Integrand size = 22, antiderivative size = 138 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=\frac {8 x}{63 c^4 \left (1-a^2 x^2\right )^{3/2}}-\frac {1}{9 a c^4 (1+a x)^3 \left (1-a^2 x^2\right )^{3/2}}-\frac {2}{21 a c^4 (1+a x)^2 \left (1-a^2 x^2\right )^{3/2}}-\frac {2}{21 a c^4 (1+a x) \left (1-a^2 x^2\right )^{3/2}}+\frac {16 x}{63 c^4 \sqrt {1-a^2 x^2}} \]
8/63*x/c^4/(-a^2*x^2+1)^(3/2)-1/9/a/c^4/(a*x+1)^3/(-a^2*x^2+1)^(3/2)-2/21/ a/c^4/(a*x+1)^2/(-a^2*x^2+1)^(3/2)-2/21/a/c^4/(a*x+1)/(-a^2*x^2+1)^(3/2)+1 6/63*x/c^4/(-a^2*x^2+1)^(1/2)
Time = 0.03 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.54 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=-\frac {19-6 a x-66 a^2 x^2-56 a^3 x^3+24 a^4 x^4+48 a^5 x^5+16 a^6 x^6}{63 a c^4 (1-a x)^{3/2} (1+a x)^{9/2}} \]
-1/63*(19 - 6*a*x - 66*a^2*x^2 - 56*a^3*x^3 + 24*a^4*x^4 + 48*a^5*x^5 + 16 *a^6*x^6)/(a*c^4*(1 - a*x)^(3/2)*(1 + a*x)^(9/2))
Time = 0.32 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {6689, 464, 461, 461, 470, 209, 208}
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.
\(\displaystyle \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx\) |
\(\Big \downarrow \) 6689 |
\(\displaystyle \frac {\int \frac {(1-a x)^3}{\left (1-a^2 x^2\right )^{11/2}}dx}{c^4}\) |
\(\Big \downarrow \) 464 |
\(\displaystyle \frac {\int \frac {1}{(a x+1)^3 \left (1-a^2 x^2\right )^{5/2}}dx}{c^4}\) |
\(\Big \downarrow \) 461 |
\(\displaystyle \frac {\frac {2}{3} \int \frac {1}{(a x+1)^2 \left (1-a^2 x^2\right )^{5/2}}dx-\frac {1}{9 a (a x+1)^3 \left (1-a^2 x^2\right )^{3/2}}}{c^4}\) |
\(\Big \downarrow \) 461 |
\(\displaystyle \frac {\frac {2}{3} \left (\frac {5}{7} \int \frac {1}{(a x+1) \left (1-a^2 x^2\right )^{5/2}}dx-\frac {1}{7 a (a x+1)^2 \left (1-a^2 x^2\right )^{3/2}}\right )-\frac {1}{9 a (a x+1)^3 \left (1-a^2 x^2\right )^{3/2}}}{c^4}\) |
\(\Big \downarrow \) 470 |
\(\displaystyle \frac {\frac {2}{3} \left (\frac {5}{7} \left (\frac {4}{5} \int \frac {1}{\left (1-a^2 x^2\right )^{5/2}}dx-\frac {1}{5 a (a x+1) \left (1-a^2 x^2\right )^{3/2}}\right )-\frac {1}{7 a (a x+1)^2 \left (1-a^2 x^2\right )^{3/2}}\right )-\frac {1}{9 a (a x+1)^3 \left (1-a^2 x^2\right )^{3/2}}}{c^4}\) |
\(\Big \downarrow \) 209 |
\(\displaystyle \frac {\frac {2}{3} \left (\frac {5}{7} \left (\frac {4}{5} \left (\frac {2}{3} \int \frac {1}{\left (1-a^2 x^2\right )^{3/2}}dx+\frac {x}{3 \left (1-a^2 x^2\right )^{3/2}}\right )-\frac {1}{5 a (a x+1) \left (1-a^2 x^2\right )^{3/2}}\right )-\frac {1}{7 a (a x+1)^2 \left (1-a^2 x^2\right )^{3/2}}\right )-\frac {1}{9 a (a x+1)^3 \left (1-a^2 x^2\right )^{3/2}}}{c^4}\) |
\(\Big \downarrow \) 208 |
