Integrand size = 33, antiderivative size = 265 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx=-\frac {2 b \left (4 a^2 A b^2-3 A b^4+2 a^4 C-a^2 b^2 C\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 (a-b)^{3/2} (a+b)^{3/2} d}+\frac {\left (6 A b^2+a^2 (A+2 C)\right ) \text {arctanh}(\sin (c+d x))}{2 a^4 d}+\frac {b \left (3 A b^2-a^2 (2 A-C)\right ) \tan (c+d x)}{a^3 \left (a^2-b^2\right ) d}-\frac {\left (3 A b^2-a^2 (A-2 C)\right ) \sec (c+d x) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac {\left (A b^2+a^2 C\right ) \sec (c+d x) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))} \]
-2*b*(4*A*a^2*b^2-3*A*b^4+2*C*a^4-C*a^2*b^2)*arctan((a-b)^(1/2)*tan(1/2*d* x+1/2*c)/(a+b)^(1/2))/a^4/(a-b)^(3/2)/(a+b)^(3/2)/d+1/2*(6*A*b^2+a^2*(A+2* C))*arctanh(sin(d*x+c))/a^4/d+b*(3*A*b^2-a^2*(2*A-C))*tan(d*x+c)/a^3/(a^2- b^2)/d-1/2*(3*A*b^2-a^2*(A-2*C))*sec(d*x+c)*tan(d*x+c)/a^2/(a^2-b^2)/d+(A* b^2+C*a^2)*sec(d*x+c)*tan(d*x+c)/a/(a^2-b^2)/d/(a+b*cos(d*x+c))
Leaf count is larger than twice the leaf count of optimal. \(712\) vs. \(2(265)=530\).
Time = 8.11 (sec) , antiderivative size = 712, normalized size of antiderivative = 2.69 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {4 b \left (4 a^2 A b^2-3 A b^4+2 a^4 C-a^2 b^2 C\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right ) \cos ^2(c+d x) \left (C+A \sec ^2(c+d x)\right )}{a^4 \left (a^2-b^2\right ) \sqrt {-a^2+b^2} d (2 A+C+C \cos (2 c+2 d x))}+\frac {\left (-a^2 A-6 A b^2-2 a^2 C\right ) \cos ^2(c+d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (C+A \sec ^2(c+d x)\right )}{a^4 d (2 A+C+C \cos (2 c+2 d x))}+\frac {\left (a^2 A+6 A b^2+2 a^2 C\right ) \cos ^2(c+d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (C+A \sec ^2(c+d x)\right )}{a^4 d (2 A+C+C \cos (2 c+2 d x))}+\frac {A \cos ^2(c+d x) \left (C+A \sec ^2(c+d x)\right )}{2 a^2 d (2 A+C+C \cos (2 c+2 d x)) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {4 A b \cos ^2(c+d x) \left (C+A \sec ^2(c+d x)\right ) \sin \left (\frac {1}{2} (c+d x)\right )}{a^3 d (2 A+C+C \cos (2 c+2 d x)) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {A \cos ^2(c+d x) \left (C+A \sec ^2(c+d x)\right )}{2 a^2 d (2 A+C+C \cos (2 c+2 d x)) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {4 A b \cos ^2(c+d x) \left (C+A \sec ^2(c+d x)\right ) \sin \left (\frac {1}{2} (c+d x)\right )}{a^3 d (2 A+C+C \cos (2 c+2 d x)) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {2 \cos ^2(c+d x) \left (C+A \sec ^2(c+d x)\right ) \left (A b^4 \sin (c+d x)+a^2 b^2 C \sin (c+d x)\right )}{a^3 (a-b) (a+b) d (a+b \cos (c+d x)) (2 A+C+C \cos (2 c+2 d x))} \]
(4*b*(4*a^2*A*b^2 - 3*A*b^4 + 2*a^4*C - a^2*b^2*C)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]]*Cos[c + d*x]^2*(C + A*Sec[c + d*x]^2))/(a^4* (a^2 - b^2)*Sqrt[-a^2 + b^2]*d*(2*A + C + C*Cos[2*c + 2*d*x])) + ((-(a^2*A ) - 6*A*b^2 - 2*a^2*C)*Cos[c + d*x]^2*Log[Cos[(c + d*x)/2] - Sin[(c + d*x) /2]]*(C + A*Sec[c + d*x]^2))/(a^4*d*(2*A + C + C*Cos[2*c + 2*d*x])) + ((a^ 2*A + 6*A*b^2 + 2*a^2*C)*Cos[c + d*x]^2*Log[Cos[(c + d*x)/2] + Sin[(c + d* x)/2]]*(C + A*Sec[c + d*x]^2))/(a^4*d*(2*A + C + C*Cos[2*c + 2*d*x])) + (A *Cos[c + d*x]^2*(C + A*Sec[c + d*x]^2))/(2*a^2*d*(2*A + C + C*Cos[2*c + 2* d*x])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) - (4*A*b*Cos[c + d*x]^2*(C + A*Sec[c + d*x]^2)*Sin[(c + d*x)/2])/(a^3*d*(2*A + C + C*Cos[2*c + 2*d*x] )*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) - (A*Cos[c + d*x]^2*(C + A*Sec[c + d*x]^2))/(2*a^2*d*(2*A + C + C*Cos[2*c + 2*d*x])*(Cos[(c + d*x)/2] + Sin [(c + d*x)/2])^2) - (4*A*b*Cos[c + d*x]^2*(C + A*Sec[c + d*x]^2)*Sin[(c + d*x)/2])/(a^3*d*(2*A + C + C*Cos[2*c + 2*d*x])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])) + (2*Cos[c + d*x]^2*(C + A*Sec[c + d*x]^2)*(A*b^4*Sin[c + d*x] + a^2*b^2*C*Sin[c + d*x]))/(a^3*(a - b)*(a + b)*d*(a + b*Cos[c + d*x])*(2 *A + C + C*Cos[2*c + 2*d*x]))
Time = 1.78 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.05, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.424, Rules used = {3042, 3535, 25, 3042, 3534, 25, 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.
