Integrand size = 22, antiderivative size = 108 \[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^2} \, dx=\frac {2}{a c^2 \sqrt {1-a^2 x^2}}-\frac {4 x}{3 c^2 \sqrt {1-a^2 x^2}}-\frac {1}{3 a c^2 (1+a x) \sqrt {1-a^2 x^2}}+\frac {\sqrt {1-a^2 x^2}}{a c^2}+\frac {\arcsin (a x)}{a c^2} \] Output:
2/a/c^2/(-a^2*x^2+1)^(1/2)-4/3*x/c^2/(-a^2*x^2+1)^(1/2)-1/3/a/c^2/(a*x+1)/ (-a^2*x^2+1)^(1/2)+(-a^2*x^2+1)^(1/2)/a/c^2+arcsin(a*x)/a/c^2
Time = 0.04 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.72 \[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^2} \, dx=\frac {8+5 a x-7 a^2 x^2-3 a^3 x^3+3 (1+a x) \sqrt {1-a^2 x^2} \arcsin (a x)}{3 a c^2 (1+a x) \sqrt {1-a^2 x^2}} \] Input:
Integrate[1/(E^ArcTanh[a*x]*(c - c/(a^2*x^2))^2),x]
Output:
(8 + 5*a*x - 7*a^2*x^2 - 3*a^3*x^3 + 3*(1 + a*x)*Sqrt[1 - a^2*x^2]*ArcSin[ a*x])/(3*a*c^2*(1 + a*x)*Sqrt[1 - a^2*x^2])
Time = 0.48 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.90, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {6707, 6699, 529, 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.
\(\displaystyle \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^2} \, dx\) |
\(\Big \downarrow \) 6707 |
\(\displaystyle \frac {a^4 \int \frac {e^{-\text {arctanh}(a x)} x^4}{\left (1-a^2 x^2\right )^2}dx}{c^2}\) |
\(\Big \downarrow \) 6699 |
\(\displaystyle \frac {a^4 \int \frac {x^4 (1-a x)}{\left (1-a^2 x^2\right )^{5/2}}dx}{c^2}\) |
\(\Big \downarrow \) 529 |
\(\displaystyle \frac {a^4 \left (-\frac {1}{3} \int \frac {-\frac {3 x^3}{a}+\frac {3 x^2}{a^2}-\frac {3 x}{a^3}+\frac {1}{a^4}}{\left (1-a^2 x^2\right )^{3/2}}dx-\frac {1-a x}{3 a^5 \left (1-a^2 x^2\right )^{3/2}}\right )}{c^2}\) |
\(\Big \downarrow \) 2345 |
\(\displaystyle \frac {a^4 \left (\frac {1}{3} \left (\int \frac {3 (1-a x)}{a^4 \sqrt {1-a^2 x^2}}dx+\frac {2 (3-2 a x)}{a^5 \sqrt {1-a^2 x^2}}\right )-\frac {1-a x}{3 a^5 \left (1-a^2 x^2\right )^{3/2}}\right )}{c^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a^4 \left (\frac {1}{3} \left (\frac {3 \int \frac {1-a x}{\sqrt {1-a^2 x^2}}dx}{a^4}+\frac {2 (3-2 a x)}{a^5 \sqrt {1-a^2 x^2}}\right )-\frac {1-a x}{3 a^5 \left (1-a^2 x^2\right )^{3/2}}\right )}{c^2}\) |
\(\Big \downarrow \) 455 |
\(\displaystyle \frac {a^4 \left (\frac {1}{3} \left (\frac {3 \left (\int \frac {1}{\sqrt {1-a^2 x^2}}dx+\frac {\sqrt {1-a^2 x^2}}{a}\right )}{a^4}+\frac {2 (3-2 a x)}{a^5 \sqrt {1-a^2 x^2}}\right )-\frac {1-a x}{3 a^5 \left (1-a^2 x^2\right )^{3/2}}\right )}{c^2}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {a^4 \left (\frac {1}{3} \left (\frac {2 (3-2 a x)}{a^5 \sqrt {1-a^2 x^2}}+\frac {3 \left (\frac {\sqrt {1-a^2 x^2}}{a}+\frac {\arcsin (a x)}{a}\right )}{a^4}\right )-\frac {1-a x}{3 a^5 \left (1-a^2 x^2\right )^{3/2}}\right )}{c^2}\) |
Input:
Int[1/(E^ArcTanh[a*x]*(c - c/(a^2*x^2))^2),x]
Output:
(a^4*(-1/3*(1 - a*x)/(a^5*(1 - a^2*x^2)^(3/2)) + ((2*(3 - 2*a*x))/(a^5*Sqr t[1 - a^2*x^2]) + (3*(Sqrt[1 - a^2*x^2]/a + ArcSin[a*x]/a))/a^4)/3))/c^2
