\(\int e^{2 \coth ^{-1}(a x)} (c-\frac {c}{a x})^5 \, dx\) [387]

Optimal result

Integrand size = 22, antiderivative size = 61 \[ \int e^{2 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^5 \, dx=-\frac {c^5}{4 a^5 x^4}+\frac {c^5}{a^4 x^3}-\frac {c^5}{a^3 x^2}-\frac {2 c^5}{a^2 x}+c^5 x-\frac {3 c^5 \log (x)}{a} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 61, 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, 76} \[ \int e^{2 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^5 \, dx=-\frac {c^5}{4 a^5 x^4}+\frac {c^5}{a^4 x^3}-\frac {c^5}{a^3 x^2}-\frac {2 c^5}{a^2 x}-\frac {3 c^5 \log (x)}{a}+c^5 x \]

[In]

Int[E^(2*ArcCoth[a*x])*(c - c/(a*x))^5,x]

[Out]

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

Rule 76

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] &&  !(ILtQ[n
+ p + 2, 0] && GtQ[n + 2*p, 0])

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

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.93 \[ \int e^{2 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^5 \, dx=-\frac {c^5 \left (\frac {5 a^4}{4}+\frac {1}{4 x^4}-\frac {a}{x^3}+\frac {a^2}{x^2}+\frac {2 a^3}{x}-a^5 x+3 a^4 \log (x)\right )}{a^5} \]

[In]

Integrate[E^(2*ArcCoth[a*x])*(c - c/(a*x))^5,x]

[Out]

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

Maple [A] (verified)

Time = 0.64 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.77

method	result	size
default	\(\frac {c^{5} \left (a^{5} x -3 a^{4} \ln \left (x \right )-\frac {1}{4 x^{4}}+\frac {a}{x^{3}}-\frac {a^{2}}{x^{2}}-\frac {2 a^{3}}{x}\right )}{a^{5}}\)	\(47\)
risch	\(c^{5} x +\frac {-2 a^{3} c^{5} x^{3}-a^{2} c^{5} x^{2}+a \,c^{5} x -\frac {1}{4} c^{5}}{a^{5} x^{4}}-\frac {3 c^{5} \ln \left (x \right )}{a}\)	\(58\)
norman	\(\frac {c^{5} x +a^{4} c^{5} x^{5}-\frac {c^{5}}{4 a}-a \,c^{5} x^{2}-2 c^{5} a^{2} x^{3}}{a^{4} x^{4}}-\frac {3 c^{5} \ln \left (x \right )}{a}\)	\(63\)
parallelrisch	\(-\frac {-4 a^{5} c^{5} x^{5}+12 c^{5} \ln \left (x \right ) a^{4} x^{4}+8 a^{3} c^{5} x^{3}+4 a^{2} c^{5} x^{2}-4 a \,c^{5} x +c^{5}}{4 a^{5} x^{4}}\)	\(66\)
meijerg	\(-\frac {c^{5} \left (-a x -\ln \left (-a x +1\right )\right )}{a}-\frac {4 c^{5} \ln \left (-a x +1\right )}{a}-\frac {5 c^{5} \left (-\ln \left (-a x +1\right )+\ln \left (x \right )+\ln \left (-a \right )\right )}{a}+\frac {5 c^{5} \left (-\ln \left (-a x +1\right )+\ln \left (x \right )+\ln \left (-a \right )-\frac {1}{2 a^{2} x^{2}}-\frac {1}{a x}\right )}{a}+\frac {4 c^{5} \left (\ln \left (-a x +1\right )-\ln \left (x \right )-\ln \left (-a \right )+\frac {1}{3 x^{3} a^{3}}+\frac {1}{2 a^{2} x^{2}}+\frac {1}{a x}\right )}{a}+\frac {c^{5} \left (-\ln \left (-a x +1\right )+\ln \left (x \right )+\ln \left (-a \right )-\frac {1}{4 a^{4} x^{4}}-\frac {1}{3 x^{3} a^{3}}-\frac {1}{2 a^{2} x^{2}}-\frac {1}{a x}\right )}{a}\)	\(207\)

[In]

int(1/(a*x-1)*(a*x+1)*(c-c/a/x)^5,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.10 \[ \int e^{2 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^5 \, dx=\frac {4 \, a^{5} c^{5} x^{5} - 12 \, a^{4} c^{5} x^{4} \log \left (x\right ) - 8 \, a^{3} c^{5} x^{3} - 4 \, a^{2} c^{5} x^{2} + 4 \, a c^{5} x - c^{5}}{4 \, a^{5} x^{4}} \]

[In]

integrate(1/(a*x-1)*(a*x+1)*(c-c/a/x)^5,x, algorithm="fricas")

[Out]

1/4*(4*a^5*c^5*x^5 - 12*a^4*c^5*x^4*log(x) - 8*a^3*c^5*x^3 - 4*a^2*c^5*x^2 + 4*a*c^5*x - c^5)/(a^5*x^4)

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.03 \[ \int e^{2 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^5 \, dx=\frac {a^{5} c^{5} x - 3 a^{4} c^{5} \log {\left (x \right )} + \frac {- 8 a^{3} c^{5} x^{3} - 4 a^{2} c^{5} x^{2} + 4 a c^{5} x - c^{5}}{4 x^{4}}}{a^{5}} \]

[In]

integrate(1/(a*x-1)*(a*x+1)*(c-c/a/x)**5,x)

[Out]

(a**5*c**5*x - 3*a**4*c**5*log(x) + (-8*a**3*c**5*x**3 - 4*a**2*c**5*x**2 + 4*a*c**5*x - c**5)/(4*x**4))/a**5

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.93 \[ \int e^{2 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^5 \, dx=c^{5} x - \frac {3 \, c^{5} \log \left (x\right )}{a} - \frac {8 \, a^{3} c^{5} x^{3} + 4 \, a^{2} c^{5} x^{2} - 4 \, a c^{5} x + c^{5}}{4 \, a^{5} x^{4}} \]

[In]

integrate(1/(a*x-1)*(a*x+1)*(c-c/a/x)^5,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.95 \[ \int e^{2 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^5 \, dx=c^{5} x - \frac {3 \, c^{5} \log \left ({\left | x \right |}\right )}{a} - \frac {8 \, a^{3} c^{5} x^{3} + 4 \, a^{2} c^{5} x^{2} - 4 \, a c^{5} x + c^{5}}{4 \, a^{5} x^{4}} \]

[In]

integrate(1/(a*x-1)*(a*x+1)*(c-c/a/x)^5,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.84 \[ \int e^{2 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^5 \, dx=-\frac {c^5\,\left (4\,a^2\,x^2-4\,a\,x+8\,a^3\,x^3-4\,a^5\,x^5+12\,a^4\,x^4\,\ln \left (x\right )+1\right )}{4\,a^5\,x^4} \]

[In]

int(((c - c/(a*x))^5*(a*x + 1))/(a*x - 1),x)

[Out]

-(c^5*(4*a^2*x^2 - 4*a*x + 8*a^3*x^3 - 4*a^5*x^5 + 12*a^4*x^4*log(x) + 1))/(4*a^5*x^4)