Integrand size = 16, antiderivative size = 65 \[ \int \frac {e^{\text {arctanh}(a x)}}{(c-a c x)^3} \, dx=\frac {\left (1-a^2 x^2\right )^{3/2}}{5 a c^3 (1-a x)^4}+\frac {\left (1-a^2 x^2\right )^{3/2}}{15 a c^3 (1-a x)^3} \] Output:
1/5*(-a^2*x^2+1)^(3/2)/a/c^3/(-a*x+1)^4+1/15*(-a^2*x^2+1)^(3/2)/a/c^3/(-a* x+1)^3
Time = 0.01 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.54 \[ \int \frac {e^{\text {arctanh}(a x)}}{(c-a c x)^3} \, dx=\frac {(4-a x) (1+a x)^{3/2}}{15 a c^3 (1-a x)^{5/2}} \] Input:
Integrate[E^ArcTanh[a*x]/(c - a*c*x)^3,x]
Output:
((4 - a*x)*(1 + a*x)^(3/2))/(15*a*c^3*(1 - a*x)^(5/2))
Time = 0.26 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.97, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6677, 27, 461, 460}
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.
\(\displaystyle \int \frac {e^{\text {arctanh}(a x)}}{(c-a c x)^3} \, dx\) |
\(\Big \downarrow \) 6677 |
\(\displaystyle c \int \frac {\sqrt {1-a^2 x^2}}{c^4 (1-a x)^4}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\sqrt {1-a^2 x^2}}{(1-a x)^4}dx}{c^3}\) |
\(\Big \downarrow \) 461 |
\(\displaystyle \frac {\frac {1}{5} \int \frac {\sqrt {1-a^2 x^2}}{(1-a x)^3}dx+\frac {\left (1-a^2 x^2\right )^{3/2}}{5 a (1-a x)^4}}{c^3}\) |
\(\Big \downarrow \) 460 |
\(\displaystyle \frac {\frac {\left (1-a^2 x^2\right )^{3/2}}{15 a (1-a x)^3}+\frac {\left (1-a^2 x^2\right )^{3/2}}{5 a (1-a x)^4}}{c^3}\) |
Input:
Int[E^ArcTanh[a*x]/(c - a*c*x)^3,x]
Output:
((1 - a^2*x^2)^(3/2)/(5*a*(1 - a*x)^4) + (1 - a^2*x^2)^(3/2)/(15*a*(1 - a* x)^3))/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[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (-d)*(c + d*x)^n*((a + b*x^2)^(p + 1)/(b*c*n)), x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && EqQ[n + 2*p + 2, 0]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (-d)*(c + d*x)^n*((a + b*x^2)^(p + 1)/(2*b*c*(n + p + 1))), x] + Simp[Simpl ify[n + 2*p + 2]/(2*c*(n + p + 1)) Int[(c + d*x)^(n + 1)*(a + b*x^2)^p, x ], x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && ILtQ[Simp lify[n + 2*p + 2], 0] && (LtQ[n, -1] || GtQ[n + p, 0])
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_))^(p_.), x_Symbol] :> S imp[c^n Int[(c + d*x)^(p - n)*(1 - a^2*x^2)^(n/2), x], x] /; FreeQ[{a, c, d, p}, x] && EqQ[a*c + d, 0] && IntegerQ[(n - 1)/2] && IntegerQ[2*p]
Time = 0.24 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.62
method | result | size |
gosper | \(-\frac {\left (a x -4\right ) \left (a x +1\right )^{2}}{15 \left (a x -1\right )^{2} c^{3} \sqrt {-a^{2} x^{2}+1}\, a}\) | \(40\) |
trager | \(\frac {\left (a^{2} x^{2}-3 a x -4\right ) \sqrt {-a^{2} x^{2}+1}}{15 c^{3} \left (a x -1\right )^{3} a}\) | \(41\) |
orering | \(\frac {\left (a x -4\right ) \left (a x -1\right ) \left (a x +1\right )^{2}}{15 a \sqrt {-a^{2} x^{2}+1}\, \left (-a c x +c \right )^{3}}\) | \(44\) |
