3.725 \(\int \frac {\cos ^8(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=141 \[ -\frac {2 \cos ^7(c+d x)}{7 a^2 d}+\frac {2 \cos ^5(c+d x)}{5 a^2 d}-\frac {\sin ^3(c+d x) \cos ^5(c+d x)}{8 a^2 d}-\frac {11 \sin (c+d x) \cos ^5(c+d x)}{48 a^2 d}+\frac {11 \sin (c+d x) \cos ^3(c+d x)}{192 a^2 d}+\frac {11 \sin (c+d x) \cos (c+d x)}{128 a^2 d}+\frac {11 x}{128 a^2} \]

[Out]

11/128*x/a^2+2/5*cos(d*x+c)^5/a^2/d-2/7*cos(d*x+c)^7/a^2/d+11/128*cos(d*x+c)*sin(d*x+c)/a^2/d+11/192*cos(d*x+c
)^3*sin(d*x+c)/a^2/d-11/48*cos(d*x+c)^5*sin(d*x+c)/a^2/d-1/8*cos(d*x+c)^5*sin(d*x+c)^3/a^2/d

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Rubi [A]  time = 0.37, antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2875, 2873, 2568, 2635, 8, 2565, 14} \[ -\frac {2 \cos ^7(c+d x)}{7 a^2 d}+\frac {2 \cos ^5(c+d x)}{5 a^2 d}-\frac {\sin ^3(c+d x) \cos ^5(c+d x)}{8 a^2 d}-\frac {11 \sin (c+d x) \cos ^5(c+d x)}{48 a^2 d}+\frac {11 \sin (c+d x) \cos ^3(c+d x)}{192 a^2 d}+\frac {11 \sin (c+d x) \cos (c+d x)}{128 a^2 d}+\frac {11 x}{128 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(11*x)/(128*a^2) + (2*Cos[c + d*x]^5)/(5*a^2*d) - (2*Cos[c + d*x]^7)/(7*a^2*d) + (11*Cos[c + d*x]*Sin[c + d*x]
)/(128*a^2*d) + (11*Cos[c + d*x]^3*Sin[c + d*x])/(192*a^2*d) - (11*Cos[c + d*x]^5*Sin[c + d*x])/(48*a^2*d) - (
Cos[c + d*x]^5*Sin[c + d*x]^3)/(8*a^2*d)

Rule 8

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

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 2565

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 2568

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(a*(b*Cos[e
+ f*x])^(n + 1)*(a*Sin[e + f*x])^(m - 1))/(b*f*(m + n)), x] + Dist[(a^2*(m - 1))/(m + n), Int[(b*Cos[e + f*x])
^n*(a*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m + n, 0] && IntegersQ[2*
m, 2*n]

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 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]

