Integrand size = 33, antiderivative size = 275 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=-\frac {\left (15 a^2 A b^4-6 A b^6-2 a^6 C-a^4 b^2 (12 A+C)\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 (a-b)^{5/2} (a+b)^{5/2} d}-\frac {3 A b \text {arctanh}(\sin (c+d x))}{a^4 d}-\frac {\left (11 a^2 A b^2-6 A b^4-a^4 (2 A-3 C)\right ) \tan (c+d x)}{2 a^3 \left (a^2-b^2\right )^2 d}+\frac {\left (A b^2+a^2 C\right ) \tan (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac {\left (3 A b^4-2 a^4 C-a^2 b^2 (6 A+C)\right ) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))} \] Output:
-(15*a^2*A*b^4-6*A*b^6-2*a^6*C-a^4*b^2*(12*A+C))*arctan((a-b)^(1/2)*tan(1/ 2*d*x+1/2*c)/(a+b)^(1/2))/a^4/(a-b)^(5/2)/(a+b)^(5/2)/d-3*A*b*arctanh(sin( d*x+c))/a^4/d-1/2*(11*A*a^2*b^2-6*A*b^4-a^4*(2*A-3*C))*tan(d*x+c)/a^3/(a^2 -b^2)^2/d+1/2*(A*b^2+C*a^2)*tan(d*x+c)/a/(a^2-b^2)/d/(a+b*cos(d*x+c))^2-1/ 2*(3*A*b^4-2*a^4*C-a^2*b^2*(6*A+C))*tan(d*x+c)/a^2/(a^2-b^2)^2/d/(a+b*cos( d*x+c))
Leaf count is larger than twice the leaf count of optimal. \(649\) vs. \(2(275)=550\).
Time = 8.10 (sec) , antiderivative size = 649, normalized size of antiderivative = 2.36 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=-\frac {2 \left (12 a^4 A b^2-15 a^2 A b^4+6 A b^6+2 a^6 C+a^4 b^2 C\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right ) \cos ^2(c+d x) \left (C+A \sec ^2(c+d x)\right )}{a^4 \left (a^2-b^2\right )^2 \sqrt {-a^2+b^2} d (2 A+C+C \cos (2 c+2 d x))}+\frac {6 A b \cos ^2(c+d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (C+A \sec ^2(c+d x)\right )}{a^4 d (2 A+C+C \cos (2 c+2 d x))}-\frac {6 A b \cos ^2(c+d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (C+A \sec ^2(c+d x)\right )}{a^4 d (2 A+C+C \cos (2 c+2 d x))}+\frac {2 A \cos ^2(c+d x) \left (C+A \sec ^2(c+d x)\right ) \sin \left (\frac {1}{2} (c+d x)\right )}{a^3 d (2 A+C+C \cos (2 c+2 d x)) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {2 A \cos ^2(c+d x) \left (C+A \sec ^2(c+d x)\right ) \sin \left (\frac {1}{2} (c+d x)\right )}{a^3 d (2 A+C+C \cos (2 c+2 d x)) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {\cos ^2(c+d x) \left (C+A \sec ^2(c+d x)\right ) \left (-A b^3 \sin (c+d x)-a^2 b C \sin (c+d x)\right )}{a^2 (a-b) (a+b) d (a+b \cos (c+d x))^2 (2 A+C+C \cos (2 c+2 d x))}+\frac {\cos ^2(c+d x) \left (C+A \sec ^2(c+d x)\right ) \left (-7 a^2 A b^3 \sin (c+d x)+4 A b^5 \sin (c+d x)-3 a^4 b C \sin (c+d x)\right )}{a^3 (a-b)^2 (a+b)^2 d (a+b \cos (c+d x)) (2 A+C+C \cos (2 c+2 d x))} \] Input:
Integrate[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^2)/(a + b*Cos[c + d*x])^3,x ]
Output:
(-2*(12*a^4*A*b^2 - 15*a^2*A*b^4 + 6*A*b^6 + 2*a^6*C + a^4*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)^2*Sqrt[-a^2 + b^2]*d*(2*A + C + C*Cos[2*c + 2*d *x])) + (6*A*b*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])) - (6*A*b*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])) + (2*A*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*A*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])) + (Cos[c + d*x]^2*(C + A*Sec[c + d*x]^2)*(-(A*b^3* Sin[c + d*x]) - a^2*b*C*Sin[c + d*x]))/(a^2*(a - b)*(a + b)*d*(a + b*Cos[c + d*x])^2*(2*A + C + C*Cos[2*c + 2*d*x])) + (Cos[c + d*x]^2*(C + A*Sec[c + d*x]^2)*(-7*a^2*A*b^3*Sin[c + d*x] + 4*A*b^5*Sin[c + d*x] - 3*a^4*b*C*Si n[c + d*x]))/(a^3*(a - b)^2*(a + b)^2*d*(a + b*Cos[c + d*x])*(2*A + C + C* Cos[2*c + 2*d*x]))
Time = 1.91 (sec) , antiderivative size = 309, normalized size of antiderivative = 1.12, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.394, Rules used = {3042, 3535, 25, 3042, 3534, 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.
