Integrand size = 19, antiderivative size = 132 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^3 (c-a c x)^2} \, dx=\frac {4 a^2 (1+a x)}{3 c^2 \left (1-a^2 x^2\right )^{3/2}}+\frac {a^2 (15+17 a x)}{3 c^2 \sqrt {1-a^2 x^2}}-\frac {\sqrt {1-a^2 x^2}}{2 c^2 x^2}-\frac {3 a \sqrt {1-a^2 x^2}}{c^2 x}-\frac {11 a^2 \text {arctanh}\left (\sqrt {1-a^2 x^2}\right )}{2 c^2} \]
4/3*a^2*(a*x+1)/c^2/(-a^2*x^2+1)^(3/2)-11/2*a^2*arctanh((-a^2*x^2+1)^(1/2) )/c^2+1/3*a^2*(17*a*x+15)/c^2/(-a^2*x^2+1)^(1/2)-1/2*(-a^2*x^2+1)^(1/2)/c^ 2/x^2-3*a*(-a^2*x^2+1)^(1/2)/c^2/x
Time = 0.04 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.78 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^3 (c-a c x)^2} \, dx=\frac {3+15 a x-59 a^2 x^2-19 a^3 x^3+52 a^4 x^4-33 a^2 x^2 (-1+a x) \sqrt {1-a^2 x^2} \text {arctanh}\left (\sqrt {1-a^2 x^2}\right )}{6 c^2 x^2 (-1+a x) \sqrt {1-a^2 x^2}} \]
(3 + 15*a*x - 59*a^2*x^2 - 19*a^3*x^3 + 52*a^4*x^4 - 33*a^2*x^2*(-1 + a*x) *Sqrt[1 - a^2*x^2]*ArcTanh[Sqrt[1 - a^2*x^2]])/(6*c^2*x^2*(-1 + a*x)*Sqrt[ 1 - a^2*x^2])
Time = 0.54 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.96, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.737, Rules used = {6678, 27, 570, 532, 25, 2336, 27, 2338, 25, 27, 534, 243, 73, 221}
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.
\(\displaystyle \int \frac {e^{\text {arctanh}(a x)}}{x^3 (c-a c x)^2} \, dx\) |
\(\Big \downarrow \) 6678 |
\(\displaystyle c \int \frac {\sqrt {1-a^2 x^2}}{c^3 x^3 (1-a x)^3}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\sqrt {1-a^2 x^2}}{x^3 (1-a x)^3}dx}{c^2}\) |
\(\Big \downarrow \) 570 |
\(\displaystyle \frac {\int \frac {(a x+1)^3}{x^3 \left (1-a^2 x^2\right )^{5/2}}dx}{c^2}\) |
\(\Big \downarrow \) 532 |
\(\displaystyle \frac {\frac {4 a^2 (a x+1)}{3 \left (1-a^2 x^2\right )^{3/2}}-\frac {1}{3} \int -\frac {8 a^3 x^3+12 a^2 x^2+9 a x+3}{x^3 \left (1-a^2 x^2\right )^{3/2}}dx}{c^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {1}{3} \int \frac {8 a^3 x^3+12 a^2 x^2+9 a x+3}{x^3 \left (1-a^2 x^2\right )^{3/2}}dx+\frac {4 a^2 (a x+1)}{3 \left (1-a^2 x^2\right )^{3/2}}}{c^2}\) |
\(\Big \downarrow \) 2336 |
\(\displaystyle \frac {\frac {1}{3} \left (\frac {a^2 (17 a x+15)}{\sqrt {1-a^2 x^2}}-\int -\frac {3 \left (5 a^2 x^2+3 a x+1\right )}{x^3 \sqrt {1-a^2 x^2}}dx\right )+\frac {4 a^2 (a x+1)}{3 \left (1-a^2 x^2\right )^{3/2}}}{c^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {1}{3} \left (3 \int \frac {5 a^2 x^2+3 a x+1}{x^3 \sqrt {1-a^2 x^2}}dx+\frac {a^2 (17 a x+15)}{\sqrt {1-a^2 x^2}}\right )+\frac {4 a^2 (a x+1)}{3 \left (1-a^2 x^2\right )^{3/2}}}{c^2}\) |
\(\Big \downarrow \) 2338 |
\(\displaystyle \frac {\frac {1}{3} \left (3 \left (-\frac {1}{2} \int -\frac {a (11 a x+6)}{x^2 \sqrt {1-a^2 x^2}}dx-\frac {\sqrt {1-a^2 x^2}}{2 x^2}\right )+\frac {a^2 (17 a x+15)}{\sqrt {1-a^2 x^2}}\right )+\frac {4 a^2 (a x+1)}{3 \left (1-a^2 x^2\right )^{3/2}}}{c^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {1}{3} \left (3 \left (\frac {1}{2} \int \frac {a (11 a x+6)}{x^2 \sqrt {1-a^2 x^2}}dx-\frac {\sqrt {1-a^2 x^2}}{2 x^2}\right )+\frac {a^2 (17 a x+15)}{\sqrt {1-a^2 x^2}}\right )+\frac {4 a^2 (a x+1)}{3 \left (1-a^2 x^2\right )^{3/2}}}{c^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {1}{3} \left (3 \left (\frac {1}{2} a \int \frac {11 a x+6}{x^2 \sqrt {1-a^2 x^2}}dx-\frac {\sqrt {1-a^2 x^2}}{2 x^2}\right )+\frac {a^2 (17 a x+15)}{\sqrt {1-a^2 x^2}}\right )+\frac {4 a^2 (a x+1)}{3 \left (1-a^2 x^2\right )^{3/2}}}{c^2}\) |
