\(\int \frac {\sec ^9(c+d x)}{a+a \sin (c+d x)} \, dx\) [907]

Optimal result

Integrand size = 21, antiderivative size = 210 \[ \int \frac {\sec ^9(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {63 \text {arctanh}(\sin (c+d x))}{256 a d}+\frac {a^3}{256 d (a-a \sin (c+d x))^4}+\frac {a^2}{64 d (a-a \sin (c+d x))^3}+\frac {21 a}{512 d (a-a \sin (c+d x))^2}+\frac {7}{64 d (a-a \sin (c+d x))}-\frac {a^4}{160 d (a+a \sin (c+d x))^5}-\frac {5 a^3}{256 d (a+a \sin (c+d x))^4}-\frac {5 a^2}{128 d (a+a \sin (c+d x))^3}-\frac {35 a}{512 d (a+a \sin (c+d x))^2}-\frac {35}{256 d (a+a \sin (c+d x))} \]

[Out]

63/256*arctanh(sin(d*x+c))/a/d+1/256*a^3/d/(a-a*sin(d*x+c))^4+1/64*a^2/d/(a-a*sin(d*x+c))^3+21/512*a/d/(a-a*si
n(d*x+c))^2+7/64/d/(a-a*sin(d*x+c))-1/160*a^4/d/(a+a*sin(d*x+c))^5-5/256*a^3/d/(a+a*sin(d*x+c))^4-5/128*a^2/d/
(a+a*sin(d*x+c))^3-35/512*a/d/(a+a*sin(d*x+c))^2-35/256/d/(a+a*sin(d*x+c))

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2746, 46, 212} \[ \int \frac {\sec ^9(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {a^4}{160 d (a \sin (c+d x)+a)^5}+\frac {a^3}{256 d (a-a \sin (c+d x))^4}-\frac {5 a^3}{256 d (a \sin (c+d x)+a)^4}+\frac {a^2}{64 d (a-a \sin (c+d x))^3}-\frac {5 a^2}{128 d (a \sin (c+d x)+a)^3}+\frac {63 \text {arctanh}(\sin (c+d x))}{256 a d}+\frac {21 a}{512 d (a-a \sin (c+d x))^2}-\frac {35 a}{512 d (a \sin (c+d x)+a)^2}+\frac {7}{64 d (a-a \sin (c+d x))}-\frac {35}{256 d (a \sin (c+d x)+a)} \]

[In]

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

[Out]

(63*ArcTanh[Sin[c + d*x]])/(256*a*d) + a^3/(256*d*(a - a*Sin[c + d*x])^4) + a^2/(64*d*(a - a*Sin[c + d*x])^3)
+ (21*a)/(512*d*(a - a*Sin[c + d*x])^2) + 7/(64*d*(a - a*Sin[c + d*x])) - a^4/(160*d*(a + a*Sin[c + d*x])^5) -
(5*a^3)/(256*d*(a + a*Sin[c + d*x])^4) - (5*a^2)/(128*d*(a + a*Sin[c + d*x])^3) - (35*a)/(512*d*(a + a*Sin[c
+ d*x])^2) - 35/(256*d*(a + a*Sin[c + d*x]))

Rule 46

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x
)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2746

