Integrand size = 22, antiderivative size = 172 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^3} \, dx=-\frac {7}{a c^3 \sqrt {1-a^2 x^2}}+\frac {181 x}{35 c^3 \sqrt {1-a^2 x^2}}+\frac {1}{7 a c^3 (1+a x)^3 \sqrt {1-a^2 x^2}}-\frac {38}{35 a c^3 (1+a x)^2 \sqrt {1-a^2 x^2}}+\frac {137}{35 a c^3 (1+a x) \sqrt {1-a^2 x^2}}-\frac {\sqrt {1-a^2 x^2}}{a c^3}-\frac {3 \arcsin (a x)}{a c^3} \] Output:
-7/a/c^3/(-a^2*x^2+1)^(1/2)+181/35*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)-38/35/a/c^3/(a*x+1)^2/(-a^2*x^2+1)^(1/2)+137/35/a /c^3/(a*x+1)/(-a^2*x^2+1)^(1/2)-(-a^2*x^2+1)^(1/2)/a/c^3-3*arcsin(a*x)/a/c ^3
Time = 0.05 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.55 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^3} \, dx=\frac {-176-423 a x-125 a^2 x^2+368 a^3 x^3+286 a^4 x^4+35 a^5 x^5-105 (1+a x)^3 \sqrt {1-a^2 x^2} \arcsin (a x)}{35 a (c+a c x)^3 \sqrt {1-a^2 x^2}} \] Input:
Integrate[1/(E^(3*ArcTanh[a*x])*(c - c/(a^2*x^2))^3),x]
Output:
(-176 - 423*a*x - 125*a^2*x^2 + 368*a^3*x^3 + 286*a^4*x^4 + 35*a^5*x^5 - 1 05*(1 + a*x)^3*Sqrt[1 - a^2*x^2]*ArcSin[a*x])/(35*a*(c + a*c*x)^3*Sqrt[1 - a^2*x^2])
Time = 0.87 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.92, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {6707, 6699, 529, 2166, 2166, 27, 2345, 27, 455, 223}
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.
\(\displaystyle \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^3} \, dx\) |
\(\Big \downarrow \) 6707 |
\(\displaystyle -\frac {a^6 \int \frac {e^{-3 \text {arctanh}(a x)} x^6}{\left (1-a^2 x^2\right )^3}dx}{c^3}\) |
\(\Big \downarrow \) 6699 |
\(\displaystyle -\frac {a^6 \int \frac {x^6 (1-a x)^3}{\left (1-a^2 x^2\right )^{9/2}}dx}{c^3}\) |
\(\Big \downarrow \) 529 |
\(\displaystyle -\frac {a^6 \left (-\frac {1}{7} \int \frac {(1-a x)^2 \left (-\frac {7 x^5}{a}+\frac {7 x^4}{a^2}-\frac {7 x^3}{a^3}+\frac {7 x^2}{a^4}-\frac {7 x}{a^5}+\frac {3}{a^6}\right )}{\left (1-a^2 x^2\right )^{7/2}}dx-\frac {(1-a x)^3}{7 a^7 \left (1-a^2 x^2\right )^{7/2}}\right )}{c^3}\) |
\(\Big \downarrow \) 2166 |
\(\displaystyle -\frac {a^6 \left (\frac {1}{7} \left (\frac {1}{5} \int \frac {(1-a x) \left (\frac {35 x^4}{a^2}-\frac {70 x^3}{a^3}+\frac {105 x^2}{a^4}-\frac {140 x}{a^5}+\frac {61}{a^6}\right )}{\left (1-a^2 x^2\right )^{5/2}}dx+\frac {38 (1-a x)^2}{5 a^7 \left (1-a^2 x^2\right )^{5/2}}\right )-\frac {(1-a x)^3}{7 a^7 \left (1-a^2 x^2\right )^{7/2}}\right )}{c^3}\) |
\(\Big \downarrow \) 2166 |
\(\displaystyle -\frac {a^6 \left (\frac {1}{7} \left (\frac {1}{5} \left (-\frac {1}{3} \int \frac {3 \left (-\frac {35 x^3}{a^3}+\frac {105 x^2}{a^4}-\frac {210 x}{a^5}+\frac {76}{a^6}\right )}{\left (1-a^2 x^2\right )^{3/2}}dx-\frac {137 (1-a x)}{a^7 \left (1-a^2 x^2\right )^{3/2}}\right )+\frac {38 (1-a x)^2}{5 a^7 \left (1-a^2 x^2\right )^{5/2}}\right )-\frac {(1-a x)^3}{7 a^7 \left (1-a^2 x^2\right )^{7/2}}\right )}{c^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {a^6 \left (\frac {1}{7} \left (\frac {1}{5} \left (-\int \frac {-\frac {35 x^3}{a^3}+\frac {105 x^2}{a^4}-\frac {210 x}{a^5}+\frac {76}{a^6}}{\left (1-a^2 x^2\right )^{3/2}}dx-\frac {137 (1-a x)}{a^7 \left (1-a^2 x^2\right )^{3/2}}\right )+\frac {38 (1-a x)^2}{5 a^7 \left (1-a^2 x^2\right )^{5/2}}\right )-\frac {(1-a x)^3}{7 a^7 \left (1-a^2 x^2\right )^{7/2}}\right )}{c^3}\) |
