]

Integrate[E^(4*ArcCoth[a*x])*(c - a*c*x)^3,x]

[Out]

-1/12*(c^3*x*(-12 - 6*a*x + 4*a^2*x^2 + 3*a^3*x^3))

Maple [A] (verified)

Time = 0.56 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.83

method	result	size
gosper	\(-\frac {x \left (3 a^{3} x^{3}+4 a^{2} x^{2}-6 a x -12\right ) c^{3}}{12}\)	\(29\)
default	\(c^{3} \left (-\frac {1}{4} a^{3} x^{4}-\frac {1}{3} a^{2} x^{3}+\frac {1}{2} a \,x^{2}+x \right )\)	\(29\)
risch	\(-\frac {1}{4} a^{3} c^{3} x^{4}-\frac {1}{3} a^{2} c^{3} x^{3}+\frac {1}{2} a \,c^{3} x^{2}+c^{3} x\)	\(38\)
parallelrisch	\(-\frac {1}{4} a^{3} c^{3} x^{4}-\frac {1}{3} a^{2} c^{3} x^{3}+\frac {1}{2} a \,c^{3} x^{2}+c^{3} x\)	\(38\)
norman	\(\frac {-c^{3} x +\frac {1}{2} a \,c^{3} x^{2}+\frac {5}{6} a^{2} c^{3} x^{3}-\frac {1}{12} a^{3} c^{3} x^{4}-\frac {1}{4} a^{4} c^{3} x^{5}}{a x -1}\)	\(58\)
meijerg	\(-\frac {c^{3} \left (\frac {a x \left (-3 a^{4} x^{4}-5 a^{3} x^{3}-10 a^{2} x^{2}-30 a x +60\right )}{-12 a x +12}+5 \ln \left (-a x +1\right )\right )}{a}-\frac {c^{3} \left (-\frac {a x \left (-5 a^{3} x^{3}-10 a^{2} x^{2}-30 a x +60\right )}{15 \left (-a x +1\right )}-4 \ln \left (-a x +1\right )\right )}{a}+\frac {2 c^{3} \left (\frac {a x \left (-2 a^{2} x^{2}-6 a x +12\right )}{-4 a x +4}+3 \ln \left (-a x +1\right )\right )}{a}+\frac {2 c^{3} \left (-\frac {a x \left (-3 a x +6\right )}{3 \left (-a x +1\right )}-2 \ln \left (-a x +1\right )\right )}{a}-\frac {c^{3} \left (\frac {a x}{-a x +1}+\ln \left (-a x +1\right )\right )}{a}+\frac {c^{3} x}{-a x +1}\)	\(234\)

[In]

int(1/(a*x-1)^2*(a*x+1)^2*(-a*c*x+c)^3,x,method=_RETURNVERBOSE)

[Out]

-1/12*x*(3*a^3*x^3+4*a^2*x^2-6*a*x-12)*c^3

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.06 \[ \int e^{4 \coth ^{-1}(a x)} (c-a c x)^3 \, dx=-\frac {1}{4} \, a^{3} c^{3} x^{4} - \frac {1}{3} \, a^{2} c^{3} x^{3} + \frac {1}{2} \, a c^{3} x^{2} + c^{3} x \]

[In]

integrate(1/(a*x-1)^2*(a*x+1)^2*(-a*c*x+c)^3,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.06 \[ \int e^{4 \coth ^{-1}(a x)} (c-a c x)^3 \, dx=- \frac {a^{3} c^{3} x^{4}}{4} - \frac {a^{2} c^{3} x^{3}}{3} + \frac {a c^{3} x^{2}}{2} + c^{3} x \]

[In]

integrate(1/(a*x-1)**2*(a*x+1)**2*(-a*c*x+c)**3,x)

[Out]

-a**3*c**3*x**4/4 - a**2*c**3*x**3/3 + a*c**3*x**2/2 + c**3*x

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.06 \[ \int e^{4 \coth ^{-1}(a x)} (c-a c x)^3 \, dx=-\frac {1}{4} \, a^{3} c^{3} x^{4} - \frac {1}{3} \, a^{2} c^{3} x^{3} + \frac {1}{2} \, a c^{3} x^{2} + c^{3} x \]

[In]

integrate(1/(a*x-1)^2*(a*x+1)^2*(-a*c*x+c)^3,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.20 \[ \int e^{4 \coth ^{-1}(a x)} (c-a c x)^3 \, dx=-\frac {{\left (3 \, c^{3} + \frac {16 \, c^{3}}{a x - 1} + \frac {24 \, c^{3}}{{\left (a x - 1\right )}^{2}}\right )} {\left (a x - 1\right )}^{4}}{12 \, a} \]

[In]

integrate(1/(a*x-1)^2*(a*x+1)^2*(-a*c*x+c)^3,x, algorithm="giac")

[Out]

-1/12*(3*c^3 + 16*c^3/(a*x - 1) + 24*c^3/(a*x - 1)^2)*(a*x - 1)^4/a

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.06 \[ \int e^{4 \coth ^{-1}(a x)} (c-a c x)^3 \, dx=-\frac {a^3\,c^3\,x^4}{4}-\frac {a^2\,c^3\,x^3}{3}+\frac {a\,c^3\,x^2}{2}+c^3\,x \]

[In]

int(((c - a*c*x)^3*(a*x + 1)^2)/(a*x - 1)^2,x)

[Out]

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