\(\int \frac {\cos ^2(c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx\) [549]

Optimal result

Integrand size = 29, antiderivative size = 47 \[ \int \frac {\cos ^2(c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {2 \csc (c+d x)}{a^2 d}-\frac {\csc ^2(c+d x)}{2 a^2 d}+\frac {\log (\sin (c+d x))}{a^2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 47, 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} \[ \int \frac {\cos ^2(c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\csc ^2(c+d x)}{2 a^2 d}+\frac {2 \csc (c+d x)}{a^2 d}+\frac {\log (\sin (c+d x))}{a^2 d} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.81 \[ \int \frac {\cos ^2(c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {4 \csc (c+d x)-\csc ^2(c+d x)+2 \log (\sin (c+d x))}{2 a^2 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.77

method	result	size
derivativedivides	\(\frac {-\frac {1}{2 \sin \left (d x +c \right )^{2}}+\ln \left (\sin \left (d x +c \right )\right )+\frac {2}{\sin \left (d x +c \right )}}{d \,a^{2}}\)	\(36\)
default	\(\frac {-\frac {1}{2 \sin \left (d x +c \right )^{2}}+\ln \left (\sin \left (d x +c \right )\right )+\frac {2}{\sin \left (d x +c \right )}}{d \,a^{2}}\)	\(36\)
parallelrisch	\(\frac {-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-8 \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+8 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{2}}\)	\(84\)
risch	\(-\frac {i x}{a^{2}}-\frac {2 i c}{d \,a^{2}}+\frac {2 i \left (-i {\mathrm e}^{2 i \left (d x +c \right )}+2 \,{\mathrm e}^{3 i \left (d x +c \right )}-2 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{a^{2} d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d \,a^{2}}\)	\(95\)
norman	\(\frac {-\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {2 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {1}{8 a d}+\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d a}+\frac {5 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}-\frac {\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d a}-\frac {23 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}-\frac {23 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}-\frac {45 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}-\frac {45 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}}-\frac {\ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}}\)	\(262\)

[In]

int(cos(d*x+c)^5*csc(d*x+c)^3/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.17 \[ \int \frac {\cos ^2(c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {2 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 4 \, \sin \left (d x + c\right ) + 1}{2 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d\right )}} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

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

[In]

integrate(cos(d*x+c)**5*csc(d*x+c)**3/(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.85 \[ \int \frac {\cos ^2(c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {2 \, \log \left (\sin \left (d x + c\right )\right )}{a^{2}} + \frac {4 \, \sin \left (d x + c\right ) - 1}{a^{2} \sin \left (d x + c\right )^{2}}}{2 \, d} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.11 \[ \int \frac {\cos ^2(c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {2 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{2}} - \frac {3 \, \sin \left (d x + c\right )^{2} - 4 \, \sin \left (d x + c\right ) + 1}{a^{2} \sin \left (d x + c\right )^{2}}}{2 \, d} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 10.04 (sec) , antiderivative size = 104, normalized size of antiderivative = 2.21 \[ \int \frac {\cos ^2(c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^2\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a^2\,d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^2\,d}-\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{a^2\,d}+\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {1}{8}\right )}{a^2\,d} \]

[In]

int(cos(c + d*x)^5/(sin(c + d*x)^3*(a + a*sin(c + d*x))^2),x)

[Out]

log(tan(c/2 + (d*x)/2))/(a^2*d) - tan(c/2 + (d*x)/2)^2/(8*a^2*d) + tan(c/2 + (d*x)/2)/(a^2*d) - log(tan(c/2 +
(d*x)/2)^2 + 1)/(a^2*d) + (cot(c/2 + (d*x)/2)^2*(tan(c/2 + (d*x)/2) - 1/8))/(a^2*d)