| \(\displaystyle \int \frac {\sec ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^3}dx\) |
| \(\Big \downarrow \) 3535 |
| \(\displaystyle \frac {\int -\frac {\left (-\left ((2 A-C) a^2\right )+2 b (A+C) \cos (c+d x) a+3 A b^2-2 \left (C a^2+A b^2\right ) \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^2}dx}{2 a \left (a^2-b^2\right )}+\frac {\left (a^2 C+A b^2\right ) \tan (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\left (a^2 C+A b^2\right ) \tan (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {\int \frac {\left (-\left ((2 A-C) a^2\right )+2 b (A+C) \cos (c+d x) a+3 A b^2-2 \left (C a^2+A b^2\right ) \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^2}dx}{2 a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\left (a^2 C+A b^2\right ) \tan (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {\int \frac {-\left ((2 A-C) a^2\right )+2 b (A+C) \sin \left (c+d x+\frac {\pi }{2}\right ) a+3 A b^2-2 \left (C a^2+A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx}{2 a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3534 |
| \(\displaystyle \frac {\left (a^2 C+A b^2\right ) \tan (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {\frac {\int \frac {\left (-\left ((2 A-3 C) a^4\right )+11 A b^2 a^2-b \left (A b^2-a^2 (4 A+3 C)\right ) \cos (c+d x) a-6 A b^4+\left (-2 C a^4-b^2 (6 A+C) a^2+3 A b^4\right ) \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)}dx}{a \left (a^2-b^2\right )}+\frac {\left (-2 a^4 C-a^2 b^2 (6 A+C)+3 A b^4\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\left (a^2 C+A b^2\right ) \tan (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {\frac {\int \frac {-\left ((2 A-3 C) a^4\right )+11 A b^2 a^2-b \left (A b^2-a^2 (4 A+3 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a-6 A b^4+\left (-2 C a^4-b^2 (6 A+C) a^2+3 A b^4\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a \left (a^2-b^2\right )}+\frac {\left (-2 a^4 C-a^2 b^2 (6 A+C)+3 A b^4\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3534 |
| \(\displaystyle \frac {\left (a^2 C+A b^2\right ) \tan (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {\frac {\frac {\int \frac {\left (6 A b \left (a^2-b^2\right )^2+a \left (-2 C a^4-b^2 (6 A+C) a^2+3 A b^4\right ) \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)}dx}{a}+\frac {\left (-\left (a^4 (2 A-3 C)\right )+11 a^2 A b^2-6 A b^4\right ) \tan (c+d x)}{a d}}{a \left (a^2-b^2\right )}+\frac {\left (-2 a^4 C-a^2 b^2 (6 A+C)+3 A b^4\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\left (a^2 C+A b^2\right ) \tan (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {\frac {\frac {\int \frac {6 A b \left (a^2-b^2\right )^2+a \left (-2 C a^4-b^2 (6 A+C) a^2+3 A b^4\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}+\frac {\left (-\left (a^4 (2 A-3 C)\right )+11 a^2 A b^2-6 A b^4\right ) \tan (c+d x)}{a d}}{a \left (a^2-b^2\right )}+\frac {\left (-2 a^4 C-a^2 b^2 (6 A+C)+3 A b^4\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3480 |
