Integrand size = 18, antiderivative size = 95 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=-\frac {a^2 \sqrt {1-\frac {1}{a^2 x^2}}}{5 c^4 \left (a-\frac {1}{x}\right )^3}+\frac {8 a \sqrt {1-\frac {1}{a^2 x^2}}}{15 c^4 \left (a-\frac {1}{x}\right )^2}-\frac {7 \sqrt {1-\frac {1}{a^2 x^2}}}{15 c^4 \left (a-\frac {1}{x}\right )} \] Output:
-1/5*a^2*(1-1/a^2/x^2)^(1/2)/c^4/(a-1/x)^3+8/15*a*(1-1/a^2/x^2)^(1/2)/c^4/ (a-1/x)^2-7/15*(1-1/a^2/x^2)^(1/2)/c^4/(a-1/x)
Time = 0.17 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.45 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=-\frac {\sqrt {1-\frac {1}{a^2 x^2}} x \left (7-6 a x+2 a^2 x^2\right )}{15 c^4 (-1+a x)^3} \] Input:
Integrate[1/(E^ArcCoth[a*x]*(c - a*c*x)^4),x]
Output:
-1/15*(Sqrt[1 - 1/(a^2*x^2)]*x*(7 - 6*a*x + 2*a^2*x^2))/(c^4*(-1 + a*x)^3)
Time = 0.57 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.37, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6724, 25, 27, 581, 25, 27, 671, 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.
\(\displaystyle \int \frac {e^{-\coth ^{-1}(a x)}}{(c-a c x)^4} \, dx\) |
\(\Big \downarrow \) 6724 |
\(\displaystyle \frac {\int -\frac {1}{c^3 \sqrt {1-\frac {1}{a^2 x^2}} \left (a-\frac {1}{x}\right )^3 x^2}d\frac {1}{x}}{a c}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {1}{c^3 \sqrt {1-\frac {1}{a^2 x^2}} \left (a-\frac {1}{x}\right )^3 x^2}d\frac {1}{x}}{a c}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int \frac {1}{\sqrt {1-\frac {1}{a^2 x^2}} \left (a-\frac {1}{x}\right )^3 x^2}d\frac {1}{x}}{a c^4}\) |
\(\Big \downarrow \) 581 |
\(\displaystyle -\frac {-\int -\frac {a \left (2 a-\frac {1}{x}\right )}{\sqrt {1-\frac {1}{a^2 x^2}} \left (a-\frac {1}{x}\right )^3}d\frac {1}{x}-\frac {a^2 \sqrt {1-\frac {1}{a^2 x^2}}}{\left (a-\frac {1}{x}\right )^2}}{a c^4}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {a \left (2 a-\frac {1}{x}\right )}{\sqrt {1-\frac {1}{a^2 x^2}} \left (a-\frac {1}{x}\right )^3}d\frac {1}{x}-\frac {a^2 \sqrt {1-\frac {1}{a^2 x^2}}}{\left (a-\frac {1}{x}\right )^2}}{a c^4}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {a \int \frac {2 a-\frac {1}{x}}{\sqrt {1-\frac {1}{a^2 x^2}} \left (a-\frac {1}{x}\right )^3}d\frac {1}{x}-\frac {a^2 \sqrt {1-\frac {1}{a^2 x^2}}}{\left (a-\frac {1}{x}\right )^2}}{a c^4}\) |
\(\Big \downarrow \) 671 |
\(\displaystyle -\frac {a \left (\frac {7}{5} \int \frac {1}{\sqrt {1-\frac {1}{a^2 x^2}} \left (a-\frac {1}{x}\right )^2}d\frac {1}{x}+\frac {a^2 \sqrt {1-\frac {1}{a^2 x^2}}}{5 \left (a-\frac {1}{x}\right )^3}\right )-\frac {a^2 \sqrt {1-\frac {1}{a^2 x^2}}}{\left (a-\frac {1}{x}\right )^2}}{a c^4}\) |
\(\Big \downarrow \) 461 |
\(\displaystyle -\frac {a \left (\frac {7}{5} \left (\frac {\int \frac {1}{\sqrt {1-\frac {1}{a^2 x^2}} \left (a-\frac {1}{x}\right )}d\frac {1}{x}}{3 a}+\frac {a \sqrt {1-\frac {1}{a^2 x^2}}}{3 \left (a-\frac {1}{x}\right )^2}\right )+\frac {a^2 \sqrt {1-\frac {1}{a^2 x^2}}}{5 \left (a-\frac {1}{x}\right )^3}\right )-\frac {a^2 \sqrt {1-\frac {1}{a^2 x^2}}}{\left (a-\frac {1}{x}\right )^2}}{a c^4}\) |
