Integrand size = 33, antiderivative size = 186 \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=2 a^4 (2 A+3 C) x+\frac {a^4 (13 A+2 C) \text {arctanh}(\sin (c+d x))}{2 d}-\frac {5 a^4 (A-2 C) \sin (c+d x)}{2 d}-\frac {(15 A-2 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{6 d}-\frac {(9 A-4 C) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{3 d}+\frac {2 a A (a+a \cos (c+d x))^3 \tan (c+d x)}{d}+\frac {A (a+a \cos (c+d x))^4 \sec (c+d x) \tan (c+d x)}{2 d} \] Output:
2*a^4*(2*A+3*C)*x+1/2*a^4*(13*A+2*C)*arctanh(sin(d*x+c))/d-5/2*a^4*(A-2*C) *sin(d*x+c)/d-1/6*(15*A-2*C)*(a^2+a^2*cos(d*x+c))^2*sin(d*x+c)/d-1/3*(9*A- 4*C)*(a^4+a^4*cos(d*x+c))*sin(d*x+c)/d+2*a*A*(a+a*cos(d*x+c))^3*tan(d*x+c) /d+1/2*A*(a+a*cos(d*x+c))^4*sec(d*x+c)*tan(d*x+c)/d
Leaf count is larger than twice the leaf count of optimal. \(756\) vs. \(2(186)=372\).
Time = 12.22 (sec) , antiderivative size = 756, normalized size of antiderivative = 4.06 \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx =\text {Too large to display} \] Input:
Integrate[(a + a*Cos[c + d*x])^4*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^3,x]
Output:
((2*A + 3*C)*x*(a + a*Cos[c + d*x])^4*Sec[c/2 + (d*x)/2]^8)/8 + ((-13*A - 2*C)*(a + a*Cos[c + d*x])^4*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]]*S ec[c/2 + (d*x)/2]^8)/(32*d) + ((13*A + 2*C)*(a + a*Cos[c + d*x])^4*Log[Cos [c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^8)/(32*d) + ((4*A + 27*C)*Cos[d*x]*(a + a*Cos[c + d*x])^4*Sec[c/2 + (d*x)/2]^8*Sin[c])/(64* d) + (C*Cos[2*d*x]*(a + a*Cos[c + d*x])^4*Sec[c/2 + (d*x)/2]^8*Sin[2*c])/( 16*d) + (C*Cos[3*d*x]*(a + a*Cos[c + d*x])^4*Sec[c/2 + (d*x)/2]^8*Sin[3*c] )/(192*d) + ((4*A + 27*C)*Cos[c]*(a + a*Cos[c + d*x])^4*Sec[c/2 + (d*x)/2] ^8*Sin[d*x])/(64*d) + (C*Cos[2*c]*(a + a*Cos[c + d*x])^4*Sec[c/2 + (d*x)/2 ]^8*Sin[2*d*x])/(16*d) + (C*Cos[3*c]*(a + a*Cos[c + d*x])^4*Sec[c/2 + (d*x )/2]^8*Sin[3*d*x])/(192*d) + (A*(a + a*Cos[c + d*x])^4*Sec[c/2 + (d*x)/2]^ 8)/(64*d*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])^2) + (A*(a + a*Cos[c + d*x])^4*Sec[c/2 + (d*x)/2]^8*Sin[(d*x)/2])/(4*d*(Cos[c/2] - Sin[c/2])*(Cos [c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])) - (A*(a + a*Cos[c + d*x])^4*Sec[c/2 + (d*x)/2]^8)/(64*d*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])^2) + (A*(a + a*Cos[c + d*x])^4*Sec[c/2 + (d*x)/2]^8*Sin[(d*x)/2])/(4*d*(Cos[c/2] + Si n[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]))
Time = 1.59 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.05, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.515, Rules used = {3042, 3523, 3042, 3454, 3042, 3455, 3042, 3455, 27, 3042, 3447, 3042, 3502, 3042, 3214, 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.
