Integrand size = 16, antiderivative size = 67 \[ \int \frac {e^{\coth ^{-1}(a x)}}{(c-a c x)^3} \, dx=\frac {a^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{5 c^3 \left (a-\frac {1}{x}\right )^4}-\frac {4 a^2 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{15 c^3 \left (a-\frac {1}{x}\right )^3} \] Output:
1/5*a^3*(1-1/a^2/x^2)^(3/2)/c^3/(a-1/x)^4-4/15*a^2*(1-1/a^2/x^2)^(3/2)/c^3 /(a-1/x)^3
Time = 0.18 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.63 \[ \int \frac {e^{\coth ^{-1}(a x)}}{(c-a c x)^3} \, dx=-\frac {\sqrt {1-\frac {1}{a^2 x^2}} x \left (-4-3 a x+a^2 x^2\right )}{15 c^3 (-1+a x)^3} \] Input:
Integrate[E^ArcCoth[a*x]/(c - a*c*x)^3,x]
Output:
-1/15*(Sqrt[1 - 1/(a^2*x^2)]*x*(-4 - 3*a*x + a^2*x^2))/(c^3*(-1 + a*x)^3)
Time = 0.43 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6724, 27, 571, 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.
\(\displaystyle \int \frac {e^{\coth ^{-1}(a x)}}{(c-a c x)^3} \, dx\) |
\(\Big \downarrow \) 6724 |
\(\displaystyle a c \int \frac {\sqrt {1-\frac {1}{a^2 x^2}}}{c^4 \left (a-\frac {1}{x}\right )^4 x}d\frac {1}{x}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a \int \frac {\sqrt {1-\frac {1}{a^2 x^2}}}{\left (a-\frac {1}{x}\right )^4 x}d\frac {1}{x}}{c^3}\) |
\(\Big \downarrow \) 571 |
\(\displaystyle \frac {a \left (\frac {a^2 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{5 \left (a-\frac {1}{x}\right )^4}-\frac {4}{5} \int \frac {\sqrt {1-\frac {1}{a^2 x^2}}}{\left (a-\frac {1}{x}\right )^3}d\frac {1}{x}\right )}{c^3}\) |
\(\Big \downarrow \) 460 |
\(\displaystyle \frac {a \left (\frac {a^2 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{5 \left (a-\frac {1}{x}\right )^4}-\frac {4 a \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{15 \left (a-\frac {1}{x}\right )^3}\right )}{c^3}\) |
Input:
Int[E^ArcCoth[a*x]/(c - a*c*x)^3,x]
Output:
(a*((a^2*(1 - 1/(a^2*x^2))^(3/2))/(5*(a - x^(-1))^4) - (4*a*(1 - 1/(a^2*x^ 2))^(3/2))/(15*(a - x^(-1))^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[(x_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c + d*x)^n*((a + b*x^2)^(p + 1)/(2*b*(n + p + 1))), x] + Simp[n/(2*d* (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] && ((LtQ[n, -1] && !IGtQ[n + p + 1, 0]) || (LtQ[n, 0] && LtQ[p, -1]) || EqQ[n + 2*p + 2, 0]) && NeQ[n + p + 1, 0]
Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_))^(p_.), x_Symbol] :> S imp[-d^n Subst[Int[(d + c*x)^(p - n)*((1 - x^2/a^2)^(n/2)/x^(p + 2)), x], x, 1/x], x] /; FreeQ[{a, c, d}, x] && EqQ[a*c + d, 0] && IntegerQ[p] && In tegerQ[n]
Time = 0.13 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.61
method | result | size |
gosper | \(-\frac {\left (a x -4\right ) \left (a x +1\right )}{15 \left (a x -1\right )^{2} c^{3} \sqrt {\frac {a x -1}{a x +1}}\, a}\) | \(41\) |
default | \(-\frac {\left (a x -4\right ) \left (a x +1\right )}{15 \left (a x -1\right )^{2} c^{3} \sqrt {\frac {a x -1}{a x +1}}\, a}\) | \(41\) |
orering | \(\frac {\left (a x -4\right ) \left (a x -1\right ) \left (a x +1\right )}{15 a \sqrt {\frac {a x -1}{a x +1}}\, \left (-a c x +c \right )^{3}}\) | \(45\) |
trager | \(-\frac {\left (a x +1\right ) \left (a^{2} x^{2}-3 a x -4\right ) \sqrt {-\frac {-a x +1}{a x +1}}}{15 a \,c^{3} \left (a x -1\right )^{3}}\) | \(51\) |
Input:
int(1/((a*x-1)/(a*x+1))^(1/2)/(-a*c*x+c)^3,x,method=_RETURNVERBOSE)
