\(\int \frac {e^{4 \coth ^{-1}(a x)}}{(c-a^2 c x^2)^4} \, dx\) [589]

Optimal result

Integrand size = 22, antiderivative size = 122 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=\frac {1}{20 a c^4 (1-a x)^5}+\frac {1}{16 a c^4 (1-a x)^4}+\frac {1}{16 a c^4 (1-a x)^3}+\frac {1}{16 a c^4 (1-a x)^2}+\frac {5}{64 a c^4 (1-a x)}-\frac {1}{64 a c^4 (1+a x)}+\frac {3 \text {arctanh}(a x)}{32 a c^4} \]

[Out]

1/20/a/c^4/(-a*x+1)^5+1/16/a/c^4/(-a*x+1)^4+1/16/a/c^4/(-a*x+1)^3+1/16/a/c^4/(-a*x+1)^2+5/64/a/c^4/(-a*x+1)-1/
64/a/c^4/(a*x+1)+3/32*arctanh(a*x)/a/c^4

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 122, 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} \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=\frac {3 \text {arctanh}(a x)}{32 a c^4}+\frac {5}{64 a c^4 (1-a x)}-\frac {1}{64 a c^4 (a x+1)}+\frac {1}{16 a c^4 (1-a x)^2}+\frac {1}{16 a c^4 (1-a x)^3}+\frac {1}{16 a c^4 (1-a x)^4}+\frac {1}{20 a c^4 (1-a x)^5} \]

[In]

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

[Out]

1/(20*a*c^4*(1 - a*x)^5) + 1/(16*a*c^4*(1 - a*x)^4) + 1/(16*a*c^4*(1 - a*x)^3) + 1/(16*a*c^4*(1 - a*x)^2) + 5/
(64*a*c^4*(1 - a*x)) - 1/(64*a*c^4*(1 + a*x)) + (3*ArcTanh[a*x])/(32*a*c^4)

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 )^4} \, dx \\ & = \frac {\int \frac {1}{(1-a x)^6 (1+a x)^2} \, dx}{c^4} \\ & = \frac {\int \left (\frac {1}{4 (-1+a x)^6}-\frac {1}{4 (-1+a x)^5}+\frac {3}{16 (-1+a x)^4}-\frac {1}{8 (-1+a x)^3}+\frac {5}{64 (-1+a x)^2}+\frac {1}{64 (1+a x)^2}-\frac {3}{32 \left (-1+a^2 x^2\right )}\right ) \, dx}{c^4} \\ & = \frac {1}{20 a c^4 (1-a x)^5}+\frac {1}{16 a c^4 (1-a x)^4}+\frac {1}{16 a c^4 (1-a x)^3}+\frac {1}{16 a c^4 (1-a x)^2}+\frac {5}{64 a c^4 (1-a x)}-\frac {1}{64 a c^4 (1+a x)}-\frac {3 \int \frac {1}{-1+a^2 x^2} \, dx}{32 c^4} \\ & = \frac {1}{20 a c^4 (1-a x)^5}+\frac {1}{16 a c^4 (1-a x)^4}+\frac {1}{16 a c^4 (1-a x)^3}+\frac {1}{16 a c^4 (1-a x)^2}+\frac {5}{64 a c^4 (1-a x)}-\frac {1}{64 a c^4 (1+a x)}+\frac {3 \text {arctanh}(a x)}{32 a c^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.66 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=\frac {-48+47 a x+20 a^2 x^2-80 a^3 x^3+60 a^4 x^4-15 a^5 x^5+15 (-1+a x)^5 (1+a x) \text {arctanh}(a x)}{160 a c^4 (-1+a x)^5 (1+a x)} \]

[In]

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

[Out]

(-48 + 47*a*x + 20*a^2*x^2 - 80*a^3*x^3 + 60*a^4*x^4 - 15*a^5*x^5 + 15*(-1 + a*x)^5*(1 + a*x)*ArcTanh[a*x])/(1
60*a*c^4*(-1 + a*x)^5*(1 + a*x))

Maple [A] (verified)

