Integrand size = 21, antiderivative size = 171 \[ \int \frac {\tan ^5(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\text {arctanh}(\sin (c+d x))}{128 a^3 d}+\frac {1}{128 a d (a-a \sin (c+d x))^2}+\frac {a^2}{40 d (a+a \sin (c+d x))^5}-\frac {7 a}{64 d (a+a \sin (c+d x))^4}+\frac {1}{6 d (a+a \sin (c+d x))^3}-\frac {5}{64 a d (a+a \sin (c+d x))^2}-\frac {1}{32 d \left (a^3-a^3 \sin (c+d x)\right )}-\frac {5}{128 d \left (a^3+a^3 \sin (c+d x)\right )} \] Output:
1/128*arctanh(sin(d*x+c))/a^3/d+1/128/a/d/(a-a*sin(d*x+c))^2+1/40*a^2/d/(a +a*sin(d*x+c))^5-7/64*a/d/(a+a*sin(d*x+c))^4+1/6/d/(a+a*sin(d*x+c))^3-5/64 /a/d/(a+a*sin(d*x+c))^2-1/32/d/(a^3-a^3*sin(d*x+c))-5/128/d/(a^3+a^3*sin(d *x+c))
Time = 0.83 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.60 \[ \int \frac {\tan ^5(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {15 \text {arctanh}(\sin (c+d x))-\frac {112+351 \sin (c+d x)+157 \sin ^2(c+d x)-540 \sin ^3(c+d x)-620 \sin ^4(c+d x)+45 \sin ^5(c+d x)+15 \sin ^6(c+d x)}{(-1+\sin (c+d x))^2 (1+\sin (c+d x))^5}}{1920 a^3 d} \] Input:
Integrate[Tan[c + d*x]^5/(a + a*Sin[c + d*x])^3,x]
Output:
(15*ArcTanh[Sin[c + d*x]] - (112 + 351*Sin[c + d*x] + 157*Sin[c + d*x]^2 - 540*Sin[c + d*x]^3 - 620*Sin[c + d*x]^4 + 45*Sin[c + d*x]^5 + 15*Sin[c + d*x]^6)/((-1 + Sin[c + d*x])^2*(1 + Sin[c + d*x])^5))/(1920*a^3*d)
Time = 0.34 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.87, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3186, 99, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\tan ^5(c+d x)}{(a \sin (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (c+d x)^5}{(a \sin (c+d x)+a)^3}dx\) |
| \(\Big \downarrow \) 3186 |
| \(\displaystyle \frac {\int \frac {a^5 \sin ^5(c+d x)}{(a-a \sin (c+d x))^3 (\sin (c+d x) a+a)^6}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \frac {\int \left (-\frac {a^2}{8 (\sin (c+d x) a+a)^6}+\frac {7 a}{16 (\sin (c+d x) a+a)^5}-\frac {1}{2 (\sin (c+d x) a+a)^4}+\frac {1}{64 (a-a \sin (c+d x))^3 a}+\frac {5}{32 (\sin (c+d x) a+a)^3 a}+\frac {1}{128 \left (a^2-a^2 \sin ^2(c+d x)\right ) a^2}-\frac {1}{32 (a-a \sin (c+d x))^2 a^2}+\frac {5}{128 (\sin (c+d x) a+a)^2 a^2}\right )d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {\text {arctanh}(\sin (c+d x))}{128 a^3}+\frac {a^2}{40 (a \sin (c+d x)+a)^5}-\frac {1}{32 a^2 (a-a \sin (c+d x))}-\frac {5}{128 a^2 (a \sin (c+d x)+a)}-\frac {7 a}{64 (a \sin (c+d x)+a)^4}+\frac {1}{6 (a \sin (c+d x)+a)^3}+\frac {1}{128 a (a-a \sin (c+d x))^2}-\frac {5}{64 a (a \sin (c+d x)+a)^2}}{d}\) |
Input:
Int[Tan[c + d*x]^5/(a + a*Sin[c + d*x])^3,x]
Output:
(ArcTanh[Sin[c + d*x]]/(128*a^3) + 1/(128*a*(a - a*Sin[c + d*x])^2) - 1/(3 2*a^2*(a - a*Sin[c + d*x])) + a^2/(40*(a + a*Sin[c + d*x])^5) - (7*a)/(64* (a + a*Sin[c + d*x])^4) + 1/(6*(a + a*Sin[c + d*x])^3) - 5/(64*a*(a + a*Si n[c + d*x])^2) - 5/(128*a^2*(a + a*Sin[c + d*x])))/d
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(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]))
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p _.), x_Symbol] :> Simp[1/f Subst[Int[x^p*((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] && E qQ[a^2 - b^2, 0] && IntegerQ[(p + 1)/2]
Time = 3.81 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.67
| method | result | size |
