Integrand size = 21, antiderivative size = 106 \[ \int \frac {\tan ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {5 \text {arctanh}(\sin (c+d x))}{16 a d}-\frac {5 \sec (c+d x) \tan (c+d x)}{16 a d}+\frac {5 \sec (c+d x) \tan ^3(c+d x)}{24 a d}-\frac {\sec (c+d x) \tan ^5(c+d x)}{6 a d}+\frac {\tan ^6(c+d x)}{6 a d} \] Output:
5/16*arctanh(sin(d*x+c))/a/d-5/16*sec(d*x+c)*tan(d*x+c)/a/d+5/24*sec(d*x+c )*tan(d*x+c)^3/a/d-1/6*sec(d*x+c)*tan(d*x+c)^5/a/d+1/6*tan(d*x+c)^6/a/d
Time = 0.37 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.79 \[ \int \frac {\tan ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {30 \text {arctanh}(\sin (c+d x))+\frac {3}{(1-\sin (c+d x))^2}-\frac {18}{1-\sin (c+d x)}+\frac {4}{(1+\sin (c+d x))^3}-\frac {21}{(1+\sin (c+d x))^2}+\frac {48}{1+\sin (c+d x)}}{96 a d} \] Input:
Integrate[Tan[c + d*x]^5/(a + a*Sin[c + d*x]),x]
Output:
(30*ArcTanh[Sin[c + d*x]] + 3/(1 - Sin[c + d*x])^2 - 18/(1 - Sin[c + d*x]) + 4/(1 + Sin[c + d*x])^3 - 21/(1 + Sin[c + d*x])^2 + 48/(1 + Sin[c + d*x] ))/(96*a*d)
Time = 0.62 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.04, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 3185, 3042, 3087, 15, 3091, 3042, 3091, 3042, 3091, 3042, 4257}
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.
| \(\displaystyle \int \frac {\tan ^5(c+d x)}{a \sin (c+d x)+a} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (c+d x)^5}{a \sin (c+d x)+a}dx\) |
| \(\Big \downarrow \) 3185 |
| \(\displaystyle \frac {\int \sec ^2(c+d x) \tan ^5(c+d x)dx}{a}-\frac {\int \sec (c+d x) \tan ^6(c+d x)dx}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \sec (c+d x)^2 \tan (c+d x)^5dx}{a}-\frac {\int \sec (c+d x) \tan (c+d x)^6dx}{a}\) |
| \(\Big \downarrow \) 3087 |
| \(\displaystyle \frac {\int \tan ^5(c+d x)d\tan (c+d x)}{a d}-\frac {\int \sec (c+d x) \tan (c+d x)^6dx}{a}\) |
| \(\Big \downarrow \) 15 |
| \(\displaystyle \frac {\tan ^6(c+d x)}{6 a d}-\frac {\int \sec (c+d x) \tan (c+d x)^6dx}{a}\) |
| \(\Big \downarrow \) 3091 |
| \(\displaystyle \frac {\tan ^6(c+d x)}{6 a d}-\frac {\frac {\tan ^5(c+d x) \sec (c+d x)}{6 d}-\frac {5}{6} \int \sec (c+d x) \tan ^4(c+d x)dx}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\tan ^6(c+d x)}{6 a d}-\frac {\frac {\tan ^5(c+d x) \sec (c+d x)}{6 d}-\frac {5}{6} \int \sec (c+d x) \tan (c+d x)^4dx}{a}\) |
| \(\Big \downarrow \) 3091 |
| \(\displaystyle \frac {\tan ^6(c+d x)}{6 a d}-\frac {\frac {\tan ^5(c+d x) \sec (c+d x)}{6 d}-\frac {5}{6} \left (\frac {\tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {3}{4} \int \sec (c+d x) \tan ^2(c+d x)dx\right )}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\tan ^6(c+d x)}{6 a d}-\frac {\frac {\tan ^5(c+d x) \sec (c+d x)}{6 d}-\frac {5}{6} \left (\frac {\tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {3}{4} \int \sec (c+d x) \tan (c+d x)^2dx\right )}{a}\) |
| \(\Big \downarrow \) 3091 |
| \(\displaystyle \frac {\tan ^6(c+d x)}{6 a d}-\frac {\frac {\tan ^5(c+d x) \sec (c+d x)}{6 d}-\frac {5}{6} \left (\frac {\tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {3}{4} \left (\frac {\tan (c+d x) \sec (c+d x)}{2 d}-\frac {1}{2} \int \sec (c+d x)dx\right )\right )}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\tan ^6(c+d x)}{6 a d}-\frac {\frac {\tan ^5(c+d x) \sec (c+d x)}{6 d}-\frac {5}{6} \left (\frac {\tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {3}{4} \left (\frac {\tan (c+d x) \sec (c+d x)}{2 d}-\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx\right )\right )}{a}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\tan ^6(c+d x)}{6 a d}-\frac {\frac {\tan ^5(c+d x) \sec (c+d x)}{6 d}-\frac {5}{6} \left (\frac {\tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {3}{4} \left (\frac {\tan (c+d x) \sec (c+d x)}{2 d}-\frac {\text {arctanh}(\sin (c+d x))}{2 d}\right )\right )}{a}\) |
