Integrand size = 24, antiderivative size = 166 \[ \int \frac {e^{n \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=-\frac {e^{n \coth ^{-1}(a x)} (n-5 a x)}{a c \left (25-n^2\right ) \left (c-a^2 c x^2\right )^{5/2}}-\frac {20 e^{n \coth ^{-1}(a x)} (n-3 a x)}{a c^2 \left (9-n^2\right ) \left (25-n^2\right ) \left (c-a^2 c x^2\right )^{3/2}}-\frac {120 e^{n \coth ^{-1}(a x)} (n-a x)}{a c^3 \left (1-n^2\right ) \left (9-n^2\right ) \left (25-n^2\right ) \sqrt {c-a^2 c x^2}} \]
-exp(n*arccoth(a*x))*(-5*a*x+n)/a/c/(-n^2+25)/(-a^2*c*x^2+c)^(5/2)-20*exp( n*arccoth(a*x))*(-3*a*x+n)/a/c^2/(n^4-34*n^2+225)/(-a^2*c*x^2+c)^(3/2)-120 *exp(n*arccoth(a*x))*(-a*x+n)/a/c^3/(-n^2+25)/(n^4-10*n^2+9)/(-a^2*c*x^2+c )^(1/2)
Time = 1.97 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.80 \[ \int \frac {e^{n \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=-\frac {a^2 e^{n \coth ^{-1}(a x)} \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^3 \left (-\frac {10 \left (225-34 n^2+n^4\right )}{\sqrt {1-\frac {1}{a^2 x^2}}}+\frac {2250 n}{a \sqrt {1-\frac {1}{a^2 x^2}} x}-\frac {340 n^3}{a \sqrt {1-\frac {1}{a^2 x^2}} x}+\frac {10 n^5}{a \sqrt {1-\frac {1}{a^2 x^2}} x}+15 \left (25-26 n^2+n^4\right ) \cosh \left (3 \coth ^{-1}(a x)\right )-45 \cosh \left (5 \coth ^{-1}(a x)\right )+50 n^2 \cosh \left (5 \coth ^{-1}(a x)\right )-5 n^4 \cosh \left (5 \coth ^{-1}(a x)\right )-125 n \sinh \left (3 \coth ^{-1}(a x)\right )+130 n^3 \sinh \left (3 \coth ^{-1}(a x)\right )-5 n^5 \sinh \left (3 \coth ^{-1}(a x)\right )+9 n \sinh \left (5 \coth ^{-1}(a x)\right )-10 n^3 \sinh \left (5 \coth ^{-1}(a x)\right )+n^5 \sinh \left (5 \coth ^{-1}(a x)\right )\right )}{16 c^2 (-5+n) (-3+n) (-1+n) (1+n) (3+n) (5+n) \left (c-a^2 c x^2\right )^{3/2}} \]
-1/16*(a^2*E^(n*ArcCoth[a*x])*(1 - 1/(a^2*x^2))^(3/2)*x^3*((-10*(225 - 34* n^2 + n^4))/Sqrt[1 - 1/(a^2*x^2)] + (2250*n)/(a*Sqrt[1 - 1/(a^2*x^2)]*x) - (340*n^3)/(a*Sqrt[1 - 1/(a^2*x^2)]*x) + (10*n^5)/(a*Sqrt[1 - 1/(a^2*x^2)] *x) + 15*(25 - 26*n^2 + n^4)*Cosh[3*ArcCoth[a*x]] - 45*Cosh[5*ArcCoth[a*x] ] + 50*n^2*Cosh[5*ArcCoth[a*x]] - 5*n^4*Cosh[5*ArcCoth[a*x]] - 125*n*Sinh[ 3*ArcCoth[a*x]] + 130*n^3*Sinh[3*ArcCoth[a*x]] - 5*n^5*Sinh[3*ArcCoth[a*x] ] + 9*n*Sinh[5*ArcCoth[a*x]] - 10*n^3*Sinh[5*ArcCoth[a*x]] + n^5*Sinh[5*Ar cCoth[a*x]]))/(c^2*(-5 + n)*(-3 + n)*(-1 + n)*(1 + n)*(3 + n)*(5 + n)*(c - a^2*c*x^2)^(3/2))
Time = 0.59 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.98, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6739, 6739, 6738}
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.