\(\Big \downarrow \) 460 |
\(\displaystyle -\frac {a \left (\frac {a^2 \sqrt {1-\frac {1}{a^2 x^2}}}{5 \left (a-\frac {1}{x}\right )^3}+\frac {7}{5} \left (\frac {a \sqrt {1-\frac {1}{a^2 x^2}}}{3 \left (a-\frac {1}{x}\right )^2}+\frac {\sqrt {1-\frac {1}{a^2 x^2}}}{3 \left (a-\frac {1}{x}\right )}\right )\right )-\frac {a^2 \sqrt {1-\frac {1}{a^2 x^2}}}{\left (a-\frac {1}{x}\right )^2}}{a c^4}\) |
Input:
Int[1/(E^ArcCoth[a*x]*(c - a*c*x)^4),x]
Output:
-((a*((7*((a*Sqrt[1 - 1/(a^2*x^2)])/(3*(a - x^(-1))^2) + Sqrt[1 - 1/(a^2*x ^2)]/(3*(a - x^(-1)))))/5 + (a^2*Sqrt[1 - 1/(a^2*x^2)])/(5*(a - x^(-1))^3) ) - (a^2*Sqrt[1 - 1/(a^2*x^2)])/(a - x^(-1))^2)/(a*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[((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[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol ] :> Simp[(c + d*x)^(m + n - 1)*((a + b*x^2)^(p + 1)/(b*d^(m - 1)*(m + n + 2*p + 1))), x] + Simp[1/(d^m*(m + n + 2*p + 1)) Int[(c + d*x)^n*(a + b*x^ 2)^p*ExpandToSum[d^m*(m + n + 2*p + 1)*x^m - (m + n + 2*p + 1)*(c + d*x)^m + c*(c + d*x)^(m - 2)*(c*(m + n - 1) + c*(m + n + 2*p + 1) + 2*d*(m + n + p )*x), x], x], x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] & & IGtQ[m, 1] && NeQ[m + n + 2*p + 1, 0] && (IntegerQ[2*p] || ILtQ[m + n, 0] )
Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_ ), x_Symbol] :> Simp[(d*g - e*f)*(d + e*x)^m*((a + c*x^2)^(p + 1)/(2*c*d*(m + p + 1))), x] + Simp[(m*(g*c*d + c*e*f) + 2*e*c*f*(p + 1))/(e*(2*c*d)*(m + p + 1)) Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && ((LtQ[m, -1] && !IGtQ[m + p + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + 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 = 50, normalized size of antiderivative = 0.53
method | result | size |
gosper | \(-\frac {\sqrt {\frac {a x -1}{a x +1}}\, \left (2 a^{2} x^{2}-6 a x +7\right ) \left (a x +1\right )}{15 \left (a x -1\right )^{3} a \,c^{4}}\) | \(50\) |
default | \(-\frac {\sqrt {\frac {a x -1}{a x +1}}\, \left (2 a^{2} x^{2}-6 a x +7\right ) \left (a x +1\right )}{15 \left (a x -1\right )^{3} a \,c^{4}}\) | \(50\) |
trager | \(-\frac {\left (2 a^{2} x^{2}-6 a x +7\right ) \left (a x +1\right ) \sqrt {-\frac {-a x +1}{a x +1}}}{15 a \,c^{4} \left (a x -1\right )^{3}}\) | \(52\) |
orering | \(-\frac {\left (2 a^{2} x^{2}-6 a x +7\right ) \left (a x -1\right ) \left (a x +1\right ) \sqrt {\frac {a x -1}{a x +1}}}{15 a \left (-a c x +c \right )^{4}}\) | \(54\) |
Input:
int(((a*x-1)/(a*x+1))^(1/2)/(-a*c*x+c)^4,x,method=_RETURNVERBOSE)
Output:
-1/15*((a*x-1)/(a*x+1))^(1/2)*(2*a^2*x^2-6*a*x+7)*(a*x+1)/(a*x-1)^3/a/c^4
