3.735 \(\int \frac {\cot ^8(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=176 \[ \frac {2 \cot ^7(c+d x)}{7 a^2 d}+\frac {2 \cot ^5(c+d x)}{5 a^2 d}-\frac {11 \tanh ^{-1}(\cos (c+d x))}{128 a^2 d}-\frac {\cot ^3(c+d x) \csc ^5(c+d x)}{8 a^2 d}-\frac {\cot ^3(c+d x) \csc ^3(c+d x)}{6 a^2 d}+\frac {\cot (c+d x) \csc ^5(c+d x)}{16 a^2 d}+\frac {7 \cot (c+d x) \csc ^3(c+d x)}{64 a^2 d}-\frac {11 \cot (c+d x) \csc (c+d x)}{128 a^2 d} \]

[Out]

-11/128*arctanh(cos(d*x+c))/a^2/d+2/5*cot(d*x+c)^5/a^2/d+2/7*cot(d*x+c)^7/a^2/d-11/128*cot(d*x+c)*csc(d*x+c)/a
^2/d+7/64*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+1/16*cot(d*x+c)*csc(d*x+c)^5/a^2/d
-1/8*cot(d*x+c)^3*csc(d*x+c)^5/a^2/d

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Rubi [A]  time = 0.41, antiderivative size = 176, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {2875, 2873, 2611, 3768, 3770, 2607, 14} \[ \frac {2 \cot ^7(c+d x)}{7 a^2 d}+\frac {2 \cot ^5(c+d x)}{5 a^2 d}-\frac {11 \tanh ^{-1}(\cos (c+d x))}{128 a^2 d}-\frac {\cot ^3(c+d x) \csc ^5(c+d x)}{8 a^2 d}-\frac {\cot ^3(c+d x) \csc ^3(c+d x)}{6 a^2 d}+\frac {\cot (c+d x) \csc ^5(c+d x)}{16 a^2 d}+\frac {7 \cot (c+d x) \csc ^3(c+d x)}{64 a^2 d}-\frac {11 \cot (c+d x) \csc (c+d x)}{128 a^2 d} \]

Antiderivative was successfully verified.

[In]

Int[(Cot[c + d*x]^8*Csc[c + d*x])/(a + a*Sin[c + d*x])^2,x]

[Out]

