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

Optimal. Leaf size=29 \[ \frac {\sin (a+b x)}{b}-\frac {\sin (a-c) \tanh ^{-1}(\cos (b x+c))}{b} \]

[Out]

-arctanh(cos(b*x+c))*sin(a-c)/b+sin(b*x+a)/b

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Rubi [A]  time = 0.02, antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {4578, 2637, 3770} \[ \frac {\sin (a+b x)}{b}-\frac {\sin (a-c) \tanh ^{-1}(\cos (b x+c))}{b} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + b*x]*Sin[a + b*x],x]

[Out]

-((ArcTanh[Cos[c + b*x]]*Sin[a - c])/b) + Sin[a + b*x]/b

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 4578

Int[Cot[w_]^(n_.)*Sin[v_], x_Symbol] :> Int[Cos[v]*Cot[w]^(n - 1), x] + Dist[Sin[v - w], Int[Csc[w]*Cot[w]^(n
- 1), x], x] /; GtQ[n, 0] && FreeQ[v - w, x] && NeQ[w, v]

Rubi steps

\begin {align*} \int \cot (c+b x) \sin (a+b x) \, dx &=\sin (a-c) \int \csc (c+b x) \, dx+\int \cos (a+b x) \, dx\\ &=-\frac {\tanh ^{-1}(\cos (c+b x)) \sin (a-c)}{b}+\frac {\sin (a+b x)}{b}\\ \end {align*}

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Mathematica [C]  time = 0.06, size = 93, normalized size = 3.21 \[ -\frac {2 i \sin (a-c) \tan ^{-1}\left (\frac {(\cos (c)-i \sin (c)) \left (\cos (c) \cos \left (\frac {b x}{2}\right )-\sin (c) \sin \left (\frac {b x}{2}\right )\right )}{\sin (c) \cos \left (\frac {b x}{2}\right )+i \cos (c) \cos \left (\frac {b x}{2}\right )}\right )}{b}+\frac {\sin (a) \cos (b x)}{b}+\frac {\cos (a) \sin (b x)}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[b*x]*Sin[a])/b - ((2*I)*ArcTan[((Cos[c] - I*Sin[c])*(Cos[c]*Cos[(b*x)/2] - Sin[c]*Sin[(b*x)/2]))/(I*Cos[c
]*Cos[(b*x)/2] + Cos[(b*x)/2]*Sin[c])]*Sin[a - c])/b + (Cos[a]*Sin[b*x])/b

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fricas [B]  time = 0.45, size = 197, normalized size = 6.79 \[ \frac {\frac {\sqrt {2} \log \left (\frac {2 \, \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) + \frac {2 \, \sqrt {2} {\left ({\left (\cos \left (-2 \, a + 2 \, c\right ) + 1\right )} \cos \left (b x + a\right ) - \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right )\right )}}{\sqrt {\cos \left (-2 \, a + 2 \, c\right ) + 1}} - \cos \left (-2 \, a + 2 \, c\right ) + 3}{2 \, \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) - \cos \left (-2 \, a + 2 \, c\right ) - 1}\right ) \sin \left (-2 \, a + 2 \, c\right )}{\sqrt {\cos \left (-2 \, a + 2 \, c\right ) + 1}} + 4 \, \sin \left (b x + a\right )}{4 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(sqrt(2)*log((2*cos(b*x + a)^2*cos(-2*a + 2*c) - 2*cos(b*x + a)*sin(b*x + a)*sin(-2*a + 2*c) + 2*sqrt(2)*(
(cos(-2*a + 2*c) + 1)*cos(b*x + a) - sin(b*x + a)*sin(-2*a + 2*c))/sqrt(cos(-2*a + 2*c) + 1) - cos(-2*a + 2*c)
+ 3)/(2*cos(b*x + a)^2*cos(-2*a + 2*c) - 2*cos(b*x + a)*sin(b*x + a)*sin(-2*a + 2*c) - cos(-2*a + 2*c) - 1))*
sin(-2*a + 2*c)/sqrt(cos(-2*a + 2*c) + 1) + 4*sin(b*x + a))/b

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)2/b*((-tan(a/2)^2*tan(c/2)+tan(a/2)*tan(c/2)^2-tan(a/2)+tan(c/
2))/(-tan(a/2)^2*tan(c/2)^2-tan(a/2)^2-tan(c/2)^2-1)*ln(abs(tan(b*x/2)+tan(c/2)))+(-tan(a/2)^2*tan(c/2)^2+tan(
a/2)*tan(c/2)^3-tan(a/2)*tan(c/2)+tan(c/2)^2)/(tan(a/2)^2*tan(c/2)^3+tan(a/2)^2*tan(c/2)+tan(c/2)^3+tan(c/2))*
ln(abs(tan(b*x/2)*tan(c/2)-1))+(-tan(b*x/2)*tan(a/2)^2+tan(b*x/2)+2*tan(a/2))/(tan(a/2)^2+1)/(tan(b*x/2)^2+1))

