\(\int \frac {\cot ^6(c+d x)}{(a+a \sin (c+d x))^2} \, dx\) [642]

Optimal result

Integrand size = 21, antiderivative size = 100 \[ \int \frac {\cot ^6(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\text {arctanh}(\cos (c+d x))}{4 a^2 d}-\frac {2 \cot ^3(c+d x)}{3 a^2 d}-\frac {\cot ^5(c+d x)}{5 a^2 d}-\frac {\cot (c+d x) \csc (c+d x)}{4 a^2 d}+\frac {\cot (c+d x) \csc ^3(c+d x)}{2 a^2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2788, 3852, 8, 3853, 3855} \[ \int \frac {\cot ^6(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\text {arctanh}(\cos (c+d x))}{4 a^2 d}-\frac {\cot ^5(c+d x)}{5 a^2 d}-\frac {2 \cot ^3(c+d x)}{3 a^2 d}+\frac {\cot (c+d x) \csc ^3(c+d x)}{2 a^2 d}-\frac {\cot (c+d x) \csc (c+d x)}{4 a^2 d} \]

[In]

Int[Cot[c + d*x]^6/(a + a*Sin[c + d*x])^2,x]

[Out]

-1/4*ArcTanh[Cos[c + d*x]]/(a^2*d) - (2*Cot[c + d*x]^3)/(3*a^2*d) - Cot[c + d*x]^5/(5*a^2*d) - (Cot[c + d*x]*C
sc[c + d*x])/(4*a^2*d) + (Cot[c + d*x]*Csc[c + d*x]^3)/(2*a^2*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2788

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^(p_), x_Symbol] :> Dist[a^p, Int[Expan
dIntegrand[Sin[e + f*x]^p*((a + b*Sin[e + f*x])^(m - p/2)/(a - b*Sin[e + f*x])^(p/2)), x], x], x] /; FreeQ[{a,
b, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p/2] && (LtQ[p, 0] || GtQ[m - p/2, 0])

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[2*n]

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \left (-a^4 \csc ^2(c+d x)+2 a^4 \csc ^3(c+d x)-2 a^4 \csc ^5(c+d x)+a^4 \csc ^6(c+d x)\right ) \, dx}{a^6} \\ & = -\frac {\int \csc ^2(c+d x) \, dx}{a^2}+\frac {\int \csc ^6(c+d x) \, dx}{a^2}+\frac {2 \int \csc ^3(c+d x) \, dx}{a^2}-\frac {2 \int \csc ^5(c+d x) \, dx}{a^2} \\ & = -\frac {\cot (c+d x) \csc (c+d x)}{a^2 d}+\frac {\cot (c+d x) \csc ^3(c+d x)}{2 a^2 d}+\frac {\int \csc (c+d x) \, dx}{a^2}-\frac {3 \int \csc ^3(c+d x) \, dx}{2 a^2}+\frac {\text {Subst}(\int 1 \, dx,x,\cot (c+d x))}{a^2 d}-\frac {\text {Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,\cot (c+d x)\right )}{a^2 d} \\ & = -\frac {\text {arctanh}(\cos (c+d x))}{a^2 d}-\frac {2 \cot ^3(c+d x)}{3 a^2 d}-\frac {\cot ^5(c+d x)}{5 a^2 d}-\frac {\cot (c+d x) \csc (c+d x)}{4 a^2 d}+\frac {\cot (c+d x) \csc ^3(c+d x)}{2 a^2 d}-\frac {3 \int \csc (c+d x) \, dx}{4 a^2} \\ & = -\frac {\text {arctanh}(\cos (c+d x))}{4 a^2 d}-\frac {2 \cot ^3(c+d x)}{3 a^2 d}-\frac {\cot ^5(c+d x)}{5 a^2 d}-\frac {\cot (c+d x) \csc (c+d x)}{4 a^2 d}+\frac {\cot (c+d x) \csc ^3(c+d x)}{2 a^2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.87 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.89 \[ \int \frac {\cot ^6(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\csc ^5(c+d x) \left (200 \cos (c+d x)+20 \cos (3 (c+d x))-28 \cos (5 (c+d x))+150 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)-150 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)-180 \sin (2 (c+d x))-75 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (3 (c+d x))+75 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (3 (c+d x))-30 \sin (4 (c+d x))+15 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (5 (c+d x))-15 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (5 (c+d x))\right )}{960 a^2 d} \]

[In]

Integrate[Cot[c + d*x]^6/(a + a*Sin[c + d*x])^2,x]

[Out]

-1/960*(Csc[c + d*x]^5*(200*Cos[c + d*x] + 20*Cos[3*(c + d*x)] - 28*Cos[5*(c + d*x)] + 150*Log[Cos[(c + d*x)/2
]]*Sin[c + d*x] - 150*Log[Sin[(c + d*x)/2]]*Sin[c + d*x] - 180*Sin[2*(c + d*x)] - 75*Log[Cos[(c + d*x)/2]]*Sin
[3*(c + d*x)] + 75*Log[Sin[(c + d*x)/2]]*Sin[3*(c + d*x)] - 30*Sin[4*(c + d*x)] + 15*Log[Cos[(c + d*x)/2]]*Sin
[5*(c + d*x)] - 15*Log[Sin[(c + d*x)/2]]*Sin[5*(c + d*x)]))/(a^2*d)

