\(\int \frac {\cos ^7(c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^2} \, dx\) [727]

Optimal result

Integrand size = 27, antiderivative size = 119 \[ \int \frac {\cos ^7(c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {3 x}{4 a^2}-\frac {\text {arctanh}(\cos (c+d x))}{a^2 d}+\frac {\cos (c+d x)}{a^2 d}+\frac {\cos ^3(c+d x)}{3 a^2 d}-\frac {\cos ^5(c+d x)}{5 a^2 d}-\frac {3 \cos (c+d x) \sin (c+d x)}{4 a^2 d}-\frac {\cos ^3(c+d x) \sin (c+d x)}{2 a^2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2954, 2952, 2715, 8, 2672, 308, 212, 2645, 30} \[ \int \frac {\cos ^7(c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\text {arctanh}(\cos (c+d x))}{a^2 d}-\frac {\cos ^5(c+d x)}{5 a^2 d}+\frac {\cos ^3(c+d x)}{3 a^2 d}+\frac {\cos (c+d x)}{a^2 d}-\frac {\sin (c+d x) \cos ^3(c+d x)}{2 a^2 d}-\frac {3 \sin (c+d x) \cos (c+d x)}{4 a^2 d}-\frac {3 x}{4 a^2} \]

[In]

Int[(Cos[c + d*x]^7*Cot[c + d*x])/(a + a*Sin[c + d*x])^2,x]

[Out]

(-3*x)/(4*a^2) - ArcTanh[Cos[c + d*x]]/(a^2*d) + Cos[c + d*x]/(a^2*d) + Cos[c + d*x]^3/(3*a^2*d) - Cos[c + d*x
]^5/(5*a^2*d) - (3*Cos[c + d*x]*Sin[c + d*x])/(4*a^2*d) - (Cos[c + d*x]^3*Sin[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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 308

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 2645

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[-(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 2672

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, a*(Sin[e + f*x]/ff)
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

Rule 2715

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

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

Mathematica [A] (verified)

Time = 1.17 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.78 \[ \int \frac {\cos ^7(c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {270 \cos (c+d x)+5 \cos (3 (c+d x))-3 \left (60 c+60 d x+\cos (5 (c+d x))+80 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-80 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+40 \sin (2 (c+d x))+5 \sin (4 (c+d x))\right )}{240 a^2 d} \]

[In]

Integrate[(Cos[c + d*x]^7*Cot[c + d*x])/(a + a*Sin[c + d*x])^2,x]

[Out]

(270*Cos[c + d*x] + 5*Cos[3*(c + d*x)] - 3*(60*c + 60*d*x + Cos[5*(c + d*x)] + 80*Log[Cos[(c + d*x)/2]] - 80*L
og[Sin[(c + d*x)/2]] + 40*Sin[2*(c + d*x)] + 5*Sin[4*(c + d*x)]))/(240*a^2*d)

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.66

method	result	size
parallelrisch	\(\frac {-180 d x -3 \cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+270 \cos \left (d x +c \right )-15 \sin \left (4 d x +4 c \right )-120 \sin \left (2 d x +2 c \right )+240 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+272}{240 d \,a^{2}}\)	\(79\)
risch	\(-\frac {3 x}{4 a^{2}}+\frac {9 \,{\mathrm e}^{i \left (d x +c \right )}}{16 d \,a^{2}}+\frac {9 \,{\mathrm e}^{-i \left (d x +c \right )}}{16 d \,a^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d \,a^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \,a^{2}}-\frac {\cos \left (5 d x +5 c \right )}{80 d \,a^{2}}-\frac {\sin \left (4 d x +4 c \right )}{16 d \,a^{2}}+\frac {\cos \left (3 d x +3 c \right )}{48 d \,a^{2}}-\frac {\sin \left (2 d x +2 c \right )}{2 d \,a^{2}}\)	\(149\)
derivativedivides	\(\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {4 \left (-\frac {5 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}-\frac {\left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}-\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}-3 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {8 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}-\frac {7 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8}-\frac {17}{30}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}-\frac {3 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}}{d \,a^{2}}\)	\(152\)
default	\(\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {4 \left (-\frac {5 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}-\frac {\left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}-\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}-3 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {8 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}-\frac {7 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8}-\frac {17}{30}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}-\frac {3 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}}{d \,a^{2}}\)	\(152\)

[In]

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

[Out]

1/240*(-180*d*x-3*cos(5*d*x+5*c)+5*cos(3*d*x+3*c)+270*cos(d*x+c)-15*sin(4*d*x+4*c)-120*sin(2*d*x+2*c)+240*ln(t
an(1/2*d*x+1/2*c))+272)/d/a^2

