\(\int \frac {\sin ^8(c+d x)}{(a+a \sec (c+d x))^3} \, dx\) [100]

Optimal result

Integrand size = 21, antiderivative size = 157 \[ \int \frac {\sin ^8(c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {29 x}{128 a^3}-\frac {29 \cos (c+d x) \sin (c+d x)}{128 a^3 d}-\frac {29 \cos ^3(c+d x) \sin (c+d x)}{192 a^3 d}+\frac {23 \cos ^5(c+d x) \sin (c+d x)}{48 a^3 d}+\frac {\cos ^7(c+d x) \sin (c+d x)}{8 a^3 d}+\frac {4 \sin ^3(c+d x)}{3 a^3 d}-\frac {7 \sin ^5(c+d x)}{5 a^3 d}+\frac {3 \sin ^7(c+d x)}{7 a^3 d} \]

[Out]

-29/128*x/a^3-29/128*cos(d*x+c)*sin(d*x+c)/a^3/d-29/192*cos(d*x+c)^3*sin(d*x+c)/a^3/d+23/48*cos(d*x+c)^5*sin(d
*x+c)/a^3/d+1/8*cos(d*x+c)^7*sin(d*x+c)/a^3/d+4/3*sin(d*x+c)^3/a^3/d-7/5*sin(d*x+c)^5/a^3/d+3/7*sin(d*x+c)^7/a
^3/d

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3957, 2954, 2952, 2644, 14, 2648, 2715, 8, 276} \[ \int \frac {\sin ^8(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {3 \sin ^7(c+d x)}{7 a^3 d}-\frac {7 \sin ^5(c+d x)}{5 a^3 d}+\frac {4 \sin ^3(c+d x)}{3 a^3 d}+\frac {\sin (c+d x) \cos ^7(c+d x)}{8 a^3 d}+\frac {23 \sin (c+d x) \cos ^5(c+d x)}{48 a^3 d}-\frac {29 \sin (c+d x) \cos ^3(c+d x)}{192 a^3 d}-\frac {29 \sin (c+d x) \cos (c+d x)}{128 a^3 d}-\frac {29 x}{128 a^3} \]

[In]

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

[Out]

(-29*x)/(128*a^3) - (29*Cos[c + d*x]*Sin[c + d*x])/(128*a^3*d) - (29*Cos[c + d*x]^3*Sin[c + d*x])/(192*a^3*d)
+ (23*Cos[c + d*x]^5*Sin[c + d*x])/(48*a^3*d) + (Cos[c + d*x]^7*Sin[c + d*x])/(8*a^3*d) + (4*Sin[c + d*x]^3)/(
3*a^3*d) - (7*Sin[c + d*x]^5)/(5*a^3*d) + (3*Sin[c + d*x]^7)/(7*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 2644

