\(\int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx\) [271]

Optimal result

Integrand size = 31, antiderivative size = 299 \[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\frac {b \left (12 a^4 A b-15 a^2 A b^3+6 A b^5-6 a^5 B+5 a^3 b^2 B-2 a b^4 B\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 (a-b)^{5/2} (a+b)^{5/2} d}-\frac {(3 A b-a B) \text {arctanh}(\sin (c+d x))}{a^4 d}+\frac {\left (2 a^4 A-11 a^2 A b^2+6 A b^4+5 a^3 b B-2 a b^3 B\right ) \tan (c+d x)}{2 a^3 \left (a^2-b^2\right )^2 d}+\frac {b (A b-a B) \tan (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac {b \left (6 a^2 A b-3 A b^3-4 a^3 B+a b^2 B\right ) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))} \] Output:

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)*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-B*a)*arctanh(sin(d*x+c))/a^4/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)*tan(d*x+c)/a^3/(a^2-b^2)^2/d+1/2*b*(A*b-B*a)*tan(d*x+c)/ 
a/(a^2-b^2)/d/(a+b*cos(d*x+c))^2+1/2*b*(6*A*a^2*b-3*A*b^3-4*B*a^3+B*a*b^2) 
*tan(d*x+c)/a^2/(a^2-b^2)^2/d/(a+b*cos(d*x+c))
 

Mathematica [A] (verified)

Time = 7.77 (sec) , antiderivative size = 418, normalized size of antiderivative = 1.40 \[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\frac {b \left (-12 a^4 A b+15 a^2 A b^3-6 A b^5+6 a^5 B-5 a^3 b^2 B+2 a b^4 B\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{a^4 \left (a^2-b^2\right )^2 \sqrt {-a^2+b^2} d}+\frac {(3 A b-a B) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{a^4 d}+\frac {(-3 A b+a B) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{a^4 d}+\frac {A \sin \left (\frac {1}{2} (c+d x)\right )}{a^3 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {A \sin \left (\frac {1}{2} (c+d x)\right )}{a^3 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {-A b^3 \sin (c+d x)+a b^2 B \sin (c+d x)}{2 a^2 (a-b) (a+b) d (a+b \cos (c+d x))^2}+\frac {-7 a^2 A b^3 \sin (c+d x)+4 A b^5 \sin (c+d x)+5 a^3 b^2 B \sin (c+d x)-2 a b^4 B \sin (c+d x)}{2 a^3 (a-b)^2 (a+b)^2 d (a+b \cos (c+d x))} \] Input:

Integrate[((A + B*Cos[c + d*x])*Sec[c + d*x]^2)/(a + b*Cos[c + d*x])^3,x]
 

Output:

(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]])/(a^4*(a^2 - b^2) 
^2*Sqrt[-a^2 + b^2]*d) + ((3*A*b - a*B)*Log[Cos[(c + d*x)/2] - Sin[(c + d* 
x)/2]])/(a^4*d) + ((-3*A*b + a*B)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] 
)/(a^4*d) + (A*Sin[(c + d*x)/2])/(a^3*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/ 
2])) + (A*Sin[(c + d*x)/2])/(a^3*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])) 
+ (-(A*b^3*Sin[c + d*x]) + a*b^2*B*Sin[c + d*x])/(2*a^2*(a - b)*(a + b)*d* 
(a + b*Cos[c + d*x])^2) + (-7*a^2*A*b^3*Sin[c + d*x] + 4*A*b^5*Sin[c + d*x 
] + 5*a^3*b^2*B*Sin[c + d*x] - 2*a*b^4*B*Sin[c + d*x])/(2*a^3*(a - b)^2*(a 
 + b)^2*d*(a + b*Cos[c + d*x]))
 

Rubi [A] (verified)

Time = 2.02 (sec) , antiderivative size = 335, normalized size of antiderivative = 1.12, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.419, Rules used = {3042, 3479, 3042, 3534, 3042, 3534, 25, 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.

\(\displaystyle \int \frac {\sec ^2(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^3}dx\)

\(\Big \downarrow \) 3479

\(\displaystyle \frac {\int \frac {\left (2 A a^2+b B a-2 (A b-a B) \cos (c+d x) a-3 A b^2+2 b (A b-a B) \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^2}dx}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \tan (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {2 A a^2+b B a-2 (A b-a B) \sin \left (c+d x+\frac {\pi }{2}\right ) a-3 A b^2+2 b (A b-a B) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \tan (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 3534

