\(\int \frac {e^{4 \coth ^{-1}(a x)}}{(c-\frac {c}{a^2 x^2})^4} \, dx\) [805]

Optimal result

Integrand size = 22, antiderivative size = 146 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^4} \, dx=\frac {x}{c^4}+\frac {1}{20 a c^4 (1-a x)^5}-\frac {7}{16 a c^4 (1-a x)^4}+\frac {83}{48 a c^4 (1-a x)^3}-\frac {67}{16 a c^4 (1-a x)^2}+\frac {501}{64 a c^4 (1-a x)}-\frac {1}{64 a c^4 (1+a x)}+\frac {261 \log (1-a x)}{64 a c^4}-\frac {5 \log (1+a x)}{64 a c^4} \]

[Out]

x/c^4+1/20/a/c^4/(-a*x+1)^5-7/16/a/c^4/(-a*x+1)^4+83/48/a/c^4/(-a*x+1)^3-67/16/a/c^4/(-a*x+1)^2+501/64/a/c^4/(
-a*x+1)-1/64/a/c^4/(a*x+1)+261/64*ln(-a*x+1)/a/c^4-5/64*ln(a*x+1)/a/c^4

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 146, 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, 6292, 6285, 90} \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^4} \, dx=\frac {501}{64 a c^4 (1-a x)}-\frac {1}{64 a c^4 (a x+1)}-\frac {67}{16 a c^4 (1-a x)^2}+\frac {83}{48 a c^4 (1-a x)^3}-\frac {7}{16 a c^4 (1-a x)^4}+\frac {1}{20 a c^4 (1-a x)^5}+\frac {261 \log (1-a x)}{64 a c^4}-\frac {5 \log (a x+1)}{64 a c^4}+\frac {x}{c^4} \]

[In]

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

[Out]

x/c^4 + 1/(20*a*c^4*(1 - a*x)^5) - 7/(16*a*c^4*(1 - a*x)^4) + 83/(48*a*c^4*(1 - a*x)^3) - 67/(16*a*c^4*(1 - a*
x)^2) + 501/(64*a*c^4*(1 - a*x)) - 1/(64*a*c^4*(1 + a*x)) + (261*Log[1 - a*x])/(64*a*c^4) - (5*Log[1 + a*x])/(
64*a*c^4)

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 6285

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[x^m*(1 -
a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p
] || GtQ[c, 0])

Rule 6292

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_.), x_Symbol] :> Dist[d^p, Int[(u/x^(2*p))*(1
- a^2*x^2)^p*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[{a, c, d, n}, x] && EqQ[c + a^2*d, 0] && IntegerQ[p]

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-\frac {c}{a^2 x^2}\right )^4} \, dx \\ & = \frac {a^8 \int \frac {e^{4 \text {arctanh}(a x)} x^8}{\left (1-a^2 x^2\right )^4} \, dx}{c^4} \\ & = \frac {a^8 \int \frac {x^8}{(1-a x)^6 (1+a x)^2} \, dx}{c^4} \\ & = \frac {a^8 \int \left (\frac {1}{a^8}+\frac {1}{4 a^8 (-1+a x)^6}+\frac {7}{4 a^8 (-1+a x)^5}+\frac {83}{16 a^8 (-1+a x)^4}+\frac {67}{8 a^8 (-1+a x)^3}+\frac {501}{64 a^8 (-1+a x)^2}+\frac {261}{64 a^8 (-1+a x)}+\frac {1}{64 a^8 (1+a x)^2}-\frac {5}{64 a^8 (1+a x)}\right ) \, dx}{c^4} \\ & = \frac {x}{c^4}+\frac {1}{20 a c^4 (1-a x)^5}-\frac {7}{16 a c^4 (1-a x)^4}+\frac {83}{48 a c^4 (1-a x)^3}-\frac {67}{16 a c^4 (1-a x)^2}+\frac {501}{64 a c^4 (1-a x)}-\frac {1}{64 a c^4 (1+a x)}+\frac {261 \log (1-a x)}{64 a c^4}-\frac {5 \log (1+a x)}{64 a c^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.67 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^4} \, dx=\frac {\frac {2 \left (-2384+7541 a x-4900 a^2 x^2-6800 a^3 x^3+9300 a^4 x^4-1365 a^5 x^5-1920 a^6 x^6+480 a^7 x^7\right )}{(-1+a x)^5 (1+a x)}+3915 \log (1-a x)-75 \log (1+a x)}{960 a c^4} \]

