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

Optimal. Leaf size=65 \[ -\frac{\csc ^2(c+d x)}{2 a^3 d}+\frac{3 \csc (c+d x)}{a^3 d}+\frac{4 \log (\sin (c+d x))}{a^3 d}-\frac{4 \log (\sin (c+d x)+1)}{a^3 d} \]

[Out]

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

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Rubi [A]  time = 0.113156, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {2836, 12, 88} \[ -\frac{\csc ^2(c+d x)}{2 a^3 d}+\frac{3 \csc (c+d x)}{a^3 d}+\frac{4 \log (\sin (c+d x))}{a^3 d}-\frac{4 \log (\sin (c+d x)+1)}{a^3 d} \]

Antiderivative was successfully verified.

[In]

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

Rule 2836

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

Rule 12

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

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

\begin{align*} \int \frac{\cos ^2(c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{a^3 (a-x)^2}{x^3 (a+x)} \, dx,x,a \sin (c+d x)\right )}{a^5 d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{(a-x)^2}{x^3 (a+x)} \, dx,x,a \sin (c+d x)\right )}{a^2 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{a}{x^3}-\frac{3}{x^2}+\frac{4}{a x}-\frac{4}{a (a+x)}\right ) \, dx,x,a \sin (c+d x)\right )}{a^2 d}\\ &=\frac{3 \csc (c+d x)}{a^3 d}-\frac{\csc ^2(c+d x)}{2 a^3 d}+\frac{4 \log (\sin (c+d x))}{a^3 d}-\frac{4 \log (1+\sin (c+d x))}{a^3 d}\\ \end{align*}

Mathematica [A]  time = 0.0819546, size = 49, normalized size = 0.75 \[ \frac{-\csc ^2(c+d x)+6 \csc (c+d x)+8 \log (\sin (c+d x))-8 \log (\sin (c+d x)+1)}{2 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*Csc[c + d*x] - Csc[c + d*x]^2 + 8*Log[Sin[c + d*x]] - 8*Log[1 + Sin[c + d*x]])/(2*a^3*d)

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Maple [A]  time = 0.155, size = 66, normalized size = 1. \begin{align*} -4\,{\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 ) }}+4\,{\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(cos(d*x+c)^5*csc(d*x+c)^3/(a+a*sin(d*x+c))^3,x)

[Out]

-4*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)+4*ln(sin(d*x+c))/a^3/d

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Maxima [A]  time = 1.10113, size = 74, normalized size = 1.14 \begin{align*} -\frac{\frac{8 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}} - \frac{8 \, \log \left (\sin \left (d x + c\right )\right )}{a^{3}} - \frac{6 \, \sin \left (d x + c\right ) - 1}{a^{3} \sin \left (d x + c\right )^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^3/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/2*(8*log(sin(d*x + c) + 1)/a^3 - 8*log(sin(d*x + c))/a^3 - (6*sin(d*x + c) - 1)/(a^3*sin(d*x + c)^2))/d

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Fricas [A]  time = 1.12308, size = 204, normalized size = 3.14 \begin{align*} \frac{8 \,{\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) - 8 \,{\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 6 \, \sin \left (d x + c\right ) + 1}{2 \,{\left (a^{3} d \cos \left (d x + c\right )^{2} - a^{3} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^3/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/2*(8*(cos(d*x + c)^2 - 1)*log(1/2*sin(d*x + c)) - 8*(cos(d*x + c)^2 - 1)*log(sin(d*x + c) + 1) - 6*sin(d*x +
c) + 1)/(a^3*d*cos(d*x + c)^2 - a^3*d)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**5*csc(d*x+c)**3/(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.21341, size = 155, normalized size = 2.38 \begin{align*} -\frac{\frac{64 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac{32 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{3}} + \frac{48 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 12 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1}{a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}} + \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(cos(d*x+c)^5*csc(d*x+c)^3/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/8*(64*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 32*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 + (48*tan(1/2*d*x + 1/
2*c)^2 - 12*tan(1/2*d*x + 1/2*c) + 1)/(a^3*tan(1/2*d*x + 1/2*c)^2) + (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