Integrand size = 41, antiderivative size = 245 \[ \int \frac {\cos ^4(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^4} \, dx=\frac {(2 A-8 B+21 C) x}{2 a^4}-\frac {8 (20 A-83 B+216 C) \sin (c+d x)}{105 a^4 d}+\frac {(2 A-8 B+21 C) \cos (c+d x) \sin (c+d x)}{2 a^4 d}-\frac {(10 A-52 B+129 C) \cos ^3(c+d x) \sin (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {4 (20 A-83 B+216 C) \cos ^2(c+d x) \sin (c+d x)}{105 a^4 d (1+\cos (c+d x))}-\frac {(A-B+C) \cos ^5(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac {(B-2 C) \cos ^4(c+d x) \sin (c+d x)}{5 a d (a+a \cos (c+d x))^3} \] Output:
1/2*(2*A-8*B+21*C)*x/a^4-8/105*(20*A-83*B+216*C)*sin(d*x+c)/a^4/d+1/2*(2*A -8*B+21*C)*cos(d*x+c)*sin(d*x+c)/a^4/d-1/105*(10*A-52*B+129*C)*cos(d*x+c)^ 3*sin(d*x+c)/a^4/d/(1+cos(d*x+c))^2-4/105*(20*A-83*B+216*C)*cos(d*x+c)^2*s in(d*x+c)/a^4/d/(1+cos(d*x+c))-1/7*(A-B+C)*cos(d*x+c)^5*sin(d*x+c)/d/(a+a* cos(d*x+c))^4+1/5*(B-2*C)*cos(d*x+c)^4*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^3
Time = 9.24 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.22 \[ \int \frac {\cos ^4(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^4} \, dx=\frac {2 \cos \left (\frac {1}{2} (c+d x)\right ) \left (15 (A-B+C) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )-6 (25 A-32 B+39 C) \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+4 (160 A-286 B+447 C) \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )-8 (260 A-764 B+1653 C) \cos ^6\left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+210 \cos ^7\left (\frac {1}{2} (c+d x)\right ) (2 (2 A-8 B+21 C) d x+4 (B-4 C) \sin (c+d x)+C \sin (2 (c+d x)))+15 (A-B+C) \cos \left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {c}{2}\right )-6 (25 A-32 B+39 C) \cos ^3\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {c}{2}\right )+4 (160 A-286 B+447 C) \cos ^5\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {c}{2}\right )\right )}{105 a^4 d (1+\cos (c+d x))^4} \] Input:
Integrate[(Cos[c + d*x]^4*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/(a + a* Cos[c + d*x])^4,x]
Output:
(2*Cos[(c + d*x)/2]*(15*(A - B + C)*Sec[c/2]*Sin[(d*x)/2] - 6*(25*A - 32*B + 39*C)*Cos[(c + d*x)/2]^2*Sec[c/2]*Sin[(d*x)/2] + 4*(160*A - 286*B + 447 *C)*Cos[(c + d*x)/2]^4*Sec[c/2]*Sin[(d*x)/2] - 8*(260*A - 764*B + 1653*C)* Cos[(c + d*x)/2]^6*Sec[c/2]*Sin[(d*x)/2] + 210*Cos[(c + d*x)/2]^7*(2*(2*A - 8*B + 21*C)*d*x + 4*(B - 4*C)*Sin[c + d*x] + C*Sin[2*(c + d*x)]) + 15*(A - B + C)*Cos[(c + d*x)/2]*Tan[c/2] - 6*(25*A - 32*B + 39*C)*Cos[(c + d*x) /2]^3*Tan[c/2] + 4*(160*A - 286*B + 447*C)*Cos[(c + d*x)/2]^5*Tan[c/2]))/( 105*a^4*d*(1 + Cos[c + d*x])^4)
Time = 1.36 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.09, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.268, Rules used = {3042, 3520, 3042, 3456, 3042, 3456, 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.
