Integrand size = 22, antiderivative size = 127 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=-\frac {16 e^{-\coth ^{-1}(a x)}}{35 a c^4}+\frac {e^{-\coth ^{-1}(a x)} (1+6 a x)}{35 a c^4 \left (1-a^2 x^2\right )^3}+\frac {2 e^{-\coth ^{-1}(a x)} (1+4 a x)}{35 a c^4 \left (1-a^2 x^2\right )^2}+\frac {8 e^{-\coth ^{-1}(a x)} (1+2 a x)}{35 a c^4 \left (1-a^2 x^2\right )} \]
-16/35/a/c^4*((a*x-1)/(a*x+1))^(1/2)+1/35*(6*a*x+1)/a/c^4*((a*x-1)/(a*x+1) )^(1/2)/(-a^2*x^2+1)^3+2/35*(4*a*x+1)/a/c^4*((a*x-1)/(a*x+1))^(1/2)/(-a^2* x^2+1)^2+8/35*(2*a*x+1)/a/c^4*((a*x-1)/(a*x+1))^(1/2)/(-a^2*x^2+1)
Time = 0.47 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.63 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=-\frac {\sqrt {1-\frac {1}{a^2 x^2}} x \left (-5+30 a x+30 a^2 x^2-40 a^3 x^3-40 a^4 x^4+16 a^5 x^5+16 a^6 x^6\right )}{35 (-1+a x)^3 (c+a c x)^4} \]
-1/35*(Sqrt[1 - 1/(a^2*x^2)]*x*(-5 + 30*a*x + 30*a^2*x^2 - 40*a^3*x^3 - 40 *a^4*x^4 + 16*a^5*x^5 + 16*a^6*x^6))/((-1 + a*x)^3*(c + a*c*x)^4)
Time = 0.57 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {6739, 27, 6739, 6739, 6737}
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.
\(\displaystyle \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx\) |
\(\Big \downarrow \) 6739 |
\(\displaystyle \frac {6 \int \frac {e^{-\coth ^{-1}(a x)}}{c^3 \left (1-a^2 x^2\right )^3}dx}{7 c}+\frac {(6 a x+1) e^{-\coth ^{-1}(a x)}}{35 a c^4 \left (1-a^2 x^2\right )^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {6 \int \frac {e^{-\coth ^{-1}(a x)}}{\left (1-a^2 x^2\right )^3}dx}{7 c^4}+\frac {(6 a x+1) e^{-\coth ^{-1}(a x)}}{35 a c^4 \left (1-a^2 x^2\right )^3}\) |
\(\Big \downarrow \) 6739 |
\(\displaystyle \frac {6 \left (\frac {4}{5} \int \frac {e^{-\coth ^{-1}(a x)}}{\left (1-a^2 x^2\right )^2}dx+\frac {(4 a x+1) e^{-\coth ^{-1}(a x)}}{15 a \left (1-a^2 x^2\right )^2}\right )}{7 c^4}+\frac {(6 a x+1) e^{-\coth ^{-1}(a x)}}{35 a c^4 \left (1-a^2 x^2\right )^3}\) |
\(\Big \downarrow \) 6739 |
\(\displaystyle \frac {6 \left (\frac {4}{5} \left (\frac {2}{3} \int \frac {e^{-\coth ^{-1}(a x)}}{1-a^2 x^2}dx+\frac {(2 a x+1) e^{-\coth ^{-1}(a x)}}{3 a \left (1-a^2 x^2\right )}\right )+\frac {(4 a x+1) e^{-\coth ^{-1}(a x)}}{15 a \left (1-a^2 x^2\right )^2}\right )}{7 c^4}+\frac {(6 a x+1) e^{-\coth ^{-1}(a x)}}{35 a c^4 \left (1-a^2 x^2\right )^3}\) |
\(\Big \downarrow \) 6737 |
\(\displaystyle \frac {(6 a x+1) e^{-\coth ^{-1}(a x)}}{35 a c^4 \left (1-a^2 x^2\right )^3}+\frac {6 \left (\frac {(4 a x+1) e^{-\coth ^{-1}(a x)}}{15 a \left (1-a^2 x^2\right )^2}+\frac {4}{5} \left (\frac {(2 a x+1) e^{-\coth ^{-1}(a x)}}{3 a \left (1-a^2 x^2\right )}-\frac {2 e^{-\coth ^{-1}(a x)}}{3 a}\right )\right )}{7 c^4}\) |
(1 + 6*a*x)/(35*a*c^4*E^ArcCoth[a*x]*(1 - a^2*x^2)^3) + (6*((1 + 4*a*x)/(1 5*a*E^ArcCoth[a*x]*(1 - a^2*x^2)^2) + (4*(-2/(3*a*E^ArcCoth[a*x]) + (1 + 2 *a*x)/(3*a*E^ArcCoth[a*x]*(1 - a^2*x^2))))/5))/(7*c^4)
3.6.97.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))/((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[ E^(n*ArcCoth[a*x])/(a*c*n), x] /; FreeQ[{a, c, d, n}, x] && EqQ[a^2*c + d, 0] && !IntegerQ[n/2]
Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(n + 2*a*(p + 1)*x)*(c + d*x^2)^(p + 1)*(E^(n*ArcCoth[a*x])/(a*c*(n^2 - 4*(p + 1)^2))), x] - Simp[2*(p + 1)*((2*p + 3)/(c*(n^2 - 4*(p + 1)^2))) Int[(c + d*x^2)^(p + 1)*E^(n*ArcCoth[a*x]), x], x] /; FreeQ[{a, c, d, n}, x] && EqQ[a^2*c + d, 0] && !IntegerQ[n/2] && LtQ[p, -1] && NeQ[p, -3/2] && NeQ[n^2 - 4*(p + 1)^2, 0] && (IntegerQ[p] || !IntegerQ[n])
