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

Optimal. Leaf size=45 \[ -\frac {4 e^{(2 i+m) x} \, _2F_1\left (2,1-\frac {i m}{2};2-\frac {i m}{2};e^{2 i x}\right )}{2 i+m} \]

[Out]

-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)

________________________________________________________________________________________

Rubi [A]
time = 0.01, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {4538} \begin {gather*} -\frac {4 e^{(m+2 i) x} \text {Hypergeometric2F1}\left (2,1-\frac {i m}{2},2-\frac {i m}{2},e^{2 i x}\right )}{m+2 i} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4538

Int[Csc[(d_.) + (e_.)*(x_)]^(n_.)*(F_)^((c_.)*((a_.) + (b_.)*(x_))), x_Symbol] :> 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*(L
og[F]/(2*e)), E^(2*I*(d + e*x))], x] /; FreeQ[{F, a, b, c, d, e}, x] && IntegerQ[n]

Rubi steps

\begin {align*} \int e^{m x} \csc ^2(x) \, dx &=-\frac {4 e^{(2 i+m) x} \, _2F_1\left (2,1-\frac {i m}{2};2-\frac {i m}{2};e^{2 i x}\right )}{2 i+m}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.15, size = 90, normalized size = 2.00 \begin {gather*} \frac {e^{m x} \left (e^{2 i x} m \, _2F_1\left (1,1-\frac {i m}{2};2-\frac {i m}{2};e^{2 i x}\right )+(2 i+m) \left (-i \cot (x)+\, _2F_1\left (1,-\frac {i m}{2};1-\frac {i m}{2};e^{2 i x}\right )\right )\right )}{-2+i m} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(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)

________________________________________________________________________________________

Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {{\mathrm e}^{m x}}{\sin \left (x \right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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))*
cos(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)*s
in(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)*sin(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)*cos(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)*sin(2*x) + 9*(m^4 + 20*m^2 + 6
4)*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) + 6
4)*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 + 20*m^2 + 64)*sin(4*x)^2 - 4*(m^4 + 20*m^2 + 64)*sin(4*x)*sin(2*x
) + 4*(m^4 + 20*m^2 + 64)*sin(2*x)^2 + 20*m^2 + 2*(m^4 + 20*m^2 - 2*(m^4 + 20*m^2 + 64)*cos(2*x) + 64)*cos(4*x
) - 4*(m^4 + 20*m^2 + 64)*cos(2*x) + 64)

________________________________________________________________________________________

Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-e^(m*x)/(cos(x)^2 - 1), x)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{m x}}{\sin ^{2}{\left (x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(exp(m*x)/sin(x)**2, x)

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(e^(m*x)/sin(x)^2, x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\mathrm {e}}^{m\,x}}{{\sin \left (x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(exp(m*x)/sin(x)^2, x)

________________________________________________________________________________________