Integrand size = 18, antiderivative size = 61 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=\frac {2}{3 a c^4 \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {1}{3 a^2 c^4 \sqrt {1-\frac {1}{a^2 x^2}} \left (a-\frac {1}{x}\right ) x^2} \] Output:
2/3/a/c^4/(1-1/a^2/x^2)^(1/2)-1/3/a^2/c^4/(1-1/a^2/x^2)^(1/2)/(a-1/x)/x^2
Time = 0.18 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.82 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=\frac {\sqrt {1-\frac {1}{a^2 x^2}} x \left (-1-2 a x+2 a^2 x^2\right )}{3 c^4 (-1+a x)^2 (1+a x)} \] Input:
Integrate[1/(E^(3*ArcCoth[a*x])*(c - a*c*x)^4),x]
Output:
(Sqrt[1 - 1/(a^2*x^2)]*x*(-1 - 2*a*x + 2*a^2*x^2))/(3*c^4*(-1 + a*x)^2*(1 + a*x))
Time = 0.44 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {6724, 25, 27, 567, 241}
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.
\(\displaystyle \int \frac {e^{-3 \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx\) |
\(\Big \downarrow \) 6724 |
\(\displaystyle \frac {\int -\frac {1}{c \left (1-\frac {1}{a^2 x^2}\right )^{3/2} \left (a-\frac {1}{x}\right ) x^2}d\frac {1}{x}}{a^3 c^3}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {1}{c \left (1-\frac {1}{a^2 x^2}\right )^{3/2} \left (a-\frac {1}{x}\right ) x^2}d\frac {1}{x}}{a^3 c^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int \frac {1}{\left (1-\frac {1}{a^2 x^2}\right )^{3/2} \left (a-\frac {1}{x}\right ) x^2}d\frac {1}{x}}{a^3 c^4}\) |
\(\Big \downarrow \) 567 |
\(\displaystyle -\frac {\frac {a}{3 x^2 \sqrt {1-\frac {1}{a^2 x^2}} \left (a-\frac {1}{x}\right )}-\frac {2}{3} \int \frac {1}{\left (1-\frac {1}{a^2 x^2}\right )^{3/2} x}d\frac {1}{x}}{a^3 c^4}\) |
\(\Big \downarrow \) 241 |
\(\displaystyle -\frac {\frac {a}{3 x^2 \sqrt {1-\frac {1}{a^2 x^2}} \left (a-\frac {1}{x}\right )}-\frac {2 a^2}{3 \sqrt {1-\frac {1}{a^2 x^2}}}}{a^3 c^4}\) |
Input:
Int[1/(E^(3*ArcCoth[a*x])*(c - a*c*x)^4),x]
Output:
-(((-2*a^2)/(3*Sqrt[1 - 1/(a^2*x^2)]) + a/(3*Sqrt[1 - 1/(a^2*x^2)]*(a - x^ (-1))*x^2))/(a^3*c^4))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(x_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(a + b*x^2)^(p + 1)/ (2*b*(p + 1)), x] /; FreeQ[{a, b, p}, x] && NeQ[p, -1]
Int[((x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_))/((c_) + (d_.)*(x_)), x_Symbol] : > Simp[c*x^m*((a + b*x^2)^(p + 1)/(2*a*d*p*(c + d*x))), x] - Simp[m/(2*d*p) Int[x^(m - 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c ^2 + a*d^2, 0] && IGtQ[m, 1] && LtQ[p, -1] && EqQ[m + 2*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 = 0.12 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.74
method | result | size |
default | \(\frac {\left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} \left (2 a^{3} x^{3}-3 a x -1\right )}{3 \left (a x -1\right )^{3} a \,c^{4}}\) | \(45\) |
trager | \(\frac {\left (2 a^{2} x^{2}-2 a x -1\right ) \sqrt {-\frac {-a x +1}{a x +1}}}{3 a \,c^{4} \left (a x -1\right )^{2}}\) | \(47\) |
gosper | \(\frac {\left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} \left (2 a^{2} x^{2}-2 a x -1\right ) \left (a x +1\right )}{3 \left (a x -1\right )^{3} a \,c^{4}}\) | \(50\) |
