Integrand size = 40, antiderivative size = 290 \[ \int \frac {\cos ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx=-\frac {(3 b B-a C) x}{a^4}+\frac {b \left (12 a^4 b B-15 a^2 b^3 B+6 b^5 B-6 a^5 C+5 a^3 b^2 C-2 a b^4 C\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 (a-b)^{5/2} (a+b)^{5/2} d}+\frac {\left (2 a^4 B-11 a^2 b^2 B+6 b^4 B+5 a^3 b C-2 a b^3 C\right ) \sin (c+d x)}{2 a^3 \left (a^2-b^2\right )^2 d}+\frac {b (b B-a C) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac {b \left (6 a^2 b B-3 b^3 B-4 a^3 C+a b^2 C\right ) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))} \]
-(3*B*b-C*a)*x/a^4+b*(12*B*a^4*b-15*B*a^2*b^3+6*B*b^5-6*C*a^5+5*C*a^3*b^2- 2*C*a*b^4)*arctanh((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+1/2*(2*B*a^4-11*B*a^2*b^2+6*B*b^4+5*C*a^3*b-2*C*a*b^3)* sin(d*x+c)/a^3/(a^2-b^2)^2/d+1/2*b*(B*b-C*a)*sin(d*x+c)/a/(a^2-b^2)/d/(a+b *sec(d*x+c))^2+1/2*b*(6*B*a^2*b-3*B*b^3-4*C*a^3+C*a*b^2)*sin(d*x+c)/a^2/(a ^2-b^2)^2/d/(a+b*sec(d*x+c))
Time = 2.37 (sec) , antiderivative size = 232, normalized size of antiderivative = 0.80 \[ \int \frac {\cos ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx=\frac {2 (-3 b B+a C) (c+d x)-\frac {2 b \left (12 a^4 b B-15 a^2 b^3 B+6 b^5 B-6 a^5 C+5 a^3 b^2 C-2 a b^4 C\right ) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+2 a B \sin (c+d x)+\frac {a b^3 (b B-a C) \sin (c+d x)}{(a-b) (a+b) (b+a \cos (c+d x))^2}+\frac {a b^2 \left (-8 a^2 b B+5 b^3 B+6 a^3 C-3 a b^2 C\right ) \sin (c+d x)}{(a-b)^2 (a+b)^2 (b+a \cos (c+d x))}}{2 a^4 d} \]
(2*(-3*b*B + a*C)*(c + d*x) - (2*b*(12*a^4*b*B - 15*a^2*b^3*B + 6*b^5*B - 6*a^5*C + 5*a^3*b^2*C - 2*a*b^4*C)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqr t[a^2 - b^2]])/(a^2 - b^2)^(5/2) + 2*a*B*Sin[c + d*x] + (a*b^3*(b*B - a*C) *Sin[c + d*x])/((a - b)*(a + b)*(b + a*Cos[c + d*x])^2) + (a*b^2*(-8*a^2*b *B + 5*b^3*B + 6*a^3*C - 3*a*b^2*C)*Sin[c + d*x])/((a - b)^2*(a + b)^2*(b + a*Cos[c + d*x])))/(2*a^4*d)
Time = 2.00 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.12, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.425, Rules used = {3042, 4560, 3042, 4518, 25, 3042, 4588, 25, 3042, 4592, 3042, 4407, 3042, 4318, 3042, 3138, 221}
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.
