\(\int e^{4 \coth ^{-1}(a x)} (c-\frac {c}{a^2 x^2})^2 \, dx\) [800]

Optimal result

Integrand size = 22, antiderivative size = 51 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^2 \, dx=-\frac {c^2}{3 a^4 x^3}-\frac {2 c^2}{a^3 x^2}-\frac {6 c^2}{a^2 x}+c^2 x+\frac {4 c^2 \log (x)}{a} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 51, 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, 45} \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^2 \, dx=-\frac {c^2}{3 a^4 x^3}-\frac {2 c^2}{a^3 x^2}-\frac {6 c^2}{a^2 x}+\frac {4 c^2 \log (x)}{a}+c^2 x \]

[In]

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

[Out]

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 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 )^2 \, dx \\ & = \frac {c^2 \int \frac {e^{4 \text {arctanh}(a x)} \left (1-a^2 x^2\right )^2}{x^4} \, dx}{a^4} \\ & = \frac {c^2 \int \frac {(1+a x)^4}{x^4} \, dx}{a^4} \\ & = \frac {c^2 \int \left (a^4+\frac {1}{x^4}+\frac {4 a}{x^3}+\frac {6 a^2}{x^2}+\frac {4 a^3}{x}\right ) \, dx}{a^4} \\ & = -\frac {c^2}{3 a^4 x^3}-\frac {2 c^2}{a^3 x^2}-\frac {6 c^2}{a^2 x}+c^2 x+\frac {4 c^2 \log (x)}{a} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^2 \, dx=-\frac {c^2}{3 a^4 x^3}-\frac {2 c^2}{a^3 x^2}-\frac {6 c^2}{a^2 x}+c^2 x+\frac {4 c^2 \log (x)}{a} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.70 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.78

method	result	size
default	\(\frac {c^{2} \left (a^{4} x +4 a^{3} \ln \left (x \right )-\frac {1}{3 x^{3}}-\frac {2 a}{x^{2}}-\frac {6 a^{2}}{x}\right )}{a^{4}}\)	\(40\)
risch	\(c^{2} x +\frac {-6 a^{2} c^{2} x^{2}-2 a \,c^{2} x -\frac {1}{3} c^{2}}{a^{4} x^{3}}+\frac {4 c^{2} \ln \left (x \right )}{a}\)	\(48\)
parallelrisch	\(\frac {3 a^{4} c^{2} x^{4}+12 c^{2} \ln \left (x \right ) a^{3} x^{3}-18 a^{2} c^{2} x^{2}-6 a \,c^{2} x -c^{2}}{3 a^{4} x^{3}}\)	\(57\)
norman	\(\frac {-7 a^{3} c^{2} x^{4}+a^{4} c^{2} x^{5}+\frac {c^{2}}{3 a}+\frac {5 c^{2} x}{3}+4 a \,c^{2} x^{2}}{\left (a x -1\right ) a^{3} x^{3}}+\frac {4 c^{2} \ln \left (x \right )}{a}\)	\(71\)
meijerg	\(-\frac {c^{2} \left (-\frac {a x \left (-3 a x +6\right )}{3 \left (-a x +1\right )}-2 \ln \left (-a x +1\right )\right )}{a}-\frac {c^{2} x}{-a x +1}+\frac {c^{2} \left (-\frac {3 a x}{-3 a x +3}+2 \ln \left (-a x +1\right )-1-2 \ln \left (x \right )-2 \ln \left (-a \right )+\frac {1}{a x}\right )}{a}+\frac {2 c^{2} \left (\frac {a x}{-a x +1}+\ln \left (-a x +1\right )\right )}{a}-\frac {4 c^{2} \left (\frac {2 a x}{-2 a x +2}-\ln \left (-a x +1\right )+1+\ln \left (x \right )+\ln \left (-a \right )\right )}{a}+\frac {2 c^{2} \left (\frac {4 a x}{-4 a x +4}-3 \ln \left (-a x +1\right )+1+3 \ln \left (x \right )+3 \ln \left (-a \right )-\frac {1}{2 a^{2} x^{2}}-\frac {2}{a x}\right )}{a}-\frac {c^{2} \left (-\frac {5 a x}{-5 a x +5}+4 \ln \left (-a x +1\right )-1-4 \ln \left (x \right )-4 \ln \left (-a \right )+\frac {1}{3 x^{3} a^{3}}+\frac {1}{a^{2} x^{2}}+\frac {3}{a x}\right )}{a}\)	\(284\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.10 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^2 \, dx=\frac {3 \, a^{4} c^{2} x^{4} + 12 \, a^{3} c^{2} x^{3} \log \left (x\right ) - 18 \, a^{2} c^{2} x^{2} - 6 \, a c^{2} x - c^{2}}{3 \, a^{4} x^{3}} \]

