3.237 \(\int \cot (c+d x) \sin (a+b x) \, dx\)

Optimal. Leaf size=139 \[ \frac{i e^{-i (a+b x)} \text{Hypergeometric2F1}\left (1,-\frac{b}{2 d},1-\frac{b}{2 d},e^{2 i (c+d x)}\right )}{b}+\frac{i e^{i (a+b x)} \text{Hypergeometric2F1}\left (1,\frac{b}{2 d},\frac{b}{2 d}+1,e^{2 i (c+d x)}\right )}{b}-\frac{i e^{-i (a+b x)}}{2 b}-\frac{i e^{i (a+b x)}}{2 b} \]

[Out]

(-I/2)/(b*E^(I*(a + b*x))) - ((I/2)*E^(I*(a + b*x)))/b + (I*Hypergeometric2F1[1, -b/(2*d), 1 - b/(2*d), E^((2*
I)*(c + d*x))])/(b*E^(I*(a + b*x))) + (I*E^(I*(a + b*x))*Hypergeometric2F1[1, b/(2*d), 1 + b/(2*d), E^((2*I)*(
c + d*x))])/b

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Rubi [A]  time = 0.110764, antiderivative size = 139, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {4559, 2194, 2251} \[ \frac{i e^{-i (a+b x)} \, _2F_1\left (1,-\frac{b}{2 d};1-\frac{b}{2 d};e^{2 i (c+d x)}\right )}{b}+\frac{i e^{i (a+b x)} \, _2F_1\left (1,\frac{b}{2 d};\frac{b}{2 d}+1;e^{2 i (c+d x)}\right )}{b}-\frac{i e^{-i (a+b x)}}{2 b}-\frac{i e^{i (a+b x)}}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]*Sin[a + b*x],x]

[Out]

(-I/2)/(b*E^(I*(a + b*x))) - ((I/2)*E^(I*(a + b*x)))/b + (I*Hypergeometric2F1[1, -b/(2*d), 1 - b/(2*d), E^((2*
I)*(c + d*x))])/(b*E^(I*(a + b*x))) + (I*E^(I*(a + b*x))*Hypergeometric2F1[1, b/(2*d), 1 + b/(2*d), E^((2*I)*(
c + d*x))])/b

Rule 4559

Int[Cot[(c_.) + (d_.)*(x_)]*Sin[(a_.) + (b_.)*(x_)], x_Symbol] :> Int[-(1/(E^(I*(a + b*x))*2)) + E^(I*(a + b*x
))/2 + 1/(E^(I*(a + b*x))*(1 - E^(2*I*(c + d*x)))) - E^(I*(a + b*x))/(1 - E^(2*I*(c + d*x))), x] /; FreeQ[{a,
b, c, d}, x] && NeQ[b^2 - d^2, 0]

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2251

Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_)*(G_)^((h_.)*((f_.) + (g_.)*(x_))), x_Symbol] :> Simp
[(a^p*G^(h*(f + g*x))*Hypergeometric2F1[-p, (g*h*Log[G])/(d*e*Log[F]), (g*h*Log[G])/(d*e*Log[F]) + 1, Simplify
[-((b*F^(e*(c + d*x)))/a)]])/(g*h*Log[G]), x] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, p}, x] && (ILtQ[p, 0] ||
GtQ[a, 0])

