Integrand size = 15, antiderivative size = 96 \[ \int \frac {A+B \sec (x)}{(a+a \cos (x))^4} \, dx=\frac {B \text {arctanh}(\sin (x))}{a^4}+\frac {(6 A-55 B) \sin (x)}{105 a^4 (1+\cos (x))^2}+\frac {2 (3 A-80 B) \sin (x)}{105 a^4 (1+\cos (x))}+\frac {(A-B) \sin (x)}{7 (a+a \cos (x))^4}+\frac {(3 A-10 B) \sin (x)}{35 a (a+a \cos (x))^3} \] Output:
B*arctanh(sin(x))/a^4+1/105*(6*A-55*B)*sin(x)/a^4/(1+cos(x))^2+2/105*(3*A- 80*B)*sin(x)/a^4/(1+cos(x))+1/7*(A-B)*sin(x)/(a+a*cos(x))^4+1/35*(3*A-10*B )*sin(x)/a/(a+a*cos(x))^3
Time = 0.69 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.08 \[ \int \frac {A+B \sec (x)}{(a+a \cos (x))^4} \, dx=\frac {-3360 B \cos ^8\left (\frac {x}{2}\right ) \left (\log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )-\log \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )\right )+(96 A-1055 B+(87 A-1480 B) \cos (x)+(24 A-535 B) \cos (2 x)+3 A \cos (3 x)-80 B \cos (3 x)) \sin (x)}{210 a^4 (1+\cos (x))^4} \] Input:
Integrate[(A + B*Sec[x])/(a + a*Cos[x])^4,x]
Output:
(-3360*B*Cos[x/2]^8*(Log[Cos[x/2] - Sin[x/2]] - Log[Cos[x/2] + Sin[x/2]]) + (96*A - 1055*B + (87*A - 1480*B)*Cos[x] + (24*A - 535*B)*Cos[2*x] + 3*A* Cos[3*x] - 80*B*Cos[3*x])*Sin[x])/(210*a^4*(1 + Cos[x])^4)
Time = 0.90 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.21, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.867, Rules used = {3042, 3307, 3042, 3457, 3042, 3457, 3042, 3457, 3042, 3457, 27, 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.
| \(\displaystyle \int \frac {A+B \sec (x)}{(a \cos (x)+a)^4} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \csc \left (x+\frac {\pi }{2}\right )}{\left (a \sin \left (x+\frac {\pi }{2}\right )+a\right )^4}dx\) |
| \(\Big \downarrow \) 3307 |
| \(\displaystyle \int \frac {\sec (x) (A \cos (x)+B)}{(a \cos (x)+a)^4}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A \sin \left (x+\frac {\pi }{2}\right )+B}{\sin \left (x+\frac {\pi }{2}\right ) \left (a \sin \left (x+\frac {\pi }{2}\right )+a\right )^4}dx\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\int \frac {(7 a B+3 a (A-B) \cos (x)) \sec (x)}{(\cos (x) a+a)^3}dx}{7 a^2}+\frac {(A-B) \sin (x)}{7 (a \cos (x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {7 a B+3 a (A-B) \sin \left (x+\frac {\pi }{2}\right )}{\sin \left (x+\frac {\pi }{2}\right ) \left (\sin \left (x+\frac {\pi }{2}\right ) a+a\right )^3}dx}{7 a^2}+\frac {(A-B) \sin (x)}{7 (a \cos (x)+a)^4}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\frac {\int \frac {\left (35 B a^2+2 (3 A-10 B) \cos (x) a^2\right ) \sec (x)}{(\cos (x) a+a)^2}dx}{5 a^2}+\frac {a (3 A-10 B) \sin (x)}{5 (a \cos (x)+a)^3}}{7 a^2}+\frac {(A-B) \sin (x)}{7 (a \cos (x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {35 B a^2+2 (3 A-10 B) \sin \left (x+\frac {\pi }{2}\right ) a^2}{\sin \left (x+\frac {\pi }{2}\right ) \left (\sin \left (x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}+\frac {a (3 A-10 B) \sin (x)}{5 (a \cos (x)+a)^3}}{7 a^2}+\frac {(A-B) \sin (x)}{7 (a \cos (x)+a)^4}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {\left (105 B a^3+(6 A-55 B) \cos (x) a^3\right ) \sec (x)}{\cos (x) a+a}dx}{3 a^2}+\frac {(6 A-55 B) \sin (x)}{3 (\cos (x)+1)^2}}{5 a^2}+\frac {a (3 A-10 B) \sin (x)}{5 (a \cos (x)+a)^3}}{7 a^2}+\frac {(A-B) \sin (x)}{7 (a \cos (x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {105 B a^3+(6 A-55 B) \sin \left (x+\frac {\pi }{2}\right ) a^3}{\sin \left (x+\frac {\pi }{2}\right ) \left (\sin \left (x+\frac {\pi }{2}\right ) a+a\right )}dx}{3 a^2}+\frac {(6 A-55 B) \sin (x)}{3 (\cos (x)+1)^2}}{5 a^2}+\frac {a (3 A-10 B) \sin (x)}{5 (a \cos (x)+a)^3}}{7 a^2}+\frac {(A-B) \sin (x)}{7 (a \cos (x)+a)^4}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\frac {\frac {\frac {\int 105 a^4 B \sec (x)dx}{a^2}+\frac {2 a^3 (3 A-80 B) \sin (x)}{a \cos (x)+a}}{3 a^2}+\frac {(6 A-55 B) \sin (x)}{3 (\cos (x)+1)^2}}{5 a^2}+\frac {a (3 A-10 B) \sin (x)}{5 (a \cos (x)+a)^3}}{7 a^2}+\frac {(A-B) \sin (x)}{7 (a \cos (x)+a)^4}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {105 a^2 B \int \sec (x)dx+\frac {2 a^3 (3 A-80 B) \sin (x)}{a \cos (x)+a}}{3 a^2}+\frac {(6 A-55 B) \sin (x)}{3 (\cos (x)+1)^2}}{5 a^2}+\frac {a (3 A-10 B) \sin (x)}{5 (a \cos (x)+a)^3}}{7 a^2}+\frac {(A-B) \sin (x)}{7 (a \cos (x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {105 a^2 B \int \csc \left (x+\frac {\pi }{2}\right )dx+\frac {2 a^3 (3 A-80 B) \sin (x)}{a \cos (x)+a}}{3 a^2}+\frac {(6 A-55 B) \sin (x)}{3 (\cos (x)+1)^2}}{5 a^2}+\frac {a (3 A-10 B) \sin (x)}{5 (a \cos (x)+a)^3}}{7 a^2}+\frac {(A-B) \sin (x)}{7 (a \cos (x)+a)^4}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {\frac {\frac {2 a^3 (3 A-80 B) \sin (x)}{a \cos (x)+a}+105 a^2 B \text {arctanh}(\sin (x))}{3 a^2}+\frac {(6 A-55 B) \sin (x)}{3 (\cos (x)+1)^2}}{5 a^2}+\frac {a (3 A-10 B) \sin (x)}{5 (a \cos (x)+a)^3}}{7 a^2}+\frac {(A-B) \sin (x)}{7 (a \cos (x)+a)^4}\) |
Input:
Int[(A + B*Sec[x])/(a + a*Cos[x])^4,x]
Output:
((A - B)*Sin[x])/(7*(a + a*Cos[x])^4) + ((a*(3*A - 10*B)*Sin[x])/(5*(a + a *Cos[x])^3) + (((6*A - 55*B)*Sin[x])/(3*(1 + Cos[x])^2) + (105*a^2*B*ArcTa nh[Sin[x]] + (2*a^3*(3*A - 80*B)*Sin[x])/(a + a*Cos[x]))/(3*a^2))/(5*a^2)) /(7*a^2)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*((d + c*Sin[e + f*x])^n/Sin[e + f*x]^n), x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IntegerQ [n]
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*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^( n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/(a*(2*m + 1)*(b*c - a*d)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b *d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*(m + n + 2)*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] && LtQ[m, -2^(-1)] && !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 0.21 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.82