\(\displaystyle \frac {\frac {2}{3} \left (\frac {5}{7} \left (\frac {4}{5} \left (\frac {2 x}{3 \sqrt {1-a^2 x^2}}+\frac {x}{3 \left (1-a^2 x^2\right )^{3/2}}\right )-\frac {1}{5 a (a x+1) \left (1-a^2 x^2\right )^{3/2}}\right )-\frac {1}{7 a (a x+1)^2 \left (1-a^2 x^2\right )^{3/2}}\right )-\frac {1}{9 a (a x+1)^3 \left (1-a^2 x^2\right )^{3/2}}}{c^4}\) |
(-1/9*1/(a*(1 + a*x)^3*(1 - a^2*x^2)^(3/2)) + (2*(-1/7*1/(a*(1 + a*x)^2*(1 - a^2*x^2)^(3/2)) + (5*(-1/5*1/(a*(1 + a*x)*(1 - a^2*x^2)^(3/2)) + (4*(x/ (3*(1 - a^2*x^2)^(3/2)) + (2*x)/(3*Sqrt[1 - a^2*x^2])))/5))/7))/3)/c^4
3.13.61.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[x/(a*Sqrt[a + b*x^2]), x] /; FreeQ[{a, b}, x]
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && ILtQ[p + 3/2, 0]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (-d)*(c + d*x)^n*((a + b*x^2)^(p + 1)/(2*b*c*(n + p + 1))), x] + Simp[Simpl ify[n + 2*p + 2]/(2*c*(n + p + 1)) Int[(c + d*x)^(n + 1)*(a + b*x^2)^p, x ], x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && ILtQ[Simp lify[n + 2*p + 2], 0] && (LtQ[n, -1] || GtQ[n + p, 0])
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[( a + b*x^2)^(n + p)/(a/c + b*(x/d))^n, x] /; FreeQ[{a, b, c, d}, x] && EqQ[b *c^2 + a*d^2, 0] && IntegerQ[n] && RationalQ[p] && (LtQ[0, -n, p] || LtQ[p, -n, 0]) && NeQ[n, 2] && NeQ[n, -1]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (-d)*(c + d*x)^n*((a + b*x^2)^(p + 1)/(2*b*c*(n + p + 1))), x] + Simp[(n + 2*p + 2)/(2*c*(n + p + 1)) Int[(c + d*x)^(n + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && EqQ[b*c^2 + a*d^2, 0] && LtQ[n, 0] && NeQ[n + p + 1, 0] && IntegerQ[2*p]
Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[c^p Int[(1 - a^2*x^2)^(p + n/2)/(1 - a*x)^n, x], x] /; FreeQ[{a, c, d, p}, x] && EqQ[a^2*c + d, 0] && IntegerQ[p] && ILtQ[(n - 1)/2, 0] && !In tegerQ[p - n/2]
Time = 0.68 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.54
method | result | size |
gosper | \(-\frac {16 a^{6} x^{6}+48 a^{5} x^{5}+24 a^{4} x^{4}-56 a^{3} x^{3}-66 a^{2} x^{2}-6 a x +19}{63 a \,c^{4} \left (a x +1\right )^{3} \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}\) | \(74\) |
trager | \(-\frac {\left (16 a^{6} x^{6}+48 a^{5} x^{5}+24 a^{4} x^{4}-56 a^{3} x^{3}-66 a^{2} x^{2}-6 a x +19\right ) \sqrt {-a^{2} x^{2}+1}}{63 c^{4} \left (a x +1\right )^{5} \left (a x -1\right )^{2} a}\) | \(81\) |
default | \(\text {Expression too large to display}\) | \(1697\) |
-1/63/a/c^4/(a*x+1)^3/(-a^2*x^2+1)^(3/2)*(16*a^6*x^6+48*a^5*x^5+24*a^4*x^4 -56*a^3*x^3-66*a^2*x^2-6*a*x+19)
