\(\int e^{-2 \coth ^{-1}(a x)} (c-a^2 c x^2)^2 \, dx\) [600]

Optimal result

Integrand size = 22, antiderivative size = 37 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^2 \, dx=\frac {c^2 (1-a x)^4}{2 a}-\frac {c^2 (1-a x)^5}{5 a} \]

[Out]

1/2*c^2*(-a*x+1)^4/a-1/5*c^2*(-a*x+1)^5/a

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {6302, 6275, 45} \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^2 \, dx=\frac {c^2 (1-a x)^4}{2 a}-\frac {c^2 (1-a x)^5}{5 a} \]

[In]

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

[Out]

(c^2*(1 - a*x)^4)/(2*a) - (c^2*(1 - a*x)^5)/(5*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 6275

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[(1 - a*x)^(p - n/2)*
(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[a^2*c + d, 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)} \left (c-a^2 c x^2\right )^2 \, dx \\ & = -\left (c^2 \int (1-a x)^3 (1+a x) \, dx\right ) \\ & = -\left (c^2 \int \left (2 (1-a x)^3-(1-a x)^4\right ) \, dx\right ) \\ & = \frac {c^2 (1-a x)^4}{2 a}-\frac {c^2 (1-a x)^5}{5 a} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.81 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^2 \, dx=\frac {1}{10} c^2 x \left (-10+10 a x-5 a^3 x^3+2 a^4 x^4\right ) \]

[In]

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

[Out]

(c^2*x*(-10 + 10*a*x - 5*a^3*x^3 + 2*a^4*x^4))/10

Maple [A] (verified)

Time = 0.63 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.78

method	result	size
gosper	\(\frac {c^{2} x \left (2 a^{4} x^{4}-5 a^{3} x^{3}+10 a x -10\right )}{10}\)	\(29\)
default	\(c^{2} \left (\frac {1}{5} a^{4} x^{5}-\frac {1}{2} a^{3} x^{4}+a \,x^{2}-x \right )\)	\(30\)
norman	\(a \,c^{2} x^{2}-c^{2} x -\frac {1}{2} a^{3} c^{2} x^{4}+\frac {1}{5} a^{4} c^{2} x^{5}\)	\(38\)
risch	\(a \,c^{2} x^{2}-c^{2} x -\frac {1}{2} a^{3} c^{2} x^{4}+\frac {1}{5} a^{4} c^{2} x^{5}\)	\(38\)
parallelrisch	\(a \,c^{2} x^{2}-c^{2} x -\frac {1}{2} a^{3} c^{2} x^{4}+\frac {1}{5} a^{4} c^{2} x^{5}\)	\(38\)
meijerg	\(\frac {c^{2} \left (\frac {a x \left (12 a^{4} x^{4}-15 a^{3} x^{3}+20 a^{2} x^{2}-30 a x +60\right )}{60}-\ln \left (a x +1\right )\right )}{a}-\frac {c^{2} \left (-\frac {a x \left (-15 a^{3} x^{3}+20 a^{2} x^{2}-30 a x +60\right )}{60}+\ln \left (a x +1\right )\right )}{a}-\frac {2 c^{2} \left (\frac {a x \left (4 a^{2} x^{2}-6 a x +12\right )}{12}-\ln \left (a x +1\right )\right )}{a}+\frac {2 c^{2} \left (-\frac {a x \left (-3 a x +6\right )}{6}+\ln \left (a x +1\right )\right )}{a}+\frac {c^{2} \left (a x -\ln \left (a x +1\right )\right )}{a}-\frac {c^{2} \ln \left (a x +1\right )}{a}\)	\(186\)

[In]

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

[Out]

1/10*c^2*x*(2*a^4*x^4-5*a^3*x^3+10*a*x-10)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^2 \, dx=\frac {1}{5} \, a^{4} c^{2} x^{5} - \frac {1}{2} \, a^{3} c^{2} x^{4} + a c^{2} x^{2} - c^{2} x \]

[In]

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

[Out]

1/5*a^4*c^2*x^5 - 1/2*a^3*c^2*x^4 + a*c^2*x^2 - c^2*x

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.97 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^2 \, dx=\frac {a^{4} c^{2} x^{5}}{5} - \frac {a^{3} c^{2} x^{4}}{2} + a c^{2} x^{2} - c^{2} x \]

[In]

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

[Out]

a**4*c**2*x**5/5 - a**3*c**2*x**4/2 + a*c**2*x**2 - c**2*x

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^2 \, dx=\frac {1}{5} \, a^{4} c^{2} x^{5} - \frac {1}{2} \, a^{3} c^{2} x^{4} + a c^{2} x^{2} - c^{2} x \]

[In]

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

[Out]

1/5*a^4*c^2*x^5 - 1/2*a^3*c^2*x^4 + a*c^2*x^2 - c^2*x

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^2 \, dx=\frac {1}{5} \, a^{4} c^{2} x^{5} - \frac {1}{2} \, a^{3} c^{2} x^{4} + a c^{2} x^{2} - c^{2} x \]

[In]

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

[Out]

1/5*a^4*c^2*x^5 - 1/2*a^3*c^2*x^4 + a*c^2*x^2 - c^2*x

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^2 \, dx=\frac {a^4\,c^2\,x^5}{5}-\frac {a^3\,c^2\,x^4}{2}+a\,c^2\,x^2-c^2\,x \]

[In]

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

[Out]

a*c^2*x^2 - c^2*x - (a^3*c^2*x^4)/2 + (a^4*c^2*x^5)/5