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

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

[Out]

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

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Rubi [A]  time = 0.0180852, antiderivative size = 45, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {4453} \[ -\frac{4 e^{(m+2 i) x} \text{Hypergeometric2F1}\left (2,1-\frac{i m}{2},2-\frac{i m}{2},e^{2 i x}\right )}{m+2 i} \]

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 4453

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 = 0.174203, size = 90, normalized size = 2. \[ \frac{e^{m x} \left (m e^{2 i x} \, _2F_1\left (1,1-\frac{i m}{2};2-\frac{i m}{2};e^{2 i x}\right )+(m+2 i) \left (\, _2F_1\left (1,-\frac{i m}{2};1-\frac{i m}{2};e^{2 i x}\right )-i \cot (x)\right )\right )}{-2+i m} \]

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

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Maple [F]  time = 0.083, size = 0, normalized size = 0. \begin{align*} \int{\frac{{{\rm e}^{mx}}}{ \left ( \sin \left ( x \right ) \right ) ^{2}}}\, dx \end{align*}

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)

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{e^{\left (m x\right )}}{\cos \left (x\right )^{2} - 1}, x\right ) \end{align*}

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)

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

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)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{e^{\left (m x\right )}}{\sin \left (x\right )^{2}}\,{d x} \end{align*}

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)