\(\int \frac {e^{4 \coth ^{-1}(a x)}}{(c-a^2 c x^2)^3} \, dx\) [588]

Optimal result

Integrand size = 22, antiderivative size = 87 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {1}{8 a c^3 (1-a x)^4}+\frac {1}{12 a c^3 (1-a x)^3}+\frac {1}{16 a c^3 (1-a x)^2}+\frac {1}{16 a c^3 (1-a x)}+\frac {\text {arctanh}(a x)}{16 a c^3} \]

[Out]

1/8/a/c^3/(-a*x+1)^4+1/12/a/c^3/(-a*x+1)^3+1/16/a/c^3/(-a*x+1)^2+1/16/a/c^3/(-a*x+1)+1/16*arctanh(a*x)/a/c^3

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {6302, 6275, 46, 213} \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {\text {arctanh}(a x)}{16 a c^3}+\frac {1}{16 a c^3 (1-a x)}+\frac {1}{16 a c^3 (1-a x)^2}+\frac {1}{12 a c^3 (1-a x)^3}+\frac {1}{8 a c^3 (1-a x)^4} \]

[In]

Int[E^(4*ArcCoth[a*x])/(c - a^2*c*x^2)^3,x]

[Out]

1/(8*a*c^3*(1 - a*x)^4) + 1/(12*a*c^3*(1 - a*x)^3) + 1/(16*a*c^3*(1 - a*x)^2) + 1/(16*a*c^3*(1 - a*x)) + ArcTa
nh[a*x]/(16*a*c^3)

Rule 46

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 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 \frac {e^{4 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx \\ & = \frac {\int \frac {1}{(1-a x)^5 (1+a x)} \, dx}{c^3} \\ & = \frac {\int \left (-\frac {1}{2 (-1+a x)^5}+\frac {1}{4 (-1+a x)^4}-\frac {1}{8 (-1+a x)^3}+\frac {1}{16 (-1+a x)^2}-\frac {1}{16 \left (-1+a^2 x^2\right )}\right ) \, dx}{c^3} \\ & = \frac {1}{8 a c^3 (1-a x)^4}+\frac {1}{12 a c^3 (1-a x)^3}+\frac {1}{16 a c^3 (1-a x)^2}+\frac {1}{16 a c^3 (1-a x)}-\frac {\int \frac {1}{-1+a^2 x^2} \, dx}{16 c^3} \\ & = \frac {1}{8 a c^3 (1-a x)^4}+\frac {1}{12 a c^3 (1-a x)^3}+\frac {1}{16 a c^3 (1-a x)^2}+\frac {1}{16 a c^3 (1-a x)}+\frac {\text {arctanh}(a x)}{16 a c^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.60 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {16-19 a x+12 a^2 x^2-3 a^3 x^3+3 (-1+a x)^4 \text {arctanh}(a x)}{48 a c^3 (-1+a x)^4} \]

[In]

Integrate[E^(4*ArcCoth[a*x])/(c - a^2*c*x^2)^3,x]

[Out]

(16 - 19*a*x + 12*a^2*x^2 - 3*a^3*x^3 + 3*(-1 + a*x)^4*ArcTanh[a*x])/(48*a*c^3*(-1 + a*x)^4)

Maple [A] (verified)

Time = 0.62 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.75

method	result	size
risch	\(\frac {-\frac {a^{2} x^{3}}{16}+\frac {a \,x^{2}}{4}-\frac {19 x}{48}+\frac {1}{3 a}}{c^{3} \left (a x -1\right )^{4}}+\frac {\ln \left (-a x -1\right )}{32 a \,c^{3}}-\frac {\ln \left (a x -1\right )}{32 a \,c^{3}}\)	\(65\)
default	\(\frac {\frac {\ln \left (a x +1\right )}{32 a}+\frac {1}{8 a \left (a x -1\right )^{4}}-\frac {1}{12 a \left (a x -1\right )^{3}}+\frac {1}{16 \left (a x -1\right )^{2} a}-\frac {1}{16 a \left (a x -1\right )}-\frac {\ln \left (a x -1\right )}{32 a}}{c^{3}}\)	\(76\)
norman	\(\frac {\frac {15 x}{16 c}+\frac {a \,x^{2}}{8 c}-\frac {31 a^{2} x^{3}}{24 c}+\frac {11 a^{3} x^{4}}{24 c}+\frac {29 a^{4} x^{5}}{48 c}-\frac {a^{5} x^{6}}{3 c}}{\left (a x +1\right )^{2} \left (a x -1\right )^{4} c^{2}}-\frac {\ln \left (a x -1\right )}{32 a \,c^{3}}+\frac {\ln \left (a x +1\right )}{32 a \,c^{3}}\)	\(108\)
parallelrisch	\(\frac {-3 \ln \left (a x -1\right ) x^{4} a^{4}+3 \ln \left (a x +1\right ) x^{4} a^{4}-32 a^{4} x^{4}+12 a^{3} \ln \left (a x -1\right ) x^{3}-12 a^{3} \ln \left (a x +1\right ) x^{3}+122 a^{3} x^{3}-18 a^{2} \ln \left (a x -1\right ) x^{2}+18 a^{2} \ln \left (a x +1\right ) x^{2}-168 a^{2} x^{2}+12 a \ln \left (a x -1\right ) x -12 a \ln \left (a x +1\right ) x +90 a x -3 \ln \left (a x -1\right )+3 \ln \left (a x +1\right )}{96 \left (a x -1\right )^{4} c^{3} a}\)	\(165\)

[In]

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

[Out]

(-1/16*a^2*x^3+1/4*a*x^2-19/48*x+1/3/a)/c^3/(a*x-1)^4+1/32/a/c^3*ln(-a*x-1)-1/32/a/c^3*ln(a*x-1)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 147 vs. \(2 (73) = 146\).