Time = 0.28 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.41 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=-\frac {19 \, a^{7} x^{7} + 57 \, a^{6} x^{6} + 19 \, a^{5} x^{5} - 95 \, a^{4} x^{4} - 95 \, a^{3} x^{3} + 19 \, a^{2} x^{2} + 57 \, a x + {\left (16 \, a^{6} x^{6} + 48 \, a^{5} x^{5} + 24 \, a^{4} x^{4} - 56 \, a^{3} x^{3} - 66 \, a^{2} x^{2} - 6 \, a x + 19\right )} \sqrt {-a^{2} x^{2} + 1} + 19}{63 \, {\left (a^{8} c^{4} x^{7} + 3 \, a^{7} c^{4} x^{6} + a^{6} c^{4} x^{5} - 5 \, a^{5} c^{4} x^{4} - 5 \, a^{4} c^{4} x^{3} + a^{3} c^{4} x^{2} + 3 \, a^{2} c^{4} x + a c^{4}\right )}} \]
-1/63*(19*a^7*x^7 + 57*a^6*x^6 + 19*a^5*x^5 - 95*a^4*x^4 - 95*a^3*x^3 + 19 *a^2*x^2 + 57*a*x + (16*a^6*x^6 + 48*a^5*x^5 + 24*a^4*x^4 - 56*a^3*x^3 - 6 6*a^2*x^2 - 6*a*x + 19)*sqrt(-a^2*x^2 + 1) + 19)/(a^8*c^4*x^7 + 3*a^7*c^4* x^6 + a^6*c^4*x^5 - 5*a^5*c^4*x^4 - 5*a^4*c^4*x^3 + a^3*c^4*x^2 + 3*a^2*c^ 4*x + a*c^4)
\[ \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=\frac {\int \frac {1}{a^{7} x^{7} \sqrt {- a^{2} x^{2} + 1} + 3 a^{6} x^{6} \sqrt {- a^{2} x^{2} + 1} + a^{5} x^{5} \sqrt {- a^{2} x^{2} + 1} - 5 a^{4} x^{4} \sqrt {- a^{2} x^{2} + 1} - 5 a^{3} x^{3} \sqrt {- a^{2} x^{2} + 1} + a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1} + 3 a x \sqrt {- a^{2} x^{2} + 1} + \sqrt {- a^{2} x^{2} + 1}}\, dx}{c^{4}} \]
Integral(1/(a**7*x**7*sqrt(-a**2*x**2 + 1) + 3*a**6*x**6*sqrt(-a**2*x**2 + 1) + a**5*x**5*sqrt(-a**2*x**2 + 1) - 5*a**4*x**4*sqrt(-a**2*x**2 + 1) - 5*a**3*x**3*sqrt(-a**2*x**2 + 1) + a**2*x**2*sqrt(-a**2*x**2 + 1) + 3*a*x* sqrt(-a**2*x**2 + 1) + sqrt(-a**2*x**2 + 1)), x)/c**4
\[ \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=\int { \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (a^{2} c x^{2} - c\right )}^{4} {\left (a x + 1\right )}^{3}} \,d x } \]
\[ \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=\int { \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (a^{2} c x^{2} - c\right )}^{4} {\left (a x + 1\right )}^{3}} \,d x } \]
Time = 4.38 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.13 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=\frac {\sqrt {1-a^2\,x^2}\,\left (\frac {197\,x}{1008\,c^4}-\frac {155}{1008\,a\,c^4}\right )}{{\left (a\,x-1\right )}^2\,{\left (a\,x+1\right )}^2}-\frac {13\,\sqrt {1-a^2\,x^2}}{252\,a\,c^4\,{\left (a\,x+1\right )}^4}-\frac {\sqrt {1-a^2\,x^2}}{36\,a\,c^4\,{\left (a\,x+1\right )}^5}-\frac {23\,\sqrt {1-a^2\,x^2}}{336\,a\,c^4\,{\left (a\,x+1\right )}^3}-\frac {16\,x\,\sqrt {1-a^2\,x^2}}{63\,c^4\,\left (a\,x-1\right )\,\left (a\,x+1\right )} \]
((1 - a^2*x^2)^(1/2)*((197*x)/(1008*c^4) - 155/(1008*a*c^4)))/((a*x - 1)^2 *(a*x + 1)^2) - (13*(1 - a^2*x^2)^(1/2))/(252*a*c^4*(a*x + 1)^4) - (1 - a^ 2*x^2)^(1/2)/(36*a*c^4*(a*x + 1)^5) - (23*(1 - a^2*x^2)^(1/2))/(336*a*c^4* (a*x + 1)^3) - (16*x*(1 - a^2*x^2)^(1/2))/(63*c^4*(a*x - 1)*(a*x + 1))