Integrand size = 22, antiderivative size = 130 \[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right )^4} \, dx=\frac {\sqrt {1-a^2 x^2}}{a c^4}+\frac {\sqrt {1-a^2 x^2}}{5 a c^4 (1-a x)^3}-\frac {6 \sqrt {1-a^2 x^2}}{5 a c^4 (1-a x)^2}+\frac {24 \sqrt {1-a^2 x^2}}{5 a c^4 (1-a x)}-\frac {3 \arcsin (a x)}{a c^4} \] Output:
(-a^2*x^2+1)^(1/2)/a/c^4+1/5*(-a^2*x^2+1)^(1/2)/a/c^4/(-a*x+1)^3-6/5*(-a^2 *x^2+1)^(1/2)/a/c^4/(-a*x+1)^2+24/5*(-a^2*x^2+1)^(1/2)/a/c^4/(-a*x+1)-3*ar csin(a*x)/a/c^4
Time = 0.21 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.55 \[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right )^4} \, dx=\frac {\frac {\sqrt {1+a x} \left (24-57 a x+39 a^2 x^2-5 a^3 x^3\right )}{(1-a x)^{5/2}}+30 \arcsin \left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )}{5 a c^4} \] Input:
Integrate[1/(E^ArcTanh[a*x]*(c - c/(a*x))^4),x]
Output:
((Sqrt[1 + a*x]*(24 - 57*a*x + 39*a^2*x^2 - 5*a^3*x^3))/(1 - a*x)^(5/2) + 30*ArcSin[Sqrt[1 - a*x]/Sqrt[2]])/(5*a*c^4)
Time = 1.09 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.96, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {6681, 6678, 570, 529, 2166, 27, 2166, 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.
\(\displaystyle \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right )^4} \, dx\) |
\(\Big \downarrow \) 6681 |
\(\displaystyle \frac {a^4 \int \frac {e^{-\text {arctanh}(a x)} x^4}{(1-a x)^4}dx}{c^4}\) |
\(\Big \downarrow \) 6678 |
\(\displaystyle \frac {a^4 \int \frac {x^4}{(1-a x)^3 \sqrt {1-a^2 x^2}}dx}{c^4}\) |
\(\Big \downarrow \) 570 |
\(\displaystyle \frac {a^4 \int \frac {x^4 (a x+1)^3}{\left (1-a^2 x^2\right )^{7/2}}dx}{c^4}\) |
\(\Big \downarrow \) 529 |
\(\displaystyle \frac {a^4 \left (\frac {(a x+1)^3}{5 a^5 \left (1-a^2 x^2\right )^{5/2}}-\frac {1}{5} \int \frac {(a x+1)^2 \left (\frac {5 x^3}{a}+\frac {5 x^2}{a^2}+\frac {5 x}{a^3}+\frac {3}{a^4}\right )}{\left (1-a^2 x^2\right )^{5/2}}dx\right )}{c^4}\) |
\(\Big \downarrow \) 2166 |
\(\displaystyle \frac {a^4 \left (\frac {1}{5} \left (\frac {1}{3} \int \frac {3 (a x+1) \left (\frac {5 x^2}{a^2}+\frac {10 x}{a^3}+\frac {9}{a^4}\right )}{\left (1-a^2 x^2\right )^{3/2}}dx-\frac {6 (a x+1)^2}{a^5 \left (1-a^2 x^2\right )^{3/2}}\right )+\frac {(a x+1)^3}{5 a^5 \left (1-a^2 x^2\right )^{5/2}}\right )}{c^4}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a^4 \left (\frac {1}{5} \left (\int \frac {(a x+1) \left (\frac {5 x^2}{a^2}+\frac {10 x}{a^3}+\frac {9}{a^4}\right )}{\left (1-a^2 x^2\right )^{3/2}}dx-\frac {6 (a x+1)^2}{a^5 \left (1-a^2 x^2\right )^{3/2}}\right )+\frac {(a x+1)^3}{5 a^5 \left (1-a^2 x^2\right )^{5/2}}\right )}{c^4}\) |
\(\Big \downarrow \) 2166 |
\(\displaystyle \frac {a^4 \left (\frac {1}{5} \left (-\int \frac {5 (a x+3)}{a^4 \sqrt {1-a^2 x^2}}dx-\frac {6 (a x+1)^2}{a^5 \left (1-a^2 x^2\right )^{3/2}}+\frac {24 (a x+1)}{a^5 \sqrt {1-a^2 x^2}}\right )+\frac {(a x+1)^3}{5 a^5 \left (1-a^2 x^2\right )^{5/2}}\right )}{c^4}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a^4 \left (\frac {1}{5} \left (-\frac {5 \int \frac {a x+3}{\sqrt {1-a^2 x^2}}dx}{a^4}-\frac {6 (a x+1)^2}{a^5 \left (1-a^2 x^2\right )^{3/2}}+\frac {24 (a x+1)}{a^5 \sqrt {1-a^2 x^2}}\right )+\frac {(a x+1)^3}{5 a^5 \left (1-a^2 x^2\right )^{5/2}}\right )}{c^4}\) |
