Integrand size = 29, antiderivative size = 290 \[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^3} \, dx=-\frac {(3 A b-a B) x}{a^4}+\frac {b \left (12 a^4 A b-15 a^2 A b^3+6 A b^5-6 a^5 B+5 a^3 b^2 B-2 a b^4 B\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 (a-b)^{5/2} (a+b)^{5/2} d}+\frac {\left (2 a^4 A-11 a^2 A b^2+6 A b^4+5 a^3 b B-2 a b^3 B\right ) \sin (c+d x)}{2 a^3 \left (a^2-b^2\right )^2 d}+\frac {b (A b-a B) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac {b \left (6 a^2 A b-3 A b^3-4 a^3 B+a b^2 B\right ) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))} \] Output:
-(3*A*b-B*a)*x/a^4+b*(12*A*a^4*b-15*A*a^2*b^3+6*A*b^5-6*B*a^5+5*B*a^3*b^2- 2*B*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*A*a^4-11*A*a^2*b^2+6*A*b^4+5*B*a^3*b-2*B*a*b^3)* sin(d*x+c)/a^3/(a^2-b^2)^2/d+1/2*b*(A*b-B*a)*sin(d*x+c)/a/(a^2-b^2)/d/(a+b *sec(d*x+c))^2+1/2*b*(6*A*a^2*b-3*A*b^3-4*B*a^3+B*a*b^2)*sin(d*x+c)/a^2/(a ^2-b^2)^2/d/(a+b*sec(d*x+c))
Time = 2.40 (sec) , antiderivative size = 306, normalized size of antiderivative = 1.06 \[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^3} \, dx=\frac {(b+a \cos (c+d x)) \sec ^2(c+d x) (A+B \sec (c+d x)) \left (2 (-3 A b+a B) (c+d x) (b+a \cos (c+d x))^2-\frac {2 b \left (12 a^4 A b-15 a^2 A b^3+6 A b^5-6 a^5 B+5 a^3 b^2 B-2 a b^4 B\right ) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right ) (b+a \cos (c+d x))^2}{\left (a^2-b^2\right )^{5/2}}+\frac {a b^3 (A b-a B) \sin (c+d x)}{(a-b) (a+b)}+\frac {a b^2 \left (-8 a^2 A b+5 A b^3+6 a^3 B-3 a b^2 B\right ) (b+a \cos (c+d x)) \sin (c+d x)}{(a-b)^2 (a+b)^2}+2 a A (b+a \cos (c+d x))^2 \sin (c+d x)\right )}{2 a^4 d (B+A \cos (c+d x)) (a+b \sec (c+d x))^3} \] Input:
Integrate[(Cos[c + d*x]*(A + B*Sec[c + d*x]))/(a + b*Sec[c + d*x])^3,x]
Output:
((b + a*Cos[c + d*x])*Sec[c + d*x]^2*(A + B*Sec[c + d*x])*(2*(-3*A*b + a*B )*(c + d*x)*(b + a*Cos[c + d*x])^2 - (2*b*(12*a^4*A*b - 15*a^2*A*b^3 + 6*A *b^5 - 6*a^5*B + 5*a^3*b^2*B - 2*a*b^4*B)*ArcTanh[((-a + b)*Tan[(c + d*x)/ 2])/Sqrt[a^2 - b^2]]*(b + a*Cos[c + d*x])^2)/(a^2 - b^2)^(5/2) + (a*b^3*(A *b - a*B)*Sin[c + d*x])/((a - b)*(a + b)) + (a*b^2*(-8*a^2*A*b + 5*A*b^3 + 6*a^3*B - 3*a*b^2*B)*(b + a*Cos[c + d*x])*Sin[c + d*x])/((a - b)^2*(a + b )^2) + 2*a*A*(b + a*Cos[c + d*x])^2*Sin[c + d*x]))/(2*a^4*d*(B + A*Cos[c + d*x])*(a + b*Sec[c + d*x])^3)
Time = 1.95 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.12, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.517, Rules used = {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.
