\(\int \frac {\cot (c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^4} \, dx\) [255]

Optimal result

Integrand size = 25, antiderivative size = 111 \[ \int \frac {\cot (c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^4} \, dx=-\frac {\csc (c+d x)}{a^4 d}-\frac {4 \log (\sin (c+d x))}{a^4 d}+\frac {4 \log (1+\sin (c+d x))}{a^4 d}-\frac {1}{3 a d (a+a \sin (c+d x))^3}-\frac {1}{d \left (a^2+a^2 \sin (c+d x)\right )^2}-\frac {3}{d \left (a^4+a^4 \sin (c+d x)\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 111, 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} \[ \int \frac {\cot (c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^4} \, dx=-\frac {3}{d \left (a^4 \sin (c+d x)+a^4\right )}-\frac {\csc (c+d x)}{a^4 d}-\frac {4 \log (\sin (c+d x))}{a^4 d}+\frac {4 \log (\sin (c+d x)+1)}{a^4 d}-\frac {1}{d \left (a^2 \sin (c+d x)+a^2\right )^2}-\frac {1}{3 a d (a \sin (c+d x)+a)^3} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.65 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.66 \[ \int \frac {\cot (c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^4} \, dx=-\frac {3 \csc (c+d x)+12 \log (\sin (c+d x))-12 \log (1+\sin (c+d x))+\frac {1}{(1+\sin (c+d x))^3}+\frac {3}{(1+\sin (c+d x))^2}+\frac {9}{1+\sin (c+d x)}}{3 a^4 d} \]

[In]

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

[Out]

-1/3*(3*Csc[c + d*x] + 12*Log[Sin[c + d*x]] - 12*Log[1 + Sin[c + d*x]] + (1 + Sin[c + d*x])^(-3) + 3/(1 + Sin[
c + d*x])^2 + 9/(1 + Sin[c + d*x]))/(a^4*d)

Maple [A] (verified)

Time = 0.53 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.57

method	result	size
derivativedivides	\(-\frac {\csc \left (d x +c \right )+\frac {2}{\left (\csc \left (d x +c \right )+1\right )^{2}}-\frac {1}{3 \left (\csc \left (d x +c \right )+1\right )^{3}}-\frac {6}{\csc \left (d x +c \right )+1}-4 \ln \left (\csc \left (d x +c \right )+1\right )}{d \,a^{4}}\)	\(63\)
default	\(-\frac {\csc \left (d x +c \right )+\frac {2}{\left (\csc \left (d x +c \right )+1\right )^{2}}-\frac {1}{3 \left (\csc \left (d x +c \right )+1\right )^{3}}-\frac {6}{\csc \left (d x +c \right )+1}-4 \ln \left (\csc \left (d x +c \right )+1\right )}{d \,a^{4}}\)	\(63\)
risch	\(-\frac {8 i \left (15 i {\mathrm e}^{6 i \left (d x +c \right )}+3 \,{\mathrm e}^{7 i \left (d x +c \right )}-36 i {\mathrm e}^{4 i \left (d x +c \right )}-31 \,{\mathrm e}^{5 i \left (d x +c \right )}+15 i {\mathrm e}^{2 i \left (d x +c \right )}+31 \,{\mathrm e}^{3 i \left (d x +c \right )}-3 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{3 \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{6} d \,a^{4}}-\frac {4 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d \,a^{4}}+\frac {8 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{4}}\)	\(160\)
parallelrisch	\(\frac {\left (144 \cos \left (2 d x +2 c \right )-360 \sin \left (d x +c \right )+24 \sin \left (3 d x +3 c \right )-240\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (-72 \cos \left (2 d x +2 c \right )+180 \sin \left (d x +c \right )-12 \sin \left (3 d x +3 c \right )+120\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (166 \cos \left (d x +c \right )+44 \cos \left (2 d x +2 c \right )-22 \cos \left (3 d x +3 c \right )-188\right ) \cot \left (\frac {d x}{2}+\frac {c}{2}\right )+6 \sec \left (\frac {d x}{2}+\frac {c}{2}\right ) \csc \left (\frac {d x}{2}+\frac {c}{2}\right )+108 \cos \left (2 d x +2 c \right )-108}{3 d \,a^{4} \left (-10+6 \cos \left (2 d x +2 c \right )+\sin \left (3 d x +3 c \right )-15 \sin \left (d x +c \right )\right )}\)	\(204\)
norman	\(\frac {-\frac {1}{2 a d}-\frac {\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d a}+\frac {51 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}+\frac {51 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}+\frac {209 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}+\frac {209 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}+\frac {1159 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d a}+\frac {1159 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d a}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}-\frac {4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{4}}+\frac {8 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{4}}\)	\(210\)

[In]

int(cos(d*x+c)*csc(d*x+c)^2/(a+a*sin(d*x+c))^4,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.81 \[ \int \frac {\cot (c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {30 \, \cos \left (d x + c\right )^{2} - 12 \, {\left (\cos \left (d x + c\right )^{4} - 5 \, \cos \left (d x + c\right )^{2} - {\left (3 \, \cos \left (d x + c\right )^{2} - 4\right )} \sin \left (d x + c\right ) + 4\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + 12 \, {\left (\cos \left (d x + c\right )^{4} - 5 \, \cos \left (d x + c\right )^{2} - {\left (3 \, \cos \left (d x + c\right )^{2} - 4\right )} \sin \left (d x + c\right ) + 4\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (6 \, \cos \left (d x + c\right )^{2} - 17\right )} \sin \left (d x + c\right ) - 33}{3 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} - 5 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d - {\left (3 \, a^{4} d \cos \left (d x + c\right )^{2} - 4 \, a^{4} d\right )} \sin \left (d x + c\right )\right )}} \]