| \(\displaystyle \int \frac {\cos ^4(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a \cos (c+d x)+a)^4} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^4 \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^4}dx\) |
| \(\Big \downarrow \) 3520 |
| \(\displaystyle \frac {\int \frac {\cos ^4(c+d x) (a (2 A+5 B-5 C)+a (2 A-2 B+9 C) \cos (c+d x))}{(\cos (c+d x) a+a)^3}dx}{7 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^5(c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^4 \left (a (2 A+5 B-5 C)+a (2 A-2 B+9 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3}dx}{7 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^5(c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3456 |
| \(\displaystyle \frac {\frac {\int \frac {\cos ^3(c+d x) \left (28 (B-2 C) a^2+(10 A-24 B+73 C) \cos (c+d x) a^2\right )}{(\cos (c+d x) a+a)^2}dx}{5 a^2}+\frac {7 a (B-2 C) \sin (c+d x) \cos ^4(c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^5(c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (28 (B-2 C) a^2+(10 A-24 B+73 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}+\frac {7 a (B-2 C) \sin (c+d x) \cos ^4(c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^5(c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3456 |
| \(\displaystyle \frac {\frac {\frac {\int -\frac {\cos ^2(c+d x) \left (3 a^3 (10 A-52 B+129 C)-a^3 (50 A-176 B+477 C) \cos (c+d x)\right )}{\cos (c+d x) a+a}dx}{3 a^2}-\frac {(10 A-52 B+129 C) \sin (c+d x) \cos ^3(c+d x)}{3 d (\cos (c+d x)+1)^2}}{5 a^2}+\frac {7 a (B-2 C) \sin (c+d x) \cos ^4(c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^5(c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {-\frac {\int \frac {\cos ^2(c+d x) \left (3 a^3 (10 A-52 B+129 C)-a^3 (50 A-176 B+477 C) \cos (c+d x)\right )}{\cos (c+d x) a+a}dx}{3 a^2}-\frac {(10 A-52 B+129 C) \sin (c+d x) \cos ^3(c+d x)}{3 d (\cos (c+d x)+1)^2}}{5 a^2}+\frac {7 a (B-2 C) \sin (c+d x) \cos ^4(c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^5(c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {-\frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (3 a^3 (10 A-52 B+129 C)-a^3 (50 A-176 B+477 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}-\frac {(10 A-52 B+129 C) \sin (c+d x) \cos ^3(c+d x)}{3 d (\cos (c+d x)+1)^2}}{5 a^2}+\frac {7 a (B-2 C) \sin (c+d x) \cos ^4(c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^5(c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3456 |
| \(\displaystyle \frac {\frac {-\frac {\frac {\int \cos (c+d x) \left (8 a^4 (20 A-83 B+216 C)-105 a^4 (2 A-8 B+21 C) \cos (c+d x)\right )dx}{a^2}+\frac {4 a^3 (20 A-83 B+216 C) \sin (c+d x) \cos ^2(c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {(10 A-52 B+129 C) \sin (c+d x) \cos ^3(c+d x)}{3 d (\cos (c+d x)+1)^2}}{5 a^2}+\frac {7 a (B-2 C) \sin (c+d x) \cos ^4(c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^5(c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {-\frac {\frac {\int \sin \left (c+d x+\frac {\pi }{2}\right ) \left (8 a^4 (20 A-83 B+216 C)-105 a^4 (2 A-8 B+21 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{a^2}+\frac {4 a^3 (20 A-83 B+216 C) \sin (c+d x) \cos ^2(c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {(10 A-52 B+129 C) \sin (c+d x) \cos ^3(c+d x)}{3 d (\cos (c+d x)+1)^2}}{5 a^2}+\frac {7 a (B-2 C) \sin (c+d x) \cos ^4(c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^5(c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3213 |
| \(\displaystyle \frac {\frac {-\frac {\frac {4 a^3 (20 A-83 B+216 C) \sin (c+d x) \cos ^2(c+d x)}{d (a \cos (c+d x)+a)}+\frac {\frac {8 a^4 (20 A-83 B+216 C) \sin (c+d x)}{d}-\frac {105 a^4 (2 A-8 B+21 C) \sin (c+d x) \cos (c+d x)}{2 d}-\frac {105}{2} a^4 x (2 A-8 B+21 C)}{a^2}}{3 a^2}-\frac {(10 A-52 B+129 C) \sin (c+d x) \cos ^3(c+d x)}{3 d (\cos (c+d x)+1)^2}}{5 a^2}+\frac {7 a (B-2 C) \sin (c+d x) \cos ^4(c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^5(c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
