Integrand size = 21, antiderivative size = 80 \[ \int \frac {e^{\text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {1+a x}{5 a^2 c^3 \left (1-a^2 x^2\right )^{5/2}}-\frac {x}{15 a c^3 \left (1-a^2 x^2\right )^{3/2}}-\frac {2 x}{15 a c^3 \sqrt {1-a^2 x^2}} \] Output:
1/5*(a*x+1)/a^2/c^3/(-a^2*x^2+1)^(5/2)-1/15*x/a/c^3/(-a^2*x^2+1)^(3/2)-2/1 5*x/a/c^3/(-a^2*x^2+1)^(1/2)
Time = 0.01 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.85 \[ \int \frac {e^{\text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {3-3 a x+3 a^2 x^2+2 a^3 x^3-2 a^4 x^4}{15 a^2 c^3 (-1+a x)^2 (1+a x) \sqrt {1-a^2 x^2}} \] Input:
Integrate[(E^ArcTanh[a*x]*x)/(c - a^2*c*x^2)^3,x]
Output:
(3 - 3*a*x + 3*a^2*x^2 + 2*a^3*x^3 - 2*a^4*x^4)/(15*a^2*c^3*(-1 + a*x)^2*( 1 + a*x)*Sqrt[1 - a^2*x^2])
Time = 0.29 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {6698, 530, 27, 209, 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.
| \(\displaystyle \int \frac {x e^{\text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx\) |
| \(\Big \downarrow \) 6698 |
| \(\displaystyle \frac {\int \frac {x (a x+1)}{\left (1-a^2 x^2\right )^{7/2}}dx}{c^3}\) |
| \(\Big \downarrow \) 530 |
| \(\displaystyle \frac {\frac {a x+1}{5 a^2 \left (1-a^2 x^2\right )^{5/2}}-\frac {1}{5} \int \frac {1}{a \left (1-a^2 x^2\right )^{5/2}}dx}{c^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {a x+1}{5 a^2 \left (1-a^2 x^2\right )^{5/2}}-\frac {\int \frac {1}{\left (1-a^2 x^2\right )^{5/2}}dx}{5 a}}{c^3}\) |
| \(\Big \downarrow \) 209 |
| \(\displaystyle \frac {\frac {a x+1}{5 a^2 \left (1-a^2 x^2\right )^{5/2}}-\frac {\frac {2}{3} \int \frac {1}{\left (1-a^2 x^2\right )^{3/2}}dx+\frac {x}{3 \left (1-a^2 x^2\right )^{3/2}}}{5 a}}{c^3}\) |
| \(\Big \downarrow \) 208 |
| \(\displaystyle \frac {\frac {a x+1}{5 a^2 \left (1-a^2 x^2\right )^{5/2}}-\frac {\frac {2 x}{3 \sqrt {1-a^2 x^2}}+\frac {x}{3 \left (1-a^2 x^2\right )^{3/2}}}{5 a}}{c^3}\) |
Input:
Int[(E^ArcTanh[a*x]*x)/(c - a^2*c*x^2)^3,x]
Output:
((1 + a*x)/(5*a^2*(1 - a^2*x^2)^(5/2)) - (x/(3*(1 - a^2*x^2)^(3/2)) + (2*x )/(3*Sqrt[1 - a^2*x^2]))/(5*a))/c^3
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[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && ILtQ[p + 3/2, 0]
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 = 0.28 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.72
| method | result | size |
| gosper | \(\frac {2 a^{4} x^{4}-2 a^{3} x^{3}-3 a^{2} x^{2}+3 a x -3}{15 a^{2} c^{3} \left (a x -1\right ) \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}\) | \(58\) |
| trager | \(\frac {\left (2 a^{4} x^{4}-2 a^{3} x^{3}-3 a^{2} x^{2}+3 a x -3\right ) \sqrt {-a^{2} x^{2}+1}}{15 c^{3} \left (a x -1\right )^{3} \left (a x +1\right )^{2} a^{2}}\) | \(65\) |
| orering | \(\frac {\left (2 a^{4} x^{4}-2 a^{3} x^{3}-3 a^{2} x^{2}+3 a x -3\right ) \left (a x -1\right ) \left (a x +1\right )^{2}}{15 a^{2} \sqrt {-a^{2} x^{2}+1}\, \left (-a^{2} c \,x^{2}+c \right )^{3}}\) | \(73\) |
| default | \(-\frac {\frac {\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{5 a \left (x -\frac {1}{a}\right )^{3}}-\frac {2 a \left (\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 a \left (x -\frac {1}{a}\right )^{2}}-\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 \left (x -\frac {1}{a}\right )}\right )}{5}}{4 a^{4}}+\frac {-\frac {\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{3 a \left (x +\frac {1}{a}\right )^{2}}-\frac {\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{3 \left (x +\frac {1}{a}\right )}}{8 a^{3}}-\frac {\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{16 a^{3} \left (x +\frac {1}{a}\right )}-\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{16 a^{3} \left (x -\frac {1}{a}\right )}}{c^{3}}\) | \(286\) |
Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*x/(-a^2*c*x^2+c)^3,x,method=_RETURNVERBOSE)
Output:
1/15/a^2/c^3/(a*x-1)/(-a^2*x^2+1)^(3/2)*(2*a^4*x^4-2*a^3*x^3-3*a^2*x^2+3*a *x-3)
Leaf count of result is larger than twice the leaf count of optimal. 145 vs. \(2 (68) = 136\).