| \(\displaystyle \int \frac {\cos ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^3}dx\) |
| \(\Big \downarrow \) 4560 |
| \(\displaystyle \int \frac {\cos (c+d x) (B+C \sec (c+d x))}{(a+b \sec (c+d x))^3}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {B+C \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^3}dx\) |
| \(\Big \downarrow \) 4518 |
| \(\displaystyle \frac {b (b B-a C) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac {\int -\frac {\cos (c+d x) \left (2 B a^2+b C a-2 (b B-a C) \sec (c+d x) a+2 b (b B-a C) \sec ^2(c+d x)-3 b^2 B\right )}{(a+b \sec (c+d x))^2}dx}{2 a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {\cos (c+d x) \left (2 B a^2+b C a-2 (b B-a C) \sec (c+d x) a+2 b (b B-a C) \sec ^2(c+d x)-3 b^2 B\right )}{(a+b \sec (c+d x))^2}dx}{2 a \left (a^2-b^2\right )}+\frac {b (b B-a C) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {2 B a^2+b C a-2 (b B-a C) \csc \left (c+d x+\frac {\pi }{2}\right ) a+2 b (b B-a C) \csc \left (c+d x+\frac {\pi }{2}\right )^2-3 b^2 B}{\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx}{2 a \left (a^2-b^2\right )}+\frac {b (b B-a C) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 4588 |
| \(\displaystyle \frac {\frac {b \left (-4 a^3 C+6 a^2 b B+a b^2 C-3 b^3 B\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\int -\frac {\cos (c+d x) \left (2 B a^4+5 b C a^3-11 b^2 B a^2-2 b^3 C a-\left (-2 C a^3+4 b B a^2-b^2 C a-b^3 B\right ) \sec (c+d x) a+b \left (-4 C a^3+6 b B a^2+b^2 C a-3 b^3 B\right ) \sec ^2(c+d x)+6 b^4 B\right )}{a+b \sec (c+d x)}dx}{a \left (a^2-b^2\right )}}{2 a \left (a^2-b^2\right )}+\frac {b (b B-a C) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {\cos (c+d x) \left (2 B a^4+5 b C a^3-11 b^2 B a^2-2 b^3 C a-\left (-2 C a^3+4 b B a^2-b^2 C a-b^3 B\right ) \sec (c+d x) a+b \left (-4 C a^3+6 b B a^2+b^2 C a-3 b^3 B\right ) \sec ^2(c+d x)+6 b^4 B\right )}{a+b \sec (c+d x)}dx}{a \left (a^2-b^2\right )}+\frac {b \left (-4 a^3 C+6 a^2 b B+a b^2 C-3 b^3 B\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b (b B-a C) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {2 B a^4+5 b C a^3-11 b^2 B a^2-2 b^3 C a-\left (-2 C a^3+4 b B a^2-b^2 C a-b^3 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right ) a+b \left (-4 C a^3+6 b B a^2+b^2 C a-3 b^3 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2+6 b^4 B}{\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a \left (a^2-b^2\right )}+\frac {b \left (-4 a^3 C+6 a^2 b B+a b^2 C-3 b^3 B\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b (b B-a C) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 4592 |