Input:
Int[(Cos[c + d*x]^4*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/(a + a*Cos[c + d*x])^4,x]
Output:
-1/7*((A - B + C)*Cos[c + d*x]^5*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^4) + ((7*a*(B - 2*C)*Cos[c + d*x]^4*Sin[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3 ) + (-1/3*((10*A - 52*B + 129*C)*Cos[c + d*x]^3*Sin[c + d*x])/(d*(1 + Cos[ c + d*x])^2) - ((4*a^3*(20*A - 83*B + 216*C)*Cos[c + d*x]^2*Sin[c + d*x])/ (d*(a + a*Cos[c + d*x])) + ((-105*a^4*(2*A - 8*B + 21*C)*x)/2 + (8*a^4*(20 *A - 83*B + 216*C)*Sin[c + d*x])/d - (105*a^4*(2*A - 8*B + 21*C)*Cos[c + d *x]*Sin[c + d*x])/(2*d))/a^2)/(3*a^2))/(5*a^2))/(7*a^2)
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_)*((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])
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[(a*A - b*B + a*C)*Cos[e + f*x]*(a + b* Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(f*(b*c - a*d)*(2*m + 1))), x ] + Simp[1/(b*(b*c - a*d)*(2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[A*(a*c*(m + 1) - b*d*(2*m + n + 2)) + B*(b*c*m + a *d*(n + 1)) - C*(a*c*m + b*d*(n + 1)) + (d*(a*A - b*B)*(m + n + 2) + C*(b*c *(2*m + 1) - a*d*(m - n - 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c ^2 - d^2, 0] && LtQ[m, -2^(-1)]
Time = 0.77 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.56
| method | result | size |
| parallelrisch | \(\frac {-2080 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\left (\frac {62 A}{13}-\frac {2741 B}{130}+\frac {3516 C}{65}\right ) \cos \left (2 d x +2 c \right )+\left (A -\frac {296 B}{65}+\frac {23619 C}{2080}\right ) \cos \left (3 d x +3 c \right )+\left (-\frac {21 B}{104}+\frac {21 C}{52}\right ) \cos \left (4 d x +4 c \right )-\frac {21 C \cos \left (5 d x +5 c \right )}{416}+\left (\frac {146 A}{13}-\frac {3124 B}{65}+\frac {128643 C}{1040}\right ) \cos \left (d x +c \right )+\frac {94 A}{13}-\frac {16171 B}{520}+\frac {20871 C}{260}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+26880 \left (-4 B +\frac {21 C}{2}+A \right ) x d}{26880 a^{4} d}\) | \(138\) |
| derivativedivides | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} A}{7}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} B}{7}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} C}{7}-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A +\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}-\frac {9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}+\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-\frac {23 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+13 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C -15 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+49 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -111 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {16 \left (B -\frac {9 C}{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+16 \left (B -\frac {7 C}{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+8 \left (2 A -8 B +21 C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{4}}\) | \(244\) |
| default | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} A}{7}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} B}{7}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} C}{7}-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A +\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}-\frac {9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}+\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-\frac {23 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+13 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C -15 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+49 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -111 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {16 \left (B -\frac {9 C}{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+16 \left (B -\frac {7 C}{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+8 \left (2 A -8 B +21 C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{4}}\) | \(244\) |
