3.1053 \(\int \cos (c+d x) \cot (c+d x) (a+b \sin (c+d x)) \, dx\)

Optimal. Leaf size=51 \[ \frac{a \cos (c+d x)}{d}-\frac{a \tanh ^{-1}(\cos (c+d x))}{d}+\frac{b \sin (c+d x) \cos (c+d x)}{2 d}+\frac{b x}{2} \]

[Out]

(b*x)/2 - (a*ArcTanh[Cos[c + d*x]])/d + (a*Cos[c + d*x])/d + (b*Cos[c + d*x]*Sin[c + d*x])/(2*d)

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Rubi [A]  time = 0.0694966, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {2838, 2592, 321, 206, 2635, 8} \[ \frac{a \cos (c+d x)}{d}-\frac{a \tanh ^{-1}(\cos (c+d x))}{d}+\frac{b \sin (c+d x) \cos (c+d x)}{2 d}+\frac{b x}{2} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]*Cot[c + d*x]*(a + b*Sin[c + d*x]),x]

[Out]

(b*x)/2 - (a*ArcTanh[Cos[c + d*x]])/d + (a*Cos[c + d*x])/d + (b*Cos[c + d*x]*Sin[c + d*x])/(2*d)

Rule 2838

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)
*(x_)]), x_Symbol] :> Dist[a, Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Dist[b/d, Int[(g*Cos[e + f*x
])^p*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x]

Rule 2592

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, (a*Sin[e + f*x])/ff
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \cos (c+d x) \cot (c+d x) (a+b \sin (c+d x)) \, dx &=a \int \cos (c+d x) \cot (c+d x) \, dx+b \int \cos ^2(c+d x) \, dx\\ &=\frac{b \cos (c+d x) \sin (c+d x)}{2 d}+\frac{1}{2} b \int 1 \, dx-\frac{a \operatorname{Subst}\left (\int \frac{x^2}{1-x^2} \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac{b x}{2}+\frac{a \cos (c+d x)}{d}+\frac{b \cos (c+d x) \sin (c+d x)}{2 d}-\frac{a \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac{b x}{2}-\frac{a \tanh ^{-1}(\cos (c+d x))}{d}+\frac{a \cos (c+d x)}{d}+\frac{b \cos (c+d x) \sin (c+d x)}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.0541021, size = 74, normalized size = 1.45 \[ \frac{a \cos (c+d x)}{d}+\frac{a \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{d}-\frac{a \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{d}+\frac{b (c+d x)}{2 d}+\frac{b \sin (2 (c+d x))}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(c + d*x))/(2*d) + (a*Cos[c + d*x])/d - (a*Log[Cos[(c + d*x)/2]])/d + (a*Log[Sin[(c + d*x)/2]])/d + (b*Sin[
2*(c + d*x)])/(4*d)

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Maple [A]  time = 0.048, size = 63, normalized size = 1.2 \begin{align*}{\frac{\cos \left ( dx+c \right ) a}{d}}+{\frac{a\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{d}}+{\frac{b\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{bx}{2}}+{\frac{cb}{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*csc(d*x+c)*(a+b*sin(d*x+c)),x)

[Out]

a*cos(d*x+c)/d+1/d*a*ln(csc(d*x+c)-cot(d*x+c))+1/2*b*cos(d*x+c)*sin(d*x+c)/d+1/2*b*x+1/2*b*c/d

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Maxima [A]  time = 1.03717, size = 77, normalized size = 1.51 \begin{align*} \frac{{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} b + 2 \, a{\left (2 \, \cos \left (d x + c\right ) - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)*(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.47206, size = 174, normalized size = 3.41 \begin{align*} \frac{b d x + b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 2 \, a \cos \left (d x + c\right ) - a \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + a \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)*(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \sin{\left (c + d x \right )}\right ) \cos ^{2}{\left (c + d x \right )} \csc{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*csc(d*x+c)*(a+b*sin(d*x+c)),x)

[Out]

Integral((a + b*sin(c + d*x))*cos(c + d*x)**2*csc(c + d*x), x)

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Giac [A]  time = 1.32054, size = 117, normalized size = 2.29 \begin{align*} \frac{{\left (d x + c\right )} b + 2 \, a \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) - \frac{2 \,{\left (b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, a\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)*(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

1/2*((d*x + c)*b + 2*a*log(abs(tan(1/2*d*x + 1/2*c))) - 2*(b*tan(1/2*d*x + 1/2*c)^3 - 2*a*tan(1/2*d*x + 1/2*c)
^2 - b*tan(1/2*d*x + 1/2*c) - 2*a)/(tan(1/2*d*x + 1/2*c)^2 + 1)^2)/d