[In]

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

[Out]

((2*(-2384 + 7541*a*x - 4900*a^2*x^2 - 6800*a^3*x^3 + 9300*a^4*x^4 - 1365*a^5*x^5 - 1920*a^6*x^6 + 480*a^7*x^7
))/((-1 + a*x)^5*(1 + a*x)) + 3915*Log[1 - a*x] - 75*Log[1 + a*x])/(960*a*c^4)

Maple [A] (verified)

Time = 0.60 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.74

method	result	size
default	\(\frac {a^{8} \left (-\frac {1}{64 a^{9} \left (a x +1\right )}-\frac {5 \ln \left (a x +1\right )}{64 a^{9}}+\frac {x}{a^{8}}-\frac {1}{20 a^{9} \left (a x -1\right )^{5}}-\frac {7}{16 a^{9} \left (a x -1\right )^{4}}-\frac {83}{48 a^{9} \left (a x -1\right )^{3}}-\frac {67}{16 a^{9} \left (a x -1\right )^{2}}-\frac {501}{64 a^{9} \left (a x -1\right )}+\frac {261 \ln \left (a x -1\right )}{64 a^{9}}\right )}{c^{4}}\)	\(108\)
risch	\(\frac {x}{c^{4}}+\frac {-\frac {251 a^{4} c^{4} x^{5}}{32}+\frac {155 a^{3} c^{4} x^{4}}{8}-\frac {55 a^{2} c^{4} x^{3}}{6}-\frac {341 a \,c^{4} x^{2}}{24}+\frac {8021 c^{4} x}{480}-\frac {149 c^{4}}{30 a}}{c^{8} \left (a x -1\right )^{4} \left (a^{2} x^{2}-1\right )}-\frac {5 \ln \left (a x +1\right )}{64 a \,c^{4}}+\frac {261 \ln \left (-a x +1\right )}{64 a \,c^{4}}\)	\(115\)
norman	\(\frac {\frac {a^{8} x^{9}}{c}-\frac {115 a^{3} x^{4}}{6 c}-\frac {133 x}{32 c}+\frac {101 a \,x^{2}}{16 c}+\frac {1049 a^{2} x^{3}}{96 c}-\frac {3869 a^{4} x^{5}}{480 c}+\frac {4709 a^{5} x^{6}}{240 c}+\frac {43 a^{6} x^{7}}{480 c}-\frac {209 a^{7} x^{8}}{30 c}}{\left (a x +1\right )^{3} \left (a x -1\right )^{5} c^{3}}+\frac {261 \ln \left (a x -1\right )}{64 a \,c^{4}}-\frac {5 \ln \left (a x +1\right )}{64 a \,c^{4}}\)	\(140\)
parallelrisch	\(\frac {-300 a \ln \left (a x +1\right ) x +375 a^{2} \ln \left (a x +1\right ) x^{2}+16342 a^{5} x^{5}-13600 a^{3} x^{3}+300 \ln \left (a x +1\right ) x^{5} a^{5}-75 \ln \left (a x +1\right ) x^{6} a^{6}-375 \ln \left (a x +1\right ) x^{4} a^{4}+3915 \ln \left (a x -1\right ) x^{6} a^{6}-15660 \ln \left (a x -1\right ) x^{5} a^{5}+19575 \ln \left (a x -1\right ) x^{4} a^{4}-8608 a^{6} x^{6}-3990 a x -19575 a^{2} \ln \left (a x -1\right ) x^{2}+15660 a \ln \left (a x -1\right ) x -5240 a^{4} x^{4}-3915 \ln \left (a x -1\right )+75 \ln \left (a x +1\right )+960 a^{7} x^{7}+14040 a^{2} x^{2}}{960 \left (a^{2} x^{2}-1\right ) \left (a x -1\right )^{4} c^{4} a}\)	\(228\)

