Integrand size = 22, antiderivative size = 121 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=\frac {1}{32 a c^4 (1-a x)^4}+\frac {1}{16 a c^4 (1-a x)^3}+\frac {3}{32 a c^4 (1-a x)^2}+\frac {5}{32 a c^4 (1-a x)}-\frac {1}{64 a c^4 (1+a x)^2}-\frac {5}{64 a c^4 (1+a x)}+\frac {15 \text {arctanh}(a x)}{64 a c^4} \]
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
Time = 0.05 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.68 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=\frac {16+17 a x-50 a^2 x^2+10 a^3 x^3+30 a^4 x^4-15 a^5 x^5+15 (-1+a x)^4 (1+a x)^2 \text {arctanh}(a x)}{64 a c^4 (-1+a x)^4 (1+a x)^2} \]
(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])/(64*a*c^4*(-1 + a*x)^4*(1 + a*x)^2)
Time = 0.31 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.86, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {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.
\(\displaystyle \int \frac {e^{2 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx\) |
\(\Big \downarrow \) 6690 |
\(\displaystyle \frac {\int \frac {1}{(1-a x)^5 (a x+1)^3}dx}{c^4}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {\int \left (\frac {5}{32 (a x-1)^2}+\frac {5}{64 (a x+1)^2}-\frac {3}{16 (a x-1)^3}+\frac {1}{32 (a x+1)^3}+\frac {3}{16 (a x-1)^4}-\frac {1}{8 (a x-1)^5}-\frac {15}{64 \left (a^2 x^2-1\right )}\right )dx}{c^4}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {15 \text {arctanh}(a x)}{64 a}+\frac {5}{32 a (1-a x)}-\frac {5}{64 a (a x+1)}+\frac {3}{32 a (1-a x)^2}-\frac {1}{64 a (a x+1)^2}+\frac {1}{16 a (1-a x)^3}+\frac {1}{32 a (1-a x)^4}}{c^4}\) |
(1/(32*a*(1 - a*x)^4) + 1/(16*a*(1 - a*x)^3) + 3/(32*a*(1 - a*x)^2) + 5/(3 2*a*(1 - a*x)) - 1/(64*a*(1 + a*x)^2) - 5/(64*a*(1 + a*x)) + (15*ArcTanh[a *x])/(64*a))/c^4
3.11.76.3.1 Defintions of rubi rules used
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])
Time = 0.20 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.76
method | result | size |
risch | \(\frac {-\frac {15 a^{4} x^{5}}{64}+\frac {15 a^{3} x^{4}}{32}+\frac {5 a^{2} x^{3}}{32}-\frac {25 a \,x^{2}}{32}+\frac {17 x}{64}+\frac {1}{4 a}}{c^{4} \left (a x -1\right )^{2} \left (a^{2} x^{2}-1\right )^{2}}+\frac {15 \ln \left (-a x -1\right )}{128 a \,c^{4}}-\frac {15 \ln \left (a x -1\right )}{128 a \,c^{4}}\) | \(92\) |
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 | \(\frac {\frac {49 x}{64 c}-\frac {73 a^{2} x^{3}}{64 c}+\frac {55 a^{4} x^{5}}{64 c}-\frac {15 a^{6} x^{7}}{64 c}+\frac {a \,x^{2}}{c}-\frac {3 a^{3} x^{4}}{2 c}+\frac {a^{5} x^{6}}{c}-\frac {a^{7} x^{8}}{4 c}}{\left (a^{2} x^{2}-1\right )^{4} c^{3}}-\frac {15 \ln \left (a x -1\right )}{128 a \,c^{4}}+\frac {15 \ln \left (a x +1\right )}{128 a \,c^{4}}\) | \(125\) |
parallelrisch | \(\frac {-30 a \ln \left (a x +1\right ) x -15 a^{2} \ln \left (a x +1\right ) x^{2}+34 a^{5} x^{5}-108 a^{3} x^{3}-30 \ln \left (a x +1\right ) x^{5} a^{5}+15 \ln \left (a x +1\right ) x^{6} a^{6}-15 \ln \left (a x +1\right ) x^{4} a^{4}-15 \ln \left (a x -1\right ) x^{6} a^{6}+30 \ln \left (a x -1\right ) x^{5} a^{5}+15 \ln \left (a x -1\right ) x^{4} a^{4}+60 a^{3} \ln \left (a x +1\right ) x^{3}-32 a^{6} x^{6}+98 a x -60 a^{3} \ln \left (a x -1\right ) x^{3}+15 a^{2} \ln \left (a x -1\right ) x^{2}+30 a \ln \left (a x -1\right ) x +92 a^{4} x^{4}-15 \ln \left (a x -1\right )+15 \ln \left (a x +1\right )-68 a^{2} x^{2}}{128 c^{4} \left (a x -1\right )^{2} \left (a^{2} x^{2}-1\right )^{2} a}\) | \(248\) |
