\(\int \frac {\cos ^2(c+d x) \cot ^6(c+d x)}{(a+a \sin (c+d x))^2} \, dx\) [732]

Optimal result

Integrand size = 29, antiderivative size = 118 \[ \int \frac {\cos ^2(c+d x) \cot ^6(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {x}{a^2}+\frac {3 \text {arctanh}(\cos (c+d x))}{4 a^2 d}+\frac {\cot (c+d x)}{a^2 d}-\frac {\cot ^3(c+d x)}{3 a^2 d}-\frac {\cot ^5(c+d x)}{5 a^2 d}-\frac {3 \cot (c+d x) \csc (c+d x)}{4 a^2 d}+\frac {\cot ^3(c+d x) \csc (c+d x)}{2 a^2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {2954, 2952, 3554, 8, 2691, 3855, 2687, 30} \[ \int \frac {\cos ^2(c+d x) \cot ^6(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {3 \text {arctanh}(\cos (c+d x))}{4 a^2 d}-\frac {\cot ^5(c+d x)}{5 a^2 d}-\frac {\cot ^3(c+d x)}{3 a^2 d}+\frac {\cot (c+d x)}{a^2 d}+\frac {\cot ^3(c+d x) \csc (c+d x)}{2 a^2 d}-\frac {3 \cot (c+d x) \csc (c+d x)}{4 a^2 d}+\frac {x}{a^2} \]

[In]

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

[Out]

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

Rule 8

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2687

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 2691

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(a*Sec[e +
f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] - Dist[b^2*((n - 1)/(m + n - 1)), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 2952

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Int[ExpandTrig[(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x]
/; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 2954

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Dist[(a/g)^(2*m), Int[(g*Cos[e + f*x])^(2*m + p)*((d*Sin[e + f*x])^n/(a - b*Sin[e +
f*x])^m), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[m, 0]

Rule 3554

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(254\) vs. \(2(118)=236\).

Time = 1.65 (sec) , antiderivative size = 254, normalized size of antiderivative = 2.15 \[ \int \frac {\cos ^2(c+d x) \cot ^6(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\csc ^5(c+d x) \left (-40 \cos (c+d x)-220 \cos (3 (c+d x))+68 \cos (5 (c+d x))+600 c \sin (c+d x)+600 d x \sin (c+d x)+450 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)-450 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)-60 \sin (2 (c+d x))-300 c \sin (3 (c+d x))-300 d x \sin (3 (c+d x))-225 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (3 (c+d x))+225 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (3 (c+d x))+150 \sin (4 (c+d x))+60 c \sin (5 (c+d x))+60 d x \sin (5 (c+d x))+45 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (5 (c+d x))-45 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (5 (c+d x))\right )}{960 a^2 d} \]

[In]

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

[Out]

(Csc[c + d*x]^5*(-40*Cos[c + d*x] - 220*Cos[3*(c + d*x)] + 68*Cos[5*(c + d*x)] + 600*c*Sin[c + d*x] + 600*d*x*
Sin[c + d*x] + 450*Log[Cos[(c + d*x)/2]]*Sin[c + d*x] - 450*Log[Sin[(c + d*x)/2]]*Sin[c + d*x] - 60*Sin[2*(c +
d*x)] - 300*c*Sin[3*(c + d*x)] - 300*d*x*Sin[3*(c + d*x)] - 225*Log[Cos[(c + d*x)/2]]*Sin[3*(c + d*x)] + 225*
Log[Sin[(c + d*x)/2]]*Sin[3*(c + d*x)] + 150*Sin[4*(c + d*x)] + 60*c*Sin[5*(c + d*x)] + 60*d*x*Sin[5*(c + d*x)
] + 45*Log[Cos[(c + d*x)/2]]*Sin[5*(c + d*x)] - 45*Log[Sin[(c + d*x)/2]]*Sin[5*(c + d*x)]))/(960*a^2*d)

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.29

method	result	size
parallelrisch	\(\frac {3 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3 \left (\cot ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-15 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+15 \left (\cot ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+5 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-5 \left (\cot ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+120 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-120 \left (\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+480 d x -360 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-270 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+270 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )}{480 d \,a^{2}}\)	\(152\)
derivativedivides	\(\frac {\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+8 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-18 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}-\frac {1}{3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}+\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}-\frac {8}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {18}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-24 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+64 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d \,a^{2}}\)	\(160\)
default	\(\frac {\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+8 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-18 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}-\frac {1}{3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}+\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}-\frac {8}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {18}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-24 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+64 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d \,a^{2}}\)	\(160\)
risch	\(\frac {x}{a^{2}}+\frac {60 i {\mathrm e}^{8 i \left (d x +c \right )}+75 \,{\mathrm e}^{9 i \left (d x +c \right )}-360 i {\mathrm e}^{6 i \left (d x +c \right )}-30 \,{\mathrm e}^{7 i \left (d x +c \right )}+320 i {\mathrm e}^{4 i \left (d x +c \right )}-280 i {\mathrm e}^{2 i \left (d x +c \right )}+30 \,{\mathrm e}^{3 i \left (d x +c \right )}+68 i-75 \,{\mathrm e}^{i \left (d x +c \right )}}{30 a^{2} d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{4 d \,a^{2}}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{4 d \,a^{2}}\)	\(163\)

