\(\int e^{3 i \arctan (a x)} x^m \, dx\) [155]

Optimal result

Integrand size = 14, antiderivative size = 118 \[ \int e^{3 i \arctan (a x)} x^m \, dx=\frac {4 x^{1+m} (1+i a x)}{\sqrt {1+a^2 x^2}}-\frac {(3+4 m) x^{1+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},-a^2 x^2\right )}{1+m}-\frac {i a (5+4 m) x^{2+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},-a^2 x^2\right )}{2+m} \] Output:

4*x^(1+m)*(1+I*a*x)/(a^2*x^2+1)^(1/2)-(3+4*m)*x^(1+m)*hypergeom([1/2, 1/2+ 
1/2*m],[3/2+1/2*m],-a^2*x^2)/(1+m)-I*a*(5+4*m)*x^(2+m)*hypergeom([1/2, 1+1 
/2*m],[2+1/2*m],-a^2*x^2)/(2+m)
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.

Time = 0.10 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.96 \[ \int e^{3 i \arctan (a x)} x^m \, dx=-\frac {i x^{1+m} \sqrt {1-i a x} \sqrt {-i+a x} \left (\operatorname {AppellF1}\left (1+m,-\frac {1}{2},\frac {1}{2},2+m,-i a x,i a x\right )-2 \operatorname {AppellF1}\left (1+m,-\frac {1}{2},\frac {3}{2},2+m,-i a x,i a x\right )\right )}{(1+m) \sqrt {1+i a x} \sqrt {i+a x}} \] Input:

Integrate[E^((3*I)*ArcTan[a*x])*x^m,x]
 

Output:

((-I)*x^(1 + m)*Sqrt[1 - I*a*x]*Sqrt[-I + a*x]*(AppellF1[1 + m, -1/2, 1/2, 
 2 + m, (-I)*a*x, I*a*x] - 2*AppellF1[1 + m, -1/2, 3/2, 2 + m, (-I)*a*x, I 
*a*x]))/((1 + m)*Sqrt[1 + I*a*x]*Sqrt[I + a*x])
 

Rubi [A] (verified)

Time = 0.86 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.36, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5583, 2355, 557, 278, 583, 557, 278}

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.

\(\displaystyle \int x^m e^{3 i \arctan (a x)} \, dx\)

\(\Big \downarrow \) 5583

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

\(\Big \downarrow \) 2355

\(\displaystyle \int \frac {x^m (-i a x-3)}{\sqrt {a^2 x^2+1}}dx+4 \int \frac {x^m}{(1-i a x) \sqrt {a^2 x^2+1}}dx\)

\(\Big \downarrow \) 557

\(\displaystyle -i a \int \frac {x^{m+1}}{\sqrt {a^2 x^2+1}}dx-3 \int \frac {x^m}{\sqrt {a^2 x^2+1}}dx+4 \int \frac {x^m}{(1-i a x) \sqrt {a^2 x^2+1}}dx\)

\(\Big \downarrow \) 278

\(\displaystyle 4 \int \frac {x^m}{(1-i a x) \sqrt {a^2 x^2+1}}dx-\frac {3 x^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},-a^2 x^2\right )}{m+1}-\frac {i a x^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},-a^2 x^2\right )}{m+2}\)

\(\Big \downarrow \) 583

\(\displaystyle 4 \int \frac {x^m (i a x+1)}{\left (a^2 x^2+1\right )^{3/2}}dx-\frac {3 x^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},-a^2 x^2\right )}{m+1}-\frac {i a x^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},-a^2 x^2\right )}{m+2}\)

\(\Big \downarrow \) 557

\(\displaystyle 4 \left (\int \frac {x^m}{\left (a^2 x^2+1\right )^{3/2}}dx+i a \int \frac {x^{m+1}}{\left (a^2 x^2+1\right )^{3/2}}dx\right )-\frac {3 x^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},-a^2 x^2\right )}{m+1}-\frac {i a x^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},-a^2 x^2\right )}{m+2}\)

\(\Big \downarrow \) 278

\(\displaystyle -\frac {3 x^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},-a^2 x^2\right )}{m+1}-\frac {i a x^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},-a^2 x^2\right )}{m+2}+4 \left (\frac {x^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {m+1}{2},\frac {m+3}{2},-a^2 x^2\right )}{m+1}+\frac {i a x^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {m+2}{2},\frac {m+4}{2},-a^2 x^2\right )}{m+2}\right )\)

