\(\int \frac {\cos ^8(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx\) [740]

Optimal result

Integrand size = 29, antiderivative size = 133 \[ \int \frac {\cos ^8(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {5 x}{16 a^3}+\frac {4 \cos ^3(c+d x)}{3 a^3 d}-\frac {\cos ^5(c+d x)}{a^3 d}+\frac {\cos ^7(c+d x)}{7 a^3 d}+\frac {5 \cos (c+d x) \sin (c+d x)}{16 a^3 d}-\frac {5 \cos ^3(c+d x) \sin (c+d x)}{8 a^3 d}-\frac {\cos ^3(c+d x) \sin ^3(c+d x)}{2 a^3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {2954, 2952, 2648, 2715, 8, 2645, 14, 276} \[ \int \frac {\cos ^8(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\cos ^7(c+d x)}{7 a^3 d}-\frac {\cos ^5(c+d x)}{a^3 d}+\frac {4 \cos ^3(c+d x)}{3 a^3 d}-\frac {\sin ^3(c+d x) \cos ^3(c+d x)}{2 a^3 d}-\frac {5 \sin (c+d x) \cos ^3(c+d x)}{8 a^3 d}+\frac {5 \sin (c+d x) \cos (c+d x)}{16 a^3 d}+\frac {5 x}{16 a^3} \]

[In]

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

[Out]

(5*x)/(16*a^3) + (4*Cos[c + d*x]^3)/(3*a^3*d) - Cos[c + d*x]^5/(a^3*d) + Cos[c + d*x]^7/(7*a^3*d) + (5*Cos[c +
d*x]*Sin[c + d*x])/(16*a^3*d) - (5*Cos[c + d*x]^3*Sin[c + d*x])/(8*a^3*d) - (Cos[c + d*x]^3*Sin[c + d*x]^3)/(
2*a^3*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 276

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 2645

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 2648

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 2715

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] && Integ
erQ[2*n]

Rule 2952

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 2954

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(429\) vs. \(2(133)=266\).

Time = 8.12 (sec) , antiderivative size = 429, normalized size of antiderivative = 3.23 \[ \int \frac {\cos ^8(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {-168 (99 c-5 d x) \cos \left (\frac {c}{2}\right )+609 \cos \left (\frac {c}{2}+d x\right )+609 \cos \left (\frac {3 c}{2}+d x\right )-63 \cos \left (\frac {3 c}{2}+2 d x\right )+63 \cos \left (\frac {5 c}{2}+2 d x\right )+91 \cos \left (\frac {5 c}{2}+3 d x\right )+91 \cos \left (\frac {7 c}{2}+3 d x\right )-105 \cos \left (\frac {7 c}{2}+4 d x\right )+105 \cos \left (\frac {9 c}{2}+4 d x\right )-63 \cos \left (\frac {9 c}{2}+5 d x\right )-63 \cos \left (\frac {11 c}{2}+5 d x\right )+21 \cos \left (\frac {11 c}{2}+6 d x\right )-21 \cos \left (\frac {13 c}{2}+6 d x\right )+3 \cos \left (\frac {13 c}{2}+7 d x\right )+3 \cos \left (\frac {15 c}{2}+7 d x\right )+16996 \sin \left (\frac {c}{2}\right )-16632 c \sin \left (\frac {c}{2}\right )+840 d x \sin \left (\frac {c}{2}\right )-609 \sin \left (\frac {c}{2}+d x\right )+609 \sin \left (\frac {3 c}{2}+d x\right )-63 \sin \left (\frac {3 c}{2}+2 d x\right )-63 \sin \left (\frac {5 c}{2}+2 d x\right )-91 \sin \left (\frac {5 c}{2}+3 d x\right )+91 \sin \left (\frac {7 c}{2}+3 d x\right )-105 \sin \left (\frac {7 c}{2}+4 d x\right )-105 \sin \left (\frac {9 c}{2}+4 d x\right )+63 \sin \left (\frac {9 c}{2}+5 d x\right )-63 \sin \left (\frac {11 c}{2}+5 d x\right )+21 \sin \left (\frac {11 c}{2}+6 d x\right )+21 \sin \left (\frac {13 c}{2}+6 d x\right )-3 \sin \left (\frac {13 c}{2}+7 d x\right )+3 \sin \left (\frac {15 c}{2}+7 d x\right )}{2688 a^3 d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right )} \]

[In]

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

[Out]