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.22

method	result	size
derivativedivides	\(\frac {\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {5 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {6}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+8 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}-\frac {1}{5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}-\frac {5}{3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}}{32 d \,a^{2}}\)	\(122\)
default	\(\frac {\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {5 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {6}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+8 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}-\frac {1}{5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}-\frac {5}{3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}}{32 d \,a^{2}}\)	\(122\)
parallelrisch	\(\frac {3 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3 \left (\cot ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-15 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+15 \left (\cot ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+25 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-25 \left (\cot ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-90 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+90 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )+120 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{480 d \,a^{2}}\)	\(122\)
risch	\(\frac {60 i {\mathrm e}^{8 i \left (d x +c \right )}+15 \,{\mathrm e}^{9 i \left (d x +c \right )}-240 i {\mathrm e}^{6 i \left (d x +c \right )}+90 \,{\mathrm e}^{7 i \left (d x +c \right )}+40 i {\mathrm e}^{4 i \left (d x +c \right )}-80 i {\mathrm e}^{2 i \left (d x +c \right )}-90 \,{\mathrm e}^{3 i \left (d x +c \right )}+28 i-15 \,{\mathrm e}^{i \left (d x +c \right )}}{30 a^{2} d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{4 d \,a^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{4 d \,a^{2}}\)	\(158\)
norman	\(\frac {-\frac {1}{160 a d}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{80 d a}+\frac {11 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{480 d a}-\frac {11 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{160 d a}+\frac {\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )}{16 d a}-\frac {\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )}{16 d a}+\frac {11 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{160 d a}-\frac {11 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{480 d a}-\frac {\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )}{80 d a}+\frac {\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )}{160 d a}+\frac {49 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{48 d a}+\frac {37 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d a}+\frac {61 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d a}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{2}}\)	\(283\)

[In]

int(cos(d*x+c)^6*csc(d*x+c)^6/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/32/d/a^2*(1/5*tan(1/2*d*x+1/2*c)^5-tan(1/2*d*x+1/2*c)^4+5/3*tan(1/2*d*x+1/2*c)^3-6*tan(1/2*d*x+1/2*c)+6/tan(
1/2*d*x+1/2*c)+8*ln(tan(1/2*d*x+1/2*c))+1/tan(1/2*d*x+1/2*c)^4-1/5/tan(1/2*d*x+1/2*c)^5-5/3/tan(1/2*d*x+1/2*c)
^3)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.67 \[ \int \frac {\cot ^6(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {56 \, \cos \left (d x + c\right )^{5} - 80 \, \cos \left (d x + c\right )^{3} - 15 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 15 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 30 \, {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} - 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d\right )} \sin \left (d x + c\right )} \]

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^6/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/120*(56*cos(d*x + c)^5 - 80*cos(d*x + c)^3 - 15*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*log(1/2*cos(d*x + c)
+ 1/2)*sin(d*x + c) + 15*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) +
30*(cos(d*x + c)^3 + cos(d*x + c))*sin(d*x + c))/((a^2*d*cos(d*x + c)^4 - 2*a^2*d*cos(d*x + c)^2 + a^2*d)*sin(
d*x + c))

Sympy [F(-1)]

Timed out. \[ \int \frac {\cot ^6(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**6*csc(d*x+c)**6/(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 195 vs. \(2 (90) = 180\).

Time = 0.23 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.95 \[ \int \frac {\cot ^6(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {\frac {90 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {25 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{2}} - \frac {120 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} - \frac {{\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {25 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {90 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - 3\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{5}}{a^{2} \sin \left (d x + c\right )^{5}}}{480 \, d} \]

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^6/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/480*((90*sin(d*x + c)/(cos(d*x + c) + 1) - 25*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 15*sin(d*x + c)^4/(cos(
d*x + c) + 1)^4 - 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^2 - 120*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^2 -
(15*sin(d*x + c)/(cos(d*x + c) + 1) - 25*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 90*sin(d*x + c)^4/(cos(d*x + c)
+ 1)^4 - 3)*(cos(d*x + c) + 1)^5/(a^2*sin(d*x + c)^5))/d

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.57 \[ \int \frac {\cot ^6(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {120 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} - \frac {274 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 90 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 25 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3}{a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}} + \frac {3 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 15 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 25 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 90 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{10}}}{480 \, d} \]

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^6/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/480*(120*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 - (274*tan(1/2*d*x + 1/2*c)^5 - 90*tan(1/2*d*x + 1/2*c)^4 + 25*t
an(1/2*d*x + 1/2*c)^2 - 15*tan(1/2*d*x + 1/2*c) + 3)/(a^2*tan(1/2*d*x + 1/2*c)^5) + (3*a^8*tan(1/2*d*x + 1/2*c
)^5 - 15*a^8*tan(1/2*d*x + 1/2*c)^4 + 25*a^8*tan(1/2*d*x + 1/2*c)^3 - 90*a^8*tan(1/2*d*x + 1/2*c))/a^10)/d

Mupad [B] (verification not implemented)

Time = 9.55 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.49 \[ \int \frac {\cot ^6(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{96\,a^2\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{32\,a^2\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160\,a^2\,d}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,a^2\,d}-\frac {3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,a^2\,d}+\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {1}{5}\right )}{32\,a^2\,d} \]

[In]

int(cos(c + d*x)^6/(sin(c + d*x)^6*(a + a*sin(c + d*x))^2),x)

[Out]

(5*tan(c/2 + (d*x)/2)^3)/(96*a^2*d) - tan(c/2 + (d*x)/2)^4/(32*a^2*d) + tan(c/2 + (d*x)/2)^5/(160*a^2*d) + log
(tan(c/2 + (d*x)/2))/(4*a^2*d) - (3*tan(c/2 + (d*x)/2))/(16*a^2*d) + (cot(c/2 + (d*x)/2)^5*(tan(c/2 + (d*x)/2)
- (5*tan(c/2 + (d*x)/2)^2)/3 + 6*tan(c/2 + (d*x)/2)^4 - 1/5))/(32*a^2*d)