Integrand size = 22, antiderivative size = 87 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {1}{8 a c^3 (1-a x)^4}+\frac {1}{12 a c^3 (1-a x)^3}+\frac {1}{16 a c^3 (1-a x)^2}+\frac {1}{16 a c^3 (1-a x)}+\frac {\text {arctanh}(a x)}{16 a c^3} \] Output:
1/8/a/c^3/(-a*x+1)^4+1/12/a/c^3/(-a*x+1)^3+1/16/a/c^3/(-a*x+1)^2+1/16/a/c^ 3/(-a*x+1)+1/16*arctanh(a*x)/a/c^3
Time = 0.05 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.60 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {16-19 a x+12 a^2 x^2-3 a^3 x^3+3 (-1+a x)^4 \text {arctanh}(a x)}{48 a c^3 (-1+a x)^4} \] Input:
Integrate[E^(4*ArcCoth[a*x])/(c - a^2*c*x^2)^3,x]
Output:
(16 - 19*a*x + 12*a^2*x^2 - 3*a^3*x^3 + 3*(-1 + a*x)^4*ArcTanh[a*x])/(48*a *c^3*(-1 + a*x)^4)
Time = 0.59 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.87, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {6717, 27, 6690, 54, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx\) |
\(\Big \downarrow \) 6717 |
\(\displaystyle \int \frac {e^{4 \text {arctanh}(a x)}}{c^3 \left (1-a^2 x^2\right )^3}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {e^{4 \text {arctanh}(a x)}}{\left (1-a^2 x^2\right )^3}dx}{c^3}\) |
\(\Big \downarrow \) 6690 |
\(\displaystyle \frac {\int \frac {1}{(1-a x)^5 (a x+1)}dx}{c^3}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {\int \left (\frac {1}{16 (a x-1)^2}-\frac {1}{8 (a x-1)^3}+\frac {1}{4 (a x-1)^4}-\frac {1}{2 (a x-1)^5}-\frac {1}{16 \left (a^2 x^2-1\right )}\right )dx}{c^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {\text {arctanh}(a x)}{16 a}+\frac {1}{16 a (1-a x)}+\frac {1}{16 a (1-a x)^2}+\frac {1}{12 a (1-a x)^3}+\frac {1}{8 a (1-a x)^4}}{c^3}\) |
Input:
Int[E^(4*ArcCoth[a*x])/(c - a^2*c*x^2)^3,x]
Output:
(1/(8*a*(1 - a*x)^4) + 1/(12*a*(1 - a*x)^3) + 1/(16*a*(1 - a*x)^2) + 1/(16 *a*(1 - a*x)) + ArcTanh[a*x]/(16*a))/c^3
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[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])
Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Simp[(-1)^(n/2) Int[ u*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[a, x] && IntegerQ[n/2]
Time = 0.17 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.75
method | result | size |
risch | \(\frac {-\frac {a^{2} x^{3}}{16}+\frac {a \,x^{2}}{4}-\frac {19 x}{48}+\frac {1}{3 a}}{c^{3} \left (a x -1\right )^{4}}-\frac {\ln \left (a x -1\right )}{32 a \,c^{3}}+\frac {\ln \left (-a x -1\right )}{32 a \,c^{3}}\) | \(65\) |
default | \(\frac {\frac {\ln \left (a x +1\right )}{32 a}+\frac {1}{8 a \left (a x -1\right )^{4}}-\frac {1}{12 a \left (a x -1\right )^{3}}+\frac {1}{16 a \left (a x -1\right )^{2}}-\frac {1}{16 \left (a x -1\right ) a}-\frac {\ln \left (a x -1\right )}{32 a}}{c^{3}}\) | \(76\) |
norman | \(\frac {\frac {15 x}{16 c}+\frac {a \,x^{2}}{8 c}-\frac {31 a^{2} x^{3}}{24 c}+\frac {11 a^{3} x^{4}}{24 c}+\frac {29 a^{4} x^{5}}{48 c}-\frac {a^{5} x^{6}}{3 c}}{c^{2} \left (a x +1\right )^{2} \left (a x -1\right )^{4}}-\frac {\ln \left (a x -1\right )}{32 a \,c^{3}}+\frac {\ln \left (a x +1\right )}{32 a \,c^{3}}\) | \(108\) |
parallelrisch | \(\frac {-3 \ln \left (a x -1\right ) x^{4} a^{4}+3 \ln \left (a x +1\right ) x^{4} a^{4}-32 a^{4} x^{4}+12 a^{3} \ln \left (a x -1\right ) x^{3}-12 \ln \left (a x +1\right ) x^{3} a^{3}+122 a^{3} x^{3}-18 a^{2} \ln \left (a x -1\right ) x^{2}+18 \ln \left (a x +1\right ) x^{2} a^{2}-168 a^{2} x^{2}+12 a \ln \left (a x -1\right ) x -12 \ln \left (a x +1\right ) x a +90 a x -3 \ln \left (a x -1\right )+3 \ln \left (a x +1\right )}{96 \left (a x -1\right )^{4} c^{3} a}\) | \(165\) |
Input:
int(1/(a*x-1)^2*(a*x+1)^2/(-a^2*c*x^2+c)^3,x,method=_RETURNVERBOSE)
Output:
(-1/16*a^2*x^3+1/4*a*x^2-19/48*x+1/3/a)/c^3/(a*x-1)^4-1/32/a/c^3*ln(a*x-1) +1/32/a/c^3*ln(-a*x-1)
Leaf count of result is larger than twice the leaf count of optimal. 147 vs. \(2 (73) = 146\).
