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

Optimal. Leaf size=86 \[ \frac{2}{d \left (a^3 \sin (c+d x)+a^3\right )}-\frac{\csc ^2(c+d x)}{2 a^3 d}+\frac{3 \csc (c+d x)}{a^3 d}+\frac{5 \log (\sin (c+d x))}{a^3 d}-\frac{5 \log (\sin (c+d x)+1)}{a^3 d} \]

[Out]

(3*Csc[c + d*x])/(a^3*d) - Csc[c + d*x]^2/(2*a^3*d) + (5*Log[Sin[c + d*x]])/(a^3*d) - (5*Log[1 + Sin[c + d*x]]
)/(a^3*d) + 2/(d*(a^3 + a^3*Sin[c + d*x]))

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Rubi [A]  time = 0.0697399, antiderivative size = 86, 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} \[ \frac{2}{d \left (a^3 \sin (c+d x)+a^3\right )}-\frac{\csc ^2(c+d x)}{2 a^3 d}+\frac{3 \csc (c+d x)}{a^3 d}+\frac{5 \log (\sin (c+d x))}{a^3 d}-\frac{5 \log (\sin (c+d x)+1)}{a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*Csc[c + d*x])/(a^3*d) - Csc[c + d*x]^2/(2*a^3*d) + (5*Log[Sin[c + d*x]])/(a^3*d) - (5*Log[1 + Sin[c + d*x]]
)/(a^3*d) + 2/(d*(a^3 + a^3*Sin[c + d*x]))

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))^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{a-x}{x^3 (a+x)^2} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{1}{a x^3}-\frac{3}{a^2 x^2}+\frac{5}{a^3 x}-\frac{2}{a^2 (a+x)^2}-\frac{5}{a^3 (a+x)}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{3 \csc (c+d x)}{a^3 d}-\frac{\csc ^2(c+d x)}{2 a^3 d}+\frac{5 \log (\sin (c+d x))}{a^3 d}-\frac{5 \log (1+\sin (c+d x))}{a^3 d}+\frac{2}{d \left (a^3+a^3 \sin (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.19094, size = 61, normalized size = 0.71 \[ \frac{\frac{4}{\sin (c+d x)+1}-\csc ^2(c+d x)+6 \csc (c+d x)+10 \log (\sin (c+d x))-10 \log (\sin (c+d x)+1)}{2 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*Csc[c + d*x] - Csc[c + d*x]^2 + 10*Log[Sin[c + d*x]] - 10*Log[1 + Sin[c + d*x]] + 4/(1 + Sin[c + d*x]))/(2*
a^3*d)

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Maple [A]  time = 0.129, size = 84, normalized size = 1. \begin{align*} 2\,{\frac{1}{d{a}^{3} \left ( 1+\sin \left ( dx+c \right ) \right ) }}-5\,{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{d{a}^{3}}}-{\frac{1}{2\,d{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}+3\,{\frac{1}{d{a}^{3}\sin \left ( dx+c \right ) }}+5\,{\frac{\ln \left ( \sin \left ( dx+c \right ) \right ) }{d{a}^{3}}} \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))^3,x)

[Out]

2/d/a^3/(1+sin(d*x+c))-5*ln(1+sin(d*x+c))/a^3/d-1/2/d/a^3/sin(d*x+c)^2+3/d/a^3/sin(d*x+c)+5*ln(sin(d*x+c))/a^3
/d

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Maxima [A]  time = 1.95102, size = 108, normalized size = 1.26 \begin{align*} \frac{\frac{10 \, \sin \left (d x + c\right )^{2} + 5 \, \sin \left (d x + c\right ) - 1}{a^{3} \sin \left (d x + c\right )^{3} + a^{3} \sin \left (d x + c\right )^{2}} - \frac{10 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}} + \frac{10 \, \log \left (\sin \left (d x + c\right )\right )}{a^{3}}}{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))^3,x, algorithm="maxima")

[Out]

1/2*((10*sin(d*x + c)^2 + 5*sin(d*x + c) - 1)/(a^3*sin(d*x + c)^3 + a^3*sin(d*x + c)^2) - 10*log(sin(d*x + c)
+ 1)/a^3 + 10*log(sin(d*x + c))/a^3)/d

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Fricas [A]  time = 1.54676, size = 393, normalized size = 4.57 \begin{align*} \frac{10 \, \cos \left (d x + c\right )^{2} + 10 \,{\left (\cos \left (d x + c\right )^{2} +{\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - 1\right )} \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) - 10 \,{\left (\cos \left (d x + c\right )^{2} +{\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 5 \, \sin \left (d x + c\right ) - 9}{2 \,{\left (a^{3} d \cos \left (d x + c\right )^{2} - a^{3} d +{\left (a^{3} d \cos \left (d x + c\right )^{2} - a^{3} d\right )} \sin \left (d x + c\right )\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))^3,x, algorithm="fricas")

[Out]

1/2*(10*cos(d*x + c)^2 + 10*(cos(d*x + c)^2 + (cos(d*x + c)^2 - 1)*sin(d*x + c) - 1)*log(1/2*sin(d*x + c)) - 1
0*(cos(d*x + c)^2 + (cos(d*x + c)^2 - 1)*sin(d*x + c) - 1)*log(sin(d*x + c) + 1) - 5*sin(d*x + c) - 9)/(a^3*d*
cos(d*x + c)^2 - a^3*d + (a^3*d*cos(d*x + c)^2 - a^3*d)*sin(d*x + c))

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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 ^{3}{\left (c + d x \right )} + 3 \sin ^{2}{\left (c + d x \right )} + 3 \sin{\left (c + d x \right )} + 1}\, dx}{a^{3}} \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))**3,x)

[Out]

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

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Giac [A]  time = 1.73369, size = 208, normalized size = 2.42 \begin{align*} -\frac{\frac{80 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac{40 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{3}} - \frac{30 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 40 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 53 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 10 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}^{2} a^{3}} + \frac{a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 12 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{6}}}{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))^3,x, algorithm="giac")

[Out]

-1/8*(80*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 40*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 - (30*tan(1/2*d*x + 1/
2*c)^4 + 40*tan(1/2*d*x + 1/2*c)^3 + 53*tan(1/2*d*x + 1/2*c)^2 + 10*tan(1/2*d*x + 1/2*c) - 1)/((tan(1/2*d*x +
1/2*c)^2 + tan(1/2*d*x + 1/2*c))^2*a^3) + (a^3*tan(1/2*d*x + 1/2*c)^2 - 12*a^3*tan(1/2*d*x + 1/2*c))/a^6)/d