Rubi steps

\begin {align*} \int \frac {\cos ^8(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac {\int \cos ^4(c+d x) \sin ^2(c+d x) (a-a \sin (c+d x))^2 \, dx}{a^4}\\ &=\frac {\int \left (a^2 \cos ^4(c+d x) \sin ^2(c+d x)-2 a^2 \cos ^4(c+d x) \sin ^3(c+d x)+a^2 \cos ^4(c+d x) \sin ^4(c+d x)\right ) \, dx}{a^4}\\ &=\frac {\int \cos ^4(c+d x) \sin ^2(c+d x) \, dx}{a^2}+\frac {\int \cos ^4(c+d x) \sin ^4(c+d x) \, dx}{a^2}-\frac {2 \int \cos ^4(c+d x) \sin ^3(c+d x) \, dx}{a^2}\\ &=-\frac {\cos ^5(c+d x) \sin (c+d x)}{6 a^2 d}-\frac {\cos ^5(c+d x) \sin ^3(c+d x)}{8 a^2 d}+\frac {\int \cos ^4(c+d x) \, dx}{6 a^2}+\frac {3 \int \cos ^4(c+d x) \sin ^2(c+d x) \, dx}{8 a^2}+\frac {2 \operatorname {Subst}\left (\int x^4 \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{a^2 d}\\ &=\frac {\cos ^3(c+d x) \sin (c+d x)}{24 a^2 d}-\frac {11 \cos ^5(c+d x) \sin (c+d x)}{48 a^2 d}-\frac {\cos ^5(c+d x) \sin ^3(c+d x)}{8 a^2 d}+\frac {\int \cos ^4(c+d x) \, dx}{16 a^2}+\frac {\int \cos ^2(c+d x) \, dx}{8 a^2}+\frac {2 \operatorname {Subst}\left (\int \left (x^4-x^6\right ) \, dx,x,\cos (c+d x)\right )}{a^2 d}\\ &=\frac {2 \cos ^5(c+d x)}{5 a^2 d}-\frac {2 \cos ^7(c+d x)}{7 a^2 d}+\frac {\cos (c+d x) \sin (c+d x)}{16 a^2 d}+\frac {11 \cos ^3(c+d x) \sin (c+d x)}{192 a^2 d}-\frac {11 \cos ^5(c+d x) \sin (c+d x)}{48 a^2 d}-\frac {\cos ^5(c+d x) \sin ^3(c+d x)}{8 a^2 d}+\frac {3 \int \cos ^2(c+d x) \, dx}{64 a^2}+\frac {\int 1 \, dx}{16 a^2}\\ &=\frac {x}{16 a^2}+\frac {2 \cos ^5(c+d x)}{5 a^2 d}-\frac {2 \cos ^7(c+d x)}{7 a^2 d}+\frac {11 \cos (c+d x) \sin (c+d x)}{128 a^2 d}+\frac {11 \cos ^3(c+d x) \sin (c+d x)}{192 a^2 d}-\frac {11 \cos ^5(c+d x) \sin (c+d x)}{48 a^2 d}-\frac {\cos ^5(c+d x) \sin ^3(c+d x)}{8 a^2 d}+\frac {3 \int 1 \, dx}{128 a^2}\\ &=\frac {11 x}{128 a^2}+\frac {2 \cos ^5(c+d x)}{5 a^2 d}-\frac {2 \cos ^7(c+d x)}{7 a^2 d}+\frac {11 \cos (c+d x) \sin (c+d x)}{128 a^2 d}+\frac {11 \cos ^3(c+d x) \sin (c+d x)}{192 a^2 d}-\frac {11 \cos ^5(c+d x) \sin (c+d x)}{48 a^2 d}-\frac {\cos ^5(c+d x) \sin ^3(c+d x)}{8 a^2 d}\\ \end {align*}