\(\Big \downarrow \) 534 |
\(\displaystyle \frac {\frac {1}{3} \left (3 \left (\frac {1}{2} a \left (11 a \int \frac {1}{x \sqrt {1-a^2 x^2}}dx-\frac {6 \sqrt {1-a^2 x^2}}{x}\right )-\frac {\sqrt {1-a^2 x^2}}{2 x^2}\right )+\frac {a^2 (17 a x+15)}{\sqrt {1-a^2 x^2}}\right )+\frac {4 a^2 (a x+1)}{3 \left (1-a^2 x^2\right )^{3/2}}}{c^2}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\frac {1}{3} \left (3 \left (\frac {1}{2} a \left (\frac {11}{2} a \int \frac {1}{x^2 \sqrt {1-a^2 x^2}}dx^2-\frac {6 \sqrt {1-a^2 x^2}}{x}\right )-\frac {\sqrt {1-a^2 x^2}}{2 x^2}\right )+\frac {a^2 (17 a x+15)}{\sqrt {1-a^2 x^2}}\right )+\frac {4 a^2 (a x+1)}{3 \left (1-a^2 x^2\right )^{3/2}}}{c^2}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {\frac {1}{3} \left (3 \left (\frac {1}{2} a \left (-\frac {11 \int \frac {1}{\frac {1}{a^2}-\frac {x^4}{a^2}}d\sqrt {1-a^2 x^2}}{a}-\frac {6 \sqrt {1-a^2 x^2}}{x}\right )-\frac {\sqrt {1-a^2 x^2}}{2 x^2}\right )+\frac {a^2 (17 a x+15)}{\sqrt {1-a^2 x^2}}\right )+\frac {4 a^2 (a x+1)}{3 \left (1-a^2 x^2\right )^{3/2}}}{c^2}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\frac {1}{3} \left (3 \left (\frac {1}{2} a \left (-11 a \text {arctanh}\left (\sqrt {1-a^2 x^2}\right )-\frac {6 \sqrt {1-a^2 x^2}}{x}\right )-\frac {\sqrt {1-a^2 x^2}}{2 x^2}\right )+\frac {a^2 (17 a x+15)}{\sqrt {1-a^2 x^2}}\right )+\frac {4 a^2 (a x+1)}{3 \left (1-a^2 x^2\right )^{3/2}}}{c^2}\) |
((4*a^2*(1 + a*x))/(3*(1 - a^2*x^2)^(3/2)) + ((a^2*(15 + 17*a*x))/Sqrt[1 - a^2*x^2] + 3*(-1/2*Sqrt[1 - a^2*x^2]/x^2 + (a*((-6*Sqrt[1 - a^2*x^2])/x - 11*a*ArcTanh[Sqrt[1 - a^2*x^2]]))/2))/3)/c^2
3.4.43.3.1 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[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo l] :> With[{Qx = PolynomialQuotient[x^m*(c + d*x)^n, a + b*x^2, x], e = Coe ff[PolynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 0], f = Coeff[Pol ynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 1]}, Simp[(a*f - b*e*x) *((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) Int[x^m *(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*(Qx/x^m) + e*((2*p + 3)/x^m), x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && ILtQ[m, 0] && LtQ[p, -1] && IntegerQ[2*p]
Int[(x_)^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-c)*x^(m + 1)*((a + b*x^2)^(p + 1)/(2*a*(p + 1))), x] + Simp[d Int[ x^(m + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, m, p}, x] && ILtQ[m, 0] && GtQ[p, -1] && EqQ[m + 2*p + 3, 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_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ {Q = PolynomialQuotient[(c*x)^m*Pq, a + b*x^2, x], f = Coeff[PolynomialRema inder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[(c*x) ^m*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)) Int[(c*x)^m*(a + b*x^2)^(p + 1)*Ex pandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x], x]] /; F reeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]
Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{ Q = PolynomialQuotient[Pq, c*x, x], R = PolynomialRemainder[Pq, c*x, x]}, S imp[R*(c*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] + Simp[1/(a*c*( m + 1)) Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*( m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && Lt Q[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])