Time = 0.25 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.69 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=-\frac {6 \, a^{3} x^{3} - 24 \, a^{2} x^{2} + 38 \, a x - 3 \, {\left (a^{4} x^{4} - 4 \, a^{3} x^{3} + 6 \, a^{2} x^{2} - 4 \, a x + 1\right )} \log \left (a x + 1\right ) + 3 \, {\left (a^{4} x^{4} - 4 \, a^{3} x^{3} + 6 \, a^{2} x^{2} - 4 \, a x + 1\right )} \log \left (a x - 1\right ) - 32}{96 \, {\left (a^{5} c^{3} x^{4} - 4 \, a^{4} c^{3} x^{3} + 6 \, a^{3} c^{3} x^{2} - 4 \, a^{2} c^{3} x + a c^{3}\right )}} \]

[In]

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

[Out]

-1/96*(6*a^3*x^3 - 24*a^2*x^2 + 38*a*x - 3*(a^4*x^4 - 4*a^3*x^3 + 6*a^2*x^2 - 4*a*x + 1)*log(a*x + 1) + 3*(a^4
*x^4 - 4*a^3*x^3 + 6*a^2*x^2 - 4*a*x + 1)*log(a*x - 1) - 32)/(a^5*c^3*x^4 - 4*a^4*c^3*x^3 + 6*a^3*c^3*x^2 - 4*
a^2*c^3*x + a*c^3)

Sympy [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.14 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=- \frac {3 a^{3} x^{3} - 12 a^{2} x^{2} + 19 a x - 16}{48 a^{5} c^{3} x^{4} - 192 a^{4} c^{3} x^{3} + 288 a^{3} c^{3} x^{2} - 192 a^{2} c^{3} x + 48 a c^{3}} - \frac {\frac {\log {\left (x - \frac {1}{a} \right )}}{32} - \frac {\log {\left (x + \frac {1}{a} \right )}}{32}}{a c^{3}} \]

[In]

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

[Out]

-(3*a**3*x**3 - 12*a**2*x**2 + 19*a*x - 16)/(48*a**5*c**3*x**4 - 192*a**4*c**3*x**3 + 288*a**3*c**3*x**2 - 192
*a**2*c**3*x + 48*a*c**3) - (log(x - 1/a)/32 - log(x + 1/a)/32)/(a*c**3)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.17 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=-\frac {3 \, a^{3} x^{3} - 12 \, a^{2} x^{2} + 19 \, a x - 16}{48 \, {\left (a^{5} c^{3} x^{4} - 4 \, a^{4} c^{3} x^{3} + 6 \, a^{3} c^{3} x^{2} - 4 \, a^{2} c^{3} x + a c^{3}\right )}} + \frac {\log \left (a x + 1\right )}{32 \, a c^{3}} - \frac {\log \left (a x - 1\right )}{32 \, a c^{3}} \]

[In]

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

[Out]

-1/48*(3*a^3*x^3 - 12*a^2*x^2 + 19*a*x - 16)/(a^5*c^3*x^4 - 4*a^4*c^3*x^3 + 6*a^3*c^3*x^2 - 4*a^2*c^3*x + a*c^
3) + 1/32*log(a*x + 1)/(a*c^3) - 1/32*log(a*x - 1)/(a*c^3)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.05 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {\log \left ({\left | -\frac {2}{a x - 1} - 1 \right |}\right )}{32 \, a c^{3}} - \frac {\frac {3 \, a^{3} c^{9}}{a x - 1} - \frac {3 \, a^{3} c^{9}}{{\left (a x - 1\right )}^{2}} + \frac {4 \, a^{3} c^{9}}{{\left (a x - 1\right )}^{3}} - \frac {6 \, a^{3} c^{9}}{{\left (a x - 1\right )}^{4}}}{48 \, a^{4} c^{12}} \]

[In]

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

[Out]

1/32*log(abs(-2/(a*x - 1) - 1))/(a*c^3) - 1/48*(3*a^3*c^9/(a*x - 1) - 3*a^3*c^9/(a*x - 1)^2 + 4*a^3*c^9/(a*x -
1)^3 - 6*a^3*c^9/(a*x - 1)^4)/(a^4*c^12)

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.95 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {\mathrm {atanh}\left (a\,x\right )}{16\,a\,c^3}-\frac {\frac {19\,x}{48}-\frac {a\,x^2}{4}-\frac {1}{3\,a}+\frac {a^2\,x^3}{16}}{a^4\,c^3\,x^4-4\,a^3\,c^3\,x^3+6\,a^2\,c^3\,x^2-4\,a\,c^3\,x+c^3} \]

[In]

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

[Out]

atanh(a*x)/(16*a*c^3) - ((19*x)/48 - (a*x^2)/4 - 1/(3*a) + (a^2*x^3)/16)/(c^3 + 6*a^2*c^3*x^2 - 4*a^3*c^3*x^3
+ a^4*c^3*x^4 - 4*a*c^3*x)