| \(\displaystyle \int \sec ^3(c+d x) (a \cos (c+d x)+a)^4 \left (A+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^4 \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx\) |
| \(\Big \downarrow \) 3523 |
| \(\displaystyle \frac {\int (\cos (c+d x) a+a)^4 (4 a A-a (3 A-2 C) \cos (c+d x)) \sec ^2(c+d x)dx}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^4}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^4 \left (4 a A-a (3 A-2 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^4}{2 d}\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle \frac {\int (\cos (c+d x) a+a)^3 \left (a^2 (13 A+2 C)-a^2 (15 A-2 C) \cos (c+d x)\right ) \sec (c+d x)dx+\frac {4 a^2 A \tan (c+d x) (a \cos (c+d x)+a)^3}{d}}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^4}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3 \left (a^2 (13 A+2 C)-a^2 (15 A-2 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {4 a^2 A \tan (c+d x) (a \cos (c+d x)+a)^3}{d}}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^4}{2 d}\) |
| \(\Big \downarrow \) 3455 |
| \(\displaystyle \frac {\frac {1}{3} \int (\cos (c+d x) a+a)^2 \left (3 a^3 (13 A+2 C)-4 a^3 (9 A-4 C) \cos (c+d x)\right ) \sec (c+d x)dx-\frac {a^3 (15 A-2 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}+\frac {4 a^2 A \tan (c+d x) (a \cos (c+d x)+a)^3}{d}}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^4}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (3 a^3 (13 A+2 C)-4 a^3 (9 A-4 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {a^3 (15 A-2 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}+\frac {4 a^2 A \tan (c+d x) (a \cos (c+d x)+a)^3}{d}}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^4}{2 d}\) |
| \(\Big \downarrow \) 3455 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {1}{2} \int 6 (\cos (c+d x) a+a) \left (a^4 (13 A+2 C)-5 a^4 (A-2 C) \cos (c+d x)\right ) \sec (c+d x)dx-\frac {2 (9 A-4 C) \sin (c+d x) \left (a^5 \cos (c+d x)+a^5\right )}{d}\right )-\frac {a^3 (15 A-2 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}+\frac {4 a^2 A \tan (c+d x) (a \cos (c+d x)+a)^3}{d}}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^4}{2 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{3} \left (3 \int (\cos (c+d x) a+a) \left (a^4 (13 A+2 C)-5 a^4 (A-2 C) \cos (c+d x)\right ) \sec (c+d x)dx-\frac {2 (9 A-4 C) \sin (c+d x) \left (a^5 \cos (c+d x)+a^5\right )}{d}\right )-\frac {a^3 (15 A-2 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}+\frac {4 a^2 A \tan (c+d x) (a \cos (c+d x)+a)^3}{d}}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^4}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \left (3 \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (a^4 (13 A+2 C)-5 a^4 (A-2 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {2 (9 A-4 C) \sin (c+d x) \left (a^5 \cos (c+d x)+a^5\right )}{d}\right )-\frac {a^3 (15 A-2 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}+\frac {4 a^2 A \tan (c+d x) (a \cos (c+d x)+a)^3}{d}}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^4}{2 d}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \frac {\frac {1}{3} \left (3 \int \left (-5 (A-2 C) \cos ^2(c+d x) a^5+(13 A+2 C) a^5+\left (a^5 (13 A+2 C)-5 a^5 (A-2 C)\right ) \cos (c+d x)\right ) \sec (c+d x)dx-\frac {2 (9 A-4 C) \sin (c+d x) \left (a^5 \cos (c+d x)+a^5\right )}{d}\right )-\frac {a^3 (15 A-2 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}+\frac {4 a^2 A \tan (c+d x) (a \cos (c+d x)+a)^3}{d}}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^4}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \left (3 \int \frac {-5 (A-2 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^5+(13 A+2 C) a^5+\left (a^5 (13 A+2 C)-5 a^5 (A-2 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {2 (9 A-4 C) \sin (c+d x) \left (a^5 \cos (c+d x)+a^5\right )}{d}\right )-\frac {a^3 (15 A-2 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}+\frac {4 a^2 A \tan (c+d x) (a \cos (c+d x)+a)^3}{d}}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^4}{2 d}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {\frac {1}{3} \left (3 \left (\int \left ((13 A+2 C) a^5+4 (2 A+3 C) \cos (c+d x) a^5\right ) \sec (c+d x)dx-\frac {5 a^5 (A-2 