\(\int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^3} \, dx\) [562]

Optimal result

Integrand size = 27, antiderivative size = 117 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {4 \csc (c+d x)}{a^3 d}+\frac {2 \csc ^2(c+d x)}{a^3 d}-\frac {4 \csc ^3(c+d x)}{3 a^3 d}+\frac {3 \csc ^4(c+d x)}{4 a^3 d}-\frac {\csc ^5(c+d x)}{5 a^3 d}-\frac {4 \log (\sin (c+d x))}{a^3 d}+\frac {4 \log (1+\sin (c+d x))}{a^3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 117, 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 = {2915, 12, 90} \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\csc ^5(c+d x)}{5 a^3 d}+\frac {3 \csc ^4(c+d x)}{4 a^3 d}-\frac {4 \csc ^3(c+d x)}{3 a^3 d}+\frac {2 \csc ^2(c+d x)}{a^3 d}-\frac {4 \csc (c+d x)}{a^3 d}-\frac {4 \log (\sin (c+d x))}{a^3 d}+\frac {4 \log (\sin (c+d x)+1)}{a^3 d} \]

[In]

Int[(Cot[c + d*x]^5*Csc[c + d*x])/(a + a*Sin[c + d*x])^3,x]

[Out]

(-4*Csc[c + d*x])/(a^3*d) + (2*Csc[c + d*x]^2)/(a^3*d) - (4*Csc[c + d*x]^3)/(3*a^3*d) + (3*Csc[c + d*x]^4)/(4*
a^3*d) - Csc[c + d*x]^5/(5*a^3*d) - (4*Log[Sin[c + d*x]])/(a^3*d) + (4*Log[1 + Sin[c + d*x]])/(a^3*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 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.68 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {240 \csc (c+d x)-120 \csc ^2(c+d x)+80 \csc ^3(c+d x)-45 \csc ^4(c+d x)+12 \csc ^5(c+d x)+240 \log (\sin (c+d x))-240 \log (1+\sin (c+d x))}{60 a^3 d} \]

[In]

Integrate[(Cot[c + d*x]^5*Csc[c + d*x])/(a + a*Sin[c + d*x])^3,x]

[Out]

-1/60*(240*Csc[c + d*x] - 120*Csc[c + d*x]^2 + 80*Csc[c + d*x]^3 - 45*Csc[c + d*x]^4 + 12*Csc[c + d*x]^5 + 240
*Log[Sin[c + d*x]] - 240*Log[1 + Sin[c + d*x]])/(a^3*d)

Maple [A] (verified)

Time = 0.33 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.58

method	result	size
derivativedivides	\(\frac {-\frac {\left (\csc ^{5}\left (d x +c \right )\right )}{5}+\frac {3 \left (\csc ^{4}\left (d x +c \right )\right )}{4}-\frac {4 \left (\csc ^{3}\left (d x +c \right )\right )}{3}+2 \left (\csc ^{2}\left (d x +c \right )\right )-4 \csc \left (d x +c \right )+4 \ln \left (\csc \left (d x +c \right )+1\right )}{d \,a^{3}}\)	\(68\)
default	\(\frac {-\frac {\left (\csc ^{5}\left (d x +c \right )\right )}{5}+\frac {3 \left (\csc ^{4}\left (d x +c \right )\right )}{4}-\frac {4 \left (\csc ^{3}\left (d x +c \right )\right )}{3}+2 \left (\csc ^{2}\left (d x +c \right )\right )-4 \csc \left (d x +c \right )+4 \ln \left (\csc \left (d x +c \right )+1\right )}{d \,a^{3}}\)	\(68\)
parallelrisch	\(\frac {-6 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-6 \left (\cot ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+45 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+45 \left (\cot ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-190 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-190 \left (\cot ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+660 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+660 \left (\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2460 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2460 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )-3840 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+7680 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{960 d \,a^{3}}\)	\(162\)
risch	\(-\frac {4 i \left (30 \,{\mathrm e}^{9 i \left (d x +c \right )}-160 \,{\mathrm e}^{7 i \left (d x +c \right )}-30 i {\mathrm e}^{8 i \left (d x +c \right )}+284 \,{\mathrm e}^{5 i \left (d x +c \right )}+135 i {\mathrm e}^{6 i \left (d x +c \right )}-160 \,{\mathrm e}^{3 i \left (d x +c \right )}-135 i {\mathrm e}^{4 i \left (d x +c \right )}+30 \,{\mathrm e}^{i \left (d x +c \right )}+30 i {\mathrm e}^{2 i \left (d x +c \right )}\right )}{15 a^{3} d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}+\frac {8 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{3}}-\frac {4 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d \,a^{3}}\)	\(169\)
norman	\(\frac {-\frac {1}{160 a d}+\frac {\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )}{64 d a}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{64 d a}-\frac {5 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{192 d a}+\frac {5 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{48 d a}-\frac {2 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}-\frac {2 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {5 \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{48 d a}-\frac {5 \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{192 d a}-\frac {\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )}{160 d a}+\frac {259 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d a}+\frac {259 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d a}+\frac {339 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}+\frac {339 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3}}+\frac {8 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{3}}\)	\(322\)

