Integrand size = 12, antiderivative size = 183 \[ \int \frac {e^{n \coth ^{-1}(a x)}}{x^5} \, dx=\frac {1}{24} a^3 \left (1-\frac {1}{a x}\right )^{1-\frac {n}{2}} \left (1+\frac {1}{a x}\right )^{\frac {2+n}{2}} \left (a \left (6+n^2\right )+\frac {2 n}{x}\right )+\frac {a^2 \left (1-\frac {1}{a x}\right )^{1-\frac {n}{2}} \left (1+\frac {1}{a x}\right )^{\frac {2+n}{2}}}{4 x^2}+\frac {2^{-2+\frac {n}{2}} a^4 n \left (8+n^2\right ) \left (1-\frac {1}{a x}\right )^{1-\frac {n}{2}} \operatorname {Hypergeometric2F1}\left (1-\frac {n}{2},-\frac {n}{2},2-\frac {n}{2},\frac {a-\frac {1}{x}}{2 a}\right )}{3 (2-n)} \]
1/24*a^3*(1-1/a/x)^(1-1/2*n)*(1+1/a/x)^(1+1/2*n)*(a*(n^2+6)+2*n/x)+1/4*a^2 *(1-1/a/x)^(1-1/2*n)*(1+1/a/x)^(1+1/2*n)/x^2+1/3*2^(-2+1/2*n)*a^4*n*(n^2+8 )*(1-1/a/x)^(1-1/2*n)*hypergeom([-1/2*n, 1-1/2*n],[2-1/2*n],1/2*(a-1/x)/a) /(2-n)
Time = 0.53 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.81 \[ \int \frac {e^{n \coth ^{-1}(a x)}}{x^5} \, dx=-\frac {1}{24} a^4 e^{n \coth ^{-1}(a x)} \left (-6-n^2+\frac {6}{a^4 x^4}+\frac {2 n}{a^3 x^3}+\frac {n^2}{a^2 x^2}+\frac {6 n}{a x}+\frac {n^3}{a x}-\frac {e^{2 \coth ^{-1}(a x)} n^2 \left (8+n^2\right ) \operatorname {Hypergeometric2F1}\left (1,1+\frac {n}{2},2+\frac {n}{2},-e^{2 \coth ^{-1}(a x)}\right )}{2+n}+n \left (8+n^2\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {n}{2},1+\frac {n}{2},-e^{2 \coth ^{-1}(a x)}\right )\right ) \]
-1/24*(a^4*E^(n*ArcCoth[a*x])*(-6 - n^2 + 6/(a^4*x^4) + (2*n)/(a^3*x^3) + n^2/(a^2*x^2) + (6*n)/(a*x) + n^3/(a*x) - (E^(2*ArcCoth[a*x])*n^2*(8 + n^2 )*Hypergeometric2F1[1, 1 + n/2, 2 + n/2, -E^(2*ArcCoth[a*x])])/(2 + n) + n *(8 + n^2)*Hypergeometric2F1[1, n/2, 1 + n/2, -E^(2*ArcCoth[a*x])]))
Time = 0.32 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6721, 111, 25, 27, 164, 79}
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.
| \(\displaystyle \int \frac {e^{n \coth ^{-1}(a x)}}{x^5} \, dx\) |
| \(\Big \downarrow \) 6721 |
| \(\displaystyle -\int \frac {\left (1-\frac {1}{a x}\right )^{-n/2} \left (1+\frac {1}{a x}\right )^{n/2}}{x^3}d\frac {1}{x}\) |
| \(\Big \downarrow \) 111 |
| \(\displaystyle \frac {1}{4} a^2 \int -\frac {\left (1-\frac {1}{a x}\right )^{-n/2} \left (1+\frac {1}{a x}\right )^{n/2} \left (2 a+\frac {n}{x}\right )}{a x}d\frac {1}{x}+\frac {a^2 \left (\frac {1}{a x}+1\right )^{\frac {n+2}{2}} \left (1-\frac {1}{a x}\right )^{1-\frac {n}{2}}}{4 x^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {a^2 \left (1-\frac {1}{a x}\right )^{1-\frac {n}{2}} \left (\frac {1}{a x}+1\right )^{\frac {n+2}{2}}}{4 x^2}-\frac {1}{4} a^2 \int \frac {\left (1-\frac {1}{a x}\right )^{-n/2} \left (1+\frac {1}{a x}\right )^{n/2} \left (2 a+\frac {n}{x}\right )}{a x}d\frac {1}{x}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {a^2 \left (1-\frac {1}{a x}\right )^{1-\frac {n}{2}} \left (\frac {1}{a x}+1\right )^{\frac {n+2}{2}}}{4 x^2}-\frac {1}{4} a \int \frac {\left (1-\frac {1}{a x}\right )^{-n/2} \left (1+\frac {1}{a x}\right )^{n/2} \left (2 a+\frac {n}{x}\right )}{x}d\frac {1}{x}\) |
| \(\Big \downarrow \) 164 |