Time = 0.08 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.81 \[ \int \frac {e^{\text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {3 \, a^{5} x^{5} - 3 \, a^{4} x^{4} - 6 \, a^{3} x^{3} + 6 \, a^{2} x^{2} + 3 \, a x + {\left (2 \, a^{4} x^{4} - 2 \, a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 3\right )} \sqrt {-a^{2} x^{2} + 1} - 3}{15 \, {\left (a^{7} c^{3} x^{5} - a^{6} c^{3} x^{4} - 2 \, a^{5} c^{3} x^{3} + 2 \, a^{4} c^{3} x^{2} + a^{3} c^{3} x - a^{2} c^{3}\right )}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x/(-a^2*c*x^2+c)^3,x, algorithm="fric as")
Output:
1/15*(3*a^5*x^5 - 3*a^4*x^4 - 6*a^3*x^3 + 6*a^2*x^2 + 3*a*x + (2*a^4*x^4 - 2*a^3*x^3 - 3*a^2*x^2 + 3*a*x - 3)*sqrt(-a^2*x^2 + 1) - 3)/(a^7*c^3*x^5 - a^6*c^3*x^4 - 2*a^5*c^3*x^3 + 2*a^4*c^3*x^2 + a^3*c^3*x - a^2*c^3)
\[ \int \frac {e^{\text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {\int \frac {x}{- a^{6} x^{6} \sqrt {- a^{2} x^{2} + 1} + 3 a^{4} x^{4} \sqrt {- a^{2} x^{2} + 1} - 3 a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1} + \sqrt {- a^{2} x^{2} + 1}}\, dx + \int \frac {a x^{2}}{- a^{6} x^{6} \sqrt {- a^{2} x^{2} + 1} + 3 a^{4} x^{4} \sqrt {- a^{2} x^{2} + 1} - 3 a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1} + \sqrt {- a^{2} x^{2} + 1}}\, dx}{c^{3}} \] Input:
integrate((a*x+1)/(-a**2*x**2+1)**(1/2)*x/(-a**2*c*x**2+c)**3,x)
Output:
(Integral(x/(-a**6*x**6*sqrt(-a**2*x**2 + 1) + 3*a**4*x**4*sqrt(-a**2*x**2 + 1) - 3*a**2*x**2*sqrt(-a**2*x**2 + 1) + sqrt(-a**2*x**2 + 1)), x) + Int egral(a*x**2/(-a**6*x**6*sqrt(-a**2*x**2 + 1) + 3*a**4*x**4*sqrt(-a**2*x** 2 + 1) - 3*a**2*x**2*sqrt(-a**2*x**2 + 1) + sqrt(-a**2*x**2 + 1)), x))/c** 3
\[ \int \frac {e^{\text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^3} \, dx=\int { -\frac {{\left (a x + 1\right )} x}{{\left (a^{2} c x^{2} - c\right )}^{3} \sqrt {-a^{2} x^{2} + 1}} \,d x } \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x/(-a^2*c*x^2+c)^3,x, algorithm="maxi ma")
Output:
-a*integrate(x^2/((a^6*c^3*x^6 - 3*a^4*c^3*x^4 + 3*a^2*c^3*x^2 - c^3)*sqrt (a*x + 1)*sqrt(-a*x + 1)), x) + 1/5/((-a^2*x^2 + 1)^(5/2)*a^2*c^3)
