\(\int \frac {\cot (c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^3} \, dx\) [246]

Optimal result

Integrand size = 25, antiderivative size = 90 \[ \int \frac {\cot (c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\csc (c+d x)}{a^3 d}-\frac {3 \log (\sin (c+d x))}{a^3 d}+\frac {3 \log (1+\sin (c+d x))}{a^3 d}-\frac {1}{2 a d (a+a \sin (c+d x))^2}-\frac {2}{d \left (a^3+a^3 \sin (c+d x)\right )} \]

[Out]

-csc(d*x+c)/a^3/d-3*ln(sin(d*x+c))/a^3/d+3*ln(1+sin(d*x+c))/a^3/d-1/2/a/d/(a+a*sin(d*x+c))^2-2/d/(a^3+a^3*sin(
d*x+c))

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2912, 12, 46} \[ \int \frac {\cot (c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {2}{d \left (a^3 \sin (c+d x)+a^3\right )}-\frac {\csc (c+d x)}{a^3 d}-\frac {3 \log (\sin (c+d x))}{a^3 d}+\frac {3 \log (\sin (c+d x)+1)}{a^3 d}-\frac {1}{2 a d (a \sin (c+d x)+a)^2} \]

[In]

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

[Out]

-(Csc[c + d*x]/(a^3*d)) - (3*Log[Sin[c + d*x]])/(a^3*d) + (3*Log[1 + Sin[c + d*x]])/(a^3*d) - 1/(2*a*d*(a + a*
Sin[c + d*x])^2) - 2/(d*(a^3 + a^3*Sin[c + d*x]))

Rule 12

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

Rule 46

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 2912

Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)
])^(n_.), x_Symbol] :> Dist[1/(b*f), Subst[Int[(a + x)^m*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[
{a, b, c, d, e, f, m, n}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {a^2}{x^2 (a+x)^3} \, dx,x,a \sin (c+d x)\right )}{a d} \\ & = \frac {a \text {Subst}\left (\int \frac {1}{x^2 (a+x)^3} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a \text {Subst}\left (\int \left (\frac {1}{a^3 x^2}-\frac {3}{a^4 x}+\frac {1}{a^2 (a+x)^3}+\frac {2}{a^3 (a+x)^2}+\frac {3}{a^4 (a+x)}\right ) \, dx,x,a \sin (c+d x)\right )}{d} \\ & = -\frac {\csc (c+d x)}{a^3 d}-\frac {3 \log (\sin (c+d x))}{a^3 d}+\frac {3 \log (1+\sin (c+d x))}{a^3 d}-\frac {1}{2 a d (a+a \sin (c+d x))^2}-\frac {2}{d \left (a^3+a^3 \sin (c+d x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.68 \[ \int \frac {\cot (c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {2 \csc (c+d x)+6 \log (\sin (c+d x))-6 \log (1+\sin (c+d x))+\frac {1}{(1+\sin (c+d x))^2}+\frac {4}{1+\sin (c+d x)}}{2 a^3 d} \]

[In]

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

[Out]

-1/2*(2*Csc[c + d*x] + 6*Log[Sin[c + d*x]] - 6*Log[1 + Sin[c + d*x]] + (1 + Sin[c + d*x])^(-2) + 4/(1 + Sin[c
+ d*x]))/(a^3*d)

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.57

method	result	size
derivativedivides	\(-\frac {\csc \left (d x +c \right )-\frac {3}{\csc \left (d x +c \right )+1}+\frac {1}{2 \left (\csc \left (d x +c \right )+1\right )^{2}}-3 \ln \left (\csc \left (d x +c \right )+1\right )}{d \,a^{3}}\)	\(51\)
default	\(-\frac {\csc \left (d x +c \right )-\frac {3}{\csc \left (d x +c \right )+1}+\frac {1}{2 \left (\csc \left (d x +c \right )+1\right )^{2}}-3 \ln \left (\csc \left (d x +c \right )+1\right )}{d \,a^{3}}\)	\(51\)
risch	\(-\frac {2 i \left (9 i {\mathrm e}^{4 i \left (d x +c \right )}+3 \,{\mathrm e}^{5 i \left (d x +c \right )}-9 i {\mathrm e}^{2 i \left (d x +c \right )}-10 \,{\mathrm e}^{3 i \left (d x +c \right )}+3 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{4} d \,a^{3}}+\frac {6 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{3}}-\frac {3 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d \,a^{3}}\)	\(137\)
parallelrisch	\(\frac {-9+12 \left (-3+\cos \left (2 d x +2 c \right )-4 \sin \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+6 \left (3-\cos \left (2 d x +2 c \right )+4 \sin \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+24 \left (-1+\cos \left (d x +c \right )\right ) \cot \left (\frac {d x}{2}+\frac {c}{2}\right )+2 \sec \left (\frac {d x}{2}+\frac {c}{2}\right ) \csc \left (\frac {d x}{2}+\frac {c}{2}\right )+9 \cos \left (2 d x +2 c \right )}{2 d \,a^{3} \left (-3+\cos \left (2 d x +2 c \right )-4 \sin \left (d x +c \right )\right )}\)	\(148\)
norman	\(\frac {\frac {13 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {13 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {1}{2 a d}-\frac {\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d a}+\frac {67 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}+\frac {67 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3}}+\frac {6 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{3}}\)	\(172\)

