Integrand size = 18, antiderivative size = 94 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=-\frac {4 \left (a+\frac {1}{x}\right )}{5 a^2 c^5 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}+\frac {\left (a+\frac {1}{x}\right )^2}{5 a^3 c^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}+\frac {5 a+\frac {2}{x}}{5 a^2 c^5 \sqrt {1-\frac {1}{a^2 x^2}}} \]
-4/5*(a+1/x)/a^2/c^5/(1-1/a^2/x^2)^(3/2)+1/5*(a+1/x)^2/a^3/c^5/(1-1/a^2/x^ 2)^(5/2)+1/5*(5*a+2/x)/a^2/c^5/(1-1/a^2/x^2)^(1/2)
Time = 0.17 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.61 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=\frac {\sqrt {1-\frac {1}{a^2 x^2}} x \left (2+a x-4 a^2 x^2+2 a^3 x^3\right )}{5 c^5 (-1+a x)^3 (1+a x)} \]
Time = 0.39 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {6724, 27, 570, 529, 2166, 27, 453}
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.
\(\displaystyle \int \frac {e^{-3 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx\) |
\(\Big \downarrow \) 6724 |
\(\displaystyle \frac {\int \frac {1}{c^2 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} \left (a-\frac {1}{x}\right )^2 x^3}d\frac {1}{x}}{a^3 c^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {1}{\left (1-\frac {1}{a^2 x^2}\right )^{3/2} \left (a-\frac {1}{x}\right )^2 x^3}d\frac {1}{x}}{a^3 c^5}\) |
\(\Big \downarrow \) 570 |
\(\displaystyle \frac {\int \frac {\left (a+\frac {1}{x}\right )^2}{\left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^3}d\frac {1}{x}}{a^7 c^5}\) |
\(\Big \downarrow \) 529 |
\(\displaystyle \frac {\frac {a^4 \left (a+\frac {1}{x}\right )^2}{5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}-\frac {1}{5} a \int \frac {\left (a+\frac {1}{x}\right ) \left (2 a^3+\frac {5 a^2}{x}+\frac {5 a}{x^2}\right )}{\left (1-\frac {1}{a^2 x^2}\right )^{5/2}}d\frac {1}{x}}{a^7 c^5}\) |
\(\Big \downarrow \) 2166 |
\(\displaystyle \frac {\frac {a^4 \left (a+\frac {1}{x}\right )^2}{5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}-\frac {1}{5} a \left (\frac {4 a^4 \left (a+\frac {1}{x}\right )}{\left (1-\frac {1}{a^2 x^2}\right )^{3/2}}-\frac {1}{3} a \int \frac {3 a^2 \left (2 a+\frac {5}{x}\right )}{\left (1-\frac {1}{a^2 x^2}\right )^{3/2}}d\frac {1}{x}\right )}{a^7 c^5}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {a^4 \left (a+\frac {1}{x}\right )^2}{5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}-\frac {1}{5} a \left (\frac {4 a^4 \left (a+\frac {1}{x}\right )}{\left (1-\frac {1}{a^2 x^2}\right )^{3/2}}-a^3 \int \frac {2 a+\frac {5}{x}}{\left (1-\frac {1}{a^2 x^2}\right )^{3/2}}d\frac {1}{x}\right )}{a^7 c^5}\) |
\(\Big \downarrow \) 453 |
\(\displaystyle \frac {\frac {a^4 \left (a+\frac {1}{x}\right )^2}{5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}-\frac {1}{5} a \left (\frac {4 a^4 \left (a+\frac {1}{x}\right )}{\left (1-\frac {1}{a^2 x^2}\right )^{3/2}}-\frac {a^4 \left (5 a+\frac {2}{x}\right )}{\sqrt {1-\frac {1}{a^2 x^2}}}\right )}{a^7 c^5}\) |
(-1/5*(a*((4*a^4*(a + x^(-1)))/(1 - 1/(a^2*x^2))^(3/2) - (a^4*(5*a + 2/x)) /Sqrt[1 - 1/(a^2*x^2)])) + (a^4*(a + x^(-1))^2)/(5*(1 - 1/(a^2*x^2))^(5/2) ))/(a^7*c^5)
3.3.24.3.1 Defintions of rubi rules used
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_))/((a_) + (b_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[-(a* d - b*c*x)/(a*b*Sqrt[a + b*x^2]), x] /; FreeQ[{a, b, c, d}, x]
Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo l] :> With[{Qx = PolynomialQuotient[x^m, a*d + b*c*x, x], R = PolynomialRem ainder[x^m, a*d + b*c*x, x]}, Simp[(-c)*R*(c + d*x)^n*((a + b*x^2)^(p + 1)/ (2*a*d*(p + 1))), x] + Simp[c/(2*a*(p + 1)) Int[(c + d*x)^(n - 1)*(a + b* x^2)^(p + 1)*ExpandToSum[2*a*d*(p + 1)*Qx + R*(n + 2*p + 2), x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && IGtQ[m, 1] && LtQ[p, -1] && EqQ[b* c^2 + a*d^2, 0]
Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c^(2*n)/a^n Int[(e*x)^m*((a + b*x^2)^(n + p)/(c - d*x)^ n), x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[b*c^2 + a*d^2, 0] && I LtQ[n, -1] && !(IGtQ[m, 0] && ILtQ[m + n, 0] && !GtQ[p, 1])
Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : > With[{Qx = PolynomialQuotient[Pq, a*e + b*d*x, x], R = PolynomialRemainde r[Pq, a*e + b*d*x, x]}, Simp[(-d)*R*(d + e*x)^m*((a + b*x^2)^(p + 1)/(2*a*e *(p + 1))), x] + Simp[d/(2*a*(p + 1)) Int[(d + e*x)^(m - 1)*(a + b*x^2)^( p + 1)*ExpandToSum[2*a*e*(p + 1)*Qx + R*(m + 2*p + 2), x], x], x]] /; FreeQ [{a, b, d, e}, x] && PolyQ[Pq, x] && EqQ[b*d^2 + a*e^2, 0] && ILtQ[p + 1/2, 0] && GtQ[m, 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 = 0.41 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.57
method | result | size |
trager | \(\frac {\left (2 a^{3} x^{3}-4 a^{2} x^{2}+a x +2\right ) \sqrt {-\frac {-a x +1}{a x +1}}}{5 a \,c^{5} \left (a x -1\right )^{3}}\) | \(54\) |
gosper | \(\frac {\left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} \left (2 a^{3} x^{3}-4 a^{2} x^{2}+a x +2\right ) \left (a x +1\right )}{5 \left (a x -1\right )^{4} c^{5} a}\) | \(57\) |
default | \(\frac {\left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} \left (2 a^{4} x^{4}-2 a^{3} x^{3}-3 a^{2} x^{2}+3 a x +2\right )}{5 \left (a x -1\right )^{4} c^{5} a}\) | \(61\) |
Time = 0.25 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.82 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=\frac {{\left (2 \, a^{3} x^{3} - 4 \, a^{2} x^{2} + a x + 2\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{5 \, {\left (a^{4} c^{5} x^{3} - 3 \, a^{3} c^{5} x^{2} + 3 \, a^{2} c^{5} x - a c^{5}\right )}} \]
1/5*(2*a^3*x^3 - 4*a^2*x^2 + a*x + 2)*sqrt((a*x - 1)/(a*x + 1))/(a^4*c^5*x ^3 - 3*a^3*c^5*x^2 + 3*a^2*c^5*x - a*c^5)
\[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=- \frac {\int \left (- \frac {\sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a^{6} x^{6} - 4 a^{5} x^{5} + 5 a^{4} x^{4} - 5 a^{2} x^{2} + 4 a x - 1}\right )\, dx + \int \frac {a x \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a^{6} x^{6} - 4 a^{5} x^{5} + 5 a^{4} x^{4} - 5 a^{2} x^{2} + 4 a x - 1}\, dx}{c^{5}} \]
-(Integral(-sqrt(a*x/(a*x + 1) - 1/(a*x + 1))/(a**6*x**6 - 4*a**5*x**5 + 5 *a**4*x**4 - 5*a**2*x**2 + 4*a*x - 1), x) + Integral(a*x*sqrt(a*x/(a*x + 1 ) - 1/(a*x + 1))/(a**6*x**6 - 4*a**5*x**5 + 5*a**4*x**4 - 5*a**2*x**2 + 4* a*x - 1), x))/c**5
Time = 0.20 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.87 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=\frac {1}{40} \, a {\left (\frac {5 \, \sqrt {\frac {a x - 1}{a x + 1}}}{a^{2} c^{5}} - \frac {\frac {5 \, {\left (a x - 1\right )}}{a x + 1} - \frac {15 \, {\left (a x - 1\right )}^{2}}{{\left (a x + 1\right )}^{2}} - 1}{a^{2} c^{5} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{2}}}\right )} \]
1/40*a*(5*sqrt((a*x - 1)/(a*x + 1))/(a^2*c^5) - (5*(a*x - 1)/(a*x + 1) - 1 5*(a*x - 1)^2/(a*x + 1)^2 - 1)/(a^2*c^5*((a*x - 1)/(a*x + 1))^(5/2)))
\[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=\int { -\frac {\left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}}{{\left (a c x - c\right )}^{5}} \,d x } \]
Time = 4.15 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.54 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=\frac {2\,a^3\,x^3-4\,a^2\,x^2+a\,x+2}{5\,a\,c^5\,{\left (a\,x+1\right )}^3\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/2}} \]