3.684 \(\int \frac{\cos ^6(c+d x) \cot (c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=99 \[ -\frac{\sin ^5(c+d x)}{5 a d}+\frac{\sin ^4(c+d x)}{4 a d}+\frac{2 \sin ^3(c+d x)}{3 a d}-\frac{\sin ^2(c+d x)}{a d}-\frac{\sin (c+d x)}{a d}+\frac{\log (\sin (c+d x))}{a d} \]

[Out]

Log[Sin[c + d*x]]/(a*d) - Sin[c + d*x]/(a*d) - Sin[c + d*x]^2/(a*d) + (2*Sin[c + d*x]^3)/(3*a*d) + Sin[c + d*x
]^4/(4*a*d) - Sin[c + d*x]^5/(5*a*d)

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Rubi [A]  time = 0.10026, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {2836, 12, 88} \[ -\frac{\sin ^5(c+d x)}{5 a d}+\frac{\sin ^4(c+d x)}{4 a d}+\frac{2 \sin ^3(c+d x)}{3 a d}-\frac{\sin ^2(c+d x)}{a d}-\frac{\sin (c+d x)}{a d}+\frac{\log (\sin (c+d x))}{a d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^6*Cot[c + d*x])/(a + a*Sin[c + d*x]),x]

[Out]

Log[Sin[c + d*x]]/(a*d) - Sin[c + d*x]/(a*d) - Sin[c + d*x]^2/(a*d) + (2*Sin[c + d*x]^3)/(3*a*d) + Sin[c + d*x
]^4/(4*a*d) - Sin[c + d*x]^5/(5*a*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 ^6(c+d x) \cot (c+d x)}{a+a \sin (c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{a (a-x)^3 (a+x)^2}{x} \, dx,x,a \sin (c+d x)\right )}{a^7 d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{(a-x)^3 (a+x)^2}{x} \, dx,x,a \sin (c+d x)\right )}{a^6 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (-a^4+\frac{a^5}{x}-2 a^3 x+2 a^2 x^2+a x^3-x^4\right ) \, dx,x,a \sin (c+d x)\right )}{a^6 d}\\ &=\frac{\log (\sin (c+d x))}{a d}-\frac{\sin (c+d x)}{a d}-\frac{\sin ^2(c+d x)}{a d}+\frac{2 \sin ^3(c+d x)}{3 a d}+\frac{\sin ^4(c+d x)}{4 a d}-\frac{\sin ^5(c+d x)}{5 a d}\\ \end{align*}

Mathematica [A]  time = 0.0722493, size = 68, normalized size = 0.69 \[ \frac{-12 \sin ^5(c+d x)+15 \sin ^4(c+d x)+40 \sin ^3(c+d x)-60 \sin ^2(c+d x)-60 \sin (c+d x)+60 \log (\sin (c+d x))}{60 a d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^6*Cot[c + d*x])/(a + a*Sin[c + d*x]),x]

[Out]

(60*Log[Sin[c + d*x]] - 60*Sin[c + d*x] - 60*Sin[c + d*x]^2 + 40*Sin[c + d*x]^3 + 15*Sin[c + d*x]^4 - 12*Sin[c
+ d*x]^5)/(60*a*d)

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Maple [A]  time = 0.101, size = 94, normalized size = 1. \begin{align*}{\frac{\ln \left ( \sin \left ( dx+c \right ) \right ) }{da}}-{\frac{\sin \left ( dx+c \right ) }{da}}-{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{da}}+{\frac{2\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{3\,da}}+{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{4\,da}}-{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{5\,da}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^7*csc(d*x+c)/(a+a*sin(d*x+c)),x)

[Out]

ln(sin(d*x+c))/a/d-sin(d*x+c)/d/a-sin(d*x+c)^2/d/a+2/3*sin(d*x+c)^3/d/a+1/4*sin(d*x+c)^4/d/a-1/5*sin(d*x+c)^5/
d/a

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Maxima [A]  time = 1.03304, size = 96, normalized size = 0.97 \begin{align*} -\frac{\frac{12 \, \sin \left (d x + c\right )^{5} - 15 \, \sin \left (d x + c\right )^{4} - 40 \, \sin \left (d x + c\right )^{3} + 60 \, \sin \left (d x + c\right )^{2} + 60 \, \sin \left (d x + c\right )}{a} - \frac{60 \, \log \left (\sin \left (d x + c\right )\right )}{a}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*csc(d*x+c)/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/60*((12*sin(d*x + c)^5 - 15*sin(d*x + c)^4 - 40*sin(d*x + c)^3 + 60*sin(d*x + c)^2 + 60*sin(d*x + c))/a - 6
0*log(sin(d*x + c))/a)/d

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Fricas [A]  time = 1.16531, size = 186, normalized size = 1.88 \begin{align*} \frac{15 \, \cos \left (d x + c\right )^{4} + 30 \, \cos \left (d x + c\right )^{2} - 4 \,{\left (3 \, \cos \left (d x + c\right )^{4} + 4 \, \cos \left (d x + c\right )^{2} + 8\right )} \sin \left (d x + c\right ) + 60 \, \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right )}{60 \, a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*csc(d*x+c)/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/60*(15*cos(d*x + c)^4 + 30*cos(d*x + c)^2 - 4*(3*cos(d*x + c)^4 + 4*cos(d*x + c)^2 + 8)*sin(d*x + c) + 60*lo
g(1/2*sin(d*x + c)))/(a*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)**7*csc(d*x+c)/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.25604, size = 119, normalized size = 1.2 \begin{align*} \frac{\frac{60 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a} - \frac{12 \, a^{4} \sin \left (d x + c\right )^{5} - 15 \, a^{4} \sin \left (d x + c\right )^{4} - 40 \, a^{4} \sin \left (d x + c\right )^{3} + 60 \, a^{4} \sin \left (d x + c\right )^{2} + 60 \, a^{4} \sin \left (d x + c\right )}{a^{5}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*csc(d*x+c)/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

1/60*(60*log(abs(sin(d*x + c)))/a - (12*a^4*sin(d*x + c)^5 - 15*a^4*sin(d*x + c)^4 - 40*a^4*sin(d*x + c)^3 + 6
0*a^4*sin(d*x + c)^2 + 60*a^4*sin(d*x + c))/a^5)/d