\(\displaystyle \frac {\frac {\int \frac {\left (2 A a^4+5 b B a^3-11 A b^2 a^2-2 b^3 B a-\left (-2 B a^3+4 A b a^2-b^2 B a-A b^3\right ) \cos (c+d x) a+6 A b^4+b \left (-4 B a^3+6 A b a^2+b^2 B a-3 A b^3\right ) \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)}dx}{a \left (a^2-b^2\right )}+\frac {b \left (-4 a^3 B+6 a^2 A b+a b^2 B-3 A b^3\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \tan (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {2 A a^4+5 b B a^3-11 A b^2 a^2-2 b^3 B a-\left (-2 B a^3+4 A b a^2-b^2 B a-A b^3\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a+6 A b^4+b \left (-4 B a^3+6 A b a^2+b^2 B a-3 A b^3\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a \left (a^2-b^2\right )}+\frac {b \left (-4 a^3 B+6 a^2 A b+a b^2 B-3 A b^3\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \tan (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 3534

\(\displaystyle \frac {\frac {\frac {\int -\frac {\left (2 \left (a^2-b^2\right )^2 (3 A b-a B)-a b \left (-4 B a^3+6 A b a^2+b^2 B a-3 A b^3\right ) \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)}dx}{a}+\frac {\left (2 a^4 A+5 a^3 b B-11 a^2 A b^2-2 a b^3 B+6 A b^4\right ) \tan (c+d x)}{a d}}{a \left (a^2-b^2\right )}+\frac {b \left (-4 a^3 B+6 a^2 A b+a b^2 B-3 A b^3\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \tan (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\frac {\left (2 a^4 A+5 a^3 b B-11 a^2 A b^2-2 a b^3 B+6 A b^4\right ) \tan (c+d x)}{a d}-\frac {\int \frac {\left (2 \left (a^2-b^2\right )^2 (3 A b-a B)-a b \left (-4 B a^3+6 A b a^2+b^2 B a-3 A b^3\right ) \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)}dx}{a}}{a \left (a^2-b^2\right )}+\frac {b \left (-4 a^3 B+6 a^2 A b+a b^2 B-3 A b^3\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \tan (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\left (2 a^4 A+5 a^3 b B-11 a^2 A b^2-2 a b^3 B+6 A b^4\right ) \tan (c+d x)}{a d}-\frac {\int \frac {2 \left (a^2-b^2\right )^2 (3 A b-a B)-a b \left (-4 B a^3+6 A b a^2+b^2 B a-3 A b^3\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}}{a \left (a^2-b^2\right )}+\frac {b \left (-4 a^3 B+6 a^2 A b+a b^2 B-3 A b^3\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \tan (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 3480

\(\displaystyle \frac {\frac {\frac {\left (2 a^4 A+5 a^3 b B-11 a^2 A b^2-2 a b^3 B+6 A b^4\right ) \tan (c+d x)}{a d}-\frac {\frac {2 \left (a^2-b^2\right )^2 (3 A b-a B) \int \sec (c+d x)dx}{a}-\frac {b \left (-6 a^5 B+12 a^4 A b+5 a^3 b^2 B-15 a^2 A b^3-2 a b^4 B+6 A b^5\right ) \int \frac {1}{a+b \cos (c+d x)}dx}{a}}{a}}{a \left (a^2-b^2\right )}+\frac {b \left (-4 a^3 B+6 a^2 A b+a b^2 B-3 A b^3\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \tan (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\left (2 a^4 A+5 a^3 b B-11 a^2 A b^2-2 a b^3 B+6 A b^4\right ) \tan (c+d x)}{a d}-\frac {\frac {2 \left (a^2-b^2\right )^2 (3 A b-a B) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {b \left (-6 a^5 B+12 a^4 A b+5 a^3 b^2 B-15 a^2 A b^3-2 a b^4 B+6 A b^5\right ) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a}}{a}}{a \left (a^2-b^2\right )}+\frac {b \left (-4 a^3 B+6 a^2 A b+a b^2 B-3 A b^3\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \tan (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 3138

