Integrand size = 33, antiderivative size = 148 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {(A-B+C) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac {(3 A+4 B-11 C) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac {(6 A+8 B+13 C) \sin (c+d x)}{105 d \left (a^2+a^2 \cos (c+d x)\right )^2}+\frac {(6 A+8 B+13 C) \sin (c+d x)}{105 d \left (a^4+a^4 \cos (c+d x)\right )} \] Output:
1/7*(A-B+C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^4+1/35*(3*A+4*B-11*C)*sin(d*x+c) /a/d/(a+a*cos(d*x+c))^3+1/105*(6*A+8*B+13*C)*sin(d*x+c)/d/(a^2+a^2*cos(d*x +c))^2+1/105*(6*A+8*B+13*C)*sin(d*x+c)/d/(a^4+a^4*cos(d*x+c))
Time = 1.80 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.41 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \left (70 (3 A+2 B+4 C) \sin \left (\frac {d x}{2}\right )-35 (4 B+5 C) \sin \left (c+\frac {d x}{2}\right )+126 A \sin \left (c+\frac {3 d x}{2}\right )+168 B \sin \left (c+\frac {3 d x}{2}\right )+168 C \sin \left (c+\frac {3 d x}{2}\right )-105 C \sin \left (2 c+\frac {3 d x}{2}\right )+42 A \sin \left (2 c+\frac {5 d x}{2}\right )+56 B \sin \left (2 c+\frac {5 d x}{2}\right )+91 C \sin \left (2 c+\frac {5 d x}{2}\right )+6 A \sin \left (3 c+\frac {7 d x}{2}\right )+8 B \sin \left (3 c+\frac {7 d x}{2}\right )+13 C \sin \left (3 c+\frac {7 d x}{2}\right )\right )}{420 a^4 d (1+\cos (c+d x))^4} \] Input:
Integrate[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(a + a*Cos[c + d*x])^4,x ]
Output:
(Cos[(c + d*x)/2]*Sec[c/2]*(70*(3*A + 2*B + 4*C)*Sin[(d*x)/2] - 35*(4*B + 5*C)*Sin[c + (d*x)/2] + 126*A*Sin[c + (3*d*x)/2] + 168*B*Sin[c + (3*d*x)/2 ] + 168*C*Sin[c + (3*d*x)/2] - 105*C*Sin[2*c + (3*d*x)/2] + 42*A*Sin[2*c + (5*d*x)/2] + 56*B*Sin[2*c + (5*d*x)/2] + 91*C*Sin[2*c + (5*d*x)/2] + 6*A* Sin[3*c + (7*d*x)/2] + 8*B*Sin[3*c + (7*d*x)/2] + 13*C*Sin[3*c + (7*d*x)/2 ]))/(420*a^4*d*(1 + Cos[c + d*x])^4)
Time = 0.60 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.97, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {3042, 3498, 25, 3042, 3229, 3042, 3129, 3042, 3127}
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.
| \(\displaystyle \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a \cos (c+d x)+a)^4} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^4}dx\) |
| \(\Big \downarrow \) 3498 |
| \(\displaystyle \frac {(A-B+C) \sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}-\frac {\int -\frac {a (3 A+4 B-4 C)+7 a C \cos (c+d x)}{(\cos (c+d x) a+a)^3}dx}{7 a^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {a (3 A+4 B-4 C)+7 a C \cos (c+d x)}{(\cos (c+d x) a+a)^3}dx}{7 a^2}+\frac {(A-B+C) \sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (3 A+4 B-4 C)+7 a C \sin \left (c+d x+\frac {\pi }{2}\right )}{\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3}dx}{7 a^2}+\frac {(A-B+C) \sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3229 |
| \(\displaystyle \frac {\frac {1}{5} (6 A+8 B+13 C) \int \frac {1}{(\cos (c+d x) a+a)^2}dx+\frac {a (3 A+4 B-11 C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}+\frac {(A-B+C) \sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} (6 A+8 B+13 C) \int \frac {1}{\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx+\frac {a (3 A+4 B-11 C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}+\frac {(A-B+C) \sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3129 |
| \(\displaystyle \frac {\frac {1}{5} (6 A+8 B+13 C) \left (\frac {\int \frac {1}{\cos (c+d x) a+a}dx}{3 a}+\frac {\sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )+\frac {a (3 A+4 B-11 C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}+\frac {(A-B+C) \sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} (6 A+8 B+13 C) \left (\frac {\int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a}+\frac {\sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )+\frac {a (3 A+4 B-11 C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}+\frac {(A-B+C) \sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3127 |
| \(\displaystyle \frac {\frac {a (3 A+4 B-11 C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}+\frac {1}{5} (6 A+8 B+13 C) \left (\frac {\sin (c+d x)}{3 a d (a \cos (c+d x)+a)}+\frac {\sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )}{7 a^2}+\frac {(A-B+C) \sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
Input:
Int[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(a + a*Cos[c + d*x])^4,x]
Output:
((A - B + C)*Sin[c + d*x])/(7*d*(a + a*Cos[c + d*x])^4) + ((a*(3*A + 4*B - 11*C)*Sin[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) + ((6*A + 8*B + 13*C)*(S in[c + d*x]/(3*d*(a + a*Cos[c + d*x])^2) + Sin[c + d*x]/(3*a*d*(a + a*Cos[ c + d*x]))))/5)/(7*a^2)