\(\Big \downarrow \) 455 |
\(\displaystyle \frac {a^4 \left (\frac {1}{5} \left (-\frac {5 \left (3 \int \frac {1}{\sqrt {1-a^2 x^2}}dx-\frac {\sqrt {1-a^2 x^2}}{a}\right )}{a^4}-\frac {6 (a x+1)^2}{a^5 \left (1-a^2 x^2\right )^{3/2}}+\frac {24 (a x+1)}{a^5 \sqrt {1-a^2 x^2}}\right )+\frac {(a x+1)^3}{5 a^5 \left (1-a^2 x^2\right )^{5/2}}\right )}{c^4}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {a^4 \left (\frac {(a x+1)^3}{5 a^5 \left (1-a^2 x^2\right )^{5/2}}+\frac {1}{5} \left (-\frac {6 (a x+1)^2}{a^5 \left (1-a^2 x^2\right )^{3/2}}+\frac {24 (a x+1)}{a^5 \sqrt {1-a^2 x^2}}-\frac {5 \left (\frac {3 \arcsin (a x)}{a}-\frac {\sqrt {1-a^2 x^2}}{a}\right )}{a^4}\right )\right )}{c^4}\) |
Input:
Int[1/(E^ArcTanh[a*x]*(c - c/(a*x))^4),x]
Output:
(a^4*((1 + a*x)^3/(5*a^5*(1 - a^2*x^2)^(5/2)) + ((-6*(1 + a*x)^2)/(a^5*(1 - a^2*x^2)^(3/2)) + (24*(1 + a*x))/(a^5*Sqrt[1 - a^2*x^2]) - (5*(-(Sqrt[1 - a^2*x^2]/a) + (3*ArcSin[a*x])/a))/a^4)/5))/c^4
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[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c^(2*n)/a^n Int[(e*x)^m*((a + b*x^2)^(n + p)/(c - d*x)^ n), x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[b*c^2 + a*d^2, 0] && I LtQ[n, -1] && !(IGtQ[m, 0] && ILtQ[m + n, 0] && !GtQ[p, 1])
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[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_))^(p_.)*((e_.) + (f_.)* (x_))^(m_.), x_Symbol] :> Simp[c^n Int[(e + f*x)^m*(c + d*x)^(p - n)*(1 - a^2*x^2)^(n/2), x], x] /; FreeQ[{a, c, d, e, f, m, p}, x] && EqQ[a*c + d, 0] && IntegerQ[(n - 1)/2] && (IntegerQ[p] || EqQ[p, n/2] || EqQ[p - n/2 - 1 , 0]) && IntegerQ[2*p]
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_))^(p_.), x_Symbol ] :> Simp[d^p Int[u*(1 + c*(x/d))^p*(E^(n*ArcTanh[a*x])/x^p), x], x] /; F reeQ[{a, c, d, n}, x] && EqQ[c^2 - a^2*d^2, 0] && IntegerQ[p]
Time = 0.24 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.48
method | result | size |
risch | \(-\frac {a^{2} x^{2}-1}{a \sqrt {-a^{2} x^{2}+1}\, c^{4}}+\frac {\left (-\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{5 a^{8} \left (x -\frac {1}{a}\right )^{3}}-\frac {6 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{5 a^{7} \left (x -\frac {1}{a}\right )^{2}}-\frac {24 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{5 a^{6} \left (x -\frac {1}{a}\right )}-\frac {3 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{a^{4} \sqrt {a^{2}}}\right ) a^{4}}{c^{4}}\) | \(192\) |
default | \(\frac {a^{4} \left (\frac {\frac {\left (-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )\right )^{\frac {3}{2}}}{5 a \left (x -\frac {1}{a}\right )^{4}}-\frac {\left (-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )\right )^{\frac {3}{2}}}{15 \left (x -\frac {1}{a}\right )^{3}}}{2 a^{8}}+\frac {7 \left (-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )\right )^{\frac {3}{2}}}{12 a^{8} \left (x -\frac {1}{a}\right )^{3}}+\frac {\frac {17 \left (-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )\right )^{\frac {3}{2}}}{8 a \left (x -\frac {1}{a}\right )^{2}}+\frac {17 a \left (\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}-\frac {a \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}\right )}{\sqrt {a^{2}}}\right )}{8}}{a^{6}}+\frac {\frac {15 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{16}-\frac {15 a \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}\right )}{16 \sqrt {a^{2}}}}{a^{5}}+\frac {\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}+\frac {a \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}\right )}{\sqrt {a^{2}}}}{16 a^{5}}\right )}{c^{4}}\) | \(397\) |
