3.685 \(\int \frac{\cos ^5(c+d x) \cot ^2(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=95 \[ -\frac{\sin ^4(c+d x)}{4 a d}+\frac{\sin ^3(c+d x)}{3 a d}+\frac{\sin ^2(c+d x)}{a d}-\frac{2 \sin (c+d x)}{a d}-\frac{\csc (c+d x)}{a d}-\frac{\log (\sin (c+d x))}{a d} \]

[Out]

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

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Rubi [A]  time = 0.119503, antiderivative size = 95, 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} \[ -\frac{\sin ^4(c+d x)}{4 a d}+\frac{\sin ^3(c+d x)}{3 a d}+\frac{\sin ^2(c+d x)}{a d}-\frac{2 \sin (c+d x)}{a d}-\frac{\csc (c+d x)}{a d}-\frac{\log (\sin (c+d x))}{a d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^5*Cot[c + d*x]^2)/(a + a*Sin[c + d*x]),x]

[Out]

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

Mathematica [A]  time = 0.128917, size = 66, normalized size = 0.69 \[ -\frac{3 \sin ^4(c+d x)-4 \sin ^3(c+d x)-12 \sin ^2(c+d x)+24 \sin (c+d x)+12 \csc (c+d x)+12 \log (\sin (c+d x))}{12 a d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^5*Cot[c + d*x]^2)/(a + a*Sin[c + d*x]),x]

[Out]

-(12*Csc[c + d*x] + 12*Log[Sin[c + d*x]] + 24*Sin[c + d*x] - 12*Sin[c + d*x]^2 - 4*Sin[c + d*x]^3 + 3*Sin[c +
d*x]^4)/(12*a*d)

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Maple [A]  time = 0.118, size = 94, normalized size = 1. \begin{align*} -{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{4\,da}}+{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{3\,da}}+{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{da}}-2\,{\frac{\sin \left ( dx+c \right ) }{da}}-{\frac{1}{da\sin \left ( dx+c \right ) }}-{\frac{\ln \left ( \sin \left ( dx+c \right ) \right ) }{da}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^7*csc(d*x+c)^2/(a+a*sin(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 1.04327, size = 100, normalized size = 1.05 \begin{align*} -\frac{\frac{3 \, \sin \left (d x + c\right )^{4} - 4 \, \sin \left (d x + c\right )^{3} - 12 \, \sin \left (d x + c\right )^{2} + 24 \, \sin \left (d x + c\right )}{a} + \frac{12 \, \log \left (\sin \left (d x + c\right )\right )}{a} + \frac{12}{a \sin \left (d x + c\right )}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*csc(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/12*((3*sin(d*x + c)^4 - 4*sin(d*x + c)^3 - 12*sin(d*x + c)^2 + 24*sin(d*x + c))/a + 12*log(sin(d*x + c))/a
+ 12/(a*sin(d*x + c)))/d

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Fricas [A]  time = 1.15926, size = 234, normalized size = 2.46 \begin{align*} \frac{32 \, \cos \left (d x + c\right )^{4} + 128 \, \cos \left (d x + c\right )^{2} - 3 \,{\left (8 \, \cos \left (d x + c\right )^{4} + 16 \, \cos \left (d x + c\right )^{2} - 11\right )} \sin \left (d x + c\right ) - 96 \, \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) - 256}{96 \, a d \sin \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*csc(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/96*(32*cos(d*x + c)^4 + 128*cos(d*x + c)^2 - 3*(8*cos(d*x + c)^4 + 16*cos(d*x + c)^2 - 11)*sin(d*x + c) - 96
*log(1/2*sin(d*x + c))*sin(d*x + c) - 256)/(a*d*sin(d*x + c))

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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)**2/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.30527, size = 128, normalized size = 1.35 \begin{align*} -\frac{\frac{12 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a} - \frac{12 \,{\left (\sin \left (d x + c\right ) - 1\right )}}{a \sin \left (d x + c\right )} + \frac{3 \, a^{3} \sin \left (d x + c\right )^{4} - 4 \, a^{3} \sin \left (d x + c\right )^{3} - 12 \, a^{3} \sin \left (d x + c\right )^{2} + 24 \, a^{3} \sin \left (d x + c\right )}{a^{4}}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*csc(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/12*(12*log(abs(sin(d*x + c)))/a - 12*(sin(d*x + c) - 1)/(a*sin(d*x + c)) + (3*a^3*sin(d*x + c)^4 - 4*a^3*si
n(d*x + c)^3 - 12*a^3*sin(d*x + c)^2 + 24*a^3*sin(d*x + c))/a^4)/d