\[ \int \frac {e^{\text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^3} \, dx=\int { -\frac {{\left (a x + 1\right )} x}{{\left (a^{2} c x^{2} - c\right )}^{3} \sqrt {-a^{2} x^{2} + 1}} \,d x } \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x/(-a^2*c*x^2+c)^3,x, algorithm="giac ")
Output:
integrate(-(a*x + 1)*x/((a^2*c*x^2 - c)^3*sqrt(-a^2*x^2 + 1)), x)
Time = 14.43 (sec) , antiderivative size = 271, normalized size of antiderivative = 3.39 \[ \int \frac {e^{\text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {\sqrt {1-a^2\,x^2}}{30\,\left (a^4\,c^3\,x^2-2\,a^3\,c^3\,x+a^2\,c^3\right )}+\frac {\sqrt {1-a^2\,x^2}}{24\,\left (a^4\,c^3\,x^2+2\,a^3\,c^3\,x+a^2\,c^3\right )}-\frac {5\,\sqrt {1-a^2\,x^2}}{48\,\left (c^3\,\sqrt {-a^2}+a\,c^3\,x\,\sqrt {-a^2}\right )\,\sqrt {-a^2}}+\frac {7\,\sqrt {1-a^2\,x^2}}{240\,\left (c^3\,\sqrt {-a^2}-a\,c^3\,x\,\sqrt {-a^2}\right )\,\sqrt {-a^2}}-\frac {\sqrt {1-a^2\,x^2}}{20\,\sqrt {-a^2}\,\left (c^3\,\sqrt {-a^2}+3\,a^2\,c^3\,x^2\,\sqrt {-a^2}-a^3\,c^3\,x^3\,\sqrt {-a^2}-3\,a\,c^3\,x\,\sqrt {-a^2}\right )} \] Input:
int((x*(a*x + 1))/((c - a^2*c*x^2)^3*(1 - a^2*x^2)^(1/2)),x)
Output:
(1 - a^2*x^2)^(1/2)/(30*(a^2*c^3 - 2*a^3*c^3*x + a^4*c^3*x^2)) + (1 - a^2* x^2)^(1/2)/(24*(a^2*c^3 + 2*a^3*c^3*x + a^4*c^3*x^2)) - (5*(1 - a^2*x^2)^( 1/2))/(48*(c^3*(-a^2)^(1/2) + a*c^3*x*(-a^2)^(1/2))*(-a^2)^(1/2)) + (7*(1 - a^2*x^2)^(1/2))/(240*(c^3*(-a^2)^(1/2) - a*c^3*x*(-a^2)^(1/2))*(-a^2)^(1 /2)) - (1 - a^2*x^2)^(1/2)/(20*(-a^2)^(1/2)*(c^3*(-a^2)^(1/2) + 3*a^2*c^3* x^2*(-a^2)^(1/2) - a^3*c^3*x^3*(-a^2)^(1/2) - 3*a*c^3*x*(-a^2)^(1/2)))
Time = 0.15 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.75 \[ \int \frac {e^{\text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {3 \sqrt {-a^{2} x^{2}+1}\, a^{3} x^{3}-3 \sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}-3 \sqrt {-a^{2} x^{2}+1}\, a x +3 \sqrt {-a^{2} x^{2}+1}-2 a^{4} x^{4}+2 a^{3} x^{3}+3 a^{2} x^{2}-3 a x +3}{15 \sqrt {-a^{2} x^{2}+1}\, a^{2} c^{3} \left (a^{3} x^{3}-a^{2} x^{2}-a x +1\right )} \] Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*x/(-a^2*c*x^2+c)^3,x)
Output:
(3*sqrt( - a**2*x**2 + 1)*a**3*x**3 - 3*sqrt( - a**2*x**2 + 1)*a**2*x**2 - 3*sqrt( - a**2*x**2 + 1)*a*x + 3*sqrt( - a**2*x**2 + 1) - 2*a**4*x**4 + 2 *a**3*x**3 + 3*a**2*x**2 - 3*a*x + 3)/(15*sqrt( - a**2*x**2 + 1)*a**2*c**3 *(a**3*x**3 - a**2*x**2 - a*x + 1))