Integrand size = 27, antiderivative size = 93 \[ \int \frac {\sec ^3(c+d x) \tan (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\sec ^3(c+d x)}{7 d (a+a \sin (c+d x))^2}-\frac {2 \sec ^3(c+d x)}{35 d \left (a^2+a^2 \sin (c+d x)\right )}+\frac {8 \tan (c+d x)}{35 a^2 d}+\frac {8 \tan ^3(c+d x)}{105 a^2 d} \] Output:
1/7*sec(d*x+c)^3/d/(a+a*sin(d*x+c))^2-2/35*sec(d*x+c)^3/d/(a^2+a^2*sin(d*x +c))+8/35*tan(d*x+c)/a^2/d+8/105*tan(d*x+c)^3/a^2/d
Time = 1.18 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.44 \[ \int \frac {\sec ^3(c+d x) \tan (c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\sec ^3(c+d x) \left (-84+\frac {21}{4} \cos (c+d x)+32 \cos (2 (c+d x))+\frac {9}{8} \cos (3 (c+d x))+16 \cos (4 (c+d x))-\frac {3}{8} \cos (5 (c+d x))-56 \sin (c+d x)+3 \sin (2 (c+d x))-12 \sin (3 (c+d x))+\frac {3}{2} \sin (4 (c+d x))+4 \sin (5 (c+d x))\right )}{420 a^2 d (1+\sin (c+d x))^2} \] Input:
Integrate[(Sec[c + d*x]^3*Tan[c + d*x])/(a + a*Sin[c + d*x])^2,x]
Output:
-1/420*(Sec[c + d*x]^3*(-84 + (21*Cos[c + d*x])/4 + 32*Cos[2*(c + d*x)] + (9*Cos[3*(c + d*x)])/8 + 16*Cos[4*(c + d*x)] - (3*Cos[5*(c + d*x)])/8 - 56 *Sin[c + d*x] + 3*Sin[2*(c + d*x)] - 12*Sin[3*(c + d*x)] + (3*Sin[4*(c + d *x)])/2 + 4*Sin[5*(c + d*x)]))/(a^2*d*(1 + Sin[c + d*x])^2)
Time = 0.50 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {3042, 3338, 3042, 3151, 3042, 4254, 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.
| \(\displaystyle \int \frac {\tan (c+d x) \sec ^3(c+d x)}{(a \sin (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)}{\cos (c+d x)^4 (a \sin (c+d x)+a)^2}dx\) |
| \(\Big \downarrow \) 3338 |
| \(\displaystyle \frac {2 \int \frac {\sec ^4(c+d x)}{\sin (c+d x) a+a}dx}{7 a}+\frac {\sec ^3(c+d x)}{7 d (a \sin (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 \int \frac {1}{\cos (c+d x)^4 (\sin (c+d x) a+a)}dx}{7 a}+\frac {\sec ^3(c+d x)}{7 d (a \sin (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3151 |
| \(\displaystyle \frac {2 \left (\frac {4 \int \sec ^4(c+d x)dx}{5 a}-\frac {\sec ^3(c+d x)}{5 d (a \sin (c+d x)+a)}\right )}{7 a}+\frac {\sec ^3(c+d x)}{7 d (a \sin (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 \left (\frac {4 \int \csc \left (c+d x+\frac {\pi }{2}\right )^4dx}{5 a}-\frac {\sec ^3(c+d x)}{5 d (a \sin (c+d x)+a)}\right )}{7 a}+\frac {\sec ^3(c+d x)}{7 d (a \sin (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {2 \left (-\frac {4 \int \left (\tan ^2(c+d x)+1\right )d(-\tan (c+d x))}{5 a d}-\frac {\sec ^3(c+d x)}{5 d (a \sin (c+d x)+a)}\right )}{7 a}+\frac {\sec ^3(c+d x)}{7 d (a \sin (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sec ^3(c+d x)}{7 d (a \sin (c+d x)+a)^2}+\frac {2 \left (-\frac {4 \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{5 a d}-\frac {\sec ^3(c+d x)}{5 d (a \sin (c+d x)+a)}\right )}{7 a}\) |
Input:
Int[(Sec[c + d*x]^3*Tan[c + d*x])/(a + a*Sin[c + d*x])^2,x]
Output:
Sec[c + d*x]^3/(7*d*(a + a*Sin[c + d*x])^2) + (2*(-1/5*Sec[c + d*x]^3/(d*( a + a*Sin[c + d*x])) - (4*(-Tan[c + d*x] - Tan[c + d*x]^3/3))/(5*a*d)))/(7 *a)
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[b*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x ])^m/(a*f*g*Simplify[2*m + p + 1])), x] + Simp[Simplify[m + p + 1]/(a*Simpl ify[2*m + p + 1]) Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1), x] , x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[Simpli fy[m + p + 1], 0] && NeQ[2*m + p + 1, 0] && !IGtQ[m, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(a*f*g*(2*m + p + 1) )), x] + Simp[(a*d*m + b*c*(m + p + 1))/(a*b*(2*m + p + 1)) Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && (LtQ[m, -1] || ILtQ[Simplify[m + p], 0 ]) && NeQ[2*m + p + 1, 0]
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Result contains complex when optimal does not.