[In]

integrate(cos(d*x+c)*csc(d*x+c)^2/(a+a*sin(d*x+c))^4,x, algorithm="fricas")

[Out]

1/3*(30*cos(d*x + c)^2 - 12*(cos(d*x + c)^4 - 5*cos(d*x + c)^2 - (3*cos(d*x + c)^2 - 4)*sin(d*x + c) + 4)*log(
1/2*sin(d*x + c)) + 12*(cos(d*x + c)^4 - 5*cos(d*x + c)^2 - (3*cos(d*x + c)^2 - 4)*sin(d*x + c) + 4)*log(sin(d
*x + c) + 1) + 2*(6*cos(d*x + c)^2 - 17)*sin(d*x + c) - 33)/(a^4*d*cos(d*x + c)^4 - 5*a^4*d*cos(d*x + c)^2 + 4
*a^4*d - (3*a^4*d*cos(d*x + c)^2 - 4*a^4*d)*sin(d*x + c))

Sympy [F]

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

[In]

integrate(cos(d*x+c)*csc(d*x+c)**2/(a+a*sin(d*x+c))**4,x)

[Out]

Integral(cos(c + d*x)*csc(c + d*x)**2/(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 = 0.20 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.03 \[ \int \frac {\cot (c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^4} \, dx=-\frac {\frac {12 \, \sin \left (d x + c\right )^{3} + 30 \, \sin \left (d x + c\right )^{2} + 22 \, \sin \left (d x + c\right ) + 3}{a^{4} \sin \left (d x + c\right )^{4} + 3 \, a^{4} \sin \left (d x + c\right )^{3} + 3 \, a^{4} \sin \left (d x + c\right )^{2} + a^{4} \sin \left (d x + c\right )} - \frac {12 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{4}} + \frac {12 \, \log \left (\sin \left (d x + c\right )\right )}{a^{4}}}{3 \, d} \]

[In]

integrate(cos(d*x+c)*csc(d*x+c)^2/(a+a*sin(d*x+c))^4,x, algorithm="maxima")

[Out]

-1/3*((12*sin(d*x + c)^3 + 30*sin(d*x + c)^2 + 22*sin(d*x + c) + 3)/(a^4*sin(d*x + c)^4 + 3*a^4*sin(d*x + c)^3
+ 3*a^4*sin(d*x + c)^2 + a^4*sin(d*x + c)) - 12*log(sin(d*x + c) + 1)/a^4 + 12*log(sin(d*x + c))/a^4)/d

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.78 \[ \int \frac {\cot (c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {\frac {12 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{4}} - \frac {12 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{4}} - \frac {12 \, \sin \left (d x + c\right )^{3} + 30 \, \sin \left (d x + c\right )^{2} + 22 \, \sin \left (d x + c\right ) + 3}{a^{4} {\left (\sin \left (d x + c\right ) + 1\right )}^{3} \sin \left (d x + c\right )}}{3 \, d} \]

[In]

integrate(cos(d*x+c)*csc(d*x+c)^2/(a+a*sin(d*x+c))^4,x, algorithm="giac")

[Out]

1/3*(12*log(abs(sin(d*x + c) + 1))/a^4 - 12*log(abs(sin(d*x + c)))/a^4 - (12*sin(d*x + c)^3 + 30*sin(d*x + c)^
2 + 22*sin(d*x + c) + 3)/(a^4*(sin(d*x + c) + 1)^3*sin(d*x + c)))/d

Mupad [B] (verification not implemented)

Time = 10.13 (sec) , antiderivative size = 251, normalized size of antiderivative = 2.26 \[ \int \frac {\cot (c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {23\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+74\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\frac {307\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}+60\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+9\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1}{d\,\left (2\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+12\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+30\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+40\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+30\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+12\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}-\frac {4\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^4\,d}+\frac {8\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{a^4\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a^4\,d} \]

[In]

int(cos(c + d*x)/(sin(c + d*x)^2*(a + a*sin(c + d*x))^4),x)

[Out]

(9*tan(c/2 + (d*x)/2)^2 - 6*tan(c/2 + (d*x)/2) + 60*tan(c/2 + (d*x)/2)^3 + (307*tan(c/2 + (d*x)/2)^4)/3 + 74*t
an(c/2 + (d*x)/2)^5 + 23*tan(c/2 + (d*x)/2)^6 - 1)/(d*(12*a^4*tan(c/2 + (d*x)/2)^2 + 30*a^4*tan(c/2 + (d*x)/2)
^3 + 40*a^4*tan(c/2 + (d*x)/2)^4 + 30*a^4*tan(c/2 + (d*x)/2)^5 + 12*a^4*tan(c/2 + (d*x)/2)^6 + 2*a^4*tan(c/2 +
(d*x)/2)^7 + 2*a^4*tan(c/2 + (d*x)/2))) - (4*log(tan(c/2 + (d*x)/2)))/(a^4*d) + (8*log(tan(c/2 + (d*x)/2) + 1
))/(a^4*d) - tan(c/2 + (d*x)/2)/(2*a^4*d)