C) \sin (c+d x)}{d}\right )-\frac {2 (9 A-4 C) \sin (c+d x) \left (a^5 \cos (c+d x)+a^5\right )}{d}\right )-\frac {a^3 (15 A-2 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}+\frac {4 a^2 A \tan (c+d x) (a \cos (c+d x)+a)^3}{d}}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^4}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \left (3 \left (\int \frac {(13 A+2 C) a^5+4 (2 A+3 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^5}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {5 a^5 (A-2 C) \sin (c+d x)}{d}\right )-\frac {2 (9 A-4 C) \sin (c+d x) \left (a^5 \cos (c+d x)+a^5\right )}{d}\right )-\frac {a^3 (15 A-2 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}+\frac {4 a^2 A \tan (c+d x) (a \cos (c+d x)+a)^3}{d}}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^4}{2 d}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {\frac {1}{3} \left (3 \left (a^5 (13 A+2 C) \int \sec (c+d x)dx-\frac {5 a^5 (A-2 C) \sin (c+d x)}{d}+4 a^5 x (2 A+3 C)\right )-\frac {2 (9 A-4 C) \sin (c+d x) \left (a^5 \cos (c+d x)+a^5\right )}{d}\right )-\frac {a^3 (15 A-2 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}+\frac {4 a^2 A \tan (c+d x) (a \cos (c+d x)+a)^3}{d}}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^4}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \left (3 \left (a^5 (13 A+2 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {5 a^5 (A-2 C) \sin (c+d x)}{d}+4 a^5 x (2 A+3 C)\right )-\frac {2 (9 A-4 C) \sin (c+d x) \left (a^5 \cos (c+d x)+a^5\right )}{d}\right )-\frac {a^3 (15 A-2 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}+\frac {4 a^2 A \tan (c+d x) (a \cos (c+d x)+a)^3}{d}}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^4}{2 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {1}{3} \left (3 \left (\frac {a^5 (13 A+2 C) \text {arctanh}(\sin (c+d x))}{d}-\frac {5 a^5 (A-2 C) \sin (c+d x)}{d}+4 a^5 x (2 A+3 C)\right )-\frac {2 (9 A-4 C) \sin (c+d x) \left (a^5 \cos (c+d x)+a^5\right )}{d}\right )-\frac {a^3 (15 A-2 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}+\frac {4 a^2 A \tan (c+d x) (a \cos (c+d x)+a)^3}{d}}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^4}{2 d}\) |
Input:
Int[(a + a*Cos[c + d*x])^4*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^3,x]
Output:
(A*(a + a*Cos[c + d*x])^4*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (-1/3*(a^3*(1 5*A - 2*C)*(a + a*Cos[c + d*x])^2*Sin[c + d*x])/d + ((-2*(9*A - 4*C)*(a^5 + a^5*Cos[c + d*x])*Sin[c + d*x])/d + 3*(4*a^5*(2*A + 3*C)*x + (a^5*(13*A + 2*C)*ArcTanh[Sin[c + d*x]])/d - (5*a^5*(A - 2*C)*Sin[c + d*x])/d))/3 + ( 4*a^2*A*(a + a*Cos[c + d*x])^3*Tan[c + d*x])/d)/(2*a)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d Int[1/(c + d *Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 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[(-b^2)*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[ e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a*d))), x] - Simp[b/(d*(n + 1)*(b*c + a*d)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp [a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n + 1) - B *(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{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] && GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || 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[(-b)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1 ) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && !LtQ[n, -1 ] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(c^2*C + A*d^2))*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(b*d*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a *d*m + b*c*(n + 1)) + c*C*(a*c*m + b*d*(n + 1)) - b*(A*d^2*(m + n + 2) + C* (c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !LtQ[m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 0])
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 12.76 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.96
| method | result | size |
