\(\int \frac {\cos ^8(c+d x)}{(a+a \sin (c+d x))^3} \, dx\) [75]

Optimal result

Integrand size = 21, antiderivative size = 103 \[ \int \frac {\cos ^8(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {7 x}{8 a^3}+\frac {7 \cos ^5(c+d x)}{15 a^3 d}+\frac {7 \cos (c+d x) \sin (c+d x)}{8 a^3 d}+\frac {7 \cos ^3(c+d x) \sin (c+d x)}{12 a^3 d}+\frac {2 \cos ^7(c+d x)}{3 a d (a+a \sin (c+d x))^2} \]

[Out]

7/8*x/a^3+7/15*cos(d*x+c)^5/a^3/d+7/8*cos(d*x+c)*sin(d*x+c)/a^3/d+7/12*cos(d*x+c)^3*sin(d*x+c)/a^3/d+2/3*cos(d
*x+c)^7/a/d/(a+a*sin(d*x+c))^2

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2759, 2761, 2715, 8} \[ \int \frac {\cos ^8(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {7 \cos ^5(c+d x)}{15 a^3 d}+\frac {7 \sin (c+d x) \cos ^3(c+d x)}{12 a^3 d}+\frac {7 \sin (c+d x) \cos (c+d x)}{8 a^3 d}+\frac {7 x}{8 a^3}+\frac {2 \cos ^7(c+d x)}{3 a d (a \sin (c+d x)+a)^2} \]

[In]

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

[Out]

(7*x)/(8*a^3) + (7*Cos[c + d*x]^5)/(15*a^3*d) + (7*Cos[c + d*x]*Sin[c + d*x])/(8*a^3*d) + (7*Cos[c + d*x]^3*Si
n[c + d*x])/(12*a^3*d) + (2*Cos[c + d*x]^7)/(3*a*d*(a + a*Sin[c + d*x])^2)

Rule 8

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

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 2759

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

Rule 2761

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[g*((g*Cos[e
+ f*x])^(p - 1)/(b*f*(p - 1))), x] + Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2), x], x] /; FreeQ[{a, b, e, f, g
}, x] && EqQ[a^2 - b^2, 0] && GtQ[p, 1] && IntegerQ[2*p]

Rubi steps \begin{align*} \text {integral}& = \frac {2 \cos ^7(c+d x)}{3 a d (a+a \sin (c+d x))^2}+\frac {7 \int \frac {\cos ^6(c+d x)}{a+a \sin (c+d x)} \, dx}{3 a^2} \\ & = \frac {7 \cos ^5(c+d x)}{15 a^3 d}+\frac {2 \cos ^7(c+d x)}{3 a d (a+a \sin (c+d x))^2}+\frac {7 \int \cos ^4(c+d x) \, dx}{3 a^3} \\ & = \frac {7 \cos ^5(c+d x)}{15 a^3 d}+\frac {7 \cos ^3(c+d x) \sin (c+d x)}{12 a^3 d}+\frac {2 \cos ^7(c+d x)}{3 a d (a+a \sin (c+d x))^2}+\frac {7 \int \cos ^2(c+d x) \, dx}{4 a^3} \\ & = \frac {7 \cos ^5(c+d x)}{15 a^3 d}+\frac {7 \cos (c+d x) \sin (c+d x)}{8 a^3 d}+\frac {7 \cos ^3(c+d x) \sin (c+d x)}{12 a^3 d}+\frac {2 \cos ^7(c+d x)}{3 a d (a+a \sin (c+d x))^2}+\frac {7 \int 1 \, dx}{8 a^3} \\ & = \frac {7 x}{8 a^3}+\frac {7 \cos ^5(c+d x)}{15 a^3 d}+\frac {7 \cos (c+d x) \sin (c+d x)}{8 a^3 d}+\frac {7 \cos ^3(c+d x) \sin (c+d x)}{12 a^3 d}+\frac {2 \cos ^7(c+d x)}{3 a d (a+a \sin (c+d x))^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.71 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.37 \[ \int \frac {\cos ^8(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\cos ^9(c+d x) \left (-210 \arcsin \left (\frac {\sqrt {1-\sin (c+d x)}}{\sqrt {2}}\right ) \sqrt {1-\sin (c+d x)}+\sqrt {1+\sin (c+d x)} \left (136-121 \sin (c+d x)-127 \sin ^2(c+d x)+202 \sin ^3(c+d x)-114 \sin ^4(c+d x)+24 \sin ^5(c+d x)\right )\right )}{120 a^3 d (-1+\sin (c+d x))^5 (1+\sin (c+d x))^{9/2}} \]

