\(\int e^{-2 \coth ^{-1}(a x)} (c-\frac {c}{a^2 x^2})^4 \, dx\) [814]

Optimal result

Integrand size = 22, antiderivative size = 90 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^4 \, dx=\frac {c^4}{7 a^8 x^7}-\frac {c^4}{3 a^7 x^6}-\frac {2 c^4}{5 a^6 x^5}+\frac {3 c^4}{2 a^5 x^4}-\frac {3 c^4}{a^3 x^2}+\frac {2 c^4}{a^2 x}+c^4 x-\frac {2 c^4 \log (x)}{a} \]

[Out]

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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 90, 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} \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^4 \, dx=\frac {c^4}{7 a^8 x^7}-\frac {c^4}{3 a^7 x^6}-\frac {2 c^4}{5 a^6 x^5}+\frac {3 c^4}{2 a^5 x^4}-\frac {3 c^4}{a^3 x^2}+\frac {2 c^4}{a^2 x}-\frac {2 c^4 \log (x)}{a}+c^4 x \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^4 \, dx=\frac {c^4}{7 a^8 x^7}-\frac {c^4}{3 a^7 x^6}-\frac {2 c^4}{5 a^6 x^5}+\frac {3 c^4}{2 a^5 x^4}-\frac {3 c^4}{a^3 x^2}+\frac {2 c^4}{a^2 x}+c^4 x-\frac {2 c^4 \log (x)}{a} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.78 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.71

method	result	size
default	\(\frac {c^{4} \left (a^{8} x -2 a^{7} \ln \left (x \right )+\frac {3 a^{3}}{2 x^{4}}-\frac {a}{3 x^{6}}-\frac {3 a^{5}}{x^{2}}+\frac {2 a^{6}}{x}+\frac {1}{7 x^{7}}-\frac {2 a^{2}}{5 x^{5}}\right )}{a^{8}}\)	\(64\)
risch	\(c^{4} x +\frac {2 a^{6} c^{4} x^{6}-3 a^{5} c^{4} x^{5}+\frac {3}{2} a^{3} c^{4} x^{3}-\frac {2}{5} a^{2} c^{4} x^{2}-\frac {1}{3} a \,c^{4} x +\frac {1}{7} c^{4}}{a^{8} x^{7}}-\frac {2 c^{4} \ln \left (x \right )}{a}\)	\(81\)
norman	\(\frac {a^{7} c^{4} x^{8}+\frac {c^{4}}{7 a}-\frac {c^{4} x}{3}-\frac {2 a \,c^{4} x^{2}}{5}+\frac {3 a^{2} c^{4} x^{3}}{2}-3 a^{4} c^{4} x^{5}+2 a^{5} c^{4} x^{6}}{a^{7} x^{7}}-\frac {2 c^{4} \ln \left (x \right )}{a}\)	\(86\)
parallelrisch	\(-\frac {-210 a^{8} c^{4} x^{8}+420 c^{4} \ln \left (x \right ) a^{7} x^{7}-420 a^{6} c^{4} x^{6}+630 a^{5} c^{4} x^{5}-315 a^{3} c^{4} x^{3}+84 a^{2} c^{4} x^{2}+70 a \,c^{4} x -30 c^{4}}{210 a^{8} x^{7}}\)	\(90\)
meijerg	\(\frac {c^{4} \left (a x -\ln \left (a x +1\right )\right )}{a}-\frac {c^{4} \ln \left (a x +1\right )}{a}-\frac {4 c^{4} \left (-\ln \left (a x +1\right )+\ln \left (x \right )+\ln \left (a \right )\right )}{a}+\frac {4 c^{4} \left (\ln \left (a x +1\right )-\ln \left (x \right )-\ln \left (a \right )-\frac {1}{a x}\right )}{a}+\frac {6 c^{4} \left (-\ln \left (a x +1\right )+\ln \left (x \right )+\ln \left (a \right )-\frac {1}{2 a^{2} x^{2}}+\frac {1}{a x}\right )}{a}-\frac {6 c^{4} \left (\ln \left (a x +1\right )-\ln \left (x \right )-\ln \left (a \right )-\frac {1}{3 x^{3} a^{3}}+\frac {1}{2 a^{2} x^{2}}-\frac {1}{a x}\right )}{a}-\frac {4 c^{4} \left (-\ln \left (a x +1\right )+\ln \left (x \right )+\ln \left (a \right )-\frac {1}{4 a^{4} x^{4}}+\frac {1}{3 x^{3} a^{3}}-\frac {1}{2 a^{2} x^{2}}+\frac {1}{a x}\right )}{a}+\frac {4 c^{4} \left (\ln \left (a x +1\right )-\ln \left (x \right )-\ln \left (a \right )-\frac {1}{5 x^{5} a^{5}}+\frac {1}{4 a^{4} x^{4}}-\frac {1}{3 x^{3} a^{3}}+\frac {1}{2 a^{2} x^{2}}-\frac {1}{a x}\right )}{a}+\frac {c^{4} \left (-\ln \left (a x +1\right )+\ln \left (x \right )+\ln \left (a \right )-\frac {1}{6 a^{6} x^{6}}+\frac {1}{5 x^{5} a^{5}}-\frac {1}{4 a^{4} x^{4}}+\frac {1}{3 x^{3} a^{3}}-\frac {1}{2 a^{2} x^{2}}+\frac {1}{a x}\right )}{a}-\frac {c^{4} \left (\ln \left (a x +1\right )-\ln \left (x \right )-\ln \left (a \right )-\frac {1}{7 x^{7} a^{7}}+\frac {1}{6 a^{6} x^{6}}-\frac {1}{5 x^{5} a^{5}}+\frac {1}{4 a^{4} x^{4}}-\frac {1}{3 x^{3} a^{3}}+\frac {1}{2 a^{2} x^{2}}-\frac {1}{a x}\right )}{a}\)	\(431\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.99 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^4 \, dx=\frac {210 \, a^{8} c^{4} x^{8} - 420 \, a^{7} c^{4} x^{7} \log \left (x\right ) + 420 \, a^{6} c^{4} x^{6} - 630 \, a^{5} c^{4} x^{5} + 315 \, a^{3} c^{4} x^{3} - 84 \, a^{2} c^{4} x^{2} - 70 \, a c^{4} x + 30 \, c^{4}}{210 \, a^{8} x^{7}} \]