default | \(-\frac {\frac {\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 a \left (x -\frac {1}{a}\right )^{2}}-\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 \left (x -\frac {1}{a}\right )}}{a^{2}}+\frac {\frac {2 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{5 a \left (x -\frac {1}{a}\right )^{3}}-\frac {4 a \left (\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 a \left (x -\frac {1}{a}\right )^{2}}-\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 \left (x -\frac {1}{a}\right )}\right )}{5}}{a^{3}}}{c^{3}}\) | \(221\) |
Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^3,x,method=_RETURNVERBOSE)
Output:
-1/15*(a*x-4)*(a*x+1)^2/(a*x-1)^2/c^3/(-a^2*x^2+1)^(1/2)/a
Time = 0.10 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.37 \[ \int \frac {e^{\text {arctanh}(a x)}}{(c-a c x)^3} \, dx=\frac {4 \, a^{3} x^{3} - 12 \, a^{2} x^{2} + 12 \, a x + {\left (a^{2} x^{2} - 3 \, a x - 4\right )} \sqrt {-a^{2} x^{2} + 1} - 4}{15 \, {\left (a^{4} c^{3} x^{3} - 3 \, a^{3} c^{3} x^{2} + 3 \, a^{2} c^{3} x - a c^{3}\right )}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^3,x, algorithm="fricas")
Output:
1/15*(4*a^3*x^3 - 12*a^2*x^2 + 12*a*x + (a^2*x^2 - 3*a*x - 4)*sqrt(-a^2*x^ 2 + 1) - 4)/(a^4*c^3*x^3 - 3*a^3*c^3*x^2 + 3*a^2*c^3*x - a*c^3)
\[ \int \frac {e^{\text {arctanh}(a x)}}{(c-a c x)^3} \, dx=- \frac {\int \frac {a x}{a^{3} x^{3} \sqrt {- a^{2} x^{2} + 1} - 3 a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1} + 3 a x \sqrt {- a^{2} x^{2} + 1} - \sqrt {- a^{2} x^{2} + 1}}\, dx + \int \frac {1}{a^{3} x^{3} \sqrt {- a^{2} x^{2} + 1} - 3 a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1} + 3 a x \sqrt {- a^{2} x^{2} + 1} - \sqrt {- a^{2} x^{2} + 1}}\, dx}{c^{3}} \] Input:
integrate((a*x+1)/(-a**2*x**2+1)**(1/2)/(-a*c*x+c)**3,x)
Output:
-(Integral(a*x/(a**3*x**3*sqrt(-a**2*x**2 + 1) - 3*a**2*x**2*sqrt(-a**2*x* *2 + 1) + 3*a*x*sqrt(-a**2*x**2 + 1) - sqrt(-a**2*x**2 + 1)), x) + Integra l(1/(a**3*x**3*sqrt(-a**2*x**2 + 1) - 3*a**2*x**2*sqrt(-a**2*x**2 + 1) + 3 *a*x*sqrt(-a**2*x**2 + 1) - sqrt(-a**2*x**2 + 1)), x))/c**3
Leaf count of result is larger than twice the leaf count of optimal. 126 vs. \(2 (55) = 110\).
Time = 0.11 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.94 \[ \int \frac {e^{\text {arctanh}(a x)}}{(c-a c x)^3} \, dx=-\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{5 \, {\left (a^{4} c^{3} x^{3} - 3 \, a^{3} c^{3} x^{2} + 3 \, a^{2} c^{3} x - a c^{3}\right )}} - \frac {\sqrt {-a^{2} x^{2} + 1}}{15 \, {\left (a^{3} c^{3} x^{2} - 2 \, a^{2} c^{3} x + a c^{3}\right )}} + \frac {\sqrt {-a^{2} x^{2} + 1}}{15 \, {\left (a^{2} c^{3} x - a c^{3}\right )}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^3,x, algorithm="maxima")
Output:
-2/5*sqrt(-a^2*x^2 + 1)/(a^4*c^3*x^3 - 3*a^3*c^3*x^2 + 3*a^2*c^3*x - a*c^3 ) - 1/15*sqrt(-a^2*x^2 + 1)/(a^3*c^3*x^2 - 2*a^2*c^3*x + a*c^3) + 1/15*sqr t(-a^2*x^2 + 1)/(a^2*c^3*x - a*c^3)
Leaf count of result is larger than twice the leaf count of optimal. 145 vs. \(2 (55) = 110\).
