3.724 \(\int \frac {\cos ^8(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=159 \[ -\frac {\cos ^9(c+d x)}{9 a^2 d}+\frac {3 \cos ^7(c+d x)}{7 a^2 d}-\frac {2 \cos ^5(c+d x)}{5 a^2 d}+\frac {\sin ^3(c+d x) \cos ^5(c+d x)}{4 a^2 d}+\frac {\sin (c+d x) \cos ^5(c+d x)}{8 a^2 d}-\frac {\sin (c+d x) \cos ^3(c+d x)}{32 a^2 d}-\frac {3 \sin (c+d x) \cos (c+d x)}{64 a^2 d}-\frac {3 x}{64 a^2} \]

[Out]

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

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Rubi [A]  time = 0.36, antiderivative size = 159, 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, 2565, 14, 2568, 2635, 8, 270} \[ -\frac {\cos ^9(c+d x)}{9 a^2 d}+\frac {3 \cos ^7(c+d x)}{7 a^2 d}-\frac {2 \cos ^5(c+d x)}{5 a^2 d}+\frac {\sin ^3(c+d x) \cos ^5(c+d x)}{4 a^2 d}+\frac {\sin (c+d x) \cos ^5(c+d x)}{8 a^2 d}-\frac {\sin (c+d x) \cos ^3(c+d x)}{32 a^2 d}-\frac {3 \sin (c+d x) \cos (c+d x)}{64 a^2 d}-\frac {3 x}{64 a^2} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^8*Sin[c + d*x]^3)/(a + a*Sin[c + d*x])^2,x]

[Out]

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

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Mathematica [B]  time = 6.91, size = 430, normalized size = 2.70 \[ -\frac {15120 d x \sin \left (\frac {c}{2}\right )-11340 \sin \left (\frac {c}{2}+d x\right )+11340 \sin \left (\frac {3 c}{2}+d x\right )-3360 \sin \left (\frac {5 c}{2}+3 d x\right )+3360 \sin \left (\frac {7 c}{2}+3 d x\right )-2520 \sin \left (\frac {7 c}{2}+4 d x\right )-2520 \sin \left (\frac {9 c}{2}+4 d x\right )+1008 \sin \left (\frac {9 c}{2}+5 d x\right )-1008 \sin \left (\frac {11 c}{2}+5 d x\right )+450 \sin \left (\frac {13 c}{2}+7 d x\right )-450 \sin \left (\frac {15 c}{2}+7 d x\right )+315 \sin \left (\frac {15 c}{2}+8 d x\right )+315 \sin \left (\frac {17 c}{2}+8 d x\right )-70 \sin \left (\frac {17 c}{2}+9 d x\right )+70 \sin \left (\frac {19 c}{2}+9 d x\right )+420 \cos \left (\frac {c}{2}\right ) (330 c+36 d x+7)+11340 \cos \left (\frac {c}{2}+d x\right )+11340 \cos \left (\frac {3 c}{2}+d x\right )+3360 \cos \left (\frac {5 c}{2}+3 d x\right )+3360 \cos \left (\frac {7 c}{2}+3 d x\right )-2520 \cos \left (\frac {7 c}{2}+4 d x\right )+2520 \cos \left (\frac {9 c}{2}+4 d x\right )-1008 \cos \left (\frac {9 c}{2}+5 d x\right )-1008 \cos \left (\frac {11 c}{2}+5 d x\right )-450 \cos \left (\frac {13 c}{2}+7 d x\right )-450 \cos \left (\frac {15 c}{2}+7 d x\right )+315 \cos \left (\frac {15 c}{2}+8 d x\right )-315 \cos \left (\frac {17 c}{2}+8 d x\right )+70 \cos \left (\frac {17 c}{2}+9 d x\right )+70 \cos \left (\frac {19 c}{2}+9 d x\right )+138600 c \sin \left (\frac {c}{2}\right )-78960 \sin \left (\frac {c}{2}\right )}{322560 a^2 d \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^8*Sin[c + d*x]^3)/(a + a*Sin[c + d*x])^2,x]

[Out]

-1/322560*(420*(7 + 330*c + 36*d*x)*Cos[c/2] + 11340*Cos[c/2 + d*x] + 11340*Cos[(3*c)/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] - 1008*Cos[(9
*c)/2 + 5*d*x] - 1008*Cos[(11*c)/2 + 5*d*x] - 450*Cos[(13*c)/2 + 7*d*x] - 450*Cos[(15*c)/2 + 7*d*x] + 315*Cos[
(15*c)/2 + 8*d*x] - 315*Cos[(17*c)/2 + 8*d*x] + 70*Cos[(17*c)/2 + 9*d*x] + 70*Cos[(19*c)/2 + 9*d*x] - 78960*Si
n[c/2] + 138600*c*Sin[c/2] + 15120*d*x*Sin[c/2] - 11340*Sin[c/2 + d*x] + 11340*Sin[(3*c)/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] + 1008*Sin
[(9*c)/2 + 5*d*x] - 1008*Sin[(11*c)/2 + 5*d*x] + 450*Sin[(13*c)/2 + 7*d*x] - 450*Sin[(15*c)/2 + 7*d*x] + 315*S
in[(15*c)/2 + 8*d*x] + 315*Sin[(17*c)/2 + 8*d*x] - 70*Sin[(17*c)/2 + 9*d*x] + 70*Sin[(19*c)/2 + 9*d*x])/(a^2*d
*(Cos[c/2] + Sin[c/2]))

