\(\int \frac {\cos ^2(c+d x) \cot (c+d x)}{a+a \sin (c+d x)} \, dx\) [360]

Optimal result

Integrand size = 27, antiderivative size = 29 \[ \int \frac {\cos ^2(c+d x) \cot (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\log (\sin (c+d x))}{a d}-\frac {\sin (c+d x)}{a d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2915, 12, 45} \[ \int \frac {\cos ^2(c+d x) \cot (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\log (\sin (c+d x))}{a d}-\frac {\sin (c+d x)}{a d} \]

[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 12

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

Rule 45

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])

Rule 2915

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/b)*x
)^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]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {a (a-x)}{x} \, dx,x,a \sin (c+d x)\right )}{a^3 d} \\ & = \frac {\text {Subst}\left (\int \frac {a-x}{x} \, dx,x,a \sin (c+d x)\right )}{a^2 d} \\ & = \frac {\text {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] (verified)

Time = 0.03 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79 \[ \int \frac {\cos ^2(c+d x) \cot (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\log (\sin (c+d x))-\sin (c+d x)}{a d} \]

[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)

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97

method	result	size
derivativedivides	\(\frac {-\ln \left (\csc \left (d x +c \right )\right )-\frac {1}{\csc \left (d x +c \right )}}{d a}\)	\(28\)
default	\(\frac {-\ln \left (\csc \left (d x +c \right )\right )-\frac {1}{\csc \left (d x +c \right )}}{d a}\)	\(28\)
parallelrisch	\(\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\sin \left (d x +c \right )}{a d}\)	\(41\)
risch	\(-\frac {i x}{a}-\frac {2 i c}{a d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d a}-\frac {\sin \left (d x +c \right )}{a d}\)	\(52\)
norman	\(\frac {\frac {2}{a d}+\frac {2 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {2 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {\ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}\)	\(136\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {\cos ^2(c+d x) \cot (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - \sin \left (d x + c\right )}{a d} \]

[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)

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^2(c+d x) \cot (c+d x)}{a+a \sin (c+d x)} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93 \[ \int \frac {\cos ^2(c+d x) \cot (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {\log \left (\sin \left (d x + c\right )\right )}{a} - \frac {\sin \left (d x + c\right )}{a}}{d} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {\cos ^2(c+d x) \cot (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {\log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a} - \frac {\sin \left (d x + c\right )}{a}}{d} \]

[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

Mupad [B] (verification not implemented)

Time = 9.45 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.45 \[ \int \frac {\cos ^2(c+d x) \cot (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a\,d}-\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a\right )}-\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{a\,d} \]

[In]

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

[Out]

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