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rubi steps \begin{align*} \text {integral}& = \frac {a^9 \text {Subst}\left (\int \frac {1}{(a-x)^5 (a+x)^6} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^9 \text {Subst}\left (\int \left (\frac {1}{64 a^6 (a-x)^5}+\frac {3}{64 a^7 (a-x)^4}+\frac {21}{256 a^8 (a-x)^3}+\frac {7}{64 a^9 (a-x)^2}+\frac {1}{32 a^5 (a+x)^6}+\frac {5}{64 a^6 (a+x)^5}+\frac {15}{128 a^7 (a+x)^4}+\frac {35}{256 a^8 (a+x)^3}+\frac {35}{256 a^9 (a+x)^2}+\frac {63}{256 a^9 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^3}{256 d (a-a \sin (c+d x))^4}+\frac {a^2}{64 d (a-a \sin (c+d x))^3}+\frac {21 a}{512 d (a-a \sin (c+d x))^2}+\frac {7}{64 d (a-a \sin (c+d x))}-\frac {a^4}{160 d (a+a \sin (c+d x))^5}-\frac {5 a^3}{256 d (a+a \sin (c+d x))^4}-\frac {5 a^2}{128 d (a+a \sin (c+d x))^3}-\frac {35 a}{512 d (a+a \sin (c+d x))^2}-\frac {35}{256 d (a+a \sin (c+d x))}+\frac {63 \text {Subst}\left (\int \frac {1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{256 d} \\ & = \frac {63 \text {arctanh}(\sin (c+d x))}{256 a d}+\frac {a^3}{256 d (a-a \sin (c+d x))^4}+\frac {a^2}{64 d (a-a \sin (c+d x))^3}+\frac {21 a}{512 d (a-a \sin (c+d x))^2}+\frac {7}{64 d (a-a \sin (c+d x))}-\frac {a^4}{160 d (a+a \sin (c+d x))^5}-\frac {5 a^3}{256 d (a+a \sin (c+d x))^4}-\frac {5 a^2}{128 d (a+a \sin (c+d x))^3}-\frac {35 a}{512 d (a+a \sin (c+d x))^2}-\frac {35}{256 d (a+a \sin (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.83 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.79 \[ \int \frac {\sec ^9(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\sec ^8(c+d x) \left (-128+315 \text {arctanh}(\sin (c+d x)) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^8 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^{10}+837 \sin (c+d x)+837 \sin ^2(c+d x)-1533 \sin ^3(c+d x)-1533 \sin ^4(c+d x)+1155 \sin ^5(c+d x)+1155 \sin ^6(c+d x)-315 \sin ^7(c+d x)-315 \sin ^8(c+d x)\right )}{1280 a d (1+\sin (c+d x))} \]

[In]

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

[Out]

(Sec[c + d*x]^8*(-128 + 315*ArcTanh[Sin[c + d*x]]*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^8*(Cos[(c + d*x)/2] +
Sin[(c + d*x)/2])^10 + 837*Sin[c + d*x] + 837*Sin[c + d*x]^2 - 1533*Sin[c + d*x]^3 - 1533*Sin[c + d*x]^4 + 115
5*Sin[c + d*x]^5 + 1155*Sin[c + d*x]^6 - 315*Sin[c + d*x]^7 - 315*Sin[c + d*x]^8))/(1280*a*d*(1 + Sin[c + d*x]
))

Maple [A] (verified)