Time = 0.07 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.81 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=-\frac {{\left (2 \, a^{3} x^{3} - 4 \, a^{2} x^{2} + a x + 7\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{15 \, {\left (a^{4} c^{4} x^{3} - 3 \, a^{3} c^{4} x^{2} + 3 \, a^{2} c^{4} x - a c^{4}\right )}} \] Input:
integrate(((a*x-1)/(a*x+1))^(1/2)/(-a*c*x+c)^4,x, algorithm="fricas")
Output:
-1/15*(2*a^3*x^3 - 4*a^2*x^2 + a*x + 7)*sqrt((a*x - 1)/(a*x + 1))/(a^4*c^4 *x^3 - 3*a^3*c^4*x^2 + 3*a^2*c^4*x - a*c^4)
\[ \int \frac {e^{-\coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=\frac {\int \frac {\sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a^{4} x^{4} - 4 a^{3} x^{3} + 6 a^{2} x^{2} - 4 a x + 1}\, dx}{c^{4}} \] Input:
integrate(((a*x-1)/(a*x+1))**(1/2)/(-a*c*x+c)**4,x)
Output:
Integral(sqrt(a*x/(a*x + 1) - 1/(a*x + 1))/(a**4*x**4 - 4*a**3*x**3 + 6*a* *2*x**2 - 4*a*x + 1), x)/c**4
Time = 0.04 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.58 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=\frac {\frac {10 \, {\left (a x - 1\right )}}{a x + 1} - \frac {15 \, {\left (a x - 1\right )}^{2}}{{\left (a x + 1\right )}^{2}} - 3}{60 \, a c^{4} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{2}}} \] Input:
integrate(((a*x-1)/(a*x+1))^(1/2)/(-a*c*x+c)^4,x, algorithm="maxima")
Output:
1/60*(10*(a*x - 1)/(a*x + 1) - 15*(a*x - 1)^2/(a*x + 1)^2 - 3)/(a*c^4*((a* x - 1)/(a*x + 1))^(5/2))
Time = 0.16 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.68 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=-\frac {4 \, {\left (10 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )}^{2} x^{2} - 5 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )} x + 1\right )}}{15 \, {\left ({\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )} x - 1\right )}^{5} a c^{4}} \] Input:
integrate(((a*x-1)/(a*x+1))^(1/2)/(-a*c*x+c)^4,x, algorithm="giac")
Output:
-4/15*(10*(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^4)
Time = 0.02 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.58 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=-\frac {\frac {{\left (a\,x-1\right )}^2}{{\left (a\,x+1\right )}^2}-\frac {2\,\left (a\,x-1\right )}{3\,\left (a\,x+1\right )}+\frac {1}{5}}{4\,a\,c^4\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/2}} \] Input:
int(((a*x - 1)/(a*x + 1))^(1/2)/(c - a*c*x)^4,x)
Output:
-((a*x - 1)^2/(a*x + 1)^2 - (2*(a*x - 1))/(3*(a*x + 1)) + 1/5)/(4*a*c^4*(( a*x - 1)/(a*x + 1))^(5/2))
Time = 0.14 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.01 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=\frac {2 \sqrt {a x -1}\, a^{2} x^{2}-4 \sqrt {a x -1}\, a x +2 \sqrt {a x -1}-2 \sqrt {a x +1}\, a^{2} x^{2}+6 \sqrt {a x +1}\, a x -7 \sqrt {a x +1}}{15 \sqrt {a x -1}\, a \,c^{4} \left (a^{2} x^{2}-2 a x +1\right )} \] Input:
int(((a*x-1)/(a*x+1))^(1/2)/(-a*c*x+c)^4,x)
Output:
(2*sqrt(a*x - 1)*a**2*x**2 - 4*sqrt(a*x - 1)*a*x + 2*sqrt(a*x - 1) - 2*sqr t(a*x + 1)*a**2*x**2 + 6*sqrt(a*x + 1)*a*x - 7*sqrt(a*x + 1))/(15*sqrt(a*x - 1)*a*c**4*(a**2*x**2 - 2*a*x + 1))