| risch | \(\frac {x A}{a^{4}}-\frac {4 x B}{a^{4}}+\frac {21 C x}{2 a^{4}}-\frac {i C \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 a^{4} d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} B}{2 a^{4} d}+\frac {2 i C \,{\mathrm e}^{i \left (d x +c \right )}}{a^{4} d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} B}{2 a^{4} d}-\frac {2 i C \,{\mathrm e}^{-i \left (d x +c \right )}}{a^{4} d}+\frac {i C \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 a^{4} d}-\frac {2 i \left (420 A \,{\mathrm e}^{6 i \left (d x +c \right )}-1050 B \,{\mathrm e}^{6 i \left (d x +c \right )}+2100 C \,{\mathrm e}^{6 i \left (d x +c \right )}+1890 A \,{\mathrm e}^{5 i \left (d x +c \right )}-5250 B \,{\mathrm e}^{5 i \left (d x +c \right )}+11025 C \,{\mathrm e}^{5 i \left (d x +c \right )}+4130 A \,{\mathrm e}^{4 i \left (d x +c \right )}-11900 B \,{\mathrm e}^{4 i \left (d x +c \right )}+25515 C \,{\mathrm e}^{4 i \left (d x +c \right )}+4970 A \,{\mathrm e}^{3 i \left (d x +c \right )}-14840 B \,{\mathrm e}^{3 i \left (d x +c \right )}+32340 C \,{\mathrm e}^{3 i \left (d x +c \right )}+3570 A \,{\mathrm e}^{2 i \left (d x +c \right )}-10794 B \,{\mathrm e}^{2 i \left (d x +c \right )}+23688 C \,{\mathrm e}^{2 i \left (d x +c \right )}+1400 A \,{\mathrm e}^{i \left (d x +c \right )}-4298 B \,{\mathrm e}^{i \left (d x +c \right )}+9471 C \,{\mathrm e}^{i \left (d x +c \right )}+260 A -764 B +1653 C \right )}{105 d \,a^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{7}}\) | \(384\) |
Input:
int(cos(d*x+c)^4*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^4,x,meth od=_RETURNVERBOSE)
Output:
1/26880*(-2080*tan(1/2*d*x+1/2*c)*((62/13*A-2741/130*B+3516/65*C)*cos(2*d* x+2*c)+(A-296/65*B+23619/2080*C)*cos(3*d*x+3*c)+(-21/104*B+21/52*C)*cos(4* d*x+4*c)-21/416*C*cos(5*d*x+5*c)+(146/13*A-3124/65*B+128643/1040*C)*cos(d* x+c)+94/13*A-16171/520*B+20871/260*C)*sec(1/2*d*x+1/2*c)^6+26880*(-4*B+21/ 2*C+A)*x*d)/a^4/d
Time = 0.08 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.08 \[ \int \frac {\cos ^4(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^4} \, dx=\frac {105 \, {\left (2 \, A - 8 \, B + 21 \, C\right )} d x \cos \left (d x + c\right )^{4} + 420 \, {\left (2 \, A - 8 \, B + 21 \, C\right )} d x \cos \left (d x + c\right )^{3} + 630 \, {\left (2 \, A - 8 \, B + 21 \, C\right )} d x \cos \left (d x + c\right )^{2} + 420 \, {\left (2 \, A - 8 \, B + 21 \, C\right )} d x \cos \left (d x + c\right ) + 105 \, {\left (2 \, A - 8 \, B + 21 \, C\right )} d x + {\left (105 \, C \cos \left (d x + c\right )^{5} + 210 \, {\left (B - 2 \, C\right )} \cos \left (d x + c\right )^{4} - 4 \, {\left (130 \, A - 592 \, B + 1509 \, C\right )} \cos \left (d x + c\right )^{3} - 4 \, {\left (310 \, A - 1318 \, B + 3411 \, C\right )} \cos \left (d x + c\right )^{2} - {\left (1070 \, A - 4472 \, B + 11619 \, C\right )} \cos \left (d x + c\right ) - 320 \, A + 1328 \, B - 3456 \, C\right )} \sin \left (d x + c\right )}{210 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \] Input:
integrate(cos(d*x+c)^4*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^4, x, algorithm="fricas")
Output:
1/210*(105*(2*A - 8*B + 21*C)*d*x*cos(d*x + c)^4 + 420*(2*A - 8*B + 21*C)* d*x*cos(d*x + c)^3 + 630*(2*A - 8*B + 21*C)*d*x*cos(d*x + c)^2 + 420*(2*A - 8*B + 21*C)*d*x*cos(d*x + c) + 105*(2*A - 8*B + 21*C)*d*x + (105*C*cos(d *x + c)^5 + 210*(B - 2*C)*cos(d*x + c)^4 - 4*(130*A - 592*B + 1509*C)*cos( d*x + c)^3 - 4*(310*A - 1318*B + 3411*C)*cos(d*x + c)^2 - (1070*A - 4472*B + 11619*C)*cos(d*x + c) - 320*A + 1328*B - 3456*C)*sin(d*x + c))/(a^4*d*c os(d*x + c)^4 + 4*a^4*d*cos(d*x + c)^3 + 6*a^4*d*cos(d*x + c)^2 + 4*a^4*d* cos(d*x + c) + a^4*d)
Leaf count of result is larger than twice the leaf count of optimal. 1624 vs. \(2 (238) = 476\).