[In]

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

[Out]

-1/120*(Cos[c + d*x]^9*(-210*ArcSin[Sqrt[1 - Sin[c + d*x]]/Sqrt[2]]*Sqrt[1 - Sin[c + d*x]] + Sqrt[1 + Sin[c +
d*x]]*(136 - 121*Sin[c + d*x] - 127*Sin[c + d*x]^2 + 202*Sin[c + d*x]^3 - 114*Sin[c + d*x]^4 + 24*Sin[c + d*x]
^5)))/(a^3*d*(-1 + Sin[c + d*x])^5*(1 + Sin[c + d*x])^(9/2))

Maple [A] (verified)

Time = 0.45 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.65

method	result	size
parallelrisch	\(\frac {420 d x +420 \cos \left (d x +c \right )-6 \cos \left (5 d x +5 c \right )-45 \sin \left (4 d x +4 c \right )+130 \cos \left (3 d x +3 c \right )+120 \sin \left (2 d x +2 c \right )+544}{480 a^{3} d}\)	\(67\)
risch	\(\frac {7 x}{8 a^{3}}+\frac {7 \cos \left (d x +c \right )}{8 a^{3} d}-\frac {\cos \left (5 d x +5 c \right )}{80 a^{3} d}-\frac {3 \sin \left (4 d x +4 c \right )}{32 a^{3} d}+\frac {13 \cos \left (3 d x +3 c \right )}{48 a^{3} d}+\frac {\sin \left (2 d x +2 c \right )}{4 a^{3} d}\)	\(90\)
derivativedivides	\(\frac {\frac {2 \left (-\frac {\left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+3 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {13 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+8 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {10 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\frac {13 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {8 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8}+\frac {17}{15}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}+\frac {7 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}}{a^{3} d}\)	\(142\)
default	\(\frac {\frac {2 \left (-\frac {\left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+3 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {13 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+8 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {10 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\frac {13 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {8 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8}+\frac {17}{15}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}+\frac {7 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}}{a^{3} d}\)	\(142\)

[In]

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

[Out]

1/480*(420*d*x+420*cos(d*x+c)-6*cos(5*d*x+5*c)-45*sin(4*d*x+4*c)+130*cos(3*d*x+3*c)+120*sin(2*d*x+2*c)+544)/a^
3/d

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.58 \[ \int \frac {\cos ^8(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {24 \, \cos \left (d x + c\right )^{5} - 160 \, \cos \left (d x + c\right )^{3} - 105 \, d x + 15 \, {\left (6 \, \cos \left (d x + c\right )^{3} - 7 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, a^{3} d} \]

[In]

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

[Out]

-1/120*(24*cos(d*x + c)^5 - 160*cos(d*x + c)^3 - 105*d*x + 15*(6*cos(d*x + c)^3 - 7*cos(d*x + c))*sin(d*x + c)
)/(a^3*d)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1717 vs. \(2 (97) = 194\).