| \(\displaystyle \int \frac {e^{n \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx\) |
| \(\Big \downarrow \) 6739 |
| \(\displaystyle \frac {20 \int \frac {e^{n \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}}dx}{c \left (25-n^2\right )}-\frac {(n-5 a x) e^{n \coth ^{-1}(a x)}}{a c \left (25-n^2\right ) \left (c-a^2 c x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 6739 |
| \(\displaystyle \frac {20 \left (\frac {6 \int \frac {e^{n \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{3/2}}dx}{c \left (9-n^2\right )}-\frac {(n-3 a x) e^{n \coth ^{-1}(a x)}}{a c \left (9-n^2\right ) \left (c-a^2 c x^2\right )^{3/2}}\right )}{c \left (25-n^2\right )}-\frac {(n-5 a x) e^{n \coth ^{-1}(a x)}}{a c \left (25-n^2\right ) \left (c-a^2 c x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 6738 |
| \(\displaystyle \frac {20 \left (-\frac {6 (n-a x) e^{n \coth ^{-1}(a x)}}{a c^2 \left (1-n^2\right ) \left (9-n^2\right ) \sqrt {c-a^2 c x^2}}-\frac {(n-3 a x) e^{n \coth ^{-1}(a x)}}{a c \left (9-n^2\right ) \left (c-a^2 c x^2\right )^{3/2}}\right )}{c \left (25-n^2\right )}-\frac {(n-5 a x) e^{n \coth ^{-1}(a x)}}{a c \left (25-n^2\right ) \left (c-a^2 c x^2\right )^{5/2}}\) |
-((E^(n*ArcCoth[a*x])*(n - 5*a*x))/(a*c*(25 - n^2)*(c - a^2*c*x^2)^(5/2))) + (20*(-((E^(n*ArcCoth[a*x])*(n - 3*a*x))/(a*c*(9 - n^2)*(c - a^2*c*x^2)^ (3/2))) - (6*E^(n*ArcCoth[a*x])*(n - a*x))/(a*c^2*(1 - n^2)*(9 - n^2)*Sqrt [c - a^2*c*x^2])))/(c*(25 - n^2))
3.8.49.3.1 Defintions of rubi rules used
Int[E^(ArcCoth[(a_.)*(x_)]*(n_))/((c_) + (d_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(n - a*x)*(E^(n*ArcCoth[a*x])/(a*c*(n^2 - 1)*Sqrt[c + d*x^2])), x] /; FreeQ[{a, c, d, n}, x] && EqQ[a^2*c + d, 0] && !IntegerQ[n]
Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(n + 2*a*(p + 1)*x)*(c + d*x^2)^(p + 1)*(E^(n*ArcCoth[a*x])/(a*c*(n^2 - 4*(p + 1)^2))), x] - Simp[2*(p + 1)*((2*p + 3)/(c*(n^2 - 4*(p + 1)^2))) Int[(c + d*x^2)^(p + 1)*E^(n*ArcCoth[a*x]), x], x] /; FreeQ[{a, c, d, n}, x] && EqQ[a^2*c + d, 0] && !IntegerQ[n/2] && LtQ[p, -1] && NeQ[p, -3/2] && NeQ[n^2 - 4*(p + 1)^2, 0] && (IntegerQ[p] || !IntegerQ[n])
Time = 0.52 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.84
| method | result | size |
| gosper | \(\frac {\left (a x -1\right ) \left (a x +1\right ) \left (120 a^{5} x^{5}-120 n \,a^{4} x^{4}+60 a^{3} n^{2} x^{3}-20 a^{2} n^{3} x^{2}-300 a^{3} x^{3}+5 a \,n^{4} x +260 n \,x^{2} a^{2}-n^{5}-110 n^{2} x a +30 n^{3}+225 a x -149 n \right ) {\mathrm e}^{n \,\operatorname {arccoth}\left (a x \right )}}{a \left (n^{6}-35 n^{4}+259 n^{2}-225\right ) \left (-a^{2} c \,x^{2}+c \right )^{\frac {7}{2}}}\) | \(140\) |
(a*x-1)*(a*x+1)*(120*a^5*x^5-120*a^4*n*x^4+60*a^3*n^2*x^3-20*a^2*n^3*x^2-3 00*a^3*x^3+5*a*n^4*x+260*a^2*n*x^2-n^5-110*a*n^2*x+30*n^3+225*a*x-149*n)*e xp(n*arccoth(a*x))/a/(n^6-35*n^4+259*n^2-225)/(-a^2*c*x^2+c)^(7/2)