Time = 10.27 (sec) , antiderivative size = 1624, normalized size of antiderivative = 6.63 \[ \int \frac {\cos ^4(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^4} \, dx=\text {Too large to display} \] Input:
integrate(cos(d*x+c)**4*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)/(a+a*cos(d*x+c))* *4,x)
Output:
Piecewise((840*A*d*x*tan(c/2 + d*x/2)**4/(840*a**4*d*tan(c/2 + d*x/2)**4 + 1680*a**4*d*tan(c/2 + d*x/2)**2 + 840*a**4*d) + 1680*A*d*x*tan(c/2 + d*x/ 2)**2/(840*a**4*d*tan(c/2 + d*x/2)**4 + 1680*a**4*d*tan(c/2 + d*x/2)**2 + 840*a**4*d) + 840*A*d*x/(840*a**4*d*tan(c/2 + d*x/2)**4 + 1680*a**4*d*tan( c/2 + d*x/2)**2 + 840*a**4*d) + 15*A*tan(c/2 + d*x/2)**11/(840*a**4*d*tan( c/2 + d*x/2)**4 + 1680*a**4*d*tan(c/2 + d*x/2)**2 + 840*a**4*d) - 75*A*tan (c/2 + d*x/2)**9/(840*a**4*d*tan(c/2 + d*x/2)**4 + 1680*a**4*d*tan(c/2 + d *x/2)**2 + 840*a**4*d) + 190*A*tan(c/2 + d*x/2)**7/(840*a**4*d*tan(c/2 + d *x/2)**4 + 1680*a**4*d*tan(c/2 + d*x/2)**2 + 840*a**4*d) - 910*A*tan(c/2 + d*x/2)**5/(840*a**4*d*tan(c/2 + d*x/2)**4 + 1680*a**4*d*tan(c/2 + d*x/2)* *2 + 840*a**4*d) - 2765*A*tan(c/2 + d*x/2)**3/(840*a**4*d*tan(c/2 + d*x/2) **4 + 1680*a**4*d*tan(c/2 + d*x/2)**2 + 840*a**4*d) - 1575*A*tan(c/2 + d*x /2)/(840*a**4*d*tan(c/2 + d*x/2)**4 + 1680*a**4*d*tan(c/2 + d*x/2)**2 + 84 0*a**4*d) - 3360*B*d*x*tan(c/2 + d*x/2)**4/(840*a**4*d*tan(c/2 + d*x/2)**4 + 1680*a**4*d*tan(c/2 + d*x/2)**2 + 840*a**4*d) - 6720*B*d*x*tan(c/2 + d* x/2)**2/(840*a**4*d*tan(c/2 + d*x/2)**4 + 1680*a**4*d*tan(c/2 + d*x/2)**2 + 840*a**4*d) - 3360*B*d*x/(840*a**4*d*tan(c/2 + d*x/2)**4 + 1680*a**4*d*t an(c/2 + d*x/2)**2 + 840*a**4*d) - 15*B*tan(c/2 + d*x/2)**11/(840*a**4*d*t an(c/2 + d*x/2)**4 + 1680*a**4*d*tan(c/2 + d*x/2)**2 + 840*a**4*d) + 117*B *tan(c/2 + d*x/2)**9/(840*a**4*d*tan(c/2 + d*x/2)**4 + 1680*a**4*d*tan(...
