\(\int \frac {(1-\cos ^2(c+d x)) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx\) [616]

Optimal result

Integrand size = 33, antiderivative size = 204 \[ \int \frac {\left (1-\cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=-\frac {\left (2 a^4-9 a^2 b^2+6 b^4\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 (a-b)^{3/2} (a+b)^{3/2} d}-\frac {3 b \text {arctanh}(\sin (c+d x))}{a^4 d}+\frac {\left (5 a^2-6 b^2\right ) \tan (c+d x)}{2 a^3 \left (a^2-b^2\right ) d}-\frac {\tan (c+d x)}{2 a d (a+b \cos (c+d x))^2}-\frac {\left (2 a^2-3 b^2\right ) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))} \] Output:

-(2*a^4-9*a^2*b^2+6*b^4)*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2) 
)/a^4/(a-b)^(3/2)/(a+b)^(3/2)/d-3*b*arctanh(sin(d*x+c))/a^4/d+1/2*(5*a^2-6 
*b^2)*tan(d*x+c)/a^3/(a^2-b^2)/d-1/2*tan(d*x+c)/a/d/(a+b*cos(d*x+c))^2-1/2 
*(2*a^2-3*b^2)*tan(d*x+c)/a^2/(a^2-b^2)/d/(a+b*cos(d*x+c))
 

Mathematica [A] (verified)

Time = 3.41 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.98 \[ \int \frac {\left (1-\cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\frac {-\frac {2 \left (2 a^4-9 a^2 b^2+6 b^4\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{3/2}}+6 b \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+\frac {a b \left (4 a^3-5 a b^2+b \left (3 a^2-4 b^2\right ) \cos (c+d x)\right ) \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))^2}+2 a \tan (c+d x)}{2 a^4 d} \] Input:

Integrate[((1 - Cos[c + d*x]^2)*Sec[c + d*x]^2)/(a + b*Cos[c + d*x])^3,x]
 

Output:

((-2*(2*a^4 - 9*a^2*b^2 + 6*b^4)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[- 
a^2 + b^2]])/(-a^2 + b^2)^(3/2) + 6*b*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x 
)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + (a*b*(4*a^3 - 5*a*b^2 
+ b*(3*a^2 - 4*b^2)*Cos[c + d*x])*Sin[c + d*x])/((a - b)*(a + b)*(a + b*Co 
s[c + d*x])^2) + 2*a*Tan[c + d*x])/(2*a^4*d)
 

Rubi [A] (verified)

Time = 1.48 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.20, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.394, Rules used = {3042, 3535, 3042, 3535, 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.

\(\displaystyle \int \frac {\left (1-\cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1-\sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^3}dx\)

\(\Big \downarrow \) 3535

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3535

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3534

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3480

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3138

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 4257

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

Input:

Int[((1 - Cos[c + d*x]^2)*Sec[c + d*x]^2)/(a + b*Cos[c + d*x])^3,x]
 

Output:

-1/2*Tan[c + d*x]/(a*d*(a + b*Cos[c + d*x])^2) + (-(((2*a^2 - 3*b^2)*Tan[c 
 + d*x])/(a*d*(a + b*Cos[c + d*x]))) + (-(((2*(2*a^6 - 11*a^4*b^2 + 15*a^2 
*b^4 - 6*b^6)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a*Sqrt[ 
a - b]*Sqrt[a + b]*d) + (6*b*(a^2 - b^2)^2*ArcTanh[Sin[c + d*x]])/(a*d))/a 
) + ((5*a^4 - 11*a^2*b^2 + 6*b^4)*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 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 3535

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])))
 

