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

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

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Rubi [A]  time = 0.02, 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 = {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*}

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Mathematica [A]  time = 0.20, size = 90, normalized size = 2.00 \[ \frac {e^{m x} \left (m e^{2 i x} \, _2F_1\left (1,1-\frac {i m}{2};2-\frac {i m}{2};e^{2 i x}\right )+(m+2 i) \left (\, _2F_1\left (1,-\frac {i m}{2};1-\frac {i m}{2};e^{2 i x}\right )-i \cot (x)\right )\right )}{-2+i m} \]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int e^{m x} \csc ^2(x) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

Could not integrate

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fricas [F]  time = 1.38, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {e^{\left (m x\right )}}{\cos \relax (x)^{2} - 1}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {e^{\left (m x\right )}}{\sin \relax (x)^{2}}\,{d x} \]

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)

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {{\mathrm e}^{m x}}{\sin \relax (x )^{2}}\, dx\]

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)

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\mathrm {e}}^{m\,x}}{{\sin \relax (x)}^2} \,d x \]

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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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {e^{m x}}{\sin ^{2}{\relax (x )}}\, dx \]

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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