Integrand size = 18, antiderivative size = 125 \[ \int \frac {e^{3 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=-\frac {152 \left (a+\frac {1}{x}\right )^5}{1155 a^6 c^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}+\frac {79 \left (a+\frac {1}{x}\right )^6}{231 a^7 c^5 \left (1-\frac {1}{a^2 x^2}\right )^{7/2}}-\frac {10 \left (a+\frac {1}{x}\right )^7}{33 a^8 c^5 \left (1-\frac {1}{a^2 x^2}\right )^{9/2}}+\frac {\left (a+\frac {1}{x}\right )^8}{11 a^9 c^5 \left (1-\frac {1}{a^2 x^2}\right )^{11/2}} \]
-152/1155*(a+1/x)^5/a^6/c^5/(1-1/a^2/x^2)^(5/2)+79/231*(a+1/x)^6/a^7/c^5/( 1-1/a^2/x^2)^(7/2)-10/33*(a+1/x)^7/a^8/c^5/(1-1/a^2/x^2)^(9/2)+1/11*(a+1/x )^8/a^9/c^5/(1-1/a^2/x^2)^(11/2)
Time = 0.18 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.46 \[ \int \frac {e^{3 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=-\frac {\sqrt {1-\frac {1}{a^2 x^2}} x (1+a x)^2 \left (-152+61 a x-16 a^2 x^2+2 a^3 x^3\right )}{1155 c^5 (-1+a x)^6} \]
-1/1155*(Sqrt[1 - 1/(a^2*x^2)]*x*(1 + a*x)^2*(-152 + 61*a*x - 16*a^2*x^2 + 2*a^3*x^3))/(c^5*(-1 + a*x)^6)
Time = 0.48 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.07, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {6724, 27, 570, 529, 2166, 27, 669, 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.
| \(\displaystyle \int \frac {e^{3 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx\) |
| \(\Big \downarrow \) 6724 |
| \(\displaystyle a^3 c^3 \int \frac {\left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{c^8 \left (a-\frac {1}{x}\right )^8 x^3}d\frac {1}{x}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {a^3 \int \frac {\left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{\left (a-\frac {1}{x}\right )^8 x^3}d\frac {1}{x}}{c^5}\) |
| \(\Big \downarrow \) 570 |
| \(\displaystyle \frac {\int \frac {\left (a+\frac {1}{x}\right )^8}{\left (1-\frac {1}{a^2 x^2}\right )^{13/2} x^3}d\frac {1}{x}}{a^{13} c^5}\) |
| \(\Big \downarrow \) 529 |
| \(\displaystyle \frac {\frac {a^4 \left (a+\frac {1}{x}\right )^8}{11 \left (1-\frac {1}{a^2 x^2}\right )^{11/2}}-\frac {1}{11} a \int \frac {\left (a+\frac {1}{x}\right )^7 \left (8 a^3+\frac {11 a^2}{x}+\frac {11 a}{x^2}\right )}{\left (1-\frac {1}{a^2 x^2}\right )^{11/2}}d\frac {1}{x}}{a^{13} c^5}\) |
| \(\Big \downarrow \) 2166 |
| \(\displaystyle \frac {\frac {a^4 \left (a+\frac {1}{x}\right )^8}{11 \left (1-\frac {1}{a^2 x^2}\right )^{11/2}}-\frac {1}{11} a \left (\frac {10 a^4 \left (a+\frac {1}{x}\right )^7}{3 \left (1-\frac {1}{a^2 x^2}\right )^{9/2}}-\frac {1}{9} a \int \frac {3 a^2 \left (a+\frac {1}{x}\right )^6 \left (46 a+\frac {33}{x}\right )}{\left (1-\frac {1}{a^2 x^2}\right )^{9/2}}d\frac {1}{x}\right )}{a^{13} c^5}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {a^4 \left (a+\frac {1}{x}\right )^8}{11 \left (1-\frac {1}{a^2 x^2}\right )^{11/2}}-\frac {1}{11} a \left (\frac {10 a^4 \left (a+\frac {1}{x}\right )^7}{3 \left (1-\frac {1}{a^2 x^2}\right )^{9/2}}-\frac {1}{3} a^3 \int \frac {\left (a+\frac {1}{x}\right )^6 \left (46 a+\frac {33}{x}\right )}{\left (1-\frac {1}{a^2 x^2}\right )^{9/2}}d\frac {1}{x}\right )}{a^{13} c^5}\) |
| \(\Big \downarrow \) 669 |