| \(\displaystyle \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^3}dx\) |
| \(\Big \downarrow \) 4518 |
| \(\displaystyle \frac {b (A b-a B) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac {\int -\frac {\cos (c+d x) \left (2 A a^2+b B a-2 (A b-a B) \sec (c+d x) a-3 A b^2+2 b (A b-a B) \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2}dx}{2 a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {\cos (c+d x) \left (2 A a^2+b B a-2 (A b-a B) \sec (c+d x) a-3 A b^2+2 b (A b-a B) \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2}dx}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {2 A a^2+b B a-2 (A b-a B) \csc \left (c+d x+\frac {\pi }{2}\right ) a-3 A b^2+2 b (A b-a B) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 4588 |
| \(\displaystyle \frac {\frac {b \left (-4 a^3 B+6 a^2 A b+a b^2 B-3 A b^3\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\int -\frac {\cos (c+d x) \left (2 A a^4+5 b B a^3-11 A b^2 a^2-2 b^3 B a-\left (-2 B a^3+4 A b a^2-b^2 B a-A b^3\right ) \sec (c+d x) a+6 A b^4+b \left (-4 B a^3+6 A b a^2+b^2 B a-3 A b^3\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)}dx}{a \left (a^2-b^2\right )}}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {\cos (c+d x) \left (2 A a^4+5 b B a^3-11 A b^2 a^2-2 b^3 B a-\left (-2 B a^3+4 A b a^2-b^2 B a-A b^3\right ) \sec (c+d x) a+6 A b^4+b \left (-4 B a^3+6 A b a^2+b^2 B a-3 A b^3\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)}dx}{a \left (a^2-b^2\right )}+\frac {b \left (-4 a^3 B+6 a^2 A b+a b^2 B-3 A b^3\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {2 A a^4+5 b B a^3-11 A b^2 a^2-2 b^3 B a-\left (-2 B a^3+4 A b a^2-b^2 B a-A b^3\right ) \csc \left (c+d x+\frac {\pi }{2}\right ) a+6 A b^4+b \left (-4 B a^3+6 A b a^2+b^2 B a-3 A b^3\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a \left (a^2-b^2\right )}+\frac {b \left (-4 a^3 B+6 a^2 A b+a b^2 B-3 A b^3\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 4592 |
| \(\displaystyle \frac {\frac {\frac {\left (2 a^4 A+5 a^3 b B-11 a^2 A b^2-2 a b^3 B+6 A b^4\right ) \sin (c+d x)}{a d}-\frac {\int \frac {2 \left (a^2-b^2\right )^2 (3 A b-a B)-a b \left (-4 B a^3+6 A b a^2+b^2 B a-3 A b^3\right ) \sec (c+d x)}{a+b \sec (c+d x)}dx}{a}}{a \left (a^2-b^2\right )}+\frac {b \left (-4 a^3 B+6 a^2 A b+a b^2 B-3 A b^3\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\left (2 a^4 A+5 a^3 b B-11 a^2 A b^2-2 a b^3 B+6 A b^4\right ) \sin (c+d x)}{a d}-\frac {\int \frac {2 \left (a^2-b^2\right )^2 (3 A b-a B)-a b \left (-4 B a^3+6 A b a^2+b^2 B a-3 A b^3\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a}}{a \left (a^2-b^2\right )}+\frac {b \left (-4 a^3 B+6 a^2 A b+a b^2 B-3 A b^3\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 4407 |
| \(\displaystyle \frac {\frac {\frac {\left (2 a^4 A+5 a^3 b B-11 a^2 A b^2-2 a b^3 B+6 A b^4\right ) \sin (c+d x)}{a d}-\frac {\frac {2 x \left (a^2-b^2\right )^2 (3 A b-a B)}{a}-\frac {b \left (-6 a^5 B+12 a^4 A b+5 a^3 b^2 B-15 a^2 A b^3-2 a b^4 B+6 A b^5\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)}dx}{a}}{a}}{a \left (a^2-b^2\right )}+\frac {b \left (-4 a^3 B+6 a^2 A b+a b^2 B-3 A b^3\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\left (2 a^4 A+5 a^3 b B-11 a^2 A b^2-2 a b^3 B+6 A b^4\right ) \sin (c+d x)}{a d}-\frac {\frac {2 x \left (a^2-b^2\right )^2 (3 A b-a B)}{a}-\frac {b \left (-6 a^5 B+12 a^4 A b+5 a^3 b^2 B-15 a^2 A b^3-2 a b^4 B+6 A b^5\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a}}{a}}{a \left (a^2-b^2\right )}+\frac {b \left (-4 a^3 B+6 a^2 A b+a b^2 B-3 A b^3\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 4318 |