Time = 0.63 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.75

method	result	size
risch	\(\frac {-\frac {3 a^{4} x^{5}}{32}+\frac {3 a^{3} x^{4}}{8}-\frac {a^{2} x^{3}}{2}+\frac {a \,x^{2}}{8}+\frac {47 x}{160}-\frac {3}{10 a}}{c^{4} \left (a x -1\right )^{4} \left (a^{2} x^{2}-1\right )}-\frac {3 \ln \left (a x -1\right )}{64 a \,c^{4}}+\frac {3 \ln \left (-a x -1\right )}{64 a \,c^{4}}\)	\(92\)
default	\(\frac {-\frac {1}{64 a \left (a x +1\right )}+\frac {3 \ln \left (a x +1\right )}{64 a}-\frac {1}{20 a \left (a x -1\right )^{5}}+\frac {1}{16 a \left (a x -1\right )^{4}}-\frac {1}{16 a \left (a x -1\right )^{3}}+\frac {1}{16 \left (a x -1\right )^{2} a}-\frac {5}{64 a \left (a x -1\right )}-\frac {3 \ln \left (a x -1\right )}{64 a}}{c^{4}}\)	\(100\)
norman	\(\frac {-\frac {a^{3} x^{4}}{2 c}-\frac {29 x}{32 c}-\frac {3 a \,x^{2}}{16 c}+\frac {59 a^{2} x^{3}}{32 c}-\frac {263 a^{4} x^{5}}{160 c}+\frac {63 a^{5} x^{6}}{80 c}+\frac {81 a^{6} x^{7}}{160 c}-\frac {3 a^{7} x^{8}}{10 c}}{\left (a x -1\right )^{5} \left (a x +1\right )^{3} c^{3}}-\frac {3 \ln \left (a x -1\right )}{64 a \,c^{4}}+\frac {3 \ln \left (a x +1\right )}{64 a \,c^{4}}\)	\(130\)
parallelrisch	\(\frac {60 a \ln \left (a x +1\right ) x -75 a^{2} \ln \left (a x +1\right ) x^{2}+354 a^{5} x^{5}-160 a^{3} x^{3}-60 \ln \left (a x +1\right ) x^{5} a^{5}+15 \ln \left (a x +1\right ) x^{6} a^{6}+75 \ln \left (a x +1\right ) x^{4} a^{4}-15 \ln \left (a x -1\right ) x^{6} a^{6}+60 \ln \left (a x -1\right ) x^{5} a^{5}-75 \ln \left (a x -1\right ) x^{4} a^{4}-96 a^{6} x^{6}-290 a x +75 a^{2} \ln \left (a x -1\right ) x^{2}-60 a \ln \left (a x -1\right ) x -360 a^{4} x^{4}+15 \ln \left (a x -1\right )-15 \ln \left (a x +1\right )+520 a^{2} x^{2}}{320 \left (a^{2} x^{2}-1\right ) \left (a x -1\right )^{4} c^{4} a}\)	\(220\)

[In]

int(1/(a*x-1)^2*(a*x+1)^2/(-a^2*c*x^2+c)^4,x,method=_RETURNVERBOSE)

[Out]

(-3/32*a^4*x^5+3/8*a^3*x^4-1/2*a^2*x^3+1/8*a*x^2+47/160*x-3/10/a)/c^4/(a*x-1)^4/(a^2*x^2-1)-3/64/a/c^4*ln(a*x-
1)+3/64/a/c^4*ln(-a*x-1)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.57 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=-\frac {30 \, a^{5} x^{5} - 120 \, a^{4} x^{4} + 160 \, a^{3} x^{3} - 40 \, a^{2} x^{2} - 94 \, a x - 15 \, {\left (a^{6} x^{6} - 4 \, a^{5} x^{5} + 5 \, a^{4} x^{4} - 5 \, a^{2} x^{2} + 4 \, a x - 1\right )} \log \left (a x + 1\right ) + 15 \, {\left (a^{6} x^{6} - 4 \, a^{5} x^{5} + 5 \, a^{4} x^{4} - 5 \, a^{2} x^{2} + 4 \, a x - 1\right )} \log \left (a x - 1\right ) + 96}{320 \, {\left (a^{7} c^{4} x^{6} - 4 \, a^{6} c^{4} x^{5} + 5 \, a^{5} c^{4} x^{4} - 5 \, a^{3} c^{4} x^{2} + 4 \, a^{2} c^{4} x - a c^{4}\right )}} \]

[In]

integrate(1/(a*x-1)^2*(a*x+1)^2/(-a^2*c*x^2+c)^4,x, algorithm="fricas")

[Out]

-1/320*(30*a^5*x^5 - 120*a^4*x^4 + 160*a^3*x^3 - 40*a^2*x^2 - 94*a*x - 15*(a^6*x^6 - 4*a^5*x^5 + 5*a^4*x^4 - 5
*a^2*x^2 + 4*a*x - 1)*log(a*x + 1) + 15*(a^6*x^6 - 4*a^5*x^5 + 5*a^4*x^4 - 5*a^2*x^2 + 4*a*x - 1)*log(a*x - 1)
+ 96)/(a^7*c^4*x^6 - 4*a^6*c^4*x^5 + 5*a^5*c^4*x^4 - 5*a^3*c^4*x^2 + 4*a^2*c^4*x - a*c^4)