| \(\displaystyle \frac {\frac {a^4 \left (a+\frac {1}{x}\right )^8}{11 \left (1-\frac {1}{a^2 x^2}\right )^{11/2}}-\frac {1}{11} a \left (\frac {10 a^4 \left (a+\frac {1}{x}\right )^7}{3 \left (1-\frac {1}{a^2 x^2}\right )^{9/2}}-\frac {1}{3} a^3 \left (\frac {79 a^2 \left (a+\frac {1}{x}\right )^6}{7 \left (1-\frac {1}{a^2 x^2}\right )^{7/2}}-\frac {152}{7} a^2 \int \frac {\left (a+\frac {1}{x}\right )^5}{\left (1-\frac {1}{a^2 x^2}\right )^{7/2}}d\frac {1}{x}\right )\right )}{a^{13} c^5}\) |
| \(\Big \downarrow \) 460 |
| \(\displaystyle \frac {\frac {a^4 \left (a+\frac {1}{x}\right )^8}{11 \left (1-\frac {1}{a^2 x^2}\right )^{11/2}}-\frac {1}{11} a \left (\frac {10 a^4 \left (a+\frac {1}{x}\right )^7}{3 \left (1-\frac {1}{a^2 x^2}\right )^{9/2}}-\frac {1}{3} a^3 \left (\frac {79 a^2 \left (a+\frac {1}{x}\right )^6}{7 \left (1-\frac {1}{a^2 x^2}\right )^{7/2}}-\frac {152 a^3 \left (a+\frac {1}{x}\right )^5}{35 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}\right )\right )}{a^{13} c^5}\) |
(-1/11*(a*(-1/3*(a^3*((-152*a^3*(a + x^(-1))^5)/(35*(1 - 1/(a^2*x^2))^(5/2 )) + (79*a^2*(a + x^(-1))^6)/(7*(1 - 1/(a^2*x^2))^(7/2)))) + (10*a^4*(a + x^(-1))^7)/(3*(1 - 1/(a^2*x^2))^(9/2)))) + (a^4*(a + x^(-1))^8)/(11*(1 - 1 /(a^2*x^2))^(11/2)))/(a^13*c^5)
3.2.86.3.1 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_))^(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_)^(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[((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* (p + 1))), x] - Simp[e*((m*(d*g + e*f) + 2*e*f*(p + 1))/(2*c*d*(p + 1))) Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && EqQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 0]
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 = 0.43 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.46
| method | result | size |
| gosper | \(-\frac {\left (2 a^{3} x^{3}-16 a^{2} x^{2}+61 a x -152\right ) \left (a x +1\right )}{1155 \left (a x -1\right )^{4} c^{5} \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} a}\) | \(58\) |
| default | \(-\frac {\left (2 a^{3} x^{3}-16 a^{2} x^{2}+61 a x -152\right ) \left (a x +1\right )}{1155 \left (a x -1\right )^{4} c^{5} \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} a}\) | \(58\) |
| trager | \(-\frac {\left (a x +1\right ) \left (2 a^{5} x^{5}-12 a^{4} x^{4}+31 a^{3} x^{3}-46 a^{2} x^{2}-243 a x -152\right ) \sqrt {-\frac {-a x +1}{a x +1}}}{1155 a \,c^{5} \left (a x -1\right )^{6}}\) | \(76\) |
Time = 0.24 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.07 \[ \int \frac {e^{3 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=-\frac {{\left (2 \, a^{6} x^{6} - 10 \, a^{5} x^{5} + 19 \, a^{4} x^{4} - 15 \, a^{3} x^{3} - 289 \, a^{2} x^{2} - 395 \, a x - 152\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{1155 \, {\left (a^{7} c^{5} x^{6} - 6 \, a^{6} c^{5} x^{5} + 15 \, a^{5} c^{5} x^{4} - 20 \, a^{4} c^{5} x^{3} + 15 \, a^{3} c^{5} x^{2} - 6 \, a^{2} c^{5} x + a c^{5}\right )}} \]
-1/1155*(2*a^6*x^6 - 10*a^5*x^5 + 19*a^4*x^4 - 15*a^3*x^3 - 289*a^2*x^2 - 395*a*x - 152)*sqrt((a*x - 1)/(a*x + 1))/(a^7*c^5*x^6 - 6*a^6*c^5*x^5 + 15 *a^5*c^5*x^4 - 20*a^4*c^5*x^3 + 15*a^3*c^5*x^2 - 6*a^2*c^5*x + a*c^5)