| \(\displaystyle \frac {\frac {\frac {\left (2 a^4 A+5 a^3 b B-11 a^2 A b^2-2 a b^3 B+6 A b^4\right ) \sin (c+d x)}{a d}-\frac {\frac {2 x \left (a^2-b^2\right )^2 (3 A b-a B)}{a}-\frac {\left (-6 a^5 B+12 a^4 A b+5 a^3 b^2 B-15 a^2 A b^3-2 a b^4 B+6 A b^5\right ) \int \frac {1}{\frac {a \cos (c+d x)}{b}+1}dx}{a}}{a}}{a \left (a^2-b^2\right )}+\frac {b \left (-4 a^3 B+6 a^2 A b+a b^2 B-3 A b^3\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\left (2 a^4 A+5 a^3 b B-11 a^2 A b^2-2 a b^3 B+6 A b^4\right ) \sin (c+d x)}{a d}-\frac {\frac {2 x \left (a^2-b^2\right )^2 (3 A b-a B)}{a}-\frac {\left (-6 a^5 B+12 a^4 A b+5 a^3 b^2 B-15 a^2 A b^3-2 a b^4 B+6 A b^5\right ) \int \frac {1}{\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{b}+1}dx}{a}}{a}}{a \left (a^2-b^2\right )}+\frac {b \left (-4 a^3 B+6 a^2 A b+a b^2 B-3 A b^3\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {\frac {\frac {\left (2 a^4 A+5 a^3 b B-11 a^2 A b^2-2 a b^3 B+6 A b^4\right ) \sin (c+d x)}{a d}-\frac {\frac {2 x \left (a^2-b^2\right )^2 (3 A b-a B)}{a}-\frac {2 \left (-6 a^5 B+12 a^4 A b+5 a^3 b^2 B-15 a^2 A b^3-2 a b^4 B+6 A b^5\right ) \int \frac {1}{\left (1-\frac {a}{b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+\frac {a+b}{b}}d\tan \left (\frac {1}{2} (c+d x)\right )}{a d}}{a}}{a \left (a^2-b^2\right )}+\frac {b \left (-4 a^3 B+6 a^2 A b+a b^2 B-3 A b^3\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {b (A b-a B) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}+\frac {\frac {b \left (-4 a^3 B+6 a^2 A b+a b^2 B-3 A b^3\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac {\frac {\left (2 a^4 A+5 a^3 b B-11 a^2 A b^2-2 a b^3 B+6 A b^4\right ) \sin (c+d x)}{a d}-\frac {\frac {2 x \left (a^2-b^2\right )^2 (3 A b-a B)}{a}-\frac {2 b \left (-6 a^5 B+12 a^4 A b+5 a^3 b^2 B-15 a^2 A b^3-2 a b^4 B+6 A b^5\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}}{a}}{a \left (a^2-b^2\right )}}{2 a \left (a^2-b^2\right )}\) |
Input:
Int[(Cos[c + d*x]*(A + B*Sec[c + d*x]))/(a + b*Sec[c + d*x])^3,x]
Output:
(b*(A*b - a*B)*Sin[c + d*x])/(2*a*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^2) + ((b*(6*a^2*A*b - 3*A*b^3 - 4*a^3*B + a*b^2*B)*Sin[c + d*x])/(a*(a^2 - b^2) *d*(a + b*Sec[c + d*x])) + (-(((2*(a^2 - b^2)^2*(3*A*b - a*B)*x)/a - (2*b* (12*a^4*A*b - 15*a^2*A*b^3 + 6*A*b^5 - 6*a^5*B + 5*a^3*b^2*B - 2*a*b^4*B)* ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*Sqrt[a + b]*d))/a) + ((2*a^4*A - 11*a^2*A*b^2 + 6*A*b^4 + 5*a^3*b*B - 2*a*b^3*B) *Sin[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)*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_.) + 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 = 0.84 (sec) , antiderivative size = 347, normalized size of antiderivative = 1.20
| method | result | size |
