Integrand size = 18, antiderivative size = 67 \[ \int \frac {e^{3 \coth ^{-1}(a x)}}{(c-a c x)^3} \, dx=\frac {a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{7 c^3 \left (a-\frac {1}{x}\right )^6}-\frac {6 a^4 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{35 c^3 \left (a-\frac {1}{x}\right )^5} \] Output:
1/7*a^5*(1-1/a^2/x^2)^(5/2)/c^3/(a-1/x)^6-6/35*a^4*(1-1/a^2/x^2)^(5/2)/c^3 /(a-1/x)^5
Time = 0.18 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.61 \[ \int \frac {e^{3 \coth ^{-1}(a x)}}{(c-a c x)^3} \, dx=-\frac {\sqrt {1-\frac {1}{a^2 x^2}} x (-6+a x) (1+a x)^2}{35 c^3 (-1+a x)^4} \] Input:
Integrate[E^(3*ArcCoth[a*x])/(c - a*c*x)^3,x]
Output:
-1/35*(Sqrt[1 - 1/(a^2*x^2)]*x*(-6 + a*x)*(1 + a*x)^2)/(c^3*(-1 + a*x)^4)
Time = 0.44 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, 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.
\(\displaystyle \int \frac {e^{3 \coth ^{-1}(a x)}}{(c-a c x)^3} \, dx\) |
\(\Big \downarrow \) 6724 |
\(\displaystyle a^3 c^3 \int \frac {\left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{c^6 \left (a-\frac {1}{x}\right )^6 x}d\frac {1}{x}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a^3 \int \frac {\left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{\left (a-\frac {1}{x}\right )^6 x}d\frac {1}{x}}{c^3}\) |
\(\Big \downarrow \) 571 |
\(\displaystyle \frac {a^3 \left (\frac {a^2 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{7 \left (a-\frac {1}{x}\right )^6}-\frac {6}{7} \int \frac {\left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{\left (a-\frac {1}{x}\right )^5}d\frac {1}{x}\right )}{c^3}\) |
\(\Big \downarrow \) 460 |
\(\displaystyle \frac {a^3 \left (\frac {a^2 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{7 \left (a-\frac {1}{x}\right )^6}-\frac {6 a \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{35 \left (a-\frac {1}{x}\right )^5}\right )}{c^3}\) |
Input:
Int[E^(3*ArcCoth[a*x])/(c - a*c*x)^3,x]
Output:
(a^3*((a^2*(1 - 1/(a^2*x^2))^(5/2))/(7*(a - x^(-1))^6) - (6*a*(1 - 1/(a^2* x^2))^(5/2))/(35*(a - x^(-1))^5)))/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 | \(-\frac {\left (a x -6\right ) \left (a x +1\right )}{35 \left (a x -1\right )^{2} c^{3} \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} a}\) | \(41\) |
default | \(-\frac {\left (a x -6\right ) \left (a x +1\right )}{35 \left (a x -1\right )^{2} c^{3} \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} a}\) | \(41\) |
orering | \(\frac {\left (a x -6\right ) \left (a x -1\right ) \left (a x +1\right )}{35 a \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} \left (-a c x +c \right )^{3}}\) | \(45\) |
trager | \(-\frac {\left (a x +1\right ) \left (a^{3} x^{3}-4 a^{2} x^{2}-11 a x -6\right ) \sqrt {-\frac {-a x +1}{a x +1}}}{35 a \,c^{3} \left (a x -1\right )^{4}}\) | \(59\) |
Input:
int(1/((a*x-1)/(a*x+1))^(3/2)/(-a*c*x+c)^3,x,method=_RETURNVERBOSE)
Output:
-1/35*(a*x-6)*(a*x+1)/(a*x-1)^2/c^3/((a*x-1)/(a*x+1))^(3/2)/a
Time = 0.07 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.42 \[ \int \frac {e^{3 \coth ^{-1}(a x)}}{(c-a c x)^3} \, dx=-\frac {{\left (a^{4} x^{4} - 3 \, a^{3} x^{3} - 15 \, a^{2} x^{2} - 17 \, a x - 6\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{35 \, {\left (a^{5} c^{3} x^{4} - 4 \, a^{4} c^{3} x^{3} + 6 \, a^{3} c^{3} x^{2} - 4 \, a^{2} c^{3} x + a c^{3}\right )}} \] Input:
integrate(1/((a*x-1)/(a*x+1))^(3/2)/(-a*c*x+c)^3,x, algorithm="fricas")
Output:
-1/35*(a^4*x^4 - 3*a^3*x^3 - 15*a^2*x^2 - 17*a*x - 6)*sqrt((a*x - 1)/(a*x + 1))/(a^5*c^3*x^4 - 4*a^4*c^3*x^3 + 6*a^3*c^3*x^2 - 4*a^2*c^3*x + a*c^3)
\[ \int \frac {e^{3 \coth ^{-1}(a x)}}{(c-a c x)^3} \, dx=- \frac {\int \frac {1}{\frac {a^{4} x^{4} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1} - \frac {4 a^{3} x^{3} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1} + \frac {6 a^{2} x^{2} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1} - \frac {4 a x \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1} + \frac {\sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1}}\, dx}{c^{3}} \] Input:
integrate(1/((a*x-1)/(a*x+1))**(3/2)/(-a*c*x+c)**3,x)
Output:
-Integral(1/(a**4*x**4*sqrt(a*x/(a*x + 1) - 1/(a*x + 1))/(a*x + 1) - 4*a** 3*x**3*sqrt(a*x/(a*x + 1) - 1/(a*x + 1))/(a*x + 1) + 6*a**2*x**2*sqrt(a*x/ (a*x + 1) - 1/(a*x + 1))/(a*x + 1) - 4*a*x*sqrt(a*x/(a*x + 1) - 1/(a*x + 1 ))/(a*x + 1) + sqrt(a*x/(a*x + 1) - 1/(a*x + 1))/(a*x + 1)), x)/c**3
Time = 0.03 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.58 \[ \int \frac {e^{3 \coth ^{-1}(a x)}}{(c-a c x)^3} \, dx=-\frac {\frac {7 \, {\left (a x - 1\right )}}{a x + 1} - 5}{70 \, a c^{3} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {7}{2}}} \] Input:
integrate(1/((a*x-1)/(a*x+1))^(3/2)/(-a*c*x+c)^3,x, algorithm="maxima")
Output:
-1/70*(7*(a*x - 1)/(a*x + 1) - 5)/(a*c^3*((a*x - 1)/(a*x + 1))^(7/2))
Leaf count of result is larger than twice the leaf count of optimal. 125 vs. \(2 (59) = 118\).
Time = 0.18 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.87 \[ \int \frac {e^{3 \coth ^{-1}(a x)}}{(c-a c x)^3} \, dx=\frac {2 \, {\left (35 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )}^{5} x^{5} + 35 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )}^{4} x^{4} + 70 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )}^{3} x^{3} + 14 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )}^{2} x^{2} + 7 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )} x - 1\right )}}{35 \, {\left ({\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )} x - 1\right )}^{7} a c^{3}} \] Input:
integrate(1/((a*x-1)/(a*x+1))^(3/2)/(-a*c*x+c)^3,x, algorithm="giac")
Output:
2/35*(35*(a + sqrt(a^2 - 1/x^2))^5*x^5 + 35*(a + sqrt(a^2 - 1/x^2))^4*x^4 + 70*(a + sqrt(a^2 - 1/x^2))^3*x^3 + 14*(a + sqrt(a^2 - 1/x^2))^2*x^2 + 7* (a + sqrt(a^2 - 1/x^2))*x - 1)/(((a + sqrt(a^2 - 1/x^2))*x - 1)^7*a*c^3)
Time = 14.16 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.58 \[ \int \frac {e^{3 \coth ^{-1}(a x)}}{(c-a c x)^3} \, dx=-\frac {\frac {a\,x-1}{5\,\left (a\,x+1\right )}-\frac {1}{7}}{2\,a\,c^3\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{7/2}} \] Input:
int(1/((c - a*c*x)^3*((a*x - 1)/(a*x + 1))^(3/2)),x)
Output:
-((a*x - 1)/(5*(a*x + 1)) - 1/7)/(2*a*c^3*((a*x - 1)/(a*x + 1))^(7/2))
Time = 0.15 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.96 \[ \int \frac {e^{3 \coth ^{-1}(a x)}}{(c-a c x)^3} \, dx=\frac {\sqrt {a x -1}\, a^{3} x^{3}-3 \sqrt {a x -1}\, a^{2} x^{2}+3 \sqrt {a x -1}\, a x -\sqrt {a x -1}-\sqrt {a x +1}\, a^{3} x^{3}+4 \sqrt {a x +1}\, a^{2} x^{2}+11 \sqrt {a x +1}\, a x +6 \sqrt {a x +1}}{35 \sqrt {a x -1}\, a \,c^{3} \left (a^{3} x^{3}-3 a^{2} x^{2}+3 a x -1\right )} \] Input:
int(1/((a*x-1)/(a*x+1))^(3/2)/(-a*c*x+c)^3,x)
Output:
(sqrt(a*x - 1)*a**3*x**3 - 3*sqrt(a*x - 1)*a**2*x**2 + 3*sqrt(a*x - 1)*a*x - sqrt(a*x - 1) - sqrt(a*x + 1)*a**3*x**3 + 4*sqrt(a*x + 1)*a**2*x**2 + 1 1*sqrt(a*x + 1)*a*x + 6*sqrt(a*x + 1))/(35*sqrt(a*x - 1)*a*c**3*(a**3*x**3 - 3*a**2*x**2 + 3*a*x - 1))