Integrand size = 23, antiderivative size = 272 \[ \int \frac {e^{n \cot ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^{7/3}} \, dx=-\frac {3 e^{n \cot ^{-1}(a x)} (3 n-8 a x)}{a c \left (64+9 n^2\right ) \left (c+a^2 c x^2\right )^{4/3}}-\frac {120 e^{n \cot ^{-1}(a x)} (3 n-2 a x)}{a c^2 \left (4+9 n^2\right ) \left (64+9 n^2\right ) \sqrt [3]{c+a^2 c x^2}}-\frac {240 \sqrt [3]{1+\frac {1}{a^2 x^2}} \left (\frac {a-\frac {i}{x}}{a+\frac {i}{x}}\right )^{\frac {1}{6} (2-3 i n)} \left (1-\frac {i}{a x}\right )^{\frac {1}{6} (-2+3 i n)} \left (1+\frac {i}{a x}\right )^{\frac {1}{6} (4-3 i n)} x \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{6} (2-3 i n),\frac {2}{3},\frac {2 i}{\left (a+\frac {i}{x}\right ) x}\right )}{c^2 \left (4+9 n^2\right ) \left (64+9 n^2\right ) \sqrt [3]{c+a^2 c x^2}} \]
-3*exp(n*arccot(a*x))*(-8*a*x+3*n)/a/c/(9*n^2+64)/(a^2*c*x^2+c)^(4/3)-120* exp(n*arccot(a*x))*(-2*a*x+3*n)/a/c^2/(81*n^4+612*n^2+256)/(a^2*c*x^2+c)^( 1/3)-240*(1+1/a^2/x^2)^(1/3)*((a-I/x)/(a+I/x))^(1/3-1/2*I*n)*(1-I/a/x)^(-1 /3+1/2*I*n)*(1+I/a/x)^(2/3-1/2*I*n)*x*hypergeom([-1/3, 1/3-1/2*I*n],[2/3], 2*I/(a+I/x)/x)/c^2/(81*n^4+612*n^2+256)/(a^2*c*x^2+c)^(1/3)
Time = 0.19 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.37 \[ \int \frac {e^{n \cot ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^{7/3}} \, dx=-\frac {3 e^{(-2 i+n) \cot ^{-1}(a x)} \left (-1+e^{2 i \cot ^{-1}(a x)}\right ) \left (c+a^2 c x^2\right )^{2/3} \operatorname {Hypergeometric2F1}\left (1,\frac {7}{3}+\frac {i n}{2},-\frac {1}{3}+\frac {i n}{2},e^{-2 i \cot ^{-1}(a x)}\right )}{a c^3 (8 i+3 n) \left (1+a^2 x^2\right )^2} \]
(-3*E^((-2*I + n)*ArcCot[a*x])*(-1 + E^((2*I)*ArcCot[a*x]))*(c + a^2*c*x^2 )^(2/3)*Hypergeometric2F1[1, 7/3 + (I/2)*n, -1/3 + (I/2)*n, E^((-2*I)*ArcC ot[a*x])])/(a*c^3*(8*I + 3*n)*(1 + a^2*x^2)^2)
Time = 0.80 (sec) , antiderivative size = 269, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {5638, 5638, 5645, 5649, 142}
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.