| \(\displaystyle \frac {\left (a^2 C+A b^2\right ) \tan (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {\frac {\frac {\frac {6 A b \left (a^2-b^2\right )^2 \int \sec (c+d x)dx}{a}+\frac {\left (-2 a^6 C-a^4 b^2 (12 A+C)+15 a^2 A b^4-6 A b^6\right ) \int \frac {1}{a+b \cos (c+d x)}dx}{a}}{a}+\frac {\left (-\left (a^4 (2 A-3 C)\right )+11 a^2 A b^2-6 A b^4\right ) \tan (c+d x)}{a d}}{a \left (a^2-b^2\right )}+\frac {\left (-2 a^4 C-a^2 b^2 (6 A+C)+3 A b^4\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\left (a^2 C+A b^2\right ) \tan (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {\frac {\frac {\frac {6 A b \left (a^2-b^2\right )^2 \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}+\frac {\left (-2 a^6 C-a^4 b^2 (12 A+C)+15 a^2 A b^4-6 A b^6\right ) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a}}{a}+\frac {\left (-\left (a^4 (2 A-3 C)\right )+11 a^2 A b^2-6 A b^4\right ) \tan (c+d x)}{a d}}{a \left (a^2-b^2\right )}+\frac {\left (-2 a^4 C-a^2 b^2 (6 A+C)+3 A b^4\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {\left (a^2 C+A b^2\right ) \tan (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {\frac {\frac {\frac {6 A b \left (a^2-b^2\right )^2 \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}+\frac {2 \left (-2 a^6 C-a^4 b^2 (12 A+C)+15 a^2 A b^4-6 A b^6\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{a d}}{a}+\frac {\left (-\left (a^4 (2 A-3 C)\right )+11 a^2 A b^2-6 A b^4\right ) \tan (c+d x)}{a d}}{a \left (a^2-b^2\right )}+\frac {\left (-2 a^4 C-a^2 b^2 (6 A+C)+3 A b^4\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\left (a^2 C+A b^2\right ) \tan (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {\frac {\frac {\frac {6 A b \left (a^2-b^2\right )^2 \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}+\frac {2 \left (-2 a^6 C-a^4 b^2 (12 A+C)+15 a^2 A b^4-6 A b^6\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}}{a}+\frac {\left (-\left (a^4 (2 A-3 C)\right )+11 a^2 A b^2-6 A b^4\right ) \tan (c+d x)}{a d}}{a \left (a^2-b^2\right )}+\frac {\left (-2 a^4 C-a^2 b^2 (6 A+C)+3 A b^4\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\left (a^2 C+A b^2\right ) \tan (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {\frac {\left (-2 a^4 C-a^2 b^2 (6 A+C)+3 A b^4\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {\frac {\left (-\left (a^4 (2 A-3 C)\right )+11 a^2 A b^2-6 A b^4\right ) \tan (c+d x)}{a d}+\frac {\frac {6 A b \left (a^2-b^2\right )^2 \text {arctanh}(\sin (c+d x))}{a d}+\frac {2 \left (-2 a^6 C-a^4 b^2 (12 A+C)+15 a^2 A b^4-6 A b^6\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}}{a}}{a \left (a^2-b^2\right )}}{2 a \left (a^2-b^2\right )}\) |
Input:
Int[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^2)/(a + b*Cos[c + d*x])^3,x]
Output:
((A*b^2 + a^2*C)*Tan[c + d*x])/(2*a*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^2) - (((3*A*b^4 - 2*a^4*C - a^2*b^2*(6*A + C))*Tan[c + d*x])/(a*(a^2 - b^2)*d *(a + b*Cos[c + d*x])) + (((2*(15*a^2*A*b^4 - 6*A*b^6 - 2*a^6*C - a^4*b^2* (12*A + C))*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*Sqrt[a + b]*d) + (6*A*b*(a^2 - b^2)^2*ArcTanh[Sin[c + d*x]])/(a*d))/a + ((11*a^2*A*b^2 - 6*A*b^4 - a^4*(2*A - 3*C))*Tan[c + d*x])/(a*d))/(a*(a^ 2 - b^2)))/(2*a*(a^2 - b^2))