[In]

int(cos(d*x+c)^8*csc(d*x+c)^6/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/480*(3*tan(1/2*d*x+1/2*c)^5-3*cot(1/2*d*x+1/2*c)^5-15*tan(1/2*d*x+1/2*c)^4+15*cot(1/2*d*x+1/2*c)^4+5*tan(1/2
*d*x+1/2*c)^3-5*cot(1/2*d*x+1/2*c)^3+120*tan(1/2*d*x+1/2*c)^2-120*cot(1/2*d*x+1/2*c)^2+480*d*x-360*ln(tan(1/2*
d*x+1/2*c))-270*tan(1/2*d*x+1/2*c)+270*cot(1/2*d*x+1/2*c))/d/a^2

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.75 \[ \int \frac {\cos ^2(c+d x) \cot ^6(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {136 \, \cos \left (d x + c\right )^{5} - 280 \, \cos \left (d x + c\right )^{3} + 45 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 45 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 30 \, {\left (4 \, d x \cos \left (d x + c\right )^{4} - 8 \, d x \cos \left (d x + c\right )^{2} + 5 \, \cos \left (d x + c\right )^{3} + 4 \, d x - 3 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) + 120 \, \cos \left (d x + c\right )}{120 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} - 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d\right )} \sin \left (d x + c\right )} \]

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)^6/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/120*(136*cos(d*x + c)^5 - 280*cos(d*x + c)^3 + 45*(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) - 45*(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*(4*d*x*cos(d*x + c)^4 - 8*d*x*cos(d*x + c)^2 + 5*cos(d*x + c)^3 + 4*d*x - 3*cos(d*x + c))*sin(d*x + c) +
120*cos(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. \[ \int \frac {\cos ^2(c+d x) \cot ^6(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**8*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. 258 vs. \(2 (108) = 216\).

Time = 0.31 (sec) , antiderivative size = 258, normalized size of antiderivative = 2.19 \[ \int \frac {\cos ^2(c+d x) \cot ^6(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {\frac {270 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {120 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{2}} - \frac {960 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} + \frac {360 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} - \frac {{\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {5 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {120 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {270 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - 3\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{5}}{a^{2} \sin \left (d x + c\right )^{5}}}{480 \, d} \]

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)^6/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/480*((270*sin(d*x + c)/(cos(d*x + c) + 1) - 120*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 5*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 - 960*a
rctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2 + 360*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^2 - (15*sin(d*x + c)/(
cos(d*x + c) + 1) - 5*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 120*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 270*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 = 0.35 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.65 \[ \int \frac {\cos ^2(c+d x) \cot ^6(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {480 \, {\left (d x + c\right )}}{a^{2}} - \frac {360 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} + \frac {822 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 270 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 120 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3}{a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}} + \frac {3 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 15 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 5 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 270 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{10}}}{480 \, d} \]

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)^6/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/480*(480*(d*x + c)/a^2 - 360*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 + (822*tan(1/2*d*x + 1/2*c)^5 + 270*tan(1/2*
d*x + 1/2*c)^4 - 120*tan(1/2*d*x + 1/2*c)^3 - 5*tan(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 + 5*a^8*tan(1/2*d*x + 1/2
*c)^3 + 120*a^8*tan(1/2*d*x + 1/2*c)^2 - 270*a^8*tan(1/2*d*x + 1/2*c))/a^10)/d

Mupad [B] (verification not implemented)

Time = 10.67 (sec) , antiderivative size = 365, normalized size of antiderivative = 3.09 \[ \int \frac {\cos ^2(c+d x) \cot ^6(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {3\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-3\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-15\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-5\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-120\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+270\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-270\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+120\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+5\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+960\,\mathrm {atan}\left (\frac {4\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )-3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+360\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{480\,a^2\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5} \]

[In]

int(cos(c + d*x)^8/(sin(c + d*x)^6*(a + a*sin(c + d*x))^2),x)

[Out]

-(3*cos(c/2 + (d*x)/2)^10 - 3*sin(c/2 + (d*x)/2)^10 + 15*cos(c/2 + (d*x)/2)*sin(c/2 + (d*x)/2)^9 - 15*cos(c/2
+ (d*x)/2)^9*sin(c/2 + (d*x)/2) - 5*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^8 - 120*cos(c/2 + (d*x)/2)^3*sin(c
/2 + (d*x)/2)^7 + 270*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^6 - 270*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2)^
4 + 120*cos(c/2 + (d*x)/2)^7*sin(c/2 + (d*x)/2)^3 + 5*cos(c/2 + (d*x)/2)^8*sin(c/2 + (d*x)/2)^2 + 960*atan((4*
cos(c/2 + (d*x)/2) - 3*sin(c/2 + (d*x)/2))/(3*cos(c/2 + (d*x)/2) + 4*sin(c/2 + (d*x)/2)))*cos(c/2 + (d*x)/2)^5
*sin(c/2 + (d*x)/2)^5 + 360*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(c/2 + (d*x)/2)^5*sin(c/2 + (d*x)/2)
^5)/(480*a^2*d*cos(c/2 + (d*x)/2)^5*sin(c/2 + (d*x)/2)^5)