\(\int \frac {e^{n \arctan (a x)} x^m}{(c+a^2 c x^2)^{5/2}} \, dx\) [384]

Optimal result

Integrand size = 26, antiderivative size = 82 \[ \int \frac {e^{n \arctan (a x)} x^m}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\frac {x^{1+m} \sqrt {1+a^2 x^2} \operatorname {AppellF1}\left (1+m,\frac {1}{2} (5-i n),\frac {1}{2} (5+i n),2+m,i a x,-i a x\right )}{c^2 (1+m) \sqrt {c+a^2 c x^2}} \] Output:

x^(1+m)*(a^2*x^2+1)^(1/2)*AppellF1(1+m,5/2+1/2*I*n,5/2-1/2*I*n,2+m,-I*a*x, 
I*a*x)/c^2/(1+m)/(a^2*c*x^2+c)^(1/2)
 

Mathematica [F]

\[ \int \frac {e^{n \arctan (a x)} x^m}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\int \frac {e^{n \arctan (a x)} x^m}{\left (c+a^2 c x^2\right )^{5/2}} \, dx \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 82, 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, 150}

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.

\(\displaystyle \int \frac {x^m e^{n \arctan (a x)}}{\left (a^2 c x^2+c\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 5608

\(\displaystyle \frac {\sqrt {a^2 x^2+1} \int \frac {e^{n \arctan (a x)} x^m}{\left (a^2 x^2+1\right )^{5/2}}dx}{c^2 \sqrt {a^2 c x^2+c}}\)

\(\Big \downarrow \) 5605

\(\displaystyle \frac {\sqrt {a^2 x^2+1} \int x^m (1-i a x)^{\frac {1}{2} (i n-5)} (i a x+1)^{\frac {1}{2} (-i n-5)}dx}{c^2 \sqrt {a^2 c x^2+c}}\)

\(\Big \downarrow \) 150

\(\displaystyle \frac {\sqrt {a^2 x^2+1} x^{m+1} \operatorname {AppellF1}\left (m+1,\frac {1}{2} (5-i n),\frac {1}{2} (i n+5),m+2,i a x,-i a x\right )}{c^2 (m+1) \sqrt {a^2 c x^2+c}}\)

Input:

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

Output:

(x^(1 + m)*Sqrt[1 + a^2*x^2]*AppellF1[1 + m, (5 - I*n)/2, (5 + I*n)/2, 2 + 
 m, I*a*x, (-I)*a*x])/(c^2*(1 + m)*Sqrt[c + a^2*c*x^2])
 

Defintions of rubi rules used

rule 150

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ 
] :> Simp[c^n*e^p*((b*x)^(m + 1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2 
, (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] &&  !In 
tegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 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]

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

Input:

int(exp(n*arctan(a*x))*x^m/(a^2*c*x^2+c)^(5/2),x)
 

Output:

int(exp(n*arctan(a*x))*x^m/(a^2*c*x^2+c)^(5/2),x)
 

Fricas [F]

\[ \int \frac {e^{n \arctan (a x)} x^m}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {x^{m} e^{\left (n \arctan \left (a x\right )\right )}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {5}{2}}} \,d x } \] Input:

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

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{n \arctan (a x)} x^m}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\text {Timed out} \] Input:

integrate(exp(n*atan(a*x))*x**m/(a**2*c*x**2+c)**(5/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {e^{n \arctan (a x)} x^m}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {x^{m} e^{\left (n \arctan \left (a x\right )\right )}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {5}{2}}} \,d x } \] Input:

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

Output:

integrate(x^m*e^(n*arctan(a*x))/(a^2*c*x^2 + c)^(5/2), x)
 

Giac [F]

\[ \int \frac {e^{n \arctan (a x)} x^m}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {x^{m} e^{\left (n \arctan \left (a x\right )\right )}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {5}{2}}} \,d x } \] Input:

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

Output:

integrate(x^m*e^(n*arctan(a*x))/(a^2*c*x^2 + c)^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{n \arctan (a x)} x^m}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\int \frac {x^m\,{\mathrm {e}}^{n\,\mathrm {atan}\left (a\,x\right )}}{{\left (c\,a^2\,x^2+c\right )}^{5/2}} \,d x \] Input:

int((x^m*exp(n*atan(a*x)))/(c + a^2*c*x^2)^(5/2),x)
 

Output:

int((x^m*exp(n*atan(a*x)))/(c + a^2*c*x^2)^(5/2), x)
 

Reduce [F]

\[ \int \frac {e^{n \arctan (a x)} x^m}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\frac {\int \frac {x^{m} e^{\mathit {atan} \left (a x \right ) n}}{\sqrt {a^{2} x^{2}+1}\, a^{4} x^{4}+2 \sqrt {a^{2} x^{2}+1}\, a^{2} x^{2}+\sqrt {a^{2} x^{2}+1}}d x}{\sqrt {c}\, c^{2}} \] Input:

int(exp(n*atan(a*x))*x^m/(a^2*c*x^2+c)^(5/2),x)
 

Output:

int((x**m*e**(atan(a*x)*n))/(sqrt(a**2*x**2 + 1)*a**4*x**4 + 2*sqrt(a**2*x 
**2 + 1)*a**2*x**2 + sqrt(a**2*x**2 + 1)),x)/(sqrt(c)*c**2)