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 = 1.79 (sec) , antiderivative size = 354, normalized size of antiderivative = 1.29
| method | result | size |
| derivativedivides | \(\frac {-\frac {A}{a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {3 A b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{4}}+\frac {\frac {2 \left (-\frac {\left (8 A \,a^{2} b^{2}+a A \,b^{3}-4 A \,b^{4}+4 a^{4} C +a^{3} b C \right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a^{2}+2 a b +b^{2}\right ) \left (a -b \right )}-\frac {\left (8 A \,a^{2} b^{2}-a A \,b^{3}-4 A \,b^{4}+4 a^{4} C -a^{3} b C \right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{2}}+\frac {\left (12 A \,a^{4} b^{2}-15 a^{2} A \,b^{4}+6 A \,b^{6}+2 a^{6} C +a^{4} b^{2} C \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{a^{4}}-\frac {A}{a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {3 A b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{4}}}{d}\) | \(354\) |
| default | \(\frac {-\frac {A}{a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {3 A b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{4}}+\frac {\frac {2 \left (-\frac {\left (8 A \,a^{2} b^{2}+a A \,b^{3}-4 A \,b^{4}+4 a^{4} C +a^{3} b C \right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a^{2}+2 a b +b^{2}\right ) \left (a -b \right )}-\frac {\left (8 A \,a^{2} b^{2}-a A \,b^{3}-4 A \,b^{4}+4 a^{4} C -a^{3} b C \right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{2}}+\frac {\left (12 A \,a^{4} b^{2}-15 a^{2} A \,b^{4}+6 A \,b^{6}+2 a^{6} C +a^{4} b^{2} C \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{a^{4}}-\frac {A}{a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {3 A b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{4}}}{d}\) | \(354\) |
| risch | \(\text {Expression too large to display}\) | \(1403\) |
Input:
int((A+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+b*cos(d*x+c))^3,x,method=_RETURNVER BOSE)
Output:
1/d*(-A/a^3/(tan(1/2*d*x+1/2*c)-1)+3*A*b/a^4*ln(tan(1/2*d*x+1/2*c)-1)+2/a^ 4*((-1/2*(8*A*a^2*b^2+A*a*b^3-4*A*b^4+4*C*a^4+C*a^3*b)*a*b/(a^2+2*a*b+b^2) /(a-b)*tan(1/2*d*x+1/2*c)^3-1/2*(8*A*a^2*b^2-A*a*b^3-4*A*b^4+4*C*a^4-C*a^3 *b)*a*b/(a+b)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c))/(tan(1/2*d*x+1/2*c)^2*a- tan(1/2*d*x+1/2*c)^2*b+a+b)^2+1/2*(12*A*a^4*b^2-15*A*a^2*b^4+6*A*b^6+2*C*a ^6+C*a^4*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)))-A/a^3/(tan(1/2*d*x+1/2*c)+1)-3*A*b/a^4*l n(tan(1/2*d*x+1/2*c)+1))
Leaf count of result is larger than twice the leaf count of optimal. 739 vs. \(2 (256) = 512\).
Time = 7.74 (sec) , antiderivative size = 1548, normalized size of antiderivative = 5.63 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Too large to display} \] Input:
integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+b*cos(d*x+c))^3,x, algorithm= "fricas")
Output:
[-1/4*(((2*C*a^6*b^2 + (12*A + C)*a^4*b^4 - 15*A*a^2*b^6 + 6*A*b^8)*cos(d* x + c)^3 + 2*(2*C*a^7*b + (12*A + C)*a^5*b^3 - 15*A*a^3*b^5 + 6*A*a*b^7)*c os(d*x + c)^2 + (2*C*a^8 + (12*A + C)*a^6*b^2 - 15*A*a^4*b^4 + 6*A*a^2*b^6 )*cos(d*x + c))*sqrt(-a^2 + b^2)*log((2*a*b*cos(d*x + c) + (2*a^2 - b^2)*c os(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)) + 6*((A*a^6*b^3 - 3*A*a^4*b^5 + 3*A*a^2*b^7 - A*b^9)*cos(d*x + c)^3 + 2*(A*a^7*b^2 - 3*A* a^5*b^4 + 3*A*a^3*b^6 - A*a*b^8)*cos(d*x + c)^2 + (A*a^8*b - 3*A*a^6*b^3 + 3*A*a^4*b^5 - A*a^2*b^7)*cos(d*x + c))*log(sin(d*x + c) + 1) - 6*((A*a^6* b^3 - 3*A*a^4*b^5 + 3*A*a^2*b^7 - A*b^9)*cos(d*x + c)^3 + 2*(A*a^7*b^2 - 3 *A*a^5*b^4 + 3*A*a^3*b^6 - A*a*b^8)*cos(d*x + c)^2 + (A*a^8*b - 3*A*a^6*b^ 3 + 3*A*a^4*b^5 - A*a^2*b^7)*cos(d*x + c))*log(-sin(d*x + c) + 1) - 2*(2*A *a^9 - 6*A*a^7*b^2 + 6*A*a^5*b^4 - 2*A*a^3*b^6 + ((2*A - 3*C)*a^7*b^2 - (1 3*A - 3*C)*a^5*b^4 + 17*A*a^3*b^6 - 6*A*a*b^8)*cos(d*x + c)^2 + (4*(A - C) *a^8*b - 5*(4*A - C)*a^6*b^3 + (25*A - C)*a^4*b^5 - 9*A*a^2*b^7)*cos(d*x + c))*sin(d*x + c))/((a^10*b^2 - 3*a^8*b^4 + 3*a^6*b^6 - a^4*b^8)*d*cos(d*x + c)^3 + 2*(a^11*b - 3*a^9*b^3 + 3*a^7*b^5 - a^5*b^7)*d*cos(d*x + c)^2 + (a^12 - 3*a^10*b^2 + 3*a^8*b^4 - a^6*b^6)*d*cos(d*x + c)), 1/2*(((2*C*a^6* b^2 + (12*A + C)*a^4*b^4 - 15*A*a^2*b^6 + 6*A*b^8)*cos(d*x + c)^3 + 2*(2*C *a^7*b + (12*A + C)*a^5*b^3 - 15*A*a^3*b^5 + 6*A*a*b^7)*cos(d*x + c)^2 ...
\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\int \frac {\left (A + C \cos ^{2}{\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}}{\left (a + b \cos {\left (c + d x \right )}\right )^{3}}\, dx \] Input:
integrate((A+C*cos(d*x+c)**2)*sec(d*x+c)**2/(a+b*cos(d*x+c))**3,x)
Output:
Integral((A + C*cos(c + d*x)**2)*sec(c + d*x)**2/(a + b*cos(c + d*x))**3, x)
Exception generated. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+b*cos(d*x+c))^3,x, algorithm= "maxima")
Output:
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. 517 vs. \(2 (256) = 512\).
Time = 0.20 (sec) , antiderivative size = 517, normalized size of antiderivative = 1.88 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx =\text {Too large to display} \] Input:
integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+b*cos(d*x+c))^3,x, algorithm= "giac")
Output:
-((2*C*a^6 + 12*A*a^4*b^2 + C*a^4*b^2 - 15*A*a^2*b^4 + 6*A*b^6)*(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^8 - 2*a^6*b^2 + a^4*b^4)*s qrt(a^2 - b^2)) + 3*A*b*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - 3*A*b*log (abs(tan(1/2*d*x + 1/2*c) - 1))/a^4 + (4*C*a^5*b*tan(1/2*d*x + 1/2*c)^3 - 3*C*a^4*b^2*tan(1/2*d*x + 1/2*c)^3 + 8*A*a^3*b^3*tan(1/2*d*x + 1/2*c)^3 - C*a^3*b^3*tan(1/2*d*x + 1/2*c)^3 - 7*A*a^2*b^4*tan(1/2*d*x + 1/2*c)^3 - 5* A*a*b^5*tan(1/2*d*x + 1/2*c)^3 + 4*A*b^6*tan(1/2*d*x + 1/2*c)^3 + 4*C*a^5* b*tan(1/2*d*x + 1/2*c) + 3*C*a^4*b^2*tan(1/2*d*x + 1/2*c) + 8*A*a^3*b^3*ta n(1/2*d*x + 1/2*c) - C*a^3*b^3*tan(1/2*d*x + 1/2*c) + 7*A*a^2*b^4*tan(1/2* d*x + 1/2*c) - 5*A*a*b^5*tan(1/2*d*x + 1/2*c) - 4*A*b^6*tan(1/2*d*x + 1/2* c))/((a^7 - 2*a^5*b^2 + a^3*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) + 2*A*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^ 2 - 1)*a^3))/d
Time = 6.08 (sec) , antiderivative size = 7211, normalized size of antiderivative = 26.22 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Too large to display} \] Input:
int((A + C*cos(c + d*x)^2)/(cos(c + d*x)^2*(a + b*cos(c + d*x))^3),x)
Output:
(A*b*atan(((A*b*((8*tan(c/2 + (d*x)/2)*(72*A^2*b^12 + 4*C^2*a^12 - 72*A^2* a*b^11 - 288*A^2*a^2*b^10 + 288*A^2*a^3*b^9 + 441*A^2*a^4*b^8 - 432*A^2*a^ 5*b^7 - 288*A^2*a^6*b^6 + 288*A^2*a^7*b^5 + 36*A^2*a^8*b^4 - 72*A^2*a^9*b^ 3 + 36*A^2*a^10*b^2 + C^2*a^8*b^4 + 4*C^2*a^10*b^2 + 12*A*C*a^4*b^8 - 6*A* C*a^6*b^6 - 36*A*C*a^8*b^4 + 48*A*C*a^10*b^2))/(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) - (3*A*b*((8*(4 *C*a^18 + 12*A*a^8*b^10 - 6*A*a^9*b^9 - 54*A*a^10*b^8 + 24*A*a^11*b^7 + 96 *A*a^12*b^6 - 42*A*a^13*b^5 - 78*A*a^14*b^4 + 36*A*a^15*b^3 + 24*A*a^16*b^ 2 - 2*C*a^11*b^7 + 2*C*a^12*b^6 + 6*C*a^15*b^3 - 6*C*a^16*b^2 - 12*A*a^17* b - 4*C*a^17*b))/(a^15*b + a^16 - a^9*b^7 - a^10*b^6 + 3*a^11*b^5 + 3*a^12 *b^4 - 3*a^13*b^3 - 3*a^14*b^2) - (24*A*b*tan(c/2 + (d*x)/2)*(8*a^17*b - 8 *a^8*b^10 + 8*a^9*b^9 + 32*a^10*b^8 - 32*a^11*b^7 - 48*a^12*b^6 + 48*a^13* b^5 + 32*a^14*b^4 - 32*a^15*b^3 - 8*a^16*b^2))/(a^4*(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))))/a^4)*3i )/a^4 + (A*b*((8*tan(c/2 + (d*x)/2)*(72*A^2*b^12 + 4*C^2*a^12 - 72*A^2*a*b ^11 - 288*A^2*a^2*b^10 + 288*A^2*a^3*b^9 + 441*A^2*a^4*b^8 - 432*A^2*a^5*b ^7 - 288*A^2*a^6*b^6 + 288*A^2*a^7*b^5 + 36*A^2*a^8*b^4 - 72*A^2*a^9*b^3 + 36*A^2*a^10*b^2 + C^2*a^8*b^4 + 4*C^2*a^10*b^2 + 12*A*C*a^4*b^8 - 6*A*C*a ^6*b^6 - 36*A*C*a^8*b^4 + 48*A*C*a^10*b^2))/(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) + (3*A*b*((8*(4...
Time = 0.19 (sec) , antiderivative size = 2778, normalized size of antiderivative = 10.10 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx =\text {Too large to display} \] Input:
int((A+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+b*cos(d*x+c))^3,x)
Output:
(4*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a **2 - b**2))*cos(c + d*x)*sin(c + d*x)**2*a**5*b**2*c + 24*sqrt(a**2 - b** 2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*cos(c + d*x)*sin(c + d*x)**2*a**4*b**4 + 2*sqrt(a**2 - b**2)*atan((tan((c + d*x )/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*cos(c + d*x)*sin(c + d*x)* *2*a**3*b**4*c - 30*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*cos(c + d*x)*sin(c + d*x)**2*a**2*b**6 + 12* sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*cos(c + d*x)*sin(c + d*x)**2*b**8 - 4*sqrt(a**2 - b**2)*atan((ta n((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*cos(c + d*x)*a** 7*c - 24*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/ sqrt(a**2 - b**2))*cos(c + d*x)*a**6*b**2 - 6*sqrt(a**2 - b**2)*atan((tan( (c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*cos(c + d*x)*a**5* b**2*c + 6*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b )/sqrt(a**2 - b**2))*cos(c + d*x)*a**4*b**4 - 2*sqrt(a**2 - b**2)*atan((ta n((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*cos(c + d*x)*a** 3*b**4*c + 18*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2 )*b)/sqrt(a**2 - b**2))*cos(c + d*x)*a**2*b**6 - 12*sqrt(a**2 - b**2)*atan ((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*cos(c + d*x) *b**8 + 8*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)...