Time = 1.76 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.66

method	result	size
derivativedivides	\(\frac {\frac {1}{256 \left (\sin \left (d x +c \right )-1\right )^{4}}-\frac {1}{64 \left (\sin \left (d x +c \right )-1\right )^{3}}+\frac {21}{512 \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {7}{64 \left (\sin \left (d x +c \right )-1\right )}-\frac {63 \ln \left (\sin \left (d x +c \right )-1\right )}{512}-\frac {1}{160 \left (1+\sin \left (d x +c \right )\right )^{5}}-\frac {5}{256 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {5}{128 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {35}{512 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {35}{256 \left (1+\sin \left (d x +c \right )\right )}+\frac {63 \ln \left (1+\sin \left (d x +c \right )\right )}{512}}{d a}\)	\(139\)
default	\(\frac {\frac {1}{256 \left (\sin \left (d x +c \right )-1\right )^{4}}-\frac {1}{64 \left (\sin \left (d x +c \right )-1\right )^{3}}+\frac {21}{512 \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {7}{64 \left (\sin \left (d x +c \right )-1\right )}-\frac {63 \ln \left (\sin \left (d x +c \right )-1\right )}{512}-\frac {1}{160 \left (1+\sin \left (d x +c \right )\right )^{5}}-\frac {5}{256 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {5}{128 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {35}{512 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {35}{256 \left (1+\sin \left (d x +c \right )\right )}+\frac {63 \ln \left (1+\sin \left (d x +c \right )\right )}{512}}{d a}\)	\(139\)
risch	\(-\frac {i \left (4830 i {\mathrm e}^{14 i \left (d x +c \right )}+315 \,{\mathrm e}^{17 i \left (d x +c \right )}+630 i {\mathrm e}^{16 i \left (d x +c \right )}+2100 \,{\mathrm e}^{15 i \left (d x +c \right )}-30318 i {\mathrm e}^{8 i \left (d x +c \right )}+5628 \,{\mathrm e}^{13 i \left (d x +c \right )}-16086 i {\mathrm e}^{6 i \left (d x +c \right )}+7116 \,{\mathrm e}^{11 i \left (d x +c \right )}+16086 i {\mathrm e}^{12 i \left (d x +c \right )}+2450 \,{\mathrm e}^{9 i \left (d x +c \right )}+30318 i {\mathrm e}^{10 i \left (d x +c \right )}+7116 \,{\mathrm e}^{7 i \left (d x +c \right )}-4830 i {\mathrm e}^{4 i \left (d x +c \right )}+5628 \,{\mathrm e}^{5 i \left (d x +c \right )}-630 i {\mathrm e}^{2 i \left (d x +c \right )}+2100 \,{\mathrm e}^{3 i \left (d x +c \right )}+315 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{640 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{10} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{8} d a}+\frac {63 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{256 a d}-\frac {63 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{256 d a}\)	\(277\)
parallelrisch	\(\frac {\left (-8820 \sin \left (3 d x +3 c \right )-6300 \sin \left (5 d x +5 c \right )-2205 \sin \left (7 d x +7 c \right )-315 \sin \left (9 d x +9 c \right )-35280 \cos \left (2 d x +2 c \right )-17640 \cos \left (4 d x +4 c \right )-5040 \cos \left (6 d x +6 c \right )-630 \cos \left (8 d x +8 c \right )-4410 \sin \left (d x +c \right )-22050\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (8820 \sin \left (3 d x +3 c \right )+6300 \sin \left (5 d x +5 c \right )+2205 \sin \left (7 d x +7 c \right )+315 \sin \left (9 d x +9 c \right )+35280 \cos \left (2 d x +2 c \right )+17640 \cos \left (4 d x +4 c \right )+5040 \cos \left (6 d x +6 c \right )+630 \cos \left (8 d x +8 c \right )+4410 \sin \left (d x +c \right )+22050\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+35756 \sin \left (3 d x +3 c \right )+12220 \sin \left (5 d x +5 c \right )+2156 \sin \left (7 d x +7 c \right )+128 \sin \left (9 d x +9 c \right )+104 \cos \left (2 d x +2 c \right )-4088 \cos \left (4 d x +4 c \right )-2152 \cos \left (6 d x +6 c \right )-374 \cos \left (8 d x +8 c \right )+62428 \sin \left (d x +c \right )+6510}{1280 a d \left (70+\sin \left (9 d x +9 c \right )+7 \sin \left (7 d x +7 c \right )+20 \sin \left (5 d x +5 c \right )+28 \sin \left (3 d x +3 c \right )+14 \sin \left (d x +c \right )+2 \cos \left (8 d x +8 c \right )+16 \cos \left (6 d x +6 c \right )+56 \cos \left (4 d x +4 c \right )+112 \cos \left (2 d x +2 c \right )\right )}\)	\(427\)

[In]

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

[Out]

