\(\int e^{2 \coth ^{-1}(a x)} (c-a^2 c x^2)^3 \, dx\) [566]

Optimal result

Integrand size = 22, antiderivative size = 52 \[ \int e^{2 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^3 \, dx=-\frac {4 c^3 (1+a x)^5}{5 a}+\frac {2 c^3 (1+a x)^6}{3 a}-\frac {c^3 (1+a x)^7}{7 a} \]

[Out]

-4/5*c^3*(a*x+1)^5/a+2/3*c^3*(a*x+1)^6/a-1/7*c^3*(a*x+1)^7/a

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 52, 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} \[ \int e^{2 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^3 \, dx=-\frac {c^3 (a x+1)^7}{7 a}+\frac {2 c^3 (a x+1)^6}{3 a}-\frac {4 c^3 (a x+1)^5}{5 a} \]

[In]

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

[Out]

(-4*c^3*(1 + a*x)^5)/(5*a) + (2*c^3*(1 + a*x)^6)/(3*a) - (c^3*(1 + a*x)^7)/(7*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 )^3 \, dx \\ & = -\left (c^3 \int (1-a x)^2 (1+a x)^4 \, dx\right ) \\ & = -\left (c^3 \int \left (4 (1+a x)^4-4 (1+a x)^5+(1+a x)^6\right ) \, dx\right ) \\ & = -\frac {4 c^3 (1+a x)^5}{5 a}+\frac {2 c^3 (1+a x)^6}{3 a}-\frac {c^3 (1+a x)^7}{7 a} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.60 \[ \int e^{2 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^3 \, dx=-\frac {c^3 (1+a x)^5 \left (29-40 a x+15 a^2 x^2\right )}{105 a} \]

[In]

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

[Out]

-1/105*(c^3*(1 + a*x)^5*(29 - 40*a*x + 15*a^2*x^2))/a

Maple [A] (verified)

Time = 0.63 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.02

method	result	size
gosper	\(-\frac {c^{3} x \left (15 a^{6} x^{6}+35 a^{5} x^{5}-21 a^{4} x^{4}-105 a^{3} x^{3}-35 a^{2} x^{2}+105 a x +105\right )}{105}\)	\(53\)
default	\(c^{3} \left (-\frac {1}{7} a^{6} x^{7}-\frac {1}{3} a^{5} x^{6}+\frac {1}{5} a^{4} x^{5}+a^{3} x^{4}+\frac {1}{3} a^{2} x^{3}-a \,x^{2}-x \right )\)	\(54\)
norman	\(a^{3} c^{3} x^{4}-c^{3} x -a \,c^{3} x^{2}+\frac {1}{3} a^{2} c^{3} x^{3}+\frac {1}{5} a^{4} c^{3} x^{5}-\frac {1}{3} a^{5} c^{3} x^{6}-\frac {1}{7} a^{6} c^{3} x^{7}\)	\(71\)
risch	\(a^{3} c^{3} x^{4}-c^{3} x -a \,c^{3} x^{2}+\frac {1}{3} a^{2} c^{3} x^{3}+\frac {1}{5} a^{4} c^{3} x^{5}-\frac {1}{3} a^{5} c^{3} x^{6}-\frac {1}{7} a^{6} c^{3} x^{7}\)	\(71\)
parallelrisch	\(a^{3} c^{3} x^{4}-c^{3} x -a \,c^{3} x^{2}+\frac {1}{3} a^{2} c^{3} x^{3}+\frac {1}{5} a^{4} c^{3} x^{5}-\frac {1}{3} a^{5} c^{3} x^{6}-\frac {1}{7} a^{6} c^{3} x^{7}\)	\(71\)
meijerg	\(\frac {c^{3} \left (-\frac {a x \left (120 a^{6} x^{6}+140 a^{5} x^{5}+168 a^{4} x^{4}+210 a^{3} x^{3}+280 a^{2} x^{2}+420 a x +840\right )}{840}-\ln \left (-a x +1\right )\right )}{a}-\frac {3 c^{3} \left (-\frac {a x \left (12 a^{4} x^{4}+15 a^{3} x^{3}+20 a^{2} x^{2}+30 a x +60\right )}{60}-\ln \left (-a x +1\right )\right )}{a}+\frac {3 c^{3} \left (-\frac {a x \left (4 a^{2} x^{2}+6 a x +12\right )}{12}-\ln \left (-a x +1\right )\right )}{a}-\frac {c^{3} \left (-a x -\ln \left (-a x +1\right )\right )}{a}-\frac {c^{3} \left (\frac {a x \left (70 a^{5} x^{5}+84 a^{4} x^{4}+105 a^{3} x^{3}+140 a^{2} x^{2}+210 a x +420\right )}{420}+\ln \left (-a x +1\right )\right )}{a}+\frac {3 c^{3} \left (\frac {a x \left (15 a^{3} x^{3}+20 a^{2} x^{2}+30 a x +60\right )}{60}+\ln \left (-a x +1\right )\right )}{a}-\frac {3 c^{3} \left (\frac {a x \left (3 a x +6\right )}{6}+\ln \left (-a x +1\right )\right )}{a}+\frac {c^{3} \ln \left (-a x +1\right )}{a}\)	\(319\)

