\(\int \frac {e^{n \cot ^{-1}(a x)}}{(c+a^2 c x^2)^{5/3}} \, dx\) [11]

Optimal result

Integrand size = 23, antiderivative size = 207 \[ \int \frac {e^{n \cot ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^{5/3}} \, dx=-\frac {3 e^{n \cot ^{-1}(a x)} (3 n-4 a x)}{a c \left (16+9 n^2\right ) \left (c+a^2 c x^2\right )^{2/3}}-\frac {12 \left (1+\frac {1}{a^2 x^2}\right )^{2/3} \left (\frac {a-\frac {i}{x}}{a+\frac {i}{x}}\right )^{\frac {1}{6} (4-3 i n)} \left (1-\frac {i}{a x}\right )^{\frac {1}{6} (-4+3 i n)} \left (1+\frac {i}{a x}\right )^{\frac {1}{6} (2-3 i n)} x \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{6} (4-3 i n),\frac {4}{3},\frac {2 i}{\left (a+\frac {i}{x}\right ) x}\right )}{c \left (16+9 n^2\right ) \left (c+a^2 c x^2\right )^{2/3}} \] Output:

-3*exp(n*arccot(a*x))*(-4*a*x+3*n)/a/c/(9*n^2+16)/(a^2*c*x^2+c)^(2/3)-12*( 
1+1/a^2/x^2)^(2/3)*((a-I/x)/(a+I/x))^(2/3-1/2*I*n)*(1-I/a/x)^(-2/3+1/2*I*n 
)*(1+I/a/x)^(1/3-1/2*I*n)*x*hypergeom([1/3, 2/3-1/2*I*n],[4/3],2*I/(a+I/x) 
/x)/c/(9*n^2+16)/(a^2*c*x^2+c)^(2/3)
 

Mathematica [A] (warning: unable to verify)

Time = 0.27 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.43 \[ \int \frac {e^{n \cot ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^{5/3}} \, dx=-\frac {3 e^{(-2 i+n) \cot ^{-1}(a x)} \left (-1+e^{2 i \cot ^{-1}(a x)}\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {5}{3}+\frac {i n}{2},\frac {1}{3}+\frac {i n}{2},e^{-2 i \cot ^{-1}(a x)}\right )}{a c (4 i+3 n) \left (c+a^2 c x^2\right )^{2/3}} \] Input:

Integrate[E^(n*ArcCot[a*x])/(c + a^2*c*x^2)^(5/3),x]
 

Output:

(-3*E^((-2*I + n)*ArcCot[a*x])*(-1 + E^((2*I)*ArcCot[a*x]))*Hypergeometric 
2F1[1, 5/3 + (I/2)*n, 1/3 + (I/2)*n, E^((-2*I)*ArcCot[a*x])])/(a*c*(4*I + 
3*n)*(c + a^2*c*x^2)^(2/3))
 

Rubi [A] (verified)

Time = 1.04 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {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.

\(\displaystyle \int \frac {e^{n \cot ^{-1}(a x)}}{\left (a^2 c x^2+c\right )^{5/3}} \, dx\)

\(\Big \downarrow \) 5638

\(\displaystyle \frac {4 \int \frac {e^{n \cot ^{-1}(a x)}}{\left (a^2 c x^2+c\right )^{2/3}}dx}{c \left (9 n^2+16\right )}-\frac {3 (3 n-4 a x) e^{n \cot ^{-1}(a x)}}{a c \left (9 n^2+16\right ) \left (a^2 c x^2+c\right )^{2/3}}\)

\(\Big \downarrow \) 5645

\(\displaystyle \frac {4 x^{4/3} \left (\frac {1}{a^2 x^2}+1\right )^{2/3} \int \frac {e^{n \cot ^{-1}(a x)}}{\left (1+\frac {1}{a^2 x^2}\right )^{2/3} x^{4/3}}dx}{c \left (9 n^2+16\right ) \left (a^2 c x^2+c\right )^{2/3}}-\frac {3 (3 n-4 a x) e^{n \cot ^{-1}(a x)}}{a c \left (9 n^2+16\right ) \left (a^2 c x^2+c\right )^{2/3}}\)

\(\Big \downarrow \) 5649

\(\displaystyle -\frac {3 (3 n-4 a x) e^{n \cot ^{-1}(a x)}}{a c \left (9 n^2+16\right ) \left (a^2 c x^2+c\right )^{2/3}}-\frac {4 \left (\frac {1}{a^2 x^2}+1\right )^{2/3} \int \frac {\left (1-\frac {i}{a x}\right )^{\frac {1}{6} (3 i n-4)} \left (1+\frac {i}{a x}\right )^{\frac {1}{6} (-3 i n-4)}}{\left (\frac {1}{x}\right )^{2/3}}d\frac {1}{x}}{c \left (9 n^2+16\right ) \left (\frac {1}{x}\right )^{4/3} \left (a^2 c x^2+c\right )^{2/3}}\)