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fricas [A]  time = 0.47, size = 90, normalized size = 0.57 \[ -\frac {2240 \, \cos \left (d x + c\right )^{9} - 8640 \, \cos \left (d x + c\right )^{7} + 8064 \, \cos \left (d x + c\right )^{5} + 945 \, d x + 315 \, {\left (16 \, \cos \left (d x + c\right )^{7} - 24 \, \cos \left (d x + c\right )^{5} + 2 \, \cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{20160 \, a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8*sin(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/20160*(2240*cos(d*x + c)^9 - 8640*cos(d*x + c)^7 + 8064*cos(d*x + c)^5 + 945*d*x + 315*(16*cos(d*x + c)^7 -
24*cos(d*x + c)^5 + 2*cos(d*x + c)^3 + 3*cos(d*x + c))*sin(d*x + c))/(a^2*d)

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giac [A]  time = 0.29, size = 231, normalized size = 1.45 \[ -\frac {\frac {945 \, {\left (d x + c\right )}}{a^{2}} + \frac {2 \, {\left (945 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{17} + 8190 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{15} + 40320 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{14} - 97650 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} + 147840 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{12} + 106470 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 120960 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{10} + 330624 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 106470 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 8064 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 97650 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 19584 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 8190 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 14976 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 945 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1664\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{9} a^{2}}}{20160 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8*sin(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/20160*(945*(d*x + c)/a^2 + 2*(945*tan(1/2*d*x + 1/2*c)^17 + 8190*tan(1/2*d*x + 1/2*c)^15 + 40320*tan(1/2*d*
x + 1/2*c)^14 - 97650*tan(1/2*d*x + 1/2*c)^13 + 147840*tan(1/2*d*x + 1/2*c)^12 + 106470*tan(1/2*d*x + 1/2*c)^1
1 - 120960*tan(1/2*d*x + 1/2*c)^10 + 330624*tan(1/2*d*x + 1/2*c)^8 - 106470*tan(1/2*d*x + 1/2*c)^7 - 8064*tan(
1/2*d*x + 1/2*c)^6 + 97650*tan(1/2*d*x + 1/2*c)^5 + 19584*tan(1/2*d*x + 1/2*c)^4 - 8190*tan(1/2*d*x + 1/2*c)^3
+ 14976*tan(1/2*d*x + 1/2*c)^2 - 945*tan(1/2*d*x + 1/2*c) + 1664)/((tan(1/2*d*x + 1/2*c)^2 + 1)^9*a^2))/d

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maple [B]  time = 0.44, size = 551, normalized size = 3.47 \[ -\frac {52}{315 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{9}}+\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{32 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{9}}-\frac {52 \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 )^{9}}+\frac {13 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{9}}-\frac {68 \left (\tan ^{4}\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 )^{9}}-\frac {155 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{9}}+\frac {4 \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 )^{9}}+\frac {169 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{9}}-\frac {164 \left (\tan ^{8}\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 )^{9}}+\frac {12 \left (\tan ^{10}\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 )^{9}}-\frac {169 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{9}}-\frac {44 \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{9}}+\frac {155 \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{9}}-\frac {4 \left (\tan ^{14}\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 )^{9}}-\frac {13 \left (\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{9}}-\frac {3 \left (\tan ^{17}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{9}}-\frac {3 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^8*sin(d*x+c)^3/(a+a*sin(d*x+c))^2,x)

[Out]

-52/315/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^9+3/32/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^9*tan(1/2*d*x+1/2*c)-52/35/d/a^2/
(1+tan(1/2*d*x+1/2*c)^2)^9*tan(1/2*d*x+1/2*c)^2+13/16/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^9*tan(1/2*d*x+1/2*c)^3-68
/35/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^9*tan(1/2*d*x+1/2*c)^4-155/16/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^9*tan(1/2*d*x+
1/2*c)^5+4/5/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^9*tan(1/2*d*x+1/2*c)^6+169/16/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^9*tan
(1/2*d*x+1/2*c)^7-164/5/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^9*tan(1/2*d*x+1/2*c)^8+12/d/a^2/(1+tan(1/2*d*x+1/2*c)^2
)^9*tan(1/2*d*x+1/2*c)^10-169/16/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^9*tan(1/2*d*x+1/2*c)^11-44/3/d/a^2/(1+tan(1/2*
d*x+1/2*c)^2)^9*tan(1/2*d*x+1/2*c)^12+155/16/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^9*tan(1/2*d*x+1/2*c)^13-4/d/a^2/(1
+tan(1/2*d*x+1/2*c)^2)^9*tan(1/2*d*x+1/2*c)^14-13/16/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^9*tan(1/2*d*x+1/2*c)^15-3/
32/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^9*tan(1/2*d*x+1/2*c)^17-3/32/d/a^2*arctan(tan(1/2*d*x+1/2*c))

