Integrand size = 41, antiderivative size = 219 \[ \int (a+a \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {1}{2} a^4 (2 A+8 B+13 C) x+\frac {a^4 (12 A+13 B+8 C) \text {arctanh}(\sin (c+d x))}{2 d}-\frac {5 a^4 (2 A+B-C) \sin (c+d x)}{2 d}-\frac {(22 A+18 B+3 C) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac {(16 A+15 B+6 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \tan (c+d x)}{6 d}+\frac {a (4 A+3 B) (a+a \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{6 d}+\frac {A (a+a \cos (c+d x))^4 \sec ^2(c+d x) \tan (c+d x)}{3 d} \]
1/2*a^4*(2*A+8*B+13*C)*x+1/2*a^4*(12*A+13*B+8*C)*arctanh(sin(d*x+c))/d-5/2 *a^4*(2*A+B-C)*sin(d*x+c)/d-1/6*(22*A+18*B+3*C)*(a^4+a^4*cos(d*x+c))*sin(d *x+c)/d+1/6*(16*A+15*B+6*C)*(a^2+a^2*cos(d*x+c))^2*tan(d*x+c)/d+1/6*a*(4*A +3*B)*(a+a*cos(d*x+c))^3*sec(d*x+c)*tan(d*x+c)/d+1/3*A*(a+a*cos(d*x+c))^4* sec(d*x+c)^2*tan(d*x+c)/d
Leaf count is larger than twice the leaf count of optimal. \(440\) vs. \(2(219)=438\).
Time = 12.75 (sec) , antiderivative size = 440, normalized size of antiderivative = 2.01 \[ \int (a+a \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=a^4 \left (\frac {(2 A+8 B+13 C) (c+d x)}{2 d}+\frac {(-12 A-13 B-8 C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 d}+\frac {(12 A+13 B+8 C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 d}+\frac {13 A+3 B}{12 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {A \sin \left (\frac {1}{2} (c+d x)\right )}{6 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {A \sin \left (\frac {1}{2} (c+d x)\right )}{6 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {-13 A-3 B}{12 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {20 A \sin \left (\frac {1}{2} (c+d x)\right )+12 B \sin \left (\frac {1}{2} (c+d x)\right )+3 C \sin \left (\frac {1}{2} (c+d x)\right )}{3 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {20 A \sin \left (\frac {1}{2} (c+d x)\right )+12 B \sin \left (\frac {1}{2} (c+d x)\right )+3 C \sin \left (\frac {1}{2} (c+d x)\right )}{3 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {(B+4 C) \sin (c+d x)}{d}+\frac {C \sin (2 (c+d x))}{4 d}\right ) \]
a^4*(((2*A + 8*B + 13*C)*(c + d*x))/(2*d) + ((-12*A - 13*B - 8*C)*Log[Cos[ (c + d*x)/2] - Sin[(c + d*x)/2]])/(2*d) + ((12*A + 13*B + 8*C)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/(2*d) + (13*A + 3*B)/(12*d*(Cos[(c + d*x)/2 ] - Sin[(c + d*x)/2])^2) + (A*Sin[(c + d*x)/2])/(6*d*(Cos[(c + d*x)/2] - S in[(c + d*x)/2])^3) + (A*Sin[(c + d*x)/2])/(6*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3) + (-13*A - 3*B)/(12*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) ^2) + (20*A*Sin[(c + d*x)/2] + 12*B*Sin[(c + d*x)/2] + 3*C*Sin[(c + d*x)/2 ])/(3*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + (20*A*Sin[(c + d*x)/2] + 12*B*Sin[(c + d*x)/2] + 3*C*Sin[(c + d*x)/2])/(3*d*(Cos[(c + d*x)/2] + Sin [(c + d*x)/2])) + ((B + 4*C)*Sin[c + d*x])/d + (C*Sin[2*(c + d*x)])/(4*d))
Time = 1.80 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.02, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.415, Rules used = {3042, 3522, 3042, 3454, 3042, 3454, 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.