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

Rule 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[e + f*x])^p*((b + a*Sin[e + f*x])^m/Sin[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rubi steps \begin{align*} \text {integral}& = -\int \frac {\cos ^3(c+d x) \sin ^8(c+d x)}{(-a-a \cos (c+d x))^3} \, dx \\ & = -\frac {\int \cos ^3(c+d x) (-a+a \cos (c+d x))^3 \sin ^2(c+d x) \, dx}{a^6} \\ & = -\frac {\int \left (-a^3 \cos ^3(c+d x) \sin ^2(c+d x)+3 a^3 \cos ^4(c+d x) \sin ^2(c+d x)-3 a^3 \cos ^5(c+d x) \sin ^2(c+d x)+a^3 \cos ^6(c+d x) \sin ^2(c+d x)\right ) \, dx}{a^6} \\ & = \frac {\int \cos ^3(c+d x) \sin ^2(c+d x) \, dx}{a^3}-\frac {\int \cos ^6(c+d x) \sin ^2(c+d x) \, dx}{a^3}-\frac {3 \int \cos ^4(c+d x) \sin ^2(c+d x) \, dx}{a^3}+\frac {3 \int \cos ^5(c+d x) \sin ^2(c+d x) \, dx}{a^3} \\ & = \frac {\cos ^5(c+d x) \sin (c+d x)}{2 a^3 d}+\frac {\cos ^7(c+d x) \sin (c+d x)}{8 a^3 d}-\frac {\int \cos ^6(c+d x) \, dx}{8 a^3}-\frac {\int \cos ^4(c+d x) \, dx}{2 a^3}+\frac {\text {Subst}\left (\int x^2 \left (1-x^2\right ) \, dx,x,\sin (c+d x)\right )}{a^3 d}+\frac {3 \text {Subst}\left (\int x^2 \left (1-x^2\right )^2 \, dx,x,\sin (c+d x)\right )}{a^3 d} \\ & = -\frac {\cos ^3(c+d x) \sin (c+d x)}{8 a^3 d}+\frac {23 \cos ^5(c+d x) \sin (c+d x)}{48 a^3 d}+\frac {\cos ^7(c+d x) \sin (c+d x)}{8 a^3 d}-\frac {5 \int \cos ^4(c+d x) \, dx}{48 a^3}-\frac {3 \int \cos ^2(c+d x) \, dx}{8 a^3}+\frac {\text {Subst}\left (\int \left (x^2-x^4\right ) \, dx,x,\sin (c+d x)\right )}{a^3 d}+\frac {3 \text {Subst}\left (\int \left (x^2-2 x^4+x^6\right ) \, dx,x,\sin (c+d x)\right )}{a^3 d} \\ & = -\frac {3 \cos (c+d x) \sin (c+d x)}{16 a^3 d}-\frac {29 \cos ^3(c+d x) \sin (c+d x)}{192 a^3 d}+\frac {23 \cos ^5(c+d x) \sin (c+d x)}{48 a^3 d}+\frac {\cos ^7(c+d x) \sin (c+d x)}{8 a^3 d}+\frac {4 \sin ^3(c+d x)}{3 a^3 d}-\frac {7 \sin ^5(c+d x)}{5 a^3 d}+\frac {3 \sin ^7(c+d x)}{7 a^3 d}-\frac {5 \int \cos ^2(c+d x) \, dx}{64 a^3}-\frac {3 \int 1 \, dx}{16 a^3} \\ & = -\frac {3 x}{16 a^3}-\frac {29 \cos (c+d x) \sin (c+d x)}{128 a^3 d}-\frac {29 \cos ^3(c+d x) \sin (c+d x)}{192 a^3 d}+\frac {23 \cos ^5(c+d x) \sin (c+d x)}{48 a^3 d}+\frac {\cos ^7(c+d x) \sin (c+d x)}{8 a^3 d}+\frac {4 \sin ^3(c+d x)}{3 a^3 d}-\frac {7 \sin ^5(c+d x)}{5 a^3 d}+\frac {3 \sin ^7(c+d x)}{7 a^3 d}-\frac {5 \int 1 \, dx}{128 a^3} \\ & = -\frac {29 x}{128 a^3}-\frac {29 \cos (c+d x) \sin (c+d x)}{128 a^3 d}-\frac {29 \cos ^3(c+d x) \sin (c+d x)}{192 a^3 d}+\frac {23 \cos ^5(c+d x) \sin (c+d x)}{48 a^3 d}+\frac {\cos ^7(c+d x) \sin (c+d x)}{8 a^3 d}+\frac {4 \sin ^3(c+d x)}{3 a^3 d}-\frac {7 \sin ^5(c+d x)}{5 a^3 d}+\frac {3 \sin ^7(c+d x)}{7 a^3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.22 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.83 \[ \int \frac {\sin ^8(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\cos ^6\left (\frac {1}{2} (c+d x)\right ) \sec ^3(c+d x) \left (-24360 d x+38640 \sin (c+d x)-6720 \sin (2 (c+d x))-3920 \sin (3 (c+d x))+5880 \sin (4 (c+d x))-4368 \sin (5 (c+d x))+2240 \sin (6 (c+d x))-720 \sin (7 (c+d x))+105 \sin (8 (c+d x))+294 \tan \left (\frac {c}{2}\right )\right )}{13440 a^3 d (1+\sec (c+d x))^3} \]

[In]

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

[Out]

(Cos[(c + d*x)/2]^6*Sec[c + d*x]^3*(-24360*d*x + 38640*Sin[c + d*x] - 6720*Sin[2*(c + d*x)] - 3920*Sin[3*(c +
d*x)] + 5880*Sin[4*(c + d*x)] - 4368*Sin[5*(c + d*x)] + 2240*Sin[6*(c + d*x)] - 720*Sin[7*(c + d*x)] + 105*Sin
[8*(c + d*x)] + 294*Tan[c/2]))/(13440*a^3*d*(1 + Sec[c + d*x])^3)