| \(\displaystyle \frac {a^2 \left (1-\frac {1}{a x}\right )^{1-\frac {n}{2}} \left (\frac {1}{a x}+1\right )^{\frac {n+2}{2}}}{4 x^2}-\frac {1}{4} a \left (\frac {1}{6} a^2 n \left (n^2+8\right ) \int \left (1-\frac {1}{a x}\right )^{-n/2} \left (1+\frac {1}{a x}\right )^{n/2}d\frac {1}{x}-\frac {1}{6} a^2 \left (1-\frac {1}{a x}\right )^{1-\frac {n}{2}} \left (\frac {1}{a x}+1\right )^{\frac {n+2}{2}} \left (a \left (n^2+6\right )+\frac {2 n}{x}\right )\right )\) |
| \(\Big \downarrow \) 79 |
| \(\displaystyle \frac {a^2 \left (1-\frac {1}{a x}\right )^{1-\frac {n}{2}} \left (\frac {1}{a x}+1\right )^{\frac {n+2}{2}}}{4 x^2}-\frac {1}{4} a \left (-\frac {a^3 2^{n/2} n \left (n^2+8\right ) \left (1-\frac {1}{a x}\right )^{1-\frac {n}{2}} \operatorname {Hypergeometric2F1}\left (1-\frac {n}{2},-\frac {n}{2},2-\frac {n}{2},\frac {a-\frac {1}{x}}{2 a}\right )}{3 (2-n)}-\frac {1}{6} a^2 \left (\frac {1}{a x}+1\right )^{\frac {n+2}{2}} \left (a \left (n^2+6\right )+\frac {2 n}{x}\right ) \left (1-\frac {1}{a x}\right )^{1-\frac {n}{2}}\right )\) |
(a^2*(1 - 1/(a*x))^(1 - n/2)*(1 + 1/(a*x))^((2 + n)/2))/(4*x^2) - (a*(-1/6 *(a^2*(1 - 1/(a*x))^(1 - n/2)*(1 + 1/(a*x))^((2 + n)/2)*(a*(6 + n^2) + (2* n)/x)) - (2^(n/2)*a^3*n*(8 + n^2)*(1 - 1/(a*x))^(1 - n/2)*Hypergeometric2F 1[1 - n/2, -1/2*n, 2 - n/2, (a - x^(-1))/(2*a)])/(3*(2 - n))))/4
3.2.56.3.1 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[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 , m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] && !IntegerQ[m] && !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] || !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[b*(a + b*x)^(m - 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 )/(d*f*(m + n + p + 1))), x] + Simp[1/(d*f*(m + n + p + 1)) Int[(a + b*x) ^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] & & GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegerQ[m]
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_ ))*((g_.) + (h_.)*(x_)), x_] :> Simp[(-(a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x))*(a + b*x)^(m + 1)*(( c + d*x)^(n + 1)/(b^2*d^2*(m + n + 2)*(m + n + 3))), x] + Simp[(a^2*d^2*f*h *(n + 1)*(n + 2) + a*b*d*(n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)) Int[( a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]
Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(x_)^(m_.), x_Symbol] :> -Subst[Int[(1 + x /a)^(n/2)/(x^(m + 2)*(1 - x/a)^(n/2)), x], x, 1/x] /; FreeQ[{a, n}, x] && !IntegerQ[n] && IntegerQ[m]
\[\int \frac {{\mathrm e}^{n \,\operatorname {arccoth}\left (a x \right )}}{x^{5}}d x\]
\[ \int \frac {e^{n \coth ^{-1}(a x)}}{x^5} \, dx=\int { \frac {\left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{x^{5}} \,d x } \]
\[ \int \frac {e^{n \coth ^{-1}(a x)}}{x^5} \, dx=\int \frac {e^{n \operatorname {acoth}{\left (a x \right )}}}{x^{5}}\, dx \]
\[ \int \frac {e^{n \coth ^{-1}(a x)}}{x^5} \, dx=\int { \frac {\left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{x^{5}} \,d x } \]
\[ \int \frac {e^{n \coth ^{-1}(a x)}}{x^5} \, dx=\int { \frac {\left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{x^{5}} \,d x } \]
Timed out. \[ \int \frac {e^{n \coth ^{-1}(a x)}}{x^5} \, dx=\int \frac {{\mathrm {e}}^{n\,\mathrm {acoth}\left (a\,x\right )}}{x^5} \,d x \]