\(\displaystyle \int \frac {e^{n \cot ^{-1}(a x)}}{\left (a^2 c x^2+c\right )^{7/3}} \, dx\) |
\(\Big \downarrow \) 5638 |
\(\displaystyle \frac {40 \int \frac {e^{n \cot ^{-1}(a x)}}{\left (a^2 c x^2+c\right )^{4/3}}dx}{c \left (9 n^2+64\right )}-\frac {3 (3 n-8 a x) e^{n \cot ^{-1}(a x)}}{a c \left (9 n^2+64\right ) \left (a^2 c x^2+c\right )^{4/3}}\) |
\(\Big \downarrow \) 5638 |
\(\displaystyle \frac {40 \left (-\frac {2 \int \frac {e^{n \cot ^{-1}(a x)}}{\sqrt [3]{a^2 c x^2+c}}dx}{c \left (9 n^2+4\right )}-\frac {3 (3 n-2 a x) e^{n \cot ^{-1}(a x)}}{a c \left (9 n^2+4\right ) \sqrt [3]{a^2 c x^2+c}}\right )}{c \left (9 n^2+64\right )}-\frac {3 (3 n-8 a x) e^{n \cot ^{-1}(a x)}}{a c \left (9 n^2+64\right ) \left (a^2 c x^2+c\right )^{4/3}}\) |
\(\Big \downarrow \) 5645 |
\(\displaystyle \frac {40 \left (-\frac {2 x^{2/3} \sqrt [3]{\frac {1}{a^2 x^2}+1} \int \frac {e^{n \cot ^{-1}(a x)}}{\sqrt [3]{1+\frac {1}{a^2 x^2}} x^{2/3}}dx}{c \left (9 n^2+4\right ) \sqrt [3]{a^2 c x^2+c}}-\frac {3 (3 n-2 a x) e^{n \cot ^{-1}(a x)}}{a c \left (9 n^2+4\right ) \sqrt [3]{a^2 c x^2+c}}\right )}{c \left (9 n^2+64\right )}-\frac {3 (3 n-8 a x) e^{n \cot ^{-1}(a x)}}{a c \left (9 n^2+64\right ) \left (a^2 c x^2+c\right )^{4/3}}\) |
\(\Big \downarrow \) 5649 |
\(\displaystyle -\frac {3 (3 n-8 a x) e^{n \cot ^{-1}(a x)}}{a c \left (9 n^2+64\right ) \left (a^2 c x^2+c\right )^{4/3}}+\frac {40 \left (-\frac {3 (3 n-2 a x) e^{n \cot ^{-1}(a x)}}{a c \left (9 n^2+4\right ) \sqrt [3]{a^2 c x^2+c}}+\frac {2 \sqrt [3]{\frac {1}{a^2 x^2}+1} \int \frac {\left (1-\frac {i}{a x}\right )^{\frac {1}{6} (3 i n-2)} \left (1+\frac {i}{a x}\right )^{\frac {1}{6} (-3 i n-2)}}{\left (\frac {1}{x}\right )^{4/3}}d\frac {1}{x}}{c \left (9 n^2+4\right ) \left (\frac {1}{x}\right )^{2/3} \sqrt [3]{a^2 c x^2+c}}\right )}{c \left (9 n^2+64\right )}\) |
\(\Big \downarrow \) 142 |
\(\displaystyle -\frac {3 (3 n-8 a x) e^{n \cot ^{-1}(a x)}}{a c \left (9 n^2+64\right ) \left (a^2 c x^2+c\right )^{4/3}}+\frac {40 \left (-\frac {3 (3 n-2 a x) e^{n \cot ^{-1}(a x)}}{a c \left (9 n^2+4\right ) \sqrt [3]{a^2 c x^2+c}}-\frac {6 x \sqrt [3]{\frac {1}{a^2 x^2}+1} \left (1-\frac {i}{a x}\right )^{\frac {1}{6} (-2+3 i n)} \left (1+\frac {i}{a x}\right )^{\frac {1}{6} (4-3 i n)} \left (\frac {a-\frac {i}{x}}{a+\frac {i}{x}}\right )^{\frac {1}{6} (2-3 i n)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{6} (2-3 i n),\frac {2}{3},\frac {2 i}{\left (a+\frac {i}{x}\right ) x}\right )}{c \left (9 n^2+4\right ) \sqrt [3]{a^2 c x^2+c}}\right )}{c \left (9 n^2+64\right )}\) |
(-3*E^(n*ArcCot[a*x])*(3*n - 8*a*x))/(a*c*(64 + 9*n^2)*(c + a^2*c*x^2)^(4/ 3)) + (40*((-3*E^(n*ArcCot[a*x])*(3*n - 2*a*x))/(a*c*(4 + 9*n^2)*(c + a^2* c*x^2)^(1/3)) - (6*(1 + 1/(a^2*x^2))^(1/3)*((a - I/x)/(a + I/x))^((2 - (3* I)*n)/6)*(1 - I/(a*x))^((-2 + (3*I)*n)/6)*(1 + I/(a*x))^((4 - (3*I)*n)/6)* x*Hypergeometric2F1[-1/3, (2 - (3*I)*n)/6, 2/3, (2*I)/((a + I/x)*x)])/(c*( 4 + 9*n^2)*(c + a^2*c*x^2)^(1/3))))/(c*(64 + 9*n^2))
