Integrand size = 40, antiderivative size = 214 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=-\frac {\left (6 a^4 b B-5 a^2 b^3 B+2 b^5 B-2 a^5 C-a^3 b^2 C\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 (a-b)^{5/2} (a+b)^{5/2} d}+\frac {B \text {arctanh}(\sin (c+d x))}{a^3 d}+\frac {b (b B-a C) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac {b \left (5 a^2 b B-2 b^3 B-3 a^3 C\right ) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))} \]
-(6*B*a^4*b-5*B*a^2*b^3+2*B*b^5-2*C*a^5-C*a^3*b^2)*arctan((a-b)^(1/2)*tan( 1/2*d*x+1/2*c)/(a+b)^(1/2))/a^3/(a-b)^(5/2)/(a+b)^(5/2)/d+B*arctanh(sin(d* x+c))/a^3/d+1/2*b*(B*b-C*a)*sin(d*x+c)/a/(a^2-b^2)/d/(a+b*cos(d*x+c))^2+1/ 2*b*(5*B*a^2*b-2*B*b^3-3*C*a^3)*sin(d*x+c)/a^2/(a^2-b^2)^2/d/(a+b*cos(d*x+ c))
Time = 1.98 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.26 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\frac {\cos (c+d x) (C+B \sec (c+d x)) \left (-\frac {2 \left (-6 a^4 b B+5 a^2 b^3 B-2 b^5 B+2 a^5 C+a^3 b^2 C\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{5/2}}-2 B \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 B \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {a^2 b (b B-a C) \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))^2}+\frac {a b \left (5 a^2 b B-2 b^3 B-3 a^3 C\right ) \sin (c+d x)}{(a-b)^2 (a+b)^2 (a+b \cos (c+d x))}\right )}{2 a^3 d (B+C \cos (c+d x))} \]
(Cos[c + d*x]*(C + B*Sec[c + d*x])*((-2*(-6*a^4*b*B + 5*a^2*b^3*B - 2*b^5* B + 2*a^5*C + a^3*b^2*C)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^ 2]])/(-a^2 + b^2)^(5/2) - 2*B*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 2 *B*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (a^2*b*(b*B - a*C)*Sin[c + d *x])/((a - b)*(a + b)*(a + b*Cos[c + d*x])^2) + (a*b*(5*a^2*b*B - 2*b^3*B - 3*a^3*C)*Sin[c + d*x])/((a - b)^2*(a + b)^2*(a + b*Cos[c + d*x]))))/(2*a ^3*d*(B + C*Cos[c + d*x]))
Time = 1.34 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.21, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3042, 3508, 3042, 3479, 3042, 3534, 3042, 3480, 3042, 3138, 218, 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.
| \(\displaystyle \int \frac {\sec ^2(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^3}dx\) |
| \(\Big \downarrow \) 3508 |
| \(\displaystyle \int \frac {\sec (c+d x) (B+C \cos (c+d x))}{(a+b \cos (c+d x))^3}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {B+C \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^3}dx\) |
| \(\Big \downarrow \) 3479 |
| \(\displaystyle \frac {\int \frac {\left (b (b B-a C) \cos ^2(c+d x)-2 a (b B-a C) \cos (c+d x)+2 \left (a^2-b^2\right ) B\right ) \sec (c+d x)}{(a+b \cos (c+d x))^2}dx}{2 a \left (a^2-b^2\right )}+\frac {b (b B-a C) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {b (b B-a C) \sin \left (c+d x+\frac {\pi }{2}\right )^2-2 a (b B-a C) \sin \left (c+d x+\frac {\pi }{2}\right )+2 \left (a^2-b^2\right ) B}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx}{2 a \left (a^2-b^2\right )}+\frac {b (b B-a C) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 3534 |
