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

Optimal. Leaf size=139 \[ \frac {i e^{-i (a+b x)} \, _2F_1\left (1,-\frac {b}{2 d};1-\frac {b}{2 d};e^{2 i (c+d x)}\right )}{b}+\frac {i e^{i (a+b x)} \, _2F_1\left (1,\frac {b}{2 d};\frac {b}{2 d}+1;e^{2 i (c+d x)}\right )}{b}-\frac {i e^{-i (a+b x)}}{2 b}-\frac {i e^{i (a+b x)}}{2 b} \]

[Out]

-1/2*I/b/exp(I*(b*x+a))-1/2*I*exp(I*(b*x+a))/b+I*hypergeom([1, -1/2*b/d],[1-1/2*b/d],exp(2*I*(d*x+c)))/b/exp(I
*(b*x+a))+I*exp(I*(b*x+a))*hypergeom([1, 1/2*b/d],[1+1/2*b/d],exp(2*I*(d*x+c)))/b

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Rubi [A]  time = 0.11, antiderivative size = 139, normalized size of antiderivative = 1.00, 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 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])

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]

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*}

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Mathematica [A]  time = 3.62, size = 260, normalized size = 1.87 \[ \frac {-\frac {i e^{-i (a+b x-2 c)} \left (b e^{2 i d x} \, _2F_1\left (1,1-\frac {b}{2 d};2-\frac {b}{2 d};e^{2 i (c+d x)}\right )-(b-2 d) \, _2F_1\left (1,-\frac {b}{2 d};1-\frac {b}{2 d};e^{2 i (c+d x)}\right )\right )}{\left (-1+e^{2 i c}\right ) (b-2 d)}-\frac {i e^{i (a+b x+2 c)} \left (b e^{2 i d x} \, _2F_1\left (1,\frac {b}{2 d}+1;\frac {b}{2 d}+2;e^{2 i (c+d x)}\right )-(b+2 d) \, _2F_1\left (1,\frac {b}{2 d};\frac {b}{2 d}+1;e^{2 i (c+d x)}\right )\right )}{\left (-1+e^{2 i c}\right ) (b+2 d)}-\cos (a) \cot (c) \cos (b x)+\sin (a) \cot (c) \sin (b x)}{b} \]

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, -1/2*b/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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fricas [F]  time = 0.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\cot \left (d x + c\right ) \sin \left (b x + a\right ), x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \cot \left (d x + c\right ) \sin \left (b x + a\right )\,{d x} \]

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)

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maple [F]  time = 1.77, size = 0, normalized size = 0.00 \[ \int \cot \left (d x +c \right ) \sin \left (b x +a \right )\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \cot \left (d x + c\right ) \sin \left (b x + a\right )\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \mathrm {cot}\left (c+d\,x\right )\,\sin \left (a+b\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)*sin(a + b*x),x)

[Out]

int(cot(c + d*x)*sin(a + b*x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sin {\left (a + b x \right )} \cot {\left (c + d x \right )}\, dx \]

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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