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

Optimal. Leaf size=168 \[ -\frac {\cot ^9(c+d x)}{9 a^2 d}-\frac {3 \cot ^7(c+d x)}{7 a^2 d}-\frac {2 \cot ^5(c+d x)}{5 a^2 d}+\frac {3 \tanh ^{-1}(\cos (c+d x))}{64 a^2 d}+\frac {\cot ^3(c+d x) \csc ^5(c+d x)}{4 a^2 d}-\frac {\cot (c+d x) \csc ^5(c+d x)}{8 a^2 d}+\frac {\cot (c+d x) \csc ^3(c+d x)}{32 a^2 d}+\frac {3 \cot (c+d x) \csc (c+d x)}{64 a^2 d} \]

[Out]

3/64*arctanh(cos(d*x+c))/a^2/d-2/5*cot(d*x+c)^5/a^2/d-3/7*cot(d*x+c)^7/a^2/d-1/9*cot(d*x+c)^9/a^2/d+3/64*cot(d
*x+c)*csc(d*x+c)/a^2/d+1/32*cot(d*x+c)*csc(d*x+c)^3/a^2/d-1/8*cot(d*x+c)*csc(d*x+c)^5/a^2/d+1/4*cot(d*x+c)^3*c
sc(d*x+c)^5/a^2/d

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Rubi [A]  time = 0.39, antiderivative size = 168, 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, 2873, 2607, 14, 2611, 3768, 3770, 270} \[ -\frac {\cot ^9(c+d x)}{9 a^2 d}-\frac {3 \cot ^7(c+d x)}{7 a^2 d}-\frac {2 \cot ^5(c+d x)}{5 a^2 d}+\frac {3 \tanh ^{-1}(\cos (c+d x))}{64 a^2 d}+\frac {\cot ^3(c+d x) \csc ^5(c+d x)}{4 a^2 d}-\frac {\cot (c+d x) \csc ^5(c+d x)}{8 a^2 d}+\frac {\cot (c+d x) \csc ^3(c+d x)}{32 a^2 d}+\frac {3 \cot (c+d x) \csc (c+d x)}{64 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*ArcTanh[Cos[c + d*x]])/(64*a^2*d) - (2*Cot[c + d*x]^5)/(5*a^2*d) - (3*Cot[c + d*x]^7)/(7*a^2*d) - Cot[c + d
*x]^9/(9*a^2*d) + (3*Cot[c + d*x]*Csc[c + d*x])/(64*a^2*d) + (Cot[c + d*x]*Csc[c + d*x]^3)/(32*a^2*d) - (Cot[c
+ d*x]*Csc[c + d*x]^5)/(8*a^2*d) + (Cot[c + d*x]^3*Csc[c + d*x]^5)/(4*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 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