[In]

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

[Out]

-1/105*c^3*x*(15*a^6*x^6+35*a^5*x^5-21*a^4*x^4-105*a^3*x^3-35*a^2*x^2+105*a*x+105)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.35 \[ \int e^{2 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^3 \, dx=-\frac {1}{7} \, a^{6} c^{3} x^{7} - \frac {1}{3} \, a^{5} c^{3} x^{6} + \frac {1}{5} \, a^{4} c^{3} x^{5} + a^{3} c^{3} x^{4} + \frac {1}{3} \, a^{2} c^{3} x^{3} - a c^{3} x^{2} - c^{3} x \]

[In]

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

[Out]

-1/7*a^6*c^3*x^7 - 1/3*a^5*c^3*x^6 + 1/5*a^4*c^3*x^5 + a^3*c^3*x^4 + 1/3*a^2*c^3*x^3 - a*c^3*x^2 - c^3*x

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.35 \[ \int e^{2 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^3 \, dx=- \frac {a^{6} c^{3} x^{7}}{7} - \frac {a^{5} c^{3} x^{6}}{3} + \frac {a^{4} c^{3} x^{5}}{5} + a^{3} c^{3} x^{4} + \frac {a^{2} c^{3} x^{3}}{3} - a c^{3} x^{2} - c^{3} x \]

[In]

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

[Out]

-a**6*c**3*x**7/7 - a**5*c**3*x**6/3 + a**4*c**3*x**5/5 + a**3*c**3*x**4 + a**2*c**3*x**3/3 - a*c**3*x**2 - c*
*3*x

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.35 \[ \int e^{2 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^3 \, dx=-\frac {1}{7} \, a^{6} c^{3} x^{7} - \frac {1}{3} \, a^{5} c^{3} x^{6} + \frac {1}{5} \, a^{4} c^{3} x^{5} + a^{3} c^{3} x^{4} + \frac {1}{3} \, a^{2} c^{3} x^{3} - a c^{3} x^{2} - c^{3} x \]

[In]

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

[Out]

-1/7*a^6*c^3*x^7 - 1/3*a^5*c^3*x^6 + 1/5*a^4*c^3*x^5 + a^3*c^3*x^4 + 1/3*a^2*c^3*x^3 - a*c^3*x^2 - c^3*x

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.35 \[ \int e^{2 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^3 \, dx=-\frac {1}{7} \, a^{6} c^{3} x^{7} - \frac {1}{3} \, a^{5} c^{3} x^{6} + \frac {1}{5} \, a^{4} c^{3} x^{5} + a^{3} c^{3} x^{4} + \frac {1}{3} \, a^{2} c^{3} x^{3} - a c^{3} x^{2} - c^{3} x \]

[In]

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

[Out]

-1/7*a^6*c^3*x^7 - 1/3*a^5*c^3*x^6 + 1/5*a^4*c^3*x^5 + a^3*c^3*x^4 + 1/3*a^2*c^3*x^3 - a*c^3*x^2 - c^3*x

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.35 \[ \int e^{2 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^3 \, dx=-\frac {a^6\,c^3\,x^7}{7}-\frac {a^5\,c^3\,x^6}{3}+\frac {a^4\,c^3\,x^5}{5}+a^3\,c^3\,x^4+\frac {a^2\,c^3\,x^3}{3}-a\,c^3\,x^2-c^3\,x \]

[In]

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

[Out]

(a^2*c^3*x^3)/3 - a*c^3*x^2 - c^3*x + a^3*c^3*x^4 + (a^4*c^3*x^5)/5 - (a^5*c^3*x^6)/3 - (a^6*c^3*x^7)/7