\(\int e^{-2 \coth ^{-1}(a x)} (c-\frac {c}{a x})^4 \, dx\) [421]

Optimal result

Integrand size = 22, antiderivative size = 65 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^4 \, dx=\frac {c^4}{3 a^4 x^3}-\frac {3 c^4}{a^3 x^2}+\frac {16 c^4}{a^2 x}+c^4 x+\frac {26 c^4 \log (x)}{a}-\frac {32 c^4 \log (1+a x)}{a} \]

[Out]

1/3*c^4/a^4/x^3-3*c^4/a^3/x^2+16*c^4/a^2/x+c^4*x+26*c^4*ln(x)/a-32*c^4*ln(a*x+1)/a

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 65, 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, 6266, 6264, 90} \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^4 \, dx=\frac {c^4}{3 a^4 x^3}-\frac {3 c^4}{a^3 x^2}+\frac {16 c^4}{a^2 x}+\frac {26 c^4 \log (x)}{a}-\frac {32 c^4 \log (a x+1)}{a}+c^4 x \]

[In]

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

[Out]

c^4/(3*a^4*x^3) - (3*c^4)/(a^3*x^2) + (16*c^4)/(a^2*x) + c^4*x + (26*c^4*Log[x])/a - (32*c^4*Log[1 + a*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 6264

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

Rule 6266

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_))^(p_.), x_Symbol] :> Dist[d^p, Int[u*(1 + c*(x/d))^
p*(E^(n*ArcTanh[a*x])/x^p), x], x] /; FreeQ[{a, c, d, n}, x] && EqQ[c^2 - a^2*d^2, 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 x}\right )^4 \, dx \\ & = -\frac {c^4 \int \frac {e^{-2 \text {arctanh}(a x)} (1-a x)^4}{x^4} \, dx}{a^4} \\ & = -\frac {c^4 \int \frac {(1-a x)^5}{x^4 (1+a x)} \, dx}{a^4} \\ & = -\frac {c^4 \int \left (-a^4+\frac {1}{x^4}-\frac {6 a}{x^3}+\frac {16 a^2}{x^2}-\frac {26 a^3}{x}+\frac {32 a^4}{1+a x}\right ) \, dx}{a^4} \\ & = \frac {c^4}{3 a^4 x^3}-\frac {3 c^4}{a^3 x^2}+\frac {16 c^4}{a^2 x}+c^4 x+\frac {26 c^4 \log (x)}{a}-\frac {32 c^4 \log (1+a x)}{a} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.83 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^4 \, dx=-\frac {c^4 \left (-\frac {1}{3 x^3}+\frac {3 a}{x^2}-\frac {16 a^2}{x}-a^4 x-26 a^3 \log (x)+32 a^3 \log (1+a x)\right )}{a^4} \]

[In]

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

[Out]

-((c^4*(-1/3*1/x^3 + (3*a)/x^2 - (16*a^2)/x - a^4*x - 26*a^3*Log[x] + 32*a^3*Log[1 + a*x]))/a^4)

Maple [A] (verified)

Time = 0.65 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.78

method	result	size
default	\(\frac {c^{4} \left (-32 a^{3} \ln \left (a x +1\right )+a^{4} x +\frac {1}{3 x^{3}}-\frac {3 a}{x^{2}}+\frac {16 a^{2}}{x}+26 a^{3} \ln \left (x \right )\right )}{a^{4}}\)	\(51\)
risch	\(c^{4} x +\frac {16 a^{2} c^{4} x^{2}-3 a \,c^{4} x +\frac {1}{3} c^{4}}{a^{4} x^{3}}+\frac {26 c^{4} \ln \left (-x \right )}{a}-\frac {32 c^{4} \ln \left (a x +1\right )}{a}\)	\(64\)
norman	\(\frac {a^{3} c^{4} x^{4}+\frac {c^{4}}{3 a}-3 c^{4} x +16 a \,c^{4} x^{2}}{a^{3} x^{3}}+\frac {26 c^{4} \ln \left (x \right )}{a}-\frac {32 c^{4} \ln \left (a x +1\right )}{a}\)	\(67\)
parallelrisch	\(\frac {3 a^{4} c^{4} x^{4}+78 c^{4} \ln \left (x \right ) a^{3} x^{3}-96 c^{4} \ln \left (a x +1\right ) a^{3} x^{3}+48 a^{2} c^{4} x^{2}-9 a \,c^{4} x +c^{4}}{3 a^{4} x^{3}}\)	\(72\)
meijerg	\(\frac {c^{4} \left (a x -\ln \left (a x +1\right )\right )}{a}-\frac {5 c^{4} \ln \left (a x +1\right )}{a}+\frac {10 c^{4} \left (-\ln \left (a x +1\right )+\ln \left (x \right )+\ln \left (a \right )\right )}{a}-\frac {10 c^{4} \left (\ln \left (a x +1\right )-\ln \left (x \right )-\ln \left (a \right )-\frac {1}{a x}\right )}{a}+\frac {5 c^{4} \left (-\ln \left (a x +1\right )+\ln \left (x \right )+\ln \left (a \right )-\frac {1}{2 a^{2} x^{2}}+\frac {1}{a x}\right )}{a}-\frac {c^{4} \left (\ln \left (a x +1\right )-\ln \left (x \right )-\ln \left (a \right )-\frac {1}{3 x^{3} a^{3}}+\frac {1}{2 a^{2} x^{2}}-\frac {1}{a x}\right )}{a}\)	\(170\)