Sympy [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.06 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=\frac {- 15 a^{5} x^{5} + 60 a^{4} x^{4} - 80 a^{3} x^{3} + 20 a^{2} x^{2} + 47 a x - 48}{160 a^{7} c^{4} x^{6} - 640 a^{6} c^{4} x^{5} + 800 a^{5} c^{4} x^{4} - 800 a^{3} c^{4} x^{2} + 640 a^{2} c^{4} x - 160 a c^{4}} + \frac {- \frac {3 \log {\left (x - \frac {1}{a} \right )}}{64} + \frac {3 \log {\left (x + \frac {1}{a} \right )}}{64}}{a c^{4}} \]

[In]

integrate(1/(a*x-1)**2*(a*x+1)**2/(-a**2*c*x**2+c)**4,x)

[Out]

(-15*a**5*x**5 + 60*a**4*x**4 - 80*a**3*x**3 + 20*a**2*x**2 + 47*a*x - 48)/(160*a**7*c**4*x**6 - 640*a**6*c**4
*x**5 + 800*a**5*c**4*x**4 - 800*a**3*c**4*x**2 + 640*a**2*c**4*x - 160*a*c**4) + (-3*log(x - 1/a)/64 + 3*log(
x + 1/a)/64)/(a*c**4)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.07 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=-\frac {15 \, a^{5} x^{5} - 60 \, a^{4} x^{4} + 80 \, a^{3} x^{3} - 20 \, a^{2} x^{2} - 47 \, a x + 48}{160 \, {\left (a^{7} c^{4} x^{6} - 4 \, a^{6} c^{4} x^{5} + 5 \, a^{5} c^{4} x^{4} - 5 \, a^{3} c^{4} x^{2} + 4 \, a^{2} c^{4} x - a c^{4}\right )}} + \frac {3 \, \log \left (a x + 1\right )}{64 \, a c^{4}} - \frac {3 \, \log \left (a x - 1\right )}{64 \, a c^{4}} \]

[In]

integrate(1/(a*x-1)^2*(a*x+1)^2/(-a^2*c*x^2+c)^4,x, algorithm="maxima")

[Out]

-1/160*(15*a^5*x^5 - 60*a^4*x^4 + 80*a^3*x^3 - 20*a^2*x^2 - 47*a*x + 48)/(a^7*c^4*x^6 - 4*a^6*c^4*x^5 + 5*a^5*
c^4*x^4 - 5*a^3*c^4*x^2 + 4*a^2*c^4*x - a*c^4) + 3/64*log(a*x + 1)/(a*c^4) - 3/64*log(a*x - 1)/(a*c^4)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.04 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=\frac {3 \, \log \left ({\left | -\frac {2}{a x - 1} - 1 \right |}\right )}{64 \, a c^{4}} + \frac {1}{128 \, a c^{4} {\left (\frac {2}{a x - 1} + 1\right )}} - \frac {\frac {25 \, a^{9} c^{16}}{a x - 1} - \frac {20 \, a^{9} c^{16}}{{\left (a x - 1\right )}^{2}} + \frac {20 \, a^{9} c^{16}}{{\left (a x - 1\right )}^{3}} - \frac {20 \, a^{9} c^{16}}{{\left (a x - 1\right )}^{4}} + \frac {16 \, a^{9} c^{16}}{{\left (a x - 1\right )}^{5}}}{320 \, a^{10} c^{20}} \]

[In]

integrate(1/(a*x-1)^2*(a*x+1)^2/(-a^2*c*x^2+c)^4,x, algorithm="giac")

[Out]

3/64*log(abs(-2/(a*x - 1) - 1))/(a*c^4) + 1/128/(a*c^4*(2/(a*x - 1) + 1)) - 1/320*(25*a^9*c^16/(a*x - 1) - 20*
a^9*c^16/(a*x - 1)^2 + 20*a^9*c^16/(a*x - 1)^3 - 20*a^9*c^16/(a*x - 1)^4 + 16*a^9*c^16/(a*x - 1)^5)/(a^10*c^20
)

Mupad [B] (verification not implemented)

Time = 3.96 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.91 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=\frac {3\,\mathrm {atanh}\left (a\,x\right )}{32\,a\,c^4}-\frac {\frac {47\,x}{160}+\frac {a\,x^2}{8}-\frac {3}{10\,a}-\frac {a^2\,x^3}{2}+\frac {3\,a^3\,x^4}{8}-\frac {3\,a^4\,x^5}{32}}{-a^6\,c^4\,x^6+4\,a^5\,c^4\,x^5-5\,a^4\,c^4\,x^4+5\,a^2\,c^4\,x^2-4\,a\,c^4\,x+c^4} \]

[In]

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

[Out]

(3*atanh(a*x))/(32*a*c^4) - ((47*x)/160 + (a*x^2)/8 - 3/(10*a) - (a^2*x^3)/2 + (3*a^3*x^4)/8 - (3*a^4*x^5)/32)
/(c^4 + 5*a^2*c^4*x^2 - 5*a^4*c^4*x^4 + 4*a^5*c^4*x^5 - a^6*c^4*x^6 - 4*a*c^4*x)