Integrand size = 23, antiderivative size = 57 \[ \int \frac {e^{\text {arctanh}(a x)} x^2}{\left (c-a^2 c x^2\right )^2} \, dx=\frac {x^2 (1+a x)}{3 a c^2 \left (1-a^2 x^2\right )^{3/2}}-\frac {2}{3 a^3 c^2 \sqrt {1-a^2 x^2}} \] Output:
1/3*x^2*(a*x+1)/a/c^2/(-a^2*x^2+1)^(3/2)-2/3/a^3/c^2/(-a^2*x^2+1)^(1/2)
Time = 0.03 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.79 \[ \int \frac {e^{\text {arctanh}(a x)} x^2}{\left (c-a^2 c x^2\right )^2} \, dx=\frac {2-2 a x-a^2 x^2}{3 a^3 c^2 (-1+a x) \sqrt {1-a^2 x^2}} \] Input:
Integrate[(E^ArcTanh[a*x]*x^2)/(c - a^2*c*x^2)^2,x]
Output:
(2 - 2*a*x - a^2*x^2)/(3*a^3*c^2*(-1 + a*x)*Sqrt[1 - a^2*x^2])
Time = 0.48 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {6698, 529, 27, 453}
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.
\(\displaystyle \int \frac {x^2 e^{\text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 6698 |
\(\displaystyle \frac {\int \frac {x^2 (a x+1)}{\left (1-a^2 x^2\right )^{5/2}}dx}{c^2}\) |
\(\Big \downarrow \) 529 |
\(\displaystyle \frac {\frac {a x+1}{3 a^3 \left (1-a^2 x^2\right )^{3/2}}-\frac {1}{3} \int \frac {3 a x+1}{a^2 \left (1-a^2 x^2\right )^{3/2}}dx}{c^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {a x+1}{3 a^3 \left (1-a^2 x^2\right )^{3/2}}-\frac {\int \frac {3 a x+1}{\left (1-a^2 x^2\right )^{3/2}}dx}{3 a^2}}{c^2}\) |
\(\Big \downarrow \) 453 |
\(\displaystyle \frac {\frac {a x+1}{3 a^3 \left (1-a^2 x^2\right )^{3/2}}-\frac {a x+3}{3 a^3 \sqrt {1-a^2 x^2}}}{c^2}\) |
Input:
Int[(E^ArcTanh[a*x]*x^2)/(c - a^2*c*x^2)^2,x]
Output:
((1 + a*x)/(3*a^3*(1 - a^2*x^2)^(3/2)) - (3 + a*x)/(3*a^3*Sqrt[1 - a^2*x^2 ]))/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[((c_) + (d_.)*(x_))/((a_) + (b_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[-(a* d - b*c*x)/(a*b*Sqrt[a + b*x^2]), x] /; FreeQ[{a, b, c, d}, x]
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^(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]) && IGtQ[(n + 1)/2, 0] && !IntegerQ[p - n/2]
Time = 0.15 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.72
method | result | size |
gosper | \(-\frac {a^{2} x^{2}+2 a x -2}{3 \left (a x -1\right ) c^{2} \sqrt {-a^{2} x^{2}+1}\, a^{3}}\) | \(41\) |
trager | \(\frac {\left (a^{2} x^{2}+2 a x -2\right ) \sqrt {-a^{2} x^{2}+1}}{3 c^{2} a^{3} \left (a x -1\right )^{2} \left (a x +1\right )}\) | \(48\) |
orering | \(-\frac {\left (a^{2} x^{2}+2 a x -2\right ) \left (a x -1\right ) \left (a x +1\right )^{2}}{3 a^{3} \sqrt {-a^{2} x^{2}+1}\, \left (-a^{2} c \,x^{2}+c \right )^{2}}\) | \(56\) |
default | \(\frac {\frac {\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 a \left (x -\frac {1}{a}\right )^{2}}-\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 \left (x -\frac {1}{a}\right )}}{2 a^{4}}-\frac {\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{4 a^{4} \left (x +\frac {1}{a}\right )}+\frac {3 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{4 a^{4} \left (x -\frac {1}{a}\right )}}{c^{2}}\) | \(167\) |
Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*x^2/(-a^2*c*x^2+c)^2,x,method=_RETURNVERBOS E)
Output:
-1/3*(a^2*x^2+2*a*x-2)/(a*x-1)/c^2/(-a^2*x^2+1)^(1/2)/a^3
