\(\int \frac {e^{n \arctan (a x)}}{x^3 \sqrt {c+a^2 c x^2}} \, dx\) [373]

Optimal result

Integrand size = 26, antiderivative size = 281 \[ \int \frac {e^{n \arctan (a x)}}{x^3 \sqrt {c+a^2 c x^2}} \, dx=-\frac {(1-i a x)^{\frac {1}{2} (1+i n)} (1+i a x)^{\frac {1}{2} (1-i n)} \sqrt {1+a^2 x^2}}{2 x^2 \sqrt {c+a^2 c x^2}}-\frac {a n (1-i a x)^{\frac {1}{2} (1+i n)} (1+i a x)^{\frac {1}{2} (1-i n)} \sqrt {1+a^2 x^2}}{2 x \sqrt {c+a^2 c x^2}}+\frac {a^2 \left (1-n^2\right ) (1-i a x)^{\frac {1}{2} (1+i n)} (1+i a x)^{\frac {1}{2} (-1-i n)} \sqrt {1+a^2 x^2} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (1+i n),\frac {1}{2} (3+i n),\frac {1-i a x}{1+i a x}\right )}{(1+i n) \sqrt {c+a^2 c x^2}} \] Output:

-1/2*(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)/x^2 
/(a^2*c*x^2+c)^(1/2)-1/2*a*n*(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)/x/(a^2*c*x^2+c)^(1/2)+a^2*(-n^2+1)*(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)*hypergeom([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 = 0.10 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.57 \[ \int \frac {e^{n \arctan (a x)}}{x^3 \sqrt {c+a^2 c x^2}} \, dx=\frac {i (1-i a x)^{\frac {1}{2}+\frac {i n}{2}} (1+i a x)^{-\frac {1}{2}-\frac {i n}{2}} \sqrt {1+a^2 x^2} \left (-((-i+n) (-i+a x) (1+a n x))+2 a^2 \left (-1+n^2\right ) x^2 \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2}+\frac {i n}{2},\frac {3}{2}+\frac {i n}{2},\frac {i+a x}{i-a x}\right )\right )}{2 (-i+n) x^2 \sqrt {c+a^2 c x^2}} \] Input:

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

Output:

((I/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]*(-((-I + n)*(-I + a*x)*(1 + a*n*x)) + 2*a^2*(-1 + n^2)*x^2*Hyperge 
ometric2F1[1, 1/2 + (I/2)*n, 3/2 + (I/2)*n, (I + a*x)/(I - a*x)]))/((-I + 
n)*x^2*Sqrt[c + a^2*c*x^2])
 

Rubi [A] (verified)

Time = 0.90 (sec) , antiderivative size = 230, normalized size of antiderivative = 0.82, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {5608, 5605, 144, 25, 27, 168, 27, 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.

\(\displaystyle \int \frac {e^{n \arctan (a x)}}{x^3 \sqrt {a^2 c x^2+c}} \, dx\)

\(\Big \downarrow \) 5608

\(\displaystyle \frac {\sqrt {a^2 x^2+1} \int \frac {e^{n \arctan (a x)}}{x^3 \sqrt {a^2 x^2+1}}dx}{\sqrt {a^2 c x^2+c}}\)

\(\Big \downarrow \) 5605

\(\displaystyle \frac {\sqrt {a^2 x^2+1} \int \frac {(1-i a x)^{\frac {1}{2} (i n-1)} (i a x+1)^{\frac {1}{2} (-i n-1)}}{x^3}dx}{\sqrt {a^2 c x^2+c}}\)

\(\Big \downarrow \) 144

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 168

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 141

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

Input:

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

Output:

(Sqrt[1 + a^2*x^2]*(-1/2*((1 - I*a*x)^((1 + I*n)/2)*(1 + I*a*x)^((1 - I*n) 
/2))/x^2 + (a*(-((n*(1 - I*a*x)^((1 + I*n)/2)*(1 + I*a*x)^((1 - I*n)/2))/x 
) + (2*a*(1 - n^2)*(1 - I*a*x)^((1 + I*n)/2)*(1 + I*a*x)^((-1 - I*n)/2)*Hy 
pergeometric2F1[1, (1 + I*n)/2, (3 + I*n)/2, (1 - I*a*x)/(1 + I*a*x)])/(1 
+ I*n)))/2))/Sqrt[c + a^2*c*x^2]
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 144