[In]

int(cos(d*x+c)^5*csc(d*x+c)^6/(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.38 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {240 \, \cos \left (d x + c\right )^{4} + 240 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) - 240 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - 560 \, \cos \left (d x + c\right )^{2} + 15 \, {\left (8 \, \cos \left (d x + c\right )^{2} - 11\right )} \sin \left (d x + c\right ) + 332}{60 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} - 2 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d\right )} \sin \left (d x + c\right )} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^6/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/60*(240*cos(d*x + c)^4 + 240*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*log(1/2*sin(d*x + c))*sin(d*x + c) - 2
40*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*log(sin(d*x + c) + 1)*sin(d*x + c) - 560*cos(d*x + c)^2 + 15*(8*cos
(d*x + c)^2 - 11)*sin(d*x + c) + 332)/((a^3*d*cos(d*x + c)^4 - 2*a^3*d*cos(d*x + c)^2 + a^3*d)*sin(d*x + c))

Sympy [F(-1)]

Timed out. \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**5*csc(d*x+c)**6/(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.73 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {240 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}} - \frac {240 \, \log \left (\sin \left (d x + c\right )\right )}{a^{3}} - \frac {240 \, \sin \left (d x + c\right )^{4} - 120 \, \sin \left (d x + c\right )^{3} + 80 \, \sin \left (d x + c\right )^{2} - 45 \, \sin \left (d x + c\right ) + 12}{a^{3} \sin \left (d x + c\right )^{5}}}{60 \, d} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^6/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/60*(240*log(sin(d*x + c) + 1)/a^3 - 240*log(sin(d*x + c))/a^3 - (240*sin(d*x + c)^4 - 120*sin(d*x + c)^3 + 8
0*sin(d*x + c)^2 - 45*sin(d*x + c) + 12)/(a^3*sin(d*x + c)^5))/d

Giac [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.74 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {7680 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {3840 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}} + \frac {8768 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 2460 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 660 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 190 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 45 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6}{a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}} - \frac {6 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 45 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 190 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 660 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2460 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{960 \, d} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^6/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/960*(7680*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 3840*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 + (8768*tan(1/2*d
*x + 1/2*c)^5 - 2460*tan(1/2*d*x + 1/2*c)^4 + 660*tan(1/2*d*x + 1/2*c)^3 - 190*tan(1/2*d*x + 1/2*c)^2 + 45*tan
(1/2*d*x + 1/2*c) - 6)/(a^3*tan(1/2*d*x + 1/2*c)^5) - (6*a^12*tan(1/2*d*x + 1/2*c)^5 - 45*a^12*tan(1/2*d*x + 1
/2*c)^4 + 190*a^12*tan(1/2*d*x + 1/2*c)^3 - 660*a^12*tan(1/2*d*x + 1/2*c)^2 + 2460*a^12*tan(1/2*d*x + 1/2*c))/
a^15)/d

Mupad [B] (verification not implemented)

Time = 10.78 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.74 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {11\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{16\,a^3\,d}-\frac {19\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{96\,a^3\,d}+\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,a^3\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160\,a^3\,d}-\frac {4\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^3\,d}+\frac {8\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{a^3\,d}-\frac {41\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,a^3\,d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (82\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-22\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {19\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}-\frac {3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}+\frac {1}{5}\right )}{32\,a^3\,d} \]

[In]

int(cos(c + d*x)^5/(sin(c + d*x)^6*(a + a*sin(c + d*x))^3),x)

[Out]

(11*tan(c/2 + (d*x)/2)^2)/(16*a^3*d) - (19*tan(c/2 + (d*x)/2)^3)/(96*a^3*d) + (3*tan(c/2 + (d*x)/2)^4)/(64*a^3
*d) - tan(c/2 + (d*x)/2)^5/(160*a^3*d) - (4*log(tan(c/2 + (d*x)/2)))/(a^3*d) + (8*log(tan(c/2 + (d*x)/2) + 1))
/(a^3*d) - (41*tan(c/2 + (d*x)/2))/(16*a^3*d) - (cot(c/2 + (d*x)/2)^5*((19*tan(c/2 + (d*x)/2)^2)/3 - (3*tan(c/
2 + (d*x)/2))/2 - 22*tan(c/2 + (d*x)/2)^3 + 82*tan(c/2 + (d*x)/2)^4 + 1/5))/(32*a^3*d)