\(\int (e x)^m \tan (a+i \log (x)) \, dx\) [150]

Optimal result

Integrand size = 15, antiderivative size = 71 \[ \int (e x)^m \tan (a+i \log (x)) \, dx=-\frac {i (e x)^{1+m}}{e (1+m)}+\frac {2 i (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (-1-m),\frac {1-m}{2},-\frac {e^{2 i a}}{x^2}\right )}{e (1+m)} \] Output:

-I*(e*x)^(1+m)/e/(1+m)+2*I*(e*x)^(1+m)*hypergeom([1, -1/2-1/2*m],[1/2-1/2* 
m],-exp(2*I*a)/x^2)/e/(1+m)
 

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.75 \[ \int (e x)^m \tan (a+i \log (x)) \, dx=\frac {x (e x)^m (\cos (a)-i \sin (a)) \left ((3+m) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-x^2 (\cos (2 a)-i \sin (2 a))\right ) (-i \cos (a)+\sin (a))+(1+m) x^2 \operatorname {Hypergeometric2F1}\left (1,\frac {3+m}{2},\frac {5+m}{2},-x^2 (\cos (2 a)-i \sin (2 a))\right ) (i \cos (a)+\sin (a))\right )}{(1+m) (3+m)} \] Input:

Integrate[(e*x)^m*Tan[a + I*Log[x]],x]
 

Output:

(x*(e*x)^m*(Cos[a] - I*Sin[a])*((3 + m)*Hypergeometric2F1[1, (1 + m)/2, (3 
 + m)/2, -(x^2*(Cos[2*a] - I*Sin[2*a]))]*((-I)*Cos[a] + Sin[a]) + (1 + m)* 
x^2*Hypergeometric2F1[1, (3 + m)/2, (5 + m)/2, -(x^2*(Cos[2*a] - I*Sin[2*a 
]))]*(I*Cos[a] + Sin[a])))/((1 + m)*(3 + m))
 

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {5006, 959, 862, 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.

\(\displaystyle \int (e x)^m \tan (a+i \log (x)) \, dx\)

\(\Big \downarrow \) 5006

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

\(\Big \downarrow \) 959

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

\(\Big \downarrow \) 862

\(\displaystyle -\frac {2 i \left (\frac {1}{x}\right )^{m+1} (e x)^{m+1} \int \frac {\left (\frac {1}{x}\right )^{-m-2}}{1+\frac {e^{2 i a}}{x^2}}d\frac {1}{x}}{e}-\frac {i (e x)^{m+1}}{e (m+1)}\)

\(\Big \downarrow \) 278

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

Input:

Int[(e*x)^m*Tan[a + I*Log[x]],x]
 

Output:

((-I)*(e*x)^(1 + m))/(e*(1 + m)) + ((2*I)*(e*x)^(1 + m)*Hypergeometric2F1[ 
1, (-1 - m)/2, (1 - m)/2, -(E^((2*I)*a)/x^2)])/(e*(1 + 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 862

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

rule 959

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n 
_)), x_Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(b*e*(m + n*(p 
+ 1) + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m + n*(p 
 + 1) + 1))   Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, 
 n, p}, x] && NeQ[b*c - a*d, 0] && NeQ[m + n*(p + 1) + 1, 0]
 

rule 5006

Int[((e_.)*(x_))^(m_.)*Tan[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.), x_Symbol] 
:> Int[(e*x)^m*((I - I*E^(2*I*a*d)*x^(2*I*b*d))/(1 + E^(2*I*a*d)*x^(2*I*b*d 
)))^p, x] /; FreeQ[{a, b, d, e, m, p}, x]
 

Maple [F]

\[\int \left (e x \right )^{m} \tan \left (a +i \ln \left (x \right )\right )d x\]

Input:

int((e*x)^m*tan(a+I*ln(x)),x)
 

Output:

int((e*x)^m*tan(a+I*ln(x)),x)
 

Fricas [F]

\[ \int (e x)^m \tan (a+i \log (x)) \, dx=\int { \left (e x\right )^{m} \tan \left (a + i \, \log \left (x\right )\right ) \,d x } \] Input:

integrate((e*x)^m*tan(a+I*log(x)),x, algorithm="fricas")
 

Output:

integral((I*x^2 - I*e^(2*I*a))*e^(m*log(e) + m*log(x))/(x^2 + e^(2*I*a)), 
x)
 

Sympy [F]

\[ \int (e x)^m \tan (a+i \log (x)) \, dx=\int \left (e x\right )^{m} \tan {\left (a + i \log {\left (x \right )} \right )}\, dx \] Input:

integrate((e*x)**m*tan(a+I*ln(x)),x)
 

Output:

Integral((e*x)**m*tan(a + I*log(x)), x)
 

Maxima [F]

\[ \int (e x)^m \tan (a+i \log (x)) \, dx=\int { \left (e x\right )^{m} \tan \left (a + i \, \log \left (x\right )\right ) \,d x } \] Input:

integrate((e*x)^m*tan(a+I*log(x)),x, algorithm="maxima")
 

Output:

integrate((e*x)^m*tan(a + I*log(x)), x)
 

Giac [F]

\[ \int (e x)^m \tan (a+i \log (x)) \, dx=\int { \left (e x\right )^{m} \tan \left (a + i \, \log \left (x\right )\right ) \,d x } \] Input:

integrate((e*x)^m*tan(a+I*log(x)),x, algorithm="giac")
 

Output:

integrate((e*x)^m*tan(a + I*log(x)), x)
 

Mupad [F(-1)]

Timed out. \[ \int (e x)^m \tan (a+i \log (x)) \, dx=\int \mathrm {tan}\left (a+\ln \left (x\right )\,1{}\mathrm {i}\right )\,{\left (e\,x\right )}^m \,d x \] Input:

int(tan(a + log(x)*1i)*(e*x)^m,x)
 

Output:

int(tan(a + log(x)*1i)*(e*x)^m, x)
 

Reduce [F]

\[ \int (e x)^m \tan (a+i \log (x)) \, dx=e^{m} \left (\int x^{m} \tan \left (\mathrm {log}\left (x \right ) i +a \right )d x \right ) \] Input:

int((e*x)^m*tan(a+I*log(x)),x)
 

Output:

e**m*int(x**m*tan(log(x)*i + a),x)