\(\displaystyle \int \sec ^4(c+d x) (a \cos (c+d x)+a)^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^4 \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx\) |
\(\Big \downarrow \) 3522 |
\(\displaystyle \frac {\int (\cos (c+d x) a+a)^4 (a (4 A+3 B)-a (2 A-3 C) \cos (c+d x)) \sec ^3(c+d x)dx}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^4}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^4 \left (a (4 A+3 B)-a (2 A-3 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^4}{3 d}\) |
\(\Big \downarrow \) 3454 |
\(\displaystyle \frac {\frac {1}{2} \int (\cos (c+d x) a+a)^3 \left (a^2 (16 A+15 B+6 C)-6 a^2 (2 A+B-C) \cos (c+d x)\right ) \sec ^2(c+d x)dx+\frac {a^2 (4 A+3 B) \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{2 d}}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^4}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{2} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3 \left (a^2 (16 A+15 B+6 C)-6 a^2 (2 A+B-C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a^2 (4 A+3 B) \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{2 d}}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^4}{3 d}\) |
\(\Big \downarrow \) 3454 |
\(\displaystyle \frac {\frac {1}{2} \left (\int (\cos (c+d x) a+a)^2 \left (3 a^3 (12 A+13 B+8 C)-2 a^3 (22 A+18 B+3 C) \cos (c+d x)\right ) \sec (c+d x)dx+\frac {a^3 (16 A+15 B+6 C) \tan (c+d x) (a \cos (c+d x)+a)^2}{d}\right )+\frac {a^2 (4 A+3 B) \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{2 d}}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^4}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{2} \left (\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (3 a^3 (12 A+13 B+8 C)-2 a^3 (22 A+18 B+3 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a^3 (16 A+15 B+6 C) \tan (c+d x) (a \cos (c+d x)+a)^2}{d}\right )+\frac {a^2 (4 A+3 B) \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{2 d}}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^4}{3 d}\) |
\(\Big \downarrow \) 3455 |
\(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{2} \int 6 (\cos (c+d x) a+a) \left (a^4 (12 A+13 B+8 C)-5 a^4 (2 A+B-C) \cos (c+d x)\right ) \sec (c+d x)dx-\frac {(22 A+18 B+3 C) \sin (c+d x) \left (a^5 \cos (c+d x)+a^5\right )}{d}+\frac {a^3 (16 A+15 B+6 C) \tan (c+d x) (a \cos (c+d x)+a)^2}{d}\right )+\frac {a^2 (4 A+3 B) \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{2 d}}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^4}{3 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {1}{2} \left (3 \int (\cos (c+d x) a+a) \left (a^4 (12 A+13 B+8 C)-5 a^4 (2 A+B-C) \cos (c+d x)\right ) \sec (c+d x)dx-\frac {(22 A+18 B+3 C) \sin (c+d x) \left (a^5 \cos (c+d x)+a^5\right )}{d}+\frac {a^3 (16 A+15 B+6 C) \tan (c+d x) (a \cos (c+d x)+a)^2}{d}\right )+\frac {a^2 (4 A+3 B) \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{2 d}}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^4}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{2} \left (3 \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (a^4 (12 A+13 B+8 C)-5 a^4 (2 A+B-C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {(22 A+18 B+3 C) \sin (c+d x) \left (a^5 \cos (c+d x)+a^5\right )}{d}+\frac {a^3 (16 A+15 B+6 C) \tan (c+d x) (a \cos (c+d x)+a)^2}{d}\right )+\frac {a^2 (4 A+3 B) \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{2 d}}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^4}{3 d}\) |
\(\Big \downarrow \) 3447 |
