\(\int \frac {\sin ^9(c+d x)}{(a+a \sec (c+d x))^2} \, dx\) [75]

Optimal result

Integrand size = 21, antiderivative size = 114 \[ \int \frac {\sin ^9(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {4 (a-a \cos (c+d x))^5}{5 a^7 d}-\frac {2 (a-a \cos (c+d x))^6}{a^8 d}+\frac {13 (a-a \cos (c+d x))^7}{7 a^9 d}-\frac {3 (a-a \cos (c+d x))^8}{4 a^{10} d}+\frac {(a-a \cos (c+d x))^9}{9 a^{11} d} \]

[Out]

4/5*(a-a*cos(d*x+c))^5/a^7/d-2*(a-a*cos(d*x+c))^6/a^8/d+13/7*(a-a*cos(d*x+c))^7/a^9/d-3/4*(a-a*cos(d*x+c))^8/a
^10/d+1/9*(a-a*cos(d*x+c))^9/a^11/d

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 114, 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 = {3957, 2915, 12, 90} \[ \int \frac {\sin ^9(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {(a-a \cos (c+d x))^9}{9 a^{11} d}-\frac {3 (a-a \cos (c+d x))^8}{4 a^{10} d}+\frac {13 (a-a \cos (c+d x))^7}{7 a^9 d}-\frac {2 (a-a \cos (c+d x))^6}{a^8 d}+\frac {4 (a-a \cos (c+d x))^5}{5 a^7 d} \]

[In]

Int[Sin[c + d*x]^9/(a + a*Sec[c + d*x])^2,x]

[Out]

(4*(a - a*Cos[c + d*x])^5)/(5*a^7*d) - (2*(a - a*Cos[c + d*x])^6)/(a^8*d) + (13*(a - a*Cos[c + d*x])^7)/(7*a^9
*d) - (3*(a - a*Cos[c + d*x])^8)/(4*a^10*d) + (a - a*Cos[c + d*x])^9/(9*a^11*d)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 2915

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x
)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2,
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 ^2(c+d x) \sin ^9(c+d x)}{(-a-a \cos (c+d x))^2} \, dx \\ & = \frac {\text {Subst}\left (\int \frac {(-a-x)^4 x^2 (-a+x)^2}{a^2} \, dx,x,-a \cos (c+d x)\right )}{a^9 d} \\ & = \frac {\text {Subst}\left (\int (-a-x)^4 x^2 (-a+x)^2 \, dx,x,-a \cos (c+d x)\right )}{a^{11} d} \\ & = \frac {\text {Subst}\left (\int \left (4 a^4 (-a-x)^4+12 a^3 (-a-x)^5+13 a^2 (-a-x)^6+6 a (-a-x)^7+(-a-x)^8\right ) \, dx,x,-a \cos (c+d x)\right )}{a^{11} d} \\ & = \frac {4 (a-a \cos (c+d x))^5}{5 a^7 d}-\frac {2 (a-a \cos (c+d x))^6}{a^8 d}+\frac {13 (a-a \cos (c+d x))^7}{7 a^9 d}-\frac {3 (a-a \cos (c+d x))^8}{4 a^{10} d}+\frac {(a-a \cos (c+d x))^9}{9 a^{11} d} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.93 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.54 \[ \int \frac {\sin ^9(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {2 (992+1615 \cos (c+d x)+970 \cos (2 (c+d x))+385 \cos (3 (c+d x))+70 \cos (4 (c+d x))) \sin ^{10}\left (\frac {1}{2} (c+d x)\right )}{315 a^2 d} \]

[In]

Integrate[Sin[c + d*x]^9/(a + a*Sec[c + d*x])^2,x]

[Out]

(2*(992 + 1615*Cos[c + d*x] + 970*Cos[2*(c + d*x)] + 385*Cos[3*(c + d*x)] + 70*Cos[4*(c + d*x)])*Sin[(c + d*x)
/2]^10)/(315*a^2*d)

Maple [A] (verified)

