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

Optimal. Leaf size=132 \[ \frac {2 \cot ^5(c+d x)}{5 a^2 d}-\frac {7 \tanh ^{-1}(\cos (c+d x))}{16 a^2 d}-\frac {\cot ^3(c+d x) \csc ^3(c+d x)}{6 a^2 d}-\frac {\cot ^3(c+d x) \csc (c+d x)}{4 a^2 d}+\frac {\cot (c+d x) \csc ^3(c+d x)}{8 a^2 d}+\frac {5 \cot (c+d x) \csc (c+d x)}{16 a^2 d} \]

[Out]

-7/16*arctanh(cos(d*x+c))/a^2/d+2/5*cot(d*x+c)^5/a^2/d+5/16*cot(d*x+c)*csc(d*x+c)/a^2/d-1/4*cot(d*x+c)^3*csc(d
*x+c)/a^2/d+1/8*cot(d*x+c)*csc(d*x+c)^3/a^2/d-1/6*cot(d*x+c)^3*csc(d*x+c)^3/a^2/d

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Rubi [A]  time = 0.34, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {2875, 2873, 2611, 3770, 2607, 30, 3768} \[ \frac {2 \cot ^5(c+d x)}{5 a^2 d}-\frac {7 \tanh ^{-1}(\cos (c+d x))}{16 a^2 d}-\frac {\cot ^3(c+d x) \csc ^3(c+d x)}{6 a^2 d}-\frac {\cot ^3(c+d x) \csc (c+d x)}{4 a^2 d}+\frac {\cot (c+d x) \csc ^3(c+d x)}{8 a^2 d}+\frac {5 \cot (c+d x) \csc (c+d x)}{16 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-7*ArcTanh[Cos[c + d*x]])/(16*a^2*d) + (2*Cot[c + d*x]^5)/(5*a^2*d) + (5*Cot[c + d*x]*Csc[c + d*x])/(16*a^2*d
) - (Cot[c + d*x]^3*Csc[c + d*x])/(4*a^2*d) + (Cot[c + d*x]*Csc[c + d*x]^3)/(8*a^2*d) - (Cot[c + d*x]^3*Csc[c
+ d*x]^3)/(6*a^2*d)

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 3768

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 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 (c+d x) \cot ^7(c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac {\int \cot ^4(c+d x) \csc ^3(c+d x) (a-a \sin (c+d x))^2 \, dx}{a^4}\\ &=\frac {\int \left (a^2 \cot ^4(c+d x) \csc (c+d x)-2 a^2 \cot ^4(c+d x) \csc ^2(c+d x)+a^2 \cot ^4(c+d x) \csc ^3(c+d x)\right ) \, dx}{a^4}\\ &=\frac {\int \cot ^4(c+d x) \csc (c+d x) \, dx}{a^2}+\frac {\int \cot ^4(c+d x) \csc ^3(c+d x) \, dx}{a^2}-\frac {2 \int \cot ^4(c+d x) \csc ^2(c+d x) \, dx}{a^2}\\ &=-\frac {\cot ^3(c+d x) \csc (c+d x)}{4 a^2 d}-\frac {\cot ^3(c+d x) \csc ^3(c+d x)}{6 a^2 d}-\frac {\int \cot ^2(c+d x) \csc ^3(c+d x) \, dx}{2 a^2}-\frac {3 \int \cot ^2(c+d x) \csc (c+d x) \, dx}{4 a^2}-\frac {2 \operatorname {Subst}\left (\int x^4 \, dx,x,-\cot (c+d x)\right )}{a^2 d}\\ &=\frac {2 \cot ^5(c+d x)}{5 a^2 d}+\frac {3 \cot (c+d x) \csc (c+d x)}{8 a^2 d}-\frac {\cot ^3(c+d x) \csc (c+d x)}{4 a^2 d}+\frac {\cot (c+d x) \csc ^3(c+d x)}{8 a^2 d}-\frac {\cot ^3(c+d x) \csc ^3(c+d x)}{6 a^2 d}+\frac {\int \csc ^3(c+d x) \, dx}{8 a^2}+\frac {3 \int \csc (c+d x) \, dx}{8 a^2}\\ &=-\frac {3 \tanh ^{-1}(\cos (c+d x))}{8 a^2 d}+\frac {2 \cot ^5(c+d x)}{5 a^2 d}+\frac {5 \cot (c+d x) \csc (c+d x)}{16 a^2 d}-\frac {\cot ^3(c+d x) \csc (c+d x)}{4 a^2 d}+\frac {\cot (c+d x) \csc ^3(c+d x)}{8 a^2 d}-\frac {\cot ^3(c+d x) \csc ^3(c+d x)}{6 a^2 d}+\frac {\int \csc (c+d x) \, dx}{16 a^2}\\ &=-\frac {7 \tanh ^{-1}(\cos (c+d x))}{16 a^2 d}+\frac {2 \cot ^5(c+d x)}{5 a^2 d}+\frac {5 \cot (c+d x) \csc (c+d x)}{16 a^2 d}-\frac {\cot ^3(c+d x) \csc (c+d x)}{4 a^2 d}+\frac {\cot (c+d x) \csc ^3(c+d x)}{8 a^2 d}-\frac {\cot ^3(c+d x) \csc ^3(c+d x)}{6 a^2 d}\\ \end {align*}

