\(\int \frac {\cot (c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^4} \, dx\) [256]

Optimal result

Integrand size = 27, antiderivative size = 131 \[ \int \frac {\cot (c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {4 \csc (c+d x)}{a^4 d}-\frac {\csc ^2(c+d x)}{2 a^4 d}+\frac {10 \log (\sin (c+d x))}{a^4 d}-\frac {10 \log (1+\sin (c+d x))}{a^4 d}+\frac {1}{3 a d (a+a \sin (c+d x))^3}+\frac {3}{2 d \left (a^2+a^2 \sin (c+d x)\right )^2}+\frac {6}{d \left (a^4+a^4 \sin (c+d x)\right )} \]

[Out]

4*csc(d*x+c)/a^4/d-1/2*csc(d*x+c)^2/a^4/d+10*ln(sin(d*x+c))/a^4/d-10*ln(1+sin(d*x+c))/a^4/d+1/3/a/d/(a+a*sin(d
*x+c))^3+3/2/d/(a^2+a^2*sin(d*x+c))^2+6/d/(a^4+a^4*sin(d*x+c))

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 131, 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 = {2912, 12, 46} \[ \int \frac {\cot (c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {6}{d \left (a^4 \sin (c+d x)+a^4\right )}-\frac {\csc ^2(c+d x)}{2 a^4 d}+\frac {4 \csc (c+d x)}{a^4 d}+\frac {10 \log (\sin (c+d x))}{a^4 d}-\frac {10 \log (\sin (c+d x)+1)}{a^4 d}+\frac {3}{2 d \left (a^2 \sin (c+d x)+a^2\right )^2}+\frac {1}{3 a d (a \sin (c+d x)+a)^3} \]

[In]

Int[(Cot[c + d*x]*Csc[c + d*x]^2)/(a + a*Sin[c + d*x])^4,x]

[Out]

(4*Csc[c + d*x])/(a^4*d) - Csc[c + d*x]^2/(2*a^4*d) + (10*Log[Sin[c + d*x]])/(a^4*d) - (10*Log[1 + Sin[c + d*x
]])/(a^4*d) + 1/(3*a*d*(a + a*Sin[c + d*x])^3) + 3/(2*d*(a^2 + a^2*Sin[c + d*x])^2) + 6/(d*(a^4 + a^4*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^3}{x^3 (a+x)^4} \, dx,x,a \sin (c+d x)\right )}{a d} \\ & = \frac {a^2 \text {Subst}\left (\int \frac {1}{x^3 (a+x)^4} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^2 \text {Subst}\left (\int \left (\frac {1}{a^4 x^3}-\frac {4}{a^5 x^2}+\frac {10}{a^6 x}-\frac {1}{a^3 (a+x)^4}-\frac {3}{a^4 (a+x)^3}-\frac {6}{a^5 (a+x)^2}-\frac {10}{a^6 (a+x)}\right ) \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {4 \csc (c+d x)}{a^4 d}-\frac {\csc ^2(c+d x)}{2 a^4 d}+\frac {10 \log (\sin (c+d x))}{a^4 d}-\frac {10 \log (1+\sin (c+d x))}{a^4 d}+\frac {1}{3 a d (a+a \sin (c+d x))^3}+\frac {3}{2 d \left (a^2+a^2 \sin (c+d x)\right )^2}+\frac {6}{d \left (a^4+a^4 \sin (c+d x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.22 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.65 \[ \int \frac {\cot (c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {24 \csc (c+d x)-3 \csc ^2(c+d x)+60 \log (\sin (c+d x))-60 \log (1+\sin (c+d x))+\frac {2}{(1+\sin (c+d x))^3}+\frac {9}{(1+\sin (c+d x))^2}+\frac {36}{1+\sin (c+d x)}}{6 a^4 d} \]

[In]

Integrate[(Cot[c + d*x]*Csc[c + d*x]^2)/(a + a*Sin[c + d*x])^4,x]

[Out]

(24*Csc[c + d*x] - 3*Csc[c + d*x]^2 + 60*Log[Sin[c + d*x]] - 60*Log[1 + Sin[c + d*x]] + 2/(1 + Sin[c + d*x])^3
+ 9/(1 + Sin[c + d*x])^2 + 36/(1 + Sin[c + d*x]))/(6*a^4*d)

Maple [A] (verified)