Maple [A] (verified)

Time = 0.97 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.63

method	result	size
parallelrisch	\(\frac {-24360 d x +38640 \sin \left (d x +c \right )+105 \sin \left (8 d x +8 c \right )+2240 \sin \left (6 d x +6 c \right )+5880 \sin \left (4 d x +4 c \right )-6720 \sin \left (2 d x +2 c \right )-720 \sin \left (7 d x +7 c \right )-4368 \sin \left (5 d x +5 c \right )-3920 \sin \left (3 d x +3 c \right )}{107520 a^{3} d}\)	\(99\)
risch	\(-\frac {29 x}{128 a^{3}}+\frac {23 \sin \left (d x +c \right )}{64 a^{3} d}+\frac {\sin \left (8 d x +8 c \right )}{1024 a^{3} d}-\frac {3 \sin \left (7 d x +7 c \right )}{448 a^{3} d}+\frac {\sin \left (6 d x +6 c \right )}{48 a^{3} d}-\frac {13 \sin \left (5 d x +5 c \right )}{320 a^{3} d}+\frac {7 \sin \left (4 d x +4 c \right )}{128 a^{3} d}-\frac {7 \sin \left (3 d x +3 c \right )}{192 a^{3} d}-\frac {\sin \left (2 d x +2 c \right )}{16 a^{3} d}\)	\(141\)
derivativedivides	\(\frac {-\frac {64 \left (-\frac {29 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4096}-\frac {667 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12288}-\frac {11107 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{61440}-\frac {146537 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{430080}-\frac {72669 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{143360}-\frac {1759 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{20480}-\frac {1143 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{4096}+\frac {29 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{4096}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{8}}-\frac {29 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64}}{a^{3} d}\)	\(142\)
default	\(\frac {-\frac {64 \left (-\frac {29 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4096}-\frac {667 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12288}-\frac {11107 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{61440}-\frac {146537 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{430080}-\frac {72669 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{143360}-\frac {1759 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{20480}-\frac {1143 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{4096}+\frac {29 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{4096}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{8}}-\frac {29 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64}}{a^{3} d}\)	\(142\)
norman	\(\frac {-\frac {29 x}{128 a}+\frac {29 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{64 a d}+\frac {667 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{192 a d}+\frac {11107 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{960 a d}+\frac {146537 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{6720 a d}+\frac {72669 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{2240 a d}+\frac {1759 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{320 a d}+\frac {1143 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{64 a d}-\frac {29 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{64 a d}-\frac {29 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{16 a}-\frac {203 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{32 a}-\frac {203 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{16 a}-\frac {1015 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{64 a}-\frac {203 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{16 a}-\frac {203 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{32 a}-\frac {29 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}}{16 a}-\frac {29 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{16}}{128 a}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{8} a^{2}}\)	\(313\)

[In]

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

[Out]

1/107520*(-24360*d*x+38640*sin(d*x+c)+105*sin(8*d*x+8*c)+2240*sin(6*d*x+6*c)+5880*sin(4*d*x+4*c)-6720*sin(2*d*
x+2*c)-720*sin(7*d*x+7*c)-4368*sin(5*d*x+5*c)-3920*sin(3*d*x+3*c))/a^3/d

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.58 \[ \int \frac {\sin ^8(c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {3045 \, d x - {\left (1680 \, \cos \left (d x + c\right )^{7} - 5760 \, \cos \left (d x + c\right )^{6} + 6440 \, \cos \left (d x + c\right )^{5} - 1536 \, \cos \left (d x + c\right )^{4} - 2030 \, \cos \left (d x + c\right )^{3} + 2432 \, \cos \left (d x + c\right )^{2} - 3045 \, \cos \left (d x + c\right ) + 4864\right )} \sin \left (d x + c\right )}{13440 \, a^{3} d} \]

[In]

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

[Out]

-1/13440*(3045*d*x - (1680*cos(d*x + c)^7 - 5760*cos(d*x + c)^6 + 6440*cos(d*x + c)^5 - 1536*cos(d*x + c)^4 -
2030*cos(d*x + c)^3 + 2432*cos(d*x + c)^2 - 3045*cos(d*x + c) + 4864)*sin(d*x + c))/(a^3*d)

Sympy [F(-1)]

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

[In]

integrate(sin(d*x+c)**8/(a+a*sec(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. 378 vs. \(2 (141) = 282\).

