Integrand size = 16, antiderivative size = 100 \[ \int \frac {e^{\coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=-\frac {a^4 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{7 c^4 \left (a-\frac {1}{x}\right )^5}+\frac {12 a^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{35 c^4 \left (a-\frac {1}{x}\right )^4}-\frac {23 a^2 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{105 c^4 \left (a-\frac {1}{x}\right )^3} \] Output:
-1/7*a^4*(1-1/a^2/x^2)^(3/2)/c^4/(a-1/x)^5+12/35*a^3*(1-1/a^2/x^2)^(3/2)/c ^4/(a-1/x)^4-23/105*a^2*(1-1/a^2/x^2)^(3/2)/c^4/(a-1/x)^3
Time = 0.18 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.51 \[ \int \frac {e^{\coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=-\frac {\sqrt {1-\frac {1}{a^2 x^2}} x \left (23+13 a x-8 a^2 x^2+2 a^3 x^3\right )}{105 c^4 (-1+a x)^4} \] Input:
Integrate[E^ArcCoth[a*x]/(c - a*c*x)^4,x]
Output:
-1/105*(Sqrt[1 - 1/(a^2*x^2)]*x*(23 + 13*a*x - 8*a^2*x^2 + 2*a^3*x^3))/(c^ 4*(-1 + a*x)^4)
Time = 0.58 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.28, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.562, 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 |
\(\displaystyle a c \int -\frac {\sqrt {1-\frac {1}{a^2 x^2}}}{c^5 \left (a-\frac {1}{x}\right )^5 x^2}d\frac {1}{x}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -a c \int \frac {\sqrt {1-\frac {1}{a^2 x^2}}}{c^5 \left (a-\frac {1}{x}\right )^5 x^2}d\frac {1}{x}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {a \int \frac {\sqrt {1-\frac {1}{a^2 x^2}}}{\left (a-\frac {1}{x}\right )^5 x^2}d\frac {1}{x}}{c^4}\) |
\(\Big \downarrow \) 581 |
\(\displaystyle -\frac {a \left (-\int -\frac {a \sqrt {1-\frac {1}{a^2 x^2}} \left (4 a-\frac {3}{x}\right )}{\left (a-\frac {1}{x}\right )^5}d\frac {1}{x}-\frac {a^2 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{\left (a-\frac {1}{x}\right )^4}\right )}{c^4}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {a \left (\int \frac {a \sqrt {1-\frac {1}{a^2 x^2}} \left (4 a-\frac {3}{x}\right )}{\left (a-\frac {1}{x}\right )^5}d\frac {1}{x}-\frac {a^2 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{\left (a-\frac {1}{x}\right )^4}\right )}{c^4}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {a \left (a \int \frac {\sqrt {1-\frac {1}{a^2 x^2}} \left (4 a-\frac {3}{x}\right )}{\left (a-\frac {1}{x}\right )^5}d\frac {1}{x}-\frac {a^2 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{\left (a-\frac {1}{x}\right )^4}\right )}{c^4}\) |
\(\Big \downarrow \) 671 |
\(\displaystyle -\frac {a \left (a \left (\frac {23}{7} \int \frac {\sqrt {1-\frac {1}{a^2 x^2}}}{\left (a-\frac {1}{x}\right )^4}d\frac {1}{x}+\frac {a^2 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{7 \left (a-\frac {1}{x}\right )^5}\right )-\frac {a^2 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{\left (a-\frac {1}{x}\right )^4}\right )}{c^4}\) |
\(\Big \downarrow \) 461 |
