\(\int \frac {\cos ^3(c+d x) \cot ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx\) [731]

Optimal result

Integrand size = 29, antiderivative size = 116 \[ \int \frac {\cos ^3(c+d x) \cot ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {2 x}{a^2}+\frac {9 \text {arctanh}(\cos (c+d x))}{8 a^2 d}-\frac {\cos (c+d x)}{a^2 d}-\frac {2 \cot (c+d x)}{a^2 d}+\frac {2 \cot ^3(c+d x)}{3 a^2 d}+\frac {\cot (c+d x) \csc (c+d x)}{8 a^2 d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a^2 d} \]

[Out]

-2*x/a^2+9/8*arctanh(cos(d*x+c))/a^2/d-cos(d*x+c)/a^2/d-2*cot(d*x+c)/a^2/d+2/3*cot(d*x+c)^3/a^2/d+1/8*cot(d*x+
c)*csc(d*x+c)/a^2/d-1/4*cot(d*x+c)*csc(d*x+c)^3/a^2/d

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2954, 2951, 3855, 3852, 8, 3853, 2718} \[ \int \frac {\cos ^3(c+d x) \cot ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {9 \text {arctanh}(\cos (c+d x))}{8 a^2 d}-\frac {\cos (c+d x)}{a^2 d}+\frac {2 \cot ^3(c+d x)}{3 a^2 d}-\frac {2 \cot (c+d x)}{a^2 d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a^2 d}+\frac {\cot (c+d x) \csc (c+d x)}{8 a^2 d}-\frac {2 x}{a^2} \]

[In]

Int[(Cos[c + d*x]^3*Cot[c + d*x]^5)/(a + a*Sin[c + d*x])^2,x]

[Out]

(-2*x)/a^2 + (9*ArcTanh[Cos[c + d*x]])/(8*a^2*d) - Cos[c + d*x]/(a^2*d) - (2*Cot[c + d*x])/(a^2*d) + (2*Cot[c
+ d*x]^3)/(3*a^2*d) + (Cot[c + d*x]*Csc[c + d*x])/(8*a^2*d) - (Cot[c + d*x]*Csc[c + d*x]^3)/(4*a^2*d)

Rule 8

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

Rule 2718

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

Rule 2951

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(
m_), x_Symbol] :> Dist[1/a^p, Int[ExpandTrig[(d*sin[e + f*x])^n*(a - b*sin[e + f*x])^(p/2)*(a + b*sin[e + f*x]
)^(m + p/2), x], x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, n, p/2] && ((GtQ[m,
0] && GtQ[p, 0] && LtQ[-m - p, n, -1]) || (GtQ[m, 2] && LtQ[p, 0] && GtQ[m + p/2, 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 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 \cot ^4(c+d x) \csc (c+d x) (a-a \sin (c+d x))^2 \, dx}{a^4} \\ & = \frac {\int \left (-2 a^6-a^6 \csc (c+d x)+4 a^6 \csc ^2(c+d x)-a^6 \csc ^3(c+d x)-2 a^6 \csc ^4(c+d x)+a^6 \csc ^5(c+d x)+a^6 \sin (c+d x)\right ) \, dx}{a^8} \\ & = -\frac {2 x}{a^2}-\frac {\int \csc (c+d x) \, dx}{a^2}-\frac {\int \csc ^3(c+d x) \, dx}{a^2}+\frac {\int \csc ^5(c+d x) \, dx}{a^2}+\frac {\int \sin (c+d x) \, dx}{a^2}-\frac {2 \int \csc ^4(c+d x) \, dx}{a^2}+\frac {4 \int \csc ^2(c+d x) \, dx}{a^2} \\ & = -\frac {2 x}{a^2}+\frac {\text {arctanh}(\cos (c+d x))}{a^2 d}-\frac {\cos (c+d x)}{a^2 d}+\frac {\cot (c+d x) \csc (c+d x)}{2 a^2 d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a^2 d}-\frac {\int \csc (c+d x) \, dx}{2 a^2}+\frac {3 \int \csc ^3(c+d x) \, dx}{4 a^2}+\frac {2 \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,\cot (c+d x)\right )}{a^2 d}-\frac {4 \text {Subst}(\int 1 \, dx,x,\cot (c+d x))}{a^2 d} \\ & = -\frac {2 x}{a^2}+\frac {3 \text {arctanh}(\cos (c+d x))}{2 a^2 d}-\frac {\cos (c+d x)}{a^2 d}-\frac {2 \cot (c+d x)}{a^2 d}+\frac {2 \cot ^3(c+d x)}{3 a^2 d}+\frac {\cot (c+d x) \csc (c+d x)}{8 a^2 d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a^2 d}+\frac {3 \int \csc (c+d x) \, dx}{8 a^2} \\ & = -\frac {2 x}{a^2}+\frac {9 \text {arctanh}(\cos (c+d x))}{8 a^2 d}-\frac {\cos (c+d x)}{a^2 d}-\frac {2 \cot (c+d x)}{a^2 d}+\frac {2 \cot ^3(c+d x)}{3 a^2 d}+\frac {\cot (c+d x) \csc (c+d x)}{8 a^2 d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a^2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.15 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.89 \[ \int \frac {\cos ^3(c+d x) \cot ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\left (\csc \left (\frac {1}{2} (c+d x)\right )+\sec \left (\frac {1}{2} (c+d x)\right )\right )^4 \left (192 \cot (c+d x)+\csc ^2\left (\frac {1}{2} (c+d x)\right ) (128-6 \csc (c+d x))+\csc ^4\left (\frac {1}{2} (c+d x)\right ) (-8+3 \csc (c+d x))+8 \left (3 \csc (c+d x) \left (16 (c+d x)-9 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+9 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )-(7+8 \cos (c+d x)) \sec ^4\left (\frac {1}{2} (c+d x)\right )+3 \csc ^3(c+d x) \sin ^2\left (\frac {1}{2} (c+d x)\right )-6 \csc ^5(c+d x) \sin ^4\left (\frac {1}{2} (c+d x)\right )\right )\right ) \sin ^5(c+d x)}{3072 a^2 d (1+\sin (c+d x))^2} \]

