Integrand size = 21, antiderivative size = 101 \[ \int (a+a \sin (c+d x))^2 \tan ^4(c+d x) \, dx=\frac {7 a^2 x}{2}-\frac {2 a^2 \cos (c+d x)}{d}+\frac {a^2 \cos (c+d x)}{3 d (1-\sin (c+d x))^2}-\frac {11 a^2 \cos (c+d x)}{3 d (1-\sin (c+d x))}-\frac {a^2 \cos (c+d x) \sin (c+d x)}{2 d} \] Output:
7/2*a^2*x-2*a^2*cos(d*x+c)/d+1/3*a^2*cos(d*x+c)/d/(1-sin(d*x+c))^2-11/3*a^ 2*cos(d*x+c)/d/(1-sin(d*x+c))-1/2*a^2*cos(d*x+c)*sin(d*x+c)/d
Time = 1.26 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.57 \[ \int (a+a \sin (c+d x))^2 \tan ^4(c+d x) \, dx=-\frac {a^2 \left (-21 (7+12 c+12 d x) \cos \left (\frac {1}{2} (c+d x)\right )+(239+84 c+84 d x) \cos \left (\frac {3}{2} (c+d x)\right )+3 \left (-5 \cos \left (\frac {5}{2} (c+d x)\right )+\cos \left (\frac {7}{2} (c+d x)\right )+2 (50+56 c+56 d x+(-27+28 c+28 d x) \cos (c+d x)-6 \cos (2 (c+d x))-\cos (3 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )\right )}{48 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3} \] Input:
Integrate[(a + a*Sin[c + d*x])^2*Tan[c + d*x]^4,x]
Output:
-1/48*(a^2*(-21*(7 + 12*c + 12*d*x)*Cos[(c + d*x)/2] + (239 + 84*c + 84*d* x)*Cos[(3*(c + d*x))/2] + 3*(-5*Cos[(5*(c + d*x))/2] + Cos[(7*(c + d*x))/2 ] + 2*(50 + 56*c + 56*d*x + (-27 + 28*c + 28*d*x)*Cos[c + d*x] - 6*Cos[2*( c + d*x)] - Cos[3*(c + d*x)])*Sin[(c + d*x)/2])))/(d*(Cos[(c + d*x)/2] - S in[(c + d*x)/2])^3)
Time = 0.62 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.27, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3187, 3042, 3244, 25, 3042, 3456, 3042, 3213}
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.
| \(\displaystyle \int \tan ^4(c+d x) (a \sin (c+d x)+a)^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \tan (c+d x)^4 (a \sin (c+d x)+a)^2dx\) |
| \(\Big \downarrow \) 3187 |
| \(\displaystyle a^4 \int \frac {\sin ^4(c+d x)}{(a-a \sin (c+d x))^2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^4 \int \frac {\sin (c+d x)^4}{(a-a \sin (c+d x))^2}dx\) |
| \(\Big \downarrow \) 3244 |
| \(\displaystyle a^4 \left (\frac {\int -\frac {\sin ^2(c+d x) (5 \sin (c+d x) a+3 a)}{a-a \sin (c+d x)}dx}{3 a^2}+\frac {\sin ^3(c+d x) \cos (c+d x)}{3 d (a-a \sin (c+d x))^2}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle a^4 \left (\frac {\sin ^3(c+d x) \cos (c+d x)}{3 d (a-a \sin (c+d x))^2}-\frac {\int \frac {\sin ^2(c+d x) (5 \sin (c+d x) a+3 a)}{a-a \sin (c+d x)}dx}{3 a^2}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^4 \left (\frac {\sin ^3(c+d x) \cos (c+d x)}{3 d (a-a \sin (c+d x))^2}-\frac {\int \frac {\sin (c+d x)^2 (5 \sin (c+d x) a+3 a)}{a-a \sin (c+d x)}dx}{3 a^2}\right )\) |
| \(\Big \downarrow \) 3456 |
| \(\displaystyle a^4 \left (\frac {\sin ^3(c+d x) \cos (c+d x)}{3 d (a-a \sin (c+d x))^2}-\frac {\frac {8 \sin ^2(c+d x) \cos (c+d x)}{d (1-\sin (c+d x))}-\frac {\int \sin (c+d x) \left (21 \sin (c+d x) a^2+16 a^2\right )dx}{a^2}}{3 a^2}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^4 \left (\frac {\sin ^3(c+d x) \cos (c+d x)}{3 d (a-a \sin (c+d x))^2}-\frac {\frac {8 \sin ^2(c+d x) \cos (c+d x)}{d (1-\sin (c+d x))}-\frac {\int \sin (c+d x) \left (21 \sin (c+d x) a^2+16 a^2\right )dx}{a^2}}{3 a^2}\right )\) |
| \(\Big \downarrow \) 3213 |
| \(\displaystyle a^4 \left (\frac {\sin ^3(c+d x) \cos (c+d x)}{3 d (a-a \sin (c+d x))^2}-\frac {\frac {8 \sin ^2(c+d x) \cos (c+d x)}{d (1-\sin (c+d x))}-\frac {-\frac {16 a^2 \cos (c+d x)}{d}-\frac {21 a^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {21 a^2 x}{2}}{a^2}}{3 a^2}\right )\) |
Input:
Int[(a + a*Sin[c + d*x])^2*Tan[c + d*x]^4,x]
Output:
a^4*((Cos[c + d*x]*Sin[c + d*x]^3)/(3*d*(a - a*Sin[c + d*x])^2) - ((8*Cos[ c + d*x]*Sin[c + d*x]^2)/(d*(1 - Sin[c + d*x])) - ((21*a^2*x)/2 - (16*a^2* Cos[c + d*x])/d - (21*a^2*Cos[c + d*x]*Sin[c + d*x])/(2*d))/a^2)/(3*a^2))
