Integrand size = 22, antiderivative size = 127 \[ \int \frac {e^{n \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {24 e^{n \coth ^{-1}(a x)}}{a c^3 n \left (64-20 n^2+n^4\right )}-\frac {e^{n \coth ^{-1}(a x)} (n-4 a x)}{a c^3 \left (16-n^2\right ) \left (1-a^2 x^2\right )^2}-\frac {12 e^{n \coth ^{-1}(a x)} (n-2 a x)}{a c^3 \left (4-n^2\right ) \left (16-n^2\right ) \left (1-a^2 x^2\right )} \]
24*exp(n*arccoth(a*x))/a/c^3/n/(n^4-20*n^2+64)-exp(n*arccoth(a*x))*(-4*a*x +n)/a/c^3/(-n^2+16)/(-a^2*x^2+1)^2-12*exp(n*arccoth(a*x))*(-2*a*x+n)/a/c^3 /(n^4-20*n^2+64)/(-a^2*x^2+1)
Time = 0.41 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.76 \[ \int \frac {e^{n \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {e^{n \coth ^{-1}(a x)} \left (n^4-4 a n^3 x+24 \left (-1+a^2 x^2\right )^2-8 a n x \left (-5+3 a^2 x^2\right )+4 n^2 \left (-4+3 a^2 x^2\right )\right )}{a c^3 n \left (-16+n^2\right ) \left (-4+n^2\right ) \left (-1+a^2 x^2\right )^2} \]
(E^(n*ArcCoth[a*x])*(n^4 - 4*a*n^3*x + 24*(-1 + a^2*x^2)^2 - 8*a*n*x*(-5 + 3*a^2*x^2) + 4*n^2*(-4 + 3*a^2*x^2)))/(a*c^3*n*(-16 + n^2)*(-4 + n^2)*(-1 + a^2*x^2)^2)
Time = 0.52 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.98, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {6739, 27, 6739, 6737}
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.
\(\displaystyle \int \frac {e^{n \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx\) |
\(\Big \downarrow \) 6739 |
\(\displaystyle \frac {12 \int \frac {e^{n \coth ^{-1}(a x)}}{c^2 \left (1-a^2 x^2\right )^2}dx}{c \left (16-n^2\right )}-\frac {(n-4 a x) e^{n \coth ^{-1}(a x)}}{a c^3 \left (16-n^2\right ) \left (1-a^2 x^2\right )^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {12 \int \frac {e^{n \coth ^{-1}(a x)}}{\left (1-a^2 x^2\right )^2}dx}{c^3 \left (16-n^2\right )}-\frac {(n-4 a x) e^{n \coth ^{-1}(a x)}}{a c^3 \left (16-n^2\right ) \left (1-a^2 x^2\right )^2}\) |
\(\Big \downarrow \) 6739 |
\(\displaystyle \frac {12 \left (\frac {2 \int \frac {e^{n \coth ^{-1}(a x)}}{1-a^2 x^2}dx}{4-n^2}-\frac {(n-2 a x) e^{n \coth ^{-1}(a x)}}{a \left (4-n^2\right ) \left (1-a^2 x^2\right )}\right )}{c^3 \left (16-n^2\right )}-\frac {(n-4 a x) e^{n \coth ^{-1}(a x)}}{a c^3 \left (16-n^2\right ) \left (1-a^2 x^2\right )^2}\) |
\(\Big \downarrow \) 6737 |
\(\displaystyle \frac {12 \left (\frac {2 e^{n \coth ^{-1}(a x)}}{a n \left (4-n^2\right )}-\frac {(n-2 a x) e^{n \coth ^{-1}(a x)}}{a \left (4-n^2\right ) \left (1-a^2 x^2\right )}\right )}{c^3 \left (16-n^2\right )}-\frac {(n-4 a x) e^{n \coth ^{-1}(a x)}}{a c^3 \left (16-n^2\right ) \left (1-a^2 x^2\right )^2}\) |
-((E^(n*ArcCoth[a*x])*(n - 4*a*x))/(a*c^3*(16 - n^2)*(1 - a^2*x^2)^2)) + ( 12*((2*E^(n*ArcCoth[a*x]))/(a*n*(4 - n^2)) - (E^(n*ArcCoth[a*x])*(n - 2*a* x))/(a*(4 - n^2)*(1 - a^2*x^2))))/(c^3*(16 - n^2))
3.8.42.3.1 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[E^(ArcCoth[(a_.)*(x_)]*(n_.))/((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[ E^(n*ArcCoth[a*x])/(a*c*n), x] /; FreeQ[{a, c, d, n}, x] && EqQ[a^2*c + d, 0] && !IntegerQ[n/2]
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 = 17.35 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.80
method | result | size |
gosper | \(\frac {\left (24 a^{4} x^{4}-24 a^{3} x^{3} n +12 a^{2} n^{2} x^{2}-4 a \,n^{3} x -48 a^{2} x^{2}+n^{4}+40 a n x -16 n^{2}+24\right ) {\mathrm e}^{n \,\operatorname {arccoth}\left (a x \right )}}{\left (a^{2} x^{2}-1\right )^{2} c^{3} a \left (n^{2}-16\right ) \left (n^{2}-4\right ) n}\) | \(101\) |