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 ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac {\int \cot ^4(c+d x) \csc ^6(c+d x) (a-a \sin (c+d x))^2 \, dx}{a^4}\\ &=\frac {\int \left (a^2 \cot ^4(c+d x) \csc ^4(c+d x)-2 a^2 \cot ^4(c+d x) \csc ^5(c+d x)+a^2 \cot ^4(c+d x) \csc ^6(c+d x)\right ) \, dx}{a^4}\\ &=\frac {\int \cot ^4(c+d x) \csc ^4(c+d x) \, dx}{a^2}+\frac {\int \cot ^4(c+d x) \csc ^6(c+d x) \, dx}{a^2}-\frac {2 \int \cot ^4(c+d x) \csc ^5(c+d x) \, dx}{a^2}\\ &=\frac {\cot ^3(c+d x) \csc ^5(c+d x)}{4 a^2 d}+\frac {3 \int \cot ^2(c+d x) \csc ^5(c+d x) \, dx}{4 a^2}+\frac {\operatorname {Subst}\left (\int x^4 \left (1+x^2\right ) \, dx,x,-\cot (c+d x)\right )}{a^2 d}+\frac {\operatorname {Subst}\left (\int x^4 \left (1+x^2\right )^2 \, dx,x,-\cot (c+d x)\right )}{a^2 d}\\ &=-\frac {\cot (c+d x) \csc ^5(c+d x)}{8 a^2 d}+\frac {\cot ^3(c+d x) \csc ^5(c+d x)}{4 a^2 d}-\frac {\int \csc ^5(c+d x) \, dx}{8 a^2}+\frac {\operatorname {Subst}\left (\int \left (x^4+x^6\right ) \, dx,x,-\cot (c+d x)\right )}{a^2 d}+\frac {\operatorname {Subst}\left (\int \left (x^4+2 x^6+x^8\right ) \, dx,x,-\cot (c+d x)\right )}{a^2 d}\\ &=-\frac {2 \cot ^5(c+d x)}{5 a^2 d}-\frac {3 \cot ^7(c+d x)}{7 a^2 d}-\frac {\cot ^9(c+d x)}{9 a^2 d}+\frac {\cot (c+d x) \csc ^3(c+d x)}{32 a^2 d}-\frac {\cot (c+d x) \csc ^5(c+d x)}{8 a^2 d}+\frac {\cot ^3(c+d x) \csc ^5(c+d x)}{4 a^2 d}-\frac {3 \int \csc ^3(c+d x) \, dx}{32 a^2}\\ &=-\frac {2 \cot ^5(c+d x)}{5 a^2 d}-\frac {3 \cot ^7(c+d x)}{7 a^2 d}-\frac {\cot ^9(c+d x)}{9 a^2 d}+\frac {3 \cot (c+d x) \csc (c+d x)}{64 a^2 d}+\frac {\cot (c+d x) \csc ^3(c+d x)}{32 a^2 d}-\frac {\cot (c+d x) \csc ^5(c+d x)}{8 a^2 d}+\frac {\cot ^3(c+d x) \csc ^5(c+d x)}{4 a^2 d}-\frac {3 \int \csc (c+d x) \, dx}{64 a^2}\\ &=\frac {3 \tanh ^{-1}(\cos (c+d x))}{64 a^2 d}-\frac {2 \cot ^5(c+d x)}{5 a^2 d}-\frac {3 \cot ^7(c+d x)}{7 a^2 d}-\frac {\cot ^9(c+d x)}{9 a^2 d}+\frac {3 \cot (c+d x) \csc (c+d x)}{64 a^2 d}+\frac {\cot (c+d x) \csc ^3(c+d x)}{32 a^2 d}-\frac {\cot (c+d x) \csc ^5(c+d x)}{8 a^2 d}+\frac {\cot ^3(c+d x) \csc ^5(c+d x)}{4 a^2 d}\\ \end {align*}

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Mathematica [A]  time = 1.87, size = 313, normalized size = 1.86 \[ \frac {\csc ^9(c+d x) \left (212940 \sin (2 (c+d x))+195300 \sin (4 (c+d x))+16380 \sin (6 (c+d x))-1890 \sin (8 (c+d x))-451584 \cos (c+d x)-155904 \cos (3 (c+d x))+20736 \cos (5 (c+d x))+14976 \cos (7 (c+d x))-1664 \cos (9 (c+d x))-119070 \sin (c+d x) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+79380 \sin (3 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-34020 \sin (5 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+8505 \sin (7 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-945 \sin (9 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+119070 \sin (c+d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-79380 \sin (3 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+34020 \sin (5 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-8505 \sin (7 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+945 \sin (9 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )}{5160960 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Csc[c + d*x]^9*(-451584*Cos[c + d*x] - 155904*Cos[3*(c + d*x)] + 20736*Cos[5*(c + d*x)] + 14976*Cos[7*(c + d*
x)] - 1664*Cos[9*(c + d*x)] + 119070*Log[Cos[(c + d*x)/2]]*Sin[c + d*x] - 119070*Log[Sin[(c + d*x)/2]]*Sin[c +
d*x] + 212940*Sin[2*(c + d*x)] - 79380*Log[Cos[(c + d*x)/2]]*Sin[3*(c + d*x)] + 79380*Log[Sin[(c + d*x)/2]]*S
in[3*(c + d*x)] + 195300*Sin[4*(c + d*x)] + 34020*Log[Cos[(c + d*x)/2]]*Sin[5*(c + d*x)] - 34020*Log[Sin[(c +
d*x)/2]]*Sin[5*(c + d*x)] + 16380*Sin[6*(c + d*x)] - 8505*Log[Cos[(c + d*x)/2]]*Sin[7*(c + d*x)] + 8505*Log[Si
n[(c + d*x)/2]]*Sin[7*(c + d*x)] - 1890*Sin[8*(c + d*x)] + 945*Log[Cos[(c + d*x)/2]]*Sin[9*(c + d*x)] - 945*Lo
g[Sin[(c + d*x)/2]]*Sin[9*(c + d*x)]))/(5160960*a^2*d)