Input:

Int[E^((3*I)*ArcTan[a*x])*x^m,x]
 

Output:

(-3*x^(1 + m)*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, -(a^2*x^2)])/(1 
 + m) - (I*a*x^(2 + m)*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, -(a^2* 
x^2)])/(2 + m) + 4*((x^(1 + m)*Hypergeometric2F1[3/2, (1 + m)/2, (3 + m)/2 
, -(a^2*x^2)])/(1 + m) + (I*a*x^(2 + m)*Hypergeometric2F1[3/2, (2 + m)/2, 
(4 + m)/2, -(a^2*x^2)])/(2 + m))
 

Defintions of rubi rules used

rule 278

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( 
c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( 
-b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && (ILtQ[p, 0 
] || GtQ[a, 0])
 

rule 557

Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Sym 
bol] :> Simp[c   Int[(e*x)^m*(a + b*x^2)^p, x], x] + Simp[d/e   Int[(e*x)^( 
m + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x]
 

rule 583

Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), 
x_Symbol] :> Simp[c^(2*n)/a^n   Int[(e*x)^m*((a + b*x^2)^(n + p)/(c - d*x)^ 
n), x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[b*c^2 + a*d^2, 0] && I 
LtQ[n, 0]
 

rule 2355

Int[(Px_)*((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2) 
^(p_.), x_Symbol] :> Int[PolynomialQuotient[Px, c + d*x, x]*(e*x)^m*(c + d* 
x)^(n + 1)*(a + b*x^2)^p, x] + Simp[PolynomialRemainder[Px, c + d*x, x]   I 
nt[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p} 
, x] && PolynomialQ[Px, x] && LtQ[n, 0]
 

rule 5583

Int[E^(ArcTan[(a_.)*(x_)]*(n_))*(x_)^(m_.), x_Symbol] :> Int[x^m*((1 - I*a* 
x)^((I*n + 1)/2)/((1 + I*a*x)^((I*n - 1)/2)*Sqrt[1 + a^2*x^2])), x] /; Free 
Q[{a, m}, x] && IntegerQ[(I*n - 1)/2]
 

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.24

method	result	size
meijerg	\(\frac {x^{1+m} \operatorname {hypergeom}\left (\left [\frac {3}{2}, \frac {m}{2}+\frac {1}{2}\right ], \left [\frac {3}{2}+\frac {m}{2}\right ], -a^{2} x^{2}\right )}{1+m}+\frac {3 i a \,x^{2+m} \operatorname {hypergeom}\left (\left [\frac {3}{2}, \frac {m}{2}+1\right ], \left [2+\frac {m}{2}\right ], -a^{2} x^{2}\right )}{2+m}-\frac {3 a^{2} x^{3+m} \operatorname {hypergeom}\left (\left [\frac {3}{2}, \frac {3}{2}+\frac {m}{2}\right ], \left [\frac {5}{2}+\frac {m}{2}\right ], -a^{2} x^{2}\right )}{3+m}-\frac {i a^{3} x^{4+m} \operatorname {hypergeom}\left (\left [\frac {3}{2}, 2+\frac {m}{2}\right ], \left [\frac {m}{2}+3\right ], -a^{2} x^{2}\right )}{4+m}\)	\(146\)

Input:

int((1+I*a*x)^3/(a^2*x^2+1)^(3/2)*x^m,x,method=_RETURNVERBOSE)
 

Output:

1/(1+m)*x^(1+m)*hypergeom([3/2,1/2*m+1/2],[3/2+1/2*m],-a^2*x^2)+3*I*a/(2+m 
)*x^(2+m)*hypergeom([3/2,1/2*m+1],[2+1/2*m],-a^2*x^2)-3*a^2/(3+m)*x^(3+m)* 
hypergeom([3/2,3/2+1/2*m],[5/2+1/2*m],-a^2*x^2)-I*a^3/(4+m)*x^(4+m)*hyperg 
eom([3/2,2+1/2*m],[1/2*m+3],-a^2*x^2)
 

Fricas [F]

\[ \int e^{3 i \arctan (a x)} x^m \, dx=\int { \frac {{\left (i \, a x + 1\right )}^{3} x^{m}}{{\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((1+I*a*x)^3/(a^2*x^2+1)^(3/2)*x^m,x, algorithm="fricas")
 

Output:

integral(sqrt(a^2*x^2 + 1)*(-I*a*x - 1)*x^m/(a^2*x^2 + 2*I*a*x - 1), x)
 

