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

Optimal. Leaf size=116 \[ -\frac {2 \cos ^3(c+d x)}{3 a^2 d}-\frac {2 \cos (c+d x)}{a^2 d}-\frac {\cot (c+d x)}{a^2 d}-\frac {\sin ^3(c+d x) \cos (c+d x)}{4 a^2 d}+\frac {\sin (c+d x) \cos (c+d x)}{8 a^2 d}+\frac {2 \tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac {9 x}{8 a^2} \]

[Out]

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

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Rubi [A]  time = 0.30, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 8, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {2875, 2872, 3770, 3767, 8, 2638, 2635, 2633} \[ -\frac {2 \cos ^3(c+d x)}{3 a^2 d}-\frac {2 \cos (c+d x)}{a^2 d}-\frac {\cot (c+d x)}{a^2 d}-\frac {\sin ^3(c+d x) \cos (c+d x)}{4 a^2 d}+\frac {\sin (c+d x) \cos (c+d x)}{8 a^2 d}+\frac {2 \tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac {9 x}{8 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

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

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 2635

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] && Integer
Q[2*n]

Rule 2638

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

Rule 2872

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 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 3767

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 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 ^6(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac {\int \cos ^2(c+d x) \cot ^2(c+d x) (a-a \sin (c+d x))^2 \, dx}{a^4}\\ &=\frac {\int \left (-a^6-2 a^6 \csc (c+d x)+a^6 \csc ^2(c+d x)+4 a^6 \sin (c+d x)-a^6 \sin ^2(c+d x)-2 a^6 \sin ^3(c+d x)+a^6 \sin ^4(c+d x)\right ) \, dx}{a^8}\\ &=-\frac {x}{a^2}+\frac {\int \csc ^2(c+d x) \, dx}{a^2}-\frac {\int \sin ^2(c+d x) \, dx}{a^2}+\frac {\int \sin ^4(c+d x) \, dx}{a^2}-\frac {2 \int \csc (c+d x) \, dx}{a^2}-\frac {2 \int \sin ^3(c+d x) \, dx}{a^2}+\frac {4 \int \sin (c+d x) \, dx}{a^2}\\ &=-\frac {x}{a^2}+\frac {2 \tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac {4 \cos (c+d x)}{a^2 d}+\frac {\cos (c+d x) \sin (c+d x)}{2 a^2 d}-\frac {\cos (c+d x) \sin ^3(c+d x)}{4 a^2 d}-\frac {\int 1 \, dx}{2 a^2}+\frac {3 \int \sin ^2(c+d x) \, dx}{4 a^2}-\frac {\operatorname {Subst}(\int 1 \, dx,x,\cot (c+d x))}{a^2 d}+\frac {2 \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{a^2 d}\\ &=-\frac {3 x}{2 a^2}+\frac {2 \tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac {2 \cos (c+d x)}{a^2 d}-\frac {2 \cos ^3(c+d x)}{3 a^2 d}-\frac {\cot (c+d x)}{a^2 d}+\frac {\cos (c+d x) \sin (c+d x)}{8 a^2 d}-\frac {\cos (c+d x) \sin ^3(c+d x)}{4 a^2 d}+\frac {3 \int 1 \, dx}{8 a^2}\\ &=-\frac {9 x}{8 a^2}+\frac {2 \tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac {2 \cos (c+d x)}{a^2 d}-\frac {2 \cos ^3(c+d x)}{3 a^2 d}-\frac {\cot (c+d x)}{a^2 d}+\frac {\cos (c+d x) \sin (c+d x)}{8 a^2 d}-\frac {\cos (c+d x) \sin ^3(c+d x)}{4 a^2 d}\\ \end {align*}

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Mathematica [A]  time = 1.52, size = 128, normalized size = 1.10 \[ \frac {\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4 \left (-108 (c+d x)+3 \sin (4 (c+d x))-240 \cos (c+d x)-16 \cos (3 (c+d x))+48 \tan \left (\frac {1}{2} (c+d x)\right )-48 \cot \left (\frac {1}{2} (c+d x)\right )-192 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+192 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )}{96 d (a \sin (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4*(-108*(c + d*x) - 240*Cos[c + d*x] - 16*Cos[3*(c + d*x)] - 48*Cot[(c
+ d*x)/2] + 192*Log[Cos[(c + d*x)/2]] - 192*Log[Sin[(c + d*x)/2]] + 3*Sin[4*(c + d*x)] + 48*Tan[(c + d*x)/2]))
/(96*d*(a + a*Sin[c + d*x])^2)

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fricas [A]  time = 0.50, size = 113, normalized size = 0.97 \[ -\frac {6 \, \cos \left (d x + c\right )^{5} - 9 \, \cos \left (d x + c\right )^{3} + {\left (16 \, \cos \left (d x + c\right )^{3} + 27 \, d x + 48 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) - 24 \, \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 24 \, \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 27 \, \cos \left (d x + c\right )}{24 \, a^{2} d \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(6*cos(d*x + c)^5 - 9*cos(d*x + c)^3 + (16*cos(d*x + c)^3 + 27*d*x + 48*cos(d*x + c))*sin(d*x + c) - 24*
log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 24*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 27*cos(d*x + c))/(a^
2*d*sin(d*x + c))

