\(\int \frac {e^{n \arctan (a x)} x}{\sqrt {c+a^2 c x^2}} \, dx\) [369]

Optimal result

Integrand size = 24, antiderivative size = 202 \[ \int \frac {e^{n \arctan (a x)} x}{\sqrt {c+a^2 c x^2}} \, dx=\frac {(1-i a x)^{\frac {1}{2} (1+i n)} (1+i a x)^{\frac {1}{2} (1-i n)} \sqrt {1+a^2 x^2}}{a^2 (1-i n) \sqrt {c+a^2 c x^2}}-\frac {i 2^{\frac {3}{2}-\frac {i n}{2}} n (1-i a x)^{\frac {1}{2} (1+i n)} \sqrt {1+a^2 x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-1+i n),\frac {1}{2} (1+i n),\frac {1}{2} (3+i n),\frac {1}{2} (1-i a x)\right )}{a^2 \left (1+n^2\right ) \sqrt {c+a^2 c x^2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.87 \[ \int \frac {e^{n \arctan (a x)} x}{\sqrt {c+a^2 c x^2}} \, dx=\frac {(1-i a x)^{\frac {1}{2}+\frac {i n}{2}} (2+2 i a x)^{-\frac {i n}{2}} \sqrt {1+a^2 x^2} \left (2^{\frac {i n}{2}} (1+i n) \sqrt {1+i a x}-2 i \sqrt {2} n (1+i a x)^{\frac {i n}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (1+i n),\frac {1}{2} i (i+n),\frac {1}{2} (3+i n),\frac {1}{2} (1-i a x)\right )\right )}{a^2 \left (1+n^2\right ) \sqrt {c+a^2 c x^2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.91, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {5608, 5605, 88, 79}

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.

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

\(\Big \downarrow \) 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

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

\(\Big \downarrow \) 88

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

\(\Big \downarrow \) 79

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

Input:

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

Output:

(Sqrt[1 + a^2*x^2]*(((1 - I*a*x)^((1 + I*n)/2)*(1 + I*a*x)^((1 - I*n)/2))/ 
(a^2*(1 - I*n)) + (2^(3/2 - (I/2)*n)*n*(1 - I*a*x)^((1 + I*n)/2)*Hypergeom 
etric2F1[(-1 + I*n)/2, (1 + I*n)/2, (3 + I*n)/2, (1 - I*a*x)/2])/(a^2*(I - 
 n)*(1 - I*n))))/Sqrt[c + a^2*c*x^2]
 

Defintions of rubi rules used

rule 79

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( 
a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 
, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] 
&&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] 
 ||  !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
 

rule 88

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^Simplify[p + 1], 
 x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] &&  !RationalQ[p] && SumSimpl 
erQ[p, 1]
 

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]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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