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fricas [A]  time = 0.49, size = 269, normalized size = 1.60 \[ -\frac {3328 \, \cos \left (d x + c\right )^{9} - 14976 \, \cos \left (d x + c\right )^{7} + 16128 \, \cos \left (d x + c\right )^{5} - 945 \, {\left (\cos \left (d x + c\right )^{8} - 4 \, \cos \left (d x + c\right )^{6} + 6 \, \cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 945 \, {\left (\cos \left (d x + c\right )^{8} - 4 \, \cos \left (d x + c\right )^{6} + 6 \, \cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 630 \, {\left (3 \, \cos \left (d x + c\right )^{7} - 11 \, \cos \left (d x + c\right )^{5} - 11 \, \cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{40320 \, {\left (a^{2} d \cos \left (d x + c\right )^{8} - 4 \, a^{2} d \cos \left (d x + c\right )^{6} + 6 \, a^{2} d \cos \left (d x + c\right )^{4} - 4 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d\right )} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)^10/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/40320*(3328*cos(d*x + c)^9 - 14976*cos(d*x + c)^7 + 16128*cos(d*x + c)^5 - 945*(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)*sin(d*x + c) + 945*(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)*sin(d*x + c) + 6
30*(3*cos(d*x + c)^7 - 11*cos(d*x + c)^5 - 11*cos(d*x + c)^3 + 3*cos(d*x + c))*sin(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)*sin(d*x + c))

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giac [A]  time = 0.30, size = 245, normalized size = 1.46 \[ -\frac {\frac {15120 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} - \frac {42774 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 11340 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 3360 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 2520 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1008 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 450 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 315 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 70}{a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9}} - \frac {70 \, a^{16} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 315 \, a^{16} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 450 \, a^{16} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 1008 \, a^{16} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 2520 \, a^{16} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 3360 \, a^{16} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 11340 \, a^{16} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{18}}}{322560 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)^10/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/322560*(15120*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 - (42774*tan(1/2*d*x + 1/2*c)^9 - 11340*tan(1/2*d*x + 1/2*
c)^8 + 3360*tan(1/2*d*x + 1/2*c)^6 - 2520*tan(1/2*d*x + 1/2*c)^5 + 1008*tan(1/2*d*x + 1/2*c)^4 - 450*tan(1/2*d
*x + 1/2*c)^2 + 315*tan(1/2*d*x + 1/2*c) - 70)/(a^2*tan(1/2*d*x + 1/2*c)^9) - (70*a^16*tan(1/2*d*x + 1/2*c)^9
- 315*a^16*tan(1/2*d*x + 1/2*c)^8 + 450*a^16*tan(1/2*d*x + 1/2*c)^7 - 1008*a^16*tan(1/2*d*x + 1/2*c)^5 + 2520*
a^16*tan(1/2*d*x + 1/2*c)^4 - 3360*a^16*tan(1/2*d*x + 1/2*c)^3 + 11340*a^16*tan(1/2*d*x + 1/2*c))/a^18)/d

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maple [A]  time = 0.71, size = 284, normalized size = 1.69 \[ \frac {\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )}{4608 d \,a^{2}}-\frac {\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )}{1024 d \,a^{2}}+\frac {5 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3584 d \,a^{2}}-\frac {\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )}{320 a^{2} d}+\frac {\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )}{128 a^{2} d}-\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{96 d \,a^{2}}+\frac {9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{256 d \,a^{2}}-\frac {9}{256 d \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 d \,a^{2}}+\frac {1}{320 a^{2} d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}-\frac {5}{3584 d \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}-\frac {1}{4608 d \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}+\frac {1}{1024 d \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}-\frac {1}{128 a^{2} d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}+\frac {1}{96 a^{2} d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^8*csc(d*x+c)^10/(a+a*sin(d*x+c))^2,x)