Sympy [F]

\[ \int e^{3 i \arctan (a x)} x^m \, dx=- i \left (\int \frac {i x^{m}}{a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} + \sqrt {a^{2} x^{2} + 1}}\, dx + \int \left (- \frac {3 a x x^{m}}{a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} + \sqrt {a^{2} x^{2} + 1}}\right )\, dx + \int \frac {a^{3} x^{3} x^{m}}{a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} + \sqrt {a^{2} x^{2} + 1}}\, dx + \int \left (- \frac {3 i a^{2} x^{2} x^{m}}{a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} + \sqrt {a^{2} x^{2} + 1}}\right )\, dx\right ) \] Input:

integrate((1+I*a*x)**3/(a**2*x**2+1)**(3/2)*x**m,x)
 

Output:

-I*(Integral(I*x**m/(a**2*x**2*sqrt(a**2*x**2 + 1) + sqrt(a**2*x**2 + 1)), 
 x) + Integral(-3*a*x*x**m/(a**2*x**2*sqrt(a**2*x**2 + 1) + sqrt(a**2*x**2 
 + 1)), x) + Integral(a**3*x**3*x**m/(a**2*x**2*sqrt(a**2*x**2 + 1) + sqrt 
(a**2*x**2 + 1)), x) + Integral(-3*I*a**2*x**2*x**m/(a**2*x**2*sqrt(a**2*x 
**2 + 1) + sqrt(a**2*x**2 + 1)), x))
 

Maxima [F]

\[ \int e^{3 i \arctan (a x)} x^m \, dx=\int { \frac {{\left (i \, a x + 1\right )}^{3} x^{m}}{{\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((1+I*a*x)^3/(a^2*x^2+1)^(3/2)*x^m,x, algorithm="maxima")
 

Output:

integrate((I*a*x + 1)^3*x^m/(a^2*x^2 + 1)^(3/2), x)
 

Giac [F(-2)]

Exception generated. \[ \int e^{3 i \arctan (a x)} x^m \, dx=\text {Exception raised: TypeError} \] Input:

integrate((1+I*a*x)^3/(a^2*x^2+1)^(3/2)*x^m,x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

Timed out. \[ \int e^{3 i \arctan (a x)} x^m \, dx=\int \frac {x^m\,{\left (1+a\,x\,1{}\mathrm {i}\right )}^3}{{\left (a^2\,x^2+1\right )}^{3/2}} \,d x \] Input:

int((x^m*(a*x*1i + 1)^3)/(a^2*x^2 + 1)^(3/2),x)
 

Output:

int((x^m*(a*x*1i + 1)^3)/(a^2*x^2 + 1)^(3/2), x)
 

Reduce [F]

\[ \int e^{3 i \arctan (a x)} x^m \, dx=\int \frac {x^{m}}{\sqrt {a^{2} x^{2}+1}\, a^{2} x^{2}+\sqrt {a^{2} x^{2}+1}}d x -\left (\int \frac {x^{m} x^{3}}{\sqrt {a^{2} x^{2}+1}\, a^{2} x^{2}+\sqrt {a^{2} x^{2}+1}}d x \right ) a^{3} i -3 \left (\int \frac {x^{m} x^{2}}{\sqrt {a^{2} x^{2}+1}\, a^{2} x^{2}+\sqrt {a^{2} x^{2}+1}}d x \right ) a^{2}+3 \left (\int \frac {x^{m} x}{\sqrt {a^{2} x^{2}+1}\, a^{2} x^{2}+\sqrt {a^{2} x^{2}+1}}d x \right ) a i \] Input:

int((1+I*a*x)^3/(a^2*x^2+1)^(3/2)*x^m,x)
 

Output:

int(x**m/(sqrt(a**2*x**2 + 1)*a**2*x**2 + sqrt(a**2*x**2 + 1)),x) - int((x 
**m*x**3)/(sqrt(a**2*x**2 + 1)*a**2*x**2 + sqrt(a**2*x**2 + 1)),x)*a**3*i 
- 3*int((x**m*x**2)/(sqrt(a**2*x**2 + 1)*a**2*x**2 + sqrt(a**2*x**2 + 1)), 
x)*a**2 + 3*int((x**m*x)/(sqrt(a**2*x**2 + 1)*a**2*x**2 + sqrt(a**2*x**2 + 
 1)),x)*a*i