\(\int e^{4 \coth ^{-1}(a x)} (c-\frac {c}{a^2 x^2})^3 \, dx\) [799]

Optimal result

Integrand size = 22, antiderivative size = 63 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^3 \, dx=\frac {c^3}{5 a^6 x^5}+\frac {c^3}{a^5 x^4}+\frac {5 c^3}{3 a^4 x^3}-\frac {5 c^3}{a^2 x}+c^3 x+\frac {4 c^3 \log (x)}{a} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 63, 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, 76} \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^3 \, dx=\frac {c^3}{5 a^6 x^5}+\frac {c^3}{a^5 x^4}+\frac {5 c^3}{3 a^4 x^3}-\frac {5 c^3}{a^2 x}+\frac {4 c^3 \log (x)}{a}+c^3 x \]

[In]

Int[E^(4*ArcCoth[a*x])*(c - c/(a^2*x^2))^3,x]

[Out]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^3 \, dx=\frac {c^3}{5 a^6 x^5}+\frac {c^3}{a^5 x^4}+\frac {5 c^3}{3 a^4 x^3}-\frac {5 c^3}{a^2 x}+c^3 x+\frac {4 c^3 \log (x)}{a} \]

[In]

Integrate[E^(4*ArcCoth[a*x])*(c - c/(a^2*x^2))^3,x]

[Out]

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

Maple [A] (verified)

Time = 0.73 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.75

method	result	size
default	\(\frac {c^{3} \left (a^{6} x +4 a^{5} \ln \left (x \right )+\frac {a}{x^{4}}+\frac {5 a^{2}}{3 x^{3}}-\frac {5 a^{4}}{x}+\frac {1}{5 x^{5}}\right )}{a^{6}}\)	\(47\)
risch	\(c^{3} x +\frac {-5 a^{4} c^{3} x^{4}+\frac {5}{3} a^{2} c^{3} x^{2}+a \,c^{3} x +\frac {1}{5} c^{3}}{a^{6} x^{5}}+\frac {4 c^{3} \ln \left (x \right )}{a}\)	\(58\)
parallelrisch	\(\frac {15 a^{6} c^{3} x^{6}+60 c^{3} \ln \left (x \right ) a^{5} x^{5}-75 a^{4} c^{3} x^{4}+25 a^{2} c^{3} x^{2}+15 a \,c^{3} x +3 c^{3}}{15 a^{6} x^{5}}\)	\(68\)
norman	\(\frac {-6 a^{5} c^{3} x^{6}+a^{6} c^{3} x^{7}-\frac {c^{3}}{5 a}-\frac {4 c^{3} x}{5}-\frac {2 a \,c^{3} x^{2}}{3}+\frac {5 a^{2} c^{3} x^{3}}{3}+5 a^{3} c^{3} x^{4}}{\left (a x -1\right ) a^{5} x^{5}}+\frac {4 c^{3} \ln \left (x \right )}{a}\)	\(93\)
meijerg	\(-\frac {c^{3} \left (-\frac {a x \left (-3 a x +6\right )}{3 \left (-a x +1\right )}-2 \ln \left (-a x +1\right )\right )}{a}-\frac {2 c^{3} x}{-a x +1}-\frac {2 c^{3} \left (-\frac {5 a x}{-5 a x +5}+4 \ln \left (-a x +1\right )-1-4 \ln \left (x \right )-4 \ln \left (-a \right )+\frac {1}{3 x^{3} a^{3}}+\frac {1}{a^{2} x^{2}}+\frac {3}{a x}\right )}{a}+\frac {2 c^{3} \left (\frac {a x}{-a x +1}+\ln \left (-a x +1\right )\right )}{a}-\frac {6 c^{3} \left (\frac {2 a x}{-2 a x +2}-\ln \left (-a x +1\right )+1+\ln \left (x \right )+\ln \left (-a \right )\right )}{a}+\frac {6 c^{3} \left (\frac {4 a x}{-4 a x +4}-3 \ln \left (-a x +1\right )+1+3 \ln \left (x \right )+3 \ln \left (-a \right )-\frac {1}{2 a^{2} x^{2}}-\frac {2}{a x}\right )}{a}-\frac {2 c^{3} \left (\frac {6 a x}{-6 a x +6}-5 \ln \left (-a x +1\right )+1+5 \ln \left (x \right )+5 \ln \left (-a \right )-\frac {1}{4 a^{4} x^{4}}-\frac {2}{3 x^{3} a^{3}}-\frac {3}{2 a^{2} x^{2}}-\frac {4}{a x}\right )}{a}+\frac {c^{3} \left (-\frac {7 a x}{-7 a x +7}+6 \ln \left (-a x +1\right )-1-6 \ln \left (x \right )-6 \ln \left (-a \right )+\frac {1}{5 x^{5} a^{5}}+\frac {1}{2 a^{4} x^{4}}+\frac {1}{x^{3} a^{3}}+\frac {2}{a^{2} x^{2}}+\frac {5}{a x}\right )}{a}\)	\(389\)