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maxima [B]  time = 0.43, size = 542, normalized size = 3.41 \[ \frac {\frac {\frac {945 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {14976 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {8190 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {19584 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {97650 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {8064 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {106470 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {330624 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {120960 \, \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} - \frac {106470 \, \sin \left (d x + c\right )^{11}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{11}} - \frac {147840 \, \sin \left (d x + c\right )^{12}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{12}} + \frac {97650 \, \sin \left (d x + c\right )^{13}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{13}} - \frac {40320 \, \sin \left (d x + c\right )^{14}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{14}} - \frac {8190 \, \sin \left (d x + c\right )^{15}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{15}} - \frac {945 \, \sin \left (d x + c\right )^{17}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{17}} - 1664}{a^{2} + \frac {9 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {36 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {84 \, a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {126 \, a^{2} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {126 \, a^{2} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} + \frac {84 \, a^{2} \sin \left (d x + c\right )^{12}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{12}} + \frac {36 \, a^{2} \sin \left (d x + c\right )^{14}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{14}} + \frac {9 \, a^{2} \sin \left (d x + c\right )^{16}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{16}} + \frac {a^{2} \sin \left (d x + c\right )^{18}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{18}}} - \frac {945 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{10080 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8*sin(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/10080*((945*sin(d*x + c)/(cos(d*x + c) + 1) - 14976*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 8190*sin(d*x + c)^
3/(cos(d*x + c) + 1)^3 - 19584*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 97650*sin(d*x + c)^5/(cos(d*x + c) + 1)^5
+ 8064*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 106470*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 330624*sin(d*x + c)
^8/(cos(d*x + c) + 1)^8 + 120960*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 - 106470*sin(d*x + c)^11/(cos(d*x + c)
+ 1)^11 - 147840*sin(d*x + c)^12/(cos(d*x + c) + 1)^12 + 97650*sin(d*x + c)^13/(cos(d*x + c) + 1)^13 - 40320*s
in(d*x + c)^14/(cos(d*x + c) + 1)^14 - 8190*sin(d*x + c)^15/(cos(d*x + c) + 1)^15 - 945*sin(d*x + c)^17/(cos(d
*x + c) + 1)^17 - 1664)/(a^2 + 9*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 36*a^2*sin(d*x + c)^4/(cos(d*x + c)
+ 1)^4 + 84*a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 126*a^2*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + 126*a^2*s
in(d*x + c)^10/(cos(d*x + c) + 1)^10 + 84*a^2*sin(d*x + c)^12/(cos(d*x + c) + 1)^12 + 36*a^2*sin(d*x + c)^14/(
cos(d*x + c) + 1)^14 + 9*a^2*sin(d*x + c)^16/(cos(d*x + c) + 1)^16 + a^2*sin(d*x + c)^18/(cos(d*x + c) + 1)^18
) - 945*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2)/d

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mupad [B]  time = 11.68, size = 225, normalized size = 1.42 \[ -\frac {3\,x}{64\,a^2}-\frac {\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{17}}{32}+\frac {13\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}}{16}+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}-\frac {155\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{16}+\frac {44\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}}{3}+\frac {169\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{16}-12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+\frac {164\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{5}-\frac {169\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{16}-\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{5}+\frac {155\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{16}+\frac {68\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{35}-\frac {13\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{16}+\frac {52\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{35}-\frac {3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{32}+\frac {52}{315}}{a^2\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^9} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^8*sin(c + d*x)^3)/(a + a*sin(c + d*x))^2,x)

[Out]

- (3*x)/(64*a^2) - ((52*tan(c/2 + (d*x)/2)^2)/35 - (3*tan(c/2 + (d*x)/2))/32 - (13*tan(c/2 + (d*x)/2)^3)/16 +
(68*tan(c/2 + (d*x)/2)^4)/35 + (155*tan(c/2 + (d*x)/2)^5)/16 - (4*tan(c/2 + (d*x)/2)^6)/5 - (169*tan(c/2 + (d*
x)/2)^7)/16 + (164*tan(c/2 + (d*x)/2)^8)/5 - 12*tan(c/2 + (d*x)/2)^10 + (169*tan(c/2 + (d*x)/2)^11)/16 + (44*t
an(c/2 + (d*x)/2)^12)/3 - (155*tan(c/2 + (d*x)/2)^13)/16 + 4*tan(c/2 + (d*x)/2)^14 + (13*tan(c/2 + (d*x)/2)^15
)/16 + (3*tan(c/2 + (d*x)/2)^17)/32 + 52/315)/(a^2*d*(tan(c/2 + (d*x)/2)^2 + 1)^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*sin(d*x+c)**3/(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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