risch | \(\frac {\left (24 a^{4} x^{4}-24 a^{3} x^{3} n +12 a^{2} n^{2} x^{2}-4 a \,n^{3} x -48 a^{2} x^{2}+n^{4}+40 a n x -16 n^{2}+24\right ) \left (a x -1\right )^{-\frac {n}{2}} \left (a x +1\right )^{\frac {n}{2}}}{\left (a^{2} x^{2}-1\right )^{2} c^{3} a \left (n^{2}-16\right ) \left (n^{2}-4\right ) n}\) | \(112\) |
parallelrisch | \(\frac {24 x^{4} {\mathrm e}^{n \,\operatorname {arccoth}\left (a x \right )} a^{4}+40 x \,{\mathrm e}^{n \,\operatorname {arccoth}\left (a x \right )} a n -48 x^{2} {\mathrm e}^{n \,\operatorname {arccoth}\left (a x \right )} a^{2}-24 a^{3} x^{3} {\mathrm e}^{n \,\operatorname {arccoth}\left (a x \right )} n +12 x^{2} {\mathrm e}^{n \,\operatorname {arccoth}\left (a x \right )} a^{2} n^{2}-4 x \,{\mathrm e}^{n \,\operatorname {arccoth}\left (a x \right )} a \,n^{3}+24 \,{\mathrm e}^{n \,\operatorname {arccoth}\left (a x \right )}+{\mathrm e}^{n \,\operatorname {arccoth}\left (a x \right )} n^{4}-16 \,{\mathrm e}^{n \,\operatorname {arccoth}\left (a x \right )} n^{2}}{c^{3} \left (a^{2} x^{2}-1\right )^{2} a \left (n^{2}-16\right ) \left (n^{2}-4\right ) n}\) | \(159\) |
(24*a^4*x^4-24*a^3*n*x^3+12*a^2*n^2*x^2-4*a*n^3*x-48*a^2*x^2+n^4+40*a*n*x- 16*n^2+24)*exp(n*arccoth(a*x))/(a^2*x^2-1)^2/c^3/a/(n^2-16)/(n^2-4)/n
Time = 0.26 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.37 \[ \int \frac {e^{n \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {{\left (24 \, a^{4} x^{4} - 24 \, a^{3} n x^{3} + n^{4} + 12 \, {\left (a^{2} n^{2} - 4 \, a^{2}\right )} x^{2} - 16 \, n^{2} - 4 \, {\left (a n^{3} - 10 \, a n\right )} x + 24\right )} \left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{a c^{3} n^{5} - 20 \, a c^{3} n^{3} + 64 \, a c^{3} n + {\left (a^{5} c^{3} n^{5} - 20 \, a^{5} c^{3} n^{3} + 64 \, a^{5} c^{3} n\right )} x^{4} - 2 \, {\left (a^{3} c^{3} n^{5} - 20 \, a^{3} c^{3} n^{3} + 64 \, a^{3} c^{3} n\right )} x^{2}} \]
(24*a^4*x^4 - 24*a^3*n*x^3 + n^4 + 12*(a^2*n^2 - 4*a^2)*x^2 - 16*n^2 - 4*( a*n^3 - 10*a*n)*x + 24)*((a*x + 1)/(a*x - 1))^(1/2*n)/(a*c^3*n^5 - 20*a*c^ 3*n^3 + 64*a*c^3*n + (a^5*c^3*n^5 - 20*a^5*c^3*n^3 + 64*a^5*c^3*n)*x^4 - 2 *(a^3*c^3*n^5 - 20*a^3*c^3*n^3 + 64*a^3*c^3*n)*x^2)
Timed out. \[ \int \frac {e^{n \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=\text {Timed out} \]
\[ \int \frac {e^{n \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=\int { -\frac {\left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{{\left (a^{2} c x^{2} - c\right )}^{3}} \,d x } \]
\[ \int \frac {e^{n \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=\int { -\frac {\left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{{\left (a^{2} c x^{2} - c\right )}^{3}} \,d x } \]
Time = 4.73 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.51 \[ \int \frac {e^{n \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {{\left (\frac {a\,x+1}{a\,x}\right )}^{n/2}\,\left (\frac {24\,x^4}{a\,c^3\,n\,\left (n^4-20\,n^2+64\right )}-\frac {4\,x\,\left (n^2-10\right )}{a^4\,c^3\,\left (n^4-20\,n^2+64\right )}-\frac {24\,x^3}{a^2\,c^3\,\left (n^4-20\,n^2+64\right )}+\frac {n^4-16\,n^2+24}{a^5\,c^3\,n\,\left (n^4-20\,n^2+64\right )}+\frac {x^2\,\left (12\,n^2-48\right )}{a^3\,c^3\,n\,\left (n^4-20\,n^2+64\right )}\right )}{{\left (\frac {a\,x-1}{a\,x}\right )}^{n/2}\,\left (\frac {1}{a^4}+x^4-\frac {2\,x^2}{a^2}\right )} \]
(((a*x + 1)/(a*x))^(n/2)*((24*x^4)/(a*c^3*n*(n^4 - 20*n^2 + 64)) - (4*x*(n ^2 - 10))/(a^4*c^3*(n^4 - 20*n^2 + 64)) - (24*x^3)/(a^2*c^3*(n^4 - 20*n^2 + 64)) + (n^4 - 16*n^2 + 24)/(a^5*c^3*n*(n^4 - 20*n^2 + 64)) + (x^2*(12*n^ 2 - 48))/(a^3*c^3*n*(n^4 - 20*n^2 + 64))))/(((a*x - 1)/(a*x))^(n/2)*(1/a^4 + x^4 - (2*x^2)/a^2))