\(\int \frac {\cos ^4(c+d x) \cot ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx\) [730]

Optimal result

Integrand size = 29, antiderivative size = 97 \[ \int \frac {\cos ^4(c+d x) \cot ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {x}{2 a^2}-\frac {3 \text {arctanh}(\cos (c+d x))}{a^2 d}+\frac {2 \cos (c+d x)}{a^2 d}-\frac {\cot ^3(c+d x)}{3 a^2 d}+\frac {\cot (c+d x) \csc (c+d x)}{a^2 d}-\frac {\cos (c+d x) \sin (c+d x)}{2 a^2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {2954, 2788, 3855, 3852, 8, 3853, 2718, 2715} \[ \int \frac {\cos ^4(c+d x) \cot ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {3 \text {arctanh}(\cos (c+d x))}{a^2 d}+\frac {2 \cos (c+d x)}{a^2 d}-\frac {\cot ^3(c+d x)}{3 a^2 d}-\frac {\sin (c+d x) \cos (c+d x)}{2 a^2 d}+\frac {\cot (c+d x) \csc (c+d x)}{a^2 d}-\frac {x}{2 a^2} \]

[In]

Int[(Cos[c + d*x]^4*Cot[c + d*x]^4)/(a + a*Sin[c + d*x])^2,x]

[Out]

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

Rule 8

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

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 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ[{c, d}, 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 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) (a-a \sin (c+d x))^2 \, dx}{a^4} \\ & = \frac {\int \left (-a^6+4 a^6 \csc (c+d x)-a^6 \csc ^2(c+d x)-2 a^6 \csc ^3(c+d x)+a^6 \csc ^4(c+d x)-2 a^6 \sin (c+d x)+a^6 \sin ^2(c+d x)\right ) \, dx}{a^8} \\ & = -\frac {x}{a^2}-\frac {\int \csc ^2(c+d x) \, dx}{a^2}+\frac {\int \csc ^4(c+d x) \, dx}{a^2}+\frac {\int \sin ^2(c+d x) \, dx}{a^2}-\frac {2 \int \csc ^3(c+d x) \, dx}{a^2}-\frac {2 \int \sin (c+d x) \, dx}{a^2}+\frac {4 \int \csc (c+d x) \, dx}{a^2} \\ & = -\frac {x}{a^2}-\frac {4 \text {arctanh}(\cos (c+d x))}{a^2 d}+\frac {2 \cos (c+d x)}{a^2 d}+\frac {\cot (c+d x) \csc (c+d x)}{a^2 d}-\frac {\cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac {\int 1 \, dx}{2 a^2}-\frac {\int \csc (c+d x) \, dx}{a^2}+\frac {\text {Subst}(\int 1 \, dx,x,\cot (c+d x))}{a^2 d}-\frac {\text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,\cot (c+d x)\right )}{a^2 d} \\ & = -\frac {x}{2 a^2}-\frac {3 \text {arctanh}(\cos (c+d x))}{a^2 d}+\frac {2 \cos (c+d x)}{a^2 d}-\frac {\cot ^3(c+d x)}{3 a^2 d}+\frac {\cot (c+d x) \csc (c+d x)}{a^2 d}-\frac {\cos (c+d x) \sin (c+d x)}{2 a^2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.45 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.90 \[ \int \frac {\cos ^4(c+d x) \cot ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\left (1+\cot \left (\frac {1}{2} (c+d x)\right )\right )^4 \sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (30 \cos (c+d x)-\cos (3 (c+d x))+3 \left (\cos (5 (c+d x))+8 \left (c+d x-6 \cos (c+d x)+2 \cos (3 (c+d x))+6 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-\cos (2 (c+d x)) \left (c+d x+6 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-6 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )-6 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right ) \sin (c+d x)\right )\right ) \tan \left (\frac {1}{2} (c+d x)\right )}{768 a^2 d (1+\sin (c+d x))^2} \]

[In]

Integrate[(Cos[c + d*x]^4*Cot[c + d*x]^4)/(a + a*Sin[c + d*x])^2,x]

[Out]

