\(\int \frac {e^{2 \coth ^{-1}(a x)}}{(c-a^2 c x^2)^4} \, dx\) [572]

Optimal result

Integrand size = 22, antiderivative size = 121 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=-\frac {1}{32 a c^4 (1-a x)^4}-\frac {1}{16 a c^4 (1-a x)^3}-\frac {3}{32 a c^4 (1-a x)^2}-\frac {5}{32 a c^4 (1-a x)}+\frac {1}{64 a c^4 (1+a x)^2}+\frac {5}{64 a c^4 (1+a x)}-\frac {15 \text {arctanh}(a x)}{64 a c^4} \]

[Out]

-1/32/a/c^4/(-a*x+1)^4-1/16/a/c^4/(-a*x+1)^3-3/32/a/c^4/(-a*x+1)^2-5/32/a/c^4/(-a*x+1)+1/64/a/c^4/(a*x+1)^2+5/
64/a/c^4/(a*x+1)-15/64*arctanh(a*x)/a/c^4

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 121, 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, 6275, 46, 213} \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=-\frac {15 \text {arctanh}(a x)}{64 a c^4}-\frac {5}{32 a c^4 (1-a x)}+\frac {5}{64 a c^4 (a x+1)}-\frac {3}{32 a c^4 (1-a x)^2}+\frac {1}{64 a c^4 (a x+1)^2}-\frac {1}{16 a c^4 (1-a x)^3}-\frac {1}{32 a c^4 (1-a x)^4} \]

[In]

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

[Out]

-1/32*1/(a*c^4*(1 - a*x)^4) - 1/(16*a*c^4*(1 - a*x)^3) - 3/(32*a*c^4*(1 - a*x)^2) - 5/(32*a*c^4*(1 - a*x)) + 1
/(64*a*c^4*(1 + a*x)^2) + 5/(64*a*c^4*(1 + a*x)) - (15*ArcTanh[a*x])/(64*a*c^4)

Rule 46

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 6275

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

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 \frac {e^{2 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx \\ & = -\frac {\int \frac {1}{(1-a x)^5 (1+a x)^3} \, dx}{c^4} \\ & = -\frac {\int \left (-\frac {1}{8 (-1+a x)^5}+\frac {3}{16 (-1+a x)^4}-\frac {3}{16 (-1+a x)^3}+\frac {5}{32 (-1+a x)^2}+\frac {1}{32 (1+a x)^3}+\frac {5}{64 (1+a x)^2}-\frac {15}{64 \left (-1+a^2 x^2\right )}\right ) \, dx}{c^4} \\ & = -\frac {1}{32 a c^4 (1-a x)^4}-\frac {1}{16 a c^4 (1-a x)^3}-\frac {3}{32 a c^4 (1-a x)^2}-\frac {5}{32 a c^4 (1-a x)}+\frac {1}{64 a c^4 (1+a x)^2}+\frac {5}{64 a c^4 (1+a x)}+\frac {15 \int \frac {1}{-1+a^2 x^2} \, dx}{64 c^4} \\ & = -\frac {1}{32 a c^4 (1-a x)^4}-\frac {1}{16 a c^4 (1-a x)^3}-\frac {3}{32 a c^4 (1-a x)^2}-\frac {5}{32 a c^4 (1-a x)}+\frac {1}{64 a c^4 (1+a x)^2}+\frac {5}{64 a c^4 (1+a x)}-\frac {15 \text {arctanh}(a x)}{64 a c^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.68 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=-\frac {16+17 a x-50 a^2 x^2+10 a^3 x^3+30 a^4 x^4-15 a^5 x^5+15 (-1+a x)^4 (1+a x)^2 \text {arctanh}(a x)}{64 a c^4 (-1+a x)^4 (1+a x)^2} \]

[In]

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

[Out]

-1/64*(16 + 17*a*x - 50*a^2*x^2 + 10*a^3*x^3 + 30*a^4*x^4 - 15*a^5*x^5 + 15*(-1 + a*x)^4*(1 + a*x)^2*ArcTanh[a
*x])/(a*c^4*(-1 + a*x)^4*(1 + a*x)^2)

Maple [A] (verified)