[In]

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

[Out]

-1/d/a^3*(csc(d*x+c)-3/(csc(d*x+c)+1)+1/2/(csc(d*x+c)+1)^2-3*ln(csc(d*x+c)+1))

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.69 \[ \int \frac {\cot (c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {6 \, \cos \left (d x + c\right )^{2} + 6 \, {\left (2 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 2\right )} \sin \left (d x + c\right ) - 2\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 6 \, {\left (2 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 2\right )} \sin \left (d x + c\right ) - 2\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 9 \, \sin \left (d x + c\right ) - 8}{2 \, {\left (2 \, a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d + {\left (a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d\right )} \sin \left (d x + c\right )\right )}} \]

[In]

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

[Out]

-1/2*(6*cos(d*x + c)^2 + 6*(2*cos(d*x + c)^2 + (cos(d*x + c)^2 - 2)*sin(d*x + c) - 2)*log(1/2*sin(d*x + c)) -
6*(2*cos(d*x + c)^2 + (cos(d*x + c)^2 - 2)*sin(d*x + c) - 2)*log(sin(d*x + c) + 1) - 9*sin(d*x + c) - 8)/(2*a^
3*d*cos(d*x + c)^2 - 2*a^3*d + (a^3*d*cos(d*x + c)^2 - 2*a^3*d)*sin(d*x + c))

Sympy [F]

\[ \int \frac {\cot (c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\int \frac {\cos {\left (c + d x \right )} \csc ^{2}{\left (c + d x \right )}}{\sin ^{3}{\left (c + d x \right )} + 3 \sin ^{2}{\left (c + d x \right )} + 3 \sin {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]

[In]

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

[Out]

Integral(cos(c + d*x)*csc(c + d*x)**2/(sin(c + d*x)**3 + 3*sin(c + d*x)**2 + 3*sin(c + d*x) + 1), x)/a**3

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.01 \[ \int \frac {\cot (c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\frac {6 \, \sin \left (d x + c\right )^{2} + 9 \, \sin \left (d x + c\right ) + 2}{a^{3} \sin \left (d x + c\right )^{3} + 2 \, a^{3} \sin \left (d x + c\right )^{2} + a^{3} \sin \left (d x + c\right )} - \frac {6 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}} + \frac {6 \, \log \left (\sin \left (d x + c\right )\right )}{a^{3}}}{2 \, d} \]

[In]

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

[Out]

-1/2*((6*sin(d*x + c)^2 + 9*sin(d*x + c) + 2)/(a^3*sin(d*x + c)^3 + 2*a^3*sin(d*x + c)^2 + a^3*sin(d*x + c)) -
6*log(sin(d*x + c) + 1)/a^3 + 6*log(sin(d*x + c))/a^3)/d

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.86 \[ \int \frac {\cot (c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {6 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3}} - \frac {6 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{3}} - \frac {6 \, \sin \left (d x + c\right )^{2} + 9 \, \sin \left (d x + c\right ) + 2}{a^{3} {\left (\sin \left (d x + c\right ) + 1\right )}^{2} \sin \left (d x + c\right )}}{2 \, d} \]

[In]

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

[Out]

1/2*(6*log(abs(sin(d*x + c) + 1))/a^3 - 6*log(abs(sin(d*x + c)))/a^3 - (6*sin(d*x + c)^2 + 9*sin(d*x + c) + 2)
/(a^3*(sin(d*x + c) + 1)^2*sin(d*x + c)))/d

Mupad [B] (verification not implemented)

Time = 9.87 (sec) , antiderivative size = 193, normalized size of antiderivative = 2.14 \[ \int \frac {\cot (c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {11\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1}{d\,\left (2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+8\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+12\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+8\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}-\frac {3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^3\,d}+\frac {6\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{a^3\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a^3\,d} \]

[In]

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

[Out]

(6*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2) + 16*tan(c/2 + (d*x)/2)^3 + 11*tan(c/2 + (d*x)/2)^4 - 1)/(d*(8*
a^3*tan(c/2 + (d*x)/2)^2 + 12*a^3*tan(c/2 + (d*x)/2)^3 + 8*a^3*tan(c/2 + (d*x)/2)^4 + 2*a^3*tan(c/2 + (d*x)/2)
^5 + 2*a^3*tan(c/2 + (d*x)/2))) - (3*log(tan(c/2 + (d*x)/2)))/(a^3*d) + (6*log(tan(c/2 + (d*x)/2) + 1))/(a^3*d
) - tan(c/2 + (d*x)/2)/(2*a^3*d)