Integrand size = 21, antiderivative size = 194 \[ \int \sec ^8(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {3 a^2 b \sec ^8(c+d x)}{8 d}+\frac {a^3 \tan (c+d x)}{d}+\frac {a \left (a^2+b^2\right ) \tan ^3(c+d x)}{d}+\frac {b^3 \tan ^4(c+d x)}{4 d}+\frac {3 a \left (a^2+3 b^2\right ) \tan ^5(c+d x)}{5 d}+\frac {b^3 \tan ^6(c+d x)}{2 d}+\frac {a \left (a^2+9 b^2\right ) \tan ^7(c+d x)}{7 d}+\frac {3 b^3 \tan ^8(c+d x)}{8 d}+\frac {a b^2 \tan ^9(c+d x)}{3 d}+\frac {b^3 \tan ^{10}(c+d x)}{10 d} \] Output:
3/8*a^2*b*sec(d*x+c)^8/d+a^3*tan(d*x+c)/d+a*(a^2+b^2)*tan(d*x+c)^3/d+1/4*b ^3*tan(d*x+c)^4/d+3/5*a*(a^2+3*b^2)*tan(d*x+c)^5/d+1/2*b^3*tan(d*x+c)^6/d+ 1/7*a*(a^2+9*b^2)*tan(d*x+c)^7/d+3/8*b^3*tan(d*x+c)^8/d+1/3*a*b^2*tan(d*x+ c)^9/d+1/10*b^3*tan(d*x+c)^10/d
Time = 1.35 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.91 \[ \int \sec ^8(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {\frac {1}{4} \left (a^2+b^2\right )^3 (a+b \tan (c+d x))^4-\frac {6}{5} a \left (a^2+b^2\right )^2 (a+b \tan (c+d x))^5+\frac {1}{2} \left (a^2+b^2\right ) \left (5 a^2+b^2\right ) (a+b \tan (c+d x))^6-\frac {4}{7} a \left (5 a^2+3 b^2\right ) (a+b \tan (c+d x))^7+\frac {3}{8} \left (5 a^2+b^2\right ) (a+b \tan (c+d x))^8-\frac {2}{3} a (a+b \tan (c+d x))^9+\frac {1}{10} (a+b \tan (c+d x))^{10}}{b^7 d} \] Input:
Integrate[Sec[c + d*x]^8*(a + b*Tan[c + d*x])^3,x]
Output:
(((a^2 + b^2)^3*(a + b*Tan[c + d*x])^4)/4 - (6*a*(a^2 + b^2)^2*(a + b*Tan[ c + d*x])^5)/5 + ((a^2 + b^2)*(5*a^2 + b^2)*(a + b*Tan[c + d*x])^6)/2 - (4 *a*(5*a^2 + 3*b^2)*(a + b*Tan[c + d*x])^7)/7 + (3*(5*a^2 + b^2)*(a + b*Tan [c + d*x])^8)/8 - (2*a*(a + b*Tan[c + d*x])^9)/3 + (a + b*Tan[c + d*x])^10 /10)/(b^7*d)
Time = 0.36 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 3987, 27, 475, 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.
| \(\displaystyle \int \sec ^8(c+d x) (a+b \tan (c+d x))^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sec (c+d x)^8 (a+b \tan (c+d x))^3dx\) |
| \(\Big \downarrow \) 3987 |
| \(\displaystyle \frac {\int \frac {(a+b \tan (c+d x))^3 \left (\tan ^2(c+d x) b^2+b^2\right )^3}{b^6}d(b \tan (c+d x))}{b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int (a+b \tan (c+d x))^3 \left (\tan ^2(c+d x) b^2+b^2\right )^3d(b \tan (c+d x))}{b^7 d}\) |
| \(\Big \downarrow \) 475 |
| \(\displaystyle \frac {\int \left (\tan ^9(c+d x) b^9+3 \tan ^7(c+d x) b^9+3 \tan ^5(c+d x) b^9+\tan ^3(c+d x) b^9+3 a \tan ^8(c+d x) b^8+a \left (a^2+9 b^2\right ) \tan ^6(c+d x) b^6+3 a \left (a^2+3 b^2\right ) \tan ^4(c+d x) b^6+a^3 b^6+3 a \left (a^2+b^2\right ) \tan ^2(c+d x) b^6\right )d(b \tan (c+d x))+\frac {3}{8} a^2 \left (b^2 \tan ^2(c+d x)+b^2\right )^4}{b^7 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^3 b^7 \tan (c+d x)+\frac {3}{8} a^2 \left (b^2 \tan ^2(c+d x)+b^2\right )^4+\frac {1}{7} a b^7 \left (a^2+9 b^2\right ) \tan ^7(c+d x)+\frac {3}{5} a b^7 \left (a^2+3 b^2\right ) \tan ^5(c+d x)+a b^7 \left (a^2+b^2\right ) \tan ^3(c+d x)+\frac {1}{3} a b^9 \tan ^9(c+d x)+\frac {1}{10} b^{10} \tan ^{10}(c+d x)+\frac {3}{8} b^{10} \tan ^8(c+d x)+\frac {1}{2} b^{10} \tan ^6(c+d x)+\frac {1}{4} b^{10} \tan ^4(c+d x)}{b^7 d}\) |