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p _), x_Symbol] :> Simp[a^p Int[Sin[e + f*x]^p/(a - b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p] && EqQ [p, 2*m]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.) *(x_)]), x_Symbol] :> Simp[(2*a*c + b*d)*(x/2), x] + (-Simp[(b*c + a*d)*(Co s[e + f*x]/f), x] - Simp[b*d*Cos[e + f*x]*(Sin[e + f*x]/(2*f)), x]) /; Free Q[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n - 1)/(a*f*(2*m + 1))), x] + Simp[1/(a*b* (2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 2)* Simp[b*(c^2*(m + 1) + d^2*(n - 1)) + a*c*d*(m - n + 1) + d*(a*d*(m - n + 1) + b*c*(m + n))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n] || (IntegerQ[m] && EqQ[c, 0]))
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/( a*f*(2*m + 1))), x] - Simp[1/(a*b*(2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], x], x], x] /; Fre eQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] & & NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] && GtQ[n, 0] && IntegerQ[2*m] && (In tegerQ[2*n] || EqQ[c, 0])
Result contains complex when optimal does not.
Time = 2.55 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.31
| method | result | size |
| risch | \(\frac {7 a^{2} x}{2}+\frac {i a^{2} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {a^{2} {\mathrm e}^{i \left (d x +c \right )}}{d}-\frac {a^{2} {\mathrm e}^{-i \left (d x +c \right )}}{d}-\frac {i a^{2} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}-\frac {2 \left (-21 i a^{2} {\mathrm e}^{i \left (d x +c \right )}+12 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-11 a^{2}\right )}{3 \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{3} d}\) | \(132\) |
| derivativedivides | \(\frac {a^{2} \left (\frac {\sin \left (d x +c \right )^{7}}{3 \cos \left (d x +c \right )^{3}}-\frac {4 \sin \left (d x +c \right )^{7}}{3 \cos \left (d x +c \right )}-\frac {4 \left (\sin \left (d x +c \right )^{5}+\frac {5 \sin \left (d x +c \right )^{3}}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{3}+\frac {5 d x}{2}+\frac {5 c}{2}\right )+2 a^{2} \left (\frac {\sin \left (d x +c \right )^{6}}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin \left (d x +c \right )^{6}}{\cos \left (d x +c \right )}-\left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )\right )+a^{2} \left (\frac {\tan \left (d x +c \right )^{3}}{3}-\tan \left (d x +c \right )+d x +c \right )}{d}\) | \(186\) |
| default | \(\frac {a^{2} \left (\frac {\sin \left (d x +c \right )^{7}}{3 \cos \left (d x +c \right )^{3}}-\frac {4 \sin \left (d x +c \right )^{7}}{3 \cos \left (d x +c \right )}-\frac {4 \left (\sin \left (d x +c \right )^{5}+\frac {5 \sin \left (d x +c \right )^{3}}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{3}+\frac {5 d x}{2}+\frac {5 c}{2}\right )+2 a^{2} \left (\frac {\sin \left (d x +c \right )^{6}}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin \left (d x +c \right )^{6}}{\cos \left (d x +c \right )}-\left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )\right )+a^{2} \left (\frac {\tan \left (d x +c \right )^{3}}{3}-\tan \left (d x +c \right )+d x +c \right )}{d}\) | \(186\) |
| parts | \(\frac {a^{2} \left (\frac {\tan \left (d x +c \right )^{3}}{3}-\tan \left (d x +c \right )+\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}+\frac {a^{2} \left (\frac {\sin \left (d x +c \right )^{7}}{3 \cos \left (d x +c \right )^{3}}-\frac {4 \sin \left (d x +c \right )^{7}}{3 \cos \left (d x +c \right )}-\frac {4 \left (\sin \left (d x +c \right )^{5}+\frac {5 \sin \left (d x +c \right )^{3}}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{3}+\frac {5 d x}{2}+\frac {5 c}{2}\right )}{d}+\frac {2 a^{2} \left (\frac {\sin \left (d x +c \right )^{6}}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin \left (d x +c \right )^{6}}{\cos \left (d x +c \right )}-\left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )\right )}{d}\) | \(194\) |
Input:
int((a+a*sin(d*x+c))^2*tan(d*x+c)^4,x,method=_RETURNVERBOSE)
Output:
7/2*a^2*x+1/8*I*a^2/d*exp(2*I*(d*x+c))-a^2/d*exp(I*(d*x+c))-a^2/d*exp(-I*( d*x+c))-1/8*I*a^2/d*exp(-2*I*(d*x+c))-2/3*(-21*I*a^2*exp(I*(d*x+c))+12*a^2 *exp(2*I*(d*x+c))-11*a^2)/(exp(I*(d*x+c))-I)^3/d
Leaf count of result is larger than twice the leaf count of optimal. 196 vs. \(2 (89) = 178\).
