\(\int \frac {e^{n \arctan (a x)}}{x \sqrt {c+a^2 c x^2}} \, dx\) [371]

Optimal result

Integrand size = 26, antiderivative size = 121 \[ \int \frac {e^{n \arctan (a x)}}{x \sqrt {c+a^2 c x^2}} \, dx=-\frac {2 (1-i a x)^{\frac {1}{2} (1+i n)} (1+i a x)^{\frac {1}{2} (-1-i n)} \sqrt {1+a^2 x^2} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (1+i n),\frac {1}{2} (3+i n),\frac {1-i a x}{1+i a x}\right )}{(1+i n) \sqrt {c+a^2 c x^2}} \] Output:

-2*(1-I*a*x)^(1/2+1/2*I*n)*(1+I*a*x)^(-1/2-1/2*I*n)*(a^2*x^2+1)^(1/2)*hype 
rgeom([1, 1/2+1/2*I*n],[3/2+1/2*I*n],(1-I*a*x)/(1+I*a*x))/(1+I*n)/(a^2*c*x 
^2+c)^(1/2)
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.99 \[ \int \frac {e^{n \arctan (a x)}}{x \sqrt {c+a^2 c x^2}} \, dx=\frac {2 (1-i a x)^{\frac {1}{2}+\frac {i n}{2}} (1+i a x)^{-\frac {1}{2}-\frac {i n}{2}} \sqrt {1+a^2 x^2} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2}+\frac {i n}{2},\frac {3}{2}+\frac {i n}{2},\frac {i+a x}{i-a x}\right )}{(-1-i n) \sqrt {c+a^2 c x^2}} \] Input:

Integrate[E^(n*ArcTan[a*x])/(x*Sqrt[c + a^2*c*x^2]),x]
 

Output:

(2*(1 - I*a*x)^(1/2 + (I/2)*n)*(1 + I*a*x)^(-1/2 - (I/2)*n)*Sqrt[1 + a^2*x 
^2]*Hypergeometric2F1[1, 1/2 + (I/2)*n, 3/2 + (I/2)*n, (I + a*x)/(I - a*x) 
])/((-1 - I*n)*Sqrt[c + a^2*c*x^2])
 

Rubi [A] (verified)

Time = 0.81 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {5608, 5605, 141}

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.

\(\displaystyle \int \frac {e^{n \arctan (a x)}}{x \sqrt {a^2 c x^2+c}} \, dx\)

\(\Big \downarrow \) 5608

\(\displaystyle \frac {\sqrt {a^2 x^2+1} \int \frac {e^{n \arctan (a x)}}{x \sqrt {a^2 x^2+1}}dx}{\sqrt {a^2 c x^2+c}}\)

\(\Big \downarrow \) 5605

\(\displaystyle \frac {\sqrt {a^2 x^2+1} \int \frac {(1-i a x)^{\frac {1}{2} (i n-1)} (i a x+1)^{\frac {1}{2} (-i n-1)}}{x}dx}{\sqrt {a^2 c x^2+c}}\)

\(\Big \downarrow \) 141

\(\displaystyle -\frac {2 \sqrt {a^2 x^2+1} (1-i a x)^{\frac {1}{2} (1+i n)} (1+i a x)^{\frac {1}{2} (-1-i n)} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (i n+1),\frac {1}{2} (i n+3),\frac {1-i a x}{i a x+1}\right )}{(1+i n) \sqrt {a^2 c x^2+c}}\)

Input:

Int[E^(n*ArcTan[a*x])/(x*Sqrt[c + a^2*c*x^2]),x]
 

Output:

(-2*(1 - I*a*x)^((1 + I*n)/2)*(1 + I*a*x)^((-1 - I*n)/2)*Sqrt[1 + a^2*x^2] 
*Hypergeometric2F1[1, (1 + I*n)/2, (3 + I*n)/2, (1 - I*a*x)/(1 + I*a*x)])/ 
((1 + I*n)*Sqrt[c + a^2*c*x^2])
 

