\(\int e^{4 \coth ^{-1}(a x)} (c-\frac {c}{a^2 x^2})^4 \, dx\) [798]

Optimal result

Integrand size = 22, antiderivative size = 100 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^4 \, dx=-\frac {c^4}{7 a^8 x^7}-\frac {2 c^4}{3 a^7 x^6}-\frac {4 c^4}{5 a^6 x^5}+\frac {c^4}{a^5 x^4}+\frac {10 c^4}{3 a^4 x^3}+\frac {2 c^4}{a^3 x^2}-\frac {4 c^4}{a^2 x}+c^4 x+\frac {4 c^4 \log (x)}{a} \]

[Out]

-1/7*c^4/a^8/x^7-2/3*c^4/a^7/x^6-4/5*c^4/a^6/x^5+c^4/a^5/x^4+10/3*c^4/a^4/x^3+2*c^4/a^3/x^2-4*c^4/a^2/x+c^4*x+
4*c^4*ln(x)/a

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 100, 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} \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^4 \, dx=-\frac {c^4}{7 a^8 x^7}-\frac {2 c^4}{3 a^7 x^6}-\frac {4 c^4}{5 a^6 x^5}+\frac {c^4}{a^5 x^4}+\frac {10 c^4}{3 a^4 x^3}+\frac {2 c^4}{a^3 x^2}-\frac {4 c^4}{a^2 x}+\frac {4 c^4 \log (x)}{a}+c^4 x \]

[In]

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

[Out]

-1/7*c^4/(a^8*x^7) - (2*c^4)/(3*a^7*x^6) - (4*c^4)/(5*a^6*x^5) + c^4/(a^5*x^4) + (10*c^4)/(3*a^4*x^3) + (2*c^4
)/(a^3*x^2) - (4*c^4)/(a^2*x) + c^4*x + (4*c^4*Log[x])/a

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 e^{4 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^4 \, dx \\ & = \frac {c^4 \int \frac {e^{4 \text {arctanh}(a x)} \left (1-a^2 x^2\right )^4}{x^8} \, dx}{a^8} \\ & = \frac {c^4 \int \frac {(1-a x)^2 (1+a x)^6}{x^8} \, dx}{a^8} \\ & = \frac {c^4 \int \left (a^8+\frac {1}{x^8}+\frac {4 a}{x^7}+\frac {4 a^2}{x^6}-\frac {4 a^3}{x^5}-\frac {10 a^4}{x^4}-\frac {4 a^5}{x^3}+\frac {4 a^6}{x^2}+\frac {4 a^7}{x}\right ) \, dx}{a^8} \\ & = -\frac {c^4}{7 a^8 x^7}-\frac {2 c^4}{3 a^7 x^6}-\frac {4 c^4}{5 a^6 x^5}+\frac {c^4}{a^5 x^4}+\frac {10 c^4}{3 a^4 x^3}+\frac {2 c^4}{a^3 x^2}-\frac {4 c^4}{a^2 x}+c^4 x+\frac {4 c^4 \log (x)}{a} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^4 \, dx=-\frac {c^4}{7 a^8 x^7}-\frac {2 c^4}{3 a^7 x^6}-\frac {4 c^4}{5 a^6 x^5}+\frac {c^4}{a^5 x^4}+\frac {10 c^4}{3 a^4 x^3}+\frac {2 c^4}{a^3 x^2}-\frac {4 c^4}{a^2 x}+c^4 x+\frac {4 c^4 \log (x)}{a} \]

[In]

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

[Out]

-1/7*c^4/(a^8*x^7) - (2*c^4)/(3*a^7*x^6) - (4*c^4)/(5*a^6*x^5) + c^4/(a^5*x^4) + (10*c^4)/(3*a^4*x^3) + (2*c^4
)/(a^3*x^2) - (4*c^4)/(a^2*x) + c^4*x + (4*c^4*Log[x])/a

Maple [A] (verified)

