\(\int \frac {\cos (c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^2} \, dx\) [311]

   Optimal result
   Rubi [A] (verified)
   Mathematica [B] (verified)
   Maple [A] (verified)
   Fricas [B] (verification not implemented)
   Sympy [F]
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 25, antiderivative size = 40 \[ \int \frac {\cos (c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\text {arctanh}(\cos (c+d x))}{a^2 d}+\frac {2 \cos (c+d x)}{a^2 d (1+\sin (c+d x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2953, 3045, 3855, 2727} \[ \int \frac {\cos (c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {2 \cos (c+d x)}{a^2 d (\sin (c+d x)+1)}-\frac {\text {arctanh}(\cos (c+d x))}{a^2 d} \]

[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 2727

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]

Rule 2953

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 3045

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 3855

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

Rubi steps \begin{align*} \text {integral}& = \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 {\text {arctanh}(\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] (verified)

Leaf count is larger than twice the leaf count of optimal. \(115\) vs. \(2(40)=80\).

Time = 0.48 (sec) , antiderivative size = 115, normalized size of antiderivative = 2.88 \[ \int \frac {\cos (c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3 \left (\cos \left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+\left (4+\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right ) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{a^2 d (1+\sin (c+d x))^2} \]

[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))

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.85

method result size
derivativedivides \(\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {4}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{d \,a^{2}}\) \(34\)
default \(\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {4}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{d \,a^{2}}\) \(34\)
parallelrisch \(\frac {\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}\) \(55\)
risch \(\frac {4}{d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d \,a^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \,a^{2}}\) \(63\)
norman \(\frac {\frac {4}{a d}+\frac {4 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}+\frac {8 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {8 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}}{a \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}}\) \(134\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 103 vs. \(2 (40) = 80\).

Time = 0.27 (sec) , antiderivative size = 103, normalized size of antiderivative = 2.58 \[ \int \frac {\cos (c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {{\left (\cos \left (d x + c\right ) + \sin \left (d x + c\right ) + 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left (\cos \left (d x + c\right ) + \sin \left (d x + c\right ) + 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 4 \, \cos \left (d x + c\right ) + 4 \, \sin \left (d x + c\right ) - 4}{2 \, {\left (a^{2} d \cos \left (d x + c\right ) + a^{2} d \sin \left (d x + c\right ) + a^{2} d\right )}} \]

[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
)

Sympy [F]

\[ \int \frac {\cos (c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\int \frac {\cos ^{2}{\left (c + d x \right )} \csc {\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.38 \[ \int \frac {\cos (c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {4}{a^{2} + \frac {a^{2} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}} + \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{d} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.95 \[ \int \frac {\cos (c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {\log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} + \frac {4}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}}}{d} \]

[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

Mupad [B] (verification not implemented)

Time = 9.69 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.98 \[ \int \frac {\cos (c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^2\,d}+\frac {4}{a^2\,d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )} \]

[In]

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

[Out]

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