[In]

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

[Out]

a^8/c^4*(-1/64/a^9/(a*x+1)-5/64/a^9*ln(a*x+1)+1/a^8*x-1/20/a^9/(a*x-1)^5-7/16/a^9/(a*x-1)^4-83/48/a^9/(a*x-1)^
3-67/16/a^9/(a*x-1)^2-501/64/a^9/(a*x-1)+261/64/a^9*ln(a*x-1))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.42 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^4} \, dx=\frac {960 \, a^{7} x^{7} - 3840 \, a^{6} x^{6} - 2730 \, a^{5} x^{5} + 18600 \, a^{4} x^{4} - 13600 \, a^{3} x^{3} - 9800 \, a^{2} x^{2} + 15082 \, a x - 75 \, {\left (a^{6} x^{6} - 4 \, a^{5} x^{5} + 5 \, a^{4} x^{4} - 5 \, a^{2} x^{2} + 4 \, a x - 1\right )} \log \left (a x + 1\right ) + 3915 \, {\left (a^{6} x^{6} - 4 \, a^{5} x^{5} + 5 \, a^{4} x^{4} - 5 \, a^{2} x^{2} + 4 \, a x - 1\right )} \log \left (a x - 1\right ) - 4768}{960 \, {\left (a^{7} c^{4} x^{6} - 4 \, a^{6} c^{4} x^{5} + 5 \, a^{5} c^{4} x^{4} - 5 \, a^{3} c^{4} x^{2} + 4 \, a^{2} c^{4} x - a c^{4}\right )}} \]

[In]

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

[Out]

1/960*(960*a^7*x^7 - 3840*a^6*x^6 - 2730*a^5*x^5 + 18600*a^4*x^4 - 13600*a^3*x^3 - 9800*a^2*x^2 + 15082*a*x -
75*(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) + 3915*(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) - 4768)/(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 = 0.48 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.99 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^4} \, dx=a^{8} \left (\frac {- 3765 a^{5} x^{5} + 9300 a^{4} x^{4} - 4400 a^{3} x^{3} - 6820 a^{2} x^{2} + 8021 a x - 2384}{480 a^{15} c^{4} x^{6} - 1920 a^{14} c^{4} x^{5} + 2400 a^{13} c^{4} x^{4} - 2400 a^{11} c^{4} x^{2} + 1920 a^{10} c^{4} x - 480 a^{9} c^{4}} + \frac {x}{a^{8} c^{4}} + \frac {\frac {261 \log {\left (x - \frac {1}{a} \right )}}{64} - \frac {5 \log {\left (x + \frac {1}{a} \right )}}{64}}{a^{9} c^{4}}\right ) \]

[In]

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

[Out]