\[ \int \frac {e^{3 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=- \frac {\int \frac {1}{\frac {a^{6} x^{6} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1} - \frac {6 a^{5} x^{5} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1} + \frac {15 a^{4} x^{4} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1} - \frac {20 a^{3} x^{3} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1} + \frac {15 a^{2} x^{2} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1} - \frac {6 a x \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1} + \frac {\sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1}}\, dx}{c^{5}} \]
-Integral(1/(a**6*x**6*sqrt(a*x/(a*x + 1) - 1/(a*x + 1))/(a*x + 1) - 6*a** 5*x**5*sqrt(a*x/(a*x + 1) - 1/(a*x + 1))/(a*x + 1) + 15*a**4*x**4*sqrt(a*x /(a*x + 1) - 1/(a*x + 1))/(a*x + 1) - 20*a**3*x**3*sqrt(a*x/(a*x + 1) - 1/ (a*x + 1))/(a*x + 1) + 15*a**2*x**2*sqrt(a*x/(a*x + 1) - 1/(a*x + 1))/(a*x + 1) - 6*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**5
Time = 0.21 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.57 \[ \int \frac {e^{3 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=-\frac {\frac {385 \, {\left (a x - 1\right )}}{a x + 1} - \frac {495 \, {\left (a x - 1\right )}^{2}}{{\left (a x + 1\right )}^{2}} + \frac {231 \, {\left (a x - 1\right )}^{3}}{{\left (a x + 1\right )}^{3}} - 105}{9240 \, a c^{5} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {11}{2}}} \]
-1/9240*(385*(a*x - 1)/(a*x + 1) - 495*(a*x - 1)^2/(a*x + 1)^2 + 231*(a*x - 1)^3/(a*x + 1)^3 - 105)/(a*c^5*((a*x - 1)/(a*x + 1))^(11/2))
Time = 0.39 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.32 \[ \int \frac {e^{3 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=\frac {4 \, {\left (1155 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )}^{7} x^{7} + 2079 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )}^{6} x^{6} + 2541 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )}^{5} x^{5} + 825 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )}^{4} x^{4} + 165 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )}^{3} x^{3} - 55 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )}^{2} x^{2} + 11 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )} x - 1\right )}}{1155 \, {\left ({\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )} x - 1\right )}^{11} a c^{5}} \]
4/1155*(1155*(a + sqrt(a^2 - 1/x^2))^7*x^7 + 2079*(a + sqrt(a^2 - 1/x^2))^ 6*x^6 + 2541*(a + sqrt(a^2 - 1/x^2))^5*x^5 + 825*(a + sqrt(a^2 - 1/x^2))^4 *x^4 + 165*(a + sqrt(a^2 - 1/x^2))^3*x^3 - 55*(a + sqrt(a^2 - 1/x^2))^2*x^ 2 + 11*(a + sqrt(a^2 - 1/x^2))*x - 1)/(((a + sqrt(a^2 - 1/x^2))*x - 1)^11* a*c^5)
Time = 0.05 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.58 \[ \int \frac {e^{3 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=\frac {\frac {3\,{\left (a\,x-1\right )}^2}{7\,{\left (a\,x+1\right )}^2}-\frac {{\left (a\,x-1\right )}^3}{5\,{\left (a\,x+1\right )}^3}-\frac {a\,x-1}{3\,\left (a\,x+1\right )}+\frac {1}{11}}{8\,a\,c^5\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{11/2}} \]