| \(\displaystyle \frac {\frac {\frac {\left (2 a^4 B+5 a^3 b C-11 a^2 b^2 B-2 a b^3 C+6 b^4 B\right ) \sin (c+d x)}{a d}-\frac {\int \frac {2 \left (a^2-b^2\right )^2 (3 b B-a C)-a b \left (-4 C a^3+6 b B a^2+b^2 C a-3 b^3 B\right ) \sec (c+d x)}{a+b \sec (c+d x)}dx}{a}}{a \left (a^2-b^2\right )}+\frac {b \left (-4 a^3 C+6 a^2 b B+a b^2 C-3 b^3 B\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b (b B-a C) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\left (2 a^4 B+5 a^3 b C-11 a^2 b^2 B-2 a b^3 C+6 b^4 B\right ) \sin (c+d x)}{a d}-\frac {\int \frac {2 \left (a^2-b^2\right )^2 (3 b B-a C)-a b \left (-4 C a^3+6 b B a^2+b^2 C a-3 b^3 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a}}{a \left (a^2-b^2\right )}+\frac {b \left (-4 a^3 C+6 a^2 b B+a b^2 C-3 b^3 B\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b (b B-a C) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 4407 |
| \(\displaystyle \frac {\frac {\frac {\left (2 a^4 B+5 a^3 b C-11 a^2 b^2 B-2 a b^3 C+6 b^4 B\right ) \sin (c+d x)}{a d}-\frac {\frac {2 x \left (a^2-b^2\right )^2 (3 b B-a C)}{a}-\frac {b \left (-6 a^5 C+12 a^4 b B+5 a^3 b^2 C-15 a^2 b^3 B-2 a b^4 C+6 b^5 B\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)}dx}{a}}{a}}{a \left (a^2-b^2\right )}+\frac {b \left (-4 a^3 C+6 a^2 b B+a b^2 C-3 b^3 B\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b (b B-a C) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\left (2 a^4 B+5 a^3 b C-11 a^2 b^2 B-2 a b^3 C+6 b^4 B\right ) \sin (c+d x)}{a d}-\frac {\frac {2 x \left (a^2-b^2\right )^2 (3 b B-a C)}{a}-\frac {b \left (-6 a^5 C+12 a^4 b B+5 a^3 b^2 C-15 a^2 b^3 B-2 a b^4 C+6 b^5 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a}}{a}}{a \left (a^2-b^2\right )}+\frac {b \left (-4 a^3 C+6 a^2 b B+a b^2 C-3 b^3 B\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b (b B-a C) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 4318 |
| \(\displaystyle \frac {\frac {\frac {\left (2 a^4 B+5 a^3 b C-11 a^2 b^2 B-2 a b^3 C+6 b^4 B\right ) \sin (c+d x)}{a d}-\frac {\frac {2 x \left (a^2-b^2\right )^2 (3 b B-a C)}{a}-\frac {\left (-6 a^5 C+12 a^4 b B+5 a^3 b^2 C-15 a^2 b^3 B-2 a b^4 C+6 b^5 B\right ) \int \frac {1}{\frac {a \cos (c+d x)}{b}+1}dx}{a}}{a}}{a \left (a^2-b^2\right )}+\frac {b \left (-4 a^3 C+6 a^2 b B+a b^2 C-3 b^3 B\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b (b B-a C) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\left (2 a^4 B+5 a^3 b C-11 a^2 b^2 B-2 a b^3 C+6 b^4 B\right ) \sin (c+d x)}{a d}-\frac {\frac {2 x \left (a^2-b^2\right )^2 (3 b B-a C)}{a}-\frac {\left (-6 a^5 C+12 a^4 b B+5 a^3 b^2 C-15 a^2 b^3 B-2 a b^4 C+6 b^5 B\right ) \int \frac {1}{\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{b}+1}dx}{a}}{a}}{a \left (a^2-b^2\right )}+\frac {b \left (-4 a^3 C+6 a^2 b B+a b^2 C-3 b^3 B\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b (b B-a C) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {\frac {\frac {\left (2 a^4 B+5 a^3 b C-11 a^2 b^2 B-2 a b^3 C+6 b^4 B\right ) \sin (c+d x)}{a d}-\frac {\frac {2 x \left (a^2-b^2\right )^2 (3 b B-a C)}{a}-\frac {2 \left (-6 a^5 C+12 a^4 b B+5 a^3 b^2 C-15 a^2 b^3 B-2 a b^4 C+6 b^5 B\right ) \int \frac {1}{\left (1-\frac {a}{b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+\frac {a+b}{b}}d\tan \left (\frac {1}{2} (c+d x)\right )}{a d}}{a}}{a \left (a^2-b^2\right )}+\frac {b \left (-4 a^3 C+6 a^2 b B+a b^2 C-3 b^3 B\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b (b B-a C) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {b (b B-a C) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}+\frac {\frac {b \left (-4 a^3 C+6 a^2 b B+a b^2 C-3 b^3 B\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac {\frac {\left (2 a^4 B+5 a^3 b C-11 a^2 b^2 B-2 a b^3 C+6 b^4 B\right ) \sin (c+d x)}{a d}-\frac {\frac {2 x \left (a^2-b^2\right )^2 (3 b B-a C)}{a}-\frac {2 b \left (-6 a^5 C+12 a^4 b B+5 a^3 b^2 C-15 a^2 b^3 B-2 a b^4 C+6 b^5 B\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}}{a}}{a \left (a^2-b^2\right )}}{2 a \left (a^2-b^2\right )}\) |
(b*(b*B - a*C)*Sin[c + d*x])/(2*a*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^2) + ((b*(6*a^2*b*B - 3*b^3*B - 4*a^3*C + a*b^2*C)*Sin[c + d*x])/(a*(a^2 - b^2) *d*(a + b*Sec[c + d*x])) + (-(((2*(a^2 - b^2)^2*(3*b*B - a*C)*x)/a - (2*b* (12*a^4*b*B - 15*a^2*b^3*B + 6*b^5*B - 6*a^5*C + 5*a^3*b^2*C - 2*a*b^4*C)* ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*Sqrt[a + b]*d))/a) + ((2*a^4*B - 11*a^2*b^2*B + 6*b^4*B + 5*a^3*b*C - 2*a*b^3*C) *Sin[c + d*x])/(a*d))/(a*(a^2 - b^2)))/(2*a*(a^2 - b^2))
3.9.13.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[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[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbo l] :> Simp[1/b Int[1/(1 + (a/b)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[c*(x/a), x] - Simp[(b*c - a*d)/a Int[Csc[e + f* x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[b*(A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d*Csc[e + f*x])^n/(a*f*( m + 1)*(a^2 - b^2))), x] + Simp[1/(a*(m + 1)*(a^2 - b^2)) Int[(a + b*Csc[ e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[A*(a^2*(m + 1) - b^2*(m + n + 1)) + a*b*B*n - a*(A*b - a*B)*(m + 1)*Csc[e + f*x] + b*(A*b - a*B)*(m + n + 2) *Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A* b - a*B, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && !(ILtQ[m + 1/2, 0] && IL tQ[n, 0])
Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_. )*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.) *(x_)]*(d_.))^(n_.), x_Symbol] :> Simp[1/b^2 Int[(a + b*Csc[e + f*x])^(m + 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))^(m_), x_Symbol] :> Simp[(A*b^2 - a*b*B + a^2*C)*Cot[e + f*x]*(a + b*Csc [e + f*x])^(m + 1)*((d*Csc[e + f*x])^n/(a*f*(m + 1)*(a^2 - b^2))), x] + Sim p[1/(a*(m + 1)*(a^2 - b^2)) Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f *x])^n*Simp[a*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C)*(m + n + 1) - a*(A*b - a*B + b*C)*(m + 1)*Csc[e + f*x] + (A*b^2 - a*b*B + a^2*C)*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x ] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && !(ILtQ[m + 1/2, 0] && ILtQ[n, 0])
