Integrand size = 20, antiderivative size = 119 \[ \int \frac {e^{\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=\frac {16 e^{\coth ^{-1}(a x)}}{35 a c^4}-\frac {e^{\coth ^{-1}(a x)} (1-6 a x)}{35 a c^4 \left (1-a^2 x^2\right )^3}-\frac {2 e^{\coth ^{-1}(a x)} (1-4 a x)}{35 a c^4 \left (1-a^2 x^2\right )^2}-\frac {8 e^{\coth ^{-1}(a x)} (1-2 a x)}{35 a c^4 \left (1-a^2 x^2\right )} \]
16/35/((a*x-1)/(a*x+1))^(1/2)/a/c^4-1/35/((a*x-1)/(a*x+1))^(1/2)*(-6*a*x+1 )/a/c^4/(-a^2*x^2+1)^3-2/35/((a*x-1)/(a*x+1))^(1/2)*(-4*a*x+1)/a/c^4/(-a^2 *x^2+1)^2-8/35/((a*x-1)/(a*x+1))^(1/2)*(-2*a*x+1)/a/c^4/(-a^2*x^2+1)
Time = 0.46 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.69 \[ \int \frac {e^{\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=\frac {\sqrt {1-\frac {1}{a^2 x^2}} x \left (-5-30 a x+30 a^2 x^2+40 a^3 x^3-40 a^4 x^4-16 a^5 x^5+16 a^6 x^6\right )}{35 c^4 (-1+a x)^4 (1+a x)^3} \]
(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))/(35*c^4*(-1 + a*x)^4*(1 + a*x)^3)
Time = 0.57 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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 |
\(\displaystyle \frac {6 \int \frac {e^{\coth ^{-1}(a x)}}{c^3 \left (1-a^2 x^2\right )^3}dx}{7 c}-\frac {(1-6 a x) e^{\coth ^{-1}(a x)}}{35 a c^4 \left (1-a^2 x^2\right )^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {6 \int \frac {e^{\coth ^{-1}(a x)}}{\left (1-a^2 x^2\right )^3}dx}{7 c^4}-\frac {(1-6 a x) e^{\coth ^{-1}(a x)}}{35 a c^4 \left (1-a^2 x^2\right )^3}\) |
\(\Big \downarrow \) 6739 |
\(\displaystyle \frac {6 \left (\frac {4}{5} \int \frac {e^{\coth ^{-1}(a x)}}{\left (1-a^2 x^2\right )^2}dx-\frac {(1-4 a x) e^{\coth ^{-1}(a x)}}{15 a \left (1-a^2 x^2\right )^2}\right )}{7 c^4}-\frac {(1-6 a x) e^{\coth ^{-1}(a x)}}{35 a c^4 \left (1-a^2 x^2\right )^3}\) |
\(\Big \downarrow \) 6739 |
\(\displaystyle \frac {6 \left (\frac {4}{5} \left (\frac {2}{3} \int \frac {e^{\coth ^{-1}(a x)}}{1-a^2 x^2}dx-\frac {(1-2 a x) e^{\coth ^{-1}(a x)}}{3 a \left (1-a^2 x^2\right )}\right )-\frac {(1-4 a x) e^{\coth ^{-1}(a x)}}{15 a \left (1-a^2 x^2\right )^2}\right )}{7 c^4}-\frac {(1-6 a x) e^{\coth ^{-1}(a x)}}{35 a c^4 \left (1-a^2 x^2\right )^3}\) |
\(\Big \downarrow \) 6737 |
\(\displaystyle \frac {6 \left (\frac {4}{5} \left (\frac {2 e^{\coth ^{-1}(a x)}}{3 a}-\frac {(1-2 a x) e^{\coth ^{-1}(a x)}}{3 a \left (1-a^2 x^2\right )}\right )-\frac {(1-4 a x) e^{\coth ^{-1}(a x)}}{15 a \left (1-a^2 x^2\right )^2}\right )}{7 c^4}-\frac {(1-6 a x) e^{\coth ^{-1}(a x)}}{35 a c^4 \left (1-a^2 x^2\right )^3}\) |
-1/35*(E^ArcCoth[a*x]*(1 - 6*a*x))/(a*c^4*(1 - a^2*x^2)^3) + (6*(-1/15*(E^ ArcCoth[a*x]*(1 - 4*a*x))/(a*(1 - a^2*x^2)^2) + (4*((2*E^ArcCoth[a*x])/(3* a) - (E^ArcCoth[a*x]*(1 - 2*a*x))/(3*a*(1 - a^2*x^2))))/5))/(7*c^4)
3.6.63.3.1 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 = 0.48 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.68
method | result | size |
gosper | \(\frac {16 a^{6} x^{6}-16 a^{5} x^{5}-40 a^{4} x^{4}+40 a^{3} x^{3}+30 a^{2} x^{2}-30 a x -5}{35 \left (a^{2} x^{2}-1\right )^{3} c^{4} \sqrt {\frac {a x -1}{a x +1}}\, a}\) | \(81\) |
default | \(\frac {16 a^{6} x^{6}-16 a^{5} x^{5}-40 a^{4} x^{4}+40 a^{3} x^{3}+30 a^{2} x^{2}-30 a x -5}{35 \sqrt {\frac {a x -1}{a x +1}}\, c^{4} \left (a x -1\right )^{3} \left (a x +1\right )^{3} a}\) | \(84\) |