| parallelrisch | \(\frac {\left (-13 \left (\cos \left (2 d x +2 c \right )+1\right ) \left (A +\frac {2 C}{13}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+13 \left (\cos \left (2 d x +2 c \right )+1\right ) \left (A +\frac {2 C}{13}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+8 \left (A +\frac {3 C}{2}\right ) x d \cos \left (2 d x +2 c \right )+\left (8 A +2 C \right ) \sin \left (2 d x +2 c \right )+\left (A +\frac {83 C}{12}\right ) \sin \left (3 d x +3 c \right )+\sin \left (4 d x +4 c \right ) C +\frac {C \sin \left (5 d x +5 c \right )}{12}+\left (3 A +\frac {41 C}{6}\right ) \sin \left (d x +c \right )+8 \left (A +\frac {3 C}{2}\right ) x d \right ) a^{4}}{2 d \left (\cos \left (2 d x +2 c \right )+1\right )}\) | \(179\) |
| parts | \(\frac {a^{4} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {\left (a^{4} A +6 a^{4} C \right ) \sin \left (d x +c \right )}{d}+\frac {\left (4 a^{4} A +4 a^{4} C \right ) \left (d x +c \right )}{d}+\frac {\left (6 a^{4} A +a^{4} C \right ) \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {a^{4} C \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3 d}+\frac {4 a^{4} A \tan \left (d x +c \right )}{d}+\frac {4 a^{4} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(186\) |
| derivativedivides | \(\frac {a^{4} A \sin \left (d x +c \right )+\frac {a^{4} C \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+4 a^{4} A \left (d x +c \right )+4 a^{4} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+6 a^{4} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+6 a^{4} C \sin \left (d x +c \right )+4 a^{4} A \tan \left (d x +c \right )+4 a^{4} C \left (d x +c \right )+a^{4} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{4} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(188\) |
| default | \(\frac {a^{4} A \sin \left (d x +c \right )+\frac {a^{4} C \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+4 a^{4} A \left (d x +c \right )+4 a^{4} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+6 a^{4} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+6 a^{4} C \sin \left (d x +c \right )+4 a^{4} A \tan \left (d x +c \right )+4 a^{4} C \left (d x +c \right )+a^{4} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{4} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(188\) |
| risch | \(4 a^{4} x A +6 a^{4} C x -\frac {i a^{4} C \,{\mathrm e}^{3 i \left (d x +c \right )}}{24 d}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )} a^{4} C}{2 d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} a^{4} A}{2 d}-\frac {27 i {\mathrm e}^{i \left (d x +c \right )} a^{4} C}{8 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} a^{4} A}{2 d}+\frac {27 i {\mathrm e}^{-i \left (d x +c \right )} a^{4} C}{8 d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} a^{4} C}{2 d}+\frac {i a^{4} C \,{\mathrm e}^{-3 i \left (d x +c \right )}}{24 d}-\frac {i a^{4} A \left ({\mathrm e}^{3 i \left (d x +c \right )}-8 \,{\mathrm e}^{2 i \left (d x +c \right )}-{\mathrm e}^{i \left (d x +c \right )}-8\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {13 a^{4} A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}-\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}+\frac {13 a^{4} A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}+\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}\) | \(311\) |
| norman | \(\frac {\left (4 a^{4} A +6 a^{4} C \right ) x +\left (-40 a^{4} A -60 a^{4} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (-16 a^{4} A -24 a^{4} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (-16 a^{4} A -24 a^{4} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (4 a^{4} A +6 a^{4} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{16}+\left (16 a^{4} A +24 a^{4} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (16 a^{4} A +24 a^{4} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (16 a^{4} A +24 a^{4} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+\left (16 a^{4} A +24 a^{4} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}+\frac {a^{4} \left (11 A +18 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {5 a^{4} \left (A -2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{d}-\frac {7 a^{4} \left (21 A -10 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{3 d}+\frac {5 a^{4} \left (39 A -46 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d}-\frac {a^{4} \left (45 A +158 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{3 d}-\frac {a^{4} \left (81 A -106 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{3 d}+\frac {a^{4} \left (159 A +130 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}+\frac {a^{4} \left (285 A -2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{6} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2}}-\frac {a^{4} \left (13 A +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a^{4} \left (13 A +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(514\) |