Time = 7.42 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.04
| method | result | size |
| risch | \(-\frac {32 \left (14 i {\mathrm e}^{4 i \left (d x +c \right )}+21 \,{\mathrm e}^{5 i \left (d x +c \right )}+3 i {\mathrm e}^{2 i \left (d x +c \right )}-8 \,{\mathrm e}^{3 i \left (d x +c \right )}-i-4 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{105 \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{7} d \,a^{2}}\) | \(97\) |
| derivativedivides | \(\frac {-\frac {1}{12 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {4}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}-\frac {2}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}+\frac {18}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {35}{12 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {11}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {4}{16 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+16}}{d \,a^{2}}\) | \(160\) |
| default | \(\frac {-\frac {1}{12 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {4}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}-\frac {2}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}+\frac {18}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {35}{12 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {11}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {4}{16 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+16}}{d \,a^{2}}\) | \(160\) |
Input:
int(sec(d*x+c)^3*tan(d*x+c)/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
-32/105*(14*I*exp(4*I*(d*x+c))+21*exp(5*I*(d*x+c))+3*I*exp(2*I*(d*x+c))-8* exp(3*I*(d*x+c))-I-4*exp(I*(d*x+c)))/(exp(I*(d*x+c))-I)^3/(exp(I*(d*x+c))+ I)^7/d/a^2
Time = 0.08 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.12 \[ \int \frac {\sec ^3(c+d x) \tan (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {32 \, \cos \left (d x + c\right )^{4} - 16 \, \cos \left (d x + c\right )^{2} + 2 \, {\left (8 \, \cos \left (d x + c\right )^{4} - 12 \, \cos \left (d x + c\right )^{2} - 5\right )} \sin \left (d x + c\right ) - 25}{105 \, {\left (a^{2} d \cos \left (d x + c\right )^{5} - 2 \, a^{2} d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) - 2 \, a^{2} d \cos \left (d x + c\right )^{3}\right )}} \] Input:
integrate(sec(d*x+c)^3*tan(d*x+c)/(a+a*sin(d*x+c))^2,x, algorithm="fricas" )
Output:
1/105*(32*cos(d*x + c)^4 - 16*cos(d*x + c)^2 + 2*(8*cos(d*x + c)^4 - 12*co s(d*x + c)^2 - 5)*sin(d*x + c) - 25)/(a^2*d*cos(d*x + c)^5 - 2*a^2*d*cos(d *x + c)^3*sin(d*x + c) - 2*a^2*d*cos(d*x + c)^3)
\[ \int \frac {\sec ^3(c+d x) \tan (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\int \frac {\tan {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin {\left (c + d x \right )} + 1}\, dx}{a^{2}} \] Input:
integrate(sec(d*x+c)**3*tan(d*x+c)/(a+a*sin(d*x+c))**2,x)
Output:
Integral(tan(c + d*x)*sec(c + d*x)**3/(sin(c + d*x)**2 + 2*sin(c + d*x) + 1), x)/a**2
Leaf count of result is larger than twice the leaf count of optimal. 376 vs. \(2 (85) = 170\).