\(\displaystyle \frac {\frac {1}{2} \left (3 \int \left (-5 (2 A+B-C) \cos ^2(c+d x) a^5+(12 A+13 B+8 C) a^5+\left (a^5 (12 A+13 B+8 C)-5 a^5 (2 A+B-C)\right ) \cos (c+d x)\right ) \sec (c+d x)dx-\frac {(22 A+18 B+3 C) \sin (c+d x) \left (a^5 \cos (c+d x)+a^5\right )}{d}+\frac {a^3 (16 A+15 B+6 C) \tan (c+d x) (a \cos (c+d x)+a)^2}{d}\right )+\frac {a^2 (4 A+3 B) \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{2 d}}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^4}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{2} \left (3 \int \frac {-5 (2 A+B-C) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^5+(12 A+13 B+8 C) a^5+\left (a^5 (12 A+13 B+8 C)-5 a^5 (2 A+B-C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {(22 A+18 B+3 C) \sin (c+d x) \left (a^5 \cos (c+d x)+a^5\right )}{d}+\frac {a^3 (16 A+15 B+6 C) \tan (c+d x) (a \cos (c+d x)+a)^2}{d}\right )+\frac {a^2 (4 A+3 B) \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{2 d}}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^4}{3 d}\) |
\(\Big \downarrow \) 3502 |
\(\displaystyle \frac {\frac {1}{2} \left (3 \left (\int \left ((12 A+13 B+8 C) a^5+(2 A+8 B+13 C) \cos (c+d x) a^5\right ) \sec (c+d x)dx-\frac {5 a^5 (2 A+B-C) \sin (c+d x)}{d}\right )-\frac {(22 A+18 B+3 C) \sin (c+d x) \left (a^5 \cos (c+d x)+a^5\right )}{d}+\frac {a^3 (16 A+15 B+6 C) \tan (c+d x) (a \cos (c+d x)+a)^2}{d}\right )+\frac {a^2 (4 A+3 B) \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{2 d}}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^4}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{2} \left (3 \left (\int \frac {(12 A+13 B+8 C) a^5+(2 A+8 B+13 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^5}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {5 a^5 (2 A+B-C) \sin (c+d x)}{d}\right )-\frac {(22 A+18 B+3 C) \sin (c+d x) \left (a^5 \cos (c+d x)+a^5\right )}{d}+\frac {a^3 (16 A+15 B+6 C) \tan (c+d x) (a \cos (c+d x)+a)^2}{d}\right )+\frac {a^2 (4 A+3 B) \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{2 d}}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^4}{3 d}\) |
\(\Big \downarrow \) 3214 |
\(\displaystyle \frac {\frac {1}{2} \left (3 \left (a^5 (12 A+13 B+8 C) \int \sec (c+d x)dx-\frac {5 a^5 (2 A+B-C) \sin (c+d x)}{d}+a^5 x (2 A+8 B+13 C)\right )-\frac {(22 A+18 B+3 C) \sin (c+d x) \left (a^5 \cos (c+d x)+a^5\right )}{d}+\frac {a^3 (16 A+15 B+6 C) \tan (c+d x) (a \cos (c+d x)+a)^2}{d}\right )+\frac {a^2 (4 A+3 B) \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{2 d}}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^4}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{2} \left (3 \left (a^5 (12 A+13 B+8 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {5 a^5 (2 A+B-C) \sin (c+d x)}{d}+a^5 x (2 A+8 B+13 C)\right )-\frac {(22 A+18 B+3 C) \sin (c+d x) \left (a^5 \cos (c+d x)+a^5\right )}{d}+\frac {a^3 (16 A+15 B+6 C) \tan (c+d x) (a \cos (c+d x)+a)^2}{d}\right )+\frac {a^2 (4 A+3 B) \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{2 d}}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^4}{3 d}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {\frac {a^2 (4 A+3 B) \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{2 d}+\frac {1}{2} \left (3 \left (\frac {a^5 (12 A+13 B+8 C) \text {arctanh}(\sin (c+d x))}{d}-\frac {5 a^5 (2 A+B-C) \sin (c+d x)}{d}+a^5 x (2 A+8 B+13 C)\right )-\frac {(22 A+18 B+3 C) \sin (c+d x) \left (a^5 \cos (c+d x)+a^5\right )}{d}+\frac {a^3 (16 A+15 B+6 C) \tan (c+d x) (a \cos (c+d x)+a)^2}{d}\right )}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^4}{3 d}\) |