Time = 0.51 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.64
method | result | size |
gosper | \(-\frac {\sqrt {\frac {a x -1}{a x +1}}\, \left (16 a^{6} x^{6}+16 a^{5} x^{5}-40 a^{4} x^{4}-40 a^{3} x^{3}+30 a^{2} x^{2}+30 a x -5\right )}{35 \left (a^{2} x^{2}-1\right )^{3} c^{4} a}\) | \(81\) |
default | \(-\frac {\sqrt {\frac {a x -1}{a x +1}}\, \left (16 a^{6} x^{6}+16 a^{5} x^{5}-40 a^{4} x^{4}-40 a^{3} x^{3}+30 a^{2} x^{2}+30 a x -5\right )}{35 c^{4} \left (a x +1\right )^{3} \left (a x -1\right )^{3} a}\) | \(84\) |
trager | \(-\frac {\left (16 a^{6} x^{6}+16 a^{5} x^{5}-40 a^{4} x^{4}-40 a^{3} x^{3}+30 a^{2} x^{2}+30 a x -5\right ) \sqrt {-\frac {-a x +1}{a x +1}}}{35 a \,c^{4} \left (a x -1\right )^{3} \left (a x +1\right )^{3}}\) | \(86\) |
-1/35*((a*x-1)/(a*x+1))^(1/2)*(16*a^6*x^6+16*a^5*x^5-40*a^4*x^4-40*a^3*x^3 +30*a^2*x^2+30*a*x-5)/(a^2*x^2-1)^3/c^4/a
Time = 0.26 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.82 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=-\frac {{\left (16 \, a^{6} x^{6} + 16 \, a^{5} x^{5} - 40 \, a^{4} x^{4} - 40 \, a^{3} x^{3} + 30 \, a^{2} x^{2} + 30 \, a x - 5\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{35 \, {\left (a^{7} c^{4} x^{6} - 3 \, a^{5} c^{4} x^{4} + 3 \, a^{3} c^{4} x^{2} - a c^{4}\right )}} \]
-1/35*(16*a^6*x^6 + 16*a^5*x^5 - 40*a^4*x^4 - 40*a^3*x^3 + 30*a^2*x^2 + 30 *a*x - 5)*sqrt((a*x - 1)/(a*x + 1))/(a^7*c^4*x^6 - 3*a^5*c^4*x^4 + 3*a^3*c ^4*x^2 - a*c^4)
\[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=\frac {\int \frac {\sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a^{8} x^{8} - 4 a^{6} x^{6} + 6 a^{4} x^{4} - 4 a^{2} x^{2} + 1}\, dx}{c^{4}} \]
Integral(sqrt(a*x/(a*x + 1) - 1/(a*x + 1))/(a**8*x**8 - 4*a**6*x**6 + 6*a* *4*x**4 - 4*a**2*x**2 + 1), x)/c**4
Time = 0.20 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.06 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=\frac {1}{2240} \, a {\left (\frac {5 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {7}{2}} - 42 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{2}} + 175 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} - 700 \, \sqrt {\frac {a x - 1}{a x + 1}}}{a^{2} c^{4}} + \frac {7 \, {\left (\frac {10 \, {\left (a x - 1\right )}}{a x + 1} - \frac {75 \, {\left (a x - 1\right )}^{2}}{{\left (a x + 1\right )}^{2}} - 1\right )}}{a^{2} c^{4} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{2}}}\right )} \]
1/2240*a*((5*((a*x - 1)/(a*x + 1))^(7/2) - 42*((a*x - 1)/(a*x + 1))^(5/2) + 175*((a*x - 1)/(a*x + 1))^(3/2) - 700*sqrt((a*x - 1)/(a*x + 1)))/(a^2*c^ 4) + 7*(10*(a*x - 1)/(a*x + 1) - 75*(a*x - 1)^2/(a*x + 1)^2 - 1)/(a^2*c^4* ((a*x - 1)/(a*x + 1))^(5/2)))
\[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=\int { \frac {\sqrt {\frac {a x - 1}{a x + 1}}}{{\left (a^{2} c x^{2} - c\right )}^{4}} \,d x } \]
Time = 4.23 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.17 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=\frac {5\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}}{64\,a\,c^4}-\frac {5\,\sqrt {\frac {a\,x-1}{a\,x+1}}}{16\,a\,c^4}-\frac {3\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/2}}{160\,a\,c^4}+\frac {{\left (\frac {a\,x-1}{a\,x+1}\right )}^{7/2}}{448\,a\,c^4}-\frac {\frac {15\,{\left (a\,x-1\right )}^2}{{\left (a\,x+1\right )}^2}-\frac {2\,\left (a\,x-1\right )}{a\,x+1}+\frac {1}{5}}{64\,a\,c^4\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/2}} \]