-1/768*((1 + Cot[(c + d*x)/2])^4*Sec[(c + d*x)/2]^2*(30*Cos[c + d*x] - Cos[3*(c + d*x)] + 3*(Cos[5*(c + d*x)]
+ 8*(c + d*x - 6*Cos[c + d*x] + 2*Cos[3*(c + d*x)] + 6*Log[Cos[(c + d*x)/2]] - Cos[2*(c + d*x)]*(c + d*x + 6*L
og[Cos[(c + d*x)/2]] - 6*Log[Sin[(c + d*x)/2]]) - 6*Log[Sin[(c + d*x)/2]])*Sin[c + d*x]))*Tan[(c + d*x)/2])/(a
^2*d*(1 + Sin[c + d*x])^2)

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.60

method	result	size
parallelrisch	\(\frac {\left (\csc ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\sec ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (-6 \left (-3 \sin \left (d x +c \right )+\sin \left (3 d x +3 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+d x \sin \left (3 d x +3 c \right )-3 d x \sin \left (d x +c \right )-5 \sin \left (3 d x +3 c \right )-2 \sin \left (4 d x +4 c \right )-\frac {5 \cos \left (d x +c \right )}{2}+\frac {\cos \left (3 d x +3 c \right )}{12}-\frac {\cos \left (5 d x +5 c \right )}{4}+15 \sin \left (d x +c \right )+8 \sin \left (2 d x +2 c \right )\right )}{64 d \,a^{2}}\)	\(155\)
derivativedivides	\(\frac {\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-2 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}+\frac {2}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+24 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {16 \left (-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}-2 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}-2\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-8 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{2}}\)	\(164\)
default	\(\frac {\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-2 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}+\frac {2}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+24 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {16 \left (-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}-2 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}-2\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-8 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{2}}\)	\(164\)
risch	\(-\frac {x}{2 a^{2}}+\frac {i {\mathrm e}^{2 i \left (d x +c \right )}}{8 d \,a^{2}}+\frac {{\mathrm e}^{i \left (d x +c \right )}}{d \,a^{2}}+\frac {{\mathrm e}^{-i \left (d x +c \right )}}{d \,a^{2}}-\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d \,a^{2}}-\frac {2 \left (-3 i {\mathrm e}^{4 i \left (d x +c \right )}+3 \,{\mathrm e}^{5 i \left (d x +c \right )}-i-3 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{3 a^{2} d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d \,a^{2}}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \,a^{2}}\)	\(174\)

[In]

int(cos(d*x+c)^8*csc(d*x+c)^4/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/64*csc(1/2*d*x+1/2*c)^3*sec(1/2*d*x+1/2*c)^3*(-6*(-3*sin(d*x+c)+sin(3*d*x+3*c))*ln(tan(1/2*d*x+1/2*c))+d*x*s
in(3*d*x+3*c)-3*d*x*sin(d*x+c)-5*sin(3*d*x+3*c)-2*sin(4*d*x+4*c)-5/2*cos(d*x+c)+1/12*cos(3*d*x+3*c)-1/4*cos(5*
d*x+5*c)+15*sin(d*x+c)+8*sin(2*d*x+2*c))/d/a^2

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.66 \[ \int \frac {\cos ^4(c+d x) \cot ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {3 \, \cos \left (d x + c\right )^{5} - 4 \, \cos \left (d x + c\right )^{3} - 9 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 9 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 3 \, {\left (d x \cos \left (d x + c\right )^{2} - 4 \, \cos \left (d x + c\right )^{3} - d x + 6 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) + 3 \, \cos \left (d x + c\right )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d\right )} \sin \left (d x + c\right )} \]

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)^4/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/6*(3*cos(d*x + c)^5 - 4*cos(d*x + c)^3 - 9*(cos(d*x + c)^2 - 1)*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 9
*(cos(d*x + c)^2 - 1)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 3*(d*x*cos(d*x + c)^2 - 4*cos(d*x + c)^3 - d
*x + 6*cos(d*x + c))*sin(d*x + c) + 3*cos(d*x + c))/((a^2*d*cos(d*x + c)^2 - a^2*d)*sin(d*x + c))

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^4(c+d x) \cot ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**8*csc(d*x+c)**4/(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. 306 vs. \(2 (91) = 182\).