Time = 0.08 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.94 \[ \int (a+a \sin (c+d x))^2 \tan ^4(c+d x) \, dx=\frac {3 \, a^{2} \cos \left (d x + c\right )^{4} - 6 \, a^{2} \cos \left (d x + c\right )^{3} - 42 \, a^{2} d x + {\left (21 \, a^{2} d x + 31 \, a^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - {\left (21 \, a^{2} d x - 38 \, a^{2}\right )} \cos \left (d x + c\right ) - {\left (3 \, a^{2} \cos \left (d x + c\right )^{3} - 42 \, a^{2} d x + 9 \, a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} - {\left (21 \, a^{2} d x - 40 \, a^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left (d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) + {\left (d \cos \left (d x + c\right ) + 2 \, d\right )} \sin \left (d x + c\right ) - 2 \, d\right )}} \] Input:
integrate((a+a*sin(d*x+c))^2*tan(d*x+c)^4,x, algorithm="fricas")
Output:
1/6*(3*a^2*cos(d*x + c)^4 - 6*a^2*cos(d*x + c)^3 - 42*a^2*d*x + (21*a^2*d* x + 31*a^2)*cos(d*x + c)^2 - 2*a^2 - (21*a^2*d*x - 38*a^2)*cos(d*x + c) - (3*a^2*cos(d*x + c)^3 - 42*a^2*d*x + 9*a^2*cos(d*x + c)^2 + 2*a^2 - (21*a^ 2*d*x - 40*a^2)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^2 - d*cos(d*x + c) + (d*cos(d*x + c) + 2*d)*sin(d*x + c) - 2*d)
\[ \int (a+a \sin (c+d x))^2 \tan ^4(c+d x) \, dx=a^{2} \left (\int 2 \sin {\left (c + d x \right )} \tan ^{4}{\left (c + d x \right )}\, dx + \int \sin ^{2}{\left (c + d x \right )} \tan ^{4}{\left (c + d x \right )}\, dx + \int \tan ^{4}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate((a+a*sin(d*x+c))**2*tan(d*x+c)**4,x)
Output:
a**2*(Integral(2*sin(c + d*x)*tan(c + d*x)**4, x) + Integral(sin(c + d*x)* *2*tan(c + d*x)**4, x) + Integral(tan(c + d*x)**4, x))
Time = 0.11 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.19 \[ \int (a+a \sin (c+d x))^2 \tan ^4(c+d x) \, dx=\frac {{\left (2 \, \tan \left (d x + c\right )^{3} + 15 \, d x + 15 \, c - \frac {3 \, \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 12 \, \tan \left (d x + c\right )\right )} a^{2} + 2 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} a^{2} - 4 \, a^{2} {\left (\frac {6 \, \cos \left (d x + c\right )^{2} - 1}{\cos \left (d x + c\right )^{3}} + 3 \, \cos \left (d x + c\right )\right )}}{6 \, d} \] Input:
integrate((a+a*sin(d*x+c))^2*tan(d*x+c)^4,x, algorithm="maxima")
Output:
1/6*((2*tan(d*x + c)^3 + 15*d*x + 15*c - 3*tan(d*x + c)/(tan(d*x + c)^2 + 1) - 12*tan(d*x + c))*a^2 + 2*(tan(d*x + c)^3 + 3*d*x + 3*c - 3*tan(d*x + c))*a^2 - 4*a^2*((6*cos(d*x + c)^2 - 1)/cos(d*x + c)^3 + 3*cos(d*x + c)))/ d
Timed out. \[ \int (a+a \sin (c+d x))^2 \tan ^4(c+d x) \, dx=\text {Timed out} \] Input:
integrate((a+a*sin(d*x+c))^2*tan(d*x+c)^4,x, algorithm="giac")
Output:
Timed out