| \(\displaystyle \frac {\frac {\int \frac {\left (2 \left (a^2-b^2\right )^2 B-a \left (-2 C a^3+4 b B a^2-b^2 C a-b^3 B\right ) \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)}dx}{a \left (a^2-b^2\right )}+\frac {b \left (-3 a^3 C+5 a^2 b B-2 b^3 B\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b (b B-a C) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {2 \left (a^2-b^2\right )^2 B-a \left (-2 C a^3+4 b B a^2-b^2 C a-b^3 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a \left (a^2-b^2\right )}+\frac {b \left (-3 a^3 C+5 a^2 b B-2 b^3 B\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b (b B-a C) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 3480 |
| \(\displaystyle \frac {\frac {\frac {2 B \left (a^2-b^2\right )^2 \int \sec (c+d x)dx}{a}-\frac {\left (-2 a^5 C+6 a^4 b B-a^3 b^2 C-5 a^2 b^3 B+2 b^5 B\right ) \int \frac {1}{a+b \cos (c+d x)}dx}{a}}{a \left (a^2-b^2\right )}+\frac {b \left (-3 a^3 C+5 a^2 b B-2 b^3 B\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b (b B-a C) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {2 B \left (a^2-b^2\right )^2 \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {\left (-2 a^5 C+6 a^4 b B-a^3 b^2 C-5 a^2 b^3 B+2 b^5 B\right ) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a}}{a \left (a^2-b^2\right )}+\frac {b \left (-3 a^3 C+5 a^2 b B-2 b^3 B\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b (b B-a C) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {\frac {\frac {2 B \left (a^2-b^2\right )^2 \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {2 \left (-2 a^5 C+6 a^4 b B-a^3 b^2 C-5 a^2 b^3 B+2 b^5 B\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{a d}}{a \left (a^2-b^2\right )}+\frac {b \left (-3 a^3 C+5 a^2 b B-2 b^3 B\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b (b B-a C) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\frac {2 B \left (a^2-b^2\right )^2 \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {2 \left (-2 a^5 C+6 a^4 b B-a^3 b^2 C-5 a^2 b^3 B+2 b^5 B\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}}{a \left (a^2-b^2\right )}+\frac {b \left (-3 a^3 C+5 a^2 b B-2 b^3 B\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b (b B-a C) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {b (b B-a C) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}+\frac {\frac {b \left (-3 a^3 C+5 a^2 b B-2 b^3 B\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {\frac {2 B \left (a^2-b^2\right )^2 \text {arctanh}(\sin (c+d x))}{a d}-\frac {2 \left (-2 a^5 C+6 a^4 b B-a^3 b^2 C-5 a^2 b^3 B+2 b^5 B\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}}{a \left (a^2-b^2\right )}}{2 a \left (a^2-b^2\right )}\) |
(b*(b*B - a*C)*Sin[c + d*x])/(2*a*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^2) + (((-2*(6*a^4*b*B - 5*a^2*b^3*B + 2*b^5*B - 2*a^5*C - a^3*b^2*C)*ArcTan[(Sq rt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*Sqrt[a + b]*d) + (2*(a^2 - b^2)^2*B*ArcTanh[Sin[c + d*x]])/(a*d))/(a*(a^2 - b^2)) + (b*(5*a ^2*b*B - 2*b^3*B - 3*a^3*C)*Sin[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Cos[c + d*x])))/(2*a*(a^2 - b^2))