Time = 0.96 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.69

method	result	size
derivativedivides	\(\frac {-\frac {\cos \left (d x +c \right )^{9}}{9}+\frac {\cos \left (d x +c \right )^{8}}{4}+\frac {\cos \left (d x +c \right )^{7}}{7}-\frac {2 \cos \left (d x +c \right )^{6}}{3}+\frac {\cos \left (d x +c \right )^{5}}{5}+\frac {\cos \left (d x +c \right )^{4}}{2}-\frac {\cos \left (d x +c \right )^{3}}{3}}{d \,a^{2}}\)	\(79\)
default	\(\frac {-\frac {\cos \left (d x +c \right )^{9}}{9}+\frac {\cos \left (d x +c \right )^{8}}{4}+\frac {\cos \left (d x +c \right )^{7}}{7}-\frac {2 \cos \left (d x +c \right )^{6}}{3}+\frac {\cos \left (d x +c \right )^{5}}{5}+\frac {\cos \left (d x +c \right )^{4}}{2}-\frac {\cos \left (d x +c \right )^{3}}{3}}{d \,a^{2}}\)	\(79\)
parallelrisch	\(\frac {-\frac {64}{315}-\frac {128 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{3}-\frac {64 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{35}-\frac {256 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{15}-\frac {128 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{5}-\frac {256 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{35}}{d \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{9} a^{2}}\)	\(91\)
norman	\(\frac {-\frac {64}{315 a d}-\frac {128 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{3 a d}-\frac {64 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{35 d a}-\frac {256 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{35 d a}-\frac {256 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{15 d a}-\frac {128 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{5 d a}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{9} a}\)	\(124\)
risch	\(-\frac {13 \cos \left (d x +c \right )}{128 a^{2} d}-\frac {\cos \left (9 d x +9 c \right )}{2304 d \,a^{2}}+\frac {\cos \left (8 d x +8 c \right )}{512 d \,a^{2}}-\frac {3 \cos \left (7 d x +7 c \right )}{1792 d \,a^{2}}-\frac {\cos \left (6 d x +6 c \right )}{192 d \,a^{2}}+\frac {\cos \left (5 d x +5 c \right )}{80 d \,a^{2}}-\frac {\cos \left (4 d x +4 c \right )}{128 d \,a^{2}}-\frac {\cos \left (3 d x +3 c \right )}{96 d \,a^{2}}+\frac {3 \cos \left (2 d x +2 c \right )}{64 d \,a^{2}}\)	\(152\)

[In]

int(sin(d*x+c)^9/(a+a*sec(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/d/a^2*(-1/9*cos(d*x+c)^9+1/4*cos(d*x+c)^8+1/7*cos(d*x+c)^7-2/3*cos(d*x+c)^6+1/5*cos(d*x+c)^5+1/2*cos(d*x+c)^
4-1/3*cos(d*x+c)^3)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.69 \[ \int \frac {\sin ^9(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {140 \, \cos \left (d x + c\right )^{9} - 315 \, \cos \left (d x + c\right )^{8} - 180 \, \cos \left (d x + c\right )^{7} + 840 \, \cos \left (d x + c\right )^{6} - 252 \, \cos \left (d x + c\right )^{5} - 630 \, \cos \left (d x + c\right )^{4} + 420 \, \cos \left (d x + c\right )^{3}}{1260 \, a^{2} d} \]

[In]

integrate(sin(d*x+c)^9/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/1260*(140*cos(d*x + c)^9 - 315*cos(d*x + c)^8 - 180*cos(d*x + c)^7 + 840*cos(d*x + c)^6 - 252*cos(d*x + c)^
5 - 630*cos(d*x + c)^4 + 420*cos(d*x + c)^3)/(a^2*d)

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin ^9(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\text {Timed out} \]

[In]

integrate(sin(d*x+c)**9/(a+a*sec(d*x+c))**2,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.69 \[ \int \frac {\sin ^9(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {140 \, \cos \left (d x + c\right )^{9} - 315 \, \cos \left (d x + c\right )^{8} - 180 \, \cos \left (d x + c\right )^{7} + 840 \, \cos \left (d x + c\right )^{6} - 252 \, \cos \left (d x + c\right )^{5} - 630 \, \cos \left (d x + c\right )^{4} + 420 \, \cos \left (d x + c\right )^{3}}{1260 \, a^{2} d} \]

[In]

integrate(sin(d*x+c)^9/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/1260*(140*cos(d*x + c)^9 - 315*cos(d*x + c)^8 - 180*cos(d*x + c)^7 + 840*cos(d*x + c)^6 - 252*cos(d*x + c)^
5 - 630*cos(d*x + c)^4 + 420*cos(d*x + c)^3)/(a^2*d)

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.24 \[ \int \frac {\sin ^9(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {64 \, {\left (\frac {9 \, {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {36 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {84 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {126 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {210 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - 1\right )}}{315 \, a^{2} d {\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1\right )}^{9}} \]

[In]

integrate(sin(d*x+c)^9/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

-64/315*(9*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 36*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 84*(cos(d*x
+ c) - 1)^3/(cos(d*x + c) + 1)^3 - 126*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4 - 210*(cos(d*x + c) - 1)^6/(c
os(d*x + c) + 1)^6 - 1)/(a^2*d*((cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 1)^9)

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.84 \[ \int \frac {\sin ^9(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\frac {{\cos \left (c+d\,x\right )}^4}{2\,a^2}-\frac {{\cos \left (c+d\,x\right )}^3}{3\,a^2}+\frac {{\cos \left (c+d\,x\right )}^5}{5\,a^2}-\frac {2\,{\cos \left (c+d\,x\right )}^6}{3\,a^2}+\frac {{\cos \left (c+d\,x\right )}^7}{7\,a^2}+\frac {{\cos \left (c+d\,x\right )}^8}{4\,a^2}-\frac {{\cos \left (c+d\,x\right )}^9}{9\,a^2}}{d} \]

[In]

int(sin(c + d*x)^9/(a + a/cos(c + d*x))^2,x)

[Out]

(cos(c + d*x)^4/(2*a^2) - cos(c + d*x)^3/(3*a^2) + cos(c + d*x)^5/(5*a^2) - (2*cos(c + d*x)^6)/(3*a^2) + cos(c
+ d*x)^7/(7*a^2) + cos(c + d*x)^8/(4*a^2) - cos(c + d*x)^9/(9*a^2))/d