Input:
Int[Sec[c + d*x]^8*(a + b*Tan[c + d*x])^3,x]
Output:
(a^3*b^7*Tan[c + d*x] + a*b^7*(a^2 + b^2)*Tan[c + d*x]^3 + (b^10*Tan[c + d *x]^4)/4 + (3*a*b^7*(a^2 + 3*b^2)*Tan[c + d*x]^5)/5 + (b^10*Tan[c + d*x]^6 )/2 + (a*b^7*(a^2 + 9*b^2)*Tan[c + d*x]^7)/7 + (3*b^10*Tan[c + d*x]^8)/8 + (a*b^9*Tan[c + d*x]^9)/3 + (b^10*Tan[c + d*x]^10)/10 + (3*a^2*(b^2 + b^2* Tan[c + d*x]^2)^4)/8)/(b^7*d)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp [d*n*c^(n - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Int[ExpandIntegran d[((c + d*x)^n - d*n*c^(n - 1)*x)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[p, 0] && IGtQ[n, 0] && LeQ[n, p]
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_ ), x_Symbol] :> Simp[1/(b*f) Subst[Int[(a + x)^n*(1 + x^2/b^2)^(m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && NeQ[a^2 + b^2, 0] && IntegerQ[m/2]
Time = 188.98 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.13
| method | result | size |
| derivativedivides | \(\frac {b^{3} \left (\frac {\sin \left (d x +c \right )^{4}}{10 \cos \left (d x +c \right )^{10}}+\frac {3 \sin \left (d x +c \right )^{4}}{40 \cos \left (d x +c \right )^{8}}+\frac {\sin \left (d x +c \right )^{4}}{20 \cos \left (d x +c \right )^{6}}+\frac {\sin \left (d x +c \right )^{4}}{40 \cos \left (d x +c \right )^{4}}\right )+3 a \,b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{9 \cos \left (d x +c \right )^{9}}+\frac {2 \sin \left (d x +c \right )^{3}}{21 \cos \left (d x +c \right )^{7}}+\frac {8 \sin \left (d x +c \right )^{3}}{105 \cos \left (d x +c \right )^{5}}+\frac {16 \sin \left (d x +c \right )^{3}}{315 \cos \left (d x +c \right )^{3}}\right )+\frac {3 a^{2} b}{8 \cos \left (d x +c \right )^{8}}-a^{3} \left (-\frac {16}{35}-\frac {\sec \left (d x +c \right )^{6}}{7}-\frac {6 \sec \left (d x +c \right )^{4}}{35}-\frac {8 \sec \left (d x +c \right )^{2}}{35}\right ) \tan \left (d x +c \right )}{d}\) | \(219\) |
| default | \(\frac {b^{3} \left (\frac {\sin \left (d x +c \right )^{4}}{10 \cos \left (d x +c \right )^{10}}+\frac {3 \sin \left (d x +c \right )^{4}}{40 \cos \left (d x +c \right )^{8}}+\frac {\sin \left (d x +c \right )^{4}}{20 \cos \left (d x +c \right )^{6}}+\frac {\sin \left (d x +c \right )^{4}}{40 \cos \left (d x +c \right )^{4}}\right )+3 a \,b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{9 \cos \left (d x +c \right )^{9}}+\frac {2 \sin \left (d x +c \right )^{3}}{21 \cos \left (d x +c \right )^{7}}+\frac {8 \sin \left (d x +c \right )^{3}}{105 \cos \left (d x +c \right )^{5}}+\frac {16 \sin \left (d x +c \right )^{3}}{315 \cos \left (d x +c \right )^{3}}\right )+\frac {3 a^{2} b}{8 \cos \left (d x +c \right )^{8}}-a^{3} \left (-\frac {16}{35}-\frac {\sec \left (d x +c \right )^{6}}{7}-\frac {6 \sec \left (d x +c \right )^{4}}{35}-\frac {8 \sec \left (d x +c \right )^{2}}{35}\right ) \tan \left (d x +c \right )}{d}\) | \(219\) |