3.9.12.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si mp[(-(A*b^2 - a*b*B))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin [e + f*x])^(1 + n)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(a*A - b*B)*(b*c - a*d)*(m + 1) + b*d*(A*b - a*B)*(m + n + 2) + (A*b - a*B)*(a*d*(m + 1) - b*c*(m + 2))*Sin[e + f*x] - b*d*(A*b - a*B) *(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n }, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Rat ionalQ[m] && m < -1 && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !(I ntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) || EqQ[a, 0]) ))
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_ .)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[(A*b - a*B)/(b*c - a*d) Int[1/(a + b*Sin[e + f*x]), x], x] + Simp[(B*c - A*d)/ (b*c - a*d) Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f , A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 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[1/b^2 Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*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*b^2 - a*b*B + a^2*C))*Cos[e + f*x ]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b* c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)) Int [(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b*c - a* d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A *b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ [n]) || !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) | | EqQ[a, 0])))
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 2.16 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.41
| method | result | size |
| derivativedivides | \(\frac {\frac {B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{3}}-\frac {2 \left (\frac {-\frac {\left (6 B \,a^{2} b +B a \,b^{2}-2 B \,b^{3}-4 C \,a^{3}-C \,a^{2} b \right ) a b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 \left (a^{2}+2 a b +b^{2}\right ) \left (a -b \right )}-\frac {b a \left (6 B \,a^{2} b -B a \,b^{2}-2 B \,b^{3}-4 C \,a^{3}+C \,a^{2} b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a -b \right )^{2}}}{{\left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b \right )}^{2}}+\frac {\left (6 B \,a^{4} b -5 B \,a^{2} b^{3}+2 B \,b^{5}-2 C \,a^{5}-C \,a^{3} b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{3}}-\frac {B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{3}}}{d}\) | \(302\) |
| default | \(\frac {\frac {B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{3}}-\frac {2 \left (\frac {-\frac {\left (6 B \,a^{2} b +B a \,b^{2}-2 B \,b^{3}-4 C \,a^{3}-C \,a^{2} b \right ) a b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 \left (a^{2}+2 a b +b^{2}\right ) \left (a -b \right )}-\frac {b a \left (6 B \,a^{2} b -B a \,b^{2}-2 B \,b^{3}-4 C \,a^{3}+C \,a^{2} b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a -b \right )^{2}}}{{\left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b \right )}^{2}}+\frac {\left (6 B \,a^{4} b -5 B \,a^{2} b^{3}+2 B \,b^{5}-2 C \,a^{5}-C \,a^{3} b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{3}}-\frac {B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{3}}}{d}\) | \(302\) |
| risch | \(\text {Expression too large to display}\) | \(1191\) |