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))^(m_), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d *Csc[e + f*x])^n/(a*f*n)), x] + Simp[1/(a*d*n) Int[(a + b*Csc[e + f*x])^m *(d*Csc[e + f*x])^(n + 1)*Simp[a*B*n - A*b*(m + n + 1) + a*(A + A*n + C*n)* Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d , e, f, A, B, C, m}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]
Time = 0.80 (sec) , antiderivative size = 347, normalized size of antiderivative = 1.20
| method | result | size |
| derivativedivides | \(\frac {-\frac {2 \left (-\frac {B a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\left (3 B b -C a \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{a^{4}}-\frac {2 b \left (\frac {-\frac {\left (8 B \,a^{2} b +B a \,b^{2}-4 B \,b^{3}-6 a^{3} C -a^{2} b C +2 C a \,b^{2}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}+\frac {b a \left (8 B \,a^{2} b -B a \,b^{2}-4 B \,b^{3}-6 a^{3} C +a^{2} b C +2 C a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a -b \right )^{2}}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )^{2}}-\frac {\left (12 B \,a^{4} b -15 B \,a^{2} b^{3}+6 B \,b^{5}-6 a^{5} C +5 C \,a^{3} b^{2}-2 C a \,b^{4}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{4}}}{d}\) | \(347\) |
| default | \(\frac {-\frac {2 \left (-\frac {B a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\left (3 B b -C a \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{a^{4}}-\frac {2 b \left (\frac {-\frac {\left (8 B \,a^{2} b +B a \,b^{2}-4 B \,b^{3}-6 a^{3} C -a^{2} b C +2 C a \,b^{2}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}+\frac {b a \left (8 B \,a^{2} b -B a \,b^{2}-4 B \,b^{3}-6 a^{3} C +a^{2} b C +2 C a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a -b \right )^{2}}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )^{2}}-\frac {\left (12 B \,a^{4} b -15 B \,a^{2} b^{3}+6 B \,b^{5}-6 a^{5} C +5 C \,a^{3} b^{2}-2 C a \,b^{4}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{4}}}{d}\) | \(347\) |
| risch | \(\text {Expression too large to display}\) | \(1399\) |
1/d*(-2/a^4*(-B*a*tan(1/2*d*x+1/2*c)/(1+tan(1/2*d*x+1/2*c)^2)+(3*B*b-C*a)* arctan(tan(1/2*d*x+1/2*c)))-2*b/a^4*((-1/2*(8*B*a^2*b+B*a*b^2-4*B*b^3-6*C* a^3-C*a^2*b+2*C*a*b^2)*a*b/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3+1/2* b*a*(8*B*a^2*b-B*a*b^2-4*B*b^3-6*C*a^3+C*a^2*b+2*C*a*b^2)/(a+b)/(a-b)^2*ta n(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*B*a^4*b-15*B*a^2*b^3+6*B*b^5-6*C*a^5+5*C*a^3*b^2-2*C*a*b^4)/(a^4-2*a ^2*b^2+b^4)/((a+b)*(a-b))^(1/2)*arctanh((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. 755 vs. \(2 (275) = 550\).
Time = 0.37 (sec) , antiderivative size = 1568, normalized size of antiderivative = 5.41 \[ \int \frac {\cos ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx=\text {Too large to display} \]
integrate(cos(d*x+c)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x, algorithm="fricas")