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giac [A]  time = 0.22, size = 186, normalized size = 1.60 \[ -\frac {\frac {27 \, {\left (d x + c\right )}}{a^{2}} + \frac {48 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} - \frac {12 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{2}} - \frac {12 \, {\left (4 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}}{a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} + \frac {2 \, {\left (3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 96 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 21 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 192 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 21 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 160 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 64\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4} a^{2}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(27*(d*x + c)/a^2 + 48*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 - 12*tan(1/2*d*x + 1/2*c)/a^2 - 12*(4*tan(1/2*
d*x + 1/2*c) - 1)/(a^2*tan(1/2*d*x + 1/2*c)) + 2*(3*tan(1/2*d*x + 1/2*c)^7 + 96*tan(1/2*d*x + 1/2*c)^6 - 21*ta
n(1/2*d*x + 1/2*c)^5 + 192*tan(1/2*d*x + 1/2*c)^4 + 21*tan(1/2*d*x + 1/2*c)^3 + 160*tan(1/2*d*x + 1/2*c)^2 - 3
*tan(1/2*d*x + 1/2*c) + 64)/((tan(1/2*d*x + 1/2*c)^2 + 1)^4*a^2))/d

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maple [B]  time = 0.54, size = 333, normalized size = 2.87 \[ \frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}-\frac {\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a^{2} d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {8 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2} d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {7 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a^{2} d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {16 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2} d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {7 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a^{2} d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {40 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a^{2} d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a^{2} d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {16}{3 a^{2} d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {9 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{2}}-\frac {1}{2 d \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d/a^2*tan(1/2*d*x+1/2*c)-1/4/a^2/d/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^7-8/a^2/d/(1+tan(1/2*d*x+
1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^6+7/4/a^2/d/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^5-16/a^2/d/(1+tan(1/2
*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^4-7/4/a^2/d/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^3-40/3/a^2/d/(1+
tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^2+1/4/a^2/d/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)-16/3/a^2/
d/(1+tan(1/2*d*x+1/2*c)^2)^4-9/4/d/a^2*arctan(tan(1/2*d*x+1/2*c))-1/2/d/a^2/tan(1/2*d*x+1/2*c)-2/d/a^2*ln(tan(
1/2*d*x+1/2*c))

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maxima [B]  time = 0.44, size = 348, normalized size = 3.00 \[ -\frac {\frac {\frac {64 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {21 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {160 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {57 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {192 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {3 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {96 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {9 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + 6}{\frac {a^{2} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {4 \, a^{2} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {6 \, a^{2} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {4 \, a^{2} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {a^{2} \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}}} + \frac {27 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} + \frac {24 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} - \frac {6 \, \sin \left (d x + c\right )}{a^{2} {\left (\cos \left (d x + c\right ) + 1\right )}}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*((64*sin(d*x + c)/(cos(d*x + c) + 1) + 21*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 160*sin(d*x + c)^3/(cos(
d*x + c) + 1)^3 + 57*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 192*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 3*sin(d*x
+ c)^6/(cos(d*x + c) + 1)^6 + 96*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 9*sin(d*x + c)^8/(cos(d*x + c) + 1)^8
+ 6)/(a^2*sin(d*x + c)/(cos(d*x + c) + 1) + 4*a^2*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 6*a^2*sin(d*x + c)^5/(
cos(d*x + c) + 1)^5 + 4*a^2*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + a^2*sin(d*x + c)^9/(cos(d*x + c) + 1)^9) + 2
7*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2 + 24*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^2 - 6*sin(d*x + c)/(
a^2*(cos(d*x + c) + 1)))/d

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mupad [B]  time = 9.21, size = 279, normalized size = 2.41 \[ \frac {9\,\mathrm {atan}\left (\frac {9\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\frac {81\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16}-9}+\frac {81}{16\,\left (\frac {81\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16}-9\right )}\right )}{4\,a^2\,d}-\frac {2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^2\,d}-\frac {\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{2}+16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{2}+32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\frac {19\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{2}+\frac {80\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+\frac {7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}+\frac {32\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}+1}{d\,\left (2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+8\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+12\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+8\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(9*atan((9*tan(c/2 + (d*x)/2))/((81*tan(c/2 + (d*x)/2))/16 - 9) + 81/(16*((81*tan(c/2 + (d*x)/2))/16 - 9))))/(
4*a^2*d) - (2*log(tan(c/2 + (d*x)/2)))/(a^2*d) - ((32*tan(c/2 + (d*x)/2))/3 + (7*tan(c/2 + (d*x)/2)^2)/2 + (80
*tan(c/2 + (d*x)/2)^3)/3 + (19*tan(c/2 + (d*x)/2)^4)/2 + 32*tan(c/2 + (d*x)/2)^5 + tan(c/2 + (d*x)/2)^6/2 + 16
*tan(c/2 + (d*x)/2)^7 + (3*tan(c/2 + (d*x)/2)^8)/2 + 1)/(d*(8*a^2*tan(c/2 + (d*x)/2)^3 + 12*a^2*tan(c/2 + (d*x
)/2)^5 + 8*a^2*tan(c/2 + (d*x)/2)^7 + 2*a^2*tan(c/2 + (d*x)/2)^9 + 2*a^2*tan(c/2 + (d*x)/2))) + tan(c/2 + (d*x
)/2)/(2*a^2*d)

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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)**2/(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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