1/d/a*(1/256/(sin(d*x+c)-1)^4-1/64/(sin(d*x+c)-1)^3+21/512/(sin(d*x+c)-1)^2-7/64/(sin(d*x+c)-1)-63/512*ln(sin(
d*x+c)-1)-1/160/(1+sin(d*x+c))^5-5/256/(1+sin(d*x+c))^4-5/128/(1+sin(d*x+c))^3-35/512/(1+sin(d*x+c))^2-35/256/
(1+sin(d*x+c))+63/512*ln(1+sin(d*x+c)))

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.89 \[ \int \frac {\sec ^9(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {630 \, \cos \left (d x + c\right )^{8} - 210 \, \cos \left (d x + c\right )^{6} - 84 \, \cos \left (d x + c\right )^{4} - 48 \, \cos \left (d x + c\right )^{2} - 315 \, {\left (\cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{8}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 315 \, {\left (\cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{8}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 6 \, {\left (105 \, \cos \left (d x + c\right )^{6} + 70 \, \cos \left (d x + c\right )^{4} + 56 \, \cos \left (d x + c\right )^{2} + 48\right )} \sin \left (d x + c\right ) - 32}{2560 \, {\left (a d \cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{8}\right )}} \]

[In]

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

[Out]

-1/2560*(630*cos(d*x + c)^8 - 210*cos(d*x + c)^6 - 84*cos(d*x + c)^4 - 48*cos(d*x + c)^2 - 315*(cos(d*x + c)^8
*sin(d*x + c) + cos(d*x + c)^8)*log(sin(d*x + c) + 1) + 315*(cos(d*x + c)^8*sin(d*x + c) + cos(d*x + c)^8)*log
(-sin(d*x + c) + 1) - 6*(105*cos(d*x + c)^6 + 70*cos(d*x + c)^4 + 56*cos(d*x + c)^2 + 48)*sin(d*x + c) - 32)/(
a*d*cos(d*x + c)^8*sin(d*x + c) + a*d*cos(d*x + c)^8)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.02 \[ \int \frac {\sec ^9(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {2 \, {\left (315 \, \sin \left (d x + c\right )^{8} + 315 \, \sin \left (d x + c\right )^{7} - 1155 \, \sin \left (d x + c\right )^{6} - 1155 \, \sin \left (d x + c\right )^{5} + 1533 \, \sin \left (d x + c\right )^{4} + 1533 \, \sin \left (d x + c\right )^{3} - 837 \, \sin \left (d x + c\right )^{2} - 837 \, \sin \left (d x + c\right ) + 128\right )}}{a \sin \left (d x + c\right )^{9} + a \sin \left (d x + c\right )^{8} - 4 \, a \sin \left (d x + c\right )^{7} - 4 \, a \sin \left (d x + c\right )^{6} + 6 \, a \sin \left (d x + c\right )^{5} + 6 \, a \sin \left (d x + c\right )^{4} - 4 \, a \sin \left (d x + c\right )^{3} - 4 \, a \sin \left (d x + c\right )^{2} + a \sin \left (d x + c\right ) + a} - \frac {315 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} + \frac {315 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a}}{2560 \, d} \]

[In]

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

[Out]

-1/2560*(2*(315*sin(d*x + c)^8 + 315*sin(d*x + c)^7 - 1155*sin(d*x + c)^6 - 1155*sin(d*x + c)^5 + 1533*sin(d*x
+ c)^4 + 1533*sin(d*x + c)^3 - 837*sin(d*x + c)^2 - 837*sin(d*x + c) + 128)/(a*sin(d*x + c)^9 + a*sin(d*x + c
)^8 - 4*a*sin(d*x + c)^7 - 4*a*sin(d*x + c)^6 + 6*a*sin(d*x + c)^5 + 6*a*sin(d*x + c)^4 - 4*a*sin(d*x + c)^3 -
4*a*sin(d*x + c)^2 + a*sin(d*x + c) + a) - 315*log(sin(d*x + c) + 1)/a + 315*log(sin(d*x + c) - 1)/a)/d