Input:
Int[Tan[c + d*x]^5/(a + a*Sin[c + d*x]),x]
Output:
Tan[c + d*x]^6/(6*a*d) - ((Sec[c + d*x]*Tan[c + d*x]^5)/(6*d) - (5*((Sec[c + d*x]*Tan[c + d*x]^3)/(4*d) - (3*(-1/2*ArcTanh[Sin[c + d*x]]/d + (Sec[c + d*x]*Tan[c + d*x])/(2*d)))/4))/6)/a
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_S ymbol] :> Simp[1/f Subst[Int[(b*x)^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] && !(IntegerQ[(n - 1) /2] && LtQ[0, n, m - 1])
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_), x_Symbol] :> Simp[b*(a*Sec[e + f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] - Simp[b^2*((n - 1)/(m + n - 1)) Int[(a*Sec[e + f*x])^m*( b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] & & NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]
Int[((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*( x_)]), x_Symbol] :> Simp[1/a Int[Sec[e + f*x]^2*(g*Tan[e + f*x])^p, x], x ] - Simp[1/(b*g) Int[Sec[e + f*x]*(g*Tan[e + f*x])^(p + 1), x], x] /; Fre eQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[p, -1]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 0.73 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.86
| method | result | size |
| derivativedivides | \(\frac {\frac {1}{24 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {7}{32 \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {1}{2+2 \sin \left (d x +c \right )}+\frac {5 \ln \left (1+\sin \left (d x +c \right )\right )}{32}+\frac {1}{32 \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {3}{16 \left (\sin \left (d x +c \right )-1\right )}-\frac {5 \ln \left (\sin \left (d x +c \right )-1\right )}{32}}{d a}\) | \(91\) |
| default | \(\frac {\frac {1}{24 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {7}{32 \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {1}{2+2 \sin \left (d x +c \right )}+\frac {5 \ln \left (1+\sin \left (d x +c \right )\right )}{32}+\frac {1}{32 \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {3}{16 \left (\sin \left (d x +c \right )-1\right )}-\frac {5 \ln \left (\sin \left (d x +c \right )-1\right )}{32}}{d a}\) | \(91\) |
| risch | \(\frac {i \left (-8 \,{\mathrm e}^{7 i \left (d x +c \right )}+2 i {\mathrm e}^{6 i \left (d x +c \right )}+78 \,{\mathrm e}^{5 i \left (d x +c \right )}+18 i {\mathrm e}^{8 i \left (d x +c \right )}+33 \,{\mathrm e}^{9 i \left (d x +c \right )}-2 i {\mathrm e}^{4 i \left (d x +c \right )}-8 \,{\mathrm e}^{3 i \left (d x +c \right )}-18 i {\mathrm e}^{2 i \left (d x +c \right )}+33 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{24 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{6} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{4} d a}-\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{16 a d}+\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{16 a d}\) | \(185\) |
Input:
int(tan(d*x+c)^5/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d/a*(1/24/(1+sin(d*x+c))^3-7/32/(1+sin(d*x+c))^2+1/2/(1+sin(d*x+c))+5/32 *ln(1+sin(d*x+c))+1/32/(sin(d*x+c)-1)^2+3/16/(sin(d*x+c)-1)-5/32*ln(sin(d* x+c)-1))
