\(\int e^{m x} \csc ^2(x) \, dx\) [550]

Optimal result

Integrand size = 10, antiderivative size = 45 \[ \int e^{m x} \csc ^2(x) \, dx=-\frac {4 e^{(2 i+m) x} \operatorname {Hypergeometric2F1}\left (2,1-\frac {i m}{2},2-\frac {i m}{2},e^{2 i x}\right )}{2 i+m} \] Output:

-4*exp((2*I+m)*x)*hypergeom([2, 1-1/2*I*m],[2-1/2*I*m],exp(2*I*x))/(2*I+m)
 

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 90, normalized size of antiderivative = 2.00 \[ \int e^{m x} \csc ^2(x) \, dx=\frac {e^{m x} \left (e^{2 i x} m \operatorname {Hypergeometric2F1}\left (1,1-\frac {i m}{2},2-\frac {i m}{2},e^{2 i x}\right )+(2 i+m) \left (-i \cot (x)+\operatorname {Hypergeometric2F1}\left (1,-\frac {i m}{2},1-\frac {i m}{2},e^{2 i x}\right )\right )\right )}{-2+i m} \] Input:

Integrate[E^(m*x)*Csc[x]^2,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {4953}

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.

\(\displaystyle \int e^{m x} \csc ^2(x) \, dx\)

\(\Big \downarrow \) 4953

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

Input:

Int[E^(m*x)*Csc[x]^2,x]
 

Output:

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

Defintions of rubi rules used

rule 4953

Int[Csc[(d_.) + (e_.)*(x_)]^(n_.)*(F_)^((c_.)*((a_.) + (b_.)*(x_))), x_Symb 
ol] :> Simp[(-2*I)^n*E^(I*n*(d + e*x))*(F^(c*(a + b*x))/(I*e*n + b*c*Log[F] 
))*Hypergeometric2F1[n, n/2 - I*b*c*(Log[F]/(2*e)), 1 + n/2 - I*b*c*(Log[F] 
/(2*e)), E^(2*I*(d + e*x))], x] /; FreeQ[{F, a, b, c, d, e}, x] && IntegerQ 
[n]
 

Maple [F]

\[\int \frac {{\mathrm e}^{m x}}{\sin \left (x \right )^{2}}d x\]

Input:

int(exp(m*x)/sin(x)^2,x)
 

Output:

int(exp(m*x)/sin(x)^2,x)
 

Fricas [F]

\[ \int e^{m x} \csc ^2(x) \, dx=\int { \frac {e^{\left (m x\right )}}{\sin \left (x\right )^{2}} \,d x } \] Input:

integrate(exp(m*x)/sin(x)^2,x, algorithm="fricas")
 

Output:

integral(-e^(m*x)/(cos(x)^2 - 1), x)
 

Sympy [F]

\[ \int e^{m x} \csc ^2(x) \, dx=\int \frac {e^{m x}}{\sin ^{2}{\left (x \right )}}\, dx \] Input:

integrate(exp(m*x)/sin(x)**2,x)
 

Output:

Integral(exp(m*x)/sin(x)**2, x)
 

Maxima [F]

\[ \int e^{m x} \csc ^2(x) \, dx=\int { \frac {e^{\left (m x\right )}}{\sin \left (x\right )^{2}} \,d x } \] Input:

integrate(exp(m*x)/sin(x)^2,x, algorithm="maxima")
 

Output:

4*(2*(m^3 + 16*m)*cos(2*x)^2*e^(m*x) + 2*(m^3 + 16*m)*e^(m*x)*sin(2*x)^2 - 
 (m^3 + 64*m)*cos(2*x)*e^(m*x) + 2*(5*m^2 - 16)*e^(m*x)*sin(2*x) - ((m^3 + 
 16*m)*cos(2*x)*e^(m*x) - 2*(m^2 + 16)*e^(m*x)*sin(2*x) - 24*m*e^(m*x))*co 
s(4*x) + 24*m*e^(m*x) - 4*(m^5 + 20*m^3 + (m^5 + 20*m^3 + 64*m)*cos(4*x)^2 
 + 4*(m^5 + 20*m^3 + 64*m)*cos(2*x)^2 + (m^5 + 20*m^3 + 64*m)*sin(4*x)^2 - 
 4*(m^5 + 20*m^3 + 64*m)*sin(4*x)*sin(2*x) + 4*(m^5 + 20*m^3 + 64*m)*sin(2 
*x)^2 + 2*(m^5 + 20*m^3 - 2*(m^5 + 20*m^3 + 64*m)*cos(2*x) + 64*m)*cos(4*x 
) - 4*(m^5 + 20*m^3 + 64*m)*cos(2*x) + 64*m)*integrate(-(6*m*cos(6*x)*e^(m 
*x) - 18*m*cos(4*x)*e^(m*x) + 18*m*cos(2*x)*e^(m*x) - (m^2 - 8)*e^(m*x)*si 
n(6*x) + 3*(m^2 - 8)*e^(m*x)*sin(4*x) - 3*(m^2 - 8)*e^(m*x)*sin(2*x) - 6*m 
*e^(m*x))/(m^4 + (m^4 + 20*m^2 + 64)*cos(6*x)^2 + 9*(m^4 + 20*m^2 + 64)*co 
s(4*x)^2 + 9*(m^4 + 20*m^2 + 64)*cos(2*x)^2 + (m^4 + 20*m^2 + 64)*sin(6*x) 
^2 + 9*(m^4 + 20*m^2 + 64)*sin(4*x)^2 - 18*(m^4 + 20*m^2 + 64)*sin(4*x)*si 
n(2*x) + 9*(m^4 + 20*m^2 + 64)*sin(2*x)^2 + 20*m^2 - 2*(m^4 + 20*m^2 + 3*( 
m^4 + 20*m^2 + 64)*cos(4*x) - 3*(m^4 + 20*m^2 + 64)*cos(2*x) + 64)*cos(6*x 
) + 6*(m^4 + 20*m^2 - 3*(m^4 + 20*m^2 + 64)*cos(2*x) + 64)*cos(4*x) - 6*(m 
^4 + 20*m^2 + 64)*cos(2*x) - 6*((m^4 + 20*m^2 + 64)*sin(4*x) - (m^4 + 20*m 
^2 + 64)*sin(2*x))*sin(6*x) + 64), x) - (2*(m^2 + 16)*cos(2*x)*e^(m*x) + ( 
m^3 + 16*m)*e^(m*x)*sin(2*x) + 4*(m^2 - 8)*e^(m*x))*sin(4*x))/(m^4 + (m^4 
+ 20*m^2 + 64)*cos(4*x)^2 + 4*(m^4 + 20*m^2 + 64)*cos(2*x)^2 + (m^4 + 2...
 

Giac [F]

\[ \int e^{m x} \csc ^2(x) \, dx=\int { \frac {e^{\left (m x\right )}}{\sin \left (x\right )^{2}} \,d x } \] Input:

integrate(exp(m*x)/sin(x)^2,x, algorithm="giac")
 

Output:

integrate(e^(m*x)/sin(x)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int e^{m x} \csc ^2(x) \, dx=\int \frac {{\mathrm {e}}^{m\,x}}{{\sin \left (x\right )}^2} \,d x \] Input:

int(exp(m*x)/sin(x)^2,x)
 

Output:

int(exp(m*x)/sin(x)^2, x)
 

Reduce [F]

\[ \int e^{m x} \csc ^2(x) \, dx=\int \frac {e^{m x}}{\sin \left (x \right )^{2}}d x \] Input:

int(exp(m*x)/sin(x)^2,x)
 

Output:

int(e**(m*x)/sin(x)**2,x)