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

Optimal. Leaf size=29 \[ \frac{\log (\sin (c+d x))}{a d}-\frac{\sin (c+d x)}{a d} \]

[Out]

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

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Rubi [A]  time = 0.0768426, antiderivative size = 29, 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, 43} \[ \frac{\log (\sin (c+d x))}{a d}-\frac{\sin (c+d x)}{a d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^2*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)

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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0332479, size = 23, normalized size = 0.79 \[ \frac{\log (\sin (c+d x))-\sin (c+d x)}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.036, size = 33, normalized size = 1.1 \begin{align*} -{\frac{1}{da\csc \left ( dx+c \right ) }}-{\frac{\ln \left ( \csc \left ( dx+c \right ) \right ) }{da}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*csc(d*x+c)/(a+a*sin(d*x+c)),x)

[Out]

-1/d/a/csc(d*x+c)-1/d/a*ln(csc(d*x+c))

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Maxima [A]  time = 0.996756, size = 36, normalized size = 1.24 \begin{align*} \frac{\frac{\log \left (\sin \left (d x + c\right )\right )}{a} - \frac{\sin \left (d x + c\right )}{a}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.52175, size = 62, normalized size = 2.14 \begin{align*} \frac{\log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) - \sin \left (d x + c\right )}{a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(1/2*sin(d*x + c)) - 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)**3*csc(d*x+c)/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.30181, size = 38, normalized size = 1.31 \begin{align*} \frac{\frac{\log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a} - \frac{\sin \left (d x + c\right )}{a}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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