Rubi steps

\begin{align*} \int \cot (c+d x) \sin (a+b x) \, dx &=\int \left (-\frac{1}{2} e^{-i (a+b x)}+\frac{1}{2} e^{i (a+b x)}+\frac{e^{-i (a+b x)}}{1-e^{2 i (c+d x)}}-\frac{e^{i (a+b x)}}{1-e^{2 i (c+d x)}}\right ) \, dx\\ &=-\left (\frac{1}{2} \int e^{-i (a+b x)} \, dx\right )+\frac{1}{2} \int e^{i (a+b x)} \, dx+\int \frac{e^{-i (a+b x)}}{1-e^{2 i (c+d x)}} \, dx-\int \frac{e^{i (a+b x)}}{1-e^{2 i (c+d x)}} \, dx\\ &=-\frac{i e^{-i (a+b x)}}{2 b}-\frac{i e^{i (a+b x)}}{2 b}+\frac{i e^{-i (a+b x)} \, _2F_1\left (1,-\frac{b}{2 d};1-\frac{b}{2 d};e^{2 i (c+d x)}\right )}{b}+\frac{i e^{i (a+b x)} \, _2F_1\left (1,\frac{b}{2 d};1+\frac{b}{2 d};e^{2 i (c+d x)}\right )}{b}\\ \end{align*}

Mathematica [A]  time = 3.59495, size = 260, normalized size = 1.87 \[ \frac{-\frac{i e^{-i (a+b x-2 c)} \left (b e^{2 i d x} \text{Hypergeometric2F1}\left (1,1-\frac{b}{2 d},2-\frac{b}{2 d},e^{2 i (c+d x)}\right )-(b-2 d) \text{Hypergeometric2F1}\left (1,-\frac{b}{2 d},1-\frac{b}{2 d},e^{2 i (c+d x)}\right )\right )}{\left (-1+e^{2 i c}\right ) (b-2 d)}-\frac{i e^{i (a+b x+2 c)} \left (b e^{2 i d x} \text{Hypergeometric2F1}\left (1,\frac{b}{2 d}+1,\frac{b}{2 d}+2,e^{2 i (c+d x)}\right )-(b+2 d) \text{Hypergeometric2F1}\left (1,\frac{b}{2 d},\frac{b}{2 d}+1,e^{2 i (c+d x)}\right )\right )}{\left (-1+e^{2 i c}\right ) (b+2 d)}-\cos (a) \cot (c) \cos (b x)+\sin (a) \cot (c) \sin (b x)}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]*Sin[a + b*x],x]

[Out]

(-(Cos[a]*Cos[b*x]*Cot[c]) - (I*(b*E^((2*I)*d*x)*Hypergeometric2F1[1, 1 - b/(2*d), 2 - b/(2*d), E^((2*I)*(c +
d*x))] - (b - 2*d)*Hypergeometric2F1[1, -b/(2*d), 1 - b/(2*d), E^((2*I)*(c + d*x))]))/((b - 2*d)*E^(I*(a - 2*c
+ b*x))*(-1 + E^((2*I)*c))) - (I*E^(I*(a + 2*c + b*x))*(b*E^((2*I)*d*x)*Hypergeometric2F1[1, 1 + b/(2*d), 2 +
b/(2*d), E^((2*I)*(c + d*x))] - (b + 2*d)*Hypergeometric2F1[1, b/(2*d), 1 + b/(2*d), E^((2*I)*(c + d*x))]))/(
(b + 2*d)*(-1 + E^((2*I)*c))) + Cot[c]*Sin[a]*Sin[b*x])/b

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Maple [F]  time = 0.208, size = 0, normalized size = 0. \begin{align*} \int \cot \left ( dx+c \right ) \sin \left ( bx+a \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)*sin(b*x+a),x)

[Out]

int(cot(d*x+c)*sin(b*x+a),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \cot \left (d x + c\right ) \sin \left (b x + a\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)*sin(b*x+a),x, algorithm="maxima")

[Out]

integrate(cot(d*x + c)*sin(b*x + a), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\cot \left (d x + c\right ) \sin \left (b x + a\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)*sin(b*x+a),x, algorithm="fricas")

[Out]

integral(cot(d*x + c)*sin(b*x + a), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sin{\left (a + b x \right )} \cot{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)*sin(b*x+a),x)

[Out]

Integral(sin(a + b*x)*cot(c + d*x), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \cot \left (d x + c\right ) \sin \left (b x + a\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)*sin(b*x+a),x, algorithm="giac")

[Out]

integrate(cot(d*x + c)*sin(b*x + a), x)