(-168*(99*c - 5*d*x)*Cos[c/2] + 609*Cos[c/2 + d*x] + 609*Cos[(3*c)/2 + d*x] - 63*Cos[(3*c)/2 + 2*d*x] + 63*Cos
[(5*c)/2 + 2*d*x] + 91*Cos[(5*c)/2 + 3*d*x] + 91*Cos[(7*c)/2 + 3*d*x] - 105*Cos[(7*c)/2 + 4*d*x] + 105*Cos[(9*
c)/2 + 4*d*x] - 63*Cos[(9*c)/2 + 5*d*x] - 63*Cos[(11*c)/2 + 5*d*x] + 21*Cos[(11*c)/2 + 6*d*x] - 21*Cos[(13*c)/
2 + 6*d*x] + 3*Cos[(13*c)/2 + 7*d*x] + 3*Cos[(15*c)/2 + 7*d*x] + 16996*Sin[c/2] - 16632*c*Sin[c/2] + 840*d*x*S
in[c/2] - 609*Sin[c/2 + d*x] + 609*Sin[(3*c)/2 + d*x] - 63*Sin[(3*c)/2 + 2*d*x] - 63*Sin[(5*c)/2 + 2*d*x] - 91
*Sin[(5*c)/2 + 3*d*x] + 91*Sin[(7*c)/2 + 3*d*x] - 105*Sin[(7*c)/2 + 4*d*x] - 105*Sin[(9*c)/2 + 4*d*x] + 63*Sin
[(9*c)/2 + 5*d*x] - 63*Sin[(11*c)/2 + 5*d*x] + 21*Sin[(11*c)/2 + 6*d*x] + 21*Sin[(13*c)/2 + 6*d*x] - 3*Sin[(13
*c)/2 + 7*d*x] + 3*Sin[(15*c)/2 + 7*d*x])/(2688*a^3*d*(Cos[c/2] + Sin[c/2]))

Maple [A] (verified)

Time = 0.31 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.67

method	result	size
parallelrisch	\(\frac {420 d x +3 \cos \left (7 d x +7 c \right )-63 \cos \left (5 d x +5 c \right )+91 \cos \left (3 d x +3 c \right )+609 \cos \left (d x +c \right )+21 \sin \left (6 d x +6 c \right )-105 \sin \left (4 d x +4 c \right )-63 \sin \left (2 d x +2 c \right )+640}{1344 d \,a^{3}}\)	\(89\)
risch	\(\frac {5 x}{16 a^{3}}+\frac {29 \cos \left (d x +c \right )}{64 a^{3} d}+\frac {\cos \left (7 d x +7 c \right )}{448 d \,a^{3}}+\frac {\sin \left (6 d x +6 c \right )}{64 d \,a^{3}}-\frac {3 \cos \left (5 d x +5 c \right )}{64 d \,a^{3}}-\frac {5 \sin \left (4 d x +4 c \right )}{64 d \,a^{3}}+\frac {13 \cos \left (3 d x +3 c \right )}{192 d \,a^{3}}-\frac {3 \sin \left (2 d x +2 c \right )}{64 d \,a^{3}}\)	\(124\)
derivativedivides	\(\frac {\frac {8 \left (\frac {5 \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64}+\frac {3 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16}+\frac {3 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}-\frac {119 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64}+\frac {23 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6}+\frac {\left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\frac {119 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64}+\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {3 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16}+\frac {5 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6}-\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{64}+\frac {5}{42}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{7}}+\frac {5 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}}{d \,a^{3}}\)	\(179\)
default	\(\frac {\frac {8 \left (\frac {5 \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64}+\frac {3 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16}+\frac {3 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}-\frac {119 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64}+\frac {23 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6}+\frac {\left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\frac {119 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64}+\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {3 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16}+\frac {5 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6}-\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{64}+\frac {5}{42}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{7}}+\frac {5 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}}{d \,a^{3}}\)	\(179\)

[In]

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

[Out]

1/1344*(420*d*x+3*cos(7*d*x+7*c)-63*cos(5*d*x+5*c)+91*cos(3*d*x+3*c)+609*cos(d*x+c)+21*sin(6*d*x+6*c)-105*sin(
4*d*x+4*c)-63*sin(2*d*x+2*c)+640)/d/a^3

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.60 \[ \int \frac {\cos ^8(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {48 \, \cos \left (d x + c\right )^{7} - 336 \, \cos \left (d x + c\right )^{5} + 448 \, \cos \left (d x + c\right )^{3} + 105 \, d x + 21 \, {\left (8 \, \cos \left (d x + c\right )^{5} - 18 \, \cos \left (d x + c\right )^{3} + 5 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{336 \, a^{3} d} \]

[In]

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

[Out]

1/336*(48*cos(d*x + c)^7 - 336*cos(d*x + c)^5 + 448*cos(d*x + c)^3 + 105*d*x + 21*(8*cos(d*x + c)^5 - 18*cos(d
*x + c)^3 + 5*cos(d*x + c))*sin(d*x + c))/(a^3*d)

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^8(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 416 vs. \(2 (121) = 242\).

