Integrand size = 22, antiderivative size = 45 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right )^3} \, dx=-\frac {1}{a c^3 \sqrt {1-a^2 x^2}}-\frac {\sqrt {1-a^2 x^2}}{a c^3} \] Output:
-1/a/c^3/(-a^2*x^2+1)^(1/2)-(-a^2*x^2+1)^(1/2)/a/c^3
Time = 0.02 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.67 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right )^3} \, dx=\frac {-2+a^2 x^2}{a c^3 \sqrt {1-a^2 x^2}} \] Input:
Integrate[1/(E^(3*ArcTanh[a*x])*(c - c/(a*x))^3),x]
Output:
(-2 + a^2*x^2)/(a*c^3*Sqrt[1 - a^2*x^2])
Time = 0.38 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.09, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {6681, 6678, 243, 53, 2009}
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.
\(\displaystyle \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right )^3} \, dx\) |
\(\Big \downarrow \) 6681 |
\(\displaystyle -\frac {a^3 \int \frac {e^{-3 \text {arctanh}(a x)} x^3}{(1-a x)^3}dx}{c^3}\) |
\(\Big \downarrow \) 6678 |
\(\displaystyle -\frac {a^3 \int \frac {x^3}{\left (1-a^2 x^2\right )^{3/2}}dx}{c^3}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle -\frac {a^3 \int \frac {x^2}{\left (1-a^2 x^2\right )^{3/2}}dx^2}{2 c^3}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle -\frac {a^3 \int \left (\frac {1}{a^2 \left (1-a^2 x^2\right )^{3/2}}-\frac {1}{a^2 \sqrt {1-a^2 x^2}}\right )dx^2}{2 c^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^3 \left (\frac {2 \sqrt {1-a^2 x^2}}{a^4}+\frac {2}{a^4 \sqrt {1-a^2 x^2}}\right )}{2 c^3}\) |
Input:
Int[1/(E^(3*ArcTanh[a*x])*(c - c/(a*x))^3),x]
Output:
-1/2*(a^3*(2/(a^4*Sqrt[1 - a^2*x^2]) + (2*Sqrt[1 - a^2*x^2])/a^4))/c^3
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_))^(p_.)*((e_.) + (f_.)* (x_))^(m_.), x_Symbol] :> Simp[c^n Int[(e + f*x)^m*(c + d*x)^(p - n)*(1 - a^2*x^2)^(n/2), x], x] /; FreeQ[{a, c, d, e, f, m, p}, x] && EqQ[a*c + d, 0] && IntegerQ[(n - 1)/2] && (IntegerQ[p] || EqQ[p, n/2] || EqQ[p - n/2 - 1 , 0]) && IntegerQ[2*p]
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_))^(p_.), x_Symbol ] :> Simp[d^p Int[u*(1 + c*(x/d))^p*(E^(n*ArcTanh[a*x])/x^p), x], x] /; F reeQ[{a, c, d, n}, x] && EqQ[c^2 - a^2*d^2, 0] && IntegerQ[p]
Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.64
method | result | size |
pseudoelliptic | \(\frac {a^{2} x^{2}-2}{\sqrt {-a^{2} x^{2}+1}\, a \,c^{3}}\) | \(29\) |
trager | \(-\frac {\left (a^{2} x^{2}-2\right ) \sqrt {-a^{2} x^{2}+1}}{a \,c^{3} \left (a^{2} x^{2}-1\right )}\) | \(41\) |
gosper | \(\frac {\left (-a^{2} x^{2}+1\right )^{\frac {3}{2}} \left (a^{2} x^{2}-2\right )}{a \left (a x -1\right )^{2} c^{3} \left (a x +1\right )^{2}}\) | \(43\) |
risch | \(\frac {a^{2} x^{2}-1}{a \sqrt {-a^{2} x^{2}+1}\, c^{3}}-\frac {1}{a \,c^{3} \sqrt {-a^{2} x^{2}+1}}\) | \(50\) |
orering | \(\frac {\left (a^{2} x^{2}-2\right ) \left (a x -1\right ) \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}{a^{4} \left (a x +1\right )^{2} x^{3} \left (c -\frac {c}{a x}\right )^{3}}\) | \(54\) |
default | \(\frac {a^{3} \left (\frac {\frac {\left (-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )\right )^{\frac {5}{2}}}{a \left (x -\frac {1}{a}\right )^{3}}+2 a \left (-\frac {\left (-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )\right )^{\frac {5}{2}}}{a \left (x -\frac {1}{a}\right )^{2}}-3 a \left (\frac {\left (-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )\right )^{\frac {3}{2}}}{3}-a \left (-\frac {\left (-2 \left (x -\frac {1}{a}\right ) a^{2}-2 a \right ) \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{4 a^{2}}+\frac {\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 )}{2 \sqrt {a^{2}}}\right )\right )\right )}{8 a^{6}}+\frac {-\frac {3 \left (-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )\right )^{\frac {5}{2}}}{16 a \left (x -\frac {1}{a}\right )^{2}}-\frac {9 a \left (\frac {\left (-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )\right )^{\frac {3}{2}}}{3}-a \left (-\frac {\left (-2 \left (x -\frac {1}{a}\right ) a^{2}-2 a \right ) \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{4 a^{2}}+\frac {\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 )}{2 \sqrt {a^{2}}}\right )\right )}{16}}{a^{5}}+\frac {-\frac {\left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {5}{2}}}{a \left (x +\frac {1}{a}\right )^{3}}-2 a \left (\frac {\left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {5}{2}}}{a \left (x +\frac {1}{a}\right )^{2}}+3 a \left (\frac {\left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {3}{2}}}{3}+a \left (-\frac {\left (-2 a^{2} \left (x +\frac {1}{a}\right )+2 a \right ) \sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{4 a^{2}}+\frac {\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 )}{2 \sqrt {a^{2}}}\right )\right )\right )}{8 a^{6}}-\frac {3 \left (\frac {\left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {5}{2}}}{a \left (x +\frac {1}{a}\right )^{2}}+3 a \left (\frac {\left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {3}{2}}}{3}+a \left (-\frac {\left (-2 a^{2} \left (x +\frac {1}{a}\right )+2 a \right ) \sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{4 a^{2}}+\frac {\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 )}{2 \sqrt {a^{2}}}\right )\right )\right )}{16 a^{5}}\right )}{c^{3}}\) | \(744\) |
Input:
int(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(c-c/a/x)^3,x,method=_RETURNVERBOSE)
Output:
(a^2*x^2-2)/(-a^2*x^2+1)^(1/2)/a/c^3
Time = 0.08 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.18 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right )^3} \, dx=-\frac {2 \, a^{2} x^{2} + {\left (a^{2} x^{2} - 2\right )} \sqrt {-a^{2} x^{2} + 1} - 2}{a^{3} c^{3} x^{2} - a c^{3}} \] Input:
integrate(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(c-c/a/x)^3,x, algorithm="fricas" )
Output:
-(2*a^2*x^2 + (a^2*x^2 - 2)*sqrt(-a^2*x^2 + 1) - 2)/(a^3*c^3*x^2 - a*c^3)
Time = 14.47 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.51 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right )^3} \, dx=- a^{3} \left (\begin {cases} \frac {2 \left (\frac {\sqrt {- a^{2} x^{2} + 1}}{2 c^{3}} + \frac {1}{2 c^{3} \sqrt {- a^{2} x^{2} + 1}}\right )}{a^{4}} & \text {for}\: a^{2} \neq 0 \\\frac {\log {\left (2 a^{4} c^{3} x^{4} + 2 c^{3} \right )}}{4 a^{4} c^{3}} & \text {otherwise} \end {cases}\right ) \] Input:
integrate(1/(a*x+1)**3*(-a**2*x**2+1)**(3/2)/(c-c/a/x)**3,x)
Output:
-a**3*Piecewise((2*(sqrt(-a**2*x**2 + 1)/(2*c**3) + 1/(2*c**3*sqrt(-a**2*x **2 + 1)))/a**4, Ne(a**2, 0)), (log(2*a**4*c**3*x**4 + 2*c**3)/(4*a**4*c** 3), True))
Time = 0.04 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right )^3} \, dx=-\frac {{\left (a^{2} x^{2} - 2\right )} \sqrt {a x + 1} \sqrt {-a x + 1}}{a^{3} c^{3} x^{2} - a c^{3}} \] Input:
integrate(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(c-c/a/x)^3,x, algorithm="maxima" )
Output:
-(a^2*x^2 - 2)*sqrt(a*x + 1)*sqrt(-a*x + 1)/(a^3*c^3*x^2 - a*c^3)
Time = 0.11 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.84 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right )^3} \, dx=-\frac {\frac {\sqrt {-a^{2} x^{2} + 1}}{c^{3}} + \frac {1}{\sqrt {-a^{2} x^{2} + 1} c^{3}}}{a} \] Input:
integrate(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(c-c/a/x)^3,x, algorithm="giac")
Output:
-(sqrt(-a^2*x^2 + 1)/c^3 + 1/(sqrt(-a^2*x^2 + 1)*c^3))/a
Time = 0.04 (sec) , antiderivative size = 116, normalized size of antiderivative = 2.58 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right )^3} \, dx=\frac {\sqrt {1-a^2\,x^2}}{2\,c^3\,\left (x\,\sqrt {-a^2}+\frac {\sqrt {-a^2}}{a}\right )\,\sqrt {-a^2}}-\frac {\sqrt {1-a^2\,x^2}}{a\,c^3}-\frac {\sqrt {1-a^2\,x^2}}{2\,c^3\,\left (x\,\sqrt {-a^2}-\frac {\sqrt {-a^2}}{a}\right )\,\sqrt {-a^2}} \] Input:
int((1 - a^2*x^2)^(3/2)/((c - c/(a*x))^3*(a*x + 1)^3),x)
Output:
(1 - a^2*x^2)^(1/2)/(2*c^3*(x*(-a^2)^(1/2) + (-a^2)^(1/2)/a)*(-a^2)^(1/2)) - (1 - a^2*x^2)^(1/2)/(a*c^3) - (1 - a^2*x^2)^(1/2)/(2*c^3*(x*(-a^2)^(1/2 ) - (-a^2)^(1/2)/a)*(-a^2)^(1/2))
Time = 0.15 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.64 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right )^3} \, dx=\frac {a^{2} x^{2}-2}{\sqrt {-a^{2} x^{2}+1}\, a \,c^{3}} \] Input:
int(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(c-c/a/x)^3,x)
Output:
(a**2*x**2 - 2)/(sqrt( - a**2*x**2 + 1)*a*c**3)