\(\int \frac {\cos ^3(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx\) [548]

Optimal result

Integrand size = 29, antiderivative size = 43 \[ \int \frac {\cos ^3(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\csc (c+d x)}{a^2 d}-\frac {2 \log (\sin (c+d x))}{a^2 d}+\frac {\sin (c+d x)}{a^2 d} \]

[Out]

-csc(d*x+c)/a^2/d-2*ln(sin(d*x+c))/a^2/d+sin(d*x+c)/a^2/d

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2915, 12, 45} \[ \int \frac {\cos ^3(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\sin (c+d x)}{a^2 d}-\frac {\csc (c+d x)}{a^2 d}-\frac {2 \log (\sin (c+d x))}{a^2 d} \]

[In]

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

[Out]

-(Csc[c + d*x]/(a^2*d)) - (2*Log[Sin[c + d*x]])/(a^2*d) + Sin[c + d*x]/(a^2*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^2 (a-x)^2}{x^2} \, dx,x,a \sin (c+d x)\right )}{a^5 d} \\ & = \frac {\text {Subst}\left (\int \frac {(a-x)^2}{x^2} \, dx,x,a \sin (c+d x)\right )}{a^3 d} \\ & = \frac {\text {Subst}\left (\int \left (1+\frac {a^2}{x^2}-\frac {2 a}{x}\right ) \, dx,x,a \sin (c+d x)\right )}{a^3 d} \\ & = -\frac {\csc (c+d x)}{a^2 d}-\frac {2 \log (\sin (c+d x))}{a^2 d}+\frac {\sin (c+d x)}{a^2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.74 \[ \int \frac {\cos ^3(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\csc (c+d x)+2 \log (\sin (c+d x))-\sin (c+d x)}{a^2 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.33 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.79

method	result	size
derivativedivides	\(\frac {\sin \left (d x +c \right )-2 \ln \left (\sin \left (d x +c \right )\right )-\frac {1}{\sin \left (d x +c \right )}}{d \,a^{2}}\)	\(34\)
default	\(\frac {\sin \left (d x +c \right )-2 \ln \left (\sin \left (d x +c \right )\right )-\frac {1}{\sin \left (d x +c \right )}}{d \,a^{2}}\)	\(34\)
parallelrisch	\(\frac {4 \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-2 \cos \left (d x +c \right )+2\right ) \cot \left (\frac {d x}{2}+\frac {c}{2}\right )-\sec \left (\frac {d x}{2}+\frac {c}{2}\right ) \csc \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}\)	\(76\)
risch	\(\frac {2 i x}{a^{2}}-\frac {i {\mathrm e}^{i \left (d x +c \right )}}{2 d \,a^{2}}+\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{2 d \,a^{2}}+\frac {4 i c}{d \,a^{2}}-\frac {2 i {\mathrm e}^{i \left (d x +c \right )}}{d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}-\frac {2 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d \,a^{2}}\)	\(106\)
norman	\(\frac {-\frac {1}{2 a d}-\frac {\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d a}+\frac {3 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {3 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {21 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {21 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {16 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {16 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {17 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}+\frac {17 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}}+\frac {2 \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}}\)	\(265\)

[In]

int(cos(d*x+c)^5*csc(d*x+c)^2/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.98 \[ \int \frac {\cos ^3(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\cos \left (d x + c\right )^{2} + 2 \, \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right )}{a^{2} d \sin \left (d x + c\right )} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^2/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^3(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**5*csc(d*x+c)**2/(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.95 \[ \int \frac {\cos ^3(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {2 \, \log \left (\sin \left (d x + c\right )\right )}{a^{2}} - \frac {\sin \left (d x + c\right )}{a^{2}} + \frac {1}{a^{2} \sin \left (d x + c\right )}}{d} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^2/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.23 \[ \int \frac {\cos ^3(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {2 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{2}} - \frac {\sin \left (d x + c\right )}{a^{2}} - \frac {2 \, \sin \left (d x + c\right ) - 1}{a^{2} \sin \left (d x + c\right )}}{d} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^2/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.46 (sec) , antiderivative size = 110, normalized size of antiderivative = 2.56 \[ \int \frac {\cos ^3(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1}{d\,\left (2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}-\frac {2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^2\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a^2\,d}+\frac {2\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{a^2\,d} \]

[In]

int(cos(c + d*x)^5/(sin(c + d*x)^2*(a + a*sin(c + d*x))^2),x)

[Out]

(3*tan(c/2 + (d*x)/2)^2 - 1)/(d*(2*a^2*tan(c/2 + (d*x)/2)^3 + 2*a^2*tan(c/2 + (d*x)/2))) - (2*log(tan(c/2 + (d
*x)/2)))/(a^2*d) - tan(c/2 + (d*x)/2)/(2*a^2*d) + (2*log(tan(c/2 + (d*x)/2)^2 + 1))/(a^2*d)