Time = 90.80 (sec) , antiderivative size = 1717, normalized size of antiderivative = 16.67 \[ \int \frac {\cos ^8(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Piecewise((105*d*x*tan(c/2 + d*x/2)**10/(120*a**3*d*tan(c/2 + d*x/2)**10 + 600*a**3*d*tan(c/2 + d*x/2)**8 + 12
00*a**3*d*tan(c/2 + d*x/2)**6 + 1200*a**3*d*tan(c/2 + d*x/2)**4 + 600*a**3*d*tan(c/2 + d*x/2)**2 + 120*a**3*d)
+ 525*d*x*tan(c/2 + d*x/2)**8/(120*a**3*d*tan(c/2 + d*x/2)**10 + 600*a**3*d*tan(c/2 + d*x/2)**8 + 1200*a**3*d
*tan(c/2 + d*x/2)**6 + 1200*a**3*d*tan(c/2 + d*x/2)**4 + 600*a**3*d*tan(c/2 + d*x/2)**2 + 120*a**3*d) + 1050*d
*x*tan(c/2 + d*x/2)**6/(120*a**3*d*tan(c/2 + d*x/2)**10 + 600*a**3*d*tan(c/2 + d*x/2)**8 + 1200*a**3*d*tan(c/2
+ d*x/2)**6 + 1200*a**3*d*tan(c/2 + d*x/2)**4 + 600*a**3*d*tan(c/2 + d*x/2)**2 + 120*a**3*d) + 1050*d*x*tan(c
/2 + d*x/2)**4/(120*a**3*d*tan(c/2 + d*x/2)**10 + 600*a**3*d*tan(c/2 + d*x/2)**8 + 1200*a**3*d*tan(c/2 + d*x/2
)**6 + 1200*a**3*d*tan(c/2 + d*x/2)**4 + 600*a**3*d*tan(c/2 + d*x/2)**2 + 120*a**3*d) + 525*d*x*tan(c/2 + d*x/
2)**2/(120*a**3*d*tan(c/2 + d*x/2)**10 + 600*a**3*d*tan(c/2 + d*x/2)**8 + 1200*a**3*d*tan(c/2 + d*x/2)**6 + 12
00*a**3*d*tan(c/2 + d*x/2)**4 + 600*a**3*d*tan(c/2 + d*x/2)**2 + 120*a**3*d) + 105*d*x/(120*a**3*d*tan(c/2 + d
*x/2)**10 + 600*a**3*d*tan(c/2 + d*x/2)**8 + 1200*a**3*d*tan(c/2 + d*x/2)**6 + 1200*a**3*d*tan(c/2 + d*x/2)**4
+ 600*a**3*d*tan(c/2 + d*x/2)**2 + 120*a**3*d) - 30*tan(c/2 + d*x/2)**9/(120*a**3*d*tan(c/2 + d*x/2)**10 + 60
0*a**3*d*tan(c/2 + d*x/2)**8 + 1200*a**3*d*tan(c/2 + d*x/2)**6 + 1200*a**3*d*tan(c/2 + d*x/2)**4 + 600*a**3*d*
tan(c/2 + d*x/2)**2 + 120*a**3*d) + 720*tan(c/2 + d*x/2)**8/(120*a**3*d*tan(c/2 + d*x/2)**10 + 600*a**3*d*tan(
c/2 + d*x/2)**8 + 1200*a**3*d*tan(c/2 + d*x/2)**6 + 1200*a**3*d*tan(c/2 + d*x/2)**4 + 600*a**3*d*tan(c/2 + d*x
/2)**2 + 120*a**3*d) - 780*tan(c/2 + d*x/2)**7/(120*a**3*d*tan(c/2 + d*x/2)**10 + 600*a**3*d*tan(c/2 + d*x/2)*
*8 + 1200*a**3*d*tan(c/2 + d*x/2)**6 + 1200*a**3*d*tan(c/2 + d*x/2)**4 + 600*a**3*d*tan(c/2 + d*x/2)**2 + 120*
a**3*d) + 1920*tan(c/2 + d*x/2)**6/(120*a**3*d*tan(c/2 + d*x/2)**10 + 600*a**3*d*tan(c/2 + d*x/2)**8 + 1200*a*
*3*d*tan(c/2 + d*x/2)**6 + 1200*a**3*d*tan(c/2 + d*x/2)**4 + 600*a**3*d*tan(c/2 + d*x/2)**2 + 120*a**3*d) + 80
0*tan(c/2 + d*x/2)**4/(120*a**3*d*tan(c/2 + d*x/2)**10 + 600*a**3*d*tan(c/2 + d*x/2)**8 + 1200*a**3*d*tan(c/2
+ d*x/2)**6 + 1200*a**3*d*tan(c/2 + d*x/2)**4 + 600*a**3*d*tan(c/2 + d*x/2)**2 + 120*a**3*d) + 780*tan(c/2 + d
*x/2)**3/(120*a**3*d*tan(c/2 + d*x/2)**10 + 600*a**3*d*tan(c/2 + d*x/2)**8 + 1200*a**3*d*tan(c/2 + d*x/2)**6 +
1200*a**3*d*tan(c/2 + d*x/2)**4 + 600*a**3*d*tan(c/2 + d*x/2)**2 + 120*a**3*d) + 640*tan(c/2 + d*x/2)**2/(120
*a**3*d*tan(c/2 + d*x/2)**10 + 600*a**3*d*tan(c/2 + d*x/2)**8 + 1200*a**3*d*tan(c/2 + d*x/2)**6 + 1200*a**3*d*
tan(c/2 + d*x/2)**4 + 600*a**3*d*tan(c/2 + d*x/2)**2 + 120*a**3*d) + 30*tan(c/2 + d*x/2)/(120*a**3*d*tan(c/2 +
d*x/2)**10 + 600*a**3*d*tan(c/2 + d*x/2)**8 + 1200*a**3*d*tan(c/2 + d*x/2)**6 + 1200*a**3*d*tan(c/2 + d*x/2)*
*4 + 600*a**3*d*tan(c/2 + d*x/2)**2 + 120*a**3*d) + 272/(120*a**3*d*tan(c/2 + d*x/2)**10 + 600*a**3*d*tan(c/2
+ d*x/2)**8 + 1200*a**3*d*tan(c/2 + d*x/2)**6 + 1200*a**3*d*tan(c/2 + d*x/2)**4 + 600*a**3*d*tan(c/2 + d*x/2)*
*2 + 120*a**3*d), Ne(d, 0)), (x*cos(c)**8/(a*sin(c) + a)**3, True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 310 vs. \(2 (93) = 186\).