\(\Big \downarrow \) 2345 |
\(\displaystyle -\frac {a^6 \left (\frac {1}{7} \left (\frac {1}{5} \left (\int \frac {35 (3-a x)}{a^6 \sqrt {1-a^2 x^2}}dx+\frac {245-181 a x}{a^7 \sqrt {1-a^2 x^2}}-\frac {137 (1-a x)}{a^7 \left (1-a^2 x^2\right )^{3/2}}\right )+\frac {38 (1-a x)^2}{5 a^7 \left (1-a^2 x^2\right )^{5/2}}\right )-\frac {(1-a x)^3}{7 a^7 \left (1-a^2 x^2\right )^{7/2}}\right )}{c^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {a^6 \left (\frac {1}{7} \left (\frac {1}{5} \left (\frac {35 \int \frac {3-a x}{\sqrt {1-a^2 x^2}}dx}{a^6}+\frac {245-181 a x}{a^7 \sqrt {1-a^2 x^2}}-\frac {137 (1-a x)}{a^7 \left (1-a^2 x^2\right )^{3/2}}\right )+\frac {38 (1-a x)^2}{5 a^7 \left (1-a^2 x^2\right )^{5/2}}\right )-\frac {(1-a x)^3}{7 a^7 \left (1-a^2 x^2\right )^{7/2}}\right )}{c^3}\) |
\(\Big \downarrow \) 455 |
\(\displaystyle -\frac {a^6 \left (\frac {1}{7} \left (\frac {1}{5} \left (\frac {35 \left (3 \int \frac {1}{\sqrt {1-a^2 x^2}}dx+\frac {\sqrt {1-a^2 x^2}}{a}\right )}{a^6}+\frac {245-181 a x}{a^7 \sqrt {1-a^2 x^2}}-\frac {137 (1-a x)}{a^7 \left (1-a^2 x^2\right )^{3/2}}\right )+\frac {38 (1-a x)^2}{5 a^7 \left (1-a^2 x^2\right )^{5/2}}\right )-\frac {(1-a x)^3}{7 a^7 \left (1-a^2 x^2\right )^{7/2}}\right )}{c^3}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle -\frac {a^6 \left (\frac {1}{7} \left (\frac {38 (1-a x)^2}{5 a^7 \left (1-a^2 x^2\right )^{5/2}}+\frac {1}{5} \left (\frac {245-181 a x}{a^7 \sqrt {1-a^2 x^2}}-\frac {137 (1-a x)}{a^7 \left (1-a^2 x^2\right )^{3/2}}+\frac {35 \left (\frac {\sqrt {1-a^2 x^2}}{a}+\frac {3 \arcsin (a x)}{a}\right )}{a^6}\right )\right )-\frac {(1-a x)^3}{7 a^7 \left (1-a^2 x^2\right )^{7/2}}\right )}{c^3}\) |
Input:
Int[1/(E^(3*ArcTanh[a*x])*(c - c/(a^2*x^2))^3),x]
Output:
-((a^6*(-1/7*(1 - a*x)^3/(a^7*(1 - a^2*x^2)^(7/2)) + ((38*(1 - a*x)^2)/(5* a^7*(1 - a^2*x^2)^(5/2)) + ((-137*(1 - a*x))/(a^7*(1 - a^2*x^2)^(3/2)) + ( 245 - 181*a*x)/(a^7*Sqrt[1 - a^2*x^2]) + (35*(Sqrt[1 - a^2*x^2]/a + (3*Arc Sin[a*x])/a))/a^6)/5)/7))/c^3)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt [a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
Int[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*(( a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Simp[c Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && !LeQ[p, -1]
Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo l] :> With[{Qx = PolynomialQuotient[x^m, a*d + b*c*x, x], R = PolynomialRem ainder[x^m, a*d + b*c*x, x]}, Simp[(-c)*R*(c + d*x)^n*((a + b*x^2)^(p + 1)/ (2*a*d*(p + 1))), x] + Simp[c/(2*a*(p + 1)) Int[(c + d*x)^(n - 1)*(a + b* x^2)^(p + 1)*ExpandToSum[2*a*d*(p + 1)*Qx + R*(n + 2*p + 2), x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && IGtQ[m, 1] && LtQ[p, -1] && EqQ[b* c^2 + a*d^2, 0]
Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : > With[{Qx = PolynomialQuotient[Pq, a*e + b*d*x, x], R = PolynomialRemainde r[Pq, a*e + b*d*x, x]}, Simp[(-d)*R*(d + e*x)^m*((a + b*x^2)^(p + 1)/(2*a*e *(p + 1))), x] + Simp[d/(2*a*(p + 1)) Int[(d + e*x)^(m - 1)*(a + b*x^2)^( p + 1)*ExpandToSum[2*a*e*(p + 1)*Qx + R*(m + 2*p + 2), x], x], x]] /; FreeQ [{a, b, d, e}, x] && PolyQ[Pq, x] && EqQ[b*d^2 + a*e^2, 0] && ILtQ[p + 1/2, 0] && GtQ[m, 0]
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuot ient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b *f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) In t[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]
Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_ Symbol] :> Simp[c^p Int[x^m*((1 - a^2*x^2)^(p + n/2)/(1 - a*x)^n), x], x] /; FreeQ[{a, c, d, m, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p] || GtQ[c , 0]) && ILtQ[(n - 1)/2, 0] && !IntegerQ[p - n/2]
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_.), x_Symb ol] :> Simp[d^p Int[(u/x^(2*p))*(1 - a^2*x^2)^p*E^(n*ArcTanh[a*x]), x], x ] /; FreeQ[{a, c, d, n}, x] && EqQ[c + a^2*d, 0] && IntegerQ[p]
Time = 0.47 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.45
method | result | size |
risch | \(\frac {a^{2} x^{2}-1}{a \sqrt {-a^{2} x^{2}+1}\, c^{3}}-\frac {\left (\frac {3 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{a^{6} \sqrt {a^{2}}}-\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{16 a^{8} \left (x -\frac {1}{a}\right )}-\frac {\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{14 a^{11} \left (x +\frac {1}{a}\right )^{4}}+\frac {71 \sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{140 a^{10} \left (x +\frac {1}{a}\right )^{3}}-\frac {477 \sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{280 a^{9} \left (x +\frac {1}{a}\right )^{2}}+\frac {2931 \sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{560 a^{8} \left (x +\frac {1}{a}\right )}\right ) a^{6}}{c^{3}}\) | \(250\) |
default | \(\text {Expression too large to display}\) | \(1325\) |
Input:
int(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(c-c/a^2/x^2)^3,x,method=_RETURNVERBOSE )
Output:
1/a*(a^2*x^2-1)/(-a^2*x^2+1)^(1/2)/c^3-(3/a^6/(a^2)^(1/2)*arctan((a^2)^(1/ 2)*x/(-a^2*x^2+1)^(1/2))-1/16/a^8/(x-1/a)*(-(x-1/a)^2*a^2-2*a*(x-1/a))^(1/ 2)-1/14/a^11/(x+1/a)^4*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2)+71/140/a^10/(x+1 /a)^3*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2)-477/280/a^9/(x+1/a)^2*(-a^2*(x+1/ a)^2+2*a*(x+1/a))^(1/2)+2931/560/a^8/(x+1/a)*(-a^2*(x+1/a)^2+2*a*(x+1/a))^ (1/2))*a^6/c^3