1/d*(B/a^3*ln(tan(1/2*d*x+1/2*c)+1)-2/a^3*((-1/2*(6*B*a^2*b+B*a*b^2-2*B*b^ 3-4*C*a^3-C*a^2*b)*a*b/(a^2+2*a*b+b^2)/(a-b)*tan(1/2*d*x+1/2*c)^3-1/2*b*a* (6*B*a^2*b-B*a*b^2-2*B*b^3-4*C*a^3+C*a^2*b)/(a+b)/(a-b)^2*tan(1/2*d*x+1/2* c))/(tan(1/2*d*x+1/2*c)^2*a-b*tan(1/2*d*x+1/2*c)^2+a+b)^2+1/2*(6*B*a^4*b-5 *B*a^2*b^3+2*B*b^5-2*C*a^5-C*a^3*b^2)/(a^4-2*a^2*b^2+b^4)/((a-b)*(a+b))^(1 /2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a-b)*(a+b))^(1/2)))-B/a^3*ln(tan(1/2 *d*x+1/2*c)-1))
Leaf count of result is larger than twice the leaf count of optimal. 665 vs. \(2 (199) = 398\).
Time = 12.83 (sec) , antiderivative size = 1400, normalized size of antiderivative = 6.54 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Too large to display} \]
integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+b*cos(d*x+c))^3,x, algorithm="fricas")
[1/4*((2*C*a^7 - 6*B*a^6*b + C*a^5*b^2 + 5*B*a^4*b^3 - 2*B*a^2*b^5 + (2*C* a^5*b^2 - 6*B*a^4*b^3 + C*a^3*b^4 + 5*B*a^2*b^5 - 2*B*b^7)*cos(d*x + c)^2 + 2*(2*C*a^6*b - 6*B*a^5*b^2 + C*a^4*b^3 + 5*B*a^3*b^4 - 2*B*a*b^6)*cos(d* x + c))*sqrt(-a^2 + b^2)*log((2*a*b*cos(d*x + c) + (2*a^2 - b^2)*cos(d*x + c)^2 - 2*sqrt(-a^2 + b^2)*(a*cos(d*x + c) + b)*sin(d*x + c) - a^2 + 2*b^2 )/(b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2)) + 2*(B*a^8 - 3*B*a^6*b^ 2 + 3*B*a^4*b^4 - B*a^2*b^6 + (B*a^6*b^2 - 3*B*a^4*b^4 + 3*B*a^2*b^6 - B*b ^8)*cos(d*x + c)^2 + 2*(B*a^7*b - 3*B*a^5*b^3 + 3*B*a^3*b^5 - B*a*b^7)*cos (d*x + c))*log(sin(d*x + c) + 1) - 2*(B*a^8 - 3*B*a^6*b^2 + 3*B*a^4*b^4 - B*a^2*b^6 + (B*a^6*b^2 - 3*B*a^4*b^4 + 3*B*a^2*b^6 - B*b^8)*cos(d*x + c)^2 + 2*(B*a^7*b - 3*B*a^5*b^3 + 3*B*a^3*b^5 - B*a*b^7)*cos(d*x + c))*log(-si n(d*x + c) + 1) - 2*(4*C*a^7*b - 6*B*a^6*b^2 - 5*C*a^5*b^3 + 9*B*a^4*b^4 + C*a^3*b^5 - 3*B*a^2*b^6 + (3*C*a^6*b^2 - 5*B*a^5*b^3 - 3*C*a^4*b^4 + 7*B* a^3*b^5 - 2*B*a*b^7)*cos(d*x + c))*sin(d*x + c))/((a^9*b^2 - 3*a^7*b^4 + 3 *a^5*b^6 - a^3*b^8)*d*cos(d*x + c)^2 + 2*(a^10*b - 3*a^8*b^3 + 3*a^6*b^5 - a^4*b^7)*d*cos(d*x + c) + (a^11 - 3*a^9*b^2 + 3*a^7*b^4 - a^5*b^6)*d), 1/ 2*((2*C*a^7 - 6*B*a^6*b + C*a^5*b^2 + 5*B*a^4*b^3 - 2*B*a^2*b^5 + (2*C*a^5 *b^2 - 6*B*a^4*b^3 + C*a^3*b^4 + 5*B*a^2*b^5 - 2*B*b^7)*cos(d*x + c)^2 + 2 *(2*C*a^6*b - 6*B*a^5*b^2 + C*a^4*b^3 + 5*B*a^3*b^4 - 2*B*a*b^6)*cos(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(...
\[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\int \frac {\left (B + C \cos {\left (c + d x \right )}\right ) \cos {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\left (a + b \cos {\left (c + d x \right )}\right )^{3}}\, dx \]
Exception generated. \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Exception raised: ValueError} \]
integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+b*cos(d*x+c))^3,x, algorithm="maxima")
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f or more de
Leaf count of result is larger than twice the leaf count of optimal. 481 vs. \(2 (199) = 398\).
Time = 0.38 (sec) , antiderivative size = 481, normalized size of antiderivative = 2.25 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\frac {\frac {{\left (2 \, C a^{5} - 6 \, B a^{4} b + C a^{3} b^{2} + 5 \, B a^{2} b^{3} - 2 \, B b^{5}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} \sqrt {a^{2} - b^{2}}} + \frac {B \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {B \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} - \frac {4 \, C a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, B