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Mathematica [A]  time = 1.89, size = 145, normalized size = 1.10 \[ \frac {\csc ^6(c+d x) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4 \left (60 (32 \sin (c+d x)-11) \cos (c+d x)+6 (32 \sin (c+d x)+45) \cos (5 (c+d x))+10 (96 \sin (c+d x)-89) \cos (3 (c+d x))+3360 \sin ^6(c+d x) \left (\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{7680 a^2 d (\sin (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Csc[c + d*x]^6*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4*(3360*(-Log[Cos[(c + d*x)/2]] + Log[Sin[(c + d*x)/2]])
*Sin[c + d*x]^6 + 60*Cos[c + d*x]*(-11 + 32*Sin[c + d*x]) + 6*Cos[5*(c + d*x)]*(45 + 32*Sin[c + d*x]) + 10*Cos
[3*(c + d*x)]*(-89 + 96*Sin[c + d*x])))/(7680*a^2*d*(1 + Sin[c + d*x])^2)

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fricas [A]  time = 0.46, size = 183, normalized size = 1.39 \[ -\frac {192 \, \cos \left (d x + c\right )^{5} \sin \left (d x + c\right ) + 270 \, \cos \left (d x + c\right )^{5} - 560 \, \cos \left (d x + c\right )^{3} + 105 \, {\left (\cos \left (d x + c\right )^{6} - 3 \, \cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{2} - 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 105 \, {\left (\cos \left (d x + c\right )^{6} - 3 \, \cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{2} - 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 210 \, \cos \left (d x + c\right )}{480 \, {\left (a^{2} d \cos \left (d x + c\right )^{6} - 3 \, a^{2} d \cos \left (d x + c\right )^{4} + 3 \, a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)^7/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/480*(192*cos(d*x + c)^5*sin(d*x + c) + 270*cos(d*x + c)^5 - 560*cos(d*x + c)^3 + 105*(cos(d*x + c)^6 - 3*co
s(d*x + c)^4 + 3*cos(d*x + c)^2 - 1)*log(1/2*cos(d*x + c) + 1/2) - 105*(cos(d*x + c)^6 - 3*cos(d*x + c)^4 + 3*
cos(d*x + c)^2 - 1)*log(-1/2*cos(d*x + c) + 1/2) + 210*cos(d*x + c))/(a^2*d*cos(d*x + c)^6 - 3*a^2*d*cos(d*x +
c)^4 + 3*a^2*d*cos(d*x + c)^2 - a^2*d)

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giac [A]  time = 0.30, size = 215, normalized size = 1.63 \[ \frac {\frac {840 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} - \frac {2058 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 240 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 255 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 120 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 24 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5}{a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6}} + \frac {5 \, a^{10} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 24 \, a^{10} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 15 \, a^{10} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 120 \, a^{10} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 255 \, a^{10} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 240 \, a^{10} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{12}}}{1920 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)^7/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/1920*(840*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 - (2058*tan(1/2*d*x + 1/2*c)^6 - 240*tan(1/2*d*x + 1/2*c)^5 - 2
55*tan(1/2*d*x + 1/2*c)^4 + 120*tan(1/2*d*x + 1/2*c)^3 + 15*tan(1/2*d*x + 1/2*c)^2 - 24*tan(1/2*d*x + 1/2*c) +
5)/(a^2*tan(1/2*d*x + 1/2*c)^6) + (5*a^10*tan(1/2*d*x + 1/2*c)^6 - 24*a^10*tan(1/2*d*x + 1/2*c)^5 + 15*a^10*t
an(1/2*d*x + 1/2*c)^4 + 120*a^10*tan(1/2*d*x + 1/2*c)^3 - 255*a^10*tan(1/2*d*x + 1/2*c)^2 - 240*a^10*tan(1/2*d
*x + 1/2*c))/a^12)/d

