\(\int \frac {\cot (c+d x) \csc ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx\) [248]

Optimal result

Integrand size = 27, antiderivative size = 126 \[ \int \frac {\cot (c+d x) \csc ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {6 \csc (c+d x)}{a^3 d}+\frac {3 \csc ^2(c+d x)}{2 a^3 d}-\frac {\csc ^3(c+d x)}{3 a^3 d}-\frac {10 \log (\sin (c+d x))}{a^3 d}+\frac {10 \log (1+\sin (c+d x))}{a^3 d}-\frac {1}{2 a d (a+a \sin (c+d x))^2}-\frac {4}{d \left (a^3+a^3 \sin (c+d x)\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 126, 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} \[ \int \frac {\cot (c+d x) \csc ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {4}{d \left (a^3 \sin (c+d x)+a^3\right )}-\frac {\csc ^3(c+d x)}{3 a^3 d}+\frac {3 \csc ^2(c+d x)}{2 a^3 d}-\frac {6 \csc (c+d x)}{a^3 d}-\frac {10 \log (\sin (c+d x))}{a^3 d}+\frac {10 \log (\sin (c+d x)+1)}{a^3 d}-\frac {1}{2 a d (a \sin (c+d x)+a)^2} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 1.57 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.64 \[ \int \frac {\cot (c+d x) \csc ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {36 \csc (c+d x)-9 \csc ^2(c+d x)+2 \csc ^3(c+d x)+60 \log (\sin (c+d x))-60 \log (1+\sin (c+d x))+\frac {3 (9+8 \sin (c+d x))}{(1+\sin (c+d x))^2}}{6 a^3 d} \]

[In]

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

[Out]

-1/6*(36*Csc[c + d*x] - 9*Csc[c + d*x]^2 + 2*Csc[c + d*x]^3 + 60*Log[Sin[c + d*x]] - 60*Log[1 + Sin[c + d*x]]
+ (3*(9 + 8*Sin[c + d*x]))/(1 + Sin[c + d*x])^2)/(a^3*d)

Maple [A] (verified)

Time = 0.60 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.58

method	result	size
derivativedivides	\(-\frac {\frac {\left (\csc ^{3}\left (d x +c \right )\right )}{3}-\frac {3 \left (\csc ^{2}\left (d x +c \right )\right )}{2}+6 \csc \left (d x +c \right )-\frac {5}{\csc \left (d x +c \right )+1}+\frac {1}{2 \left (\csc \left (d x +c \right )+1\right )^{2}}-10 \ln \left (\csc \left (d x +c \right )+1\right )}{d \,a^{3}}\)	\(73\)
default	\(-\frac {\frac {\left (\csc ^{3}\left (d x +c \right )\right )}{3}-\frac {3 \left (\csc ^{2}\left (d x +c \right )\right )}{2}+6 \csc \left (d x +c \right )-\frac {5}{\csc \left (d x +c \right )+1}+\frac {1}{2 \left (\csc \left (d x +c \right )+1\right )^{2}}-10 \ln \left (\csc \left (d x +c \right )+1\right )}{d \,a^{3}}\)	\(73\)
risch	\(-\frac {4 i \left (45 i {\mathrm e}^{8 i \left (d x +c \right )}+15 \,{\mathrm e}^{9 i \left (d x +c \right )}-125 i {\mathrm e}^{6 i \left (d x +c \right )}-80 \,{\mathrm e}^{7 i \left (d x +c \right )}+125 i {\mathrm e}^{4 i \left (d x +c \right )}+138 \,{\mathrm e}^{5 i \left (d x +c \right )}-45 i {\mathrm e}^{2 i \left (d x +c \right )}-80 \,{\mathrm e}^{3 i \left (d x +c \right )}+15 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{3 \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{4} d \,a^{3}}+\frac {20 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{3}}-\frac {10 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d \,a^{3}}\)	\(183\)
norman	\(\frac {-\frac {1}{24 a d}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{6 d a}-\frac {5 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}-\frac {5 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )}{6 d a}-\frac {\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )}{24 d a}+\frac {325 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}+\frac {325 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}+\frac {425 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d a}+\frac {425 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d a}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {10 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3}}+\frac {20 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{3}}\)	\(246\)
parallelrisch	\(\frac {3840 \left (-3+\cos \left (2 d x +2 c \right )-4 \sin \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+1920 \left (3-\cos \left (2 d x +2 c \right )+4 \sin \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+16 \left (\sec ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )+40 \sec \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\csc ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+5 \left (\left (-17 \cos \left (\frac {9 d x}{2}+\frac {9 c}{2}\right )-17 \cos \left (\frac {7 d x}{2}+\frac {7 c}{2}\right )+64 \cos \left (\frac {5 d x}{2}+\frac {5 c}{2}\right )+64 \cos \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )-110 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\sec ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2944 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\csc ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-7040 \left (\cot ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\cos \left (d x +c \right )-3\right )}{192 d \,a^{3} \left (-3+\cos \left (2 d x +2 c \right )-4 \sin \left (d x +c \right )\right )}\)	\(247\)

[In]

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

[Out]

-1/d/a^3*(1/3*csc(d*x+c)^3-3/2*csc(d*x+c)^2+6*csc(d*x+c)-5/(csc(d*x+c)+1)+1/2/(csc(d*x+c)+1)^2-10*ln(csc(d*x+c
)+1))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 242 vs. \(2 (120) = 240\).