\(\displaystyle -\frac {a \left (a \left (\frac {23}{7} \left (\frac {\int \frac {\sqrt {1-\frac {1}{a^2 x^2}}}{\left (a-\frac {1}{x}\right )^3}d\frac {1}{x}}{5 a}+\frac {a \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{5 \left (a-\frac {1}{x}\right )^4}\right )+\frac {a^2 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{7 \left (a-\frac {1}{x}\right )^5}\right )-\frac {a^2 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{\left (a-\frac {1}{x}\right )^4}\right )}{c^4}\) |
\(\Big \downarrow \) 460 |
\(\displaystyle -\frac {a \left (a \left (\frac {a^2 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{7 \left (a-\frac {1}{x}\right )^5}+\frac {23}{7} \left (\frac {\left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{15 \left (a-\frac {1}{x}\right )^3}+\frac {a \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{5 \left (a-\frac {1}{x}\right )^4}\right )\right )-\frac {a^2 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{\left (a-\frac {1}{x}\right )^4}\right )}{c^4}\) |
Input:
Int[E^ArcCoth[a*x]/(c - a*c*x)^4,x]
Output:
-((a*(a*((23*((a*(1 - 1/(a^2*x^2))^(3/2))/(5*(a - x^(-1))^4) + (1 - 1/(a^2 *x^2))^(3/2)/(15*(a - x^(-1))^3)))/7 + (a^2*(1 - 1/(a^2*x^2))^(3/2))/(7*(a - x^(-1))^5)) - (a^2*(1 - 1/(a^2*x^2))^(3/2))/(a - x^(-1))^4))/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 = 0.13 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.50
method | result | size |
gosper | \(-\frac {\left (2 a^{2} x^{2}-10 a x +23\right ) \left (a x +1\right )}{105 \left (a x -1\right )^{3} c^{4} \sqrt {\frac {a x -1}{a x +1}}\, a}\) | \(50\) |
default | \(-\frac {\left (2 a^{2} x^{2}-10 a x +23\right ) \left (a x +1\right )}{105 \left (a x -1\right )^{3} c^{4} \sqrt {\frac {a x -1}{a x +1}}\, a}\) | \(50\) |
orering | \(-\frac {\left (2 a^{2} x^{2}-10 a x +23\right ) \left (a x -1\right ) \left (a x +1\right )}{105 a \sqrt {\frac {a x -1}{a x +1}}\, \left (-a c x +c \right )^{4}}\) | \(54\) |
trager | \(-\frac {\left (a x +1\right ) \left (2 a^{3} x^{3}-8 a^{2} x^{2}+13 a x +23\right ) \sqrt {-\frac {-a x +1}{a x +1}}}{105 a \,c^{4} \left (a x -1\right )^{4}}\) | \(60\) |
Input:
int(1/((a*x-1)/(a*x+1))^(1/2)/(-a*c*x+c)^4,x,method=_RETURNVERBOSE)
Output:
-1/105*(2*a^2*x^2-10*a*x+23)*(a*x+1)/(a*x-1)^3/c^4/((a*x-1)/(a*x+1))^(1/2) /a
Time = 0.10 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=-\frac {{\left (2 \, a^{4} x^{4} - 6 \, a^{3} x^{3} + 5 \, a^{2} x^{2} + 36 \, a x + 23\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{105 \, {\left (a^{5} c^{4} x^{4} - 4 \, a^{4} c^{4} x^{3} + 6 \, a^{3} c^{4} x^{2} - 4 \, a^{2} c^{4} x + a c^{4}\right )}} \] Input:
integrate(1/((a*x-1)/(a*x+1))^(1/2)/(-a*c*x+c)^4,x, algorithm="fricas")
Output:
-1/105*(2*a^4*x^4 - 6*a^3*x^3 + 5*a^2*x^2 + 36*a*x + 23)*sqrt((a*x - 1)/(a *x + 1))/(a^5*c^4*x^4 - 4*a^4*c^4*x^3 + 6*a^3*c^4*x^2 - 4*a^2*c^4*x + a*c^ 4)
\[ \int \frac {e^{\coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=\frac {\int \frac {1}{a^{4} x^{4} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}} - 4 a^{3} x^{3} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}} + 6 a^{2} x^{2} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}} - 4 a x \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}} + \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}\, dx}{c^{4}} \] Input:
integrate(1/((a*x-1)/(a*x+1))**(1/2)/(-a*c*x+c)**4,x)
Output:
Integral(1/(a**4*x**4*sqrt(a*x/(a*x + 1) - 1/(a*x + 1)) - 4*a**3*x**3*sqrt (a*x/(a*x + 1) - 1/(a*x + 1)) + 6*a**2*x**2*sqrt(a*x/(a*x + 1) - 1/(a*x + 1)) - 4*a*x*sqrt(a*x/(a*x + 1) - 1/(a*x + 1)) + sqrt(a*x/(a*x + 1) - 1/(a* x + 1))), x)/c**4
Time = 0.03 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.55 \[ \int \frac {e^{\coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=\frac {\frac {42 \, {\left (a x - 1\right )}}{a x + 1} - \frac {35 \, {\left (a x - 1\right )}^{2}}{{\left (a x + 1\right )}^{2}} - 15}{420 \, a c^{4} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {7}{2}}} \] Input:
integrate(1/((a*x-1)/(a*x+1))^(1/2)/(-a*c*x+c)^4,x, algorithm="maxima")
Output:
1/420*(42*(a*x - 1)/(a*x + 1) - 35*(a*x - 1)^2/(a*x + 1)^2 - 15)/(a*c^4*(( a*x - 1)/(a*x + 1))^(7/2))
Time = 0.17 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.05 \[ \int \frac {e^{\coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=-\frac {4 \, {\left (70 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )}^{4} x^{4} + 35 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )}^{3} x^{3} + 21 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )}^{2} x^{2} - 7 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )} x + 1\right )}}{105 \, {\left ({\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )} x - 1\right )}^{7} a c^{4}} \] Input:
integrate(1/((a*x-1)/(a*x+1))^(1/2)/(-a*c*x+c)^4,x, algorithm="giac")
Output:
-4/105*(70*(a + sqrt(a^2 - 1/x^2))^4*x^4 + 35*(a + sqrt(a^2 - 1/x^2))^3*x^ 3 + 21*(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^4)
Time = 13.32 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.56 \[ \int \frac {e^{\coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=-\frac {\frac {{\left (a\,x-1\right )}^2}{3\,{\left (a\,x+1\right )}^2}-\frac {2\,\left (a\,x-1\right )}{5\,\left (a\,x+1\right )}+\frac {1}{7}}{4\,a\,c^4\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{7/2}} \] Input:
int(1/((c - a*c*x)^4*((a*x - 1)/(a*x + 1))^(1/2)),x)
Output:
-((a*x - 1)^2/(3*(a*x + 1)^2) - (2*(a*x - 1))/(5*(a*x + 1)) + 1/7)/(4*a*c^ 4*((a*x - 1)/(a*x + 1))^(7/2))
Time = 0.15 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.32 \[ \int \frac {e^{\coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=\frac {2 \sqrt {a x -1}\, a^{3} x^{3}-6 \sqrt {a x -1}\, a^{2} x^{2}+6 \sqrt {a x -1}\, a x -2 \sqrt {a x -1}-2 \sqrt {a x +1}\, a^{3} x^{3}+8 \sqrt {a x +1}\, a^{2} x^{2}-13 \sqrt {a x +1}\, a x -23 \sqrt {a x +1}}{105 \sqrt {a x -1}\, a \,c^{4} \left (a^{3} x^{3}-3 a^{2} x^{2}+3 a x -1\right )} \] Input:
int(1/((a*x-1)/(a*x+1))^(1/2)/(-a*c*x+c)^4,x)
Output:
(2*sqrt(a*x - 1)*a**3*x**3 - 6*sqrt(a*x - 1)*a**2*x**2 + 6*sqrt(a*x - 1)*a *x - 2*sqrt(a*x - 1) - 2*sqrt(a*x + 1)*a**3*x**3 + 8*sqrt(a*x + 1)*a**2*x* *2 - 13*sqrt(a*x + 1)*a*x - 23*sqrt(a*x + 1))/(105*sqrt(a*x - 1)*a*c**4*(a **3*x**3 - 3*a**2*x**2 + 3*a*x - 1))