[1/4*(4*(C*a^9 - 3*B*a^8*b - 3*C*a^7*b^2 + 9*B*a^6*b^3 + 3*C*a^5*b^4 - 9*B *a^4*b^5 - C*a^3*b^6 + 3*B*a^2*b^7)*d*x*cos(d*x + c)^2 + 8*(C*a^8*b - 3*B* a^7*b^2 - 3*C*a^6*b^3 + 9*B*a^5*b^4 + 3*C*a^4*b^5 - 9*B*a^3*b^6 - C*a^2*b^ 7 + 3*B*a*b^8)*d*x*cos(d*x + c) + 4*(C*a^7*b^2 - 3*B*a^6*b^3 - 3*C*a^5*b^4 + 9*B*a^4*b^5 + 3*C*a^3*b^6 - 9*B*a^2*b^7 - C*a*b^8 + 3*B*b^9)*d*x - (6*C *a^5*b^3 - 12*B*a^4*b^4 - 5*C*a^3*b^5 + 15*B*a^2*b^6 + 2*C*a*b^7 - 6*B*b^8 + (6*C*a^7*b - 12*B*a^6*b^2 - 5*C*a^5*b^3 + 15*B*a^4*b^4 + 2*C*a^3*b^5 - 6*B*a^2*b^6)*cos(d*x + c)^2 + 2*(6*C*a^6*b^2 - 12*B*a^5*b^3 - 5*C*a^4*b^4 + 15*B*a^3*b^5 + 2*C*a^2*b^6 - 6*B*a*b^7)*cos(d*x + c))*sqrt(a^2 - b^2)*lo g((2*a*b*cos(d*x + c) - (a^2 - 2*b^2)*cos(d*x + c)^2 + 2*sqrt(a^2 - b^2)*( b*cos(d*x + c) + a)*sin(d*x + c) + 2*a^2 - b^2)/(a^2*cos(d*x + c)^2 + 2*a* b*cos(d*x + c) + b^2)) + 2*(2*B*a^7*b^2 + 5*C*a^6*b^3 - 13*B*a^5*b^4 - 7*C *a^4*b^5 + 17*B*a^3*b^6 + 2*C*a^2*b^7 - 6*B*a*b^8 + 2*(B*a^9 - 3*B*a^7*b^2 + 3*B*a^5*b^4 - B*a^3*b^6)*cos(d*x + c)^2 + (4*B*a^8*b + 6*C*a^7*b^2 - 20 *B*a^6*b^3 - 9*C*a^5*b^4 + 25*B*a^4*b^5 + 3*C*a^3*b^6 - 9*B*a^2*b^7)*cos(d *x + c))*sin(d*x + c))/((a^12 - 3*a^10*b^2 + 3*a^8*b^4 - a^6*b^6)*d*cos(d* x + c)^2 + 2*(a^11*b - 3*a^9*b^3 + 3*a^7*b^5 - a^5*b^7)*d*cos(d*x + c) + ( a^10*b^2 - 3*a^8*b^4 + 3*a^6*b^6 - a^4*b^8)*d), 1/2*(2*(C*a^9 - 3*B*a^8*b - 3*C*a^7*b^2 + 9*B*a^6*b^3 + 3*C*a^5*b^4 - 9*B*a^4*b^5 - C*a^3*b^6 + 3*B* a^2*b^7)*d*x*cos(d*x + c)^2 + 4*(C*a^8*b - 3*B*a^7*b^2 - 3*C*a^6*b^3 + ...
\[ \int \frac {\cos ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx=\int \frac {\left (B + C \sec {\left (c + d x \right )}\right ) \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{3}}\, dx \]
Exception generated. \[ \int \frac {\cos ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx=\text {Exception raised: ValueError} \]
integrate(cos(d*x+c)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(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*a^2-4*b^2>0)', see `assume?` f or more de
Time = 0.38 (sec) , antiderivative size = 546, normalized size of antiderivative = 1.88 \[ \int \frac {\cos ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx=-\frac {\frac {{\left (6 \, C a^{5} b - 12 \, B a^{4} b^{2} - 5 \, C a^{3} b^{3} + 15 \, B a^{2} b^{4} + 2 \, C a b^{5} - 6 \, B b^{6}\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^{8} - 2 \, a^{6} b^{2} + a^{4} b^{4}\right )} \sqrt {-a^{2} + b^{2}}} + \frac {6 \, C a^{4} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 8 \, B a^{3} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, C a^{3} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 7 \, B a^{2} b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, C a^{2} b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 5 \, B a b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, C a b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, B b^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, C a^{4} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 8 \, B a^{3} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 5 \, C a^{3} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 7 \, B