| derivativedivides | \(\frac {-\frac {2 b \left (\frac {-\frac {\left (8 A \,a^{2} b +A a \,b^{2}-4 A \,b^{3}-6 B \,a^{3}-B \,a^{2} b +2 B a \,b^{2}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}+\frac {b a \left (8 A \,a^{2} b -A a \,b^{2}-4 A \,b^{3}-6 B \,a^{3}+B \,a^{2} b +2 B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a -b \right )^{2}}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )^{2}}-\frac {\left (12 A \,a^{4} b -15 A \,a^{2} b^{3}+6 A \,b^{5}-6 B \,a^{5}+5 B \,a^{3} b^{2}-2 B a \,b^{4}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{4}}-\frac {2 \left (-\frac {A a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\left (3 A b -B a \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{a^{4}}}{d}\) | \(347\) |
| default | \(\frac {-\frac {2 b \left (\frac {-\frac {\left (8 A \,a^{2} b +A a \,b^{2}-4 A \,b^{3}-6 B \,a^{3}-B \,a^{2} b +2 B a \,b^{2}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}+\frac {b a \left (8 A \,a^{2} b -A a \,b^{2}-4 A \,b^{3}-6 B \,a^{3}+B \,a^{2} b +2 B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a -b \right )^{2}}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )^{2}}-\frac {\left (12 A \,a^{4} b -15 A \,a^{2} b^{3}+6 A \,b^{5}-6 B \,a^{5}+5 B \,a^{3} b^{2}-2 B a \,b^{4}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{4}}-\frac {2 \left (-\frac {A a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\left (3 A b -B a \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{a^{4}}}{d}\) | \(347\) |
| risch | \(\text {Expression too large to display}\) | \(1411\) |
Input:
int(cos(d*x+c)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^3,x,method=_RETURNVERBOSE )
Output:
1/d*(-2*b/a^4*((-1/2*(8*A*a^2*b+A*a*b^2-4*A*b^3-6*B*a^3-B*a^2*b+2*B*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*A*a^2*b-A*a*b^2 -4*A*b^3-6*B*a^3+B*a^2*b+2*B*a*b^2)/(a+b)/(a-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-15*A*a^2 *b^3+6*A*b^5-6*B*a^5+5*B*a^3*b^2-2*B*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)))-2/a^4*(-A *a*tan(1/2*d*x+1/2*c)/(1+tan(1/2*d*x+1/2*c)^2)+(3*A*b-B*a)*arctan(tan(1/2* d*x+1/2*c))))
Leaf count of result is larger than twice the leaf count of optimal. 755 vs. \(2 (275) = 550\).
Time = 0.19 (sec) , antiderivative size = 1568, normalized size of antiderivative = 5.41 \[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^3} \, dx=\text {Too large to display} \] Input:
integrate(cos(d*x+c)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^3,x, algorithm="fri cas")
Output:
[1/4*(4*(B*a^9 - 3*A*a^8*b - 3*B*a^7*b^2 + 9*A*a^6*b^3 + 3*B*a^5*b^4 - 9*A *a^4*b^5 - B*a^3*b^6 + 3*A*a^2*b^7)*d*x*cos(d*x + c)^2 + 8*(B*a^8*b - 3*A* a^7*b^2 - 3*B*a^6*b^3 + 9*A*a^5*b^4 + 3*B*a^4*b^5 - 9*A*a^3*b^6 - B*a^2*b^ 7 + 3*A*a*b^8)*d*x*cos(d*x + c) + 4*(B*a^7*b^2 - 3*A*a^6*b^3 - 3*B*a^5*b^4 + 9*A*a^4*b^5 + 3*B*a^3*b^6 - 9*A*a^2*b^7 - B*a*b^8 + 3*A*b^9)*d*x - (6*B *a^5*b^3 - 12*A*a^4*b^4 - 5*B*a^3*b^5 + 15*A*a^2*b^6 + 2*B*a*b^7 - 6*A*b^8 + (6*B*a^7*b - 12*A*a^6*b^2 - 5*B*a^5*b^3 + 15*A*a^4*b^4 + 2*B*a^3*b^5 - 6*A*a^2*b^6)*cos(d*x + c)^2 + 2*(6*B*a^6*b^2 - 12*A*a^5*b^3 - 5*B*a^4*b^4 + 15*A*a^3*b^5 + 2*B*a^2*b^6 - 6*A*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*A*a^7*b^2 + 5*B*a^6*b^3 - 13*A*a^5*b^4 - 7*B *a^4*b^5 + 17*A*a^3*b^6 + 2*B*a^2*b^7 - 6*A*a*b^8 + 2*(A*a^9 - 3*A*a^7*b^2 + 3*A*a^5*b^4 - A*a^3*b^6)*cos(d*x + c)^2 + (4*A*a^8*b + 6*B*a^7*b^2 - 20 *A*a^6*b^3 - 9*B*a^5*b^4 + 25*A*a^4*b^5 + 3*B*a^3*b^6 - 9*A*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*(B*a^9 - 3*A*a^8*b - 3*B*a^7*b^2 + 9*A*a^6*b^3 + 3*B*a^5*b^4 - 9*A*a^4*b^5 - B*a^3*b^6 + 3*A* a^2*b^7)*d*x*cos(d*x + c)^2 + 4*(B*a^8*b - 3*A*a^7*b^2 - 3*B*a^6*b^3 + ...