Input:
int((a+a*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^3,x,method=_RETURNVER BOSE)
Output:
1/2*(-13*(cos(2*d*x+2*c)+1)*(A+2/13*C)*ln(tan(1/2*d*x+1/2*c)-1)+13*(cos(2* d*x+2*c)+1)*(A+2/13*C)*ln(tan(1/2*d*x+1/2*c)+1)+8*(A+3/2*C)*x*d*cos(2*d*x+ 2*c)+(8*A+2*C)*sin(2*d*x+2*c)+(A+83/12*C)*sin(3*d*x+3*c)+sin(4*d*x+4*c)*C+ 1/12*C*sin(5*d*x+5*c)+(3*A+41/6*C)*sin(d*x+c)+8*(A+3/2*C)*x*d)*a^4/d/(cos( 2*d*x+2*c)+1)
Time = 0.14 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.92 \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {24 \, {\left (2 \, A + 3 \, C\right )} a^{4} d x \cos \left (d x + c\right )^{2} + 3 \, {\left (13 \, A + 2 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (13 \, A + 2 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, C a^{4} \cos \left (d x + c\right )^{4} + 12 \, C a^{4} \cos \left (d x + c\right )^{3} + 2 \, {\left (3 \, A + 20 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 24 \, A a^{4} \cos \left (d x + c\right ) + 3 \, A a^{4}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{2}} \] Input:
integrate((a+a*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^3,x, algorithm= "fricas")
Output:
1/12*(24*(2*A + 3*C)*a^4*d*x*cos(d*x + c)^2 + 3*(13*A + 2*C)*a^4*cos(d*x + c)^2*log(sin(d*x + c) + 1) - 3*(13*A + 2*C)*a^4*cos(d*x + c)^2*log(-sin(d *x + c) + 1) + 2*(2*C*a^4*cos(d*x + c)^4 + 12*C*a^4*cos(d*x + c)^3 + 2*(3* A + 20*C)*a^4*cos(d*x + c)^2 + 24*A*a^4*cos(d*x + c) + 3*A*a^4)*sin(d*x + c))/(d*cos(d*x + c)^2)
Timed out. \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\text {Timed out} \] Input:
integrate((a+a*cos(d*x+c))**4*(A+C*cos(d*x+c)**2)*sec(d*x+c)**3,x)
Output:
Timed out
Time = 0.05 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.13 \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {48 \, {\left (d x + c\right )} A a^{4} - 4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{4} + 12 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{4} + 48 \, {\left (d x + c\right )} C a^{4} - 3 \, A a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 36 \, A a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, C a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, A a^{4} \sin \left (d x + c\right ) + 72 \, C a^{4} \sin \left (d x + c\right ) + 48 \, A a^{4} \tan \left (d x + c\right )}{12 \, d} \] Input:
integrate((a+a*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^3,x, algorithm= "maxima")
Output:
1/12*(48*(d*x + c)*A*a^4 - 4*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a^4 + 12* (2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^4 + 48*(d*x + c)*C*a^4 - 3*A*a^4*(2*s in(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c ) - 1)) + 36*A*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 6*C*a ^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 12*A*a^4*sin(d*x + c) + 72*C*a^4*sin(d*x + c) + 48*A*a^4*tan(d*x + c))/d
Time = 0.40 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.33 \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {12 \, {\left (2 \, A a^{4} + 3 \, C a^{4}\right )} {\left (d x + c\right )} + 3 \, {\left (13 \, A a^{4} + 2 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (13 \, A a^{4} + 2 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {6 \, {\left (7 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}} + \frac {4 \, {\left (3 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 15 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 38 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 27 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \] Input:
integrate((a+a*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^3,x, algorithm= "giac")
Output:
1/6*(12*(2*A*a^4 + 3*C*a^4)*(d*x + c) + 3*(13*A*a^4 + 2*C*a^4)*log(abs(tan (1/2*d*x + 1/2*c) + 1)) - 3*(13*A*a^4 + 2*C*a^4)*log(abs(tan(1/2*d*x + 1/2 *c) - 1)) - 6*(7*A*a^4*tan(1/2*d*x + 1/2*c)^3 - 9*A*a^4*tan(1/2*d*x + 1/2* c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^2 + 4*(3*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 1 5*C*a^4*tan(1/2*d*x + 1/2*c)^5 + 6*A*a^4*tan(1/2*d*x + 1/2*c)^3 + 38*C*a^4 *tan(1/2*d*x + 1/2*c)^3 + 3*A*a^4*tan(1/2*d*x + 1/2*c) + 27*C*a^4*tan(1/2* d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d
Time = 0.27 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.31 \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {A\,a^4\,\sin \left (c+d\,x\right )}{d}+\frac {20\,C\,a^4\,\sin \left (c+d\,x\right )}{3\,d}+\frac {8\,A\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {13\,A\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {12\,C\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,C\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {4\,A\,a^4\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {A\,a^4\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2}+\frac {C\,a^4\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{3\,d}+\frac {2\,C\,a^4\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{d} \] Input:
int(((A + C*cos(c + d*x)^2)*(a + a*cos(c + d*x))^4)/cos(c + d*x)^3,x)
Output:
(A*a^4*sin(c + d*x))/d + (20*C*a^4*sin(c + d*x))/(3*d) + (8*A*a^4*atan(sin (c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (13*A*a^4*atanh(sin(c/2 + (d*x)/2 )/cos(c/2 + (d*x)/2)))/d + (12*C*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d* x)/2)))/d + (2*C*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (4* A*a^4*sin(c + d*x))/(d*cos(c + d*x)) + (A*a^4*sin(c + d*x))/(2*d*cos(c + d *x)^2) + (C*a^4*cos(c + d*x)^2*sin(c + d*x))/(3*d) + (2*C*a^4*cos(c + d*x) *sin(c + d*x))/d
Time = 0.20 (sec) , antiderivative size = 426, normalized size of antiderivative = 2.29 \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {a^{4} \left (12 \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )^{3} c -12 \cos \left (d x +c \right )^{2} \sin \left (d x +c \right ) c -39 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a -6 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} c +39 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a +6 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) c +39 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a +6 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} c -39 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a -6 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) c -2 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5} c +6 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a +44 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} c +24 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a d x +36 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} c d x -9 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a -42 \cos \left (d x +c \right ) \sin \left (d x +c \right ) c -24 \cos \left (d x +c \right ) a d x -36 \cos \left (d x +c \right ) c d x +24 \sin \left (d x +c \right )^{3} a -24 \sin \left (d x +c \right ) a \right )}{6 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:
int((a+a*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^3,x)
Output:
(a**4*(12*cos(c + d*x)**2*sin(c + d*x)**3*c - 12*cos(c + d*x)**2*sin(c + d *x)*c - 39*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a - 6*co s(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*c + 39*cos(c + d*x)*l og(tan((c + d*x)/2) - 1)*a + 6*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*c + 39*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a + 6*cos(c + d* x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*c - 39*cos(c + d*x)*log(tan(( c + d*x)/2) + 1)*a - 6*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*c - 2*cos(c + d*x)*sin(c + d*x)**5*c + 6*cos(c + d*x)*sin(c + d*x)**3*a + 44*cos(c + d *x)*sin(c + d*x)**3*c + 24*cos(c + d*x)*sin(c + d*x)**2*a*d*x + 36*cos(c + d*x)*sin(c + d*x)**2*c*d*x - 9*cos(c + d*x)*sin(c + d*x)*a - 42*cos(c + d *x)*sin(c + d*x)*c - 24*cos(c + d*x)*a*d*x - 36*cos(c + d*x)*c*d*x + 24*si n(c + d*x)**3*a - 24*sin(c + d*x)*a))/(6*cos(c + d*x)*d*(sin(c + d*x)**2 - 1))