Input:
int(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(c-c/a/x)^4,x,method=_RETURNVERBOSE)
Output:
-1/a*(a^2*x^2-1)/(-a^2*x^2+1)^(1/2)/c^4+(-1/5/a^8/(x-1/a)^3*(-(x-1/a)^2*a^ 2-2*a*(x-1/a))^(1/2)-6/5/a^7/(x-1/a)^2*(-(x-1/a)^2*a^2-2*a*(x-1/a))^(1/2)- 24/5/a^6/(x-1/a)*(-(x-1/a)^2*a^2-2*a*(x-1/a))^(1/2)-3/a^4/(a^2)^(1/2)*arct an((a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2)))*a^4/c^4
Time = 0.09 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.10 \[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right )^4} \, dx=\frac {24 \, a^{3} x^{3} - 72 \, a^{2} x^{2} + 72 \, a x + 30 \, {\left (a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + {\left (5 \, a^{3} x^{3} - 39 \, a^{2} x^{2} + 57 \, a x - 24\right )} \sqrt {-a^{2} x^{2} + 1} - 24}{5 \, {\left (a^{4} c^{4} x^{3} - 3 \, a^{3} c^{4} x^{2} + 3 \, a^{2} c^{4} x - a c^{4}\right )}} \] Input:
integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(c-c/a/x)^4,x, algorithm="fricas")
Output:
1/5*(24*a^3*x^3 - 72*a^2*x^2 + 72*a*x + 30*(a^3*x^3 - 3*a^2*x^2 + 3*a*x - 1)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + (5*a^3*x^3 - 39*a^2*x^2 + 57*a *x - 24)*sqrt(-a^2*x^2 + 1) - 24)/(a^4*c^4*x^3 - 3*a^3*c^4*x^2 + 3*a^2*c^4 *x - a*c^4)
\[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right )^4} \, dx=\frac {a^{4} \int \frac {x^{4} \sqrt {- a^{2} x^{2} + 1}}{a^{5} x^{5} - 3 a^{4} x^{4} + 2 a^{3} x^{3} + 2 a^{2} x^{2} - 3 a x + 1}\, dx}{c^{4}} \] Input:
integrate(1/(a*x+1)*(-a**2*x**2+1)**(1/2)/(c-c/a/x)**4,x)
Output:
a**4*Integral(x**4*sqrt(-a**2*x**2 + 1)/(a**5*x**5 - 3*a**4*x**4 + 2*a**3* x**3 + 2*a**2*x**2 - 3*a*x + 1), x)/c**4
\[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right )^4} \, dx=\int { \frac {\sqrt {-a^{2} x^{2} + 1}}{{\left (a x + 1\right )} {\left (c - \frac {c}{a x}\right )}^{4}} \,d x } \] Input:
integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(c-c/a/x)^4,x, algorithm="maxima")
Output:
integrate(sqrt(-a^2*x^2 + 1)/((a*x + 1)*(c - c/(a*x))^4), x)
Time = 0.14 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.38 \[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right )^4} \, dx=-\frac {3 \, \arcsin \left (a x\right ) \mathrm {sgn}\left (a\right )}{c^{4} {\left | a \right |}} + \frac {\sqrt {-a^{2} x^{2} + 1}}{a c^{4}} - \frac {2 \, {\left (\frac {80 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}}{a^{2} x} - \frac {120 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2}}{a^{4} x^{2}} + \frac {70 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{3}}{a^{6} x^{3}} - \frac {15 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{4}}{a^{8} x^{4}} - 19\right )}}{5 \, c^{4} {\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{a^{2} x} - 1\right )}^{5} {\left | a \right |}} \] Input:
integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(c-c/a/x)^4,x, algorithm="giac")