Time = 0.52 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.57

method	result	size
derivativedivides	\(-\frac {\frac {\left (\csc ^{2}\left (d x +c \right )\right )}{2}-4 \csc \left (d x +c \right )-\frac {5}{2 \left (\csc \left (d x +c \right )+1\right )^{2}}+\frac {1}{3 \left (\csc \left (d x +c \right )+1\right )^{3}}+10 \ln \left (\csc \left (d x +c \right )+1\right )+\frac {10}{\csc \left (d x +c \right )+1}}{d \,a^{4}}\)	\(75\)
default	\(-\frac {\frac {\left (\csc ^{2}\left (d x +c \right )\right )}{2}-4 \csc \left (d x +c \right )-\frac {5}{2 \left (\csc \left (d x +c \right )+1\right )^{2}}+\frac {1}{3 \left (\csc \left (d x +c \right )+1\right )^{3}}+10 \ln \left (\csc \left (d x +c \right )+1\right )+\frac {10}{\csc \left (d x +c \right )+1}}{d \,a^{4}}\)	\(75\)
risch	\(\frac {4 i \left (75 i {\mathrm e}^{8 i \left (d x +c \right )}+15 \,{\mathrm e}^{9 i \left (d x +c \right )}-255 i {\mathrm e}^{6 i \left (d x +c \right )}-170 \,{\mathrm e}^{7 i \left (d x +c \right )}+255 i {\mathrm e}^{4 i \left (d x +c \right )}+298 \,{\mathrm e}^{5 i \left (d x +c \right )}-75 i {\mathrm e}^{2 i \left (d x +c \right )}-170 \,{\mathrm e}^{3 i \left (d x +c \right )}+15 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{3 \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{6} d \,a^{4}}-\frac {20 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{4}}+\frac {10 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d \,a^{4}}\)	\(183\)
norman	\(\frac {-\frac {60 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {60 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {1}{8 a d}+\frac {9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d a}+\frac {9 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}-\frac {\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d a}-\frac {1995 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}-\frac {1995 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}-\frac {11095 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d a}-\frac {11095 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d a}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}+\frac {10 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{4}}-\frac {20 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{4}}\)	\(246\)
parallelrisch	\(\frac {240 \left (-\sin \left (3 d x +3 c \right )+15 \sin \left (d x +c \right )+10-6 \cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+120 \left (-10+6 \cos \left (2 d x +2 c \right )+\sin \left (3 d x +3 c \right )-15 \sin \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-5\right ) \left (\csc ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-275 \left (\cos \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )+\cos \left (\frac {5 d x}{2}+\frac {5 c}{2}\right )-\frac {17 \cos \left (\frac {7 d x}{2}+\frac {7 c}{2}\right )}{176}-\frac {17 \cos \left (\frac {9 d x}{2}+\frac {9 c}{2}\right )}{176}-\frac {699 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )}{440}\right ) \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \csc \left (\frac {d x}{2}+\frac {c}{2}\right )-4140 \left (\cos \left (d x +c \right )-\frac {\cos \left (2 d x +2 c \right )}{4}-\frac {209}{276}\right ) \left (\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d \,a^{4} \left (-10+6 \cos \left (2 d x +2 c \right )+\sin \left (3 d x +3 c \right )-15 \sin \left (d x +c \right )\right )}\)	\(262\)

[In]

int(cos(d*x+c)*csc(d*x+c)^3/(a+a*sin(d*x+c))^4,x,method=_RETURNVERBOSE)

[Out]

-1/d/a^4*(1/2*csc(d*x+c)^2-4*csc(d*x+c)-5/2/(csc(d*x+c)+1)^2+1/3/(csc(d*x+c)+1)^3+10*ln(csc(d*x+c)+1)+10/(csc(
d*x+c)+1))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.85 \[ \int \frac {\cot (c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {60 \, \cos \left (d x + c\right )^{4} - 230 \, \cos \left (d x + c\right )^{2} + 60 \, {\left (3 \, \cos \left (d x + c\right )^{4} - 7 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{4} - 5 \, \cos \left (d x + c\right )^{2} + 4\right )} \sin \left (d x + c\right ) + 4\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 60 \, {\left (3 \, \cos \left (d x + c\right )^{4} - 7 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{4} - 5 \, \cos \left (d x + c\right )^{2} + 4\right )} \sin \left (d x + c\right ) + 4\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (10 \, \cos \left (d x + c\right )^{2} - 11\right )} \sin \left (d x + c\right ) + 167}{6 \, {\left (3 \, a^{4} d \cos \left (d x + c\right )^{4} - 7 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d + {\left (a^{4} d \cos \left (d x + c\right )^{4} - 5 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d\right )} \sin \left (d x + c\right )\right )}} \]

[In]

integrate(cos(d*x+c)*csc(d*x+c)^3/(a+a*sin(d*x+c))^4,x, algorithm="fricas")

[Out]

1/6*(60*cos(d*x + c)^4 - 230*cos(d*x + c)^2 + 60*(3*cos(d*x + c)^4 - 7*cos(d*x + c)^2 + (cos(d*x + c)^4 - 5*co
s(d*x + c)^2 + 4)*sin(d*x + c) + 4)*log(1/2*sin(d*x + c)) - 60*(3*cos(d*x + c)^4 - 7*cos(d*x + c)^2 + (cos(d*x
+ c)^4 - 5*cos(d*x + c)^2 + 4)*sin(d*x + c) + 4)*log(sin(d*x + c) + 1) - 15*(10*cos(d*x + c)^2 - 11)*sin(d*x
+ c) + 167)/(3*a^4*d*cos(d*x + c)^4 - 7*a^4*d*cos(d*x + c)^2 + 4*a^4*d + (a^4*d*cos(d*x + c)^4 - 5*a^4*d*cos(d
*x + c)^2 + 4*a^4*d)*sin(d*x + c))