[In]

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

[Out]

1/210*(210*a^8*c^4*x^8 - 420*a^7*c^4*x^7*log(x) + 420*a^6*c^4*x^6 - 630*a^5*c^4*x^5 + 315*a^3*c^4*x^3 - 84*a^2
*c^4*x^2 - 70*a*c^4*x + 30*c^4)/(a^8*x^7)

Sympy [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.98 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^4 \, dx=\frac {a^{8} c^{4} x - 2 a^{7} c^{4} \log {\left (x \right )} + \frac {420 a^{6} c^{4} x^{6} - 630 a^{5} c^{4} x^{5} + 315 a^{3} c^{4} x^{3} - 84 a^{2} c^{4} x^{2} - 70 a c^{4} x + 30 c^{4}}{210 x^{7}}}{a^{8}} \]

[In]

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

[Out]

(a**8*c**4*x - 2*a**7*c**4*log(x) + (420*a**6*c**4*x**6 - 630*a**5*c**4*x**5 + 315*a**3*c**4*x**3 - 84*a**2*c*
*4*x**2 - 70*a*c**4*x + 30*c**4)/(210*x**7))/a**8

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.90 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^4 \, dx=c^{4} x - \frac {2 \, c^{4} \log \left (x\right )}{a} + \frac {420 \, a^{6} c^{4} x^{6} - 630 \, a^{5} c^{4} x^{5} + 315 \, a^{3} c^{4} x^{3} - 84 \, a^{2} c^{4} x^{2} - 70 \, a c^{4} x + 30 \, c^{4}}{210 \, a^{8} x^{7}} \]

[In]

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

[Out]

c^4*x - 2*c^4*log(x)/a + 1/210*(420*a^6*c^4*x^6 - 630*a^5*c^4*x^5 + 315*a^3*c^4*x^3 - 84*a^2*c^4*x^2 - 70*a*c^
4*x + 30*c^4)/(a^8*x^7)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.91 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^4 \, dx=c^{4} x - \frac {2 \, c^{4} \log \left ({\left | x \right |}\right )}{a} + \frac {420 \, a^{6} c^{4} x^{6} - 630 \, a^{5} c^{4} x^{5} + 315 \, a^{3} c^{4} x^{3} - 84 \, a^{2} c^{4} x^{2} - 70 \, a c^{4} x + 30 \, c^{4}}{210 \, a^{8} x^{7}} \]

[In]

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

[Out]

c^4*x - 2*c^4*log(abs(x))/a + 1/210*(420*a^6*c^4*x^6 - 630*a^5*c^4*x^5 + 315*a^3*c^4*x^3 - 84*a^2*c^4*x^2 - 70
*a*c^4*x + 30*c^4)/(a^8*x^7)

Mupad [B] (verification not implemented)

Time = 4.03 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.74 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^4 \, dx=-\frac {c^4\,\left (\frac {a\,x}{3}+\frac {2\,a^2\,x^2}{5}-\frac {3\,a^3\,x^3}{2}+3\,a^5\,x^5-2\,a^6\,x^6-a^8\,x^8+2\,a^7\,x^7\,\ln \left (x\right )-\frac {1}{7}\right )}{a^8\,x^7} \]

[In]

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

[Out]

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