[In]

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

[Out]

c^4/a^4*(-32*a^3*ln(a*x+1)+a^4*x+1/3/x^3-3*a/x^2+16*a^2/x+26*a^3*ln(x))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.09 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^4 \, dx=\frac {3 \, a^{4} c^{4} x^{4} - 96 \, a^{3} c^{4} x^{3} \log \left (a x + 1\right ) + 78 \, a^{3} c^{4} x^{3} \log \left (x\right ) + 48 \, a^{2} c^{4} x^{2} - 9 \, a c^{4} x + c^{4}}{3 \, a^{4} x^{3}} \]

[In]

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

[Out]

1/3*(3*a^4*c^4*x^4 - 96*a^3*c^4*x^3*log(a*x + 1) + 78*a^3*c^4*x^3*log(x) + 48*a^2*c^4*x^2 - 9*a*c^4*x + c^4)/(
a^4*x^3)

Sympy [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.86 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^4 \, dx=c^{4} x + \frac {2 c^{4} \cdot \left (13 \log {\left (x \right )} - 16 \log {\left (x + \frac {1}{a} \right )}\right )}{a} + \frac {48 a^{2} c^{4} x^{2} - 9 a c^{4} x + c^{4}}{3 a^{4} x^{3}} \]

[In]

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

[Out]

c**4*x + 2*c**4*(13*log(x) - 16*log(x + 1/a))/a + (48*a**2*c**4*x**2 - 9*a*c**4*x + c**4)/(3*a**4*x**3)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.92 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^4 \, dx=c^{4} x - \frac {32 \, c^{4} \log \left (a x + 1\right )}{a} + \frac {26 \, c^{4} \log \left (x\right )}{a} + \frac {48 \, a^{2} c^{4} x^{2} - 9 \, a c^{4} x + c^{4}}{3 \, a^{4} x^{3}} \]

[In]

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

[Out]

c^4*x - 32*c^4*log(a*x + 1)/a + 26*c^4*log(x)/a + 1/3*(48*a^2*c^4*x^2 - 9*a*c^4*x + c^4)/(a^4*x^3)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.95 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^4 \, dx=c^{4} x - \frac {32 \, c^{4} \log \left ({\left | a x + 1 \right |}\right )}{a} + \frac {26 \, c^{4} \log \left ({\left | x \right |}\right )}{a} + \frac {48 \, a^{2} c^{4} x^{2} - 9 \, a c^{4} x + c^{4}}{3 \, a^{4} x^{3}} \]

[In]

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

[Out]

c^4*x - 32*c^4*log(abs(a*x + 1))/a + 26*c^4*log(abs(x))/a + 1/3*(48*a^2*c^4*x^2 - 9*a*c^4*x + c^4)/(a^4*x^3)

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.94 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^4 \, dx=c^4\,x+\frac {16\,a^2\,c^4\,x^2-3\,a\,c^4\,x+\frac {c^4}{3}}{a^4\,x^3}+\frac {26\,c^4\,\ln \left (x\right )}{a}-\frac {32\,c^4\,\ln \left (a\,x+1\right )}{a} \]

[In]

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

[Out]

c^4*x + (c^4/3 + 16*a^2*c^4*x^2 - 3*a*c^4*x)/(a^4*x^3) + (26*c^4*log(x))/a - (32*c^4*log(a*x + 1))/a