Time = 0.12 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.39 \[ \int \frac {\tan ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {66 \, \cos \left (d x + c\right )^{4} - 70 \, \cos \left (d x + c\right )^{2} + 15 \, {\left (\cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{4}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (\cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{4}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (9 \, \cos \left (d x + c\right )^{2} - 2\right )} \sin \left (d x + c\right ) + 20}{96 \, {\left (a d \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{4}\right )}} \] Input:
integrate(tan(d*x+c)^5/(a+a*sin(d*x+c)),x, algorithm="fricas")
Output:
1/96*(66*cos(d*x + c)^4 - 70*cos(d*x + c)^2 + 15*(cos(d*x + c)^4*sin(d*x + c) + cos(d*x + c)^4)*log(sin(d*x + c) + 1) - 15*(cos(d*x + c)^4*sin(d*x + c) + cos(d*x + c)^4)*log(-sin(d*x + c) + 1) - 2*(9*cos(d*x + c)^2 - 2)*si n(d*x + c) + 20)/(a*d*cos(d*x + c)^4*sin(d*x + c) + a*d*cos(d*x + c)^4)
\[ \int \frac {\tan ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {\tan ^{5}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \] Input:
integrate(tan(d*x+c)**5/(a+a*sin(d*x+c)),x)
Output:
Integral(tan(c + d*x)**5/(sin(c + d*x) + 1), x)/a
Time = 0.03 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.23 \[ \int \frac {\tan ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {2 \, {\left (33 \, \sin \left (d x + c\right )^{4} + 9 \, \sin \left (d x + c\right )^{3} - 31 \, \sin \left (d x + c\right )^{2} - 7 \, \sin \left (d x + c\right ) + 8\right )}}{a \sin \left (d x + c\right )^{5} + a \sin \left (d x + c\right )^{4} - 2 \, a \sin \left (d x + c\right )^{3} - 2 \, a \sin \left (d x + c\right )^{2} + a \sin \left (d x + c\right ) + a} + \frac {15 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} - \frac {15 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a}}{96 \, d} \] Input:
integrate(tan(d*x+c)^5/(a+a*sin(d*x+c)),x, algorithm="maxima")
Output:
1/96*(2*(33*sin(d*x + c)^4 + 9*sin(d*x + c)^3 - 31*sin(d*x + c)^2 - 7*sin( d*x + c) + 8)/(a*sin(d*x + c)^5 + a*sin(d*x + c)^4 - 2*a*sin(d*x + c)^3 - 2*a*sin(d*x + c)^2 + a*sin(d*x + c) + a) + 15*log(sin(d*x + c) + 1)/a - 15 *log(sin(d*x + c) - 1)/a)/d
Time = 0.15 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.99 \[ \int \frac {\tan ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {5 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{32 \, a d} - \frac {5 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{32 \, a d} + \frac {33 \, \sin \left (d x + c\right )^{4} + 9 \, \sin \left (d x + c\right )^{3} - 31 \, \sin \left (d x + c\right )^{2} - 7 \, \sin \left (d x + c\right ) + 8}{48 \, a d {\left (\sin \left (d x + c\right ) + 1\right )}^{3} {\left (\sin \left (d x + c\right ) - 1\right )}^{2}} \] Input:
integrate(tan(d*x+c)^5/(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
5/32*log(abs(sin(d*x + c) + 1))/(a*d) - 5/32*log(abs(sin(d*x + c) - 1))/(a *d) + 1/48*(33*sin(d*x + c)^4 + 9*sin(d*x + c)^3 - 31*sin(d*x + c)^2 - 7*s in(d*x + c) + 8)/(a*d*(sin(d*x + c) + 1)^3*(sin(d*x + c) - 1)^2)