Time = 0.11 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.24 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^3} \, dx=-\frac {176 \, a^{5} x^{5} + 528 \, a^{4} x^{4} + 352 \, a^{3} x^{3} - 352 \, a^{2} x^{2} - 528 \, a x - 210 \, {\left (a^{5} x^{5} + 3 \, a^{4} x^{4} + 2 \, a^{3} x^{3} - 2 \, a^{2} x^{2} - 3 \, a x - 1\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + {\left (35 \, a^{5} x^{5} + 286 \, a^{4} x^{4} + 368 \, a^{3} x^{3} - 125 \, a^{2} x^{2} - 423 \, a x - 176\right )} \sqrt {-a^{2} x^{2} + 1} - 176}{35 \, {\left (a^{6} c^{3} x^{5} + 3 \, a^{5} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{3} - 2 \, a^{3} c^{3} x^{2} - 3 \, a^{2} c^{3} x - a c^{3}\right )}} \] Input:
integrate(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(c-c/a^2/x^2)^3,x, algorithm="fri cas")
Output:
-1/35*(176*a^5*x^5 + 528*a^4*x^4 + 352*a^3*x^3 - 352*a^2*x^2 - 528*a*x - 2 10*(a^5*x^5 + 3*a^4*x^4 + 2*a^3*x^3 - 2*a^2*x^2 - 3*a*x - 1)*arctan((sqrt( -a^2*x^2 + 1) - 1)/(a*x)) + (35*a^5*x^5 + 286*a^4*x^4 + 368*a^3*x^3 - 125* a^2*x^2 - 423*a*x - 176)*sqrt(-a^2*x^2 + 1) - 176)/(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)
\[ \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^3} \, dx=\frac {a^{6} \left (\int \frac {x^{6} \sqrt {- a^{2} x^{2} + 1}}{a^{9} x^{9} + 3 a^{8} x^{8} - 8 a^{6} x^{6} - 6 a^{5} x^{5} + 6 a^{4} x^{4} + 8 a^{3} x^{3} - 3 a x - 1}\, dx + \int \left (- \frac {a^{2} x^{8} \sqrt {- a^{2} x^{2} + 1}}{a^{9} x^{9} + 3 a^{8} x^{8} - 8 a^{6} x^{6} - 6 a^{5} x^{5} + 6 a^{4} x^{4} + 8 a^{3} x^{3} - 3 a x - 1}\right )\, dx\right )}{c^{3}} \] Input:
integrate(1/(a*x+1)**3*(-a**2*x**2+1)**(3/2)/(c-c/a**2/x**2)**3,x)
Output:
a**6*(Integral(x**6*sqrt(-a**2*x**2 + 1)/(a**9*x**9 + 3*a**8*x**8 - 8*a**6 *x**6 - 6*a**5*x**5 + 6*a**4*x**4 + 8*a**3*x**3 - 3*a*x - 1), x) + Integra l(-a**2*x**8*sqrt(-a**2*x**2 + 1)/(a**9*x**9 + 3*a**8*x**8 - 8*a**6*x**6 - 6*a**5*x**5 + 6*a**4*x**4 + 8*a**3*x**3 - 3*a*x - 1), x))/c**3
\[ \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^3} \, dx=\int { \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (a x + 1\right )}^{3} {\left (c - \frac {c}{a^{2} x^{2}}\right )}^{3}} \,d x } \] Input:
integrate(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(c-c/a^2/x^2)^3,x, algorithm="max ima")
Output:
integrate((-a^2*x^2 + 1)^(3/2)/((a*x + 1)^3*(c - c/(a^2*x^2))^3), x)
\[ \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^3} \, dx=\int { \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (a x + 1\right )}^{3} {\left (c - \frac {c}{a^{2} x^{2}}\right )}^{3}} \,d x } \] Input:
integrate(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(c-c/a^2/x^2)^3,x, algorithm="gia c")
Output:
integrate((-a^2*x^2 + 1)^(3/2)/((a*x + 1)^3*(c - c/(a^2*x^2))^3), x)
