Integrand size = 21, antiderivative size = 55 \[ \int \frac {e^{\text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^2} \, dx=\frac {1+a x}{3 a^2 c^2 \left (1-a^2 x^2\right )^{3/2}}-\frac {x}{3 a c^2 \sqrt {1-a^2 x^2}} \] Output:
1/3*(a*x+1)/a^2/c^2/(-a^2*x^2+1)^(3/2)-1/3*x/a/c^2/(-a^2*x^2+1)^(1/2)
Time = 0.03 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.80 \[ \int \frac {e^{\text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^2} \, dx=\frac {-1+a x-a^2 x^2}{3 a^2 c^2 (-1+a x) \sqrt {1-a^2 x^2}} \] Input:
Integrate[(E^ArcTanh[a*x]*x)/(c - a^2*c*x^2)^2,x]
Output:
(-1 + a*x - a^2*x^2)/(3*a^2*c^2*(-1 + a*x)*Sqrt[1 - a^2*x^2])
Time = 0.42 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.96, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {6698, 530, 27, 208}
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.
\(\displaystyle \int \frac {x e^{\text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 6698 |
\(\displaystyle \frac {\int \frac {x (a x+1)}{\left (1-a^2 x^2\right )^{5/2}}dx}{c^2}\) |
\(\Big \downarrow \) 530 |
\(\displaystyle \frac {\frac {a x+1}{3 a^2 \left (1-a^2 x^2\right )^{3/2}}-\frac {1}{3} \int \frac {1}{a \left (1-a^2 x^2\right )^{3/2}}dx}{c^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {a x+1}{3 a^2 \left (1-a^2 x^2\right )^{3/2}}-\frac {\int \frac {1}{\left (1-a^2 x^2\right )^{3/2}}dx}{3 a}}{c^2}\) |
\(\Big \downarrow \) 208 |
\(\displaystyle \frac {\frac {a x+1}{3 a^2 \left (1-a^2 x^2\right )^{3/2}}-\frac {x}{3 a \sqrt {1-a^2 x^2}}}{c^2}\) |
Input:
Int[(E^ArcTanh[a*x]*x)/(c - a^2*c*x^2)^2,x]
Output:
((1 + a*x)/(3*a^2*(1 - a^2*x^2)^(3/2)) - x/(3*a*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[((a_) + (b_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[x/(a*Sqrt[a + b*x^2]), x] /; FreeQ[{a, b}, x]
Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symb ol] :> With[{Qx = PolynomialQuotient[x^m*(c + d*x)^n, a + b*x^2, x], e = Co eff[PolynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 0], f = Coeff[Po lynomialRemainder[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[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Qx + e*(2*p + 3), x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && IGtQ[m, 0] && LtQ[p, -1] && EqQ[n, 1] && IntegerQ[2*p]
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 = 0.75
method | result | size |
gosper | \(-\frac {a^{2} x^{2}-a x +1}{3 \left (a x -1\right ) c^{2} \sqrt {-a^{2} x^{2}+1}\, a^{2}}\) | \(41\) |
trager | \(\frac {\left (a^{2} x^{2}-a x +1\right ) \sqrt {-a^{2} x^{2}+1}}{3 c^{2} a^{2} \left (a x -1\right )^{2} \left (a x +1\right )}\) | \(48\) |
orering | \(-\frac {\left (a^{2} x^{2}-a x +1\right ) \left (a x -1\right ) \left (a x +1\right )^{2}}{3 a^{2} \sqrt {-a^{2} x^{2}+1}\, \left (-a^{2} c \,x^{2}+c \right )^{2}}\) | \(56\) |
default | \(\frac {\frac {\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 a \left (x -\frac {1}{a}\right )^{2}}-\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 \left (x -\frac {1}{a}\right )}}{2 a^{3}}+\frac {\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{4 a^{3} \left (x +\frac {1}{a}\right )}+\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{4 a^{3} \left (x -\frac {1}{a}\right )}}{c^{2}}\) | \(167\) |
Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*x/(-a^2*c*x^2+c)^2,x,method=_RETURNVERBOSE)
Output:
-1/3*(a^2*x^2-a*x+1)/(a*x-1)/c^2/(-a^2*x^2+1)^(1/2)/a^2
Time = 0.07 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.62 \[ \int \frac {e^{\text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^2} \, dx=\frac {a^{3} x^{3} - a^{2} x^{2} - a x + {\left (a^{2} x^{2} - a x + 1\right )} \sqrt {-a^{2} x^{2} + 1} + 1}{3 \, {\left (a^{5} c^{2} x^{3} - a^{4} c^{2} x^{2} - a^{3} c^{2} x + a^{2} c^{2}\right )}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x/(-a^2*c*x^2+c)^2,x, algorithm="fric as")
Output:
1/3*(a^3*x^3 - a^2*x^2 - a*x + (a^2*x^2 - a*x + 1)*sqrt(-a^2*x^2 + 1) + 1) /(a^5*c^2*x^3 - a^4*c^2*x^2 - a^3*c^2*x + a^2*c^2)
\[ \int \frac {e^{\text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^2} \, dx=\frac {\int \frac {x}{a^{4} x^{4} \sqrt {- a^{2} x^{2} + 1} - 2 a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1} + \sqrt {- a^{2} x^{2} + 1}}\, dx + \int \frac {a x^{2}}{a^{4} x^{4} \sqrt {- a^{2} x^{2} + 1} - 2 a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1} + \sqrt {- a^{2} x^{2} + 1}}\, dx}{c^{2}} \] Input:
integrate((a*x+1)/(-a**2*x**2+1)**(1/2)*x/(-a**2*c*x**2+c)**2,x)
Output:
(Integral(x/(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**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))/c **2
\[ \int \frac {e^{\text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^2} \, dx=\int { \frac {{\left (a x + 1\right )} x}{{\left (a^{2} c x^{2} - c\right )}^{2} \sqrt {-a^{2} x^{2} + 1}} \,d x } \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x/(-a^2*c*x^2+c)^2,x, algorithm="maxi ma")
Output:
a*integrate(x^2/((a^4*c^2*x^4 - 2*a^2*c^2*x^2 + c^2)*sqrt(a*x + 1)*sqrt(-a *x + 1)), x) + 1/3/((-a^2*x^2 + 1)^(3/2)*a^2*c^2)
\[ \int \frac {e^{\text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^2} \, dx=\int { \frac {{\left (a x + 1\right )} x}{{\left (a^{2} c x^{2} - c\right )}^{2} \sqrt {-a^{2} x^{2} + 1}} \,d x } \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x/(-a^2*c*x^2+c)^2,x, algorithm="giac ")
Output:
integrate((a*x + 1)*x/((a^2*c*x^2 - c)^2*sqrt(-a^2*x^2 + 1)), x)
Time = 23.47 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.15 \[ \int \frac {e^{\text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^2} \, dx=\frac {1}{3\,a^2\,c^2\,{\left (1-a^2\,x^2\right )}^{3/2}}-\frac {x}{3\,a\,c^2\,\sqrt {1-a^2\,x^2}}+\frac {x}{3\,a\,c^2\,{\left (1-a^2\,x^2\right )}^{3/2}} \] Input:
int((x*(a*x + 1))/((c - a^2*c*x^2)^2*(1 - a^2*x^2)^(1/2)),x)
Output:
1/(3*a^2*c^2*(1 - a^2*x^2)^(3/2)) - x/(3*a*c^2*(1 - a^2*x^2)^(1/2)) + x/(3 *a*c^2*(1 - a^2*x^2)^(3/2))
Time = 0.15 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.24 \[ \int \frac {e^{\text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^2} \, dx=\frac {\sqrt {-a^{2} x^{2}+1}\, a x -\sqrt {-a^{2} x^{2}+1}-a^{2} x^{2}+a x -1}{3 \sqrt {-a^{2} x^{2}+1}\, a^{2} c^{2} \left (a x -1\right )} \] Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*x/(-a^2*c*x^2+c)^2,x)
Output:
(sqrt( - a**2*x**2 + 1)*a*x - sqrt( - a**2*x**2 + 1) - a**2*x**2 + a*x - 1 )/(3*sqrt( - a**2*x**2 + 1)*a**2*c**2*(a*x - 1))