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]
Time = 0.16 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.37
method | result | size |
default | \(\frac {-\frac {\sqrt {-a^{2} x^{2}+1}}{2 x^{2}}-\frac {11 a^{2} \operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )}{2}-\frac {3 a \sqrt {-a^{2} x^{2}+1}}{x}+2 a \left (\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 \left (x -\frac {1}{a}\right ) a}}{3 a \left (x -\frac {1}{a}\right )^{2}}-\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 \left (x -\frac {1}{a}\right ) a}}{3 \left (x -\frac {1}{a}\right )}\right )-\frac {5 a \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 \left (x -\frac {1}{a}\right ) a}}{x -\frac {1}{a}}}{c^{2}}\) | \(181\) |
risch | \(\frac {6 a^{3} x^{3}+a^{2} x^{2}-6 a x -1}{2 x^{2} \sqrt {-a^{2} x^{2}+1}\, c^{2}}+\frac {a^{2} \left (-11 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )+\frac {\frac {4 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 \left (x -\frac {1}{a}\right ) a}}{3 a \left (x -\frac {1}{a}\right )^{2}}-\frac {4 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 \left (x -\frac {1}{a}\right ) a}}{3 \left (x -\frac {1}{a}\right )}}{a}-\frac {10 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 \left (x -\frac {1}{a}\right ) a}}{a \left (x -\frac {1}{a}\right )}\right )}{2 c^{2}}\) | \(193\) |
1/c^2*(-1/2*(-a^2*x^2+1)^(1/2)/x^2-11/2*a^2*arctanh(1/(-a^2*x^2+1)^(1/2))- 3*a*(-a^2*x^2+1)^(1/2)/x+2*a*(1/3/a/(x-1/a)^2*(-(x-1/a)^2*a^2-2*(x-1/a)*a) ^(1/2)-1/3/(x-1/a)*(-(x-1/a)^2*a^2-2*(x-1/a)*a)^(1/2))-5*a/(x-1/a)*(-(x-1/ a)^2*a^2-2*(x-1/a)*a)^(1/2))
Time = 0.28 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.03 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^3 (c-a c x)^2} \, dx=\frac {38 \, a^{4} x^{4} - 76 \, a^{3} x^{3} + 38 \, a^{2} x^{2} + 33 \, {\left (a^{4} x^{4} - 2 \, a^{3} x^{3} + a^{2} x^{2}\right )} \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) - {\left (52 \, a^{3} x^{3} - 71 \, a^{2} x^{2} + 12 \, a x + 3\right )} \sqrt {-a^{2} x^{2} + 1}}{6 \, {\left (a^{2} c^{2} x^{4} - 2 \, a c^{2} x^{3} + c^{2} x^{2}\right )}} \]
1/6*(38*a^4*x^4 - 76*a^3*x^3 + 38*a^2*x^2 + 33*(a^4*x^4 - 2*a^3*x^3 + a^2* x^2)*log((sqrt(-a^2*x^2 + 1) - 1)/x) - (52*a^3*x^3 - 71*a^2*x^2 + 12*a*x + 3)*sqrt(-a^2*x^2 + 1))/(a^2*c^2*x^4 - 2*a*c^2*x^3 + c^2*x^2)
\[ \int \frac {e^{\text {arctanh}(a x)}}{x^3 (c-a c x)^2} \, dx=\frac {\int \frac {a x}{a^{2} x^{5} \sqrt {- a^{2} x^{2} + 1} - 2 a x^{4} \sqrt {- a^{2} x^{2} + 1} + x^{3} \sqrt {- a^{2} x^{2} + 1}}\, dx + \int \frac {1}{a^{2} x^{5} \sqrt {- a^{2} x^{2} + 1} - 2 a x^{4} \sqrt {- a^{2} x^{2} + 1} + x^{3} \sqrt {- a^{2} x^{2} + 1}}\, dx}{c^{2}} \]
(Integral(a*x/(a**2*x**5*sqrt(-a**2*x**2 + 1) - 2*a*x**4*sqrt(-a**2*x**2 + 1) + x**3*sqrt(-a**2*x**2 + 1)), x) + Integral(1/(a**2*x**5*sqrt(-a**2*x* *2 + 1) - 2*a*x**4*sqrt(-a**2*x**2 + 1) + x**3*sqrt(-a**2*x**2 + 1)), x))/ c**2
\[ \int \frac {e^{\text {arctanh}(a x)}}{x^3 (c-a c x)^2} \, dx=\int { \frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1} {\left (a c x - c\right )}^{2} x^{3}} \,d x } \]
Result contains complex when optimal does not.