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Mathematica [B]  time = 3.88, size = 481, normalized size = 3.41 \[ \frac {18480 d x \sin \left (\frac {c}{2}\right )-10080 \sin \left (\frac {c}{2}+d x\right )+10080 \sin \left (\frac {3 c}{2}+d x\right )+1680 \sin \left (\frac {3 c}{2}+2 d x\right )+1680 \sin \left (\frac {5 c}{2}+2 d x\right )-3360 \sin \left (\frac {5 c}{2}+3 d x\right )+3360 \sin \left (\frac {7 c}{2}+3 d x\right )-2520 \sin \left (\frac {7 c}{2}+4 d x\right )-2520 \sin \left (\frac {9 c}{2}+4 d x\right )+672 \sin \left (\frac {9 c}{2}+5 d x\right )-672 \sin \left (\frac {11 c}{2}+5 d x\right )-560 \sin \left (\frac {11 c}{2}+6 d x\right )-560 \sin \left (\frac {13 c}{2}+6 d x\right )+480 \sin \left (\frac {13 c}{2}+7 d x\right )-480 \sin \left (\frac {15 c}{2}+7 d x\right )+105 \sin \left (\frac {15 c}{2}+8 d x\right )+105 \sin \left (\frac {17 c}{2}+8 d x\right )+9240 \cos \left (\frac {c}{2}\right ) (15 c+2 d x)+10080 \cos \left (\frac {c}{2}+d x\right )+10080 \cos \left (\frac {3 c}{2}+d x\right )+1680 \cos \left (\frac {3 c}{2}+2 d x\right )-1680 \cos \left (\frac {5 c}{2}+2 d x\right )+3360 \cos \left (\frac {5 c}{2}+3 d x\right )+3360 \cos \left (\frac {7 c}{2}+3 d x\right )-2520 \cos \left (\frac {7 c}{2}+4 d x\right )+2520 \cos \left (\frac {9 c}{2}+4 d x\right )-672 \cos \left (\frac {9 c}{2}+5 d x\right )-672 \cos \left (\frac {11 c}{2}+5 d x\right )-560 \cos \left (\frac {11 c}{2}+6 d x\right )+560 \cos \left (\frac {13 c}{2}+6 d x\right )-480 \cos \left (\frac {13 c}{2}+7 d x\right )-480 \cos \left (\frac {15 c}{2}+7 d x\right )+105 \cos \left (\frac {15 c}{2}+8 d x\right )-105 \cos \left (\frac {17 c}{2}+8 d x\right )+138600 c \sin \left (\frac {c}{2}\right )-79800 \sin \left (\frac {c}{2}\right )}{215040 a^2 d \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(9240*(15*c + 2*d*x)*Cos[c/2] + 10080*Cos[c/2 + d*x] + 10080*Cos[(3*c)/2 + d*x] + 1680*Cos[(3*c)/2 + 2*d*x] -
1680*Cos[(5*c)/2 + 2*d*x] + 3360*Cos[(5*c)/2 + 3*d*x] + 3360*Cos[(7*c)/2 + 3*d*x] - 2520*Cos[(7*c)/2 + 4*d*x]
+ 2520*Cos[(9*c)/2 + 4*d*x] - 672*Cos[(9*c)/2 + 5*d*x] - 672*Cos[(11*c)/2 + 5*d*x] - 560*Cos[(11*c)/2 + 6*d*x]
+ 560*Cos[(13*c)/2 + 6*d*x] - 480*Cos[(13*c)/2 + 7*d*x] - 480*Cos[(15*c)/2 + 7*d*x] + 105*Cos[(15*c)/2 + 8*d*
x] - 105*Cos[(17*c)/2 + 8*d*x] - 79800*Sin[c/2] + 138600*c*Sin[c/2] + 18480*d*x*Sin[c/2] - 10080*Sin[c/2 + d*x
] + 10080*Sin[(3*c)/2 + d*x] + 1680*Sin[(3*c)/2 + 2*d*x] + 1680*Sin[(5*c)/2 + 2*d*x] - 3360*Sin[(5*c)/2 + 3*d*
x] + 3360*Sin[(7*c)/2 + 3*d*x] - 2520*Sin[(7*c)/2 + 4*d*x] - 2520*Sin[(9*c)/2 + 4*d*x] + 672*Sin[(9*c)/2 + 5*d
*x] - 672*Sin[(11*c)/2 + 5*d*x] - 560*Sin[(11*c)/2 + 6*d*x] - 560*Sin[(13*c)/2 + 6*d*x] + 480*Sin[(13*c)/2 + 7
*d*x] - 480*Sin[(15*c)/2 + 7*d*x] + 105*Sin[(15*c)/2 + 8*d*x] + 105*Sin[(17*c)/2 + 8*d*x])/(215040*a^2*d*(Cos[
c/2] + Sin[c/2]))