Time = 0.32 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.92 \[ \int \frac {\cot (c+d x) \csc ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {60 \, \cos \left (d x + c\right )^{4} - 140 \, \cos \left (d x + c\right )^{2} + 60 \, {\left (2 \, \cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{4} - 3 \, \cos \left (d x + c\right )^{2} + 2\right )} \sin \left (d x + c\right ) + 2\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 60 \, {\left (2 \, \cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{4} - 3 \, \cos \left (d x + c\right )^{2} + 2\right )} \sin \left (d x + c\right ) + 2\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 5 \, {\left (18 \, \cos \left (d x + c\right )^{2} - 17\right )} \sin \left (d x + c\right ) + 82}{6 \, {\left (2 \, a^{3} d \cos \left (d x + c\right )^{4} - 4 \, a^{3} d \cos \left (d x + c\right )^{2} + 2 \, a^{3} d + {\left (a^{3} d \cos \left (d x + c\right )^{4} - 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 2 \, a^{3} d\right )} \sin \left (d x + c\right )\right )}} \]

[In]

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

[Out]

-1/6*(60*cos(d*x + c)^4 - 140*cos(d*x + c)^2 + 60*(2*cos(d*x + c)^4 - 4*cos(d*x + c)^2 + (cos(d*x + c)^4 - 3*c
os(d*x + c)^2 + 2)*sin(d*x + c) + 2)*log(1/2*sin(d*x + c)) - 60*(2*cos(d*x + c)^4 - 4*cos(d*x + c)^2 + (cos(d*
x + c)^4 - 3*cos(d*x + c)^2 + 2)*sin(d*x + c) + 2)*log(sin(d*x + c) + 1) - 5*(18*cos(d*x + c)^2 - 17)*sin(d*x
+ c) + 82)/(2*a^3*d*cos(d*x + c)^4 - 4*a^3*d*cos(d*x + c)^2 + 2*a^3*d + (a^3*d*cos(d*x + c)^4 - 3*a^3*d*cos(d*
x + c)^2 + 2*a^3*d)*sin(d*x + c))

Sympy [F]

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

[In]

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

[Out]

Integral(cos(c + d*x)*csc(c + d*x)**4/(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 = 0.24 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.90 \[ \int \frac {\cot (c+d x) \csc ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\frac {60 \, \sin \left (d x + c\right )^{4} + 90 \, \sin \left (d x + c\right )^{3} + 20 \, \sin \left (d x + c\right )^{2} - 5 \, \sin \left (d x + c\right ) + 2}{a^{3} \sin \left (d x + c\right )^{5} + 2 \, a^{3} \sin \left (d x + c\right )^{4} + a^{3} \sin \left (d x + c\right )^{3}} - \frac {60 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}} + \frac {60 \, \log \left (\sin \left (d x + c\right )\right )}{a^{3}}}{6 \, d} \]

[In]

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

[Out]

-1/6*((60*sin(d*x + c)^4 + 90*sin(d*x + c)^3 + 20*sin(d*x + c)^2 - 5*sin(d*x + c) + 2)/(a^3*sin(d*x + c)^5 + 2
*a^3*sin(d*x + c)^4 + a^3*sin(d*x + c)^3) - 60*log(sin(d*x + c) + 1)/a^3 + 60*log(sin(d*x + c))/a^3)/d

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.77 \[ \int \frac {\cot (c+d x) \csc ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {60 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3}} - \frac {60 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{3}} - \frac {60 \, \sin \left (d x + c\right )^{4} + 90 \, \sin \left (d x + c\right )^{3} + 20 \, \sin \left (d x + c\right )^{2} - 5 \, \sin \left (d x + c\right ) + 2}{a^{3} {\left (\sin \left (d x + c\right ) + 1\right )}^{2} \sin \left (d x + c\right )^{3}}}{6 \, d} \]

[In]

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

[Out]

1/6*(60*log(abs(sin(d*x + c) + 1))/a^3 - 60*log(abs(sin(d*x + c)))/a^3 - (60*sin(d*x + c)^4 + 90*sin(d*x + c)^
3 + 20*sin(d*x + c)^2 - 5*sin(d*x + c) + 2)/(a^3*(sin(d*x + c) + 1)^2*sin(d*x + c)^3))/d

Mupad [B] (verification not implemented)

Time = 10.20 (sec) , antiderivative size = 260, normalized size of antiderivative = 2.06 \[ \int \frac {\cot (c+d x) \csc ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a^3\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,a^3\,d}-\frac {10\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^3\,d}+\frac {20\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{a^3\,d}-\frac {25\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,a^3\,d}-\frac {-55\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-47\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\frac {175\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}+\frac {250\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-\frac {5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}+\frac {1}{3}}{d\,\left (8\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+32\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+48\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+32\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+8\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\right )} \]

[In]

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

[Out]

(3*tan(c/2 + (d*x)/2)^2)/(8*a^3*d) - tan(c/2 + (d*x)/2)^3/(24*a^3*d) - (10*log(tan(c/2 + (d*x)/2)))/(a^3*d) +
(20*log(tan(c/2 + (d*x)/2) + 1))/(a^3*d) - (25*tan(c/2 + (d*x)/2))/(8*a^3*d) - (15*tan(c/2 + (d*x)/2)^2 - (5*t
an(c/2 + (d*x)/2))/3 + (250*tan(c/2 + (d*x)/2)^3)/3 + (175*tan(c/2 + (d*x)/2)^4)/3 - 47*tan(c/2 + (d*x)/2)^5 -
55*tan(c/2 + (d*x)/2)^6 + 1/3)/(d*(8*a^3*tan(c/2 + (d*x)/2)^3 + 32*a^3*tan(c/2 + (d*x)/2)^4 + 48*a^3*tan(c/2
+ (d*x)/2)^5 + 32*a^3*tan(c/2 + (d*x)/2)^6 + 8*a^3*tan(c/2 + (d*x)/2)^7))