\(\displaystyle \frac {\frac {\frac {\left (2 a^4 A+5 a^3 b B-11 a^2 A b^2-2 a b^3 B+6 A b^4\right ) \tan (c+d x)}{a d}-\frac {\frac {2 \left (a^2-b^2\right )^2 (3 A b-a B) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {2 b \left (-6 a^5 B+12 a^4 A b+5 a^3 b^2 B-15 a^2 A b^3-2 a b^4 B+6 A b^5\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{a d}}{a}}{a \left (a^2-b^2\right )}+\frac {b \left (-4 a^3 B+6 a^2 A b+a b^2 B-3 A b^3\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \tan (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {\frac {\left (2 a^4 A+5 a^3 b B-11 a^2 A b^2-2 a b^3 B+6 A b^4\right ) \tan (c+d x)}{a d}-\frac {\frac {2 \left (a^2-b^2\right )^2 (3 A b-a B) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {2 b \left (-6 a^5 B+12 a^4 A b+5 a^3 b^2 B-15 a^2 A b^3-2 a b^4 B+6 A b^5\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}}{a}}{a \left (a^2-b^2\right )}+\frac {b \left (-4 a^3 B+6 a^2 A b+a b^2 B-3 A b^3\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \tan (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {b (A b-a B) \tan (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}+\frac {\frac {b \left (-4 a^3 B+6 a^2 A b+a b^2 B-3 A b^3\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {\frac {\left (2 a^4 A+5 a^3 b B-11 a^2 A b^2-2 a b^3 B+6 A b^4\right ) \tan (c+d x)}{a d}-\frac {\frac {2 \left (a^2-b^2\right )^2 (3 A b-a B) \text {arctanh}(\sin (c+d x))}{a d}-\frac {2 b \left (-6 a^5 B+12 a^4 A b+5 a^3 b^2 B-15 a^2 A b^3-2 a b^4 B+6 A b^5\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}}{a}}{a \left (a^2-b^2\right )}}{2 a \left (a^2-b^2\right )}\)

Input:

Int[((A + B*Cos[c + d*x])*Sec[c + d*x]^2)/(a + b*Cos[c + d*x])^3,x]
 

Output:

(b*(A*b - a*B)*Tan[c + d*x])/(2*a*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^2) + 
((b*(6*a^2*A*b - 3*A*b^3 - 4*a^3*B + a*b^2*B)*Tan[c + d*x])/(a*(a^2 - b^2) 
*d*(a + b*Cos[c + d*x])) + (-(((-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)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2]) 
/Sqrt[a + b]])/(a*Sqrt[a - b]*Sqrt[a + b]*d) + (2*(a^2 - b^2)^2*(3*A*b - a 
*B)*ArcTanh[Sin[c + d*x]])/(a*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)*Tan[c + d*x])/(a*d))/(a*(a^2 - b^2)))/(2*a*(a^2 - 
 b^2))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 218

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]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3138

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]
 

rule 3479

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
 (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si 
mp[(-(A*b^2 - a*b*B))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin 
[e + f*x])^(1 + n)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 
1)*(b*c - a*d)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e 
 + f*x])^n*Simp[(a*A - b*B)*(b*c - a*d)*(m + 1) + b*d*(A*b - a*B)*(m + n + 
2) + (A*b - a*B)*(a*d*(m + 1) - b*c*(m + 2))*Sin[e + f*x] - b*d*(A*b - a*B) 
*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n 
}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Rat 
ionalQ[m] && m < -1 && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(I 
ntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0]) 
))
 

rule 3480

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]
 

rule 3534

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])))
 

rule 4257

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 

Maple [A] (verified)

Time = 3.09 (sec) , antiderivative size = 376, normalized size of antiderivative = 1.26

method	result	size
derivativedivides	\(\frac {-\frac {A}{a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (-3 A b +B a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{4}}-\frac {A}{a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (3 A b -B a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{4}}+\frac {2 b \left (\frac {-\frac {\left (8 A \,a^{2} b +A a \,b^{2}-4 A \,b^{3}-6 a^{3} B -B \,a^{2} b +2 B a \,b^{2}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {b a \left (8 A \,a^{2} b -A a \,b^{2}-4 A \,b^{3}-6 a^{3} B +B \,a^{2} b +2 B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a -b \right )^{2}}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{2}}+\frac {\left (12 A \,a^{4} b -15 A \,a^{2} b^{3}+6 A \,b^{5}-6 B \,a^{5}+5 B \,a^{3} b^{2}-2 B a \,b^{4}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{4}}}{d}\)	\(376\)
default	\(\frac {-\frac {A}{a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (-3 A b +B a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{4}}-\frac {A}{a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (3 A b -B a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{4}}+\frac {2 b \left (\frac {-\frac {\left (8 A \,a^{2} b +A a \,b^{2}-4 A \,b^{3}-6 a^{3} B -B \,a^{2} b +2 B a \,b^{2}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {b a \left (8 A \,a^{2} b -A a \,b^{2}-4 A \,b^{3}-6 a^{3} B +B \,a^{2} b +2 B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a -b \right )^{2}}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{2}}+\frac {\left (12 A \,a^{4} b -15 A \,a^{2} b^{3}+6 A \,b^{5}-6 B \,a^{5}+5 B \,a^{3} b^{2}-2 B a \,b^{4}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{4}}}{d}\)	\(376\)
risch	\(\text {Expression too large to display}\)	\(1691\)

Input:

int((A+B*cos(d*x+c))*sec(d*x+c)^2/(a+cos(d*x+c)*b)^3,x,method=_RETURNVERBO 
SE)
 

Output:

1/d*(-A/a^3/(tan(1/2*d*x+1/2*c)+1)+1/a^4*(-3*A*b+B*a)*ln(tan(1/2*d*x+1/2*c 
)+1)-A/a^3/(tan(1/2*d*x+1/2*c)-1)+(3*A*b-B*a)/a^4*ln(tan(1/2*d*x+1/2*c)-1) 
+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)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a-b)*(a+b))^(1/2))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1015 vs. \(2 (284) = 568\).