| \(\displaystyle \int \frac {\sec ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 3535 |
| \(\displaystyle \frac {\int -\frac {\left (-\left ((A-2 C) a^2\right )+b (A+C) \cos (c+d x) a+3 A b^2-2 \left (C a^2+A b^2\right ) \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)}dx}{a \left (a^2-b^2\right )}+\frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\int \frac {\left (-\left ((A-2 C) a^2\right )+b (A+C) \cos (c+d x) a+3 A b^2-2 \left (C a^2+A b^2\right ) \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)}dx}{a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\int \frac {-\left ((A-2 C) a^2\right )+b (A+C) \sin \left (c+d x+\frac {\pi }{2}\right ) a+3 A b^2-2 \left (C a^2+A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3534 |
| \(\displaystyle \frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\int -\frac {\left (-b \left (3 A b^2-a^2 (A-2 C)\right ) \cos ^2(c+d x)+a \left ((A+2 C) a^2+A b^2\right ) \cos (c+d x)+2 \left (3 A b^3-\frac {1}{2} a^2 (4 A b-2 b C)\right )\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)}dx}{2 a}+\frac {\left (3 A b^2-a^2 (A-2 C)\right ) \tan (c+d x) \sec (c+d x)}{2 a d}}{a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\left (3 A b^2-a^2 (A-2 C)\right ) \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\int \frac {\left (-b \left (3 A b^2-a^2 (A-2 C)\right ) \cos ^2(c+d x)+a \left ((A+2 C) a^2+A b^2\right ) \cos (c+d x)+2 b \left (3 A b^2-a^2 (2 A-C)\right )\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)}dx}{2 a}}{a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\left (3 A b^2-a^2 (A-2 C)\right ) \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\int \frac {-b \left (3 A b^2-a^2 (A-2 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2+a \left ((A+2 C) a^2+A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+2 b \left (3 A b^2-a^2 (2 A-C)\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 a}}{a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3534 |
| \(\displaystyle \frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\left (3 A b^2-a^2 (A-2 C)\right ) \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {\int \frac {\left (\left (a^2-b^2\right ) \left ((A+2 C) a^2+6 A b^2\right )-a b \left (3 A b^2-a^2 (A-2 C)\right ) \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)}dx}{a}+\frac {2 b \left (3 A b^2-a^2 (2 A-C)\right ) \tan (c+d x)}{a d}}{2 a}}{a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\left (3 A b^2-a^2 (A-2 C)\right ) \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {\int \frac {\left (a^2-b^2\right ) \left ((A+2 C) a^2+6 A b^2\right )-a b \left (3 A b^2-a^2 (A-2 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}+\frac {2 b \left (3 A b^2-a^2 (2 A-C)\right ) \tan (c+d x)}{a d}}{2 a}}{a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3480 |
| \(\displaystyle \frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\left (3 A b^2-a^2 (A-2 C)\right ) \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {\frac {\left (a^2-b^2\right ) \left (a^2 (A+2 C)+6 A b^2\right ) \int \sec (c+d x)dx}{a}+\frac {2 b \left (-2 a^4 C-a^2 b^2 (4 A-C)+3 A b^4\right ) \int \frac {1}{a+b \cos (c+d x)}dx}{a}}{a}+\frac {2 b \left (3 A b^2-a^2 (2 A-C)\right ) \tan (c+d x)}{a d}}{2 a}}{a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\left (3 A b^2-a^2 (A-2 C)\right ) \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {\frac {\left (a^2-b^2\right ) \left (a^2 (A+2 C)+6 A b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}+\frac {2 b \left (-2 a^4 C-a^2 b^2 (4 A-C)+3 A b^4\right ) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a}}{a}+\frac {2 b \left (3 A b^2-a^2 (2 A-C)\right ) \tan (c+d x)}{a d}}{2 a}}{a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\left (3 A b^2-a^2 (A-2 C)\right ) \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {\frac {\left (a^2-b^2\right ) \left (a^2 (A+2 C)+6 A b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}+\frac {4 b \left (-2 a^4 C-a^2 b^2 (4 A-C)+3 A b^4\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{a d}}{a}+\frac {2 b \left (3 A b^2-a^2 (2 A-C)\right ) \tan (c+d x)}{a d}}{2 a}}{a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\left (3 A b^2-a^2 (A-2 C)\right ) \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {\frac {\left (a^2-b^2\right ) \left (a^2 (A+2 C)+6 A b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}+\frac {4 b \left (-2 a^4 C-a^2 b^2 (4 A-C)+3 A b^4\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}}{a}+\frac {2 b \left (3 A b^2-a^2 (2 A-C)\right ) \tan (c+d x)}{a d}}{2 a}}{a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\left (3 A b^2-a^2 (A-2 C)\right ) \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 b \left (3 A b^2-a^2 (2 A-C)\right ) \tan (c+d x)}{a d}+\frac {\frac {\left (a^2-b^2\right ) \left (a^2 (A+2 C)+6 A b^2\right ) \text {arctanh}(\sin (c+d x))}{a d}+\frac {4 b \left (-2 a^4 C-a^2 b^2 (4 A-C)+3 A b^4\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}}{a}}{2 a}}{a \left (a^2-b^2\right )}\) |
((A*b^2 + a^2*C)*Sec[c + d*x]*Tan[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Cos[c + d*x])) - (((3*A*b^2 - a^2*(A - 2*C))*Sec[c + d*x]*Tan[c + d*x])/(2*a*d) - (((4*b*(3*A*b^4 - a^2*b^2*(4*A - C) - 2*a^4*C)*ArcTan[(Sqrt[a - b]*Tan[( c + d*x)/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*Sqrt[a + b]*d) + ((a^2 - b^2)*(6 *A*b^2 + a^2*(A + 2*C))*ArcTanh[Sin[c + d*x]])/(a*d))/a + (2*b*(3*A*b^2 - a^2*(2*A - C))*Tan[c + d*x])/(a*d))/(2*a))/(a*(a^2 - b^2))
3.6.76.3.1 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_)])/(((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[(-(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[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 + a^2*C))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*S in[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[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d *(A*b^2 + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, 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 = 2.94 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.20
| method | result | size |
| derivativedivides | \(\frac {-\frac {A}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {\left (A \,a^{2}+6 A \,b^{2}+2 a^{2} C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{4}}+\frac {A \left (a +4 b \right )}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {A}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {\left (-A \,a^{2}-6 A \,b^{2}-2 a^{2} C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{4}}+\frac {A \left (a +4 b \right )}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {2 b \left (-\frac {a \left (A \,b^{2}+a^{2} C \right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b \right )}+\frac {\left (4 A \,a^{2} b^{2}-3 A \,b^{4}+2 C \,a^{4}-C \,a^{2} b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{4}}}{d}\) | \(319\) |
| default | \(\frac {-\frac {A}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {\left (A \,a^{2}+6 A \,b^{2}+2 a^{2} C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{4}}+\frac {A \left (a +4 b \right )}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {A}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {\left (-A \,a^{2}-6 A \,b^{2}-2 a^{2} C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{4}}+\frac {A \left (a +4 b \right )}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {2 b \left (-\frac {a \left (A \,b^{2}+a^{2} C \right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b \right )}+\frac {\left (4 A \,a^{2} b^{2}-3 A \,b^{4}+2 C \,a^{4}-C \,a^{2} b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{4}}}{d}\) | \(319\) |
| risch | \(\text {Expression too large to display}\) | \(1184\) |
1/d*(-1/2*A/a^2/(tan(1/2*d*x+1/2*c)+1)^2+1/2*(A*a^2+6*A*b^2+2*C*a^2)/a^4*l n(tan(1/2*d*x+1/2*c)+1)+1/2*A*(a+4*b)/a^3/(tan(1/2*d*x+1/2*c)+1)+1/2*A/a^2 /(tan(1/2*d*x+1/2*c)-1)^2+1/2/a^4*(-A*a^2-6*A*b^2-2*C*a^2)*ln(tan(1/2*d*x+ 1/2*c)-1)+1/2*A*(a+4*b)/a^3/(tan(1/2*d*x+1/2*c)-1)-2*b/a^4*(-a*(A*b^2+C*a^ 2)*b/(a^2-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)+(4*A*a^2*b^2-3*A*b^4+2*C*a^4-C*a^2*b^2)/(a-b)/(a+b)/((a-b)*(a+ b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a-b)*(a+b))^(1/2))))
Leaf count of result is larger than twice the leaf count of optimal. 540 vs. \(2 (251) = 502\).
