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

Optimal. Leaf size=62 \[ \frac{\sin ^2(c+d x)}{2 a d}-\frac{\sin (c+d x)}{a d}-\frac{\csc (c+d x)}{a d}-\frac{\log (\sin (c+d x))}{a d} \]

[Out]

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

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Rubi [A]  time = 0.108435, antiderivative size = 62, 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, 75} \[ \frac{\sin ^2(c+d x)}{2 a d}-\frac{\sin (c+d x)}{a d}-\frac{\csc (c+d x)}{a d}-\frac{\log (\sin (c+d x))}{a d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^3*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) - Sin[c + d*x]/(a*d) + Sin[c + d*x]^2/(2*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 75

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] &&  !(ILtQ[n
+ p + 2, 0] && GtQ[n + 2*p, 0])

Rubi steps

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

Mathematica [A]  time = 0.0648701, size = 45, normalized size = 0.73 \[ \frac{\sin ^2(c+d x)-2 \sin (c+d x)-2 \csc (c+d x)-2 \log (\sin (c+d x))+6}{2 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.107, size = 63, normalized size = 1. \begin{align*}{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{2\,da}}-{\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)^5*csc(d*x+c)^2/(a+a*sin(d*x+c)),x)

[Out]

1/2*sin(d*x+c)^2/d/a-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 = 1.16758, size = 70, normalized size = 1.13 \begin{align*} \frac{\frac{\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right )}{a} - \frac{2 \, \log \left (\sin \left (d x + c\right )\right )}{a} - \frac{2}{a \sin \left (d x + c\right )}}{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)^2/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.14821, size = 167, normalized size = 2.69 \begin{align*} \frac{4 \, \cos \left (d x + c\right )^{2} -{\left (2 \, \cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - 4 \, \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) - 8}{4 \, 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)^5*csc(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/4*(4*cos(d*x + c)^2 - (2*cos(d*x + c)^2 - 1)*sin(d*x + c) - 4*log(1/2*sin(d*x + c))*sin(d*x + c) - 8)/(a*d*s
in(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)**5*csc(d*x+c)**2/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.26467, size = 88, normalized size = 1.42 \begin{align*} -\frac{\frac{2 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a} - \frac{a \sin \left (d x + c\right )^{2} - 2 \, a \sin \left (d x + c\right )}{a^{2}} - \frac{2 \,{\left (\sin \left (d x + c\right ) - 1\right )}}{a \sin \left (d x + c\right )}}{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)^2/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

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