Time = 0.13 (sec) , antiderivative size = 145, normalized size of antiderivative = 2.23 \[ \int \frac {e^{\text {arctanh}(a x)}}{(c-a c x)^3} \, dx=-\frac {2 \, {\left (\frac {5 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}}{a^{2} x} - \frac {25 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2}}{a^{4} x^{2}} + \frac {15 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{3}}{a^{6} x^{3}} - \frac {15 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{4}}{a^{8} x^{4}} - 4\right )}}{15 \, c^{3} {\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{a^{2} x} - 1\right )}^{5} {\left | a \right |}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^3,x, algorithm="giac")
Output:
-2/15*(5*(sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) - 25*(sqrt(-a^2*x^2 + 1)* abs(a) + a)^2/(a^4*x^2) + 15*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^3/(a^6*x^3) - 15*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^4/(a^8*x^4) - 4)/(c^3*((sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) - 1)^5*abs(a))
Time = 13.98 (sec) , antiderivative size = 183, normalized size of antiderivative = 2.82 \[ \int \frac {e^{\text {arctanh}(a x)}}{(c-a c x)^3} \, dx=\frac {2\,\sqrt {1-a^2\,x^2}}{5\,\sqrt {-a^2}\,\left (3\,c^3\,x\,\sqrt {-a^2}-\frac {c^3\,\sqrt {-a^2}}{a}+a^2\,c^3\,x^3\,\sqrt {-a^2}-3\,a\,c^3\,x^2\,\sqrt {-a^2}\right )}-\frac {\sqrt {1-a^2\,x^2}}{15\,\sqrt {-a^2}\,\left (c^3\,x\,\sqrt {-a^2}-\frac {c^3\,\sqrt {-a^2}}{a}\right )}-\frac {a\,\sqrt {1-a^2\,x^2}}{15\,\left (a^4\,c^3\,x^2-2\,a^3\,c^3\,x+a^2\,c^3\right )} \] Input:
int((a*x + 1)/((1 - a^2*x^2)^(1/2)*(c - a*c*x)^3),x)
Output:
(2*(1 - a^2*x^2)^(1/2))/(5*(-a^2)^(1/2)*(3*c^3*x*(-a^2)^(1/2) - (c^3*(-a^2 )^(1/2))/a + a^2*c^3*x^3*(-a^2)^(1/2) - 3*a*c^3*x^2*(-a^2)^(1/2))) - (1 - a^2*x^2)^(1/2)/(15*(-a^2)^(1/2)*(c^3*x*(-a^2)^(1/2) - (c^3*(-a^2)^(1/2))/a )) - (a*(1 - a^2*x^2)^(1/2))/(15*(a^2*c^3 - 2*a^3*c^3*x + a^4*c^3*x^2))
Time = 0.15 (sec) , antiderivative size = 142, normalized size of antiderivative = 2.18 \[ \int \frac {e^{\text {arctanh}(a x)}}{(c-a c x)^3} \, dx=\frac {-\sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}+\sqrt {-a^{2} x^{2}+1}\, a x -6 \sqrt {-a^{2} x^{2}+1}-3 a^{3} x^{3}+8 a^{2} x^{2}+a x +6}{15 a \,c^{3} \left (\sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}-2 \sqrt {-a^{2} x^{2}+1}\, a x +\sqrt {-a^{2} x^{2}+1}+a^{3} x^{3}-3 a^{2} x^{2}+3 a x -1\right )} \] Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^3,x)
Output:
( - sqrt( - a**2*x**2 + 1)*a**2*x**2 + sqrt( - a**2*x**2 + 1)*a*x - 6*sqrt ( - a**2*x**2 + 1) - 3*a**3*x**3 + 8*a**2*x**2 + a*x + 6)/(15*a*c**3*(sqrt ( - a**2*x**2 + 1)*a**2*x**2 - 2*sqrt( - a**2*x**2 + 1)*a*x + sqrt( - a**2 *x**2 + 1) + a**3*x**3 - 3*a**2*x**2 + 3*a*x - 1))