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[-Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b ^2, 0]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*Cos[c + d*x]*((a + b*Sin[c + d*x])^n/(a*d*(2*n + 1))), x] + Simp[(n + 1)/(a*(2*n + 1)) Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)*Cos[e + f*x]*((a + b*Sin[e + f* x])^m/(a*f*(2*m + 1))), x] + Simp[(a*d*m + b*c*(m + 1))/(a*b*(2*m + 1)) I nt[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*b - a* B + b*C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/(a*f*(2*m + 1))), x] + Simp[1 /(a^2*(2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[a*A*(m + 1) + m*(b *B - a*C) + b*C*(2*m + 1)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && EqQ[a^2 - b^2, 0]
Time = 0.34 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.64
| method | result | size |
| parallelrisch | \(\frac {\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\left (8 A +\frac {32 B}{3}+\frac {52 C}{3}\right ) \cos \left (2 d x +2 c \right )+\left (A +\frac {4 B}{3}+\frac {13 C}{6}\right ) \cos \left (3 d x +3 c \right )+\left (29 A +\frac {116 B}{3}+\frac {167 C}{6}\right ) \cos \left (d x +c \right )+32 A +\frac {58 B}{3}+\frac {68 C}{3}\right )}{560 a^{4} d}\) | \(94\) |
| derivativedivides | \(\frac {\frac {\left (A -B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{7}+\frac {\left (3 A -C -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}+\frac {\left (3 A +B -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}+A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}\) | \(106\) |
| default | \(\frac {\frac {\left (A -B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{7}+\frac {\left (3 A -C -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}+\frac {\left (3 A +B -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}+A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}\) | \(106\) |
| risch | \(\frac {2 i \left (105 C \,{\mathrm e}^{5 i \left (d x +c \right )}+140 B \,{\mathrm e}^{4 i \left (d x +c \right )}+175 C \,{\mathrm e}^{4 i \left (d x +c \right )}+210 A \,{\mathrm e}^{3 i \left (d x +c \right )}+140 B \,{\mathrm e}^{3 i \left (d x +c \right )}+280 C \,{\mathrm e}^{3 i \left (d x +c \right )}+126 A \,{\mathrm e}^{2 i \left (d x +c \right )}+168 B \,{\mathrm e}^{2 i \left (d x +c \right )}+168 C \,{\mathrm e}^{2 i \left (d x +c \right )}+42 A \,{\mathrm e}^{i \left (d x +c \right )}+56 B \,{\mathrm e}^{i \left (d x +c \right )}+91 C \,{\mathrm e}^{i \left (d x +c \right )}+6 A +8 B +13 C \right )}{105 d \,a^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{7}}\) | \(177\) |
| norman | \(\frac {\frac {\left (A -B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{56 a d}+\frac {\left (A +B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d a}+\frac {\left (9 A +7 B +5 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{24 d a}+\frac {\left (27 A +11 B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{60 d a}+\frac {\left (31 A -17 B +3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{280 d a}+\frac {\left (123 A -11 B -31 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{420 d a}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} a^{3}}\) | \(181\) |
Input:
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^4,x,method=_RETURNVER BOSE)
Output:
1/560*sec(1/2*d*x+1/2*c)^6*tan(1/2*d*x+1/2*c)*((8*A+32/3*B+52/3*C)*cos(2*d *x+2*c)+(A+4/3*B+13/6*C)*cos(3*d*x+3*c)+(29*A+116/3*B+167/6*C)*cos(d*x+c)+ 32*A+58/3*B+68/3*C)/a^4/d
Time = 0.09 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.91 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {{\left ({\left (6 \, A + 8 \, B + 13 \, C\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (6 \, A + 8 \, B + 13 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (39 \, A + 52 \, B + 32 \, C\right )} \cos \left (d x + c\right ) + 36 \, A + 13 \, B + 8 \, C\right )} \sin \left (d x + c\right )}{105 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \] Input:
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^4,x, algorithm= "fricas")
Output:
1/105*((6*A + 8*B + 13*C)*cos(d*x + c)^3 + 4*(6*A + 8*B + 13*C)*cos(d*x + c)^2 + (39*A + 52*B + 32*C)*cos(d*x + c) + 36*A + 13*B + 8*C)*sin(d*x + c) /(a^4*d*cos(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)
Time = 2.14 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.78 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\begin {cases} \frac {A \tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{56 a^{4} d} + \frac {3 A \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{40 a^{4} d} + \frac {A \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{8 a^{4} d} + \frac {A \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{8 a^{4} d} - \frac {B \tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{56 a^{4} d} - \frac {B \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{40 a^{4} d} + \frac {B \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{24 a^{4} d} + \frac {B \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{8 a^{4} d} + \frac {C \tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{56 a^{4} d} - \frac {C \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{40 a^{4} d} - \frac {C \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{24 a^{4} d} + \frac {C \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{8 a^{4} d} & \text {for}\: d \neq 0 \\\frac {x \left (A + B \cos {\left (c \right )} + C \cos ^{2}{\left (c \right )}\right )}{\left (a \cos {\left (c \right )} + a\right )^{4}} & \text {otherwise} \end {cases} \] Input:
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)**2)/(a+a*cos(d*x+c))**4,x)
Output:
Piecewise((A*tan(c/2 + d*x/2)**7/(56*a**4*d) + 3*A*tan(c/2 + d*x/2)**5/(40 *a**4*d) + A*tan(c/2 + d*x/2)**3/(8*a**4*d) + A*tan(c/2 + d*x/2)/(8*a**4*d ) - B*tan(c/2 + d*x/2)**7/(56*a**4*d) - B*tan(c/2 + d*x/2)**5/(40*a**4*d) + B*tan(c/2 + d*x/2)**3/(24*a**4*d) + B*tan(c/2 + d*x/2)/(8*a**4*d) + C*ta n(c/2 + d*x/2)**7/(56*a**4*d) - C*tan(c/2 + d*x/2)**5/(40*a**4*d) - C*tan( c/2 + d*x/2)**3/(24*a**4*d) + C*tan(c/2 + d*x/2)/(8*a**4*d), Ne(d, 0)), (x *(A + B*cos(c) + C*cos(c)**2)/(a*cos(c) + a)**4, True))
Time = 0.05 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.75 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {\frac {B {\left (\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}} + \frac {C {\left (\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}} + \frac {3 \, A {\left (\frac {35 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {5 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}}}{840 \, d} \] Input:
integrate((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*(B*(105*sin(d*x + c)/(cos(d*x + c) + 1) + 35*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 21*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 + C*(105*sin(d*x + c)/(cos(d*x + c) + 1) - 35*si n(d*x + c)^3/(cos(d*x + c) + 1)^3 - 21*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 + 3*A*(35*sin(d*x + c)/(cos (d*x + c) + 1) + 35*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 21*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)/d
Time = 0.15 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.16 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {15 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 15 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 63 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 21 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 21 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 105 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 35 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 35 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 105 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 105 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{840 \, a^{4} d} \] Input:
integrate((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*(15*A*tan(1/2*d*x + 1/2*c)^7 - 15*B*tan(1/2*d*x + 1/2*c)^7 + 15*C*ta n(1/2*d*x + 1/2*c)^7 + 63*A*tan(1/2*d*x + 1/2*c)^5 - 21*B*tan(1/2*d*x + 1/ 2*c)^5 - 21*C*tan(1/2*d*x + 1/2*c)^5 + 105*A*tan(1/2*d*x + 1/2*c)^3 + 35*B *tan(1/2*d*x + 1/2*c)^3 - 35*C*tan(1/2*d*x + 1/2*c)^3 + 105*A*tan(1/2*d*x + 1/2*c) + 105*B*tan(1/2*d*x + 1/2*c) + 105*C*tan(1/2*d*x + 1/2*c))/(a^4*d )
Time = 0.32 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.67 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (A-B+C\right )}{56\,a^4\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (B-3\,A+C\right )}{40\,a^4\,d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A+B+C\right )}{8\,a^4\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (3\,A+B-C\right )}{24\,a^4\,d} \] Input:
int((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)^7*(A - B + C))/(56*a^4*d) - (tan(c/2 + (d*x)/2)^5*(B - 3*A + C))/(40*a^4*d) + (tan(c/2 + (d*x)/2)*(A + B + C))/(8*a^4*d) + (tan( c/2 + (d*x)/2)^3*(3*A + B - C))/(24*a^4*d)
Time = 0.17 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.03 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} a -15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} b +15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} c +63 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a -21 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} b -21 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} c +105 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +35 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -35 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} c +105 a +105 b +105 c \right )}{840 a^{4} d} \] Input:
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^4,x)
Output:
(tan((c + d*x)/2)*(15*tan((c + d*x)/2)**6*a - 15*tan((c + d*x)/2)**6*b + 1 5*tan((c + d*x)/2)**6*c + 63*tan((c + d*x)/2)**4*a - 21*tan((c + d*x)/2)** 4*b - 21*tan((c + d*x)/2)**4*c + 105*tan((c + d*x)/2)**2*a + 35*tan((c + d *x)/2)**2*b - 35*tan((c + d*x)/2)**2*c + 105*a + 105*b + 105*c))/(840*a**4 *d)