Time = 0.35 (sec) , antiderivative size = 332, normalized size of antiderivative = 2.52 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^3 (c-a c x)^2} \, dx=\frac {{\left (33 \, a^{3} \log \left (2\right ) - 66 \, a^{3} \log \left (i + 1\right ) + 104 i \, a^{3}\right )} \mathrm {sgn}\left (\frac {1}{a c x - c}\right ) \mathrm {sgn}\left (a\right ) \mathrm {sgn}\left (c\right ) + \frac {66 \, a^{3} \log \left (\sqrt {-\frac {2 \, c}{a c x - c} - 1} + 1\right )}{\mathrm {sgn}\left (\frac {1}{a c x - c}\right ) \mathrm {sgn}\left (a\right ) \mathrm {sgn}\left (c\right )} - \frac {66 \, a^{3} \log \left ({\left | \sqrt {-\frac {2 \, c}{a c x - c} - 1} - 1 \right |}\right )}{\mathrm {sgn}\left (\frac {1}{a c x - c}\right ) \mathrm {sgn}\left (a\right ) \mathrm {sgn}\left (c\right )} + \frac {3 \, {\left (7 \, a^{3} {\left (-\frac {2 \, c}{a c x - c} - 1\right )}^{\frac {3}{2}} - 5 \, a^{3} \sqrt {-\frac {2 \, c}{a c x - c} - 1}\right )}}{{\left (\frac {c}{a c x - c} + 1\right )}^{2} \mathrm {sgn}\left (\frac {1}{a c x - c}\right ) \mathrm {sgn}\left (a\right ) \mathrm {sgn}\left (c\right )} - \frac {4 \, {\left (a^{3} {\left (-\frac {2 \, c}{a c x - c} - 1\right )}^{\frac {3}{2}} \mathrm {sgn}\left (\frac {1}{a c x - c}\right )^{2} \mathrm {sgn}\left (a\right )^{2} \mathrm {sgn}\left (c\right )^{2} + 18 \, a^{3} \sqrt {-\frac {2 \, c}{a c x - c} - 1} \mathrm {sgn}\left (\frac {1}{a c x - c}\right )^{2} \mathrm {sgn}\left (a\right )^{2} \mathrm {sgn}\left (c\right )^{2}\right )}}{\mathrm {sgn}\left (\frac {1}{a c x - c}\right )^{3} \mathrm {sgn}\left (a\right )^{3} \mathrm {sgn}\left (c\right )^{3}}}{12 \, c^{2} {\left | a \right |}} \]
1/12*((33*a^3*log(2) - 66*a^3*log(I + 1) + 104*I*a^3)*sgn(1/(a*c*x - c))*s gn(a)*sgn(c) + 66*a^3*log(sqrt(-2*c/(a*c*x - c) - 1) + 1)/(sgn(1/(a*c*x - c))*sgn(a)*sgn(c)) - 66*a^3*log(abs(sqrt(-2*c/(a*c*x - c) - 1) - 1))/(sgn( 1/(a*c*x - c))*sgn(a)*sgn(c)) + 3*(7*a^3*(-2*c/(a*c*x - c) - 1)^(3/2) - 5* a^3*sqrt(-2*c/(a*c*x - c) - 1))/((c/(a*c*x - c) + 1)^2*sgn(1/(a*c*x - c))* sgn(a)*sgn(c)) - 4*(a^3*(-2*c/(a*c*x - c) - 1)^(3/2)*sgn(1/(a*c*x - c))^2* sgn(a)^2*sgn(c)^2 + 18*a^3*sqrt(-2*c/(a*c*x - c) - 1)*sgn(1/(a*c*x - c))^2 *sgn(a)^2*sgn(c)^2)/(sgn(1/(a*c*x - c))^3*sgn(a)^3*sgn(c)^3))/(c^2*abs(a))
Time = 0.06 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.28 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^3 (c-a c x)^2} \, dx=\frac {2\,a^4\,\sqrt {1-a^2\,x^2}}{3\,\left (a^4\,c^2\,x^2-2\,a^3\,c^2\,x+a^2\,c^2\right )}-\frac {\sqrt {1-a^2\,x^2}}{2\,c^2\,x^2}-\frac {3\,a\,\sqrt {1-a^2\,x^2}}{c^2\,x}+\frac {17\,a^3\,\sqrt {1-a^2\,x^2}}{3\,\sqrt {-a^2}\,\left (c^2\,x\,\sqrt {-a^2}-\frac {c^2\,\sqrt {-a^2}}{a}\right )}+\frac {a^2\,\mathrm {atan}\left (\sqrt {1-a^2\,x^2}\,1{}\mathrm {i}\right )\,11{}\mathrm {i}}{2\,c^2} \]