\(\int e^{4 \coth ^{-1}(a x)} (c-a^2 c x^2) \, dx\) [585]

Optimal result

Integrand size = 20, antiderivative size = 46 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right ) \, dx=-4 c x-\frac {c (1+a x)^2}{a}-\frac {c (1+a x)^3}{3 a}-\frac {8 c \log (1-a x)}{a} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {6302, 6275, 45} \[ \int e^{4 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right ) \, dx=-\frac {c (a x+1)^3}{3 a}-\frac {c (a x+1)^2}{a}-\frac {8 c \log (1-a x)}{a}-4 c x \]

[In]

Int[E^(4*ArcCoth[a*x])*(c - a^2*c*x^2),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.83 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right ) \, dx=-\frac {c \left (4+21 a x+6 a^2 x^2+a^3 x^3+24 \log (1-a x)\right )}{3 a} \]

[In]

Integrate[E^(4*ArcCoth[a*x])*(c - a^2*c*x^2),x]

[Out]

-1/3*(c*(4 + 21*a*x + 6*a^2*x^2 + a^3*x^3 + 24*Log[1 - a*x]))/a

Maple [A] (verified)

Time = 0.60 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.70

method	result	size
default	\(c \left (-\frac {a^{2} x^{3}}{3}-2 a \,x^{2}-7 x -\frac {8 \ln \left (a x -1\right )}{a}\right )\)	\(32\)
risch	\(-\frac {a^{2} c \,x^{3}}{3}-2 a c \,x^{2}-7 c x -\frac {8 c \ln \left (a x -1\right )}{a}\)	\(34\)
parallelrisch	\(-\frac {a^{3} c \,x^{3}+6 a^{2} c \,x^{2}+21 a c x +24 c \ln \left (a x -1\right )}{3 a}\)	\(38\)
norman	\(\frac {7 c x -5 a c \,x^{2}-\frac {5}{3} a^{2} c \,x^{3}-\frac {1}{3} a^{3} c \,x^{4}}{a x -1}-\frac {8 c \ln \left (a x -1\right )}{a}\)	\(52\)
meijerg	\(\frac {c \left (-\frac {a x \left (-5 a^{3} x^{3}-10 a^{2} x^{2}-30 a x +60\right )}{15 \left (-a x +1\right )}-4 \ln \left (-a x +1\right )\right )}{a}-\frac {2 c \left (\frac {a x \left (-2 a^{2} x^{2}-6 a x +12\right )}{-4 a x +4}+3 \ln \left (-a x +1\right )\right )}{a}+\frac {2 c \left (\frac {a x}{-a x +1}+\ln \left (-a x +1\right )\right )}{a}+\frac {c x}{-a x +1}\)	\(129\)

[In]

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

[Out]

c*(-1/3*a^2*x^3-2*a*x^2-7*x-8/a*ln(a*x-1))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.80 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right ) \, dx=-\frac {a^{3} c x^{3} + 6 \, a^{2} c x^{2} + 21 \, a c x + 24 \, c \log \left (a x - 1\right )}{3 \, a} \]

[In]

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

[Out]

-1/3*(a^3*c*x^3 + 6*a^2*c*x^2 + 21*a*c*x + 24*c*log(a*x - 1))/a

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.78 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right ) \, dx=- \frac {a^{2} c x^{3}}{3} - 2 a c x^{2} - 7 c x - \frac {8 c \log {\left (a x - 1 \right )}}{a} \]

[In]

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

[Out]

-a**2*c*x**3/3 - 2*a*c*x**2 - 7*c*x - 8*c*log(a*x - 1)/a

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.72 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right ) \, dx=-\frac {1}{3} \, a^{2} c x^{3} - 2 \, a c x^{2} - 7 \, c x - \frac {8 \, c \log \left (a x - 1\right )}{a} \]

[In]

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

[Out]

-1/3*a^2*c*x^3 - 2*a*c*x^2 - 7*c*x - 8*c*log(a*x - 1)/a

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.30 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right ) \, dx=-\frac {{\left (a x - 1\right )}^{3} {\left (c + \frac {9 \, c}{a x - 1} + \frac {36 \, c}{{\left (a x - 1\right )}^{2}}\right )}}{3 \, a} + \frac {8 \, c \log \left (\frac {{\left | a x - 1 \right |}}{{\left (a x - 1\right )}^{2} {\left | a \right |}}\right )}{a} \]

[In]

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

[Out]

-1/3*(a*x - 1)^3*(c + 9*c/(a*x - 1) + 36*c/(a*x - 1)^2)/a + 8*c*log(abs(a*x - 1)/((a*x - 1)^2*abs(a)))/a

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.72 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right ) \, dx=-7\,c\,x-\frac {a^2\,c\,x^3}{3}-\frac {8\,c\,\ln \left (a\,x-1\right )}{a}-2\,a\,c\,x^2 \]

[In]

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

[Out]

- 7*c*x - (a^2*c*x^3)/3 - (8*c*log(a*x - 1))/a - 2*a*c*x^2