[In]

Integrate[(Cos[c + d*x]^3*Cot[c + d*x]^5)/(a + a*Sin[c + d*x])^2,x]

[Out]

-1/3072*((Csc[(c + d*x)/2] + Sec[(c + d*x)/2])^4*(192*Cot[c + d*x] + Csc[(c + d*x)/2]^2*(128 - 6*Csc[c + d*x])
+ Csc[(c + d*x)/2]^4*(-8 + 3*Csc[c + d*x]) + 8*(3*Csc[c + d*x]*(16*(c + d*x) - 9*Log[Cos[(c + d*x)/2]] + 9*Lo
g[Sin[(c + d*x)/2]]) - (7 + 8*Cos[c + d*x])*Sec[(c + d*x)/2]^4 + 3*Csc[c + d*x]^3*Sin[(c + d*x)/2]^2 - 6*Csc[c
+ d*x]^5*Sin[(c + d*x)/2]^4))*Sin[c + d*x]^5)/(a^2*d*(1 + Sin[c + d*x])^2)

Maple [A] (verified)

Time = 0.35 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.09

method	result	size
derivativedivides	\(\frac {\frac {\left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}-\frac {4 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+20 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}+\frac {4}{3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {20}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-18 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {32}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}-64 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d \,a^{2}}\)	\(127\)
default	\(\frac {\frac {\left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}-\frac {4 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+20 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}+\frac {4}{3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {20}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-18 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {32}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}-64 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d \,a^{2}}\)	\(127\)
parallelrisch	\(\frac {\left (-216 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-216\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-16 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+224 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-384 d x +384\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3 \left (\cot ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+16 \left (\cot ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-384 d x -3 \left (\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-224 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )}{192 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\)	\(175\)
risch	\(-\frac {2 x}{a^{2}}-\frac {{\mathrm e}^{i \left (d x +c \right )}}{2 d \,a^{2}}-\frac {{\mathrm e}^{-i \left (d x +c \right )}}{2 d \,a^{2}}-\frac {96 i {\mathrm e}^{6 i \left (d x +c \right )}+3 \,{\mathrm e}^{7 i \left (d x +c \right )}-192 i {\mathrm e}^{4 i \left (d x +c \right )}+21 \,{\mathrm e}^{5 i \left (d x +c \right )}+160 i {\mathrm e}^{2 i \left (d x +c \right )}+21 \,{\mathrm e}^{3 i \left (d x +c \right )}-64 i+3 \,{\mathrm e}^{i \left (d x +c \right )}}{12 d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}+\frac {9 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{8 d \,a^{2}}-\frac {9 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{8 d \,a^{2}}\)	\(186\)

[In]

int(cos(d*x+c)^8*csc(d*x+c)^5/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/16/d/a^2*(1/4*tan(1/2*d*x+1/2*c)^4-4/3*tan(1/2*d*x+1/2*c)^3+20*tan(1/2*d*x+1/2*c)-1/4/tan(1/2*d*x+1/2*c)^4+4
/3/tan(1/2*d*x+1/2*c)^3-20/tan(1/2*d*x+1/2*c)-18*ln(tan(1/2*d*x+1/2*c))-32/(1+tan(1/2*d*x+1/2*c)^2)-64*arctan(
tan(1/2*d*x+1/2*c)))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.61 \[ \int \frac {\cos ^3(c+d x) \cot ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {96 \, d x \cos \left (d x + c\right )^{4} + 48 \, \cos \left (d x + c\right )^{5} - 192 \, d x \cos \left (d x + c\right )^{2} - 90 \, \cos \left (d x + c\right )^{3} + 96 \, d x - 27 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 27 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 32 \, {\left (4 \, \cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) + 54 \, \cos \left (d x + c\right )}{48 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} - 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d\right )}} \]