Time = 36.92 (sec) , antiderivative size = 287, normalized size of antiderivative = 2.84 \[ \int (a+a \sin (c+d x))^2 \tan ^4(c+d x) \, dx=\frac {7\,a^2\,x}{2}+\frac {\frac {7\,a^2\,\left (c+d\,x\right )}{2}-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {21\,a^2\,\left (c+d\,x\right )}{2}-\frac {a^2\,\left (63\,c+63\,d\,x-150\right )}{6}\right )-\frac {a^2\,\left (21\,c+21\,d\,x-64\right )}{6}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {21\,a^2\,\left (c+d\,x\right )}{2}-\frac {a^2\,\left (63\,c+63\,d\,x-42\right )}{6}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {35\,a^2\,\left (c+d\,x\right )}{2}-\frac {a^2\,\left (105\,c+105\,d\,x-126\right )}{6}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {35\,a^2\,\left (c+d\,x\right )}{2}-\frac {a^2\,\left (105\,c+105\,d\,x-194\right )}{6}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {49\,a^2\,\left (c+d\,x\right )}{2}-\frac {a^2\,\left (147\,c+147\,d\,x-196\right )}{6}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {49\,a^2\,\left (c+d\,x\right )}{2}-\frac {a^2\,\left (147\,c+147\,d\,x-252\right )}{6}\right )}{d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}^3\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^2} \] Input:
int(tan(c + d*x)^4*(a + a*sin(c + d*x))^2,x)
Output:
(7*a^2*x)/2 + ((7*a^2*(c + d*x))/2 - tan(c/2 + (d*x)/2)*((21*a^2*(c + d*x) )/2 - (a^2*(63*c + 63*d*x - 150))/6) - (a^2*(21*c + 21*d*x - 64))/6 + tan( c/2 + (d*x)/2)^6*((21*a^2*(c + d*x))/2 - (a^2*(63*c + 63*d*x - 42))/6) - t an(c/2 + (d*x)/2)^5*((35*a^2*(c + d*x))/2 - (a^2*(105*c + 105*d*x - 126))/ 6) + tan(c/2 + (d*x)/2)^2*((35*a^2*(c + d*x))/2 - (a^2*(105*c + 105*d*x - 194))/6) + tan(c/2 + (d*x)/2)^4*((49*a^2*(c + d*x))/2 - (a^2*(147*c + 147* d*x - 196))/6) - tan(c/2 + (d*x)/2)^3*((49*a^2*(c + d*x))/2 - (a^2*(147*c + 147*d*x - 252))/6))/(d*(tan(c/2 + (d*x)/2) - 1)^3*(tan(c/2 + (d*x)/2)^2 + 1)^2)
Time = 0.19 (sec) , antiderivative size = 232, normalized size of antiderivative = 2.30 \[ \int (a+a \sin (c+d x))^2 \tan ^4(c+d x) \, dx=\frac {a^{2} \left (2 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} \tan \left (d x +c \right )^{3}-6 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} \tan \left (d x +c \right )+15 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} c +21 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} d x +32 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}-2 \cos \left (d x +c \right ) \tan \left (d x +c \right )^{3}+6 \cos \left (d x +c \right ) \tan \left (d x +c \right )-15 \cos \left (d x +c \right ) c -21 \cos \left (d x +c \right ) d x -32 \cos \left (d x +c \right )+3 \sin \left (d x +c \right )^{5}+12 \sin \left (d x +c \right )^{4}-20 \sin \left (d x +c \right )^{3}-48 \sin \left (d x +c \right )^{2}+15 \sin \left (d x +c \right )+32\right )}{6 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:
int((a+a*sin(d*x+c))^2*tan(d*x+c)^4,x)
Output:
(a**2*(2*cos(c + d*x)*sin(c + d*x)**2*tan(c + d*x)**3 - 6*cos(c + d*x)*sin (c + d*x)**2*tan(c + d*x) + 15*cos(c + d*x)*sin(c + d*x)**2*c + 21*cos(c + d*x)*sin(c + d*x)**2*d*x + 32*cos(c + d*x)*sin(c + d*x)**2 - 2*cos(c + d* x)*tan(c + d*x)**3 + 6*cos(c + d*x)*tan(c + d*x) - 15*cos(c + d*x)*c - 21* cos(c + d*x)*d*x - 32*cos(c + d*x) + 3*sin(c + d*x)**5 + 12*sin(c + d*x)** 4 - 20*sin(c + d*x)**3 - 48*sin(c + d*x)**2 + 15*sin(c + d*x) + 32))/(6*co s(c + d*x)*d*(sin(c + d*x)**2 - 1))