Time = 20.22 (sec) , antiderivative size = 281, normalized size of antiderivative = 2.65 \[ \int \frac {\tan ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {5\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{8\,a\,d}-\frac {\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{8}+\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{4}-\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{3}-\frac {55\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{12}+\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}-\frac {55\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{12}-\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4}+\frac {5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-8\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+12\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-8\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )} \] Input:
int(tan(c + d*x)^5/(a + a*sin(c + d*x)),x)
Output:
(5*atanh(tan(c/2 + (d*x)/2)))/(8*a*d) - ((5*tan(c/2 + (d*x)/2))/8 + (5*tan (c/2 + (d*x)/2)^2)/4 - (5*tan(c/2 + (d*x)/2)^3)/3 - (55*tan(c/2 + (d*x)/2) ^4)/12 + (3*tan(c/2 + (d*x)/2)^5)/4 - (55*tan(c/2 + (d*x)/2)^6)/12 - (5*ta n(c/2 + (d*x)/2)^7)/3 + (5*tan(c/2 + (d*x)/2)^8)/4 + (5*tan(c/2 + (d*x)/2) ^9)/8)/(d*(a + 2*a*tan(c/2 + (d*x)/2) - 3*a*tan(c/2 + (d*x)/2)^2 - 8*a*tan (c/2 + (d*x)/2)^3 + 2*a*tan(c/2 + (d*x)/2)^4 + 12*a*tan(c/2 + (d*x)/2)^5 + 2*a*tan(c/2 + (d*x)/2)^6 - 8*a*tan(c/2 + (d*x)/2)^7 - 3*a*tan(c/2 + (d*x) /2)^8 + 2*a*tan(c/2 + (d*x)/2)^9 + a*tan(c/2 + (d*x)/2)^10))
Time = 0.20 (sec) , antiderivative size = 340, normalized size of antiderivative = 3.21 \[ \int \frac {\tan ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {-15 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{5}-15 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4}+30 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{3}+30 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2}-15 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )-15 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+15 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{5}+15 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4}-30 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{3}-30 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2}+15 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )+15 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+7 \sin \left (d x +c \right )^{5}+40 \sin \left (d x +c \right )^{4}-5 \sin \left (d x +c \right )^{3}-45 \sin \left (d x +c \right )^{2}+15}{48 a d \left (\sin \left (d x +c \right )^{5}+\sin \left (d x +c \right )^{4}-2 \sin \left (d x +c \right )^{3}-2 \sin \left (d x +c \right )^{2}+\sin \left (d x +c \right )+1\right )} \] Input:
int(tan(d*x+c)^5/(a+a*sin(d*x+c)),x)
Output:
( - 15*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**5 - 15*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4 + 30*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**3 + 30* log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2 - 15*log(tan((c + d*x)/2) - 1)*s in(c + d*x) - 15*log(tan((c + d*x)/2) - 1) + 15*log(tan((c + d*x)/2) + 1)* sin(c + d*x)**5 + 15*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4 - 30*log(ta n((c + d*x)/2) + 1)*sin(c + d*x)**3 - 30*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2 + 15*log(tan((c + d*x)/2) + 1)*sin(c + d*x) + 15*log(tan((c + d*x )/2) + 1) + 7*sin(c + d*x)**5 + 40*sin(c + d*x)**4 - 5*sin(c + d*x)**3 - 4 5*sin(c + d*x)**2 + 15)/(48*a*d*(sin(c + d*x)**5 + sin(c + d*x)**4 - 2*sin (c + d*x)**3 - 2*sin(c + d*x)**2 + sin(c + d*x) + 1))