Time = 0.62 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.76

method	result	size
risch	\(\frac {\frac {15 a^{4} x^{5}}{64}-\frac {15 a^{3} x^{4}}{32}-\frac {5 a^{2} x^{3}}{32}+\frac {25 a \,x^{2}}{32}-\frac {17 x}{64}-\frac {1}{4 a}}{\left (a x -1\right )^{2} \left (a^{2} x^{2}-1\right )^{2} c^{4}}+\frac {15 \ln \left (-a x +1\right )}{128 a \,c^{4}}-\frac {15 \ln \left (a x +1\right )}{128 a \,c^{4}}\)	\(92\)
default	\(\frac {\frac {1}{64 a \left (a x +1\right )^{2}}+\frac {5}{64 a \left (a x +1\right )}-\frac {15 \ln \left (a x +1\right )}{128 a}-\frac {1}{32 a \left (a x -1\right )^{4}}+\frac {1}{16 a \left (a x -1\right )^{3}}-\frac {3}{32 \left (a x -1\right )^{2} a}+\frac {5}{32 a \left (a x -1\right )}+\frac {15 \ln \left (a x -1\right )}{128 a}}{c^{4}}\)	\(100\)
norman	\(\frac {-\frac {49 x}{64 c}-\frac {15 a \,x^{2}}{64 c}+\frac {11 a^{2} x^{3}}{8 c}+\frac {a^{3} x^{4}}{8 c}-\frac {63 a^{4} x^{5}}{64 c}-\frac {a^{5} x^{6}}{64 c}+\frac {a^{6} x^{7}}{4 c}}{\left (a x +1\right )^{3} c^{3} \left (a x -1\right )^{4}}+\frac {15 \ln \left (a x -1\right )}{128 a \,c^{4}}-\frac {15 \ln \left (a x +1\right )}{128 a \,c^{4}}\)	\(119\)
parallelrisch	\(\frac {30 a \ln \left (a x +1\right ) x +15 a^{2} \ln \left (a x +1\right ) x^{2}-34 a^{5} x^{5}+108 a^{3} x^{3}+30 \ln \left (a x +1\right ) x^{5} a^{5}-15 \ln \left (a x +1\right ) x^{6} a^{6}+15 \ln \left (a x +1\right ) x^{4} a^{4}+15 \ln \left (a x -1\right ) x^{6} a^{6}-30 \ln \left (a x -1\right ) x^{5} a^{5}-15 \ln \left (a x -1\right ) x^{4} a^{4}-60 a^{3} \ln \left (a x +1\right ) x^{3}+32 a^{6} x^{6}-98 a x +60 a^{3} \ln \left (a x -1\right ) x^{3}-15 a^{2} \ln \left (a x -1\right ) x^{2}-30 a \ln \left (a x -1\right ) x -92 a^{4} x^{4}+15 \ln \left (a x -1\right )-15 \ln \left (a x +1\right )+68 a^{2} x^{2}}{128 c^{4} \left (a x -1\right )^{2} \left (a^{2} x^{2}-1\right )^{2} a}\)	\(248\)

[In]

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

[Out]

(15/64*a^4*x^5-15/32*a^3*x^4-5/32*a^2*x^3+25/32*a*x^2-17/64*x-1/4/a)/(a*x-1)^2/(a^2*x^2-1)^2/c^4+15/128*ln(-a*
x+1)/a/c^4-15/128*ln(a*x+1)/a/c^4

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 217 vs. \(2 (103) = 206\).

Time = 0.25 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.79 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=\frac {30 \, a^{5} x^{5} - 60 \, a^{4} x^{4} - 20 \, a^{3} x^{3} + 100 \, a^{2} x^{2} - 34 \, a x - 15 \, {\left (a^{6} x^{6} - 2 \, a^{5} x^{5} - a^{4} x^{4} + 4 \, a^{3} x^{3} - a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (a x + 1\right ) + 15 \, {\left (a^{6} x^{6} - 2 \, a^{5} x^{5} - a^{4} x^{4} + 4 \, a^{3} x^{3} - a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (a x - 1\right ) - 32}{128 \, {\left (a^{7} c^{4} x^{6} - 2 \, a^{6} c^{4} x^{5} - a^{5} c^{4} x^{4} + 4 \, a^{4} c^{4} x^{3} - a^{3} c^{4} x^{2} - 2 \, a^{2} c^{4} x + a c^{4}\right )}} \]

[In]

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

[Out]

1/128*(30*a^5*x^5 - 60*a^4*x^4 - 20*a^3*x^3 + 100*a^2*x^2 - 34*a*x - 15*(a^6*x^6 - 2*a^5*x^5 - a^4*x^4 + 4*a^3
*x^3 - a^2*x^2 - 2*a*x + 1)*log(a*x + 1) + 15*(a^6*x^6 - 2*a^5*x^5 - a^4*x^4 + 4*a^3*x^3 - a^2*x^2 - 2*a*x + 1
)*log(a*x - 1) - 32)/(a^7*c^4*x^6 - 2*a^6*c^4*x^5 - a^5*c^4*x^4 + 4*a^4*c^4*x^3 - a^3*c^4*x^2 - 2*a^2*c^4*x +
a*c^4)