(A*(a + a*Cos[c + d*x])^4*Sec[c + d*x]^2*Tan[c + d*x])/(3*d) + ((a^2*(4*A + 3*B)*(a + a*Cos[c + d*x])^3*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (-(((22*A + 18*B + 3*C)*(a^5 + a^5*Cos[c + d*x])*Sin[c + d*x])/d) + 3*(a^5*(2*A + 8 *B + 13*C)*x + (a^5*(12*A + 13*B + 8*C)*ArcTanh[Sin[c + d*x]])/d - (5*a^5* (2*A + B - C)*Sin[c + d*x])/d) + (a^3*(16*A + 15*B + 6*C)*(a + a*Cos[c + d *x])^2*Tan[c + d*x])/d)/2)/(3*a)
3.4.34.3.1 Defintions of rubi rules used
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_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(c^2*C - B*c*d + 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 - B*d)*( a*c*m + b*d*(n + 1)) + b*(d*(B*c - A*d)*(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, B, 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 = 9.31 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.00
method | result | size |
parts | \(-\frac {a^{4} A \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (4 a^{4} A +B \,a^{4}\right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {\left (B \,a^{4}+4 C \,a^{4}\right ) \sin \left (d x +c \right )}{d}+\frac {\left (a^{4} A +4 B \,a^{4}+6 C \,a^{4}\right ) \left (d x +c \right )}{d}+\frac {\left (4 a^{4} A +6 B \,a^{4}+4 C \,a^{4}\right ) \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {\left (6 a^{4} A +4 B \,a^{4}+C \,a^{4}\right ) \tan \left (d x +c \right )}{d}+\frac {C \,a^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(220\) |
parallelrisch | \(\frac {8 a^{4} \left (-\frac {9 \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \left (A +\frac {13 B}{12}+\frac {2 C}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{4}+\frac {9 \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \left (A +\frac {13 B}{12}+\frac {2 C}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4}+\frac {x \left (A +4 B +\frac {13 C}{2}\right ) d \cos \left (3 d x +3 c \right )}{8}+\frac {\left (\frac {B}{2}+C +A \right ) \sin \left (2 d x +2 c \right )}{2}+\frac {\left (\frac {11 C}{32}+\frac {5 A}{3}+B \right ) \sin \left (3 d x +3 c \right )}{2}+\frac {\left (\frac {B}{4}+C \right ) \sin \left (4 d x +4 c \right )}{4}+\frac {\sin \left (5 d x +5 c \right ) C}{64}+\frac {3 x \left (A +4 B +\frac {13 C}{2}\right ) d \cos \left (d x +c \right )}{8}+\sin \left (d x +c \right ) \left (\frac {5 C}{32}+\frac {B}{2}+A \right )\right )}{d \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\) | \(228\) |
derivativedivides | \(\frac {a^{4} A \left (d x +c \right )+B \,a^{4} \sin \left (d x +c \right )+C \,a^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 a^{4} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 B \,a^{4} \left (d x +c \right )+4 C \,a^{4} \sin \left (d x +c \right )+6 a^{4} A \tan \left (d x +c \right )+6 B \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+6 C \,a^{4} \left (d x +c \right )+4 a^{4} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+4 B \,a^{4} \tan \left (d x +c \right )+4 C \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )-a^{4} A \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+B \,a^{4} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+C \,a^{4} \tan \left (d x +c \right )}{d}\) | \(280\) |