Leaf count of result is larger than twice the leaf count of optimal. 474 vs. \(2 (231) = 462\).
Time = 0.15 (sec) , antiderivative size = 474, normalized size of antiderivative = 1.93 \[ \int \frac {\cos ^4(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^4} \, dx =\text {Too large to display} \] Input:
integrate(cos(d*x+c)^4*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^4, x, algorithm="maxima")
Output:
-1/840*(3*C*(280*(7*sin(d*x + c)/(cos(d*x + c) + 1) + 9*sin(d*x + c)^3/(co s(d*x + c) + 1)^3)/(a^4 + 2*a^4*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a^4* sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (3885*sin(d*x + c)/(cos(d*x + c) + 1) - 455*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 63*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 5*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4 - 5880*arctan(sin (d*x + c)/(cos(d*x + c) + 1))/a^4) - B*(1680*sin(d*x + c)/((a^4 + a^4*sin( d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) + (5145*sin(d*x + c)/ (cos(d*x + c) + 1) - 805*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 147*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 15*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4 - 6720*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^4) + 5*A*((315*sin(d*x + c)/(cos(d*x + c) + 1) - 77*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 21*sin(d *x + c)^5/(cos(d*x + c) + 1)^5 - 3*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^ 4 - 336*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^4))/d
Time = 0.15 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.23 \[ \int \frac {\cos ^4(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^4} \, dx=\frac {\frac {420 \, {\left (d x + c\right )} {\left (2 \, A - 8 \, B + 21 \, C\right )}}{a^{4}} + \frac {840 \, {\left (2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 7 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a^{4}} + \frac {15 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 15 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 105 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 147 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 189 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 385 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 805 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1365 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 1575 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5145 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 11655 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{28}}}{840 \, d} \] Input:
integrate(cos(d*x+c)^4*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^4, x, algorithm="giac")
Output:
1/840*(420*(d*x + c)*(2*A - 8*B + 21*C)/a^4 + 840*(2*B*tan(1/2*d*x + 1/2*c )^3 - 9*C*tan(1/2*d*x + 1/2*c)^3 + 2*B*tan(1/2*d*x + 1/2*c) - 7*C*tan(1/2* d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^2*a^4) + (15*A*a^24*tan(1/2*d* x + 1/2*c)^7 - 15*B*a^24*tan(1/2*d*x + 1/2*c)^7 + 15*C*a^24*tan(1/2*d*x + 1/2*c)^7 - 105*A*a^24*tan(1/2*d*x + 1/2*c)^5 + 147*B*a^24*tan(1/2*d*x + 1/ 2*c)^5 - 189*C*a^24*tan(1/2*d*x + 1/2*c)^5 + 385*A*a^24*tan(1/2*d*x + 1/2* c)^3 - 805*B*a^24*tan(1/2*d*x + 1/2*c)^3 + 1365*C*a^24*tan(1/2*d*x + 1/2*c )^3 - 1575*A*a^24*tan(1/2*d*x + 1/2*c) + 5145*B*a^24*tan(1/2*d*x + 1/2*c) - 11655*C*a^24*tan(1/2*d*x + 1/2*c))/a^28)/d