Time = 0.09 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.69 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=-\frac {6 \, a^{3} x^{3} - 24 \, a^{2} x^{2} + 38 \, a x - 3 \, {\left (a^{4} x^{4} - 4 \, a^{3} x^{3} + 6 \, a^{2} x^{2} - 4 \, a x + 1\right )} \log \left (a x + 1\right ) + 3 \, {\left (a^{4} x^{4} - 4 \, a^{3} x^{3} + 6 \, a^{2} x^{2} - 4 \, a x + 1\right )} \log \left (a x - 1\right ) - 32}{96 \, {\left (a^{5} c^{3} x^{4} - 4 \, a^{4} c^{3} x^{3} + 6 \, a^{3} c^{3} x^{2} - 4 \, a^{2} c^{3} x + a c^{3}\right )}} \] Input:
integrate(1/(a*x-1)^2*(a*x+1)^2/(-a^2*c*x^2+c)^3,x, algorithm="fricas")
Output:
-1/96*(6*a^3*x^3 - 24*a^2*x^2 + 38*a*x - 3*(a^4*x^4 - 4*a^3*x^3 + 6*a^2*x^ 2 - 4*a*x + 1)*log(a*x + 1) + 3*(a^4*x^4 - 4*a^3*x^3 + 6*a^2*x^2 - 4*a*x + 1)*log(a*x - 1) - 32)/(a^5*c^3*x^4 - 4*a^4*c^3*x^3 + 6*a^3*c^3*x^2 - 4*a^ 2*c^3*x + a*c^3)
Time = 0.24 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.14 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=- \frac {3 a^{3} x^{3} - 12 a^{2} x^{2} + 19 a x - 16}{48 a^{5} c^{3} x^{4} - 192 a^{4} c^{3} x^{3} + 288 a^{3} c^{3} x^{2} - 192 a^{2} c^{3} x + 48 a c^{3}} - \frac {\frac {\log {\left (x - \frac {1}{a} \right )}}{32} - \frac {\log {\left (x + \frac {1}{a} \right )}}{32}}{a c^{3}} \] Input:
integrate(1/(a*x-1)**2*(a*x+1)**2/(-a**2*c*x**2+c)**3,x)
Output:
-(3*a**3*x**3 - 12*a**2*x**2 + 19*a*x - 16)/(48*a**5*c**3*x**4 - 192*a**4* c**3*x**3 + 288*a**3*c**3*x**2 - 192*a**2*c**3*x + 48*a*c**3) - (log(x - 1 /a)/32 - log(x + 1/a)/32)/(a*c**3)
Time = 0.03 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.17 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=-\frac {3 \, a^{3} x^{3} - 12 \, a^{2} x^{2} + 19 \, a x - 16}{48 \, {\left (a^{5} c^{3} x^{4} - 4 \, a^{4} c^{3} x^{3} + 6 \, a^{3} c^{3} x^{2} - 4 \, a^{2} c^{3} x + a c^{3}\right )}} + \frac {\log \left (a x + 1\right )}{32 \, a c^{3}} - \frac {\log \left (a x - 1\right )}{32 \, a c^{3}} \] Input:
integrate(1/(a*x-1)^2*(a*x+1)^2/(-a^2*c*x^2+c)^3,x, algorithm="maxima")
Output:
-1/48*(3*a^3*x^3 - 12*a^2*x^2 + 19*a*x - 16)/(a^5*c^3*x^4 - 4*a^4*c^3*x^3 + 6*a^3*c^3*x^2 - 4*a^2*c^3*x + a*c^3) + 1/32*log(a*x + 1)/(a*c^3) - 1/32* log(a*x - 1)/(a*c^3)
