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

Optimal. Leaf size=118 \[ -\frac {\cot ^5(c+d x)}{5 a^2 d}-\frac {\cot ^3(c+d x)}{3 a^2 d}+\frac {\cot (c+d x)}{a^2 d}+\frac {3 \tanh ^{-1}(\cos (c+d x))}{4 a^2 d}+\frac {\cot ^3(c+d x) \csc (c+d x)}{2 a^2 d}-\frac {3 \cot (c+d x) \csc (c+d x)}{4 a^2 d}+\frac {x}{a^2} \]

[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

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Rubi [A]  time = 0.32, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {2875, 2873, 3473, 8, 2611, 3770, 2607, 30} \[ -\frac {\cot ^5(c+d x)}{5 a^2 d}-\frac {\cot ^3(c+d x)}{3 a^2 d}+\frac {\cot (c+d x)}{a^2 d}+\frac {3 \tanh ^{-1}(\cos (c+d x))}{4 a^2 d}+\frac {\cot ^3(c+d x) \csc (c+d x)}{2 a^2 d}-\frac {3 \cot (c+d x) \csc (c+d x)}{4 a^2 d}+\frac {x}{a^2} \]

Antiderivative was successfully verified.

[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 2607

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 2611

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 2873

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 2875

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 3473

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 3770

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

Rubi steps

\begin {align*} \int \frac {\cos ^2(c+d x) \cot ^6(c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\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 {\operatorname {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 \tanh ^{-1}(\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*}

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Mathematica [B]  time = 1.12, size = 254, normalized size = 2.15 \[ \frac {\csc ^5(c+d x) \left (600 c \sin (c+d x)+600 d x \sin (c+d x)-60 \sin (2 (c+d x))-300 c \sin (3 (c+d x))-300 d x \sin (3 (c+d x))+150 \sin (4 (c+d x))+60 c \sin (5 (c+d x))+60 d x \sin (5 (c+d x))-40 \cos (c+d x)-220 \cos (3 (c+d x))+68 \cos (5 (c+d x))-450 \sin (c+d x) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+225 \sin (3 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-45 \sin (5 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+450 \sin (c+d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-225 \sin (3 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+45 \sin (5 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )}{960 a^2 d} \]

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.48, size = 207, normalized size = 1.75 \[ \frac {136 \, \cos \left (d x + c\right )^{5} - 280 \, \cos \left (d x + c\right )^{3} + 45 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 45 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 30 \, {\left (4 \, d x \cos \left (d x + c\right )^{4} - 8 \, d x \cos \left (d x + c\right )^{2} + 5 \, \cos \left (d x + c\right )^{3} + 4 \, d x - 3 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) + 120 \, \cos \left (d x + c\right )}{120 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} - 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d\right )} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [A]  time = 0.26, size = 195, normalized size = 1.65 \[ \frac {\frac {480 \, {\left (d x + c\right )}}{a^{2}} - \frac {360 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} + \frac {822 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 270 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 120 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3}{a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}} + \frac {3 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 15 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 5 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 270 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{10}}}{480 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [B]  time = 0.74, size = 226, normalized size = 1.92 \[ \frac {\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )}{160 a^{2} d}-\frac {\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )}{32 a^{2} d}+\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{96 d \,a^{2}}+\frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a^{2} d}-\frac {9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16 d \,a^{2}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}}-\frac {1}{160 a^{2} d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}-\frac {1}{96 a^{2} d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}+\frac {1}{32 a^{2} d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}-\frac {1}{4 a^{2} d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {9}{16 d \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/160/d/a^2*tan(1/2*d*x+1/2*c)^5-1/32/d/a^2*tan(1/2*d*x+1/2*c)^4+1/96/d/a^2*tan(1/2*d*x+1/2*c)^3+1/4/d/a^2*tan
(1/2*d*x+1/2*c)^2-9/16/d/a^2*tan(1/2*d*x+1/2*c)+2/d/a^2*arctan(tan(1/2*d*x+1/2*c))-1/160/a^2/d/tan(1/2*d*x+1/2
*c)^5-1/96/a^2/d/tan(1/2*d*x+1/2*c)^3+1/32/a^2/d/tan(1/2*d*x+1/2*c)^4-1/4/a^2/d/tan(1/2*d*x+1/2*c)^2+9/16/d/a^
2/tan(1/2*d*x+1/2*c)-3/4/d/a^2*ln(tan(1/2*d*x+1/2*c))

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maxima [B]  time = 0.43, size = 258, normalized size = 2.19 \[ -\frac {\frac {\frac {270 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {120 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{2}} - \frac {960 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} + \frac {360 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} - \frac {{\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {5 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {120 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {270 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - 3\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{5}}{a^{2} \sin \left (d x + c\right )^{5}}}{480 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 9.85, size = 365, normalized size = 3.09 \[ -\frac {3\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-3\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-15\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-5\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-120\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+270\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-270\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+120\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+5\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+960\,\mathrm {atan}\left (\frac {4\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )-3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+360\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{480\,a^2\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**8*csc(d*x+c)**6/(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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