Time = 0.35 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.16 \[ \int \frac {\cos ^4(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^4} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (A+5\,B-15\,C\right )}{8\,a^4}-\frac {3\,\left (2\,A-4\,B+6\,C\right )}{4\,a^4}-\frac {5\,\left (A-B+C\right )}{4\,a^4}+\frac {4\,A-20\,C}{8\,a^4}\right )}{d}+\frac {\left (2\,B-9\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,B-7\,C\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^4\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {2\,A-4\,B+6\,C}{40\,a^4}+\frac {3\,\left (A-B+C\right )}{40\,a^4}\right )}{d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {2\,A-4\,B+6\,C}{8\,a^4}-\frac {A+5\,B-15\,C}{24\,a^4}+\frac {A-B+C}{4\,a^4}\right )}{d}+\frac {x\,\left (2\,A-8\,B+21\,C\right )}{2\,a^4}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (A-B+C\right )}{56\,a^4\,d} \] Input:
int((cos(c + d*x)^4*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/(a + a*cos(c + d*x))^4,x)
Output:
(tan(c/2 + (d*x)/2)*((3*(A + 5*B - 15*C))/(8*a^4) - (3*(2*A - 4*B + 6*C))/ (4*a^4) - (5*(A - B + C))/(4*a^4) + (4*A - 20*C)/(8*a^4)))/d + (tan(c/2 + (d*x)/2)^3*(2*B - 9*C) + tan(c/2 + (d*x)/2)*(2*B - 7*C))/(d*(2*a^4*tan(c/2 + (d*x)/2)^2 + a^4*tan(c/2 + (d*x)/2)^4 + a^4)) - (tan(c/2 + (d*x)/2)^5*( (2*A - 4*B + 6*C)/(40*a^4) + (3*(A - B + C))/(40*a^4)))/d + (tan(c/2 + (d* x)/2)^3*((2*A - 4*B + 6*C)/(8*a^4) - (A + 5*B - 15*C)/(24*a^4) + (A - B + C)/(4*a^4)))/d + (x*(2*A - 8*B + 21*C))/(2*a^4) + (tan(c/2 + (d*x)/2)^7*(A - B + C))/(56*a^4*d)
Time = 0.16 (sec) , antiderivative size = 394, normalized size of antiderivative = 1.61 \[ \int \frac {\cos ^4(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^4} \, dx=\frac {-3360 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} b d x -910 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} a -9114 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} c -2765 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a -29505 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} c +17640 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} c d x +117 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9} b -75 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9} a -159 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9} c +840 a d x +8820 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} c d x -17535 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) c +15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11} a +15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11} c +190 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} a +1002 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} c +8820 c d x +11165 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} b -15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11} b +3682 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} b +840 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a d x +1680 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a d x -1575 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +6825 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b -3360 b d x -526 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} b -6720 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b d x}{840 a^{4} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right )} \] Input:
int(cos(d*x+c)^4*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^4,x)
Output:
(15*tan((c + d*x)/2)**11*a - 15*tan((c + d*x)/2)**11*b + 15*tan((c + d*x)/ 2)**11*c - 75*tan((c + d*x)/2)**9*a + 117*tan((c + d*x)/2)**9*b - 159*tan( (c + d*x)/2)**9*c + 190*tan((c + d*x)/2)**7*a - 526*tan((c + d*x)/2)**7*b + 1002*tan((c + d*x)/2)**7*c - 910*tan((c + d*x)/2)**5*a + 3682*tan((c + d *x)/2)**5*b - 9114*tan((c + d*x)/2)**5*c + 840*tan((c + d*x)/2)**4*a*d*x - 3360*tan((c + d*x)/2)**4*b*d*x + 8820*tan((c + d*x)/2)**4*c*d*x - 2765*ta n((c + d*x)/2)**3*a + 11165*tan((c + d*x)/2)**3*b - 29505*tan((c + d*x)/2) **3*c + 1680*tan((c + d*x)/2)**2*a*d*x - 6720*tan((c + d*x)/2)**2*b*d*x + 17640*tan((c + d*x)/2)**2*c*d*x - 1575*tan((c + d*x)/2)*a + 6825*tan((c + d*x)/2)*b - 17535*tan((c + d*x)/2)*c + 840*a*d*x - 3360*b*d*x + 8820*c*d*x )/(840*a**4*d*(tan((c + d*x)/2)**4 + 2*tan((c + d*x)/2)**2 + 1))