orering | \(\frac {\left (2 a^{2} x^{2}-2 a x -1\right ) \left (a x -1\right ) \left (a x +1\right ) \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}{3 a \left (-a c x +c \right )^{4}}\) | \(54\) |
Input:
int(((a*x-1)/(a*x+1))^(3/2)/(-a*c*x+c)^4,x,method=_RETURNVERBOSE)
Output:
1/3*((a*x-1)/(a*x+1))^(3/2)*(2*a^3*x^3-3*a*x-1)/(a*x-1)^3/a/c^4
Time = 0.09 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.95 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=\frac {{\left (2 \, a^{2} x^{2} - 2 \, a x - 1\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{3 \, {\left (a^{3} c^{4} x^{2} - 2 \, a^{2} c^{4} x + a c^{4}\right )}} \] Input:
integrate(((a*x-1)/(a*x+1))^(3/2)/(-a*c*x+c)^4,x, algorithm="fricas")
Output:
1/3*(2*a^2*x^2 - 2*a*x - 1)*sqrt((a*x - 1)/(a*x + 1))/(a^3*c^4*x^2 - 2*a^2 *c^4*x + a*c^4)
\[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=\frac {\int \left (- \frac {\sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a^{5} x^{5} - 3 a^{4} x^{4} + 2 a^{3} x^{3} + 2 a^{2} x^{2} - 3 a x + 1}\right )\, dx + \int \frac {a x \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a^{5} x^{5} - 3 a^{4} x^{4} + 2 a^{3} x^{3} + 2 a^{2} x^{2} - 3 a x + 1}\, dx}{c^{4}} \] Input:
integrate(((a*x-1)/(a*x+1))**(3/2)/(-a*c*x+c)**4,x)
Output:
(Integral(-sqrt(a*x/(a*x + 1) - 1/(a*x + 1))/(a**5*x**5 - 3*a**4*x**4 + 2* a**3*x**3 + 2*a**2*x**2 - 3*a*x + 1), x) + Integral(a*x*sqrt(a*x/(a*x + 1) - 1/(a*x + 1))/(a**5*x**5 - 3*a**4*x**4 + 2*a**3*x**3 + 2*a**2*x**2 - 3*a *x + 1), x))/c**4
Time = 0.03 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.07 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=\frac {1}{12} \, a {\left (\frac {3 \, \sqrt {\frac {a x - 1}{a x + 1}}}{a^{2} c^{4}} + \frac {\frac {6 \, {\left (a x - 1\right )}}{a x + 1} - 1}{a^{2} c^{4} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}}\right )} \] Input:
integrate(((a*x-1)/(a*x+1))^(3/2)/(-a*c*x+c)^4,x, algorithm="maxima")
Output:
1/12*a*(3*sqrt((a*x - 1)/(a*x + 1))/(a^2*c^4) + (6*(a*x - 1)/(a*x + 1) - 1 )/(a^2*c^4*((a*x - 1)/(a*x + 1))^(3/2)))
\[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=\int { \frac {\left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}}{{\left (a c x - c\right )}^{4}} \,d x } \] Input:
integrate(((a*x-1)/(a*x+1))^(3/2)/(-a*c*x+c)^4,x, algorithm="giac")
Output:
integrate(((a*x - 1)/(a*x + 1))^(3/2)/(a*c*x - c)^4, x)
Time = 13.29 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.82 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=\frac {-2\,a^2\,x^2+2\,a\,x+1}{\left (3\,a\,c^4-3\,a^3\,c^4\,x^2\right )\,\sqrt {\frac {a\,x-1}{a\,x+1}}} \] Input:
int(((a*x - 1)/(a*x + 1))^(3/2)/(c - a*c*x)^4,x)
Output:
(2*a*x - 2*a^2*x^2 + 1)/((3*a*c^4 - 3*a^3*c^4*x^2)*((a*x - 1)/(a*x + 1))^( 1/2))
Time = 0.14 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.34 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=\frac {-2 \sqrt {a x -1}\, a^{2} x^{2}+2 \sqrt {a x -1}+2 \sqrt {a x +1}\, a^{2} x^{2}-2 \sqrt {a x +1}\, a x -\sqrt {a x +1}}{3 \sqrt {a x -1}\, a \,c^{4} \left (a^{2} x^{2}-1\right )} \] Input:
int(((a*x-1)/(a*x+1))^(3/2)/(-a*c*x+c)^4,x)
Output:
( - 2*sqrt(a*x - 1)*a**2*x**2 + 2*sqrt(a*x - 1) + 2*sqrt(a*x + 1)*a**2*x** 2 - 2*sqrt(a*x + 1)*a*x - sqrt(a*x + 1))/(3*sqrt(a*x - 1)*a*c**4*(a**2*x** 2 - 1))