Time = 0.84 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.71

method	result	size
default	\(\frac {c^{4} \left (a^{8} x +4 a^{7} \ln \left (x \right )+\frac {a^{3}}{x^{4}}+\frac {10 a^{4}}{3 x^{3}}-\frac {2 a}{3 x^{6}}+\frac {2 a^{5}}{x^{2}}-\frac {4 a^{6}}{x}-\frac {1}{7 x^{7}}-\frac {4 a^{2}}{5 x^{5}}\right )}{a^{8}}\)	\(71\)
risch	\(c^{4} x +\frac {-4 a^{6} c^{4} x^{6}+2 a^{5} c^{4} x^{5}+\frac {10}{3} a^{4} c^{4} x^{4}+a^{3} c^{4} x^{3}-\frac {4}{5} a^{2} c^{4} x^{2}-\frac {2}{3} a \,c^{4} x -\frac {1}{7} c^{4}}{a^{8} x^{7}}+\frac {4 c^{4} \ln \left (x \right )}{a}\)	\(91\)
parallelrisch	\(\frac {105 a^{8} c^{4} x^{8}+420 c^{4} \ln \left (x \right ) a^{7} x^{7}-420 a^{6} c^{4} x^{6}+210 a^{5} c^{4} x^{5}+350 a^{4} c^{4} x^{4}+105 a^{3} c^{4} x^{3}-84 a^{2} c^{4} x^{2}-70 a \,c^{4} x -15 c^{4}}{105 a^{8} x^{7}}\)	\(101\)
norman	\(\frac {-5 a^{7} c^{4} x^{8}+a^{8} c^{4} x^{9}+\frac {c^{4}}{7 a}+\frac {11 c^{4} x}{21}+\frac {2 a \,c^{4} x^{2}}{15}-\frac {9 a^{2} c^{4} x^{3}}{5}-\frac {7 a^{3} c^{4} x^{4}}{3}+\frac {4 a^{4} c^{4} x^{5}}{3}+6 a^{5} c^{4} x^{6}}{\left (a x -1\right ) a^{7} x^{7}}+\frac {4 c^{4} \ln \left (x \right )}{a}\)	\(115\)
meijerg	\(-\frac {c^{4} \left (-\frac {a x \left (-3 a x +6\right )}{3 \left (-a x +1\right )}-2 \ln \left (-a x +1\right )\right )}{a}-\frac {3 c^{4} x}{-a x +1}-\frac {2 c^{4} \left (-\frac {3 a x}{-3 a x +3}+2 \ln \left (-a x +1\right )-1-2 \ln \left (x \right )-2 \ln \left (-a \right )+\frac {1}{a x}\right )}{a}-\frac {2 c^{4} \left (-\frac {5 a x}{-5 a x +5}+4 \ln \left (-a x +1\right )-1-4 \ln \left (x \right )-4 \ln \left (-a \right )+\frac {1}{3 x^{3} a^{3}}+\frac {1}{a^{2} x^{2}}+\frac {3}{a x}\right )}{a}+\frac {3 c^{4} \left (-\frac {7 a x}{-7 a x +7}+6 \ln \left (-a x +1\right )-1-6 \ln \left (x \right )-6 \ln \left (-a \right )+\frac {1}{5 x^{5} a^{5}}+\frac {1}{2 a^{4} x^{4}}+\frac {1}{x^{3} a^{3}}+\frac {2}{a^{2} x^{2}}+\frac {5}{a x}\right )}{a}+\frac {2 c^{4} \left (\frac {a x}{-a x +1}+\ln \left (-a x +1\right )\right )}{a}-\frac {8 c^{4} \left (\frac {2 a x}{-2 a x +2}-\ln \left (-a x +1\right )+1+\ln \left (x \right )+\ln \left (-a \right )\right )}{a}+\frac {12 c^{4} \left (\frac {4 a x}{-4 a x +4}-3 \ln \left (-a x +1\right )+1+3 \ln \left (x \right )+3 \ln \left (-a \right )-\frac {1}{2 a^{2} x^{2}}-\frac {2}{a x}\right )}{a}-\frac {8 c^{4} \left (\frac {6 a x}{-6 a x +6}-5 \ln \left (-a x +1\right )+1+5 \ln \left (x \right )+5 \ln \left (-a \right )-\frac {1}{4 a^{4} x^{4}}-\frac {2}{3 x^{3} a^{3}}-\frac {3}{2 a^{2} x^{2}}-\frac {4}{a x}\right )}{a}+\frac {2 c^{4} \left (\frac {8 a x}{-8 a x +8}-7 \ln \left (-a x +1\right )+1+7 \ln \left (x \right )+7 \ln \left (-a \right )-\frac {1}{6 a^{6} x^{6}}-\frac {2}{5 x^{5} a^{5}}-\frac {3}{4 a^{4} x^{4}}-\frac {4}{3 x^{3} a^{3}}-\frac {5}{2 a^{2} x^{2}}-\frac {6}{a x}\right )}{a}-\frac {c^{4} \left (-\frac {9 a x}{-9 a x +9}+8 \ln \left (-a x +1\right )-1-8 \ln \left (x \right )-8 \ln \left (-a \right )+\frac {1}{7 x^{7} a^{7}}+\frac {1}{3 a^{6} x^{6}}+\frac {3}{5 x^{5} a^{5}}+\frac {1}{a^{4} x^{4}}+\frac {5}{3 x^{3} a^{3}}+\frac {3}{a^{2} x^{2}}+\frac {7}{a x}\right )}{a}\)	\(623\)

[In]

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

[Out]

c^4/a^8*(a^8*x+4*a^7*ln(x)+a^3/x^4+10/3*a^4/x^3-2/3*a/x^6+2*a^5/x^2-4*a^6/x-1/7/x^7-4/5*a^2/x^5)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^4 \, dx=\frac {105 \, a^{8} c^{4} x^{8} + 420 \, a^{7} c^{4} x^{7} \log \left (x\right ) - 420 \, a^{6} c^{4} x^{6} + 210 \, a^{5} c^{4} x^{5} + 350 \, a^{4} c^{4} x^{4} + 105 \, a^{3} c^{4} x^{3} - 84 \, a^{2} c^{4} x^{2} - 70 \, a c^{4} x - 15 \, c^{4}}{105 \, a^{8} x^{7}} \]