Time = 0.07 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.60 \[ \int \frac {e^{\text {arctanh}(a x)} x^2}{\left (c-a^2 c x^2\right )^2} \, dx=-\frac {2 \, a^{3} x^{3} - 2 \, a^{2} x^{2} - 2 \, a x - {\left (a^{2} x^{2} + 2 \, a x - 2\right )} \sqrt {-a^{2} x^{2} + 1} + 2}{3 \, {\left (a^{6} c^{2} x^{3} - a^{5} c^{2} x^{2} - a^{4} c^{2} x + a^{3} c^{2}\right )}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^2/(-a^2*c*x^2+c)^2,x, algorithm="fr icas")
Output:
-1/3*(2*a^3*x^3 - 2*a^2*x^2 - 2*a*x - (a^2*x^2 + 2*a*x - 2)*sqrt(-a^2*x^2 + 1) + 2)/(a^6*c^2*x^3 - a^5*c^2*x^2 - a^4*c^2*x + a^3*c^2)
\[ \int \frac {e^{\text {arctanh}(a x)} x^2}{\left (c-a^2 c x^2\right )^2} \, dx=\frac {\int \frac {x^{2}}{a^{4} x^{4} \sqrt {- a^{2} x^{2} + 1} - 2 a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1} + \sqrt {- a^{2} x^{2} + 1}}\, dx + \int \frac {a x^{3}}{a^{4} x^{4} \sqrt {- a^{2} x^{2} + 1} - 2 a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1} + \sqrt {- a^{2} x^{2} + 1}}\, dx}{c^{2}} \] Input:
integrate((a*x+1)/(-a**2*x**2+1)**(1/2)*x**2/(-a**2*c*x**2+c)**2,x)
Output:
(Integral(x**2/(a**4*x**4*sqrt(-a**2*x**2 + 1) - 2*a**2*x**2*sqrt(-a**2*x* *2 + 1) + sqrt(-a**2*x**2 + 1)), x) + Integral(a*x**3/(a**4*x**4*sqrt(-a** 2*x**2 + 1) - 2*a**2*x**2*sqrt(-a**2*x**2 + 1) + sqrt(-a**2*x**2 + 1)), x) )/c**2
\[ \int \frac {e^{\text {arctanh}(a x)} x^2}{\left (c-a^2 c x^2\right )^2} \, dx=\int { \frac {{\left (a x + 1\right )} x^{2}}{{\left (a^{2} c x^{2} - c\right )}^{2} \sqrt {-a^{2} x^{2} + 1}} \,d x } \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^2/(-a^2*c*x^2+c)^2,x, algorithm="ma xima")
Output:
integrate((a*x + 1)*x^2/((a^2*c*x^2 - c)^2*sqrt(-a^2*x^2 + 1)), x)
\[ \int \frac {e^{\text {arctanh}(a x)} x^2}{\left (c-a^2 c x^2\right )^2} \, dx=\int { \frac {{\left (a x + 1\right )} x^{2}}{{\left (a^{2} c x^{2} - c\right )}^{2} \sqrt {-a^{2} x^{2} + 1}} \,d x } \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^2/(-a^2*c*x^2+c)^2,x, algorithm="gi ac")
Output:
integrate((a*x + 1)*x^2/((a^2*c*x^2 - c)^2*sqrt(-a^2*x^2 + 1)), x)
Time = 23.50 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.79 \[ \int \frac {e^{\text {arctanh}(a x)} x^2}{\left (c-a^2 c x^2\right )^2} \, dx=-\frac {12\,a^7\,c^2\,{\left (1-a^2\,x^2\right )}^{3/2}-36\,a^7\,c^2\,{\left (1-a^2\,x^2\right )}^{5/2}+12\,a^8\,c^2\,x\,{\left (1-a^2\,x^2\right )}^{3/2}-12\,a^8\,c^2\,x\,{\left (1-a^2\,x^2\right )}^{5/2}}{36\,a^{10}\,c^4\,{\left (a^2\,x^2-1\right )}^3} \] Input:
int((x^2*(a*x + 1))/((c - a^2*c*x^2)^2*(1 - a^2*x^2)^(1/2)),x)
Output:
-(12*a^7*c^2*(1 - a^2*x^2)^(3/2) - 36*a^7*c^2*(1 - a^2*x^2)^(5/2) + 12*a^8 *c^2*x*(1 - a^2*x^2)^(3/2) - 12*a^8*c^2*x*(1 - a^2*x^2)^(5/2))/(36*a^10*c^ 4*(a^2*x^2 - 1)^3)
Time = 0.17 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.23 \[ \int \frac {e^{\text {arctanh}(a x)} x^2}{\left (c-a^2 c x^2\right )^2} \, dx=\frac {-2 \sqrt {-a^{2} x^{2}+1}\, a x +2 \sqrt {-a^{2} x^{2}+1}-a^{2} x^{2}-2 a x +2}{3 \sqrt {-a^{2} x^{2}+1}\, a^{3} c^{2} \left (a x -1\right )} \] Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*x^2/(-a^2*c*x^2+c)^2,x)
Output:
( - 2*sqrt( - a**2*x**2 + 1)*a*x + 2*sqrt( - a**2*x**2 + 1) - a**2*x**2 - 2*a*x + 2)/(3*sqrt( - a**2*x**2 + 1)*a**3*c**2*(a*x - 1))