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maple [B]  time = 0.66, size = 246, normalized size = 1.86 \[ \frac {\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )}{384 d \,a^{2}}-\frac {\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )}{80 a^{2} d}+\frac {\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )}{128 a^{2} d}+\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{16 d \,a^{2}}-\frac {17 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{128 a^{2} d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{2}}-\frac {1}{384 d \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}+\frac {1}{8 d \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {7 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d \,a^{2}}+\frac {1}{80 a^{2} d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}+\frac {17}{128 a^{2} d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-\frac {1}{128 a^{2} d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}-\frac {1}{16 a^{2} d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^8*csc(d*x+c)^7/(a+a*sin(d*x+c))^2,x)

[Out]

1/384/d/a^2*tan(1/2*d*x+1/2*c)^6-1/80/d/a^2*tan(1/2*d*x+1/2*c)^5+1/128/d/a^2*tan(1/2*d*x+1/2*c)^4+1/16/d/a^2*t
an(1/2*d*x+1/2*c)^3-17/128/d/a^2*tan(1/2*d*x+1/2*c)^2-1/8/d/a^2*tan(1/2*d*x+1/2*c)-1/384/d/a^2/tan(1/2*d*x+1/2
*c)^6+1/8/d/a^2/tan(1/2*d*x+1/2*c)+7/16/d/a^2*ln(tan(1/2*d*x+1/2*c))+1/80/a^2/d/tan(1/2*d*x+1/2*c)^5+17/128/a^
2/d/tan(1/2*d*x+1/2*c)^2-1/128/a^2/d/tan(1/2*d*x+1/2*c)^4-1/16/a^2/d/tan(1/2*d*x+1/2*c)^3

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maxima [B]  time = 0.33, size = 275, normalized size = 2.08 \[ -\frac {\frac {\frac {240 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {255 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {120 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {15 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {24 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {5 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}}{a^{2}} - \frac {840 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} - \frac {{\left (\frac {24 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {15 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {120 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {255 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {240 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - 5\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{6}}{a^{2} \sin \left (d x + c\right )^{6}}}{1920 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)^7/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/1920*((240*sin(d*x + c)/(cos(d*x + c) + 1) + 255*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 120*sin(d*x + c)^3/(
cos(d*x + c) + 1)^3 - 15*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 24*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 5*sin(
d*x + c)^6/(cos(d*x + c) + 1)^6)/a^2 - 840*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^2 - (24*sin(d*x + c)/(cos(d*
x + c) + 1) - 15*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 120*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 255*sin(d*x +
c)^4/(cos(d*x + c) + 1)^4 + 240*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 5)*(cos(d*x + c) + 1)^6/(a^2*sin(d*x +
c)^6))/d

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mupad [B]  time = 10.34, size = 339, normalized size = 2.57 \[ \frac {5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-5\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-24\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+24\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+15\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+120\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-255\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-240\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+240\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+255\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-120\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-15\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+840\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{1920\,a^2\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^8/(sin(c + d*x)^7*(a + a*sin(c + d*x))^2),x)

[Out]

(5*sin(c/2 + (d*x)/2)^12 - 5*cos(c/2 + (d*x)/2)^12 - 24*cos(c/2 + (d*x)/2)*sin(c/2 + (d*x)/2)^11 + 24*cos(c/2
+ (d*x)/2)^11*sin(c/2 + (d*x)/2) + 15*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^10 + 120*cos(c/2 + (d*x)/2)^3*si
n(c/2 + (d*x)/2)^9 - 255*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^8 - 240*cos(c/2 + (d*x)/2)^5*sin(c/2 + (d*x)/
2)^7 + 240*cos(c/2 + (d*x)/2)^7*sin(c/2 + (d*x)/2)^5 + 255*cos(c/2 + (d*x)/2)^8*sin(c/2 + (d*x)/2)^4 - 120*cos
(c/2 + (d*x)/2)^9*sin(c/2 + (d*x)/2)^3 - 15*cos(c/2 + (d*x)/2)^10*sin(c/2 + (d*x)/2)^2 + 840*log(sin(c/2 + (d*
x)/2)/cos(c/2 + (d*x)/2))*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2)^6)/(1920*a^2*d*cos(c/2 + (d*x)/2)^6*sin(c/2
+ (d*x)/2)^6)

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

[Out]

Timed out

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