Time = 7.38 (sec) , antiderivative size = 1149, normalized size of antiderivative = 4.34 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Too large to display} \]
[-1/4*(2*((2*C*a^4*b^2 + (4*A - C)*a^2*b^4 - 3*A*b^6)*cos(d*x + c)^3 + (2* C*a^5*b + (4*A - C)*a^3*b^3 - 3*A*a*b^5)*cos(d*x + c)^2)*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)) - (((A + 2*C)*a^6*b + 4*(A - C)*a^4*b^3 - (11*A - 2*C)*a^2*b^5 + 6*A*b^7)*cos(d*x + c)^3 + ((A + 2*C)*a^7 + 4*(A - C)*a^5* b^2 - (11*A - 2*C)*a^3*b^4 + 6*A*a*b^6)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) + (((A + 2*C)*a^6*b + 4*(A - C)*a^4*b^3 - (11*A - 2*C)*a^2*b^5 + 6*A*b ^7)*cos(d*x + c)^3 + ((A + 2*C)*a^7 + 4*(A - C)*a^5*b^2 - (11*A - 2*C)*a^3 *b^4 + 6*A*a*b^6)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*(A*a^7 - 2*A* a^5*b^2 + A*a^3*b^4 - 2*((2*A - C)*a^5*b^2 - (5*A - C)*a^3*b^4 + 3*A*a*b^6 )*cos(d*x + c)^2 - 3*(A*a^6*b - 2*A*a^4*b^3 + A*a^2*b^5)*cos(d*x + c))*sin (d*x + c))/((a^8*b - 2*a^6*b^3 + a^4*b^5)*d*cos(d*x + c)^3 + (a^9 - 2*a^7* b^2 + a^5*b^4)*d*cos(d*x + c)^2), -1/4*(4*((2*C*a^4*b^2 + (4*A - C)*a^2*b^ 4 - 3*A*b^6)*cos(d*x + c)^3 + (2*C*a^5*b + (4*A - C)*a^3*b^3 - 3*A*a*b^5)* cos(d*x + c)^2)*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b ^2)*sin(d*x + c))) - (((A + 2*C)*a^6*b + 4*(A - C)*a^4*b^3 - (11*A - 2*C)* a^2*b^5 + 6*A*b^7)*cos(d*x + c)^3 + ((A + 2*C)*a^7 + 4*(A - C)*a^5*b^2 - ( 11*A - 2*C)*a^3*b^4 + 6*A*a*b^6)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) + ( ((A + 2*C)*a^6*b + 4*(A - C)*a^4*b^3 - (11*A - 2*C)*a^2*b^5 + 6*A*b^7)*...