a^{2} b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, C a^{2} b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 5 \, B a b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, C a b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, B b^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} 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}} - \frac {{\left (C a - 3 \, B b\right )} {\left (d x + c\right )}}{a^{4}} - \frac {2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a^{3}}}{d} \]
-((6*C*a^5*b - 12*B*a^4*b^2 - 5*C*a^3*b^3 + 15*B*a^2*b^4 + 2*C*a*b^5 - 6*B *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)*sqrt(-a^2 + b^2)) + (6*C*a^4*b^2*tan(1/2*d*x + 1/2*c)^3 - 8*B*a^3*b^3*tan(1/2*d*x + 1/2*c)^3 - 5*C*a^3*b^3*tan(1/2*d*x + 1/2*c)^3 + 7*B*a^2*b^4*tan(1/2*d*x + 1/2*c)^3 - 3*C*a^2*b^4*tan(1/2*d*x + 1/2*c)^3 + 5*B*a*b^5*tan(1/2*d*x + 1/2*c)^3 + 2*C*a*b^5*tan(1/2*d*x + 1/2*c)^3 - 4*B* b^6*tan(1/2*d*x + 1/2*c)^3 - 6*C*a^4*b^2*tan(1/2*d*x + 1/2*c) + 8*B*a^3*b^ 3*tan(1/2*d*x + 1/2*c) - 5*C*a^3*b^3*tan(1/2*d*x + 1/2*c) + 7*B*a^2*b^4*ta n(1/2*d*x + 1/2*c) + 3*C*a^2*b^4*tan(1/2*d*x + 1/2*c) - 5*B*a*b^5*tan(1/2* d*x + 1/2*c) + 2*C*a*b^5*tan(1/2*d*x + 1/2*c) - 4*B*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) - (C*a - 3*B*b)*(d*x + c)/a^4 - 2*B*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 + 1)*a^3))/d
Time = 25.08 (sec) , antiderivative size = 5530, normalized size of antiderivative = 19.07 \[ \int \frac {\cos ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx=\text {Too large to display} \]
((tan(c/2 + (d*x)/2)^5*(6*B*b^5 - 2*B*a^5 - 12*B*a^2*b^3 + 4*B*a^3*b^2 + C *a^2*b^3 + 6*C*a^3*b^2 - 3*B*a*b^4 + 2*B*a^4*b - 2*C*a*b^4))/((a^3*b - a^4 )*(a + b)^2) + (tan(c/2 + (d*x)/2)*(2*B*a^5 + 6*B*b^5 - 12*B*a^2*b^3 - 4*B *a^3*b^2 - C*a^2*b^3 + 6*C*a^3*b^2 + 3*B*a*b^4 + 2*B*a^4*b - 2*C*a*b^4))/( (a + b)*(a^5 - 2*a^4*b + a^3*b^2)) + (2*tan(c/2 + (d*x)/2)^3*(2*B*a^6 - 6* B*b^6 + 13*B*a^2*b^4 - 6*B*a^4*b^2 - 5*C*a^3*b^3 + 2*C*a*b^5))/(a*(a^2*b - a^3)*(a + b)^2*(a - b)))/(d*(2*a*b + tan(c/2 + (d*x)/2)^2*(2*a*b - a^2 + 3*b^2) + tan(c/2 + (d*x)/2)^6*(a^2 - 2*a*b + b^2) + a^2 + b^2 - tan(c/2 + (d*x)/2)^4*(2*a*b + a^2 - 3*b^2))) + (log(tan(c/2 + (d*x)/2) - 1i)*(3*B*b - C*a)*1i)/(a^4*d) - (log(tan(c/2 + (d*x)/2) + 1i)*(B*b*3i - C*a*1i))/(a^4 *d) - (b*atan(((b*((a + b)^5*(a - b)^5)^(1/2)*((8*tan(c/2 + (d*x)/2)*(72*B ^2*b^12 + 4*C^2*a^12 - 72*B^2*a*b^11 - 8*C^2*a^11*b - 288*B^2*a^2*b^10 + 2 88*B^2*a^3*b^9 + 441*B^2*a^4*b^8 - 432*B^2*a^5*b^7 - 288*B^2*a^6*b^6 + 288 *B^2*a^7*b^5 + 36*B^2*a^8*b^4 - 72*B^2*a^9*b^3 + 36*B^2*a^10*b^2 + 8*C^2*a ^2*b^10 - 8*C^2*a^3*b^9 - 32*C^2*a^4*b^8 + 32*C^2*a^5*b^7 + 57*C^2*a^6*b^6 - 48*C^2*a^7*b^5 - 52*C^2*a^8*b^4 + 32*C^2*a^9*b^3 + 24*C^2*a^10*b^2 - 48 *B*C*a*b^11 - 24*B*C*a^11*b + 48*B*C*a^2*b^10 + 192*B*C*a^3*b^9 - 192*B*C* a^4*b^8 - 318*B*C*a^5*b^7 + 288*B*C*a^6*b^6 + 252*B*C*a^7*b^5 - 192*B*C*a^ 8*b^4 - 72*B*C*a^9*b^3 + 48*B*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) + (b*((8*(4*C*a^...