| method | result | size |
| parallelrisch | \(\frac {-56 B \ln \left (\tan \left (\frac {x}{2}\right )-1\right )+56 B \ln \left (\tan \left (\frac {x}{2}\right )+1\right )+\left (\left (A -B \right ) \tan \left (\frac {x}{2}\right )^{6}+7 \left (\frac {3 A}{5}-B \right ) \tan \left (\frac {x}{2}\right )^{4}+7 \left (A -\frac {11 B}{3}\right ) \tan \left (\frac {x}{2}\right )^{2}+7 A -105 B \right ) \tan \left (\frac {x}{2}\right )}{56 a^{4}}\) | \(79\) |
| default | \(\frac {A \tan \left (\frac {x}{2}\right )^{3}-\frac {11 B \tan \left (\frac {x}{2}\right )^{3}}{3}-8 B \ln \left (\tan \left (\frac {x}{2}\right )-1\right )+A \tan \left (\frac {x}{2}\right )-15 B \tan \left (\frac {x}{2}\right )+8 B \ln \left (\tan \left (\frac {x}{2}\right )+1\right )+\frac {\tan \left (\frac {x}{2}\right )^{7} A}{7}-\frac {\tan \left (\frac {x}{2}\right )^{7} B}{7}+\frac {3 A \tan \left (\frac {x}{2}\right )^{5}}{5}-B \tan \left (\frac {x}{2}\right )^{5}}{8 a^{4}}\) | \(93\) |
| norman | \(\frac {\frac {\left (A -15 B \right ) \tan \left (\frac {x}{2}\right )}{8 a}+\frac {\left (A -B \right ) \tan \left (\frac {x}{2}\right )^{7}}{56 a}+\frac {\left (3 A -11 B \right ) \tan \left (\frac {x}{2}\right )^{3}}{24 a}+\frac {\left (3 A -5 B \right ) \tan \left (\frac {x}{2}\right )^{5}}{40 a}}{a^{3}}+\frac {B \ln \left (\tan \left (\frac {x}{2}\right )+1\right )}{a^{4}}-\frac {B \ln \left (\tan \left (\frac {x}{2}\right )-1\right )}{a^{4}}\) | \(98\) |
| risch | \(-\frac {2 i \left (105 B \,{\mathrm e}^{6 i x}+735 B \,{\mathrm e}^{5 i x}+2170 B \,{\mathrm e}^{4 i x}-210 A \,{\mathrm e}^{3 i x}+3430 B \,{\mathrm e}^{3 i x}-126 A \,{\mathrm e}^{2 i x}+2625 B \,{\mathrm e}^{2 i x}-42 A \,{\mathrm e}^{i x}+1015 B \,{\mathrm e}^{i x}-6 A +160 B \right )}{105 \left ({\mathrm e}^{i x}+1\right )^{7} a^{4}}+\frac {B \ln \left ({\mathrm e}^{i x}+i\right )}{a^{4}}-\frac {B \ln \left ({\mathrm e}^{i x}-i\right )}{a^{4}}\) | \(125\) |
Input:
int((A+B*sec(x))/(a+a*cos(x))^4,x,method=_RETURNVERBOSE)
Output:
1/56*(-56*B*ln(tan(1/2*x)-1)+56*B*ln(tan(1/2*x)+1)+((A-B)*tan(1/2*x)^6+7*( 3/5*A-B)*tan(1/2*x)^4+7*(A-11/3*B)*tan(1/2*x)^2+7*A-105*B)*tan(1/2*x))/a^4
Time = 0.09 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.65 \[ \int \frac {A+B \sec (x)}{(a+a \cos (x))^4} \, dx=\frac {105 \, {\left (B \cos \left (x\right )^{4} + 4 \, B \cos \left (x\right )^{3} + 6 \, B \cos \left (x\right )^{2} + 4 \, B \cos \left (x\right ) + B\right )} \log \left (\sin \left (x\right ) + 1\right ) - 105 \, {\left (B \cos \left (x\right )^{4} + 4 \, B \cos \left (x\right )^{3} + 6 \, B \cos \left (x\right )^{2} + 4 \, B \cos \left (x\right ) + B\right )} \log \left (-\sin \left (x\right ) + 1\right ) + 2 \, {\left (2 \, {\left (3 \, A - 80 \, B\right )} \cos \left (x\right )^{3} + {\left (24 \, A - 535 \, B\right )} \cos \left (x\right )^{2} + {\left (39 \, A - 620 \, B\right )} \cos \left (x\right ) + 36 \, A - 260 \, B\right )} \sin \left (x\right )}{210 \, {\left (a^{4} \cos \left (x\right )^{4} + 4 \, a^{4} \cos \left (x\right )^{3} + 6 \, a^{4} \cos \left (x\right )^{2} + 4 \, a^{4} \cos \left (x\right ) + a^{4}\right )}} \] Input:
integrate((A+B*sec(x))/(a+a*cos(x))^4,x, algorithm="fricas")
Output:
1/210*(105*(B*cos(x)^4 + 4*B*cos(x)^3 + 6*B*cos(x)^2 + 4*B*cos(x) + B)*log (sin(x) + 1) - 105*(B*cos(x)^4 + 4*B*cos(x)^3 + 6*B*cos(x)^2 + 4*B*cos(x) + B)*log(-sin(x) + 1) + 2*(2*(3*A - 80*B)*cos(x)^3 + (24*A - 535*B)*cos(x) ^2 + (39*A - 620*B)*cos(x) + 36*A - 260*B)*sin(x))/(a^4*cos(x)^4 + 4*a^4*c os(x)^3 + 6*a^4*cos(x)^2 + 4*a^4*cos(x) + a^4)
\[ \int \frac {A+B \sec (x)}{(a+a \cos (x))^4} \, dx=\frac {\int \frac {A}{\cos ^{4}{\left (x \right )} + 4 \cos ^{3}{\left (x \right )} + 6 \cos ^{2}{\left (x \right )} + 4 \cos {\left (x \right )} + 1}\, dx + \int \frac {B \sec {\left (x \right )}}{\cos ^{4}{\left (x \right )} + 4 \cos ^{3}{\left (x \right )} + 6 \cos ^{2}{\left (x \right )} + 4 \cos {\left (x \right )} + 1}\, dx}{a^{4}} \] Input:
integrate((A+B*sec(x))/(a+a*cos(x))**4,x)
Output:
(Integral(A/(cos(x)**4 + 4*cos(x)**3 + 6*cos(x)**2 + 4*cos(x) + 1), x) + I ntegral(B*sec(x)/(cos(x)**4 + 4*cos(x)**3 + 6*cos(x)**2 + 4*cos(x) + 1), x ))/a**4
Time = 0.04 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.49 \[ \int \frac {A+B \sec (x)}{(a+a \cos (x))^4} \, dx=-\frac {1}{168} \, B {\left (\frac {\frac {315 \, \sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac {77 \, \sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (x\right )^{5}}{{\left (\cos \left (x\right ) + 1\right )}^{5}} + \frac {3 \, \sin \left (x\right )^{7}}{{\left (\cos \left (x\right ) + 1\right )}^{7}}}{a^{4}} - \frac {168 \, \log \left (\frac {\sin \left (x\right )}{\cos \left (x\right ) + 1} + 1\right )}{a^{4}} + \frac {168 \, \log \left (\frac {\sin \left (x\right )}{\cos \left (x\right ) + 1} - 1\right )}{a^{4}}\right )} + \frac {A {\left (\frac {35 \, \sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac {35 \, \sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (x\right )^{5}}{{\left (\cos \left (x\right ) + 1\right )}^{5}} + \frac {5 \, \sin \left (x\right )^{7}}{{\left (\cos \left (x\right ) + 1\right )}^{7}}\right )}}{280 \, a^{4}} \] Input:
integrate((A+B*sec(x))/(a+a*cos(x))^4,x, algorithm="maxima")
Output:
-1/168*B*((315*sin(x)/(cos(x) + 1) + 77*sin(x)^3/(cos(x) + 1)^3 + 21*sin(x )^5/(cos(x) + 1)^5 + 3*sin(x)^7/(cos(x) + 1)^7)/a^4 - 168*log(sin(x)/(cos( x) + 1) + 1)/a^4 + 168*log(sin(x)/(cos(x) + 1) - 1)/a^4) + 1/280*A*(35*sin (x)/(cos(x) + 1) + 35*sin(x)^3/(cos(x) + 1)^3 + 21*sin(x)^5/(cos(x) + 1)^5 + 5*sin(x)^7/(cos(x) + 1)^7)/a^4