Sympy [F]

\[ \int \frac {\cot (c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {\int \frac {\cos {\left (c + d x \right )} \csc ^{3}{\left (c + d x \right )}}{\sin ^{4}{\left (c + d x \right )} + 4 \sin ^{3}{\left (c + d x \right )} + 6 \sin ^{2}{\left (c + d x \right )} + 4 \sin {\left (c + d x \right )} + 1}\, dx}{a^{4}} \]

[In]

integrate(cos(d*x+c)*csc(d*x+c)**3/(a+a*sin(d*x+c))**4,x)

[Out]

Integral(cos(c + d*x)*csc(c + d*x)**3/(sin(c + d*x)**4 + 4*sin(c + d*x)**3 + 6*sin(c + d*x)**2 + 4*sin(c + d*x
) + 1), x)/a**4

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.96 \[ \int \frac {\cot (c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {\frac {60 \, \sin \left (d x + c\right )^{4} + 150 \, \sin \left (d x + c\right )^{3} + 110 \, \sin \left (d x + c\right )^{2} + 15 \, \sin \left (d x + c\right ) - 3}{a^{4} \sin \left (d x + c\right )^{5} + 3 \, a^{4} \sin \left (d x + c\right )^{4} + 3 \, a^{4} \sin \left (d x + c\right )^{3} + a^{4} \sin \left (d x + c\right )^{2}} - \frac {60 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{4}} + \frac {60 \, \log \left (\sin \left (d x + c\right )\right )}{a^{4}}}{6 \, d} \]

[In]

integrate(cos(d*x+c)*csc(d*x+c)^3/(a+a*sin(d*x+c))^4,x, algorithm="maxima")

[Out]

1/6*((60*sin(d*x + c)^4 + 150*sin(d*x + c)^3 + 110*sin(d*x + c)^2 + 15*sin(d*x + c) - 3)/(a^4*sin(d*x + c)^5 +
3*a^4*sin(d*x + c)^4 + 3*a^4*sin(d*x + c)^3 + a^4*sin(d*x + c)^2) - 60*log(sin(d*x + c) + 1)/a^4 + 60*log(sin
(d*x + c))/a^4)/d

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.74 \[ \int \frac {\cot (c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^4} \, dx=-\frac {\frac {60 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{4}} - \frac {60 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{4}} - \frac {60 \, \sin \left (d x + c\right )^{4} + 150 \, \sin \left (d x + c\right )^{3} + 110 \, \sin \left (d x + c\right )^{2} + 15 \, \sin \left (d x + c\right ) - 3}{a^{4} {\left (\sin \left (d x + c\right ) + 1\right )}^{3} \sin \left (d x + c\right )^{2}}}{6 \, d} \]

[In]

integrate(cos(d*x+c)*csc(d*x+c)^3/(a+a*sin(d*x+c))^4,x, algorithm="giac")

[Out]

-1/6*(60*log(abs(sin(d*x + c) + 1))/a^4 - 60*log(abs(sin(d*x + c)))/a^4 - (60*sin(d*x + c)^4 + 150*sin(d*x + c
)^3 + 110*sin(d*x + c)^2 + 15*sin(d*x + c) - 3)/(a^4*(sin(d*x + c) + 1)^3*sin(d*x + c)^2))/d

Mupad [B] (verification not implemented)

Time = 10.76 (sec) , antiderivative size = 286, normalized size of antiderivative = 2.18 \[ \int \frac {\cot (c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {10\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^4\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a^4\,d}-\frac {72\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\frac {465\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{2}+\frac {881\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{3}+\frac {255\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{2}-30\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-\frac {81\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}-5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {1}{2}}{d\,\left (4\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+24\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+60\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+80\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+60\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+24\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+4\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}-\frac {20\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{a^4\,d}+\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^4\,d} \]

[In]

int(cos(c + d*x)/(sin(c + d*x)^3*(a + a*sin(c + d*x))^4),x)

[Out]

(10*log(tan(c/2 + (d*x)/2)))/(a^4*d) - tan(c/2 + (d*x)/2)^2/(8*a^4*d) - ((255*tan(c/2 + (d*x)/2)^4)/2 - (81*ta
n(c/2 + (d*x)/2)^2)/2 - 30*tan(c/2 + (d*x)/2)^3 - 5*tan(c/2 + (d*x)/2) + (881*tan(c/2 + (d*x)/2)^5)/3 + (465*t
an(c/2 + (d*x)/2)^6)/2 + 72*tan(c/2 + (d*x)/2)^7 + 1/2)/(d*(4*a^4*tan(c/2 + (d*x)/2)^2 + 24*a^4*tan(c/2 + (d*x
)/2)^3 + 60*a^4*tan(c/2 + (d*x)/2)^4 + 80*a^4*tan(c/2 + (d*x)/2)^5 + 60*a^4*tan(c/2 + (d*x)/2)^6 + 24*a^4*tan(
c/2 + (d*x)/2)^7 + 4*a^4*tan(c/2 + (d*x)/2)^8)) - (20*log(tan(c/2 + (d*x)/2) + 1))/(a^4*d) + (2*tan(c/2 + (d*x
)/2))/(a^4*d)