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_)*((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 = 0.32 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.66
method | result | size |
risch | \(-\frac {a^{2} x^{2}-1}{a \sqrt {-a^{2} x^{2}+1}\, c^{2}}+\frac {\left (\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{a^{4} \sqrt {a^{2}}}+\frac {19 \sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{12 a^{6} \left (x +\frac {1}{a}\right )}-\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{4 a^{6} \left (x -\frac {1}{a}\right )}-\frac {\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{6 a^{7} \left (x +\frac {1}{a}\right )^{2}}\right ) a^{4}}{c^{2}}\) | \(179\) |
default | \(\frac {a^{4} \left (\frac {\frac {\left (-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )\right )^{\frac {3}{2}}}{a \left (x -\frac {1}{a}\right )^{2}}+a \left (\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}-\frac {a \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}\right )}{\sqrt {a^{2}}}\right )}{8 a^{6}}+\frac {\frac {5 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{16}-\frac {5 a \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}\right )}{16 \sqrt {a^{2}}}}{a^{5}}-\frac {\left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {3}{2}}}{12 a^{8} \left (x +\frac {1}{a}\right )^{3}}-\frac {3 \left (-\frac {\left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {3}{2}}}{a \left (x +\frac {1}{a}\right )^{2}}-a \left (\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}+\frac {a \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}\right )}{\sqrt {a^{2}}}\right )\right )}{4 a^{6}}+\frac {\frac {11 \sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{16}+\frac {11 a \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}\right )}{16 \sqrt {a^{2}}}}{a^{5}}\right )}{c^{2}}\) | \(412\) |
Input:
int(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(c-c/a^2/x^2)^2,x,method=_RETURNVERBOSE)
Output:
-1/a*(a^2*x^2-1)/(-a^2*x^2+1)^(1/2)/c^2+(1/a^4/(a^2)^(1/2)*arctan((a^2)^(1 /2)*x/(-a^2*x^2+1)^(1/2))+19/12/a^6/(x+1/a)*(-a^2*(x+1/a)^2+2*a*(x+1/a))^( 1/2)-1/4/a^6/(x-1/a)*(-(x-1/a)^2*a^2-2*a*(x-1/a))^(1/2)-1/6/a^7/(x+1/a)^2* (-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2))*a^4/c^2
Time = 0.09 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.31 \[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^2} \, dx=\frac {8 \, a^{3} x^{3} + 8 \, a^{2} x^{2} - 8 \, a x - 6 \, {\left (a^{3} x^{3} + a^{2} x^{2} - a x - 1\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + {\left (3 \, a^{3} x^{3} + 7 \, a^{2} x^{2} - 5 \, a x - 8\right )} \sqrt {-a^{2} x^{2} + 1} - 8}{3 \, {\left (a^{4} c^{2} x^{3} + a^{3} c^{2} x^{2} - a^{2} c^{2} x - a c^{2}\right )}} \] Input:
integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(c-c/a^2/x^2)^2,x, algorithm="frica s")
Output:
1/3*(8*a^3*x^3 + 8*a^2*x^2 - 8*a*x - 6*(a^3*x^3 + a^2*x^2 - a*x - 1)*arcta n((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + (3*a^3*x^3 + 7*a^2*x^2 - 5*a*x - 8)*sq rt(-a^2*x^2 + 1) - 8)/(a^4*c^2*x^3 + a^3*c^2*x^2 - a^2*c^2*x - a*c^2)
\[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^2} \, dx=\frac {a^{4} \int \frac {x^{4} \sqrt {- a^{2} x^{2} + 1}}{a^{5} x^{5} + a^{4} x^{4} - 2 a^{3} x^{3} - 2 a^{2} x^{2} + a x + 1}\, dx}{c^{2}} \] Input:
integrate(1/(a*x+1)*(-a**2*x**2+1)**(1/2)/(c-c/a**2/x**2)**2,x)
Output:
a**4*Integral(x**4*sqrt(-a**2*x**2 + 1)/(a**5*x**5 + a**4*x**4 - 2*a**3*x* *3 - 2*a**2*x**2 + a*x + 1), x)/c**2
\[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^2} \, dx=\int { \frac {\sqrt {-a^{2} x^{2} + 1}}{{\left (a x + 1\right )} {\left (c - \frac {c}{a^{2} x^{2}}\right )}^{2}} \,d x } \] Input:
integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(c-c/a^2/x^2)^2,x, algorithm="maxim a")
Output:
integrate(sqrt(-a^2*x^2 + 1)/((a*x + 1)*(c - c/(a^2*x^2))^2), x)
\[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^2} \, dx=\int { \frac {\sqrt {-a^{2} x^{2} + 1}}{{\left (a x + 1\right )} {\left (c - \frac {c}{a^{2} x^{2}}\right )}^{2}} \,d x } \] Input:
integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(c-c/a^2/x^2)^2,x, algorithm="giac" )
Output:
integrate(sqrt(-a^2*x^2 + 1)/((a*x + 1)*(c - c/(a^2*x^2))^2), x)
Time = 14.29 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.74 \[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^2} \, dx=\frac {\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{c^2\,\sqrt {-a^2}}-\frac {a\,\sqrt {1-a^2\,x^2}}{6\,\left (a^4\,c^2\,x^2+2\,a^3\,c^2\,x+a^2\,c^2\right )}+\frac {\sqrt {1-a^2\,x^2}}{a\,c^2}-\frac {19\,\sqrt {1-a^2\,x^2}}{12\,\sqrt {-a^2}\,\left (c^2\,x\,\sqrt {-a^2}+\frac {c^2\,\sqrt {-a^2}}{a}\right )}+\frac {\sqrt {1-a^2\,x^2}}{4\,\sqrt {-a^2}\,\left (c^2\,x\,\sqrt {-a^2}-\frac {c^2\,\sqrt {-a^2}}{a}\right )} \] Input:
int((1 - a^2*x^2)^(1/2)/((c - c/(a^2*x^2))^2*(a*x + 1)),x)
Output:
asinh(x*(-a^2)^(1/2))/(c^2*(-a^2)^(1/2)) - (a*(1 - a^2*x^2)^(1/2))/(6*(a^2 *c^2 + 2*a^3*c^2*x + a^4*c^2*x^2)) + (1 - a^2*x^2)^(1/2)/(a*c^2) - (19*(1 - a^2*x^2)^(1/2))/(12*(-a^2)^(1/2)*(c^2*x*(-a^2)^(1/2) + (c^2*(-a^2)^(1/2) )/a)) + (1 - a^2*x^2)^(1/2)/(4*(-a^2)^(1/2)*(c^2*x*(-a^2)^(1/2) - (c^2*(-a ^2)^(1/2))/a))
Time = 0.15 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.06 \[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^2} \, dx=\frac {3 \sqrt {-a^{2} x^{2}+1}\, \mathit {asin} \left (a x \right ) a x +3 \sqrt {-a^{2} x^{2}+1}\, \mathit {asin} \left (a x \right )+5 \sqrt {-a^{2} x^{2}+1}\, a x +5 \sqrt {-a^{2} x^{2}+1}-3 a^{3} x^{3}-7 a^{2} x^{2}+5 a x +8}{3 \sqrt {-a^{2} x^{2}+1}\, a \,c^{2} \left (a x +1\right )} \] Input:
int(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(c-c/a^2/x^2)^2,x)
Output:
(3*sqrt( - a**2*x**2 + 1)*asin(a*x)*a*x + 3*sqrt( - a**2*x**2 + 1)*asin(a* x) + 5*sqrt( - a**2*x**2 + 1)*a*x + 5*sqrt( - a**2*x**2 + 1) - 3*a**3*x**3 - 7*a**2*x**2 + 5*a*x + 8)/(3*sqrt( - a**2*x**2 + 1)*a*c**2*(a*x + 1))