Time = 0.12 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.31 \[ \int \frac {A+B \sec (x)}{(a+a \cos (x))^4} \, dx=\frac {B \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) + 1 \right |}\right )}{a^{4}} - \frac {B \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) - 1 \right |}\right )}{a^{4}} + \frac {15 \, A a^{24} \tan \left (\frac {1}{2} \, x\right )^{7} - 15 \, B a^{24} \tan \left (\frac {1}{2} \, x\right )^{7} + 63 \, A a^{24} \tan \left (\frac {1}{2} \, x\right )^{5} - 105 \, B a^{24} \tan \left (\frac {1}{2} \, x\right )^{5} + 105 \, A a^{24} \tan \left (\frac {1}{2} \, x\right )^{3} - 385 \, B a^{24} \tan \left (\frac {1}{2} \, x\right )^{3} + 105 \, A a^{24} \tan \left (\frac {1}{2} \, x\right ) - 1575 \, B a^{24} \tan \left (\frac {1}{2} \, x\right )}{840 \, a^{28}} \] Input:
integrate((A+B*sec(x))/(a+a*cos(x))^4,x, algorithm="giac")
Output:
B*log(abs(tan(1/2*x) + 1))/a^4 - B*log(abs(tan(1/2*x) - 1))/a^4 + 1/840*(1 5*A*a^24*tan(1/2*x)^7 - 15*B*a^24*tan(1/2*x)^7 + 63*A*a^24*tan(1/2*x)^5 - 105*B*a^24*tan(1/2*x)^5 + 105*A*a^24*tan(1/2*x)^3 - 385*B*a^24*tan(1/2*x)^ 3 + 105*A*a^24*tan(1/2*x) - 1575*B*a^24*tan(1/2*x))/a^28
Time = 15.58 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.46 \[ \int \frac {A+B \sec (x)}{(a+a \cos (x))^4} \, dx=\mathrm {tan}\left (\frac {x}{2}\right )\,\left (\frac {A-B}{8\,a^4}-\frac {3\,B}{4\,a^4}+\frac {2\,A-4\,B}{8\,a^4}-\frac {2\,A+4\,B}{8\,a^4}\right )+{\mathrm {tan}\left (\frac {x}{2}\right )}^5\,\left (\frac {A-B}{40\,a^4}+\frac {2\,A-4\,B}{40\,a^4}\right )+{\mathrm {tan}\left (\frac {x}{2}\right )}^3\,\left (\frac {A-B}{24\,a^4}-\frac {B}{4\,a^4}+\frac {2\,A-4\,B}{24\,a^4}\right )+\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^7\,\left (A-B\right )}{56\,a^4}+\frac {2\,B\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {x}{2}\right )\right )}{a^4} \] Input:
int((A + B/cos(x))/(a + a*cos(x))^4,x)
Output:
tan(x/2)*((A - B)/(8*a^4) - (3*B)/(4*a^4) + (2*A - 4*B)/(8*a^4) - (2*A + 4 *B)/(8*a^4)) + tan(x/2)^5*((A - B)/(40*a^4) + (2*A - 4*B)/(40*a^4)) + tan( x/2)^3*((A - B)/(24*a^4) - B/(4*a^4) + (2*A - 4*B)/(24*a^4)) + (tan(x/2)^7 *(A - B))/(56*a^4) + (2*B*atanh(tan(x/2)))/a^4
Time = 0.16 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.98 \[ \int \frac {A+B \sec (x)}{(a+a \cos (x))^4} \, dx=\frac {-840 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )-1\right ) b +840 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )+1\right ) b +15 \tan \left (\frac {x}{2}\right )^{7} a -15 \tan \left (\frac {x}{2}\right )^{7} b +63 \tan \left (\frac {x}{2}\right )^{5} a -105 \tan \left (\frac {x}{2}\right )^{5} b +105 \tan \left (\frac {x}{2}\right )^{3} a -385 \tan \left (\frac {x}{2}\right )^{3} b +105 \tan \left (\frac {x}{2}\right ) a -1575 \tan \left (\frac {x}{2}\right ) b}{840 a^{4}} \] Input:
int((A+B*sec(x))/(a+a*cos(x))^4,x)
Output:
( - 840*log(tan(x/2) - 1)*b + 840*log(tan(x/2) + 1)*b + 15*tan(x/2)**7*a - 15*tan(x/2)**7*b + 63*tan(x/2)**5*a - 105*tan(x/2)**5*b + 105*tan(x/2)**3 *a - 385*tan(x/2)**3*b + 105*tan(x/2)*a - 1575*tan(x/2)*b)/(840*a**4)