[In]

int(1/(a*x-1)^2*(a*x+1)^2*(c-c/a^2/x^2)^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.06 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^3 \, dx=\frac {15 \, a^{6} c^{3} x^{6} + 60 \, a^{5} c^{3} x^{5} \log \left (x\right ) - 75 \, a^{4} c^{3} x^{4} + 25 \, a^{2} c^{3} x^{2} + 15 \, a c^{3} x + 3 \, c^{3}}{15 \, a^{6} x^{5}} \]

[In]

integrate(1/(a*x-1)^2*(a*x+1)^2*(c-c/a^2/x^2)^3,x, algorithm="fricas")

[Out]

1/15*(15*a^6*c^3*x^6 + 60*a^5*c^3*x^5*log(x) - 75*a^4*c^3*x^4 + 25*a^2*c^3*x^2 + 15*a*c^3*x + 3*c^3)/(a^6*x^5)

Sympy [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.03 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^3 \, dx=\frac {a^{6} c^{3} x + 4 a^{5} c^{3} \log {\left (x \right )} + \frac {- 75 a^{4} c^{3} x^{4} + 25 a^{2} c^{3} x^{2} + 15 a c^{3} x + 3 c^{3}}{15 x^{5}}}{a^{6}} \]

[In]

integrate(1/(a*x-1)**2*(a*x+1)**2*(c-c/a**2/x**2)**3,x)

[Out]

(a**6*c**3*x + 4*a**5*c**3*log(x) + (-75*a**4*c**3*x**4 + 25*a**2*c**3*x**2 + 15*a*c**3*x + 3*c**3)/(15*x**5))
/a**6

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.94 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^3 \, dx=c^{3} x + \frac {4 \, c^{3} \log \left (x\right )}{a} - \frac {75 \, a^{4} c^{3} x^{4} - 25 \, a^{2} c^{3} x^{2} - 15 \, a c^{3} x - 3 \, c^{3}}{15 \, a^{6} x^{5}} \]

[In]

integrate(1/(a*x-1)^2*(a*x+1)^2*(c-c/a^2/x^2)^3,x, algorithm="maxima")

[Out]

c^3*x + 4*c^3*log(x)/a - 1/15*(75*a^4*c^3*x^4 - 25*a^2*c^3*x^2 - 15*a*c^3*x - 3*c^3)/(a^6*x^5)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 136 vs. \(2 (59) = 118\).

Time = 0.27 (sec) , antiderivative size = 136, normalized size of antiderivative = 2.16 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^3 \, dx=-\frac {4 \, c^{3} \log \left (\frac {{\left | a x - 1 \right |}}{{\left (a x - 1\right )}^{2} {\left | a \right |}}\right )}{a} + \frac {4 \, c^{3} \log \left ({\left | -\frac {1}{a x - 1} - 1 \right |}\right )}{a} + \frac {{\left (15 \, c^{3} + \frac {107 \, c^{3}}{a x - 1} + \frac {235 \, c^{3}}{{\left (a x - 1\right )}^{2}} + \frac {170 \, c^{3}}{{\left (a x - 1\right )}^{3}} - \frac {30 \, c^{3}}{{\left (a x - 1\right )}^{4}} - \frac {60 \, c^{3}}{{\left (a x - 1\right )}^{5}}\right )} {\left (a x - 1\right )}}{15 \, a {\left (\frac {1}{a x - 1} + 1\right )}^{5}} \]

[In]

integrate(1/(a*x-1)^2*(a*x+1)^2*(c-c/a^2/x^2)^3,x, algorithm="giac")

[Out]

-4*c^3*log(abs(a*x - 1)/((a*x - 1)^2*abs(a)))/a + 4*c^3*log(abs(-1/(a*x - 1) - 1))/a + 1/15*(15*c^3 + 107*c^3/
(a*x - 1) + 235*c^3/(a*x - 1)^2 + 170*c^3/(a*x - 1)^3 - 30*c^3/(a*x - 1)^4 - 60*c^3/(a*x - 1)^5)*(a*x - 1)/(a*
(1/(a*x - 1) + 1)^5)

Mupad [B] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.76 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^3 \, dx=\frac {c^3\,\left (a\,x+\frac {5\,a^2\,x^2}{3}-5\,a^4\,x^4+a^6\,x^6+4\,a^5\,x^5\,\ln \left (x\right )+\frac {1}{5}\right )}{a^6\,x^5} \]

[In]

int(((c - c/(a^2*x^2))^3*(a*x + 1)^2)/(a*x - 1)^2,x)

[Out]

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