Integrand size = 22, antiderivative size = 116 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {8 x}{35 c^3 \sqrt {1-a^2 x^2}}-\frac {1}{7 a c^3 (1+a x)^3 \sqrt {1-a^2 x^2}}-\frac {4}{35 a c^3 (1+a x)^2 \sqrt {1-a^2 x^2}}-\frac {4}{35 a c^3 (1+a x) \sqrt {1-a^2 x^2}} \] Output:
8/35*x/c^3/(-a^2*x^2+1)^(1/2)-1/7/a/c^3/(a*x+1)^3/(-a^2*x^2+1)^(1/2)-4/35/ a/c^3/(a*x+1)^2/(-a^2*x^2+1)^(1/2)-4/35/a/c^3/(a*x+1)/(-a^2*x^2+1)^(1/2)
Time = 0.04 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.51 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {-13-4 a x+20 a^2 x^2+24 a^3 x^3+8 a^4 x^4}{35 a c^3 \sqrt {1-a x} (1+a x)^{7/2}} \] Input:
Integrate[1/(E^(3*ArcTanh[a*x])*(c - a^2*c*x^2)^3),x]
Output:
(-13 - 4*a*x + 20*a^2*x^2 + 24*a^3*x^3 + 8*a^4*x^4)/(35*a*c^3*Sqrt[1 - a*x ]*(1 + a*x)^(7/2))
Time = 0.50 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.02, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {6689, 464, 461, 461, 470, 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.
\(\displaystyle \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx\) |
\(\Big \downarrow \) 6689 |
\(\displaystyle \frac {\int \frac {(1-a x)^3}{\left (1-a^2 x^2\right )^{9/2}}dx}{c^3}\) |
\(\Big \downarrow \) 464 |
\(\displaystyle \frac {\int \frac {1}{(a x+1)^3 \left (1-a^2 x^2\right )^{3/2}}dx}{c^3}\) |
\(\Big \downarrow \) 461 |
\(\displaystyle \frac {\frac {4}{7} \int \frac {1}{(a x+1)^2 \left (1-a^2 x^2\right )^{3/2}}dx-\frac {1}{7 a (a x+1)^3 \sqrt {1-a^2 x^2}}}{c^3}\) |
\(\Big \downarrow \) 461 |
\(\displaystyle \frac {\frac {4}{7} \left (\frac {3}{5} \int \frac {1}{(a x+1) \left (1-a^2 x^2\right )^{3/2}}dx-\frac {1}{5 a (a x+1)^2 \sqrt {1-a^2 x^2}}\right )-\frac {1}{7 a (a x+1)^3 \sqrt {1-a^2 x^2}}}{c^3}\) |
\(\Big \downarrow \) 470 |
\(\displaystyle \frac {\frac {4}{7} \left (\frac {3}{5} \left (\frac {2}{3} \int \frac {1}{\left (1-a^2 x^2\right )^{3/2}}dx-\frac {1}{3 a (a x+1) \sqrt {1-a^2 x^2}}\right )-\frac {1}{5 a (a x+1)^2 \sqrt {1-a^2 x^2}}\right )-\frac {1}{7 a (a x+1)^3 \sqrt {1-a^2 x^2}}}{c^3}\) |
\(\Big \downarrow \) 208 |
\(\displaystyle \frac {\frac {4}{7} \left (\frac {3}{5} \left (\frac {2 x}{3 \sqrt {1-a^2 x^2}}-\frac {1}{3 a (a x+1) \sqrt {1-a^2 x^2}}\right )-\frac {1}{5 a (a x+1)^2 \sqrt {1-a^2 x^2}}\right )-\frac {1}{7 a (a x+1)^3 \sqrt {1-a^2 x^2}}}{c^3}\) |
Input:
Int[1/(E^(3*ArcTanh[a*x])*(c - a^2*c*x^2)^3),x]
Output:
(-1/7*1/(a*(1 + a*x)^3*Sqrt[1 - a^2*x^2]) + (4*(-1/5*1/(a*(1 + a*x)^2*Sqrt [1 - a^2*x^2]) + (3*((2*x)/(3*Sqrt[1 - a^2*x^2]) - 1/(3*a*(1 + a*x)*Sqrt[1 - a^2*x^2])))/5))/7)/c^3
Int[((a_) + (b_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[x/(a*Sqrt[a + b*x^2]), x] /; FreeQ[{a, b}, x]
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 = 0.40 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.50
method | result | size |
gosper | \(\frac {8 a^{4} x^{4}+24 a^{3} x^{3}+20 a^{2} x^{2}-4 a x -13}{35 a \,c^{3} \left (a x +1\right )^{3} \sqrt {-a^{2} x^{2}+1}}\) | \(58\) |
trager | \(-\frac {\left (8 a^{4} x^{4}+24 a^{3} x^{3}+20 a^{2} x^{2}-4 a x -13\right ) \sqrt {-a^{2} x^{2}+1}}{35 c^{3} \left (a x +1\right )^{4} a \left (a x -1\right )}\) | \(65\) |
orering | \(-\frac {\left (8 a^{4} x^{4}+24 a^{3} x^{3}+20 a^{2} x^{2}-4 a x -13\right ) \left (a x -1\right ) \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}{35 a \left (a x +1\right )^{2} \left (-a^{2} c \,x^{2}+c \right )^{3}}\) | \(73\) |
default | \(\text {Expression too large to display}\) | \(1323\) |
Input:
int(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(-a^2*c*x^2+c)^3,x,method=_RETURNVERBOS E)
Output:
1/35/a/c^3/(a*x+1)^3/(-a^2*x^2+1)^(1/2)*(8*a^4*x^4+24*a^3*x^3+20*a^2*x^2-4 *a*x-13)