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 = 1.08 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.26

method	result	size
derivativedivides	\(\frac {-\frac {1}{a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {3 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{4}}-\frac {2 \left (\frac {-\frac {\left (4 a^{2}+a b -4 b^{2}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a +b \right )}-\frac {\left (4 a^{2}-a b -4 b^{2}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a -b \right )}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{2}}+\frac {\left (2 a^{4}-9 a^{2} b^{2}+6 b^{4}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{2 \left (a^{2}-b^{2}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{4}}-\frac {1}{a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {3 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{4}}}{d}\)	\(257\)
default	\(\frac {-\frac {1}{a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {3 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{4}}-\frac {2 \left (\frac {-\frac {\left (4 a^{2}+a b -4 b^{2}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a +b \right )}-\frac {\left (4 a^{2}-a b -4 b^{2}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a -b \right )}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{2}}+\frac {\left (2 a^{4}-9 a^{2} b^{2}+6 b^{4}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{2 \left (a^{2}-b^{2}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{4}}-\frac {1}{a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {3 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{4}}}{d}\)	\(257\)
risch	\(\frac {i \left (2 a^{3} {\mathrm e}^{5 i \left (d x +c \right )} b -3 a \,{\mathrm e}^{5 i \left (d x +c \right )} b^{3}+6 a^{4} {\mathrm e}^{4 i \left (d x +c \right )}-3 a^{2} {\mathrm e}^{4 i \left (d x +c \right )} b^{2}-6 \,{\mathrm e}^{4 i \left (d x +c \right )} b^{4}+20 a^{3} {\mathrm e}^{3 i \left (d x +c \right )} b -24 a \,{\mathrm e}^{3 i \left (d x +c \right )} b^{3}+14 a^{4} {\mathrm e}^{2 i \left (d x +c \right )}-6 a^{2} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-12 \,{\mathrm e}^{2 i \left (d x +c \right )} b^{4}+18 \,{\mathrm e}^{i \left (d x +c \right )} a^{3} b -21 \,{\mathrm e}^{i \left (d x +c \right )} a \,b^{3}+5 a^{2} b^{2}-6 b^{4}\right )}{a^{3} d \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )^{2} \left (a^{2}-b^{2}\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {3 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{4} d}-\frac {3 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a^{4} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}+\frac {9 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) b^{2}}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{2}}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) b^{4}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{4}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}-\frac {9 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) b^{2}}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{2}}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) b^{4}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{4}}\)	\(809\)

Input:

int((1-cos(d*x+c)^2)*sec(d*x+c)^2/(a+b*cos(d*x+c))^3,x,method=_RETURNVERBO 
SE)
 

Output:

1/d*(-1/a^3/(tan(1/2*d*x+1/2*c)-1)+3*b/a^4*ln(tan(1/2*d*x+1/2*c)-1)-2/a^4* 
((-1/2*(4*a^2+a*b-4*b^2)*a*b/(a+b)*tan(1/2*d*x+1/2*c)^3-1/2*(4*a^2-a*b-4*b 
^2)*a*b/(a-b)*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*(2*a^4-9*a^2*b^2+6*b^4)/(a^2-b^2)/((a+b)*(a-b))^(1/2)*ar 
ctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2)))-1/a^3/(tan(1/2*d*x+1/2 
*c)+1)-3*b/a^4*ln(tan(1/2*d*x+1/2*c)+1))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 526 vs. \(2 (189) = 378\).

Time = 0.37 (sec) , antiderivative size = 1121, normalized size of antiderivative = 5.50 \[ \int \frac {\left (1-\cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Too large to display} \] Input:

integrate((1-cos(d*x+c)^2)*sec(d*x+c)^2/(a+b*cos(d*x+c))^3,x, algorithm="f 
ricas")
 

Output:

