3.311 \(\int \frac{\cos (c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=40 \[ \frac{2 \cos (c+d x)}{a^2 d (\sin (c+d x)+1)}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^2 d} \]

[Out]

-(ArcTanh[Cos[c + d*x]]/(a^2*d)) + (2*Cos[c + d*x])/(a^2*d*(1 + Sin[c + d*x]))

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Rubi [A]  time = 0.157168, antiderivative size = 40, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {2874, 2966, 3770, 2648} \[ \frac{2 \cos (c+d x)}{a^2 d (\sin (c+d x)+1)}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^2 d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]*Cot[c + d*x])/(a + a*Sin[c + d*x])^2,x]

[Out]

-(ArcTanh[Cos[c + d*x]]/(a^2*d)) + (2*Cos[c + d*x])/(a^2*d*(1 + Sin[c + d*x]))

Rule 2874

Int[cos[(e_.) + (f_.)*(x_)]^2*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Dist[1/b^2, Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x])^(m + 1)*(a - b*Sin[e + f*x]), x], x] /;
FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && (ILtQ[m, 0] ||  !IGtQ[n, 0])

Rule 2966

Int[sin[(e_.) + (f_.)*(x_)]^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.
)*(x_)]), x_Symbol] :> Int[ExpandTrig[sin[e + f*x]^n*(a + b*sin[e + f*x])^m*(A + B*sin[e + f*x]), x], x] /; Fr
eeQ[{a, b, e, f, A, B}, x] && EqQ[A*b + a*B, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && IntegerQ[n]

Rule 3770

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

Rule 2648

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

Rubi steps

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

Mathematica [B]  time = 0.148496, size = 115, normalized size = 2.88 \[ -\frac{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3 \left (\cos \left (\frac{1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )\right )+\sin \left (\frac{1}{2} (c+d x)\right ) \left (-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+\log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+4\right )\right )}{a^2 d (\sin (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]*Cot[c + d*x])/(a + a*Sin[c + d*x])^2,x]

[Out]

-(((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3*(Cos[(c + d*x)/2]*(Log[Cos[(c + d*x)/2]] - Log[Sin[(c + d*x)/2]]) +
(4 + Log[Cos[(c + d*x)/2]] - Log[Sin[(c + d*x)/2]])*Sin[(c + d*x)/2]))/(a^2*d*(1 + Sin[c + d*x])^2))

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Maple [A]  time = 0.128, size = 40, normalized size = 1. \begin{align*} 4\,{\frac{1}{d{a}^{2} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }}+{\frac{1}{d{a}^{2}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/d/a^2/(tan(1/2*d*x+1/2*c)+1)+1/d/a^2*ln(tan(1/2*d*x+1/2*c))

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Maxima [A]  time = 1.11075, size = 74, normalized size = 1.85 \begin{align*} \frac{\frac{4}{a^{2} + \frac{a^{2} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}} + \frac{\log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{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+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.62081, size = 301, normalized size = 7.52 \begin{align*} -\frac{{\left (\cos \left (d x + c\right ) + \sin \left (d x + c\right ) + 1\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) -{\left (\cos \left (d x + c\right ) + \sin \left (d x + c\right ) + 1\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 4 \, \cos \left (d x + c\right ) + 4 \, \sin \left (d x + c\right ) - 4}{2 \,{\left (a^{2} d \cos \left (d x + c\right ) + a^{2} d \sin \left (d x + c\right ) + a^{2} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\cos ^{2}{\left (c + d x \right )} \csc{\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin{\left (c + d x \right )} + 1}\, dx}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(cos(c + d*x)**2*csc(c + d*x)/(sin(c + d*x)**2 + 2*sin(c + d*x) + 1), x)/a**2

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Giac [A]  time = 1.34909, size = 51, normalized size = 1.27 \begin{align*} \frac{\frac{\log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{2}} + \frac{4}{a^{2}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}}}{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+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

(log(abs(tan(1/2*d*x + 1/2*c)))/a^2 + 4/(a^2*(tan(1/2*d*x + 1/2*c) + 1)))/d