| risch | \(-\frac {32 \left (-30 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-105 i a \,b^{2} {\mathrm e}^{8 i \left (d x +c \right )}-315 a^{2} b \,{\mathrm e}^{12 i \left (d x +c \right )}+105 b^{3} {\mathrm e}^{12 i \left (d x +c \right )}-135 i a^{3} {\mathrm e}^{4 i \left (d x +c \right )}-3 i a^{3}-630 a^{2} b \,{\mathrm e}^{10 i \left (d x +c \right )}-126 b^{3} {\mathrm e}^{10 i \left (d x +c \right )}-525 i a^{3} {\mathrm e}^{8 i \left (d x +c \right )}+45 i a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-315 a^{2} b \,{\mathrm e}^{8 i \left (d x +c \right )}+105 b^{3} {\mathrm e}^{8 i \left (d x +c \right )}+315 i a \,b^{2} {\mathrm e}^{12 i \left (d x +c \right )}+i a \,b^{2}-378 i a^{3} {\mathrm e}^{10 i \left (d x +c \right )}-105 i a^{3} {\mathrm e}^{12 i \left (d x +c \right )}+120 i a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-360 i a^{3} {\mathrm e}^{6 i \left (d x +c \right )}+126 i a \,b^{2} {\mathrm e}^{10 i \left (d x +c \right )}+10 i a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}\right )}{105 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{10}}\) | \(306\) |
Input:
int(sec(d*x+c)^8*(a+b*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/d*(b^3*(1/10*sin(d*x+c)^4/cos(d*x+c)^10+3/40*sin(d*x+c)^4/cos(d*x+c)^8+1 /20*sin(d*x+c)^4/cos(d*x+c)^6+1/40*sin(d*x+c)^4/cos(d*x+c)^4)+3*a*b^2*(1/9 *sin(d*x+c)^3/cos(d*x+c)^9+2/21*sin(d*x+c)^3/cos(d*x+c)^7+8/105*sin(d*x+c) ^3/cos(d*x+c)^5+16/315*sin(d*x+c)^3/cos(d*x+c)^3)+3/8*a^2*b/cos(d*x+c)^8-a ^3*(-16/35-1/7*sec(d*x+c)^6-6/35*sec(d*x+c)^4-8/35*sec(d*x+c)^2)*tan(d*x+c ))
Time = 0.10 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.77 \[ \int \sec ^8(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {84 \, b^{3} + 105 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} + 8 \, {\left (16 \, {\left (3 \, a^{3} - a b^{2}\right )} \cos \left (d x + c\right )^{9} + 8 \, {\left (3 \, a^{3} - a b^{2}\right )} \cos \left (d x + c\right )^{7} + 6 \, {\left (3 \, a^{3} - a b^{2}\right )} \cos \left (d x + c\right )^{5} + 35 \, a b^{2} \cos \left (d x + c\right ) + 5 \, {\left (3 \, a^{3} - a b^{2}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{840 \, d \cos \left (d x + c\right )^{10}} \] Input:
integrate(sec(d*x+c)^8*(a+b*tan(d*x+c))^3,x, algorithm="fricas")
Output:
1/840*(84*b^3 + 105*(3*a^2*b - b^3)*cos(d*x + c)^2 + 8*(16*(3*a^3 - a*b^2) *cos(d*x + c)^9 + 8*(3*a^3 - a*b^2)*cos(d*x + c)^7 + 6*(3*a^3 - a*b^2)*cos (d*x + c)^5 + 35*a*b^2*cos(d*x + c) + 5*(3*a^3 - a*b^2)*cos(d*x + c)^3)*si n(d*x + c))/(d*cos(d*x + c)^10)
\[ \int \sec ^8(c+d x) (a+b \tan (c+d x))^3 \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right )^{3} \sec ^{8}{\left (c + d x \right )}\, dx \] Input:
integrate(sec(d*x+c)**8*(a+b*tan(d*x+c))**3,x)
Output:
Integral((a + b*tan(c + d*x))**3*sec(c + d*x)**8, x)