\[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^3} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )}\right ) \cos {\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{3}}\, dx \] Input:
integrate(cos(d*x+c)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))**3,x)
Output:
Integral((A + B*sec(c + d*x))*cos(c + d*x)/(a + b*sec(c + d*x))**3, x)
Exception generated. \[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^3} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(cos(d*x+c)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^3,x, algorithm="max ima")
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*a^2-4*b^2>0)', see `assume?` f or more de
Time = 0.25 (sec) , antiderivative size = 546, normalized size of antiderivative = 1.88 \[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^3} \, dx =\text {Too large to display} \] Input:
integrate(cos(d*x+c)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^3,x, algorithm="gia c")
Output:
-((6*B*a^5*b - 12*A*a^4*b^2 - 5*B*a^3*b^3 + 15*A*a^2*b^4 + 2*B*a*b^5 - 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)*sqrt(-a^2 + b^2)) + (6*B*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 - 5*B*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 - 3*B*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 + 2*B*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 - 6*B*a^4*b^2*tan(1/2*d*x + 1/2*c) + 8*A*a^3*b^ 3*tan(1/2*d*x + 1/2*c) - 5*B*a^3*b^3*tan(1/2*d*x + 1/2*c) + 7*A*a^2*b^4*ta n(1/2*d*x + 1/2*c) + 3*B*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) + 2*B*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) - (B*a - 3*A*b)*(d*x + c)/a^4 - 2*A*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 + 1)*a^3))/d
Time = 18.84 (sec) , antiderivative size = 5530, normalized size of antiderivative = 19.07 \[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^3} \, dx=\text {Too large to display} \] Input:
int((cos(c + d*x)*(A + B/cos(c + d*x)))/(a + b/cos(c + d*x))^3,x)
Output:
((tan(c/2 + (d*x)/2)^5*(6*A*b^5 - 2*A*a^5 - 12*A*a^2*b^3 + 4*A*a^3*b^2 + B *a^2*b^3 + 6*B*a^3*b^2 - 3*A*a*b^4 + 2*A*a^4*b - 2*B*a*b^4))/((a^3*b - a^4 )*(a + b)^2) + (tan(c/2 + (d*x)/2)*(2*A*a^5 + 6*A*b^5 - 12*A*a^2*b^3 - 4*A *a^3*b^2 - B*a^2*b^3 + 6*B*a^3*b^2 + 3*A*a*b^4 + 2*A*a^4*b - 2*B*a*b^4))/( (a + b)*(a^5 - 2*a^4*b + a^3*b^2)) + (2*tan(c/2 + (d*x)/2)^3*(2*A*a^6 - 6* A*b^6 + 13*A*a^2*b^4 - 6*A*a^4*b^2 - 5*B*a^3*b^3 + 2*B*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*A*b - B*a)*1i)/(a^4*d) - (log(tan(c/2 + (d*x)/2) + 1i)*(A*b*3i - B*a*1i))/(a^4 *d) - (b*atan(((b*((8*tan(c/2 + (d*x)/2)*(72*A^2*b^12 + 4*B^2*a^12 - 72*A^ 2*a*b^11 - 8*B^2*a^11*b - 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 + 8*B^2*a^2*b^10 - 8*B^2*a^3*b^9 - 32 *B^2*a^4*b^8 + 32*B^2*a^5*b^7 + 57*B^2*a^6*b^6 - 48*B^2*a^7*b^5 - 52*B^2*a ^8*b^4 + 32*B^2*a^9*b^3 + 24*B^2*a^10*b^2 - 48*A*B*a*b^11 - 24*A*B*a^11*b + 48*A*B*a^2*b^10 + 192*A*B*a^3*b^9 - 192*A*B*a^4*b^8 - 318*A*B*a^5*b^7 + 288*A*B*a^6*b^6 + 252*A*B*a^7*b^5 - 192*A*B*a^8*b^4 - 72*A*B*a^9*b^3 + 48* A*B*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*B*a^18 + 12*A*a^8*b^10 - 6*A*a^9...