a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, C a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 5 \, B a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, B a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, C a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, B a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, C a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 5 \, B a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - C a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, B b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )}^{2}}}{d} \]
((2*C*a^5 - 6*B*a^4*b + C*a^3*b^2 + 5*B*a^2*b^3 - 2*B*b^5)*(pi*floor(1/2*( d*x + c)/pi + 1/2)*sgn(2*a - 2*b) + arctan((a*tan(1/2*d*x + 1/2*c) - b*tan (1/2*d*x + 1/2*c))/sqrt(a^2 - b^2)))/((a^7 - 2*a^5*b^2 + a^3*b^4)*sqrt(a^2 - b^2)) + B*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - B*log(abs(tan(1/2*d* x + 1/2*c) - 1))/a^3 - (4*C*a^4*b*tan(1/2*d*x + 1/2*c)^3 - 6*B*a^3*b^2*tan (1/2*d*x + 1/2*c)^3 - 3*C*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 + 5*B*a^2*b^3*tan (1/2*d*x + 1/2*c)^3 - C*a^2*b^3*tan(1/2*d*x + 1/2*c)^3 + 3*B*a*b^4*tan(1/2 *d*x + 1/2*c)^3 - 2*B*b^5*tan(1/2*d*x + 1/2*c)^3 + 4*C*a^4*b*tan(1/2*d*x + 1/2*c) - 6*B*a^3*b^2*tan(1/2*d*x + 1/2*c) + 3*C*a^3*b^2*tan(1/2*d*x + 1/2 *c) - 5*B*a^2*b^3*tan(1/2*d*x + 1/2*c) - C*a^2*b^3*tan(1/2*d*x + 1/2*c) + 3*B*a*b^4*tan(1/2*d*x + 1/2*c) + 2*B*b^5*tan(1/2*d*x + 1/2*c))/((a^6 - 2*a ^4*b^2 + a^2*b^4)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 + a + b)^2))/d
Time = 11.66 (sec) , antiderivative size = 6911, normalized size of antiderivative = 32.29 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Too large to display} \]
((tan(c/2 + (d*x)/2)^3*(2*B*b^4 - 6*B*a^2*b^2 + C*a^2*b^2 - B*a*b^3 + 4*C* a^3*b))/((a^2*b - a^3)*(a + b)^2) - (tan(c/2 + (d*x)/2)*(2*B*b^4 - 6*B*a^2 *b^2 - C*a^2*b^2 + B*a*b^3 + 4*C*a^3*b))/((a + b)*(a^4 - 2*a^3*b + a^2*b^2 )))/(d*(2*a*b + tan(c/2 + (d*x)/2)^2*(2*a^2 - 2*b^2) + tan(c/2 + (d*x)/2)^ 4*(a^2 - 2*a*b + b^2) + a^2 + b^2)) - (B*atan(-((B*((B*((8*(4*B*a^15 + 4*C *a^15 - 4*B*a^6*b^9 + 2*B*a^7*b^8 + 18*B*a^8*b^7 - 4*B*a^9*b^6 - 36*B*a^10 *b^5 + 6*B*a^11*b^4 + 34*B*a^12*b^3 - 8*B*a^13*b^2 - 2*C*a^8*b^7 + 2*C*a^9 *b^6 + 6*C*a^12*b^3 - 6*C*a^13*b^2 - 12*B*a^14*b - 4*C*a^14*b))/(a^12*b + a^13 - a^6*b^7 - a^7*b^6 + 3*a^8*b^5 + 3*a^9*b^4 - 3*a^10*b^3 - 3*a^11*b^2 ) - (8*B*tan(c/2 + (d*x)/2)*(8*a^15*b - 8*a^6*b^10 + 8*a^7*b^9 + 32*a^8*b^ 8 - 32*a^9*b^7 - 48*a^10*b^6 + 48*a^11*b^5 + 32*a^12*b^4 - 32*a^13*b^3 - 8 *a^14*b^2))/(a^3*(a^10*b + a^11 - a^4*b^7 - a^5*b^6 + 3*a^6*b^5 + 3*a^7*b^ 4 - 3*a^8*b^3 - 3*a^9*b^2))))/a^3 - (8*tan(c/2 + (d*x)/2)*(4*B^2*a^10 + 8* B^2*b^10 + 4*C^2*a^10 - 8*B^2*a*b^9 - 8*B^2*a^9*b - 32*B^2*a^2*b^8 + 32*B^ 2*a^3*b^7 + 57*B^2*a^4*b^6 - 48*B^2*a^5*b^5 - 52*B^2*a^6*b^4 + 32*B^2*a^7* b^3 + 24*B^2*a^8*b^2 + C^2*a^6*b^4 + 4*C^2*a^8*b^2 - 24*B*C*a^9*b - 4*B*C* a^3*b^7 + 2*B*C*a^5*b^5 + 8*B*C*a^7*b^3))/(a^10*b + a^11 - a^4*b^7 - a^5*b ^6 + 3*a^6*b^5 + 3*a^7*b^4 - 3*a^8*b^3 - 3*a^9*b^2))*1i)/a^3 - (B*((B*((8* (4*B*a^15 + 4*C*a^15 - 4*B*a^6*b^9 + 2*B*a^7*b^8 + 18*B*a^8*b^7 - 4*B*a^9* b^6 - 36*B*a^10*b^5 + 6*B*a^11*b^4 + 34*B*a^12*b^3 - 8*B*a^13*b^2 - 2*C...