| derivativedivides | \(\frac {\frac {1}{40 \left (1+\sin \left (d x +c \right )\right )^{5}}-\frac {7}{64 \left (1+\sin \left (d x +c \right )\right )^{4}}+\frac {1}{6 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {5}{64 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {5}{128 \left (1+\sin \left (d x +c \right )\right )}+\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{256}+\frac {1}{128 \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {1}{32 \sin \left (d x +c \right )-32}-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{256}}{d \,a^{3}}\) | \(115\) |
| default | \(\frac {\frac {1}{40 \left (1+\sin \left (d x +c \right )\right )^{5}}-\frac {7}{64 \left (1+\sin \left (d x +c \right )\right )^{4}}+\frac {1}{6 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {5}{64 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {5}{128 \left (1+\sin \left (d x +c \right )\right )}+\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{256}+\frac {1}{128 \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {1}{32 \sin \left (d x +c \right )-32}-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{256}}{d \,a^{3}}\) | \(115\) |
| risch | \(-\frac {i \left (-7183 \,{\mathrm e}^{5 i \left (d x +c \right )}+2390 \,{\mathrm e}^{3 i \left (d x +c \right )}+15 \,{\mathrm e}^{i \left (d x +c \right )}+15 \,{\mathrm e}^{13 i \left (d x +c \right )}+2390 \,{\mathrm e}^{11 i \left (d x +c \right )}-7183 \,{\mathrm e}^{9 i \left (d x +c \right )}+2388 \,{\mathrm e}^{7 i \left (d x +c \right )}+3870 i {\mathrm e}^{10 i \left (d x +c \right )}-90 i {\mathrm e}^{2 i \left (d x +c \right )}+90 i {\mathrm e}^{12 i \left (d x +c \right )}-828 i {\mathrm e}^{8 i \left (d x +c \right )}+828 i {\mathrm e}^{6 i \left (d x +c \right )}-3870 i {\mathrm e}^{4 i \left (d x +c \right )}\right )}{960 \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{10} d \,a^{3}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{128 d \,a^{3}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{128 d \,a^{3}}\) | \(231\) |
Input:
int(tan(d*x+c)^5/(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/d/a^3*(1/40/(1+sin(d*x+c))^5-7/64/(1+sin(d*x+c))^4+1/6/(1+sin(d*x+c))^3- 5/64/(1+sin(d*x+c))^2-5/128/(1+sin(d*x+c))+1/256*ln(1+sin(d*x+c))+1/128/(s in(d*x+c)-1)^2+1/32/(sin(d*x+c)-1)-1/256*ln(sin(d*x+c)-1))
Time = 0.10 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.45 \[ \int \frac {\tan ^5(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {30 \, \cos \left (d x + c\right )^{6} + 1150 \, \cos \left (d x + c\right )^{4} - 2076 \, \cos \left (d x + c\right )^{2} - 15 \, {\left (3 \, \cos \left (d x + c\right )^{6} - 4 \, \cos \left (d x + c\right )^{4} + {\left (\cos \left (d x + c\right )^{6} - 4 \, \cos \left (d x + c\right )^{4}\right )} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, {\left (3 \, \cos \left (d x + c\right )^{6} - 4 \, \cos \left (d x + c\right )^{4} + {\left (\cos \left (d x + c\right )^{6} - 4 \, \cos \left (d x + c\right )^{4}\right )} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 18 \, {\left (5 \, \cos \left (d x + c\right )^{4} + 50 \, \cos \left (d x + c\right )^{2} - 16\right )} \sin \left (d x + c\right ) + 672}{3840 \, {\left (3 \, a^{3} d \cos \left (d x + c\right )^{6} - 4 \, a^{3} d \cos \left (d x + c\right )^{4} + {\left (a^{3} d \cos \left (d x + c\right )^{6} - 4 \, a^{3} d \cos \left (d x + c\right )^{4}\right )} \sin \left (d x + c\right )\right )}} \] Input:
integrate(tan(d*x+c)^5/(a+a*sin(d*x+c))^3,x, algorithm="fricas")
Output:
-1/3840*(30*cos(d*x + c)^6 + 1150*cos(d*x + c)^4 - 2076*cos(d*x + c)^2 - 1 5*(3*cos(d*x + c)^6 - 4*cos(d*x + c)^4 + (cos(d*x + c)^6 - 4*cos(d*x + c)^ 4)*sin(d*x + c))*log(sin(d*x + c) + 1) + 15*(3*cos(d*x + c)^6 - 4*cos(d*x + c)^4 + (cos(d*x + c)^6 - 4*cos(d*x + c)^4)*sin(d*x + c))*log(-sin(d*x + c) + 1) - 18*(5*cos(d*x + c)^4 + 50*cos(d*x + c)^2 - 16)*sin(d*x + c) + 67 2)/(3*a^3*d*cos(d*x + c)^6 - 4*a^3*d*cos(d*x + c)^4 + (a^3*d*cos(d*x + c)^ 6 - 4*a^3*d*cos(d*x + c)^4)*sin(d*x + c))
\[ \int \frac {\tan ^5(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\int \frac {\tan ^{5}{\left (c + d x \right )}}{\sin ^{3}{\left (c + d x \right )} + 3 \sin ^{2}{\left (c + d x \right )} + 3 \sin {\left (c + d x \right )} + 1}\, dx}{a^{3}} \] Input:
integrate(tan(d*x+c)**5/(a+a*sin(d*x+c))**3,x)
Output:
Integral(tan(c + d*x)**5/(sin(c + d*x)**3 + 3*sin(c + d*x)**2 + 3*sin(c + d*x) + 1), x)/a**3
Time = 0.03 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.10 \[ \int \frac {\tan ^5(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{6} + 45 \, \sin \left (d x + c\right )^{5} - 620 \, \sin \left (d x + c\right )^{4} - 540 \, \sin \left (d x + c\right )^{3} + 157 \, \sin \left (d x + c\right )^{2} + 351 \, \sin \left (d x + c\right ) + 112\right )}}{a^{3} \sin \left (d x + c\right )^{7} + 3 \, a^{3} \sin \left (d x + c\right )^{6} + a^{3} \sin \left (d x + c\right )^{5} - 5 \, a^{3} \sin \left (d x + c\right )^{4} - 5 \, a^{3} \sin \left (d x + c\right )^{3} + a^{3} \sin \left (d x + c\right )^{2} + 3 \, a^{3} \sin \left (d x + c\right ) + a^{3}} - \frac {15 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}} + \frac {15 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3}}}{3840 \, d} \] Input:
integrate(tan(d*x+c)^5/(a+a*sin(d*x+c))^3,x, algorithm="maxima")
Output:
-1/3840*(2*(15*sin(d*x + c)^6 + 45*sin(d*x + c)^5 - 620*sin(d*x + c)^4 - 5 40*sin(d*x + c)^3 + 157*sin(d*x + c)^2 + 351*sin(d*x + c) + 112)/(a^3*sin( d*x + c)^7 + 3*a^3*sin(d*x + c)^6 + a^3*sin(d*x + c)^5 - 5*a^3*sin(d*x + c )^4 - 5*a^3*sin(d*x + c)^3 + a^3*sin(d*x + c)^2 + 3*a^3*sin(d*x + c) + a^3 ) - 15*log(sin(d*x + c) + 1)/a^3 + 15*log(sin(d*x + c) - 1)/a^3)/d
Time = 0.16 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.73 \[ \int \frac {\tan ^5(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{256 \, a^{3} d} - \frac {\log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{256 \, a^{3} d} - \frac {15 \, \sin \left (d x + c\right )^{6} + 45 \, \sin \left (d x + c\right )^{5} - 620 \, \sin \left (d x + c\right )^{4} - 540 \, \sin \left (d x + c\right )^{3} + 157 \, \sin \left (d x + c\right )^{2} + 351 \, \sin \left (d x + c\right ) + 112}{1920 \, a^{3} d {\left (\sin \left (d x + c\right ) + 1\right )}^{5} {\left (\sin \left (d x + c\right ) - 1\right )}^{2}} \] Input:
integrate(tan(d*x+c)^5/(a+a*sin(d*x+c))^3,x, algorithm="giac")
Output:
1/256*log(abs(sin(d*x + c) + 1))/(a^3*d) - 1/256*log(abs(sin(d*x + c) - 1) )/(a^3*d) - 1/1920*(15*sin(d*x + c)^6 + 45*sin(d*x + c)^5 - 620*sin(d*x + c)^4 - 540*sin(d*x + c)^3 + 157*sin(d*x + c)^2 + 351*sin(d*x + c) + 112)/( a^3*d*(sin(d*x + c) + 1)^5*(sin(d*x + c) - 1)^2)
Time = 20.36 (sec) , antiderivative size = 418, normalized size of antiderivative = 2.44 \[ \int \frac {\tan ^5(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{64}-\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}}{32}-\frac {17\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{96}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{32}+\frac {527\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{960}+\frac {901\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{80}+\frac {711\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{80}+\frac {901\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{80}+\frac {527\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{960}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{32}-\frac {17\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{96}-\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{32}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64}}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}+6\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}+11\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-4\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}-39\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-38\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+27\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+72\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+27\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-38\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-39\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+11\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+6\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a^3\right )}+\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{64\,a^3\,d} \] Input:
int(tan(c + d*x)^5/(a + a*sin(c + d*x))^3,x)
Output:
(tan(c/2 + (d*x)/2)^4/32 - (3*tan(c/2 + (d*x)/2)^2)/32 - (17*tan(c/2 + (d* x)/2)^3)/96 - tan(c/2 + (d*x)/2)/64 + (527*tan(c/2 + (d*x)/2)^5)/960 + (90 1*tan(c/2 + (d*x)/2)^6)/80 + (711*tan(c/2 + (d*x)/2)^7)/80 + (901*tan(c/2 + (d*x)/2)^8)/80 + (527*tan(c/2 + (d*x)/2)^9)/960 + tan(c/2 + (d*x)/2)^10/ 32 - (17*tan(c/2 + (d*x)/2)^11)/96 - (3*tan(c/2 + (d*x)/2)^12)/32 - tan(c/ 2 + (d*x)/2)^13/64)/(d*(11*a^3*tan(c/2 + (d*x)/2)^2 - 4*a^3*tan(c/2 + (d*x )/2)^3 - 39*a^3*tan(c/2 + (d*x)/2)^4 - 38*a^3*tan(c/2 + (d*x)/2)^5 + 27*a^ 3*tan(c/2 + (d*x)/2)^6 + 72*a^3*tan(c/2 + (d*x)/2)^7 + 27*a^3*tan(c/2 + (d *x)/2)^8 - 38*a^3*tan(c/2 + (d*x)/2)^9 - 39*a^3*tan(c/2 + (d*x)/2)^10 - 4* a^3*tan(c/2 + (d*x)/2)^11 + 11*a^3*tan(c/2 + (d*x)/2)^12 + 6*a^3*tan(c/2 + (d*x)/2)^13 + a^3*tan(c/2 + (d*x)/2)^14 + a^3 + 6*a^3*tan(c/2 + (d*x)/2)) ) + atanh(tan(c/2 + (d*x)/2))/(64*a^3*d)
Time = 0.18 (sec) , antiderivative size = 468, normalized size of antiderivative = 2.74 \[ \int \frac {\tan ^5(c+d x)}{(a+a \sin (c+d x))^3} \, dx =\text {Too large to display} \] Input:
int(tan(d*x+c)^5/(a+a*sin(d*x+c))^3,x)
Output:
( - 15*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**7 - 45*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**6 - 15*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**5 + 75* log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4 + 75*log(tan((c + d*x)/2) - 1)*s in(c + d*x)**3 - 15*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2 - 45*log(tan ((c + d*x)/2) - 1)*sin(c + d*x) - 15*log(tan((c + d*x)/2) - 1) + 15*log(ta n((c + d*x)/2) + 1)*sin(c + d*x)**7 + 45*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**6 + 15*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**5 - 75*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4 - 75*log(tan((c + d*x)/2) + 1)*sin(c + d*x)** 3 + 15*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2 + 45*log(tan((c + d*x)/2) + 1)*sin(c + d*x) + 15*log(tan((c + d*x)/2) + 1) + 117*sin(c + d*x)**7 + 336*sin(c + d*x)**6 + 72*sin(c + d*x)**5 + 35*sin(c + d*x)**4 - 45*sin(c + d*x)**3 - 40*sin(c + d*x)**2 + 5)/(1920*a**3*d*(sin(c + d*x)**7 + 3*sin(c + d*x)**6 + sin(c + d*x)**5 - 5*sin(c + d*x)**4 - 5*sin(c + d*x)**3 + sin (c + d*x)**2 + 3*sin(c + d*x) + 1))