default | \(\frac {a^{4} A \left (d x +c \right )+B \,a^{4} \sin \left (d x +c \right )+C \,a^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 a^{4} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 B \,a^{4} \left (d x +c \right )+4 C \,a^{4} \sin \left (d x +c \right )+6 a^{4} A \tan \left (d x +c \right )+6 B \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+6 C \,a^{4} \left (d x +c \right )+4 a^{4} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+4 B \,a^{4} \tan \left (d x +c \right )+4 C \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )-a^{4} A \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+B \,a^{4} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+C \,a^{4} \tan \left (d x +c \right )}{d}\) | \(280\) |
risch | \(a^{4} x A +4 a^{4} B x +\frac {13 a^{4} C x}{2}-\frac {i {\mathrm e}^{i \left (d x +c \right )} B \,a^{4}}{2 d}+\frac {2 i {\mathrm e}^{-i \left (d x +c \right )} C \,a^{4}}{d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} B \,a^{4}}{2 d}-\frac {i a^{4} \left (12 A \,{\mathrm e}^{5 i \left (d x +c \right )}+3 B \,{\mathrm e}^{5 i \left (d x +c \right )}-36 A \,{\mathrm e}^{4 i \left (d x +c \right )}-24 B \,{\mathrm e}^{4 i \left (d x +c \right )}-6 C \,{\mathrm e}^{4 i \left (d x +c \right )}-84 A \,{\mathrm e}^{2 i \left (d x +c \right )}-48 B \,{\mathrm e}^{2 i \left (d x +c \right )}-12 C \,{\mathrm e}^{2 i \left (d x +c \right )}-12 A \,{\mathrm e}^{i \left (d x +c \right )}-3 B \,{\mathrm e}^{i \left (d x +c \right )}-40 A -24 B -6 C \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}-\frac {2 i {\mathrm e}^{i \left (d x +c \right )} C \,a^{4}}{d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} C \,a^{4}}{8 d}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )} C \,a^{4}}{8 d}-\frac {6 a^{4} A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}-\frac {13 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{2 d}-\frac {4 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}+\frac {6 a^{4} A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}+\frac {13 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{2 d}+\frac {4 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}\) | \(420\) |
-a^4*A/d*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+(4*A*a^4+B*a^4)/d*(1/2*sec(d*x +c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+(B*a^4+4*C*a^4)/d*sin(d*x+c) +(A*a^4+4*B*a^4+6*C*a^4)/d*(d*x+c)+(4*A*a^4+6*B*a^4+4*C*a^4)/d*ln(sec(d*x+ c)+tan(d*x+c))+(6*A*a^4+4*B*a^4+C*a^4)/d*tan(d*x+c)+C*a^4/d*(1/2*cos(d*x+c )*sin(d*x+c)+1/2*d*x+1/2*c)
Time = 0.27 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.87 \[ \int (a+a \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {6 \, {\left (2 \, A + 8 \, B + 13 \, C\right )} a^{4} d x \cos \left (d x + c\right )^{3} + 3 \, {\left (12 \, A + 13 \, B + 8 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (12 \, A + 13 \, B + 8 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (3 \, C a^{4} \cos \left (d x + c\right )^{4} + 6 \, {\left (B + 4 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} + 2 \, {\left (20 \, A + 12 \, B + 3 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 3 \, {\left (4 \, A + B\right )} a^{4} \cos \left (d x + c\right ) + 2 \, A a^{4}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \]
integrate((a+a*cos(d*x+c))^4*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4, x, algorithm="fricas")
1/12*(6*(2*A + 8*B + 13*C)*a^4*d*x*cos(d*x + c)^3 + 3*(12*A + 13*B + 8*C)* a^4*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(12*A + 13*B + 8*C)*a^4*cos(d *x + c)^3*log(-sin(d*x + c) + 1) + 2*(3*C*a^4*cos(d*x + c)^4 + 6*(B + 4*C) *a^4*cos(d*x + c)^3 + 2*(20*A + 12*B + 3*C)*a^4*cos(d*x + c)^2 + 3*(4*A + B)*a^4*cos(d*x + c) + 2*A*a^4)*sin(d*x + c))/(d*cos(d*x + c)^3)