Output:
-3*arcsin(a*x)*sgn(a)/(c^4*abs(a)) + sqrt(-a^2*x^2 + 1)/(a*c^4) - 2/5*(80* (sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) - 120*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2/(a^4*x^2) + 70*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^3/(a^6*x^3) - 15*(sqr t(-a^2*x^2 + 1)*abs(a) + a)^4/(a^8*x^4) - 19)/(c^4*((sqrt(-a^2*x^2 + 1)*ab s(a) + a)/(a^2*x) - 1)^5*abs(a))
Time = 23.21 (sec) , antiderivative size = 272, normalized size of antiderivative = 2.09 \[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right )^4} \, dx=\frac {2\,a^4\,\sqrt {1-a^2\,x^2}}{15\,\left (a^7\,c^4\,x^2-2\,a^6\,c^4\,x+a^5\,c^4\right )}-\frac {3\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{c^4\,\sqrt {-a^2}}-\frac {4\,a\,\sqrt {1-a^2\,x^2}}{3\,\left (a^4\,c^4\,x^2-2\,a^3\,c^4\,x+a^2\,c^4\right )}+\frac {\sqrt {1-a^2\,x^2}}{a\,c^4}+\frac {24\,\sqrt {1-a^2\,x^2}}{5\,\sqrt {-a^2}\,\left (c^4\,x\,\sqrt {-a^2}-\frac {c^4\,\sqrt {-a^2}}{a}\right )}+\frac {\sqrt {1-a^2\,x^2}}{5\,\sqrt {-a^2}\,\left (3\,c^4\,x\,\sqrt {-a^2}-\frac {c^4\,\sqrt {-a^2}}{a}+a^2\,c^4\,x^3\,\sqrt {-a^2}-3\,a\,c^4\,x^2\,\sqrt {-a^2}\right )} \] Input:
int((1 - a^2*x^2)^(1/2)/((c - c/(a*x))^4*(a*x + 1)),x)
Output:
(2*a^4*(1 - a^2*x^2)^(1/2))/(15*(a^5*c^4 - 2*a^6*c^4*x + a^7*c^4*x^2)) - ( 3*asinh(x*(-a^2)^(1/2)))/(c^4*(-a^2)^(1/2)) - (4*a*(1 - a^2*x^2)^(1/2))/(3 *(a^2*c^4 - 2*a^3*c^4*x + a^4*c^4*x^2)) + (1 - a^2*x^2)^(1/2)/(a*c^4) + (2 4*(1 - a^2*x^2)^(1/2))/(5*(-a^2)^(1/2)*(c^4*x*(-a^2)^(1/2) - (c^4*(-a^2)^( 1/2))/a)) + (1 - a^2*x^2)^(1/2)/(5*(-a^2)^(1/2)*(3*c^4*x*(-a^2)^(1/2) - (c ^4*(-a^2)^(1/2))/a + a^2*c^4*x^3*(-a^2)^(1/2) - 3*a*c^4*x^2*(-a^2)^(1/2)))
Time = 0.15 (sec) , antiderivative size = 268, normalized size of antiderivative = 2.06 \[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right )^4} \, dx=\frac {-15 \sqrt {-a^{2} x^{2}+1}\, \mathit {asin} \left (a x \right ) a^{2} x^{2}+30 \sqrt {-a^{2} x^{2}+1}\, \mathit {asin} \left (a x \right ) a x -15 \sqrt {-a^{2} x^{2}+1}\, \mathit {asin} \left (a x \right )-15 \mathit {asin} \left (a x \right ) a^{3} x^{3}+45 \mathit {asin} \left (a x \right ) a^{2} x^{2}-45 \mathit {asin} \left (a x \right ) a x +15 \mathit {asin} \left (a x \right )+5 \sqrt {-a^{2} x^{2}+1}\, a^{3} x^{3}-21 \sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}+21 \sqrt {-a^{2} x^{2}+1}\, a x -6 \sqrt {-a^{2} x^{2}+1}-5 a^{4} x^{4}+52 a^{3} x^{3}-72 a^{2} x^{2}+21 a x +6}{5 a \,c^{4} \left (\sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}-2 \sqrt {-a^{2} x^{2}+1}\, a x +\sqrt {-a^{2} x^{2}+1}+a^{3} x^{3}-3 a^{2} x^{2}+3 a x -1\right )} \] Input:
int(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(c-c/a/x)^4,x)
Output:
( - 15*sqrt( - a**2*x**2 + 1)*asin(a*x)*a**2*x**2 + 30*sqrt( - a**2*x**2 + 1)*asin(a*x)*a*x - 15*sqrt( - a**2*x**2 + 1)*asin(a*x) - 15*asin(a*x)*a** 3*x**3 + 45*asin(a*x)*a**2*x**2 - 45*asin(a*x)*a*x + 15*asin(a*x) + 5*sqrt ( - a**2*x**2 + 1)*a**3*x**3 - 21*sqrt( - a**2*x**2 + 1)*a**2*x**2 + 21*sq rt( - a**2*x**2 + 1)*a*x - 6*sqrt( - a**2*x**2 + 1) - 5*a**4*x**4 + 52*a** 3*x**3 - 72*a**2*x**2 + 21*a*x + 6)/(5*a*c**4*(sqrt( - a**2*x**2 + 1)*a**2 *x**2 - 2*sqrt( - a**2*x**2 + 1)*a*x + sqrt( - a**2*x**2 + 1) + a**3*x**3 - 3*a**2*x**2 + 3*a*x - 1))