Time = 24.39 (sec) , antiderivative size = 2100, normalized size of antiderivative = 7.02 \[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Too large to display} \] Input:

integrate((A+B*cos(d*x+c))*sec(d*x+c)^2/(a+b*cos(d*x+c))^3,x, algorithm="f 
ricas")
 

Output:

[1/4*(((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)*cos(d*x + c)^3 + 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)^2 + (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))*sqrt(-a^2 + b^2)*log((2*a*b*cos(d*x + c) + (2*a^2 - b^2)*cos(d*x 
+ c)^2 + 2*sqrt(-a^2 + b^2)*(a*cos(d*x + c) + b)*sin(d*x + c) - a^2 + 2*b^ 
2)/(b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2)) + 2*((B*a^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)*cos(d*x + c)^3 + 2*(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)*cos(d*x + c)^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)*cos(d*x + c))*log(sin(d*x + c) + 1) - 2*((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)*cos(d*x + c)^3 + 2*(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) 
*cos(d*x + c)^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)*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*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)*cos(d*x + c)^2 + (4*A*a^8*b + 6*B*a^7*b^2 - 20*A*a^6*b^...
 

Sympy [F]

\[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\int \frac {\left (A + B \cos {\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}}{\left (a + b \cos {\left (c + d x \right )}\right )^{3}}\, dx \] Input:

integrate((A+B*cos(d*x+c))*sec(d*x+c)**2/(a+b*cos(d*x+c))**3,x)
 

Output:

Integral((A + B*cos(c + d*x))*sec(c + d*x)**2/(a + b*cos(c + d*x))**3, x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((A+B*cos(d*x+c))*sec(d*x+c)^2/(a+b*cos(d*x+c))^3,x, algorithm="m 
axima")
 

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
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 574 vs. \(2 (284) = 568\).

Time = 0.24 (sec) , antiderivative size = 574, normalized size of antiderivative = 1.92 \[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx =\text {Too large to display} \] Input:

integrate((A+B*cos(d*x+c))*sec(d*x+c)^2/(a+b*cos(d*x+c))^3,x, algorithm="g 
iac")
 

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*t 
an(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*tan(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)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^ 
4 - (B*a - 3*A*b)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/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
 

Mupad [B] (verification not implemented)

Time = 34.52 (sec) , antiderivative size = 9312, normalized size of antiderivative = 31.14 \[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Too large to display} \] Input:

int((A + B*cos(c + d*x))/(cos(c + d*x)^2*(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))) + (atan(((((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 + 28 
8*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) + (((8*(4*B*a^18 + 1 
2*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 - 4*B*...
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 977, normalized size of antiderivative = 3.27 \[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx =\text {Too large to display} \] Input:

int((A+B*cos(d*x+c))*sec(d*x+c)^2/(a+b*cos(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)/sqrt(a 
**2 - b**2))*cos(c + d*x)*a**3*b**2 - 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)*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))*sin(c + d*x)**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))*sin(c + d*x)**2*b**5 + 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 + 2*cos(c + d*x)*log(tan((c + d*x) 
/2) - 1)*a**5*b - 4*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a**3*b**3 + 2*c 
os(c + d*x)*log(tan((c + d*x)/2) - 1)*a*b**5 - 2*cos(c + d*x)*log(tan((c + 
 d*x)/2) + 1)*a**5*b + 4*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a**3*b**3 
- 2*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a*b**5 + cos(c + d*x)*sin(c + d 
*x)*a**5*b - 3*cos(c + d*x)*sin(c + d*x)*a**3*b**3 + 2*cos(c + d*x)*sin(c 
+ d*x)*a*b**5 - 2*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**4*b**2 + 4* 
log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**2*b**4 - 2*log(tan((c + d*x)/ 
2) - 1)*sin(c + d*x)**2*b**6 + 2*log(tan((c + d*x)/2) - 1)*a**4*b**2 - 4*l 
og(tan((c + d*x)/2) - 1)*a**2*b**4 + 2*log(tan((c + d*x)/2) - 1)*b**6 + 2* 
log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**4*b**2 - 4*log(tan((c + d*x)/ 
2) + 1)*sin(c + d*x)**2*a**2*b**4 + 2*log(tan((c + d*x)/2) + 1)*sin(c +...