[In]

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

[Out]

1/105*(105*a^8*c^4*x^8 + 420*a^7*c^4*x^7*log(x) - 420*a^6*c^4*x^6 + 210*a^5*c^4*x^5 + 350*a^4*c^4*x^4 + 105*a^
3*c^4*x^3 - 84*a^2*c^4*x^2 - 70*a*c^4*x - 15*c^4)/(a^8*x^7)

Sympy [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^4 \, dx=\frac {a^{8} c^{4} x + 4 a^{7} c^{4} \log {\left (x \right )} + \frac {- 420 a^{6} c^{4} x^{6} + 210 a^{5} c^{4} x^{5} + 350 a^{4} c^{4} x^{4} + 105 a^{3} c^{4} x^{3} - 84 a^{2} c^{4} x^{2} - 70 a c^{4} x - 15 c^{4}}{105 x^{7}}}{a^{8}} \]

[In]

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

[Out]

(a**8*c**4*x + 4*a**7*c**4*log(x) + (-420*a**6*c**4*x**6 + 210*a**5*c**4*x**5 + 350*a**4*c**4*x**4 + 105*a**3*
c**4*x**3 - 84*a**2*c**4*x**2 - 70*a*c**4*x - 15*c**4)/(105*x**7))/a**8

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.92 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^4 \, dx=c^{4} x + \frac {4 \, c^{4} \log \left (x\right )}{a} - \frac {420 \, a^{6} c^{4} x^{6} - 210 \, a^{5} c^{4} x^{5} - 350 \, a^{4} c^{4} x^{4} - 105 \, a^{3} c^{4} x^{3} + 84 \, a^{2} c^{4} x^{2} + 70 \, a c^{4} x + 15 \, c^{4}}{105 \, a^{8} x^{7}} \]

[In]

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

[Out]

c^4*x + 4*c^4*log(x)/a - 1/105*(420*a^6*c^4*x^6 - 210*a^5*c^4*x^5 - 350*a^4*c^4*x^4 - 105*a^3*c^4*x^3 + 84*a^2
*c^4*x^2 + 70*a*c^4*x + 15*c^4)/(a^8*x^7)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.60 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^4 \, dx=-\frac {4 \, c^{4} \log \left (\frac {{\left | a x - 1 \right |}}{{\left (a x - 1\right )}^{2} {\left | a \right |}}\right )}{a} + \frac {4 \, c^{4} \log \left ({\left | -\frac {1}{a x - 1} - 1 \right |}\right )}{a} + \frac {{\left (105 \, c^{4} + \frac {659 \, c^{4}}{a x - 1} + \frac {1253 \, c^{4}}{{\left (a x - 1\right )}^{2}} - \frac {231 \, c^{4}}{{\left (a x - 1\right )}^{3}} - \frac {3885 \, c^{4}}{{\left (a x - 1\right )}^{4}} - \frac {5250 \, c^{4}}{{\left (a x - 1\right )}^{5}} - \frac {2730 \, c^{4}}{{\left (a x - 1\right )}^{6}} - \frac {420 \, c^{4}}{{\left (a x - 1\right )}^{7}}\right )} {\left (a x - 1\right )}}{105 \, a {\left (\frac {1}{a x - 1} + 1\right )}^{7}} \]

[In]

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

[Out]

-4*c^4*log(abs(a*x - 1)/((a*x - 1)^2*abs(a)))/a + 4*c^4*log(abs(-1/(a*x - 1) - 1))/a + 1/105*(105*c^4 + 659*c^
4/(a*x - 1) + 1253*c^4/(a*x - 1)^2 - 231*c^4/(a*x - 1)^3 - 3885*c^4/(a*x - 1)^4 - 5250*c^4/(a*x - 1)^5 - 2730*
c^4/(a*x - 1)^6 - 420*c^4/(a*x - 1)^7)*(a*x - 1)/(a*(1/(a*x - 1) + 1)^7)

Mupad [B] (verification not implemented)

Time = 4.24 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.72 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^4 \, dx=\frac {c^4\,\left (a^3\,x^3-\frac {4\,a^2\,x^2}{5}-\frac {2\,a\,x}{3}+\frac {10\,a^4\,x^4}{3}+2\,a^5\,x^5-4\,a^6\,x^6+a^8\,x^8+4\,a^7\,x^7\,\ln \left (x\right )-\frac {1}{7}\right )}{a^8\,x^7} \]

[In]

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

[Out]

(c^4*(a^3*x^3 - (4*a^2*x^2)/5 - (2*a*x)/3 + (10*a^4*x^4)/3 + 2*a^5*x^5 - 4*a^6*x^6 + a^8*x^8 + 4*a^7*x^7*log(x
) - 1/7))/(a^8*x^7)