Timed out. \[ \int (a+a \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\text {Timed out} \]
Time = 0.23 (sec) , antiderivative size = 320, normalized size of antiderivative = 1.46 \[ \int (a+a \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{4} + 12 \, {\left (d x + c\right )} A a^{4} + 48 \, {\left (d x + c\right )} B a^{4} + 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{4} + 72 \, {\left (d x + c\right )} C a^{4} - 12 \, A a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 3 \, B a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, A a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 36 \, B a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, C a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, B a^{4} \sin \left (d x + c\right ) + 48 \, C a^{4} \sin \left (d x + c\right ) + 72 \, A a^{4} \tan \left (d x + c\right ) + 48 \, B a^{4} \tan \left (d x + c\right ) + 12 \, C a^{4} \tan \left (d x + c\right )}{12 \, d} \]
integrate((a+a*cos(d*x+c))^4*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4, x, algorithm="maxima")
1/12*(4*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^4 + 12*(d*x + c)*A*a^4 + 48* (d*x + c)*B*a^4 + 3*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^4 + 72*(d*x + c)* C*a^4 - 12*A*a^4*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 3*B*a^4*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1 ) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 24*A*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 36*B*a^4*(log(sin(d*x + c) + 1) - lo g(sin(d*x + c) - 1)) + 24*C*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 12*B*a^4*sin(d*x + c) + 48*C*a^4*sin(d*x + c) + 72*A*a^4*tan(d*x + c) + 48*B*a^4*tan(d*x + c) + 12*C*a^4*tan(d*x + c))/d
Time = 0.39 (sec) , antiderivative size = 347, normalized size of antiderivative = 1.58 \[ \int (a+a \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {3 \, {\left (2 \, A a^{4} + 8 \, B a^{4} + 13 \, C a^{4}\right )} {\left (d x + c\right )} + 3 \, {\left (12 \, A a^{4} + 13 \, B a^{4} + 8 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (12 \, A a^{4} + 13 \, B a^{4} + 8 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {6 \, {\left (2 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 7 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}} - \frac {2 \, {\left (30 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 21 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 76 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 48 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 54 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 27 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \]
integrate((a+a*cos(d*x+c))^4*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4, x, algorithm="giac")
1/6*(3*(2*A*a^4 + 8*B*a^4 + 13*C*a^4)*(d*x + c) + 3*(12*A*a^4 + 13*B*a^4 + 8*C*a^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(12*A*a^4 + 13*B*a^4 + 8* C*a^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 6*(2*B*a^4*tan(1/2*d*x + 1/2*c )^3 + 7*C*a^4*tan(1/2*d*x + 1/2*c)^3 + 2*B*a^4*tan(1/2*d*x + 1/2*c) + 9*C* a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^2 - 2*(30*A*a^4*tan (1/2*d*x + 1/2*c)^5 + 21*B*a^4*tan(1/2*d*x + 1/2*c)^5 + 6*C*a^4*tan(1/2*d* x + 1/2*c)^5 - 76*A*a^4*tan(1/2*d*x + 1/2*c)^3 - 48*B*a^4*tan(1/2*d*x + 1/ 2*c)^3 - 12*C*a^4*tan(1/2*d*x + 1/2*c)^3 + 54*A*a^4*tan(1/2*d*x + 1/2*c) + 27*B*a^4*tan(1/2*d*x + 1/2*c) + 6*C*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d* x + 1/2*c)^2 - 1)^3)/d