Time = 0.28 (sec) , antiderivative size = 378, normalized size of antiderivative = 2.41 \[ \int \frac {\sin ^8(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\frac {\frac {3045 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {23345 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {77749 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {146537 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {218007 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} + \frac {36939 \, \sin \left (d x + c\right )^{11}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{11}} + \frac {120015 \, \sin \left (d x + c\right )^{13}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{13}} - \frac {3045 \, \sin \left (d x + c\right )^{15}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{15}}}{a^{3} + \frac {8 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {28 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {56 \, a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {70 \, a^{3} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {56 \, a^{3} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} + \frac {28 \, a^{3} \sin \left (d x + c\right )^{12}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{12}} + \frac {8 \, a^{3} \sin \left (d x + c\right )^{14}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{14}} + \frac {a^{3} \sin \left (d x + c\right )^{16}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{16}}} - \frac {3045 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}}{6720 \, d} \]

[In]

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

[Out]

1/6720*((3045*sin(d*x + c)/(cos(d*x + c) + 1) + 23345*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 77749*sin(d*x + c)
^5/(cos(d*x + c) + 1)^5 + 146537*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 218007*sin(d*x + c)^9/(cos(d*x + c) + 1
)^9 + 36939*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 + 120015*sin(d*x + c)^13/(cos(d*x + c) + 1)^13 - 3045*sin(d*
x + c)^15/(cos(d*x + c) + 1)^15)/(a^3 + 8*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 28*a^3*sin(d*x + c)^4/(cos
(d*x + c) + 1)^4 + 56*a^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 70*a^3*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + 5
6*a^3*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 + 28*a^3*sin(d*x + c)^12/(cos(d*x + c) + 1)^12 + 8*a^3*sin(d*x + c
)^14/(cos(d*x + c) + 1)^14 + a^3*sin(d*x + c)^16/(cos(d*x + c) + 1)^16) - 3045*arctan(sin(d*x + c)/(cos(d*x +
c) + 1))/a^3)/d

Giac [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.89 \[ \int \frac {\sin ^8(c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {\frac {3045 \, {\left (d x + c\right )}}{a^{3}} + \frac {2 \, {\left (3045 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{15} - 120015 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} - 36939 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 218007 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 146537 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 77749 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 23345 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3045 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{8} a^{3}}}{13440 \, d} \]

[In]

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

[Out]

-1/13440*(3045*(d*x + c)/a^3 + 2*(3045*tan(1/2*d*x + 1/2*c)^15 - 120015*tan(1/2*d*x + 1/2*c)^13 - 36939*tan(1/
2*d*x + 1/2*c)^11 - 218007*tan(1/2*d*x + 1/2*c)^9 - 146537*tan(1/2*d*x + 1/2*c)^7 - 77749*tan(1/2*d*x + 1/2*c)
^5 - 23345*tan(1/2*d*x + 1/2*c)^3 - 3045*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^8*a^3))/d

Mupad [B] (verification not implemented)

Time = 16.58 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.84 \[ \int \frac {\sin ^8(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {-\frac {29\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}}{64}+\frac {1143\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{64}+\frac {1759\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{320}+\frac {72669\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{2240}+\frac {146537\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{6720}+\frac {11107\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{960}+\frac {667\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{192}+\frac {29\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64}}{a^3\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^8}-\frac {29\,x}{128\,a^3} \]

[In]

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

[Out]

((29*tan(c/2 + (d*x)/2))/64 + (667*tan(c/2 + (d*x)/2)^3)/192 + (11107*tan(c/2 + (d*x)/2)^5)/960 + (146537*tan(
c/2 + (d*x)/2)^7)/6720 + (72669*tan(c/2 + (d*x)/2)^9)/2240 + (1759*tan(c/2 + (d*x)/2)^11)/320 + (1143*tan(c/2
+ (d*x)/2)^13)/64 - (29*tan(c/2 + (d*x)/2)^15)/64)/(a^3*d*(tan(c/2 + (d*x)/2)^2 + 1)^8) - (29*x)/(128*a^3)