Time = 0.12 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.24 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=-\frac {13 \, a^{5} x^{5} + 39 \, a^{4} x^{4} + 26 \, a^{3} x^{3} - 26 \, a^{2} x^{2} - 39 \, a x + {\left (8 \, a^{4} x^{4} + 24 \, a^{3} x^{3} + 20 \, a^{2} x^{2} - 4 \, a x - 13\right )} \sqrt {-a^{2} x^{2} + 1} - 13}{35 \, {\left (a^{6} c^{3} x^{5} + 3 \, a^{5} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{3} - 2 \, a^{3} c^{3} x^{2} - 3 \, a^{2} c^{3} x - a c^{3}\right )}} \] Input:
integrate(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(-a^2*c*x^2+c)^3,x, algorithm="fr icas")
Output:
-1/35*(13*a^5*x^5 + 39*a^4*x^4 + 26*a^3*x^3 - 26*a^2*x^2 - 39*a*x + (8*a^4 *x^4 + 24*a^3*x^3 + 20*a^2*x^2 - 4*a*x - 13)*sqrt(-a^2*x^2 + 1) - 13)/(a^6 *c^3*x^5 + 3*a^5*c^3*x^4 + 2*a^4*c^3*x^3 - 2*a^3*c^3*x^2 - 3*a^2*c^3*x - a *c^3)
\[ \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {\int \frac {1}{- a^{5} x^{5} \sqrt {- a^{2} x^{2} + 1} - 3 a^{4} x^{4} \sqrt {- a^{2} x^{2} + 1} - 2 a^{3} x^{3} \sqrt {- a^{2} x^{2} + 1} + 2 a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1} + 3 a x \sqrt {- a^{2} x^{2} + 1} + \sqrt {- a^{2} x^{2} + 1}}\, dx}{c^{3}} \] Input:
integrate(1/(a*x+1)**3*(-a**2*x**2+1)**(3/2)/(-a**2*c*x**2+c)**3,x)
Output:
Integral(1/(-a**5*x**5*sqrt(-a**2*x**2 + 1) - 3*a**4*x**4*sqrt(-a**2*x**2 + 1) - 2*a**3*x**3*sqrt(-a**2*x**2 + 1) + 2*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**3
\[ \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=\int { -\frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (a^{2} c x^{2} - c\right )}^{3} {\left (a x + 1\right )}^{3}} \,d x } \] Input:
integrate(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(-a^2*c*x^2+c)^3,x, algorithm="ma xima")
Output:
-integrate((-a^2*x^2 + 1)^(3/2)/((a^2*c*x^2 - c)^3*(a*x + 1)^3), x)
\[ \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=\int { -\frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (a^{2} c x^{2} - c\right )}^{3} {\left (a x + 1\right )}^{3}} \,d x } \] Input:
integrate(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(-a^2*c*x^2+c)^3,x, algorithm="gi ac")
Output:
integrate(-(-a^2*x^2 + 1)^(3/2)/((a^2*c*x^2 - c)^3*(a*x + 1)^3), x)
Time = 23.90 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.08 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=-\frac {29\,\sqrt {1-a^2\,x^2}}{280\,a\,c^3\,{\left (a\,x+1\right )}^2}-\frac {13\,\sqrt {1-a^2\,x^2}}{140\,a\,c^3\,{\left (a\,x+1\right )}^3}-\frac {\sqrt {1-a^2\,x^2}}{14\,a\,c^3\,{\left (a\,x+1\right )}^4}-\frac {\sqrt {1-a^2\,x^2}\,\left (\frac {8\,x}{35\,c^3}-\frac {29}{280\,a\,c^3}\right )}{\left (a\,x-1\right )\,\left (a\,x+1\right )} \] Input:
int((1 - a^2*x^2)^(3/2)/((c - a^2*c*x^2)^3*(a*x + 1)^3),x)
Output:
- (29*(1 - a^2*x^2)^(1/2))/(280*a*c^3*(a*x + 1)^2) - (13*(1 - a^2*x^2)^(1/ 2))/(140*a*c^3*(a*x + 1)^3) - (1 - a^2*x^2)^(1/2)/(14*a*c^3*(a*x + 1)^4) - ((1 - a^2*x^2)^(1/2)*((8*x)/(35*c^3) - 29/(280*a*c^3)))/((a*x - 1)*(a*x + 1))
Time = 0.16 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.21 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {-4 \sqrt {-a^{2} x^{2}+1}\, a^{3} x^{3}-12 \sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}-12 \sqrt {-a^{2} x^{2}+1}\, a x -4 \sqrt {-a^{2} x^{2}+1}+24 a^{4} x^{4}+72 a^{3} x^{3}+60 a^{2} x^{2}-12 a x -39}{105 \sqrt {-a^{2} x^{2}+1}\, a \,c^{3} \left (a^{3} x^{3}+3 a^{2} x^{2}+3 a x +1\right )} \] Input:
int(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(-a^2*c*x^2+c)^3,x)
Output:
( - 4*sqrt( - a**2*x**2 + 1)*a**3*x**3 - 12*sqrt( - a**2*x**2 + 1)*a**2*x* *2 - 12*sqrt( - a**2*x**2 + 1)*a*x - 4*sqrt( - a**2*x**2 + 1) + 24*a**4*x* *4 + 72*a**3*x**3 + 60*a**2*x**2 - 12*a*x - 39)/(105*sqrt( - a**2*x**2 + 1 )*a*c**3*(a**3*x**3 + 3*a**2*x**2 + 3*a*x + 1))