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.79 \[ \int \frac {\cos ^7(c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {12 \, \cos \left (d x + c\right )^{5} - 20 \, \cos \left (d x + c\right )^{3} + 45 \, d x + 15 \, {\left (2 \, \cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) - 60 \, \cos \left (d x + c\right ) + 30 \, \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 30 \, \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{60 \, a^{2} d} \]

[In]

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

[Out]

-1/60*(12*cos(d*x + c)^5 - 20*cos(d*x + c)^3 + 45*d*x + 15*(2*cos(d*x + c)^3 + 3*cos(d*x + c))*sin(d*x + c) -
60*cos(d*x + c) + 30*log(1/2*cos(d*x + c) + 1/2) - 30*log(-1/2*cos(d*x + c) + 1/2))/(a^2*d)

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^7(c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**8*csc(d*x+c)/(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. 333 vs. \(2 (109) = 218\).

Time = 0.32 (sec) , antiderivative size = 333, normalized size of antiderivative = 2.80 \[ \int \frac {\cos ^7(c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {\frac {75 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {280 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {30 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {320 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {360 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {30 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {60 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {75 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - 68}{a^{2} + \frac {5 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {10 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {10 \, a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {5 \, a^{2} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {a^{2} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}}} + \frac {45 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} - \frac {30 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{30 \, d} \]

[In]

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

[Out]

-1/30*((75*sin(d*x + c)/(cos(d*x + c) + 1) - 280*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 30*sin(d*x + c)^3/(cos(
d*x + c) + 1)^3 - 320*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 360*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 30*sin(d
*x + c)^7/(cos(d*x + c) + 1)^7 - 60*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 75*sin(d*x + c)^9/(cos(d*x + c) + 1)
^9 - 68)/(a^2 + 5*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 10*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 10*a^
2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 5*a^2*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + a^2*sin(d*x + c)^10/(cos(d
*x + c) + 1)^10) + 45*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2 - 30*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^
2)/d

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.31 \[ \int \frac {\cos ^7(c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {45 \, {\left (d x + c\right )}}{a^{2}} - \frac {60 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} - \frac {2 \, {\left (75 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 60 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 30 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 360 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 320 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 30 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 280 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 75 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 68\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5} a^{2}}}{60 \, d} \]

[In]

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

[Out]

-1/60*(45*(d*x + c)/a^2 - 60*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 - 2*(75*tan(1/2*d*x + 1/2*c)^9 + 60*tan(1/2*d*
x + 1/2*c)^8 + 30*tan(1/2*d*x + 1/2*c)^7 + 360*tan(1/2*d*x + 1/2*c)^6 + 320*tan(1/2*d*x + 1/2*c)^4 - 30*tan(1/
2*d*x + 1/2*c)^3 + 280*tan(1/2*d*x + 1/2*c)^2 - 75*tan(1/2*d*x + 1/2*c) + 68)/((tan(1/2*d*x + 1/2*c)^2 + 1)^5*
a^2))/d

Mupad [B] (verification not implemented)

Time = 11.38 (sec) , antiderivative size = 262, normalized size of antiderivative = 2.20 \[ \int \frac {\cos ^7(c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {3\,\mathrm {atan}\left (\frac {9}{4\,\left (\frac {9\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}+3\right )}-\frac {3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\frac {9\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}+3}\right )}{2\,a^2\,d}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^2\,d}+\frac {\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{2}+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\frac {32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {28\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}-\frac {5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}+\frac {34}{15}}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^2\right )} \]

[In]

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

[Out]

(3*atan(9/(4*((9*tan(c/2 + (d*x)/2))/4 + 3)) - (3*tan(c/2 + (d*x)/2))/((9*tan(c/2 + (d*x)/2))/4 + 3)))/(2*a^2*
d) + log(tan(c/2 + (d*x)/2))/(a^2*d) + ((28*tan(c/2 + (d*x)/2)^2)/3 - (5*tan(c/2 + (d*x)/2))/2 - tan(c/2 + (d*
x)/2)^3 + (32*tan(c/2 + (d*x)/2)^4)/3 + 12*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^7 + 2*tan(c/2 + (d*x)/2)^
8 + (5*tan(c/2 + (d*x)/2)^9)/2 + 34/15)/(d*(5*a^2*tan(c/2 + (d*x)/2)^2 + 10*a^2*tan(c/2 + (d*x)/2)^4 + 10*a^2*
tan(c/2 + (d*x)/2)^6 + 5*a^2*tan(c/2 + (d*x)/2)^8 + a^2*tan(c/2 + (d*x)/2)^10 + a^2))