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)^5/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/48*(96*d*x*cos(d*x + c)^4 + 48*cos(d*x + c)^5 - 192*d*x*cos(d*x + c)^2 - 90*cos(d*x + c)^3 + 96*d*x - 27*(c
os(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*log(1/2*cos(d*x + c) + 1/2) + 27*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)
*log(-1/2*cos(d*x + c) + 1/2) - 32*(4*cos(d*x + c)^3 - 3*cos(d*x + c))*sin(d*x + c) + 54*cos(d*x + c))/(a^2*d*
cos(d*x + c)^4 - 2*a^2*d*cos(d*x + c)^2 + a^2*d)

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^3(c+d x) \cot ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**8*csc(d*x+c)**5/(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. 263 vs. \(2 (108) = 216\).

Time = 0.30 (sec) , antiderivative size = 263, normalized size of antiderivative = 2.27 \[ \int \frac {\cos ^3(c+d x) \cot ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {\frac {16 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {3 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {224 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {384 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {240 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - 3}{\frac {a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac {\frac {240 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {16 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}}{a^{2}} - \frac {768 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} - \frac {216 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{192 \, d} \]

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)^5/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/192*((16*sin(d*x + c)/(cos(d*x + c) + 1) - 3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 224*sin(d*x + c)^3/(cos(d
*x + c) + 1)^3 - 384*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 240*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 3)/(a^2*s
in(d*x + c)^4/(cos(d*x + c) + 1)^4 + a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6) + (240*sin(d*x + c)/(cos(d*x + c
) + 1) - 16*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4)/a^2 - 768*arctan(sin(
d*x + c)/(cos(d*x + c) + 1))/a^2 - 216*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^2)/d

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.37 \[ \int \frac {\cos ^3(c+d x) \cot ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {384 \, {\left (d x + c\right )}}{a^{2}} + \frac {216 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} + \frac {384}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a^{2}} - \frac {450 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 240 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 16 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3}{a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}} - \frac {3 \, a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 16 \, a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 240 \, a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{8}}}{192 \, d} \]

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)^5/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/192*(384*(d*x + c)/a^2 + 216*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 + 384/((tan(1/2*d*x + 1/2*c)^2 + 1)*a^2) -
(450*tan(1/2*d*x + 1/2*c)^4 - 240*tan(1/2*d*x + 1/2*c)^3 + 16*tan(1/2*d*x + 1/2*c) - 3)/(a^2*tan(1/2*d*x + 1/2
*c)^4) - (3*a^6*tan(1/2*d*x + 1/2*c)^4 - 16*a^6*tan(1/2*d*x + 1/2*c)^3 + 240*a^6*tan(1/2*d*x + 1/2*c))/a^8)/d

Mupad [B] (verification not implemented)

Time = 9.95 (sec) , antiderivative size = 232, normalized size of antiderivative = 2.00 \[ \int \frac {\cos ^3(c+d x) \cot ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,a^2\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{12\,a^2\,d}+\frac {4\,\mathrm {atan}\left (\frac {9\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-9}+\frac {16}{16\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-9}\right )}{a^2\,d}-\frac {9\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{8\,a^2\,d}-\frac {20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {56\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4}-\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}+\frac {1}{4}}{d\,\left (16\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+16\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\right )}+\frac {5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4\,a^2\,d} \]

[In]

int(cos(c + d*x)^8/(sin(c + d*x)^5*(a + a*sin(c + d*x))^2),x)

[Out]

tan(c/2 + (d*x)/2)^4/(64*a^2*d) - tan(c/2 + (d*x)/2)^3/(12*a^2*d) + (4*atan((9*tan(c/2 + (d*x)/2))/(16*tan(c/2
+ (d*x)/2) - 9) + 16/(16*tan(c/2 + (d*x)/2) - 9)))/(a^2*d) - (9*log(tan(c/2 + (d*x)/2)))/(8*a^2*d) - (tan(c/2
+ (d*x)/2)^2/4 - (4*tan(c/2 + (d*x)/2))/3 + (56*tan(c/2 + (d*x)/2)^3)/3 + 32*tan(c/2 + (d*x)/2)^4 + 20*tan(c/
2 + (d*x)/2)^5 + 1/4)/(d*(16*a^2*tan(c/2 + (d*x)/2)^4 + 16*a^2*tan(c/2 + (d*x)/2)^6)) + (5*tan(c/2 + (d*x)/2))
/(4*a^2*d)