(-11*ArcTanh[Cos[c + d*x]])/(128*a^2*d) + (2*Cot[c + d*x]^5)/(5*a^2*d) + (2*Cot[c + d*x]^7)/(7*a^2*d) - (11*Co
t[c + d*x]*Csc[c + d*x])/(128*a^2*d) + (7*Cot[c + d*x]*Csc[c + d*x]^3)/(64*a^2*d) - (Cot[c + d*x]^3*Csc[c + d*
x]^3)/(6*a^2*d) + (Cot[c + d*x]*Csc[c + d*x]^5)/(16*a^2*d) - (Cot[c + d*x]^3*Csc[c + d*x]^5)/(8*a^2*d)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 {\cot ^8(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac {\int \cot ^4(c+d x) \csc ^5(c+d x) (a-a \sin (c+d x))^2 \, dx}{a^4}\\ &=\frac {\int \left (a^2 \cot ^4(c+d x) \csc ^3(c+d x)-2 a^2 \cot ^4(c+d x) \csc ^4(c+d x)+a^2 \cot ^4(c+d x) \csc ^5(c+d x)\right ) \, dx}{a^4}\\ &=\frac {\int \cot ^4(c+d x) \csc ^3(c+d x) \, dx}{a^2}+\frac {\int \cot ^4(c+d x) \csc ^5(c+d x) \, dx}{a^2}-\frac {2 \int \cot ^4(c+d x) \csc ^4(c+d x) \, dx}{a^2}\\ &=-\frac {\cot ^3(c+d x) \csc ^3(c+d x)}{6 a^2 d}-\frac {\cot ^3(c+d x) \csc ^5(c+d x)}{8 a^2 d}-\frac {3 \int \cot ^2(c+d x) \csc ^5(c+d x) \, dx}{8 a^2}-\frac {\int \cot ^2(c+d x) \csc ^3(c+d x) \, dx}{2 a^2}-\frac {2 \operatorname {Subst}\left (\int x^4 \left (1+x^2\right ) \, dx,x,-\cot (c+d x)\right )}{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 {\cot (c+d x) \csc ^5(c+d x)}{16 a^2 d}-\frac {\cot ^3(c+d x) \csc ^5(c+d x)}{8 a^2 d}+\frac {\int \csc ^5(c+d x) \, dx}{16 a^2}+\frac {\int \csc ^3(c+d x) \, dx}{8 a^2}-\frac {2 \operatorname {Subst}\left (\int \left (x^4+x^6\right ) \, dx,x,-\cot (c+d x)\right )}{a^2 d}\\ &=\frac {2 \cot ^5(c+d x)}{5 a^2 d}+\frac {2 \cot ^7(c+d x)}{7 a^2 d}-\frac {\cot (c+d x) \csc (c+d x)}{16 a^2 d}+\frac {7 \cot (c+d x) \csc ^3(c+d x)}{64 a^2 d}-\frac {\cot ^3(c+d x) \csc ^3(c+d x)}{6 a^2 d}+\frac {\cot (c+d x) \csc ^5(c+d x)}{16 a^2 d}-\frac {\cot ^3(c+d x) \csc ^5(c+d x)}{8 a^2 d}+\frac {3 \int \csc ^3(c+d x) \, dx}{64 a^2}+\frac {\int \csc (c+d x) \, dx}{16 a^2}\\ &=-\frac {\tanh ^{-1}(\cos (c+d x))}{16 a^2 d}+\frac {2 \cot ^5(c+d x)}{5 a^2 d}+\frac {2 \cot ^7(c+d x)}{7 a^2 d}-\frac {11 \cot (c+d x) \csc (c+d x)}{128 a^2 d}+\frac {7 \cot (c+d x) \csc ^3(c+d x)}{64 a^2 d}-\frac {\cot ^3(c+d x) \csc ^3(c+d x)}{6 a^2 d}+\frac {\cot (c+d x) \csc ^5(c+d x)}{16 a^2 d}-\frac {\cot ^3(c+d x) \csc ^5(c+d x)}{8 a^2 d}+\frac {3 \int \csc (c+d x) \, dx}{128 a^2}\\ &=-\frac {11 \tanh ^{-1}(\cos (c+d x))}{128 a^2 d}+\frac {2 \cot ^5(c+d x)}{5 a^2 d}+\frac {2 \cot ^7(c+d x)}{7 a^2 d}-\frac {11 \cot (c+d x) \csc (c+d x)}{128 a^2 d}+\frac {7 \cot (c+d x) \csc ^3(c+d x)}{64 a^2 d}-\frac {\cot ^3(c+d x) \csc ^3(c+d x)}{6 a^2 d}+\frac {\cot (c+d x) \csc ^5(c+d x)}{16 a^2 d}-\frac {\cot ^3(c+d x) \csc ^5(c+d x)}{8 a^2 d}\\ \end {align*}

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Mathematica [A]  time = 0.95, size = 291, normalized size = 1.65 \[ -\frac {\csc ^8(c+d x) \left (-86016 \sin (2 (c+d x))-64512 \sin (4 (c+d x))-12288 \sin (6 (c+d x))+1536 \sin (8 (c+d x))+158270 \cos (c+d x)+77210 \cos (3 (c+d x))-18130 \cos (5 (c+d x))-2310 \cos (7 (c+d x))-40425 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-64680 \cos (2 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+32340 \cos (4 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-9240 \cos (6 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+1155 \cos (8 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+40425 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+64680 \cos (2 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-32340 \cos (4 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+9240 \cos (6 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-1155 \cos (8 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )}{1720320 a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cot[c + d*x]^8*Csc[c + d*x])/(a + a*Sin[c + d*x])^2,x]

[Out]

-1/1720320*(Csc[c + d*x]^8*(158270*Cos[c + d*x] + 77210*Cos[3*(c + d*x)] - 18130*Cos[5*(c + d*x)] - 2310*Cos[7
*(c + d*x)] + 40425*Log[Cos[(c + d*x)/2]] - 64680*Cos[2*(c + d*x)]*Log[Cos[(c + d*x)/2]] + 32340*Cos[4*(c + d*
x)]*Log[Cos[(c + d*x)/2]] - 9240*Cos[6*(c + d*x)]*Log[Cos[(c + d*x)/2]] + 1155*Cos[8*(c + d*x)]*Log[Cos[(c + d
*x)/2]] - 40425*Log[Sin[(c + d*x)/2]] + 64680*Cos[2*(c + d*x)]*Log[Sin[(c + d*x)/2]] - 32340*Cos[4*(c + d*x)]*
Log[Sin[(c + d*x)/2]] + 9240*Cos[6*(c + d*x)]*Log[Sin[(c + d*x)/2]] - 1155*Cos[8*(c + d*x)]*Log[Sin[(c + d*x)/
2]] - 86016*Sin[2*(c + d*x)] - 64512*Sin[4*(c + d*x)] - 12288*Sin[6*(c + d*x)] + 1536*Sin[8*(c + d*x)]))/(a^2*
d)