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> With[{mnp = Simplify[m + n + p]}, Simp[b*(a + b*x)^(m + 1)*( 
c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] 
+ Simp[1/((m + 1)*(b*c - a*d)*(b*e - a*f))   Int[(a + b*x)^(m + 1)*(c + d*x 
)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) + c*f*(m + p + 2)) 
- b*d*f*(mnp + 3)*x, x], x], x] /; ILtQ[mnp + 2, 0] && (SumSimplerQ[m, 1] | 
| ( !SumSimplerQ[n, 1] &&  !SumSimplerQ[p, 1]))] /; FreeQ[{a, b, c, d, e, f 
, m, n, p}, x] && NeQ[m, -1]
 

rule 168

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + 
 d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + S 
imp[1/((m + 1)*(b*c - a*d)*(b*e - a*f))   Int[(a + b*x)^(m + 1)*(c + d*x)^n 
*(e + f*x)^p*Simp[(a*d*f*g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a* 
h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p + 3)*x, x], x], 
 x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m, -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]

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

Input:

int(exp(n*arctan(a*x))/x^3/(a^2*c*x^2+c)^(1/2),x)
 

Output:

int(exp(n*arctan(a*x))/x^3/(a^2*c*x^2+c)^(1/2),x)
 

Fricas [F]

\[ \int \frac {e^{n \arctan (a x)}}{x^3 \sqrt {c+a^2 c x^2}} \, dx=\int { \frac {e^{\left (n \arctan \left (a x\right )\right )}}{\sqrt {a^{2} c x^{2} + c} x^{3}} \,d x } \] Input:

integrate(exp(n*arctan(a*x))/x^3/(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^5 + c*x^3), x)
 

Sympy [F]

\[ \int \frac {e^{n \arctan (a x)}}{x^3 \sqrt {c+a^2 c x^2}} \, dx=\int \frac {e^{n \operatorname {atan}{\left (a x \right )}}}{x^{3} \sqrt {c \left (a^{2} x^{2} + 1\right )}}\, dx \] Input:

integrate(exp(n*atan(a*x))/x**3/(a**2*c*x**2+c)**(1/2),x)
 

Output:

Integral(exp(n*atan(a*x))/(x**3*sqrt(c*(a**2*x**2 + 1))), x)
 

Maxima [F]

\[ \int \frac {e^{n \arctan (a x)}}{x^3 \sqrt {c+a^2 c x^2}} \, dx=\int { \frac {e^{\left (n \arctan \left (a x\right )\right )}}{\sqrt {a^{2} c x^{2} + c} x^{3}} \,d x } \] Input:

integrate(exp(n*arctan(a*x))/x^3/(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^3), x)
 

Giac [F]

\[ \int \frac {e^{n \arctan (a x)}}{x^3 \sqrt {c+a^2 c x^2}} \, dx=\int { \frac {e^{\left (n \arctan \left (a x\right )\right )}}{\sqrt {a^{2} c x^{2} + c} x^{3}} \,d x } \] Input:

integrate(exp(n*arctan(a*x))/x^3/(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^3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{n \arctan (a x)}}{x^3 \sqrt {c+a^2 c x^2}} \, dx=\int \frac {{\mathrm {e}}^{n\,\mathrm {atan}\left (a\,x\right )}}{x^3\,\sqrt {c\,a^2\,x^2+c}} \,d x \] Input:

int(exp(n*atan(a*x))/(x^3*(c + a^2*c*x^2)^(1/2)),x)
 

Output:

int(exp(n*atan(a*x))/(x^3*(c + a^2*c*x^2)^(1/2)), x)
 

Reduce [F]

\[ \int \frac {e^{n \arctan (a x)}}{x^3 \sqrt {c+a^2 c x^2}} \, dx=\frac {\int \frac {e^{\mathit {atan} \left (a x \right ) n}}{\sqrt {a^{2} x^{2}+1}\, x^{3}}d x}{\sqrt {c}} \] Input:

int(exp(n*atan(a*x))/x^3/(a^2*c*x^2+c)^(1/2),x)
 

Output:

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