[Out]

1/4608/d/a^2*tan(1/2*d*x+1/2*c)^9-1/1024/d/a^2*tan(1/2*d*x+1/2*c)^8+5/3584/d/a^2*tan(1/2*d*x+1/2*c)^7-1/320/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)^4-1/96/d/a^2*tan(1/2*d*x+1/2*c)^3+9/256/d/a^2*tan(1/2*
d*x+1/2*c)-9/256/d/a^2/tan(1/2*d*x+1/2*c)-3/64/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)^5-5
/3584/d/a^2/tan(1/2*d*x+1/2*c)^7-1/4608/d/a^2/tan(1/2*d*x+1/2*c)^9+1/1024/d/a^2/tan(1/2*d*x+1/2*c)^8-1/128/a^2
/d/tan(1/2*d*x+1/2*c)^4+1/96/a^2/d/tan(1/2*d*x+1/2*c)^3

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maxima [B]  time = 0.34, size = 314, normalized size = 1.87 \[ \frac {\frac {\frac {11340 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {3360 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {2520 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {1008 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {450 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {315 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {70 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}}}{a^{2}} - \frac {15120 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} + \frac {{\left (\frac {315 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {450 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {1008 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {2520 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {3360 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {11340 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - 70\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{9}}{a^{2} \sin \left (d x + c\right )^{9}}}{322560 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)^10/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/322560*((11340*sin(d*x + c)/(cos(d*x + c) + 1) - 3360*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 2520*sin(d*x + c
)^4/(cos(d*x + c) + 1)^4 - 1008*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 450*sin(d*x + c)^7/(cos(d*x + c) + 1)^7
- 315*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + 70*sin(d*x + c)^9/(cos(d*x + c) + 1)^9)/a^2 - 15120*log(sin(d*x +
c)/(cos(d*x + c) + 1))/a^2 + (315*sin(d*x + c)/(cos(d*x + c) + 1) - 450*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 +
1008*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 2520*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 3360*sin(d*x + c)^6/(cos
(d*x + c) + 1)^6 - 11340*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 70)*(cos(d*x + c) + 1)^9/(a^2*sin(d*x + c)^9))/
d

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mupad [B]  time = 12.56, size = 387, normalized size = 2.30 \[ -\frac {70\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{18}-70\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{18}+315\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{17}-315\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{17}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-450\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{16}+1008\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}-2520\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}+3360\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-11340\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+11340\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-3360\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+2520\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-1008\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+450\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{16}\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+15120\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{322560\,a^2\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^8/(sin(c + d*x)^10*(a + a*sin(c + d*x))^2),x)

[Out]

-(70*cos(c/2 + (d*x)/2)^18 - 70*sin(c/2 + (d*x)/2)^18 + 315*cos(c/2 + (d*x)/2)*sin(c/2 + (d*x)/2)^17 - 315*cos
(c/2 + (d*x)/2)^17*sin(c/2 + (d*x)/2) - 450*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^16 + 1008*cos(c/2 + (d*x)/
2)^4*sin(c/2 + (d*x)/2)^14 - 2520*cos(c/2 + (d*x)/2)^5*sin(c/2 + (d*x)/2)^13 + 3360*cos(c/2 + (d*x)/2)^6*sin(c
/2 + (d*x)/2)^12 - 11340*cos(c/2 + (d*x)/2)^8*sin(c/2 + (d*x)/2)^10 + 11340*cos(c/2 + (d*x)/2)^10*sin(c/2 + (d
*x)/2)^8 - 3360*cos(c/2 + (d*x)/2)^12*sin(c/2 + (d*x)/2)^6 + 2520*cos(c/2 + (d*x)/2)^13*sin(c/2 + (d*x)/2)^5 -
1008*cos(c/2 + (d*x)/2)^14*sin(c/2 + (d*x)/2)^4 + 450*cos(c/2 + (d*x)/2)^16*sin(c/2 + (d*x)/2)^2 + 15120*log(
sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(c/2 + (d*x)/2)^9*sin(c/2 + (d*x)/2)^9)/(322560*a^2*d*cos(c/2 + (d*x
)/2)^9*sin(c/2 + (d*x)/2)^9)

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

[Out]

Timed out

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