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fricas [A]  time = 0.48, size = 239, normalized size = 1.36 \[ \frac {2310 \, \cos \left (d x + c\right )^{7} + 490 \, \cos \left (d x + c\right )^{5} - 8470 \, \cos \left (d x + c\right )^{3} - 1155 \, {\left (\cos \left (d x + c\right )^{8} - 4 \, \cos \left (d x + c\right )^{6} + 6 \, \cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 1155 \, {\left (\cos \left (d x + c\right )^{8} - 4 \, \cos \left (d x + c\right )^{6} + 6 \, \cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 1536 \, {\left (2 \, \cos \left (d x + c\right )^{7} - 7 \, \cos \left (d x + c\right )^{5}\right )} \sin \left (d x + c\right ) + 2310 \, \cos \left (d x + c\right )}{26880 \, {\left (a^{2} d \cos \left (d x + c\right )^{8} - 4 \, a^{2} d \cos \left (d x + c\right )^{6} + 6 \, a^{2} d \cos \left (d x + c\right )^{4} - 4 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)^9/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/26880*(2310*cos(d*x + c)^7 + 490*cos(d*x + c)^5 - 8470*cos(d*x + c)^3 - 1155*(cos(d*x + c)^8 - 4*cos(d*x + c
)^6 + 6*cos(d*x + c)^4 - 4*cos(d*x + c)^2 + 1)*log(1/2*cos(d*x + c) + 1/2) + 1155*(cos(d*x + c)^8 - 4*cos(d*x
+ c)^6 + 6*cos(d*x + c)^4 - 4*cos(d*x + c)^2 + 1)*log(-1/2*cos(d*x + c) + 1/2) - 1536*(2*cos(d*x + c)^7 - 7*co
s(d*x + c)^5)*sin(d*x + c) + 2310*cos(d*x + c))/(a^2*d*cos(d*x + c)^8 - 4*a^2*d*cos(d*x + c)^6 + 6*a^2*d*cos(d
*x + c)^4 - 4*a^2*d*cos(d*x + c)^2 + a^2*d)

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giac [A]  time = 0.31, size = 273, normalized size = 1.55 \[ \frac {\frac {18480 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} - \frac {50226 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 10080 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 1680 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 3360 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 2520 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 672 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 560 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 480 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 105}{a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8}} + \frac {105 \, a^{14} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 480 \, a^{14} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 560 \, a^{14} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 672 \, a^{14} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 2520 \, a^{14} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 3360 \, a^{14} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 1680 \, a^{14} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 10080 \, a^{14} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{16}}}{215040 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)^9/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/215040*(18480*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 - (50226*tan(1/2*d*x + 1/2*c)^8 - 10080*tan(1/2*d*x + 1/2*c
)^7 - 1680*tan(1/2*d*x + 1/2*c)^6 + 3360*tan(1/2*d*x + 1/2*c)^5 - 2520*tan(1/2*d*x + 1/2*c)^4 + 672*tan(1/2*d*
x + 1/2*c)^3 + 560*tan(1/2*d*x + 1/2*c)^2 - 480*tan(1/2*d*x + 1/2*c) + 105)/(a^2*tan(1/2*d*x + 1/2*c)^8) + (10
5*a^14*tan(1/2*d*x + 1/2*c)^8 - 480*a^14*tan(1/2*d*x + 1/2*c)^7 + 560*a^14*tan(1/2*d*x + 1/2*c)^6 + 672*a^14*t
an(1/2*d*x + 1/2*c)^5 - 2520*a^14*tan(1/2*d*x + 1/2*c)^4 + 3360*a^14*tan(1/2*d*x + 1/2*c)^3 - 1680*a^14*tan(1/
2*d*x + 1/2*c)^2 - 10080*a^14*tan(1/2*d*x + 1/2*c))/a^16)/d

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maple [B]  time = 0.70, size = 322, normalized size = 1.83 \[ \frac {\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )}{2048 d \,a^{2}}-\frac {\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )}{448 d \,a^{2}}+\frac {\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )}{384 d \,a^{2}}+\frac {\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )}{320 a^{2} d}-\frac {3 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{256 a^{2} d}+\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{64 d \,a^{2}}-\frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{128 a^{2} d}-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{64 d \,a^{2}}-\frac {1}{384 d \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}+\frac {3}{64 d \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {11 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{128 d \,a^{2}}-\frac {1}{320 a^{2} d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}+\frac {1}{448 d \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}+\frac {1}{128 a^{2} d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-\frac {1}{2048 d \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}+\frac {3}{256 a^{2} d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}-\frac {1}{64 a^{2} d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^8*csc(d*x+c)^9/(a+a*sin(d*x+c))^2,x)

