3.66 \(\int \frac{\cot ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=65 \[ -\frac{\csc ^2(c+d x)}{2 a^2 d}+\frac{2 \csc (c+d x)}{a^2 d}+\frac{2 \log (\sin (c+d x))}{a^2 d}-\frac{2 \log (\sin (c+d x)+1)}{a^2 d} \]

[Out]

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

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Rubi [A]  time = 0.0593806, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2707, 77} \[ -\frac{\csc ^2(c+d x)}{2 a^2 d}+\frac{2 \csc (c+d x)}{a^2 d}+\frac{2 \log (\sin (c+d x))}{a^2 d}-\frac{2 \log (\sin (c+d x)+1)}{a^2 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^3/(a + a*Sin[c + d*x])^2,x]

[Out]

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

Rule 2707

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(a + x)^(m - (p + 1)/2))/(a - x)^((p + 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& EqQ[a^2 - b^2, 0] && IntegerQ[(p + 1)/2]

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

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

Mathematica [A]  time = 0.0701507, size = 49, normalized size = 0.75 \[ \frac{-\csc ^2(c+d x)+4 \csc (c+d x)+4 \log (\sin (c+d x))-4 \log (\sin (c+d x)+1)}{2 a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^3/(a + a*Sin[c + d*x])^2,x]

[Out]

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

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Maple [A]  time = 0.106, size = 66, normalized size = 1. \begin{align*} -2\,{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{{a}^{2}d}}-{\frac{1}{2\,{a}^{2}d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}+2\,{\frac{1}{{a}^{2}d\sin \left ( dx+c \right ) }}+2\,{\frac{\ln \left ( \sin \left ( dx+c \right ) \right ) }{{a}^{2}d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^3/(a+a*sin(d*x+c))^2,x)

[Out]

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

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Maxima [A]  time = 1.38376, size = 74, normalized size = 1.14 \begin{align*} -\frac{\frac{4 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2}} - \frac{4 \, \log \left (\sin \left (d x + c\right )\right )}{a^{2}} - \frac{4 \, \sin \left (d x + c\right ) - 1}{a^{2} \sin \left (d x + c\right )^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.54804, size = 204, normalized size = 3.14 \begin{align*} \frac{4 \,{\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) - 4 \,{\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 4 \, \sin \left (d x + c\right ) + 1}{2 \,{\left (a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\cot ^{3}{\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(cot(d*x+c)**3/(a+a*sin(d*x+c))**2,x)

[Out]

Integral(cot(c + d*x)**3/(sin(c + d*x)**2 + 2*sin(c + d*x) + 1), x)/a**2

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Giac [A]  time = 2.10221, size = 155, normalized size = 2.38 \begin{align*} -\frac{\frac{32 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac{16 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{2}} + \frac{a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 8 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{4}} + \frac{24 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 8 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1}{a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/8*(32*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 - 16*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 + (a^2*tan(1/2*d*x + 1
/2*c)^2 - 8*a^2*tan(1/2*d*x + 1/2*c))/a^4 + (24*tan(1/2*d*x + 1/2*c)^2 - 8*tan(1/2*d*x + 1/2*c) + 1)/(a^2*tan(
1/2*d*x + 1/2*c)^2))/d