Sympy [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.17 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=\frac {15 a^{5} x^{5} - 30 a^{4} x^{4} - 10 a^{3} x^{3} + 50 a^{2} x^{2} - 17 a x - 16}{64 a^{7} c^{4} x^{6} - 128 a^{6} c^{4} x^{5} - 64 a^{5} c^{4} x^{4} + 256 a^{4} c^{4} x^{3} - 64 a^{3} c^{4} x^{2} - 128 a^{2} c^{4} x + 64 a c^{4}} + \frac {\frac {15 \log {\left (x - \frac {1}{a} \right )}}{128} - \frac {15 \log {\left (x + \frac {1}{a} \right )}}{128}}{a c^{4}} \]

[In]

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

[Out]

(15*a**5*x**5 - 30*a**4*x**4 - 10*a**3*x**3 + 50*a**2*x**2 - 17*a*x - 16)/(64*a**7*c**4*x**6 - 128*a**6*c**4*x
**5 - 64*a**5*c**4*x**4 + 256*a**4*c**4*x**3 - 64*a**3*c**4*x**2 - 128*a**2*c**4*x + 64*a*c**4) + (15*log(x -
1/a)/128 - 15*log(x + 1/a)/128)/(a*c**4)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.16 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=\frac {15 \, a^{5} x^{5} - 30 \, a^{4} x^{4} - 10 \, a^{3} x^{3} + 50 \, a^{2} x^{2} - 17 \, a x - 16}{64 \, {\left (a^{7} c^{4} x^{6} - 2 \, a^{6} c^{4} x^{5} - a^{5} c^{4} x^{4} + 4 \, a^{4} c^{4} x^{3} - a^{3} c^{4} x^{2} - 2 \, a^{2} c^{4} x + a c^{4}\right )}} - \frac {15 \, \log \left (a x + 1\right )}{128 \, a c^{4}} + \frac {15 \, \log \left (a x - 1\right )}{128 \, a c^{4}} \]

[In]

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

[Out]

1/64*(15*a^5*x^5 - 30*a^4*x^4 - 10*a^3*x^3 + 50*a^2*x^2 - 17*a*x - 16)/(a^7*c^4*x^6 - 2*a^6*c^4*x^5 - a^5*c^4*
x^4 + 4*a^4*c^4*x^3 - a^3*c^4*x^2 - 2*a^2*c^4*x + a*c^4) - 15/128*log(a*x + 1)/(a*c^4) + 15/128*log(a*x - 1)/(
a*c^4)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.75 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=-\frac {15 \, \log \left ({\left | a x + 1 \right |}\right )}{128 \, a c^{4}} + \frac {15 \, \log \left ({\left | a x - 1 \right |}\right )}{128 \, a c^{4}} + \frac {15 \, a^{5} x^{5} - 30 \, a^{4} x^{4} - 10 \, a^{3} x^{3} + 50 \, a^{2} x^{2} - 17 \, a x - 16}{64 \, {\left (a x + 1\right )}^{2} {\left (a x - 1\right )}^{4} a c^{4}} \]

[In]

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

[Out]

-15/128*log(abs(a*x + 1))/(a*c^4) + 15/128*log(abs(a*x - 1))/(a*c^4) + 1/64*(15*a^5*x^5 - 30*a^4*x^4 - 10*a^3*
x^3 + 50*a^2*x^2 - 17*a*x - 16)/((a*x + 1)^2*(a*x - 1)^4*a*c^4)

Mupad [B] (verification not implemented)

Time = 4.14 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=\frac {\frac {17\,x}{64}-\frac {25\,a\,x^2}{32}+\frac {1}{4\,a}+\frac {5\,a^2\,x^3}{32}+\frac {15\,a^3\,x^4}{32}-\frac {15\,a^4\,x^5}{64}}{-a^6\,c^4\,x^6+2\,a^5\,c^4\,x^5+a^4\,c^4\,x^4-4\,a^3\,c^4\,x^3+a^2\,c^4\,x^2+2\,a\,c^4\,x-c^4}-\frac {15\,\mathrm {atanh}\left (a\,x\right )}{64\,a\,c^4} \]

[In]

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

[Out]

((17*x)/64 - (25*a*x^2)/32 + 1/(4*a) + (5*a^2*x^3)/32 + (15*a^3*x^4)/32 - (15*a^4*x^5)/64)/(a^2*c^4*x^2 - c^4
- 4*a^3*c^4*x^3 + a^4*c^4*x^4 + 2*a^5*c^4*x^5 - a^6*c^4*x^6 + 2*a*c^4*x) - (15*atanh(a*x))/(64*a*c^4)