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

Optimal. Leaf size=68 \[ -\frac{1}{d \left (a^2 \sin (c+d x)+a^2\right )}-\frac{\csc (c+d x)}{a^2 d}-\frac{2 \log (\sin (c+d x))}{a^2 d}+\frac{2 \log (\sin (c+d x)+1)}{a^2 d} \]

[Out]

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

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Rubi [A]  time = 0.0741999, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {2833, 12, 44} \[ -\frac{1}{d \left (a^2 \sin (c+d x)+a^2\right )}-\frac{\csc (c+d x)}{a^2 d}-\frac{2 \log (\sin (c+d x))}{a^2 d}+\frac{2 \log (\sin (c+d x)+1)}{a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2833

Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)
])^(n_.), x_Symbol] :> Dist[1/(b*f), Subst[Int[(a + x)^m*(c + (d*x)/b)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[
{a, b, c, d, e, f, m, n}, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{\cot (c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{a^2}{x^2 (a+x)^2} \, dx,x,a \sin (c+d x)\right )}{a d}\\ &=\frac{a \operatorname{Subst}\left (\int \frac{1}{x^2 (a+x)^2} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{a \operatorname{Subst}\left (\int \left (\frac{1}{a^2 x^2}-\frac{2}{a^3 x}+\frac{1}{a^2 (a+x)^2}+\frac{2}{a^3 (a+x)}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac{\csc (c+d x)}{a^2 d}-\frac{2 \log (\sin (c+d x))}{a^2 d}+\frac{2 \log (1+\sin (c+d x))}{a^2 d}-\frac{1}{d \left (a^2+a^2 \sin (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.153453, size = 45, normalized size = 0.66 \[ -\frac{\frac{1}{\sin (c+d x)+1}+\csc (c+d x)+2 \log (\sin (c+d x))-2 \log (\sin (c+d x)+1)}{a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.046, size = 68, normalized size = 1. \begin{align*} -{\frac{1}{d{a}^{2} \left ( 1+\sin \left ( dx+c \right ) \right ) }}+2\,{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{d{a}^{2}}}-{\frac{1}{d{a}^{2}\sin \left ( dx+c \right ) }}-2\,{\frac{\ln \left ( \sin \left ( dx+c \right ) \right ) }{d{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.09386, size = 92, normalized size = 1.35 \begin{align*} -\frac{\frac{2 \, \sin \left (d x + c\right ) + 1}{a^{2} \sin \left (d x + c\right )^{2} + a^{2} \sin \left (d x + c\right )} - \frac{2 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2}} + \frac{2 \, \log \left (\sin \left (d x + c\right )\right )}{a^{2}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.41018, size = 269, normalized size = 3.96 \begin{align*} -\frac{2 \,{\left (\cos \left (d x + c\right )^{2} - \sin \left (d x + c\right ) - 1\right )} \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) - 2 \,{\left (\cos \left (d x + c\right )^{2} - \sin \left (d x + c\right ) - 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 2 \, \sin \left (d x + c\right ) - 1}{a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d \sin \left (d x + c\right ) - a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*(cos(d*x + c)^2 - sin(d*x + c) - 1)*log(1/2*sin(d*x + c)) - 2*(cos(d*x + c)^2 - sin(d*x + c) - 1)*log(sin(
d*x + c) + 1) - 2*sin(d*x + c) - 1)/(a^2*d*cos(d*x + c)^2 - 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{\left (c + d x \right )} \csc ^{2}{\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)*csc(d*x+c)**2/(a+a*sin(d*x+c))**2,x)

[Out]

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

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Giac [A]  time = 1.28909, size = 93, normalized size = 1.37 \begin{align*} -\frac{\frac{2 \, \log \left ({\left | -\frac{a}{a \sin \left (d x + c\right ) + a} + 1 \right |}\right )}{a^{2}} + \frac{1}{{\left (a \sin \left (d x + c\right ) + a\right )} a} - \frac{1}{a^{2}{\left (\frac{a}{a \sin \left (d x + c\right ) + a} - 1\right )}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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