Time = 0.26 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.75 \[ \int \frac {e^{n \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=-\frac {{\left (120 \, a^{5} x^{5} - 120 \, a^{4} n x^{4} - n^{5} + 60 \, {\left (a^{3} n^{2} - 5 \, a^{3}\right )} x^{3} + 30 \, n^{3} - 20 \, {\left (a^{2} n^{3} - 13 \, a^{2} n\right )} x^{2} + 5 \, {\left (a n^{4} - 22 \, a n^{2} + 45 \, a\right )} x - 149 \, n\right )} \sqrt {-a^{2} c x^{2} + c} \left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{a c^{4} n^{6} - 35 \, a c^{4} n^{4} + 259 \, a c^{4} n^{2} - {\left (a^{7} c^{4} n^{6} - 35 \, a^{7} c^{4} n^{4} + 259 \, a^{7} c^{4} n^{2} - 225 \, a^{7} c^{4}\right )} x^{6} - 225 \, a c^{4} + 3 \, {\left (a^{5} c^{4} n^{6} - 35 \, a^{5} c^{4} n^{4} + 259 \, a^{5} c^{4} n^{2} - 225 \, a^{5} c^{4}\right )} x^{4} - 3 \, {\left (a^{3} c^{4} n^{6} - 35 \, a^{3} c^{4} n^{4} + 259 \, a^{3} c^{4} n^{2} - 225 \, a^{3} c^{4}\right )} x^{2}} \]
-(120*a^5*x^5 - 120*a^4*n*x^4 - n^5 + 60*(a^3*n^2 - 5*a^3)*x^3 + 30*n^3 - 20*(a^2*n^3 - 13*a^2*n)*x^2 + 5*(a*n^4 - 22*a*n^2 + 45*a)*x - 149*n)*sqrt( -a^2*c*x^2 + c)*((a*x + 1)/(a*x - 1))^(1/2*n)/(a*c^4*n^6 - 35*a*c^4*n^4 + 259*a*c^4*n^2 - (a^7*c^4*n^6 - 35*a^7*c^4*n^4 + 259*a^7*c^4*n^2 - 225*a^7* c^4)*x^6 - 225*a*c^4 + 3*(a^5*c^4*n^6 - 35*a^5*c^4*n^4 + 259*a^5*c^4*n^2 - 225*a^5*c^4)*x^4 - 3*(a^3*c^4*n^6 - 35*a^3*c^4*n^4 + 259*a^3*c^4*n^2 - 22 5*a^3*c^4)*x^2)
Timed out. \[ \int \frac {e^{n \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=\text {Timed out} \]
\[ \int \frac {e^{n \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=\int { \frac {\left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {7}{2}}} \,d x } \]
\[ \int \frac {e^{n \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=\int { \frac {\left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {7}{2}}} \,d x } \]
Time = 4.81 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.74 \[ \int \frac {e^{n \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=-\frac {{\left (\frac {a\,x+1}{a\,x}\right )}^{n/2}\,\left (\frac {120\,x^5}{c^3\,\left (n^6-35\,n^4+259\,n^2-225\right )}-\frac {120\,n\,x^4}{a\,c^3\,\left (n^6-35\,n^4+259\,n^2-225\right )}+\frac {x^3\,\left (60\,n^2-300\right )}{a^2\,c^3\,\left (n^6-35\,n^4+259\,n^2-225\right )}-\frac {n\,\left (n^4-30\,n^2+149\right )}{a^5\,c^3\,\left (n^6-35\,n^4+259\,n^2-225\right )}+\frac {5\,x\,\left (n^4-22\,n^2+45\right )}{a^4\,c^3\,\left (n^6-35\,n^4+259\,n^2-225\right )}-\frac {20\,n\,x^2\,\left (n^2-13\right )}{a^3\,c^3\,\left (n^6-35\,n^4+259\,n^2-225\right )}\right )}{\left (\frac {\sqrt {c-a^2\,c\,x^2}}{a^4}+x^4\,\sqrt {c-a^2\,c\,x^2}-\frac {2\,x^2\,\sqrt {c-a^2\,c\,x^2}}{a^2}\right )\,{\left (\frac {a\,x-1}{a\,x}\right )}^{n/2}} \]
-(((a*x + 1)/(a*x))^(n/2)*((120*x^5)/(c^3*(259*n^2 - 35*n^4 + n^6 - 225)) - (120*n*x^4)/(a*c^3*(259*n^2 - 35*n^4 + n^6 - 225)) + (x^3*(60*n^2 - 300) )/(a^2*c^3*(259*n^2 - 35*n^4 + n^6 - 225)) - (n*(n^4 - 30*n^2 + 149))/(a^5 *c^3*(259*n^2 - 35*n^4 + n^6 - 225)) + (5*x*(n^4 - 22*n^2 + 45))/(a^4*c^3* (259*n^2 - 35*n^4 + n^6 - 225)) - (20*n*x^2*(n^2 - 13))/(a^3*c^3*(259*n^2 - 35*n^4 + n^6 - 225))))/(((c - a^2*c*x^2)^(1/2)/a^4 + x^4*(c - a^2*c*x^2) ^(1/2) - (2*x^2*(c - a^2*c*x^2)^(1/2))/a^2)*((a*x - 1)/(a*x))^(n/2))