Time = 0.04 (sec) , antiderivative size = 376, normalized size of antiderivative = 4.04 \[ \int \frac {\sec ^3(c+d x) \tan (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {2 \, {\left (\frac {36 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {132 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {68 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {14 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {84 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {140 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {140 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {105 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + 9\right )}}{105 \, {\left (a^{2} + \frac {4 \, a^{2} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {3 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {8 \, a^{2} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {14 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {14 \, a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {8 \, a^{2} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {3 \, a^{2} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {4 \, a^{2} \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac {a^{2} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}}\right )} d} \] Input:
integrate(sec(d*x+c)^3*tan(d*x+c)/(a+a*sin(d*x+c))^2,x, algorithm="maxima" )
Output:
2/105*(36*sin(d*x + c)/(cos(d*x + c) + 1) + 132*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 68*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 14*sin(d*x + c)^4/(co s(d*x + c) + 1)^4 - 84*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 140*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 140*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 105 *sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + 9)/((a^2 + 4*a^2*sin(d*x + c)/(cos( d*x + c) + 1) + 3*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 8*a^2*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 14*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 14*a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 8*a^2*sin(d*x + c)^7/(cos(d* x + c) + 1)^7 - 3*a^2*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 4*a^2*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - a^2*sin(d*x + c)^10/(cos(d*x + c) + 1)^10)*d )
Time = 0.36 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.57 \[ \int \frac {\sec ^3(c+d x) \tan (c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {35 \, {\left (6 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 9 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5\right )}}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}} - \frac {210 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 105 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 175 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 910 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 756 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 427 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 31}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{7}}}{840 \, d} \] Input:
integrate(sec(d*x+c)^3*tan(d*x+c)/(a+a*sin(d*x+c))^2,x, algorithm="giac")
Output:
-1/840*(35*(6*tan(1/2*d*x + 1/2*c)^2 - 9*tan(1/2*d*x + 1/2*c) + 5)/(a^2*(t an(1/2*d*x + 1/2*c) - 1)^3) - (210*tan(1/2*d*x + 1/2*c)^6 + 105*tan(1/2*d* x + 1/2*c)^5 - 175*tan(1/2*d*x + 1/2*c)^4 - 910*tan(1/2*d*x + 1/2*c)^3 - 7 56*tan(1/2*d*x + 1/2*c)^2 - 427*tan(1/2*d*x + 1/2*c) - 31)/(a^2*(tan(1/2*d *x + 1/2*c) + 1)^7))/d
Time = 35.25 (sec) , antiderivative size = 254, normalized size of antiderivative = 2.73 \[ \int \frac {\sec ^3(c+d x) \tan (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {2\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (9\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+36\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+132\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+68\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+14\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-84\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+140\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+140\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+105\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\right )}{105\,a^2\,d\,{\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}^3\,{\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}^7} \] Input:
int(tan(c + d*x)/(cos(c + d*x)^3*(a + a*sin(c + d*x))^2),x)
Output:
(2*cos(c/2 + (d*x)/2)^2*(9*cos(c/2 + (d*x)/2)^8 + 105*sin(c/2 + (d*x)/2)^8 + 140*cos(c/2 + (d*x)/2)*sin(c/2 + (d*x)/2)^7 + 36*cos(c/2 + (d*x)/2)^7*s in(c/2 + (d*x)/2) + 140*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^6 - 84*cos (c/2 + (d*x)/2)^3*sin(c/2 + (d*x)/2)^5 + 14*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^4 + 68*cos(c/2 + (d*x)/2)^5*sin(c/2 + (d*x)/2)^3 + 132*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2)^2))/(105*a^2*d*(cos(c/2 + (d*x)/2) - sin(c/ 2 + (d*x)/2))^3*(cos(c/2 + (d*x)/2) + sin(c/2 + (d*x)/2))^7)
Time = 0.23 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.61 \[ \int \frac {\sec ^3(c+d x) \tan (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {9 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}+18 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}-18 \cos \left (d x +c \right ) \sin \left (d x +c \right )-9 \cos \left (d x +c \right )+16 \sin \left (d x +c \right )^{5}+32 \sin \left (d x +c \right )^{4}-8 \sin \left (d x +c \right )^{3}-48 \sin \left (d x +c \right )^{2}-18 \sin \left (d x +c \right )-9}{105 \cos \left (d x +c \right ) a^{2} d \left (\sin \left (d x +c \right )^{4}+2 \sin \left (d x +c \right )^{3}-2 \sin \left (d x +c \right )-1\right )} \] Input:
int(sec(d*x+c)^3*tan(d*x+c)/(a+a*sin(d*x+c))^2,x)
Output:
(9*cos(c + d*x)*sin(c + d*x)**4 + 18*cos(c + d*x)*sin(c + d*x)**3 - 18*cos (c + d*x)*sin(c + d*x) - 9*cos(c + d*x) + 16*sin(c + d*x)**5 + 32*sin(c + d*x)**4 - 8*sin(c + d*x)**3 - 48*sin(c + d*x)**2 - 18*sin(c + d*x) - 9)/(1 05*cos(c + d*x)*a**2*d*(sin(c + d*x)**4 + 2*sin(c + d*x)**3 - 2*sin(c + d* x) - 1))