Defintions of rubi rules used

rule 141

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[(b*c - a*d)^n*((a + b*x)^(m + 1)/((m + 1)*(b*e - a*f)^( 
n + 1)*(e + f*x)^(m + 1)))*Hypergeometric2F1[m + 1, -n, m + 2, (-(d*e - c*f 
))*((a + b*x)/((b*c - a*d)*(e + f*x)))], x] /; FreeQ[{a, b, c, d, e, f, m, 
p}, x] && EqQ[m + n + p + 2, 0] && ILtQ[n, 0] && (SumSimplerQ[m, 1] ||  !Su 
mSimplerQ[p, 1]) &&  !ILtQ[m, 0]
 

rule 5605

Int[E^(ArcTan[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_ 
Symbol] :> Simp[c^p   Int[x^m*(1 - I*a*x)^(p + I*(n/2))*(1 + I*a*x)^(p - I* 
(n/2)), x], x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[d, a^2*c] && (Integer 
Q[p] || GtQ[c, 0])
 

rule 5608

Int[E^(ArcTan[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[c^IntPart[p]*((c + d*x^2)^FracPart[p]/(1 + a^2*x^2)^FracPart 
[p])   Int[x^m*(1 + a^2*x^2)^p*E^(n*ArcTan[a*x]), x], x] /; FreeQ[{a, c, d, 
 m, n, p}, x] && EqQ[d, a^2*c] &&  !(IntegerQ[p] || GtQ[c, 0])
 

Maple [F]

\[\int \frac {{\mathrm e}^{n \arctan \left (a x \right )}}{x \sqrt {a^{2} c \,x^{2}+c}}d x\]

Input:

int(exp(n*arctan(a*x))/x/(a^2*c*x^2+c)^(1/2),x)
 

Output:

int(exp(n*arctan(a*x))/x/(a^2*c*x^2+c)^(1/2),x)
 

Fricas [F]

\[ \int \frac {e^{n \arctan (a x)}}{x \sqrt {c+a^2 c x^2}} \, dx=\int { \frac {e^{\left (n \arctan \left (a x\right )\right )}}{\sqrt {a^{2} c x^{2} + c} x} \,d x } \] Input:

integrate(exp(n*arctan(a*x))/x/(a^2*c*x^2+c)^(1/2),x, algorithm="fricas")
 

Output:

integral(sqrt(a^2*c*x^2 + c)*e^(n*arctan(a*x))/(a^2*c*x^3 + c*x), x)
 

Sympy [F]

\[ \int \frac {e^{n \arctan (a x)}}{x \sqrt {c+a^2 c x^2}} \, dx=\int \frac {e^{n \operatorname {atan}{\left (a x \right )}}}{x \sqrt {c \left (a^{2} x^{2} + 1\right )}}\, dx \] Input:

integrate(exp(n*atan(a*x))/x/(a**2*c*x**2+c)**(1/2),x)
 

Output:

Integral(exp(n*atan(a*x))/(x*sqrt(c*(a**2*x**2 + 1))), x)
 

Maxima [F]

\[ \int \frac {e^{n \arctan (a x)}}{x \sqrt {c+a^2 c x^2}} \, dx=\int { \frac {e^{\left (n \arctan \left (a x\right )\right )}}{\sqrt {a^{2} c x^{2} + c} x} \,d x } \] Input:

integrate(exp(n*arctan(a*x))/x/(a^2*c*x^2+c)^(1/2),x, algorithm="maxima")
 

Output:

integrate(e^(n*arctan(a*x))/(sqrt(a^2*c*x^2 + c)*x), x)
 

Giac [F]

\[ \int \frac {e^{n \arctan (a x)}}{x \sqrt {c+a^2 c x^2}} \, dx=\int { \frac {e^{\left (n \arctan \left (a x\right )\right )}}{\sqrt {a^{2} c x^{2} + c} x} \,d x } \] Input:

integrate(exp(n*arctan(a*x))/x/(a^2*c*x^2+c)^(1/2),x, algorithm="giac")
 

Output:

integrate(e^(n*arctan(a*x))/(sqrt(a^2*c*x^2 + c)*x), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{n \arctan (a x)}}{x \sqrt {c+a^2 c x^2}} \, dx=\int \frac {{\mathrm {e}}^{n\,\mathrm {atan}\left (a\,x\right )}}{x\,\sqrt {c\,a^2\,x^2+c}} \,d x \] Input:

int(exp(n*atan(a*x))/(x*(c + a^2*c*x^2)^(1/2)),x)
 

Output:

int(exp(n*atan(a*x))/(x*(c + a^2*c*x^2)^(1/2)), x)
 

Reduce [F]

\[ \int \frac {e^{n \arctan (a x)}}{x \sqrt {c+a^2 c x^2}} \, dx=\frac {\int \frac {e^{\mathit {atan} \left (a x \right ) n}}{\sqrt {a^{2} x^{2}+1}\, x}d x}{\sqrt {c}} \] Input:

int(exp(n*atan(a*x))/x/(a^2*c*x^2+c)^(1/2),x)
 

Output:

int(e**(atan(a*x)*n)/(sqrt(a**2*x**2 + 1)*x),x)/sqrt(c)