[In]

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

[Out]

1/3*(3*a^4*c^2*x^4 + 12*a^3*c^2*x^3*log(x) - 18*a^2*c^2*x^2 - 6*a*c^2*x - c^2)/(a^4*x^3)

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.04 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^2 \, dx=\frac {a^{4} c^{2} x + 4 a^{3} c^{2} \log {\left (x \right )} + \frac {- 18 a^{2} c^{2} x^{2} - 6 a c^{2} x - c^{2}}{3 x^{3}}}{a^{4}} \]

[In]

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

[Out]

(a**4*c**2*x + 4*a**3*c**2*log(x) + (-18*a**2*c**2*x**2 - 6*a*c**2*x - c**2)/(3*x**3))/a**4

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.90 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^2 \, dx=c^{2} x + \frac {4 \, c^{2} \log \left (x\right )}{a} - \frac {18 \, a^{2} c^{2} x^{2} + 6 \, a c^{2} x + c^{2}}{3 \, a^{4} x^{3}} \]

[In]

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

[Out]

c^2*x + 4*c^2*log(x)/a - 1/3*(18*a^2*c^2*x^2 + 6*a*c^2*x + c^2)/(a^4*x^3)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 112 vs. \(2 (49) = 98\).

Time = 0.26 (sec) , antiderivative size = 112, normalized size of antiderivative = 2.20 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^2 \, dx=-\frac {4 \, c^{2} \log \left (\frac {{\left | a x - 1 \right |}}{{\left (a x - 1\right )}^{2} {\left | a \right |}}\right )}{a} + \frac {4 \, c^{2} \log \left ({\left | -\frac {1}{a x - 1} - 1 \right |}\right )}{a} + \frac {{\left (3 \, c^{2} + \frac {34 \, c^{2}}{a x - 1} + \frac {66 \, c^{2}}{{\left (a x - 1\right )}^{2}} + \frac {36 \, c^{2}}{{\left (a x - 1\right )}^{3}}\right )} {\left (a x - 1\right )}}{3 \, a {\left (\frac {1}{a x - 1} + 1\right )}^{3}} \]

[In]

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

[Out]

-4*c^2*log(abs(a*x - 1)/((a*x - 1)^2*abs(a)))/a + 4*c^2*log(abs(-1/(a*x - 1) - 1))/a + 1/3*(3*c^2 + 34*c^2/(a*
x - 1) + 66*c^2/(a*x - 1)^2 + 36*c^2/(a*x - 1)^3)*(a*x - 1)/(a*(1/(a*x - 1) + 1)^3)

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.84 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^2 \, dx=-\frac {c^2\,\left (6\,a\,x+18\,a^2\,x^2-3\,a^4\,x^4-12\,a^3\,x^3\,\ln \left (x\right )+1\right )}{3\,a^4\,x^3} \]

[In]

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

[Out]

-(c^2*(6*a*x + 18*a^2*x^2 - 3*a^4*x^4 - 12*a^3*x^3*log(x) + 1))/(3*a^4*x^3)