Time = 0.28 (sec) , antiderivative size = 310, normalized size of antiderivative = 3.01 \[ \int \frac {\cos ^8(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {320 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {390 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {400 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {960 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {390 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {360 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {15 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} + 136}{a^{3} + \frac {5 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {10 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {10 \, a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {5 \, a^{3} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {a^{3} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}}} + \frac {105 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}}{60 \, d} \]

[In]

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

[Out]

1/60*((15*sin(d*x + c)/(cos(d*x + c) + 1) + 320*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 390*sin(d*x + c)^3/(cos(
d*x + c) + 1)^3 + 400*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 960*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 390*sin(
d*x + c)^7/(cos(d*x + c) + 1)^7 + 360*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 15*sin(d*x + c)^9/(cos(d*x + c) +
1)^9 + 136)/(a^3 + 5*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 10*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 10
*a^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 5*a^3*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + a^3*sin(d*x + c)^10/(co
s(d*x + c) + 1)^10) + 105*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3)/d

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.36 \[ \int \frac {\cos ^8(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {105 \, {\left (d x + c\right )}}{a^{3}} - \frac {2 \, {\left (15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 360 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 390 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 960 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 400 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 390 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 320 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 136\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5} a^{3}}}{120 \, d} \]

[In]

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

[Out]

1/120*(105*(d*x + c)/a^3 - 2*(15*tan(1/2*d*x + 1/2*c)^9 - 360*tan(1/2*d*x + 1/2*c)^8 + 390*tan(1/2*d*x + 1/2*c
)^7 - 960*tan(1/2*d*x + 1/2*c)^6 - 400*tan(1/2*d*x + 1/2*c)^4 - 390*tan(1/2*d*x + 1/2*c)^3 - 320*tan(1/2*d*x +
1/2*c)^2 - 15*tan(1/2*d*x + 1/2*c) - 136)/((tan(1/2*d*x + 1/2*c)^2 + 1)^5*a^3))/d

Mupad [B] (verification not implemented)

Time = 6.05 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.79 \[ \int \frac {\cos ^8(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {7\,x}{8\,a^3}+\frac {4\,{\cos \left (c+d\,x\right )}^3}{3\,a^3\,d}-\frac {{\cos \left (c+d\,x\right )}^5}{5\,a^3\,d}-\frac {3\,{\cos \left (c+d\,x\right )}^3\,\sin \left (c+d\,x\right )}{4\,a^3\,d}+\frac {7\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{8\,a^3\,d} \]

[In]

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

[Out]

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