trager | \(\frac {\left (16 a^{6} x^{6}-16 a^{5} x^{5}-40 a^{4} x^{4}+40 a^{3} x^{3}+30 a^{2} x^{2}-30 a x -5\right ) \sqrt {-\frac {-a x +1}{a x +1}}}{35 a \,c^{4} \left (a x +1\right )^{2} \left (a x -1\right )^{4}}\) | \(86\) |
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)/(a^ 2*x^2-1)^3/c^4/((a*x-1)/(a*x+1))^(1/2)/a
Time = 0.24 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.13 \[ \int \frac {e^{\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=\frac {{\left (16 \, a^{6} x^{6} - 16 \, a^{5} x^{5} - 40 \, a^{4} x^{4} + 40 \, a^{3} x^{3} + 30 \, a^{2} x^{2} - 30 \, a x - 5\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{35 \, {\left (a^{7} c^{4} x^{6} - 2 \, a^{6} c^{4} x^{5} - a^{5} c^{4} x^{4} + 4 \, a^{4} c^{4} x^{3} - a^{3} c^{4} x^{2} - 2 \, a^{2} c^{4} x + a c^{4}\right )}} \]
1/35*(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 - 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)
\[ \int \frac {e^{\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=\frac {\int \frac {1}{a^{8} x^{8} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}} - 4 a^{6} x^{6} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}} + 6 a^{4} x^{4} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}} - 4 a^{2} x^{2} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}} + \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}\, dx}{c^{4}} \]
Integral(1/(a**8*x**8*sqrt(a*x/(a*x + 1) - 1/(a*x + 1)) - 4*a**6*x**6*sqrt (a*x/(a*x + 1) - 1/(a*x + 1)) + 6*a**4*x**4*sqrt(a*x/(a*x + 1) - 1/(a*x + 1)) - 4*a**2*x**2*sqrt(a*x/(a*x + 1) - 1/(a*x + 1)) + sqrt(a*x/(a*x + 1) - 1/(a*x + 1))), x)/c**4
Time = 0.21 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.11 \[ \int \frac {e^{\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=\frac {1}{2240} \, a {\left (\frac {7 \, {\left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{2}} - 10 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} + 75 \, \sqrt {\frac {a x - 1}{a x + 1}}\right )}}{a^{2} c^{4}} + \frac {\frac {42 \, {\left (a x - 1\right )}}{a x + 1} - \frac {175 \, {\left (a x - 1\right )}^{2}}{{\left (a x + 1\right )}^{2}} + \frac {700 \, {\left (a x - 1\right )}^{3}}{{\left (a x + 1\right )}^{3}} - 5}{a^{2} c^{4} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {7}{2}}}\right )} \]
1/2240*a*(7*(((a*x - 1)/(a*x + 1))^(5/2) - 10*((a*x - 1)/(a*x + 1))^(3/2) + 75*sqrt((a*x - 1)/(a*x + 1)))/(a^2*c^4) + (42*(a*x - 1)/(a*x + 1) - 175* (a*x - 1)^2/(a*x + 1)^2 + 700*(a*x - 1)^3/(a*x + 1)^3 - 5)/(a^2*c^4*((a*x - 1)/(a*x + 1))^(7/2)))
\[ \int \frac {e^{\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=\int { \frac {1}{{\left (a^{2} c x^{2} - c\right )}^{4} \sqrt {\frac {a x - 1}{a x + 1}}} \,d x } \]
Time = 0.07 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.19 \[ \int \frac {e^{\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=\frac {15\,\sqrt {\frac {a\,x-1}{a\,x+1}}}{64\,a\,c^4}-\frac {{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}}{32\,a\,c^4}+\frac {{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/2}}{320\,a\,c^4}-\frac {\frac {5\,{\left (a\,x-1\right )}^2}{{\left (a\,x+1\right )}^2}-\frac {20\,{\left (a\,x-1\right )}^3}{{\left (a\,x+1\right )}^3}-\frac {6\,\left (a\,x-1\right )}{5\,\left (a\,x+1\right )}+\frac {1}{7}}{64\,a\,c^4\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{7/2}} \]