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maple [C]  time = 0.80, size = 95, normalized size = 3.28 \[ -\frac {i {\mathrm e}^{i \left (b x +a \right )}}{2 b}+\frac {i {\mathrm e}^{-i \left (b x +a \right )}}{2 b}-\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}+{\mathrm e}^{i \left (a -c \right )}\right ) \sin \left (a -c \right )}{b}+\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}-{\mathrm e}^{i \left (a -c \right )}\right ) \sin \left (a -c \right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*I*exp(I*(b*x+a))/b+1/2*I/b*exp(-I*(b*x+a))-ln(exp(I*(b*x+a))+exp(I*(a-c)))/b*sin(a-c)+ln(exp(I*(b*x+a))-e
xp(I*(a-c)))/b*sin(a-c)

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maxima [B]  time = 0.35, size = 105, normalized size = 3.62 \[ \frac {\log \left (\cos \left (b x\right )^{2} + 2 \, \cos \left (b x\right ) \cos \relax (c) + \cos \relax (c)^{2} + \sin \left (b x\right )^{2} - 2 \, \sin \left (b x\right ) \sin \relax (c) + \sin \relax (c)^{2}\right ) \sin \left (-a + c\right ) - \log \left (\cos \left (b x\right )^{2} - 2 \, \cos \left (b x\right ) \cos \relax (c) + \cos \relax (c)^{2} + \sin \left (b x\right )^{2} + 2 \, \sin \left (b x\right ) \sin \relax (c) + \sin \relax (c)^{2}\right ) \sin \left (-a + c\right ) + 2 \, \sin \left (b x + a\right )}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(log(cos(b*x)^2 + 2*cos(b*x)*cos(c) + cos(c)^2 + sin(b*x)^2 - 2*sin(b*x)*sin(c) + sin(c)^2)*sin(-a + c) -
log(cos(b*x)^2 - 2*cos(b*x)*cos(c) + cos(c)^2 + sin(b*x)^2 + 2*sin(b*x)*sin(c) + sin(c)^2)*sin(-a + c) + 2*sin
(b*x + a))/b

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mupad [B]  time = 4.85, size = 233, normalized size = 8.03 \[ \frac {{\mathrm {e}}^{-a\,1{}\mathrm {i}-b\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2\,b}-\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2\,b}-\frac {\ln \left (-{\mathrm {e}}^{a\,1{}\mathrm {i}}\,{\mathrm {e}}^{b\,x\,1{}\mathrm {i}}\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\mathrm {e}}^{-c\,2{}\mathrm {i}}-1\right )-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\mathrm {e}}^{-c\,2{}\mathrm {i}}\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\mathrm {e}}^{-c\,2{}\mathrm {i}}-1\right )\,1{}\mathrm {i}}{\sqrt {-{\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\mathrm {e}}^{-c\,2{}\mathrm {i}}}}\right )\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}-c\,2{}\mathrm {i}}-1\right )}{2\,b\,\sqrt {-{\mathrm {e}}^{a\,2{}\mathrm {i}-c\,2{}\mathrm {i}}}}+\frac {\ln \left (-{\mathrm {e}}^{a\,1{}\mathrm {i}}\,{\mathrm {e}}^{b\,x\,1{}\mathrm {i}}\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\mathrm {e}}^{-c\,2{}\mathrm {i}}-1\right )+\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\mathrm {e}}^{-c\,2{}\mathrm {i}}\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\mathrm {e}}^{-c\,2{}\mathrm {i}}-1\right )\,1{}\mathrm {i}}{\sqrt {-{\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\mathrm {e}}^{-c\,2{}\mathrm {i}}}}\right )\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}-c\,2{}\mathrm {i}}-1\right )}{2\,b\,\sqrt {-{\mathrm {e}}^{a\,2{}\mathrm {i}-c\,2{}\mathrm {i}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + b*x)*sin(a + b*x),x)

[Out]

(exp(- a*1i - b*x*1i)*1i)/(2*b) - (exp(a*1i + b*x*1i)*1i)/(2*b) - (log(- exp(a*1i)*exp(b*x*1i)*(exp(a*2i)*exp(
-c*2i) - 1) - (exp(a*2i)*exp(-c*2i)*(exp(a*2i)*exp(-c*2i) - 1)*1i)/(-exp(a*2i)*exp(-c*2i))^(1/2))*(exp(a*2i -
c*2i) - 1))/(2*b*(-exp(a*2i - c*2i))^(1/2)) + (log((exp(a*2i)*exp(-c*2i)*(exp(a*2i)*exp(-c*2i) - 1)*1i)/(-exp(
a*2i)*exp(-c*2i))^(1/2) - exp(a*1i)*exp(b*x*1i)*(exp(a*2i)*exp(-c*2i) - 1))*(exp(a*2i - c*2i) - 1))/(2*b*(-exp
(a*2i - c*2i))^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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