Time = 0.13 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.05 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {\log \left ({\left | -\frac {2}{a x - 1} - 1 \right |}\right )}{32 \, a c^{3}} - \frac {\frac {3 \, a^{3} c^{9}}{a x - 1} - \frac {3 \, a^{3} c^{9}}{{\left (a x - 1\right )}^{2}} + \frac {4 \, a^{3} c^{9}}{{\left (a x - 1\right )}^{3}} - \frac {6 \, a^{3} c^{9}}{{\left (a x - 1\right )}^{4}}}{48 \, a^{4} c^{12}} \] Input:
integrate(1/(a*x-1)^2*(a*x+1)^2/(-a^2*c*x^2+c)^3,x, algorithm="giac")
Output:
1/32*log(abs(-2/(a*x - 1) - 1))/(a*c^3) - 1/48*(3*a^3*c^9/(a*x - 1) - 3*a^ 3*c^9/(a*x - 1)^2 + 4*a^3*c^9/(a*x - 1)^3 - 6*a^3*c^9/(a*x - 1)^4)/(a^4*c^ 12)
Time = 0.05 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.95 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {\mathrm {atanh}\left (a\,x\right )}{16\,a\,c^3}-\frac {\frac {19\,x}{48}-\frac {a\,x^2}{4}-\frac {1}{3\,a}+\frac {a^2\,x^3}{16}}{a^4\,c^3\,x^4-4\,a^3\,c^3\,x^3+6\,a^2\,c^3\,x^2-4\,a\,c^3\,x+c^3} \] Input:
int((a*x + 1)^2/((c - a^2*c*x^2)^3*(a*x - 1)^2),x)
Output:
atanh(a*x)/(16*a*c^3) - ((19*x)/48 - (a*x^2)/4 - 1/(3*a) + (a^2*x^3)/16)/( c^3 + 6*a^2*c^3*x^2 - 4*a^3*c^3*x^3 + a^4*c^3*x^4 - 4*a*c^3*x)
Time = 0.14 (sec) , antiderivative size = 181, normalized size of antiderivative = 2.08 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {-6 \,\mathrm {log}\left (a x -1\right ) a^{4} x^{4}+24 \,\mathrm {log}\left (a x -1\right ) a^{3} x^{3}-36 \,\mathrm {log}\left (a x -1\right ) a^{2} x^{2}+24 \,\mathrm {log}\left (a x -1\right ) a x -6 \,\mathrm {log}\left (a x -1\right )+6 \,\mathrm {log}\left (a x +1\right ) a^{4} x^{4}-24 \,\mathrm {log}\left (a x +1\right ) a^{3} x^{3}+36 \,\mathrm {log}\left (a x +1\right ) a^{2} x^{2}-24 \,\mathrm {log}\left (a x +1\right ) a x +6 \,\mathrm {log}\left (a x +1\right )-3 a^{4} x^{4}+30 a^{2} x^{2}-64 a x +61}{192 a \,c^{3} \left (a^{4} x^{4}-4 a^{3} x^{3}+6 a^{2} x^{2}-4 a x +1\right )} \] Input:
int(1/(a*x-1)^2*(a*x+1)^2/(-a^2*c*x^2+c)^3,x)
Output:
( - 6*log(a*x - 1)*a**4*x**4 + 24*log(a*x - 1)*a**3*x**3 - 36*log(a*x - 1) *a**2*x**2 + 24*log(a*x - 1)*a*x - 6*log(a*x - 1) + 6*log(a*x + 1)*a**4*x* *4 - 24*log(a*x + 1)*a**3*x**3 + 36*log(a*x + 1)*a**2*x**2 - 24*log(a*x + 1)*a*x + 6*log(a*x + 1) - 3*a**4*x**4 + 30*a**2*x**2 - 64*a*x + 61)/(192*a *c**3*(a**4*x**4 - 4*a**3*x**3 + 6*a**2*x**2 - 4*a*x + 1))