Integrand size = 18, antiderivative size = 65 \[ \int \frac {e^{-\text {arctanh}(a x)}}{(c-a c x)^3} \, dx=\frac {\sqrt {1-a^2 x^2}}{3 a c^3 (1-a x)^2}+\frac {\sqrt {1-a^2 x^2}}{3 a c^3 (1-a x)} \] Output:
1/3*(-a^2*x^2+1)^(1/2)/a/c^3/(-a*x+1)^2+1/3*(-a^2*x^2+1)^(1/2)/a/c^3/(-a*x +1)
Time = 0.01 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.52 \[ \int \frac {e^{-\text {arctanh}(a x)}}{(c-a c x)^3} \, dx=-\frac {(-2+a x) \sqrt {1+a x}}{3 a c^3 (1-a x)^{3/2}} \] Input:
Integrate[1/(E^ArcTanh[a*x]*(c - a*c*x)^3),x]
Output:
-1/3*((-2 + a*x)*Sqrt[1 + a*x])/(a*c^3*(1 - a*x)^(3/2))
Time = 0.25 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.97, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, 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 |
\(\displaystyle \frac {\int \frac {1}{c^2 (1-a x)^2 \sqrt {1-a^2 x^2}}dx}{c}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {1}{(1-a x)^2 \sqrt {1-a^2 x^2}}dx}{c^3}\) |
\(\Big \downarrow \) 461 |
\(\displaystyle \frac {\frac {1}{3} \int \frac {1}{(1-a x) \sqrt {1-a^2 x^2}}dx+\frac {\sqrt {1-a^2 x^2}}{3 a (1-a x)^2}}{c^3}\) |
\(\Big \downarrow \) 460 |
\(\displaystyle \frac {\frac {\sqrt {1-a^2 x^2}}{3 a (1-a x)}+\frac {\sqrt {1-a^2 x^2}}{3 a (1-a x)^2}}{c^3}\) |
Input:
Int[1/(E^ArcTanh[a*x]*(c - a*c*x)^3),x]
Output:
(Sqrt[1 - a^2*x^2]/(3*a*(1 - a*x)^2) + Sqrt[1 - a^2*x^2]/(3*a*(1 - a*x)))/ 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 = 0.26 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.51
method | result | size |
gosper | \(-\frac {\sqrt {-a^{2} x^{2}+1}\, \left (a x -2\right )}{3 \left (a x -1\right )^{2} a \,c^{3}}\) | \(33\) |
trager | \(-\frac {\sqrt {-a^{2} x^{2}+1}\, \left (a x -2\right )}{3 \left (a x -1\right )^{2} a \,c^{3}}\) | \(33\) |
orering | \(\frac {\left (a x -2\right ) \left (a x -1\right ) \sqrt {-a^{2} x^{2}+1}}{3 a \left (-a c x +c \right )^{3}}\) | \(37\) |
default | \(-\frac {\frac {\left (-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )\right )^{\frac {3}{2}}}{6 a^{4} \left (x -\frac {1}{a}\right )^{3}}-\frac {\frac {\left (-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )\right )^{\frac {3}{2}}}{a \left (x -\frac {1}{a}\right )^{2}}+a \left (\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}-\frac {a \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}\right )}{\sqrt {a^{2}}}\right )}{4 a^{2}}-\frac {\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}+\frac {a \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}\right )}{\sqrt {a^{2}}}}{8 a}+\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}-\frac {a \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}\right )}{\sqrt {a^{2}}}}{8 a}}{c^{3}}\) | \(310\) |
Input:
int(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^3,x,method=_RETURNVERBOSE)
Output:
-1/3*(-a^2*x^2+1)^(1/2)*(a*x-2)/(a*x-1)^2/a/c^3
Time = 0.08 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.95 \[ \int \frac {e^{-\text {arctanh}(a x)}}{(c-a c x)^3} \, dx=\frac {2 \, a^{2} x^{2} - 4 \, a x - \sqrt {-a^{2} x^{2} + 1} {\left (a x - 2\right )} + 2}{3 \, {\left (a^{3} c^{3} x^{2} - 2 \, a^{2} c^{3} x + a c^{3}\right )}} \] Input:
integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^3,x, algorithm="fricas")
Output:
1/3*(2*a^2*x^2 - 4*a*x - sqrt(-a^2*x^2 + 1)*(a*x - 2) + 2)/(a^3*c^3*x^2 - 2*a^2*c^3*x + a*c^3)
\[ \int \frac {e^{-\text {arctanh}(a x)}}{(c-a c x)^3} \, dx=- \frac {\int \frac {\sqrt {- a^{2} x^{2} + 1}}{a^{4} x^{4} - 2 a^{3} x^{3} + 2 a x - 1}\, dx}{c^{3}} \] Input:
integrate(1/(a*x+1)*(-a**2*x**2+1)**(1/2)/(-a*c*x+c)**3,x)
Output:
-Integral(sqrt(-a**2*x**2 + 1)/(a**4*x**4 - 2*a**3*x**3 + 2*a*x - 1), x)/c **3
\[ \int \frac {e^{-\text {arctanh}(a x)}}{(c-a c x)^3} \, dx=\int { -\frac {\sqrt {-a^{2} x^{2} + 1}}{{\left (a c x - c\right )}^{3} {\left (a x + 1\right )}} \,d x } \] Input:
integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^3,x, algorithm="maxima")
Output:
-integrate(sqrt(-a^2*x^2 + 1)/((a*c*x - c)^3*(a*x + 1)), x)
Time = 0.12 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.40 \[ \int \frac {e^{-\text {arctanh}(a x)}}{(c-a c x)^3} \, dx=-\frac {2 \, {\left (\frac {3 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}}{a^{2} x} - \frac {3 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2}}{a^{4} x^{2}} - 2\right )}}{3 \, c^{3} {\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{a^{2} x} - 1\right )}^{3} {\left | a \right |}} \] Input:
integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^3,x, algorithm="giac")
Output:
-2/3*(3*(sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) - 3*(sqrt(-a^2*x^2 + 1)*ab s(a) + a)^2/(a^4*x^2) - 2)/(c^3*((sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) - 1)^3*abs(a))
Time = 14.47 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.49 \[ \int \frac {e^{-\text {arctanh}(a x)}}{(c-a c x)^3} \, dx=-\frac {\sqrt {1-a^2\,x^2}\,\left (a\,x-2\right )}{3\,a\,c^3\,{\left (a\,x-1\right )}^2} \] Input:
int((1 - a^2*x^2)^(1/2)/((c - a*c*x)^3*(a*x + 1)),x)
Output:
-((1 - a^2*x^2)^(1/2)*(a*x - 2))/(3*a*c^3*(a*x - 1)^2)
Time = 0.15 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.83 \[ \int \frac {e^{-\text {arctanh}(a x)}}{(c-a c x)^3} \, dx=\frac {-2 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{3}-2}{3 a \,c^{3} \left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{3}-3 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{2}+3 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )-1\right )} \] Input:
int(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^3,x)
Output:
( - 2*(tan(asin(a*x)/2)**3 + 1))/(3*a*c**3*(tan(asin(a*x)/2)**3 - 3*tan(as in(a*x)/2)**2 + 3*tan(asin(a*x)/2) - 1))