Time = 0.31 (sec) , antiderivative size = 306, normalized size of antiderivative = 3.15 \[ \int \frac {\cos ^4(c+d x) \cot ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {\frac {6 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {108 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {19 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {102 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {27 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - 1}{\frac {a^{2} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {2 \, a^{2} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {a^{2} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}} - \frac {\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {6 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {24 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} + \frac {72 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{24 \, d} \]

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)^4/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/24*((6*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 108*sin(d*x + c)^3/(cos(d*x +
c) + 1)^3 - 19*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 102*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 27*sin(d*x + c
)^6/(cos(d*x + c) + 1)^6 - 1)/(a^2*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 2*a^2*sin(d*x + c)^5/(cos(d*x + c) +
1)^5 + a^2*sin(d*x + c)^7/(cos(d*x + c) + 1)^7) - (3*sin(d*x + c)/(cos(d*x + c) + 1) + 6*sin(d*x + c)^2/(cos(d
*x + c) + 1)^2 - sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 24*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2 + 7
2*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^2)/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 194 vs. \(2 (91) = 182\).

Time = 0.35 (sec) , antiderivative size = 194, normalized size of antiderivative = 2.00 \[ \int \frac {\cos ^4(c+d x) \cot ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {12 \, {\left (d x + c\right )}}{a^{2}} - \frac {72 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} - \frac {24 \, {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a^{2}} + \frac {132 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 6 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1}{a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}} - \frac {a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{24 \, d} \]

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)^4/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/24*(12*(d*x + c)/a^2 - 72*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 - 24*(tan(1/2*d*x + 1/2*c)^3 + 4*tan(1/2*d*x +
1/2*c)^2 - tan(1/2*d*x + 1/2*c) + 4)/((tan(1/2*d*x + 1/2*c)^2 + 1)^2*a^2) + (132*tan(1/2*d*x + 1/2*c)^3 - 3*t
an(1/2*d*x + 1/2*c)^2 - 6*tan(1/2*d*x + 1/2*c) + 1)/(a^2*tan(1/2*d*x + 1/2*c)^3) - (a^4*tan(1/2*d*x + 1/2*c)^3
- 6*a^4*tan(1/2*d*x + 1/2*c)^2 - 3*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d

Mupad [B] (verification not implemented)

Time = 9.82 (sec) , antiderivative size = 253, normalized size of antiderivative = 2.61 \[ \int \frac {\cos ^4(c+d x) \cot ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,a^2\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4\,a^2\,d}+\frac {3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^2\,d}+\frac {9\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+34\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-\frac {19\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}+36\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {1}{3}}{d\,\left (8\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+16\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+8\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,a^2\,d}+\frac {\mathrm {atan}\left (\frac {1}{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+6}-\frac {6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+6}\right )}{a^2\,d} \]

[In]

int(cos(c + d*x)^8/(sin(c + d*x)^4*(a + a*sin(c + d*x))^2),x)

[Out]

tan(c/2 + (d*x)/2)^3/(24*a^2*d) - tan(c/2 + (d*x)/2)^2/(4*a^2*d) + (3*log(tan(c/2 + (d*x)/2)))/(a^2*d) + (2*ta
n(c/2 + (d*x)/2) + tan(c/2 + (d*x)/2)^2/3 + 36*tan(c/2 + (d*x)/2)^3 - (19*tan(c/2 + (d*x)/2)^4)/3 + 34*tan(c/2
+ (d*x)/2)^5 + 9*tan(c/2 + (d*x)/2)^6 - 1/3)/(d*(8*a^2*tan(c/2 + (d*x)/2)^3 + 16*a^2*tan(c/2 + (d*x)/2)^5 + 8
*a^2*tan(c/2 + (d*x)/2)^7)) - tan(c/2 + (d*x)/2)/(8*a^2*d) + atan(1/(tan(c/2 + (d*x)/2) + 6) - (6*tan(c/2 + (d
*x)/2))/(tan(c/2 + (d*x)/2) + 6))/(a^2*d)