a**8*((-3765*a**5*x**5 + 9300*a**4*x**4 - 4400*a**3*x**3 - 6820*a**2*x**2 + 8021*a*x - 2384)/(480*a**15*c**4*x
**6 - 1920*a**14*c**4*x**5 + 2400*a**13*c**4*x**4 - 2400*a**11*c**4*x**2 + 1920*a**10*c**4*x - 480*a**9*c**4)
+ x/(a**8*c**4) + (261*log(x - 1/a)/64 - 5*log(x + 1/a)/64)/(a**9*c**4))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.92 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^4} \, dx=-\frac {3765 \, a^{5} x^{5} - 9300 \, a^{4} x^{4} + 4400 \, a^{3} x^{3} + 6820 \, a^{2} x^{2} - 8021 \, a x + 2384}{480 \, {\left (a^{7} c^{4} x^{6} - 4 \, a^{6} c^{4} x^{5} + 5 \, a^{5} c^{4} x^{4} - 5 \, a^{3} c^{4} x^{2} + 4 \, a^{2} c^{4} x - a c^{4}\right )}} + \frac {x}{c^{4}} - \frac {5 \, \log \left (a x + 1\right )}{64 \, a c^{4}} + \frac {261 \, \log \left (a x - 1\right )}{64 \, a c^{4}} \]

[In]

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

[Out]

-1/480*(3765*a^5*x^5 - 9300*a^4*x^4 + 4400*a^3*x^3 + 6820*a^2*x^2 - 8021*a*x + 2384)/(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) + x/c^4 - 5/64*log(a*x + 1)/(a*c^4) + 261/64*log(a*
x - 1)/(a*c^4)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.16 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^4} \, dx=\frac {{\left (a x - 1\right )} {\left (\frac {257}{a x - 1} + 128\right )}}{128 \, a c^{4} {\left (\frac {2}{a x - 1} + 1\right )}} - \frac {4 \, \log \left (\frac {{\left | a x - 1 \right |}}{{\left (a x - 1\right )}^{2} {\left | a \right |}}\right )}{a c^{4}} - \frac {5 \, \log \left ({\left | -\frac {2}{a x - 1} - 1 \right |}\right )}{64 \, a c^{4}} - \frac {\frac {7515 \, a^{19} c^{16}}{a x - 1} + \frac {4020 \, a^{19} c^{16}}{{\left (a x - 1\right )}^{2}} + \frac {1660 \, a^{19} c^{16}}{{\left (a x - 1\right )}^{3}} + \frac {420 \, a^{19} c^{16}}{{\left (a x - 1\right )}^{4}} + \frac {48 \, a^{19} c^{16}}{{\left (a x - 1\right )}^{5}}}{960 \, a^{20} c^{20}} \]

[In]

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

[Out]

1/128*(a*x - 1)*(257/(a*x - 1) + 128)/(a*c^4*(2/(a*x - 1) + 1)) - 4*log(abs(a*x - 1)/((a*x - 1)^2*abs(a)))/(a*
c^4) - 5/64*log(abs(-2/(a*x - 1) - 1))/(a*c^4) - 1/960*(7515*a^19*c^16/(a*x - 1) + 4020*a^19*c^16/(a*x - 1)^2
+ 1660*a^19*c^16/(a*x - 1)^3 + 420*a^19*c^16/(a*x - 1)^4 + 48*a^19*c^16/(a*x - 1)^5)/(a^20*c^20)

Mupad [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.90 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^4} \, dx=\frac {\frac {341\,a\,x^2}{24}-\frac {8021\,x}{480}+\frac {149}{30\,a}+\frac {55\,a^2\,x^3}{6}-\frac {155\,a^3\,x^4}{8}+\frac {251\,a^4\,x^5}{32}}{-a^6\,c^4\,x^6+4\,a^5\,c^4\,x^5-5\,a^4\,c^4\,x^4+5\,a^2\,c^4\,x^2-4\,a\,c^4\,x+c^4}+\frac {x}{c^4}+\frac {261\,\ln \left (a\,x-1\right )}{64\,a\,c^4}-\frac {5\,\ln \left (a\,x+1\right )}{64\,a\,c^4} \]

[In]

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

[Out]

((341*a*x^2)/24 - (8021*x)/480 + 149/(30*a) + (55*a^2*x^3)/6 - (155*a^3*x^4)/8 + (251*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) + x/c^4 + (261*log(a*x - 1))/(64*a*c^4) -
(5*log(a*x + 1))/(64*a*c^4)