(-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)/c^4/ (a*x-1)^2/(a^2*x^2-1)^2+15/128/a/c^4*ln(-a*x-1)-15/128/a/c^4*ln(a*x-1)
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 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=-\frac {30 \, a^{5} x^{5} - 60 \, a^{4} x^{4} - 20 \, a^{3} x^{3} + 100 \, a^{2} x^{2} - 34 \, a x - 15 \, {\left (a^{6} x^{6} - 2 \, a^{5} x^{5} - a^{4} x^{4} + 4 \, a^{3} x^{3} - a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (a x + 1\right ) + 15 \, {\left (a^{6} x^{6} - 2 \, a^{5} x^{5} - a^{4} x^{4} + 4 \, a^{3} x^{3} - a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (a x - 1\right ) - 32}{128 \, {\left (a^{7} c^{4} x^{6} - 2 \, a^{6} c^{4} x^{5} - a^{5} c^{4} x^{4} + 4 \, a^{4} c^{4} x^{3} - a^{3} c^{4} x^{2} - 2 \, a^{2} c^{4} x + a c^{4}\right )}} \]
-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)
Time = 0.34 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.18 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=- \frac {15 a^{5} x^{5} - 30 a^{4} x^{4} - 10 a^{3} x^{3} + 50 a^{2} x^{2} - 17 a x - 16}{64 a^{7} c^{4} x^{6} - 128 a^{6} c^{4} x^{5} - 64 a^{5} c^{4} x^{4} + 256 a^{4} c^{4} x^{3} - 64 a^{3} c^{4} x^{2} - 128 a^{2} c^{4} x + 64 a c^{4}} - \frac {\frac {15 \log {\left (x - \frac {1}{a} \right )}}{128} - \frac {15 \log {\left (x + \frac {1}{a} \right )}}{128}}{a c^{4}} \]
-(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)
Time = 0.19 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.16 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=-\frac {15 \, a^{5} x^{5} - 30 \, a^{4} x^{4} - 10 \, a^{3} x^{3} + 50 \, a^{2} x^{2} - 17 \, a x - 16}{64 \, {\left (a^{7} c^{4} x^{6} - 2 \, a^{6} c^{4} x^{5} - a^{5} c^{4} x^{4} + 4 \, a^{4} c^{4} x^{3} - a^{3} c^{4} x^{2} - 2 \, a^{2} c^{4} x + a c^{4}\right )}} + \frac {15 \, \log \left (a x + 1\right )}{128 \, a c^{4}} - \frac {15 \, \log \left (a x - 1\right )}{128 \, a c^{4}} \]
-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)
Time = 0.26 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.75 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=\frac {15 \, \log \left ({\left | a x + 1 \right |}\right )}{128 \, a c^{4}} - \frac {15 \, \log \left ({\left | a x - 1 \right |}\right )}{128 \, a c^{4}} - \frac {15 \, a^{5} x^{5} - 30 \, a^{4} x^{4} - 10 \, a^{3} x^{3} + 50 \, a^{2} x^{2} - 17 \, a x - 16}{64 \, {\left (a x + 1\right )}^{2} {\left (a x - 1\right )}^{4} a c^{4}} \]
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)
Time = 0.15 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.01 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=\frac {15\,\mathrm {atanh}\left (a\,x\right )}{64\,a\,c^4}-\frac {\frac {17\,x}{64}-\frac {25\,a\,x^2}{32}+\frac {1}{4\,a}+\frac {5\,a^2\,x^3}{32}+\frac {15\,a^3\,x^4}{32}-\frac {15\,a^4\,x^5}{64}}{-a^6\,c^4\,x^6+2\,a^5\,c^4\,x^5+a^4\,c^4\,x^4-4\,a^3\,c^4\,x^3+a^2\,c^4\,x^2+2\,a\,c^4\,x-c^4} \]