Time = 0.04 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.91 \[ \int \sec ^8(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {84 \, b^{3} \tan \left (d x + c\right )^{10} + 280 \, a b^{2} \tan \left (d x + c\right )^{9} + 315 \, {\left (a^{2} b + b^{3}\right )} \tan \left (d x + c\right )^{8} + 120 \, {\left (a^{3} + 9 \, a b^{2}\right )} \tan \left (d x + c\right )^{7} + 420 \, {\left (3 \, a^{2} b + b^{3}\right )} \tan \left (d x + c\right )^{6} + 504 \, {\left (a^{3} + 3 \, a b^{2}\right )} \tan \left (d x + c\right )^{5} + 1260 \, a^{2} b \tan \left (d x + c\right )^{2} + 210 \, {\left (9 \, a^{2} b + b^{3}\right )} \tan \left (d x + c\right )^{4} + 840 \, a^{3} \tan \left (d x + c\right ) + 840 \, {\left (a^{3} + a b^{2}\right )} \tan \left (d x + c\right )^{3}}{840 \, d} \] Input:
integrate(sec(d*x+c)^8*(a+b*tan(d*x+c))^3,x, algorithm="maxima")
Output:
1/840*(84*b^3*tan(d*x + c)^10 + 280*a*b^2*tan(d*x + c)^9 + 315*(a^2*b + b^ 3)*tan(d*x + c)^8 + 120*(a^3 + 9*a*b^2)*tan(d*x + c)^7 + 420*(3*a^2*b + b^ 3)*tan(d*x + c)^6 + 504*(a^3 + 3*a*b^2)*tan(d*x + c)^5 + 1260*a^2*b*tan(d* x + c)^2 + 210*(9*a^2*b + b^3)*tan(d*x + c)^4 + 840*a^3*tan(d*x + c) + 840 *(a^3 + a*b^2)*tan(d*x + c)^3)/d
Time = 0.27 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.13 \[ \int \sec ^8(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {84 \, b^{3} \tan \left (d x + c\right )^{10} + 280 \, a b^{2} \tan \left (d x + c\right )^{9} + 315 \, a^{2} b \tan \left (d x + c\right )^{8} + 315 \, b^{3} \tan \left (d x + c\right )^{8} + 120 \, a^{3} \tan \left (d x + c\right )^{7} + 1080 \, a b^{2} \tan \left (d x + c\right )^{7} + 1260 \, a^{2} b \tan \left (d x + c\right )^{6} + 420 \, b^{3} \tan \left (d x + c\right )^{6} + 504 \, a^{3} \tan \left (d x + c\right )^{5} + 1512 \, a b^{2} \tan \left (d x + c\right )^{5} + 1890 \, a^{2} b \tan \left (d x + c\right )^{4} + 210 \, b^{3} \tan \left (d x + c\right )^{4} + 840 \, a^{3} \tan \left (d x + c\right )^{3} + 840 \, a b^{2} \tan \left (d x + c\right )^{3} + 1260 \, a^{2} b \tan \left (d x + c\right )^{2} + 840 \, a^{3} \tan \left (d x + c\right )}{840 \, d} \] Input:
integrate(sec(d*x+c)^8*(a+b*tan(d*x+c))^3,x, algorithm="giac")
Output:
1/840*(84*b^3*tan(d*x + c)^10 + 280*a*b^2*tan(d*x + c)^9 + 315*a^2*b*tan(d *x + c)^8 + 315*b^3*tan(d*x + c)^8 + 120*a^3*tan(d*x + c)^7 + 1080*a*b^2*t an(d*x + c)^7 + 1260*a^2*b*tan(d*x + c)^6 + 420*b^3*tan(d*x + c)^6 + 504*a ^3*tan(d*x + c)^5 + 1512*a*b^2*tan(d*x + c)^5 + 1890*a^2*b*tan(d*x + c)^4 + 210*b^3*tan(d*x + c)^4 + 840*a^3*tan(d*x + c)^3 + 840*a*b^2*tan(d*x + c) ^3 + 1260*a^2*b*tan(d*x + c)^2 + 840*a^3*tan(d*x + c))/d
Time = 0.75 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.90 \[ \int \sec ^8(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {{\mathrm {tan}\left (c+d\,x\right )}^5\,\left (\frac {3\,a^3}{5}+\frac {9\,a\,b^2}{5}\right )+{\mathrm {tan}\left (c+d\,x\right )}^7\,\left (\frac {a^3}{7}+\frac {9\,a\,b^2}{7}\right )+{\mathrm {tan}\left (c+d\,x\right )}^6\,\left (\frac {3\,a^2\,b}{2}+\frac {b^3}{2}\right )+{\mathrm {tan}\left (c+d\,x\right )}^4\,\left (\frac {9\,a^2\,b}{4}+\frac {b^3}{4}\right )+a^3\,\mathrm {tan}\left (c+d\,x\right )+\frac {b^3\,{\mathrm {tan}\left (c+d\,x\right )}^{10}}{10}+\frac {3\,a^2\,b\,{\mathrm {tan}\left (c+d\,x\right )}^2}{2}+\frac {a\,b^2\,{\mathrm {tan}\left (c+d\,x\right )}^9}{3}+a\,{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (a^2+b^2\right )+\frac {3\,b\,{\mathrm {tan}\left (c+d\,x\right )}^8\,\left (a^2+b^2\right )}{8}}{d} \] Input:
int((a + b*tan(c + d*x))^3/cos(c + d*x)^8,x)
Output:
(tan(c + d*x)^5*((9*a*b^2)/5 + (3*a^3)/5) + tan(c + d*x)^7*((9*a*b^2)/7 + a^3/7) + tan(c + d*x)^6*((3*a^2*b)/2 + b^3/2) + tan(c + d*x)^4*((9*a^2*b)/ 4 + b^3/4) + a^3*tan(c + d*x) + (b^3*tan(c + d*x)^10)/10 + (3*a^2*b*tan(c + d*x)^2)/2 + (a*b^2*tan(c + d*x)^9)/3 + a*tan(c + d*x)^3*(a^2 + b^2) + (3 *b*tan(c + d*x)^8*(a^2 + b^2))/8)/d
Time = 0.17 (sec) , antiderivative size = 351, normalized size of antiderivative = 1.81 \[ \int \sec ^8(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {\sin \left (d x +c \right ) \left (-384 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{8} a^{3}+128 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{8} a \,b^{2}+1728 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{6} a^{3}-576 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{6} a \,b^{2}-3024 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} a^{3}+1008 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} a \,b^{2}+2520 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{3}-840 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a \,b^{2}-840 \cos \left (d x +c \right ) a^{3}-315 \sin \left (d x +c \right )^{9} a^{2} b +21 \sin \left (d x +c \right )^{9} b^{3}+1575 \sin \left (d x +c \right )^{7} a^{2} b -105 \sin \left (d x +c \right )^{7} b^{3}-3150 \sin \left (d x +c \right )^{5} a^{2} b +210 \sin \left (d x +c \right )^{5} b^{3}+3150 \sin \left (d x +c \right )^{3} a^{2} b -210 \sin \left (d x +c \right )^{3} b^{3}-1260 \sin \left (d x +c \right ) a^{2} b \right )}{840 d \left (\sin \left (d x +c \right )^{10}-5 \sin \left (d x +c \right )^{8}+10 \sin \left (d x +c \right )^{6}-10 \sin \left (d x +c \right )^{4}+5 \sin \left (d x +c \right )^{2}-1\right )} \] Input:
int(sec(d*x+c)^8*(a+b*tan(d*x+c))^3,x)
Output:
(sin(c + d*x)*( - 384*cos(c + d*x)*sin(c + d*x)**8*a**3 + 128*cos(c + d*x) *sin(c + d*x)**8*a*b**2 + 1728*cos(c + d*x)*sin(c + d*x)**6*a**3 - 576*cos (c + d*x)*sin(c + d*x)**6*a*b**2 - 3024*cos(c + d*x)*sin(c + d*x)**4*a**3 + 1008*cos(c + d*x)*sin(c + d*x)**4*a*b**2 + 2520*cos(c + d*x)*sin(c + d*x )**2*a**3 - 840*cos(c + d*x)*sin(c + d*x)**2*a*b**2 - 840*cos(c + d*x)*a** 3 - 315*sin(c + d*x)**9*a**2*b + 21*sin(c + d*x)**9*b**3 + 1575*sin(c + d* x)**7*a**2*b - 105*sin(c + d*x)**7*b**3 - 3150*sin(c + d*x)**5*a**2*b + 21 0*sin(c + d*x)**5*b**3 + 3150*sin(c + d*x)**3*a**2*b - 210*sin(c + d*x)**3 *b**3 - 1260*sin(c + d*x)*a**2*b))/(840*d*(sin(c + d*x)**10 - 5*sin(c + d* x)**8 + 10*sin(c + d*x)**6 - 10*sin(c + d*x)**4 + 5*sin(c + d*x)**2 - 1))