Time = 0.34 (sec) , antiderivative size = 416, normalized size of antiderivative = 3.13 \[ \int \frac {\cos ^8(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {1120 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {252 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {1344 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {2499 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {448 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {5152 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {2499 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac {2016 \, \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} - \frac {252 \, \sin \left (d x + c\right )^{11}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{11}} - \frac {105 \, \sin \left (d x + c\right )^{13}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{13}} - 160}{a^{3} + \frac {7 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {21 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {35 \, a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {35 \, a^{3} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {21 \, a^{3} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} + \frac {7 \, a^{3} \sin \left (d x + c\right )^{12}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{12}} + \frac {a^{3} \sin \left (d x + c\right )^{14}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{14}}} - \frac {105 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}}{168 \, d} \]

[In]

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

[Out]

-1/168*((105*sin(d*x + c)/(cos(d*x + c) + 1) - 1120*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 252*sin(d*x + c)^3/(
cos(d*x + c) + 1)^3 - 1344*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 2499*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 44
8*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 5152*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + 2499*sin(d*x + c)^9/(cos(d*
x + c) + 1)^9 - 2016*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 - 252*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 - 105*s
in(d*x + c)^13/(cos(d*x + c) + 1)^13 - 160)/(a^3 + 7*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 21*a^3*sin(d*x
+ c)^4/(cos(d*x + c) + 1)^4 + 35*a^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 35*a^3*sin(d*x + c)^8/(cos(d*x + c)
+ 1)^8 + 21*a^3*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 + 7*a^3*sin(d*x + c)^12/(cos(d*x + c) + 1)^12 + a^3*sin
(d*x + c)^14/(cos(d*x + c) + 1)^14) - 105*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3)/d

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.35 \[ \int \frac {\cos ^8(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {105 \, {\left (d x + c\right )}}{a^{3}} + \frac {2 \, {\left (105 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} + 252 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 2016 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{10} - 2499 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 5152 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 448 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 2499 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1344 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 252 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1120 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 105 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 160\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{7} a^{3}}}{336 \, d} \]

[In]

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

[Out]

1/336*(105*(d*x + c)/a^3 + 2*(105*tan(1/2*d*x + 1/2*c)^13 + 252*tan(1/2*d*x + 1/2*c)^11 + 2016*tan(1/2*d*x + 1
/2*c)^10 - 2499*tan(1/2*d*x + 1/2*c)^9 + 5152*tan(1/2*d*x + 1/2*c)^8 + 448*tan(1/2*d*x + 1/2*c)^6 + 2499*tan(1
/2*d*x + 1/2*c)^5 + 1344*tan(1/2*d*x + 1/2*c)^4 - 252*tan(1/2*d*x + 1/2*c)^3 + 1120*tan(1/2*d*x + 1/2*c)^2 - 1
05*tan(1/2*d*x + 1/2*c) + 160)/((tan(1/2*d*x + 1/2*c)^2 + 1)^7*a^3))/d

Mupad [B] (verification not implemented)

Time = 13.45 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.29 \[ \int \frac {\cos ^8(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {5\,x}{16\,a^3}+\frac {\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{8}+\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{2}+12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-\frac {119\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{8}+\frac {92\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{3}+\frac {8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{3}+\frac {119\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{8}+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2}+\frac {20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}-\frac {5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}+\frac {20}{21}}{a^3\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^7} \]

[In]

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

[Out]

(5*x)/(16*a^3) + ((20*tan(c/2 + (d*x)/2)^2)/3 - (5*tan(c/2 + (d*x)/2))/8 - (3*tan(c/2 + (d*x)/2)^3)/2 + 8*tan(
c/2 + (d*x)/2)^4 + (119*tan(c/2 + (d*x)/2)^5)/8 + (8*tan(c/2 + (d*x)/2)^6)/3 + (92*tan(c/2 + (d*x)/2)^8)/3 - (
119*tan(c/2 + (d*x)/2)^9)/8 + 12*tan(c/2 + (d*x)/2)^10 + (3*tan(c/2 + (d*x)/2)^11)/2 + (5*tan(c/2 + (d*x)/2)^1
3)/8 + 20/21)/(a^3*d*(tan(c/2 + (d*x)/2)^2 + 1)^7)