3.1.12.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((b*e - a*f)*(m + 1)))*Hypergeometric2F1[m + 1, -n, m + 2, (-(d*e - c*f))*((a + b*x)/((b*c - a*d)*(e + f*x)))])/((b*e - a*f)*((c + d*x)/((b*c - a*d)*(e + f *x))))^n, x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[m + n + p + 2, 0] && !IntegerQ[n]
Int[E^(ArcCot[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> S imp[(-(n + 2*a*(p + 1)*x))*(c + d*x^2)^(p + 1)*(E^(n*ArcCot[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*ArcCot[a*x]), x], x] /; FreeQ[{a, c, d, n}, x] && EqQ[d, a^2*c] && LtQ[p, -1] && NeQ[p, -3/2] && NeQ[n^2 + 4*(p + 1)^2, 0] && !(IntegerQ[p] && IntegerQ[I*(n/2)]) && !( !IntegerQ[p] && IntegerQ [(I*n - 1)/2])
Int[E^(ArcCot[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol ] :> Simp[(c + d*x^2)^p/(x^(2*p)*(1 + 1/(a^2*x^2))^p) Int[u*x^(2*p)*(1 + 1/(a^2*x^2))^p*E^(n*ArcCot[a*x]), x], x] /; FreeQ[{a, c, d, n, p}, x] && Eq Q[d, a^2*c] && !IntegerQ[I*(n/2)] && !IntegerQ[p]
Int[E^(ArcCot[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_)^2)^(p_.)*(x_)^(m_), x_S ymbol] :> Simp[(-c^p)*x^m*(1/x)^m Subst[Int[(1 - I*(x/a))^(p + I*(n/2))*( (1 + I*(x/a))^(p - I*(n/2))/x^(m + 2)), x], x, 1/x], x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[c, a^2*d] && !IntegerQ[I*(n/2)] && (IntegerQ[p] || GtQ [c, 0]) && !(IntegerQ[2*p] && IntegerQ[p + I*(n/2)]) && !IntegerQ[m]
\[\int \frac {{\mathrm e}^{n \,\operatorname {arccot}\left (a x \right )}}{\left (a^{2} c \,x^{2}+c \right )^{\frac {7}{3}}}d x\]
\[ \int \frac {e^{n \cot ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^{7/3}} \, dx=\int { \frac {e^{\left (n \operatorname {arccot}\left (a x\right )\right )}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {7}{3}}} \,d x } \]
integral((a^2*c*x^2 + c)^(2/3)*e^(n*arccot(a*x))/(a^6*c^3*x^6 + 3*a^4*c^3* x^4 + 3*a^2*c^3*x^2 + c^3), x)
\[ \int \frac {e^{n \cot ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^{7/3}} \, dx=\int \frac {e^{n \operatorname {acot}{\left (a x \right )}}}{\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {7}{3}}}\, dx \]
\[ \int \frac {e^{n \cot ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^{7/3}} \, dx=\int { \frac {e^{\left (n \operatorname {arccot}\left (a x\right )\right )}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {7}{3}}} \,d x } \]
\[ \int \frac {e^{n \cot ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^{7/3}} \, dx=\int { \frac {e^{\left (n \operatorname {arccot}\left (a x\right )\right )}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {7}{3}}} \,d x } \]
Timed out. \[ \int \frac {e^{n \cot ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^{7/3}} \, dx=\int \frac {{\mathrm {e}}^{n\,\mathrm {acot}\left (a\,x\right )}}{{\left (c\,a^2\,x^2+c\right )}^{7/3}} \,d x \]