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fricas [A]  time = 0.49, size = 80, normalized size = 0.57 \[ -\frac {3840 \, \cos \left (d x + c\right )^{7} - 5376 \, \cos \left (d x + c\right )^{5} - 1155 \, d x - 35 \, {\left (48 \, \cos \left (d x + c\right )^{7} - 136 \, \cos \left (d x + c\right )^{5} + 22 \, \cos \left (d x + c\right )^{3} + 33 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{13440 \, a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/13440*(3840*cos(d*x + c)^7 - 5376*cos(d*x + c)^5 - 1155*d*x - 35*(48*cos(d*x + c)^7 - 136*cos(d*x + c)^5 +
22*cos(d*x + c)^3 + 33*cos(d*x + c))*sin(d*x + c))/(a^2*d)

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giac [A]  time = 0.24, size = 205, normalized size = 1.45 \[ \frac {\frac {1155 \, {\left (d x + c\right )}}{a^{2}} + \frac {2 \, {\left (1155 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{15} - 9065 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} + 53760 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{12} - 38605 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 79135 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 53760 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 79135 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 86016 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 38605 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 10752 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 9065 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12288 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1155 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1536\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{8} a^{2}}}{13440 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/13440*(1155*(d*x + c)/a^2 + 2*(1155*tan(1/2*d*x + 1/2*c)^15 - 9065*tan(1/2*d*x + 1/2*c)^13 + 53760*tan(1/2*d
*x + 1/2*c)^12 - 38605*tan(1/2*d*x + 1/2*c)^11 + 79135*tan(1/2*d*x + 1/2*c)^9 + 53760*tan(1/2*d*x + 1/2*c)^8 -
79135*tan(1/2*d*x + 1/2*c)^7 + 86016*tan(1/2*d*x + 1/2*c)^6 + 38605*tan(1/2*d*x + 1/2*c)^5 - 10752*tan(1/2*d*
x + 1/2*c)^4 + 9065*tan(1/2*d*x + 1/2*c)^3 + 12288*tan(1/2*d*x + 1/2*c)^2 - 1155*tan(1/2*d*x + 1/2*c) + 1536)/
((tan(1/2*d*x + 1/2*c)^2 + 1)^8*a^2))/d

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maple [B]  time = 0.38, size = 483, normalized size = 3.43 \[ \frac {8}{35 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{8}}-\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{64 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{8}}+\frac {64 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{35 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{8}}+\frac {259 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{192 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{8}}-\frac {8 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{8}}+\frac {1103 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{192 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{8}}+\frac {64 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{8}}-\frac {2261 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{192 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{8}}+\frac {8 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{8}}+\frac {2261 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{192 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{8}}-\frac {1103 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{192 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{8}}+\frac {8 \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{8}}-\frac {259 \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{192 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{8}}+\frac {11 \left (\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{8}}+\frac {11 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8/35/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^8-11/64/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^8*tan(1/2*d*x+1/2*c)+64/35/d/a^2/(1
+tan(1/2*d*x+1/2*c)^2)^8*tan(1/2*d*x+1/2*c)^2+259/192/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^8*tan(1/2*d*x+1/2*c)^3-8/
5/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^8*tan(1/2*d*x+1/2*c)^4+1103/192/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^8*tan(1/2*d*x+
1/2*c)^5+64/5/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^8*tan(1/2*d*x+1/2*c)^6-2261/192/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^8*
tan(1/2*d*x+1/2*c)^7+8/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^8*tan(1/2*d*x+1/2*c)^8+2261/192/d/a^2/(1+tan(1/2*d*x+1/2
*c)^2)^8*tan(1/2*d*x+1/2*c)^9-1103/192/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^8*tan(1/2*d*x+1/2*c)^11+8/d/a^2/(1+tan(1
/2*d*x+1/2*c)^2)^8*tan(1/2*d*x+1/2*c)^12-259/192/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^8*tan(1/2*d*x+1/2*c)^13+11/64/
d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^8*tan(1/2*d*x+1/2*c)^15+11/64/d/a^2*arctan(tan(1/2*d*x+1/2*c))