[1/4*(((2*a^4*b^2 - 9*a^2*b^4 + 6*b^6)*cos(d*x + c)^3 + 2*(2*a^5*b - 9*a^3 
*b^3 + 6*a*b^5)*cos(d*x + c)^2 + (2*a^6 - 9*a^4*b^2 + 6*a^2*b^4)*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)) - 6*((a^4*b^3 - 2*a^2*b^5 
+ b^7)*cos(d*x + c)^3 + 2*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*cos(d*x + c)^2 + ( 
a^6*b - 2*a^4*b^3 + a^2*b^5)*cos(d*x + c))*log(sin(d*x + c) + 1) + 6*((a^4 
*b^3 - 2*a^2*b^5 + b^7)*cos(d*x + c)^3 + 2*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*c 
os(d*x + c)^2 + (a^6*b - 2*a^4*b^3 + a^2*b^5)*cos(d*x + c))*log(-sin(d*x + 
 c) + 1) + 2*(2*a^7 - 4*a^5*b^2 + 2*a^3*b^4 + (5*a^5*b^2 - 11*a^3*b^4 + 6* 
a*b^6)*cos(d*x + c)^2 + (8*a^6*b - 17*a^4*b^3 + 9*a^2*b^5)*cos(d*x + c))*s 
in(d*x + c))/((a^8*b^2 - 2*a^6*b^4 + a^4*b^6)*d*cos(d*x + c)^3 + 2*(a^9*b 
- 2*a^7*b^3 + a^5*b^5)*d*cos(d*x + c)^2 + (a^10 - 2*a^8*b^2 + a^6*b^4)*d*c 
os(d*x + c)), -1/2*(((2*a^4*b^2 - 9*a^2*b^4 + 6*b^6)*cos(d*x + c)^3 + 2*(2 
*a^5*b - 9*a^3*b^3 + 6*a*b^5)*cos(d*x + c)^2 + (2*a^6 - 9*a^4*b^2 + 6*a^2* 
b^4)*cos(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 
- b^2)*sin(d*x + c))) + 3*((a^4*b^3 - 2*a^2*b^5 + b^7)*cos(d*x + c)^3 + 2* 
(a^5*b^2 - 2*a^3*b^4 + a*b^6)*cos(d*x + c)^2 + (a^6*b - 2*a^4*b^3 + a^2*b^ 
5)*cos(d*x + c))*log(sin(d*x + c) + 1) - 3*((a^4*b^3 - 2*a^2*b^5 + b^7)*co 
s(d*x + c)^3 + 2*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*cos(d*x + c)^2 + (a^6*b ...
 

Sympy [F]

\[ \int \frac {\left (1-\cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=- \int \left (- \frac {\sec ^{2}{\left (c + d x \right )}}{a^{3} + 3 a^{2} b \cos {\left (c + d x \right )} + 3 a b^{2} \cos ^{2}{\left (c + d x \right )} + b^{3} \cos ^{3}{\left (c + d x \right )}}\right )\, dx - \int \frac {\cos ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{a^{3} + 3 a^{2} b \cos {\left (c + d x \right )} + 3 a b^{2} \cos ^{2}{\left (c + d x \right )} + b^{3} \cos ^{3}{\left (c + d x \right )}}\, dx \] Input:

integrate((1-cos(d*x+c)**2)*sec(d*x+c)**2/(a+b*cos(d*x+c))**3,x)
 

Output:

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (1-\cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((1-cos(d*x+c)^2)*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 [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 357, normalized size of antiderivative = 1.75 \[ \int \frac {\left (1-\cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\frac {\frac {{\left (2 \, a^{4} - 9 \, a^{2} b^{2} + 6 \, b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{6} - a^{4} b^{2}\right )} \sqrt {a^{2} - b^{2}}} + \frac {4 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 5 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{5} - a^{3} b^{2}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )}^{2}} - \frac {3 \, b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} + \frac {3 \, b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} - \frac {2 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{3}}}{d} \] Input:

integrate((1-cos(d*x+c)^2)*sec(d*x+c)^2/(a+b*cos(d*x+c))^3,x, algorithm="g 
iac")
 

Output:

((2*a^4 - 9*a^2*b^2 + 6*b^4)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 
2*b) + arctan(-(a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(a^2 
- b^2)))/((a^6 - a^4*b^2)*sqrt(a^2 - b^2)) + (4*a^3*b*tan(1/2*d*x + 1/2*c) 
^3 - 3*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 - 5*a*b^3*tan(1/2*d*x + 1/2*c)^3 + 4 
*b^4*tan(1/2*d*x + 1/2*c)^3 + 4*a^3*b*tan(1/2*d*x + 1/2*c) + 3*a^2*b^2*tan 
(1/2*d*x + 1/2*c) - 5*a*b^3*tan(1/2*d*x + 1/2*c) - 4*b^4*tan(1/2*d*x + 1/2 
*c))/((a^5 - a^3*b^2)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 
 + a + b)^2) - 3*b*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 + 3*b*log(abs(ta 
n(1/2*d*x + 1/2*c) - 1))/a^4 - 2*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 = 4.65 (sec) , antiderivative size = 3376, normalized size of antiderivative = 16.55 \[ \int \frac {\left (1-\cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Too large to display} \] Input:

int(-(cos(c + d*x)^2 - 1)/(cos(c + d*x)^2*(a + b*cos(c + d*x))^3),x)
 