Giac [A] (verification not implemented)

none

Time = 0.65 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.74 \[ \int \frac {\sec ^9(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {1260 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} - \frac {1260 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a} + \frac {5 \, {\left (525 \, \sin \left (d x + c\right )^{4} - 2324 \, \sin \left (d x + c\right )^{3} + 3906 \, \sin \left (d x + c\right )^{2} - 2972 \, \sin \left (d x + c\right ) + 873\right )}}{a {\left (\sin \left (d x + c\right ) - 1\right )}^{4}} - \frac {2877 \, \sin \left (d x + c\right )^{5} + 15785 \, \sin \left (d x + c\right )^{4} + 35070 \, \sin \left (d x + c\right )^{3} + 39670 \, \sin \left (d x + c\right )^{2} + 23085 \, \sin \left (d x + c\right ) + 5641}{a {\left (\sin \left (d x + c\right ) + 1\right )}^{5}}}{10240 \, d} \]

[In]

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

[Out]

1/10240*(1260*log(abs(sin(d*x + c) + 1))/a - 1260*log(abs(sin(d*x + c) - 1))/a + 5*(525*sin(d*x + c)^4 - 2324*
sin(d*x + c)^3 + 3906*sin(d*x + c)^2 - 2972*sin(d*x + c) + 873)/(a*(sin(d*x + c) - 1)^4) - (2877*sin(d*x + c)^
5 + 15785*sin(d*x + c)^4 + 35070*sin(d*x + c)^3 + 39670*sin(d*x + c)^2 + 23085*sin(d*x + c) + 5641)/(a*(sin(d*
x + c) + 1)^5))/d

Mupad [B] (verification not implemented)

Time = 9.56 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.95 \[ \int \frac {\sec ^9(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {63\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )}{256\,a\,d}-\frac {\frac {63\,{\sin \left (c+d\,x\right )}^8}{256}+\frac {63\,{\sin \left (c+d\,x\right )}^7}{256}-\frac {231\,{\sin \left (c+d\,x\right )}^6}{256}-\frac {231\,{\sin \left (c+d\,x\right )}^5}{256}+\frac {1533\,{\sin \left (c+d\,x\right )}^4}{1280}+\frac {1533\,{\sin \left (c+d\,x\right )}^3}{1280}-\frac {837\,{\sin \left (c+d\,x\right )}^2}{1280}-\frac {837\,\sin \left (c+d\,x\right )}{1280}+\frac {1}{10}}{d\,\left (a\,{\sin \left (c+d\,x\right )}^9+a\,{\sin \left (c+d\,x\right )}^8-4\,a\,{\sin \left (c+d\,x\right )}^7-4\,a\,{\sin \left (c+d\,x\right )}^6+6\,a\,{\sin \left (c+d\,x\right )}^5+6\,a\,{\sin \left (c+d\,x\right )}^4-4\,a\,{\sin \left (c+d\,x\right )}^3-4\,a\,{\sin \left (c+d\,x\right )}^2+a\,\sin \left (c+d\,x\right )+a\right )} \]

[In]

int(1/(cos(c + d*x)^9*(a + a*sin(c + d*x))),x)

[Out]

(63*atanh(sin(c + d*x)))/(256*a*d) - ((1533*sin(c + d*x)^3)/1280 - (837*sin(c + d*x)^2)/1280 - (837*sin(c + d*
x))/1280 + (1533*sin(c + d*x)^4)/1280 - (231*sin(c + d*x)^5)/256 - (231*sin(c + d*x)^6)/256 + (63*sin(c + d*x)
^7)/256 + (63*sin(c + d*x)^8)/256 + 1/10)/(d*(a + a*sin(c + d*x) - 4*a*sin(c + d*x)^2 - 4*a*sin(c + d*x)^3 + 6
*a*sin(c + d*x)^4 + 6*a*sin(c + d*x)^5 - 4*a*sin(c + d*x)^6 - 4*a*sin(c + d*x)^7 + a*sin(c + d*x)^8 + a*sin(c
+ d*x)^9))