Time = 14.48 (sec) , antiderivative size = 436, normalized size of antiderivative = 2.53 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^3} \, dx=\frac {49\,a\,\sqrt {1-a^2\,x^2}}{24\,\left (a^4\,c^3\,x^2+2\,a^3\,c^3\,x+a^2\,c^3\right )}-\frac {3\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{c^3\,\sqrt {-a^2}}+\frac {a^3\,\sqrt {1-a^2\,x^2}}{35\,\left (a^6\,c^3\,x^2+2\,a^5\,c^3\,x+a^4\,c^3\right )}-\frac {11\,a^6\,\sqrt {1-a^2\,x^2}}{30\,\left (a^9\,c^3\,x^2+2\,a^8\,c^3\,x+a^7\,c^3\right )}-\frac {\sqrt {1-a^2\,x^2}}{a\,c^3}+\frac {a\,\sqrt {1-a^2\,x^2}}{14\,\left (a^6\,c^3\,x^4+4\,a^5\,c^3\,x^3+6\,a^4\,c^3\,x^2+4\,a^3\,c^3\,x+a^2\,c^3\right )}+\frac {2931\,\sqrt {1-a^2\,x^2}}{560\,\sqrt {-a^2}\,\left (c^3\,x\,\sqrt {-a^2}+\frac {c^3\,\sqrt {-a^2}}{a}\right )}-\frac {\sqrt {1-a^2\,x^2}}{16\,\sqrt {-a^2}\,\left (c^3\,x\,\sqrt {-a^2}-\frac {c^3\,\sqrt {-a^2}}{a}\right )}+\frac {71\,\sqrt {1-a^2\,x^2}}{140\,\sqrt {-a^2}\,\left (3\,c^3\,x\,\sqrt {-a^2}+\frac {c^3\,\sqrt {-a^2}}{a}+a^2\,c^3\,x^3\,\sqrt {-a^2}+3\,a\,c^3\,x^2\,\sqrt {-a^2}\right )} \] Input:
int((1 - a^2*x^2)^(3/2)/((c - c/(a^2*x^2))^3*(a*x + 1)^3),x)
Output:
(49*a*(1 - a^2*x^2)^(1/2))/(24*(a^2*c^3 + 2*a^3*c^3*x + a^4*c^3*x^2)) - (3 *asinh(x*(-a^2)^(1/2)))/(c^3*(-a^2)^(1/2)) + (a^3*(1 - a^2*x^2)^(1/2))/(35 *(a^4*c^3 + 2*a^5*c^3*x + a^6*c^3*x^2)) - (11*a^6*(1 - a^2*x^2)^(1/2))/(30 *(a^7*c^3 + 2*a^8*c^3*x + a^9*c^3*x^2)) - (1 - a^2*x^2)^(1/2)/(a*c^3) + (a *(1 - a^2*x^2)^(1/2))/(14*(a^2*c^3 + 4*a^3*c^3*x + 6*a^4*c^3*x^2 + 4*a^5*c ^3*x^3 + a^6*c^3*x^4)) + (2931*(1 - a^2*x^2)^(1/2))/(560*(-a^2)^(1/2)*(c^3 *x*(-a^2)^(1/2) + (c^3*(-a^2)^(1/2))/a)) - (1 - a^2*x^2)^(1/2)/(16*(-a^2)^ (1/2)*(c^3*x*(-a^2)^(1/2) - (c^3*(-a^2)^(1/2))/a)) + (71*(1 - a^2*x^2)^(1/ 2))/(140*(-a^2)^(1/2)*(3*c^3*x*(-a^2)^(1/2) + (c^3*(-a^2)^(1/2))/a + a^2*c ^3*x^3*(-a^2)^(1/2) + 3*a*c^3*x^2*(-a^2)^(1/2)))
Time = 0.15 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.34 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^3} \, dx=\frac {-105 \sqrt {-a^{2} x^{2}+1}\, \mathit {asin} \left (a x \right ) a^{3} x^{3}-315 \sqrt {-a^{2} x^{2}+1}\, \mathit {asin} \left (a x \right ) a^{2} x^{2}-315 \sqrt {-a^{2} x^{2}+1}\, \mathit {asin} \left (a x \right ) a x -105 \sqrt {-a^{2} x^{2}+1}\, \mathit {asin} \left (a x \right )-141 \sqrt {-a^{2} x^{2}+1}\, a^{3} x^{3}-423 \sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}-423 \sqrt {-a^{2} x^{2}+1}\, a x -141 \sqrt {-a^{2} x^{2}+1}+35 a^{5} x^{5}+286 a^{4} x^{4}+368 a^{3} x^{3}-125 a^{2} x^{2}-423 a x -176}{35 \sqrt {-a^{2} x^{2}+1}\, a \,c^{3} \left (a^{3} x^{3}+3 a^{2} x^{2}+3 a x +1\right )} \] Input:
int(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(c-c/a^2/x^2)^3,x)
Output:
( - 105*sqrt( - a**2*x**2 + 1)*asin(a*x)*a**3*x**3 - 315*sqrt( - a**2*x**2 + 1)*asin(a*x)*a**2*x**2 - 315*sqrt( - a**2*x**2 + 1)*asin(a*x)*a*x - 105 *sqrt( - a**2*x**2 + 1)*asin(a*x) - 141*sqrt( - a**2*x**2 + 1)*a**3*x**3 - 423*sqrt( - a**2*x**2 + 1)*a**2*x**2 - 423*sqrt( - a**2*x**2 + 1)*a*x - 1 41*sqrt( - a**2*x**2 + 1) + 35*a**5*x**5 + 286*a**4*x**4 + 368*a**3*x**3 - 125*a**2*x**2 - 423*a*x - 176)/(35*sqrt( - a**2*x**2 + 1)*a*c**3*(a**3*x* *3 + 3*a**2*x**2 + 3*a*x + 1))