\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\int \frac {\left (A + C \cos ^{2}{\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{\left (a + b \cos {\left (c + d x \right )}\right )^{2}}\, dx \]
Exception generated. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Exception raised: ValueError} \]
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
Time = 0.37 (sec) , antiderivative size = 353, normalized size of antiderivative = 1.33 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {\frac {4 \, {\left (2 \, C a^{4} b + 4 \, A a^{2} b^{3} - C a^{2} b^{3} - 3 \, A b^{5}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{6} - a^{4} b^{2}\right )} \sqrt {a^{2} - b^{2}}} + \frac {4 \, {\left (C a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + A b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (a^{5} - a^{3} b^{2}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )}} + \frac {{\left (A a^{2} + 2 \, C a^{2} + 6 \, A b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac {{\left (A a^{2} + 2 \, C a^{2} + 6 \, A b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} + \frac {2 \, {\left (A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{3}}}{2 \, d} \]
1/2*(4*(2*C*a^4*b + 4*A*a^2*b^3 - C*a^2*b^3 - 3*A*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^6 - a^4*b^2)*sqrt(a^2 - b^2)) + 4*( C*a^2*b^2*tan(1/2*d*x + 1/2*c) + A*b^4*tan(1/2*d*x + 1/2*c))/((a^5 - a^3*b ^2)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 + a + b)) + (A*a^ 2 + 2*C*a^2 + 6*A*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - (A*a^2 + 2 *C*a^2 + 6*A*b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^4 + 2*(A*a*tan(1/2* d*x + 1/2*c)^3 + 4*A*b*tan(1/2*d*x + 1/2*c)^3 + A*a*tan(1/2*d*x + 1/2*c) - 4*A*b*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*a^3))/d
Time = 11.25 (sec) , antiderivative size = 6465, normalized size of antiderivative = 24.40 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Too large to display} \]
- ((tan(c/2 + (d*x)/2)*(A*a^4 + 6*A*b^4 - 5*A*a^2*b^2 + 2*C*a^2*b^2 + 3*A* a*b^3 - 3*A*a^3*b))/((a^3*b - a^4)*(a + b)) + (tan(c/2 + (d*x)/2)^5*(A*a^4 + 6*A*b^4 - 5*A*a^2*b^2 + 2*C*a^2*b^2 - 3*A*a*b^3 + 3*A*a^3*b))/((a^3*b - a^4)*(a + b)) + (2*tan(c/2 + (d*x)/2)^3*(A*a^4 - 6*A*b^4 + 3*A*a^2*b^2 - 2*C*a^2*b^2))/(a*(a^2*b - a^3)*(a + b)))/(d*(a + b - tan(c/2 + (d*x)/2)^2* (a + 3*b) - tan(c/2 + (d*x)/2)^4*(a - 3*b) + tan(c/2 + (d*x)/2)^6*(a - b)) ) - (atan(-(((3*A*b^2 + a^2*(A/2 + C))*(((3*A*b^2 + a^2*(A/2 + C))*((8*(2* A*a^15 + 4*C*a^15 - 12*A*a^8*b^7 + 6*A*a^9*b^6 + 28*A*a^10*b^5 - 14*A*a^11 *b^4 - 16*A*a^12*b^3 + 6*A*a^13*b^2 - 4*C*a^10*b^5 + 12*C*a^12*b^3 - 4*C*a ^13*b^2 - 8*C*a^14*b))/(a^11*b + a^12 - a^9*b^3 - a^10*b^2) - (8*tan(c/2 + (d*x)/2)*(3*A*b^2 + a^2*(A/2 + C))*(8*a^13*b - 8*a^8*b^6 + 8*a^9*b^5 + 16 *a^10*b^4 - 16*a^11*b^3 - 8*a^12*b^2))/(a^4*(a^8*b + a^9 - a^6*b^3 - a^7*b ^2))))/a^4 - (8*tan(c/2 + (d*x)/2)*(A^2*a^10 + 72*A^2*b^10 + 4*C^2*a^10 - 72*A^2*a*b^9 - 2*A^2*a^9*b - 8*C^2*a^9*b - 120*A^2*a^2*b^8 + 120*A^2*a^3*b ^7 + 17*A^2*a^4*b^6 - 26*A^2*a^5*b^5 + 23*A^2*a^6*b^4 - 20*A^2*a^7*b^3 + 1 1*A^2*a^8*b^2 + 8*C^2*a^4*b^6 - 8*C^2*a^5*b^5 - 20*C^2*a^6*b^4 + 16*C^2*a^ 7*b^3 + 12*C^2*a^8*b^2 + 4*A*C*a^10 - 8*A*C*a^9*b + 48*A*C*a^2*b^8 - 48*A* C*a^3*b^7 - 100*A*C*a^4*b^6 + 88*A*C*a^5*b^5 + 36*A*C*a^6*b^4 - 32*A*C*a^7 *b^3 + 20*A*C*a^8*b^2))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2))*1i)/a^4 - ((3*A *b^2 + a^2*(A/2 + C))*(((3*A*b^2 + a^2*(A/2 + C))*((8*(2*A*a^15 + 4*C*a...