Time = 0.17 (sec) , antiderivative size = 521, normalized size of antiderivative = 1.80 \[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^3} \, dx=\frac {6 \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {-a^{2}+b^{2}}}\right ) \cos \left (d x +c \right ) a^{3} b^{2}-4 \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {-a^{2}+b^{2}}}\right ) \cos \left (d x +c \right ) a \,b^{4}+6 \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {-a^{2}+b^{2}}}\right ) a^{2} b^{3}-4 \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {-a^{2}+b^{2}}}\right ) b^{5}+\cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{6}-2 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{4} b^{2}+\cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2} b^{4}-2 \cos \left (d x +c \right ) a^{5} b c -2 \cos \left (d x +c \right ) a^{5} b d x +4 \cos \left (d x +c \right ) a^{3} b^{3} c +4 \cos \left (d x +c \right ) a^{3} b^{3} d x -2 \cos \left (d x +c \right ) a \,b^{5} c -2 \cos \left (d x +c \right ) a \,b^{5} d x +\sin \left (d x +c \right ) a^{5} b -3 \sin \left (d x +c \right ) a^{3} b^{3}+2 \sin \left (d x +c \right ) a \,b^{5}-2 a^{4} b^{2} c -2 a^{4} b^{2} d x +4 a^{2} b^{4} c +4 a^{2} b^{4} d x -2 b^{6} c -2 b^{6} d x}{a^{3} d \left (\cos \left (d x +c \right ) a^{5}-2 \cos \left (d x +c \right ) a^{3} b^{2}+\cos \left (d x +c \right ) a \,b^{4}+a^{4} b -2 a^{2} b^{3}+b^{5}\right )} \] Input:
int(cos(d*x+c)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^3,x)
Output:
(6*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqr t( - a**2 + b**2))*cos(c + d*x)*a**3*b**2 - 4*sqrt( - a**2 + b**2)*atan((t an((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b**2))*cos(c + d*x) *a*b**4 + 6*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/ 2)*b)/sqrt( - a**2 + b**2))*a**2*b**3 - 4*sqrt( - a**2 + b**2)*atan((tan(( c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b**2))*b**5 + cos(c + d *x)*sin(c + d*x)*a**6 - 2*cos(c + d*x)*sin(c + d*x)*a**4*b**2 + cos(c + d* x)*sin(c + d*x)*a**2*b**4 - 2*cos(c + d*x)*a**5*b*c - 2*cos(c + d*x)*a**5* b*d*x + 4*cos(c + d*x)*a**3*b**3*c + 4*cos(c + d*x)*a**3*b**3*d*x - 2*cos( c + d*x)*a*b**5*c - 2*cos(c + d*x)*a*b**5*d*x + sin(c + d*x)*a**5*b - 3*si n(c + d*x)*a**3*b**3 + 2*sin(c + d*x)*a*b**5 - 2*a**4*b**2*c - 2*a**4*b**2 *d*x + 4*a**2*b**4*c + 4*a**2*b**4*d*x - 2*b**6*c - 2*b**6*d*x)/(a**3*d*(c os(c + d*x)*a**5 - 2*cos(c + d*x)*a**3*b**2 + cos(c + d*x)*a*b**4 + a**4*b - 2*a**2*b**3 + b**5))