\(\Big \downarrow \) 142

\(\displaystyle -\frac {3 (3 n-4 a x) e^{n \cot ^{-1}(a x)}}{a c \left (9 n^2+16\right ) \left (a^2 c x^2+c\right )^{2/3}}-\frac {12 x \left (\frac {1}{a^2 x^2}+1\right )^{2/3} \left (\frac {a-\frac {i}{x}}{a+\frac {i}{x}}\right )^{\frac {1}{6} (4-3 i n)} \left (1-\frac {i}{a x}\right )^{\frac {1}{6} (-4+3 i n)} \left (1+\frac {i}{a x}\right )^{\frac {1}{6} (2-3 i n)} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{6} (4-3 i n),\frac {4}{3},\frac {2 i}{\left (a+\frac {i}{x}\right ) x}\right )}{c \left (9 n^2+16\right ) \left (a^2 c x^2+c\right )^{2/3}}\)

Input:

Int[E^(n*ArcCot[a*x])/(c + a^2*c*x^2)^(5/3),x]
 

Output:

(-3*E^(n*ArcCot[a*x])*(3*n - 4*a*x))/(a*c*(16 + 9*n^2)*(c + a^2*c*x^2)^(2/ 
3)) - (12*(1 + 1/(a^2*x^2))^(2/3)*((a - I/x)/(a + I/x))^((4 - (3*I)*n)/6)* 
(1 - I/(a*x))^((-4 + (3*I)*n)/6)*(1 + I/(a*x))^((2 - (3*I)*n)/6)*x*Hyperge 
ometric2F1[1/3, (4 - (3*I)*n)/6, 4/3, (2*I)/((a + I/x)*x)])/(c*(16 + 9*n^2 
)*(c + a^2*c*x^2)^(2/3))
 

Defintions of rubi rules used

rule 142

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]
 

rule 5638

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])
 

rule 5645

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]
 

rule 5649

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]
 

Maple [F]

\[\int \frac {{\mathrm e}^{n \,\operatorname {arccot}\left (a x \right )}}{\left (a^{2} c \,x^{2}+c \right )^{\frac {5}{3}}}d x\]

Input:

int(exp(n*arccot(a*x))/(a^2*c*x^2+c)^(5/3),x)
 

Output:

int(exp(n*arccot(a*x))/(a^2*c*x^2+c)^(5/3),x)
 

Fricas [F]

\[ \int \frac {e^{n \cot ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^{5/3}} \, dx=\int { \frac {e^{\left (n \operatorname {arccot}\left (a x\right )\right )}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {5}{3}}} \,d x } \] Input:

integrate(exp(n*arccot(a*x))/(a^2*c*x^2+c)^(5/3),x, algorithm="fricas")
 

Output:

integral((a^2*c*x^2 + c)^(1/3)*e^(n*arccot(a*x))/(a^4*c^2*x^4 + 2*a^2*c^2* 
x^2 + c^2), x)
 

Sympy [F]

\[ \int \frac {e^{n \cot ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^{5/3}} \, dx=\int \frac {e^{n \operatorname {acot}{\left (a x \right )}}}{\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {5}{3}}}\, dx \] Input:

integrate(exp(n*acot(a*x))/(a**2*c*x**2+c)**(5/3),x)
 

Output:

Integral(exp(n*acot(a*x))/(c*(a**2*x**2 + 1))**(5/3), x)
 

Maxima [F]

\[ \int \frac {e^{n \cot ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^{5/3}} \, dx=\int { \frac {e^{\left (n \operatorname {arccot}\left (a x\right )\right )}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {5}{3}}} \,d x } \] Input:

integrate(exp(n*arccot(a*x))/(a^2*c*x^2+c)^(5/3),x, algorithm="maxima")
 

Output:

integrate(e^(n*arccot(a*x))/(a^2*c*x^2 + c)^(5/3), x)
 

Giac [F]

\[ \int \frac {e^{n \cot ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^{5/3}} \, dx=\int { \frac {e^{\left (n \operatorname {arccot}\left (a x\right )\right )}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {5}{3}}} \,d x } \] Input:

integrate(exp(n*arccot(a*x))/(a^2*c*x^2+c)^(5/3),x, algorithm="giac")
 

Output:

integrate(e^(n*arccot(a*x))/(a^2*c*x^2 + c)^(5/3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{n \cot ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^{5/3}} \, dx=\int \frac {{\mathrm {e}}^{n\,\mathrm {acot}\left (a\,x\right )}}{{\left (c\,a^2\,x^2+c\right )}^{5/3}} \,d x \] Input:

int(exp(n*acot(a*x))/(c + a^2*c*x^2)^(5/3),x)
 

Output:

int(exp(n*acot(a*x))/(c + a^2*c*x^2)^(5/3), x)
 

Reduce [F]

\[ \int \frac {e^{n \cot ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^{5/3}} \, dx=\frac {\int \frac {e^{\mathit {acot} \left (a x \right ) n}}{\left (a^{2} x^{2}+1\right )^{\frac {2}{3}} a^{2} x^{2}+\left (a^{2} x^{2}+1\right )^{\frac {2}{3}}}d x}{c^{\frac {5}{3}}} \] Input:

int(exp(n*acot(a*x))/(a^2*c*x^2+c)^(5/3),x)
 

Output:

int(e**(acot(a*x)*n)/((a**2*x**2 + 1)**(2/3)*a**2*x**2 + (a**2*x**2 + 1)** 
(2/3)),x)/(c**(2/3)*c)