Output:
-1/15*(a*x-4)*(a*x+1)/(a*x-1)^2/c^3/((a*x-1)/(a*x+1))^(1/2)/a
Time = 0.08 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.15 \[ \int \frac {e^{\coth ^{-1}(a x)}}{(c-a c x)^3} \, dx=-\frac {{\left (a^{3} x^{3} - 2 \, a^{2} x^{2} - 7 \, a x - 4\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{15 \, {\left (a^{4} c^{3} x^{3} - 3 \, a^{3} c^{3} x^{2} + 3 \, a^{2} c^{3} x - a c^{3}\right )}} \] Input:
integrate(1/((a*x-1)/(a*x+1))^(1/2)/(-a*c*x+c)^3,x, algorithm="fricas")
Output:
-1/15*(a^3*x^3 - 2*a^2*x^2 - 7*a*x - 4)*sqrt((a*x - 1)/(a*x + 1))/(a^4*c^3 *x^3 - 3*a^3*c^3*x^2 + 3*a^2*c^3*x - a*c^3)
\[ \int \frac {e^{\coth ^{-1}(a x)}}{(c-a c x)^3} \, dx=- \frac {\int \frac {1}{a^{3} x^{3} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}} - 3 a^{2} x^{2} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}} + 3 a x \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}} - \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}\, dx}{c^{3}} \] Input:
integrate(1/((a*x-1)/(a*x+1))**(1/2)/(-a*c*x+c)**3,x)
Output:
-Integral(1/(a**3*x**3*sqrt(a*x/(a*x + 1) - 1/(a*x + 1)) - 3*a**2*x**2*sqr t(a*x/(a*x + 1) - 1/(a*x + 1)) + 3*a*x*sqrt(a*x/(a*x + 1) - 1/(a*x + 1)) - sqrt(a*x/(a*x + 1) - 1/(a*x + 1))), x)/c**3
Time = 0.04 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.58 \[ \int \frac {e^{\coth ^{-1}(a x)}}{(c-a c x)^3} \, dx=-\frac {\frac {5 \, {\left (a x - 1\right )}}{a x + 1} - 3}{30 \, a c^{3} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{2}}} \] Input:
integrate(1/((a*x-1)/(a*x+1))^(1/2)/(-a*c*x+c)^3,x, algorithm="maxima")
Output:
-1/30*(5*(a*x - 1)/(a*x + 1) - 3)/(a*c^3*((a*x - 1)/(a*x + 1))^(5/2))
Time = 0.14 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.27 \[ \int \frac {e^{\coth ^{-1}(a x)}}{(c-a c x)^3} \, dx=\frac {2 \, {\left (15 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )}^{3} x^{3} + 5 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )}^{2} x^{2} + 5 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )} x - 1\right )}}{15 \, {\left ({\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )} x - 1\right )}^{5} a c^{3}} \] Input:
integrate(1/((a*x-1)/(a*x+1))^(1/2)/(-a*c*x+c)^3,x, algorithm="giac")
Output:
2/15*(15*(a + sqrt(a^2 - 1/x^2))^3*x^3 + 5*(a + sqrt(a^2 - 1/x^2))^2*x^2 + 5*(a + sqrt(a^2 - 1/x^2))*x - 1)/(((a + sqrt(a^2 - 1/x^2))*x - 1)^5*a*c^3 )
Time = 13.29 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.58 \[ \int \frac {e^{\coth ^{-1}(a x)}}{(c-a c x)^3} \, dx=-\frac {\frac {a\,x-1}{3\,\left (a\,x+1\right )}-\frac {1}{5}}{2\,a\,c^3\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/2}} \] Input:
int(1/((c - a*c*x)^3*((a*x - 1)/(a*x + 1))^(1/2)),x)
Output:
-((a*x - 1)/(3*(a*x + 1)) - 1/5)/(2*a*c^3*((a*x - 1)/(a*x + 1))^(5/2))
Time = 0.16 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.39 \[ \int \frac {e^{\coth ^{-1}(a x)}}{(c-a c x)^3} \, dx=\frac {\sqrt {a x -1}\, a^{2} x^{2}-2 \sqrt {a x -1}\, a x +\sqrt {a x -1}-\sqrt {a x +1}\, a^{2} x^{2}+3 \sqrt {a x +1}\, a x +4 \sqrt {a x +1}}{15 \sqrt {a x -1}\, a \,c^{3} \left (a^{2} x^{2}-2 a x +1\right )} \] Input:
int(1/((a*x-1)/(a*x+1))^(1/2)/(-a*c*x+c)^3,x)
Output:
(sqrt(a*x - 1)*a**2*x**2 - 2*sqrt(a*x - 1)*a*x + sqrt(a*x - 1) - sqrt(a*x + 1)*a**2*x**2 + 3*sqrt(a*x + 1)*a*x + 4*sqrt(a*x + 1))/(15*sqrt(a*x - 1)* a*c**3*(a**2*x**2 - 2*a*x + 1))