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maxima [B]  time = 0.44, size = 479, normalized size = 3.40 \[ -\frac {\frac {\frac {1155 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {12288 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {9065 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {10752 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {38605 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {86016 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {79135 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {53760 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {79135 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} + \frac {38605 \, \sin \left (d x + c\right )^{11}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{11}} - \frac {53760 \, \sin \left (d x + c\right )^{12}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{12}} + \frac {9065 \, \sin \left (d x + c\right )^{13}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{13}} - \frac {1155 \, \sin \left (d x + c\right )^{15}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{15}} - 1536}{a^{2} + \frac {8 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {28 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {56 \, a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {70 \, a^{2} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {56 \, a^{2} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} + \frac {28 \, a^{2} \sin \left (d x + c\right )^{12}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{12}} + \frac {8 \, a^{2} \sin \left (d x + c\right )^{14}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{14}} + \frac {a^{2} \sin \left (d x + c\right )^{16}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{16}}} - \frac {1155 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{6720 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6720*((1155*sin(d*x + c)/(cos(d*x + c) + 1) - 12288*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 9065*sin(d*x + c)
^3/(cos(d*x + c) + 1)^3 + 10752*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 38605*sin(d*x + c)^5/(cos(d*x + c) + 1)^
5 - 86016*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 79135*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 53760*sin(d*x + c)
^8/(cos(d*x + c) + 1)^8 - 79135*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 + 38605*sin(d*x + c)^11/(cos(d*x + c) + 1)
^11 - 53760*sin(d*x + c)^12/(cos(d*x + c) + 1)^12 + 9065*sin(d*x + c)^13/(cos(d*x + c) + 1)^13 - 1155*sin(d*x
+ c)^15/(cos(d*x + c) + 1)^15 - 1536)/(a^2 + 8*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 28*a^2*sin(d*x + c)^4
/(cos(d*x + c) + 1)^4 + 56*a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 70*a^2*sin(d*x + c)^8/(cos(d*x + c) + 1)^
8 + 56*a^2*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 + 28*a^2*sin(d*x + c)^12/(cos(d*x + c) + 1)^12 + 8*a^2*sin(d*
x + c)^14/(cos(d*x + c) + 1)^14 + a^2*sin(d*x + c)^16/(cos(d*x + c) + 1)^16) - 1155*arctan(sin(d*x + c)/(cos(d
*x + c) + 1))/a^2)/d

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mupad [B]  time = 11.70, size = 198, normalized size = 1.40 \[ \frac {11\,x}{128\,a^2}+\frac {\frac {11\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}}{64}-\frac {259\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{192}+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-\frac {1103\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{192}+\frac {2261\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{192}+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-\frac {2261\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{192}+\frac {64\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{5}+\frac {1103\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{192}-\frac {8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{5}+\frac {259\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{192}+\frac {64\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{35}-\frac {11\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64}+\frac {8}{35}}{a^2\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^8} \]

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]

(11*x)/(128*a^2) + ((64*tan(c/2 + (d*x)/2)^2)/35 - (11*tan(c/2 + (d*x)/2))/64 + (259*tan(c/2 + (d*x)/2)^3)/192
- (8*tan(c/2 + (d*x)/2)^4)/5 + (1103*tan(c/2 + (d*x)/2)^5)/192 + (64*tan(c/2 + (d*x)/2)^6)/5 - (2261*tan(c/2
+ (d*x)/2)^7)/192 + 8*tan(c/2 + (d*x)/2)^8 + (2261*tan(c/2 + (d*x)/2)^9)/192 - (1103*tan(c/2 + (d*x)/2)^11)/19
2 + 8*tan(c/2 + (d*x)/2)^12 - (259*tan(c/2 + (d*x)/2)^13)/192 + (11*tan(c/2 + (d*x)/2)^15)/64 + 8/35)/(a^2*d*(
tan(c/2 + (d*x)/2)^2 + 1)^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*sin(d*x+c)**2/(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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