[Out]

1/2048/d/a^2*tan(1/2*d*x+1/2*c)^8-1/448/d/a^2*tan(1/2*d*x+1/2*c)^7+1/384/d/a^2*tan(1/2*d*x+1/2*c)^6+1/320/d/a^
2*tan(1/2*d*x+1/2*c)^5-3/256/d/a^2*tan(1/2*d*x+1/2*c)^4+1/64/d/a^2*tan(1/2*d*x+1/2*c)^3-1/128/d/a^2*tan(1/2*d*
x+1/2*c)^2-3/64/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+3/64/d/a^2/tan(1/2*d*x+1/2*c)+11/128
/d/a^2*ln(tan(1/2*d*x+1/2*c))-1/320/a^2/d/tan(1/2*d*x+1/2*c)^5+1/448/d/a^2/tan(1/2*d*x+1/2*c)^7+1/128/a^2/d/ta
n(1/2*d*x+1/2*c)^2-1/2048/d/a^2/tan(1/2*d*x+1/2*c)^8+3/256/a^2/d/tan(1/2*d*x+1/2*c)^4-1/64/a^2/d/tan(1/2*d*x+1
/2*c)^3

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maxima [B]  time = 0.33, size = 355, normalized size = 2.02 \[ -\frac {\frac {\frac {10080 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {1680 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {3360 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {2520 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {672 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {560 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {480 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {105 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}}{a^{2}} - \frac {18480 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} - \frac {{\left (\frac {480 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {560 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {672 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {2520 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {3360 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {1680 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {10080 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - 105\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{8}}{a^{2} \sin \left (d x + c\right )^{8}}}{215040 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)^9/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/215040*((10080*sin(d*x + c)/(cos(d*x + c) + 1) + 1680*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 3360*sin(d*x +
c)^3/(cos(d*x + c) + 1)^3 + 2520*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 672*sin(d*x + c)^5/(cos(d*x + c) + 1)^5
- 560*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 480*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 105*sin(d*x + c)^8/(cos
(d*x + c) + 1)^8)/a^2 - 18480*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^2 - (480*sin(d*x + c)/(cos(d*x + c) + 1)
- 560*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 672*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 2520*sin(d*x + c)^4/(cos
(d*x + c) + 1)^4 - 3360*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 1680*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 10080
*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 105)*(cos(d*x + c) + 1)^8/(a^2*sin(d*x + c)^8))/d

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mupad [B]  time = 11.96, size = 435, normalized size = 2.47 \[ \frac {105\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{16}-105\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{16}-480\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}+480\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+560\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}+672\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}-2520\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+3360\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}-1680\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-10080\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+10080\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+1680\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3360\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+2520\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-672\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-560\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+18480\,\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 )}^8\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{215040\,a^2\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^8/(sin(c + d*x)^9*(a + a*sin(c + d*x))^2),x)

[Out]

(105*sin(c/2 + (d*x)/2)^16 - 105*cos(c/2 + (d*x)/2)^16 - 480*cos(c/2 + (d*x)/2)*sin(c/2 + (d*x)/2)^15 + 480*co
s(c/2 + (d*x)/2)^15*sin(c/2 + (d*x)/2) + 560*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^14 + 672*cos(c/2 + (d*x)/
2)^3*sin(c/2 + (d*x)/2)^13 - 2520*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^12 + 3360*cos(c/2 + (d*x)/2)^5*sin(c
/2 + (d*x)/2)^11 - 1680*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2)^10 - 10080*cos(c/2 + (d*x)/2)^7*sin(c/2 + (d*x
)/2)^9 + 10080*cos(c/2 + (d*x)/2)^9*sin(c/2 + (d*x)/2)^7 + 1680*cos(c/2 + (d*x)/2)^10*sin(c/2 + (d*x)/2)^6 - 3
360*cos(c/2 + (d*x)/2)^11*sin(c/2 + (d*x)/2)^5 + 2520*cos(c/2 + (d*x)/2)^12*sin(c/2 + (d*x)/2)^4 - 672*cos(c/2
+ (d*x)/2)^13*sin(c/2 + (d*x)/2)^3 - 560*cos(c/2 + (d*x)/2)^14*sin(c/2 + (d*x)/2)^2 + 18480*log(sin(c/2 + (d*
x)/2)/cos(c/2 + (d*x)/2))*cos(c/2 + (d*x)/2)^8*sin(c/2 + (d*x)/2)^8)/(215040*a^2*d*cos(c/2 + (d*x)/2)^8*sin(c/
2 + (d*x)/2)^8)

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

[Out]

Timed out

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