Output:

((tan(c/2 + (d*x)/2)*(3*a*b^2 - 6*a^2*b - 2*a^3 + 6*b^3))/(a^3*b - a^4) - 
(tan(c/2 + (d*x)/2)^5*(3*a*b^2 + 6*a^2*b - 2*a^3 - 6*b^3))/(a^3*(a + b)) + 
 (2*tan(c/2 + (d*x)/2)^3*(2*a^4 + 6*b^4 - 7*a^2*b^2))/(a^3*(a + b)*(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))) - (b*atan(((b*((8*tan(c/2 + (d*x)/2)*(4*a^8 - 72*a*b^7 + 72*b^ 
8 - 144*a^2*b^6 + 144*a^3*b^5 + 69*a^4*b^4 - 72*a^5*b^3))/(a^8*b + a^9 - a 
^6*b^3 - a^7*b^2) - (3*b*((8*(8*a^13*b + 4*a^14 - 12*a^8*b^6 + 6*a^9*b^5 + 
 30*a^10*b^4 - 14*a^11*b^3 - 22*a^12*b^2))/(a^11*b + a^12 - a^9*b^3 - a^10 
*b^2) - (24*b*tan(c/2 + (d*x)/2)*(8*a^13*b - 8*a^8*b^6 + 8*a^9*b^5 + 16*a^ 
10*b^4 - 16*a^11*b^3 - 8*a^12*b^2))/(a^4*(a^8*b + a^9 - a^6*b^3 - a^7*b^2) 
)))/a^4)*3i)/a^4 + (b*((8*tan(c/2 + (d*x)/2)*(4*a^8 - 72*a*b^7 + 72*b^8 - 
144*a^2*b^6 + 144*a^3*b^5 + 69*a^4*b^4 - 72*a^5*b^3))/(a^8*b + a^9 - a^6*b 
^3 - a^7*b^2) + (3*b*((8*(8*a^13*b + 4*a^14 - 12*a^8*b^6 + 6*a^9*b^5 + 30* 
a^10*b^4 - 14*a^11*b^3 - 22*a^12*b^2))/(a^11*b + a^12 - a^9*b^3 - a^10*b^2 
) + (24*b*tan(c/2 + (d*x)/2)*(8*a^13*b - 8*a^8*b^6 + 8*a^9*b^5 + 16*a^10*b 
^4 - 16*a^11*b^3 - 8*a^12*b^2))/(a^4*(a^8*b + a^9 - a^6*b^3 - a^7*b^2))))/ 
a^4)*3i)/a^4)/((16*(54*a*b^7 - 12*a^7*b - 108*b^8 + 270*a^2*b^6 - 117*a^3* 
b^5 - 198*a^4*b^4 + 72*a^5*b^3 + 36*a^6*b^2))/(a^11*b + a^12 - a^9*b^3 - a 
^10*b^2) + (3*b*((8*tan(c/2 + (d*x)/2)*(4*a^8 - 72*a*b^7 + 72*b^8 - 144...
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 1817, normalized size of antiderivative = 8.91 \[ \int \frac {\left (1-\cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx =\text {Too large to display} \] Input:

int((1-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)/sqr 
t(a**2 - b**2))*cos(c + d*x)*sin(c + d*x)**2*a**4*b**2 + 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)*sin(c + d*x)**2*a**2*b**4 - 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**6 + 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**6 - 14*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**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**2*b**4 + 12*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)*b** 
6 - 8*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqr 
t(a**2 - b**2))*sin(c + d*x)**2*a**5*b + 36*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**3 
*b**3 - 24*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*b**5 + 8*sqrt(a**2 - b**2)*atan((ta 
n((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*a**5*b - 36*sqrt 
(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b 
**2))*a**3*b**3 + 24*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + 
 d*x)/2)*b)/sqrt(a**2 - b**2))*a*b**5 + 6*cos(c + d*x)*log(tan((c + d*x...