Time = 3.77 (sec) , antiderivative size = 625, normalized size of antiderivative = 2.85 \[ \int (a+a \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {3\,A\,a^4\,\sin \left (2\,c+2\,d\,x\right )+5\,A\,a^4\,\sin \left (3\,c+3\,d\,x\right )+\frac {3\,B\,a^4\,\sin \left (2\,c+2\,d\,x\right )}{2}+3\,B\,a^4\,\sin \left (3\,c+3\,d\,x\right )+\frac {3\,B\,a^4\,\sin \left (4\,c+4\,d\,x\right )}{8}+3\,C\,a^4\,\sin \left (2\,c+2\,d\,x\right )+\frac {33\,C\,a^4\,\sin \left (3\,c+3\,d\,x\right )}{32}+\frac {3\,C\,a^4\,\sin \left (4\,c+4\,d\,x\right )}{2}+\frac {3\,C\,a^4\,\sin \left (5\,c+5\,d\,x\right )}{32}+6\,A\,a^4\,\sin \left (c+d\,x\right )+3\,B\,a^4\,\sin \left (c+d\,x\right )+\frac {15\,C\,a^4\,\sin \left (c+d\,x\right )}{16}+\frac {9\,A\,a^4\,\cos \left (c+d\,x\right )\,\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 )}{2}+27\,A\,a^4\,\cos \left (c+d\,x\right )\,\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 )+18\,B\,a^4\,\cos \left (c+d\,x\right )\,\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 )+\frac {117\,B\,a^4\,\cos \left (c+d\,x\right )\,\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 )}{4}+\frac {117\,C\,a^4\,\cos \left (c+d\,x\right )\,\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 )}{4}+18\,C\,a^4\,\cos \left (c+d\,x\right )\,\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 )+\frac {3\,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 )\,\cos \left (3\,c+3\,d\,x\right )}{2}+9\,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 )\,\cos \left (3\,c+3\,d\,x\right )+6\,B\,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 )\,\cos \left (3\,c+3\,d\,x\right )+\frac {39\,B\,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 )\,\cos \left (3\,c+3\,d\,x\right )}{4}+\frac {39\,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 )\,\cos \left (3\,c+3\,d\,x\right )}{4}+6\,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 )\,\cos \left (3\,c+3\,d\,x\right )}{3\,d\,\left (\frac {3\,\cos \left (c+d\,x\right )}{4}+\frac {\cos \left (3\,c+3\,d\,x\right )}{4}\right )} \]
(3*A*a^4*sin(2*c + 2*d*x) + 5*A*a^4*sin(3*c + 3*d*x) + (3*B*a^4*sin(2*c + 2*d*x))/2 + 3*B*a^4*sin(3*c + 3*d*x) + (3*B*a^4*sin(4*c + 4*d*x))/8 + 3*C* a^4*sin(2*c + 2*d*x) + (33*C*a^4*sin(3*c + 3*d*x))/32 + (3*C*a^4*sin(4*c + 4*d*x))/2 + (3*C*a^4*sin(5*c + 5*d*x))/32 + 6*A*a^4*sin(c + d*x) + 3*B*a^ 4*sin(c + d*x) + (15*C*a^4*sin(c + d*x))/16 + (9*A*a^4*cos(c + d*x)*atan(s in(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/2 + 27*A*a^4*cos(c + d*x)*atanh(sin (c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + 18*B*a^4*cos(c + d*x)*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + (117*B*a^4*cos(c + d*x)*atanh(sin(c/2 + (d *x)/2)/cos(c/2 + (d*x)/2)))/4 + (117*C*a^4*cos(c + d*x)*atan(sin(c/2 + (d* x)/2)/cos(c/2 + (d*x)/2)))/4 + 18*C*a^4*cos(c + d*x)*atanh(sin(c/2 + (d*x) /2)/cos(c/2 + (d*x)/2)) + (3*A*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x) /2))*cos(3*c + 3*d*x))/2 + 9*A*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x )/2))*cos(3*c + 3*d*x) + 6*B*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2 ))*cos(3*c + 3*d*x) + (39*B*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2 ))*cos(3*c + 3*d*x))/4 + (39*C*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x) /2))*cos(3*c + 3*d*x))/4 + 6*C*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x )/2))*cos(3*c + 3*d*x))/(3*d*((3*cos(c + d*x))/4 + cos(3*c + 3*d*x)/4))