\(\int \frac {\sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx\) [476]

Optimal result

Integrand size = 21, antiderivative size = 232 \[ \int \frac {\sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\frac {3 b^2 \left (4 a^4-5 a^2 b^2+2 b^4\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 (a-b)^{5/2} (a+b)^{5/2} d}-\frac {3 b \text {arctanh}(\sin (c+d x))}{a^4 d}+\frac {\left (2 a^4-11 a^2 b^2+6 b^4\right ) \tan (c+d x)}{2 a^3 \left (a^2-b^2\right )^2 d}+\frac {b^2 \tan (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac {3 b^2 \left (2 a^2-b^2\right ) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))} \] Output:

3*b^2*(4*a^4-5*a^2*b^2+2*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*b*arctanh(sin(d*x+c))/a^4/d+1/2*(2* 
a^4-11*a^2*b^2+6*b^4)*tan(d*x+c)/a^3/(a^2-b^2)^2/d+1/2*b^2*tan(d*x+c)/a/(a 
^2-b^2)/d/(a+b*cos(d*x+c))^2+3/2*b^2*(2*a^2-b^2)*tan(d*x+c)/a^2/(a^2-b^2)^ 
2/d/(a+b*cos(d*x+c))
 

Mathematica [A] (verified)

Time = 4.92 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.88 \[ \int \frac {\sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=-\frac {\frac {6 b^2 \left (4 a^4-5 a^2 b^2+2 b^4\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{5/2}}-6 b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+6 b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {a b^3 \left (8 a^3-5 a b^2+b \left (7 a^2-4 b^2\right ) \cos (c+d x)\right ) \sin (c+d x)}{(a-b)^2 (a+b)^2 (a+b \cos (c+d x))^2}-2 a \tan (c+d x)}{2 a^4 d} \] Input:

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

Output:

-1/2*((6*b^2*(4*a^4 - 5*a^2*b^2 + 2*b^4)*ArcTanh[((a - b)*Tan[(c + d*x)/2] 
)/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(5/2) - 6*b*Log[Cos[(c + d*x)/2] - Sin[( 
c + d*x)/2]] + 6*b*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (a*b^3*(8*a^ 
3 - 5*a*b^2 + b*(7*a^2 - 4*b^2)*Cos[c + d*x])*Sin[c + d*x])/((a - b)^2*(a 
+ b)^2*(a + b*Cos[c + d*x])^2) - 2*a*Tan[c + d*x])/(a^4*d)
 

Rubi [A] (verified)

Time = 1.58 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.16, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.619, Rules used = {3042, 3281, 3042, 3534, 3042, 3534, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3281

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3534

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3534

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3480

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3138

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 4257

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

Input:

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

Output:

(b^2*Tan[c + d*x])/(2*a*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^2) + ((3*b^2*(2 
*a^2 - b^2)*Tan[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Cos[c + d*x])) + ((-3*(( 
-2*b^2*(4*a^4 - 5*a^2*b^2 + 2*b^4)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/S 
qrt[a + b]])/(a*Sqrt[a - b]*Sqrt[a + b]*d) + (2*b*(a^2 - b^2)^2*ArcTanh[Si 
n[c + d*x]])/(a*d)))/a + ((2*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 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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 3281

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*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[a*(b*c - a*d)*(m + 1) + b^2*d*(m + n 
 + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m + n + 3)*Si 
n[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a* 
d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && IntegersQ[ 
2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[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 = 2.01 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.26

method	result	size
derivativedivides	\(\frac {\frac {2 b^{2} \left (\frac {-\frac {\left (8 a^{2}+a b -4 b^{2}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {\left (8 a^{2}-a b -4 b^{2}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{2}}+\frac {3 \left (4 a^{4}-5 a^{2} b^{2}+2 b^{4}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{4}}-\frac {1}{a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {3 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{4}}-\frac {1}{a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {3 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{4}}}{d}\)	\(292\)
default	\(\frac {\frac {2 b^{2} \left (\frac {-\frac {\left (8 a^{2}+a b -4 b^{2}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {\left (8 a^{2}-a b -4 b^{2}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{2}}+\frac {3 \left (4 a^{4}-5 a^{2} b^{2}+2 b^{4}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{4}}-\frac {1}{a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {3 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{4}}-\frac {1}{a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {3 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{4}}}{d}\)	\(292\)
risch	\(\frac {i \left (-6 a^{3} b^{3} {\mathrm e}^{5 i \left (d x +c \right )}+3 a \,b^{5} {\mathrm e}^{5 i \left (d x +c \right )}-12 a^{4} b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-3 a^{2} b^{4} {\mathrm e}^{4 i \left (d x +c \right )}+6 b^{6} {\mathrm e}^{4 i \left (d x +c \right )}+8 a^{5} b \,{\mathrm e}^{3 i \left (d x +c \right )}-44 a^{3} b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+24 a \,b^{5} {\mathrm e}^{3 i \left (d x +c \right )}+8 a^{6} {\mathrm e}^{2 i \left (d x +c \right )}-26 a^{4} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-6 a^{2} b^{4} {\mathrm e}^{2 i \left (d x +c \right )}+12 b^{6} {\mathrm e}^{2 i \left (d x +c \right )}+8 a^{5} b \,{\mathrm e}^{i \left (d x +c \right )}-38 a^{3} b^{3} {\mathrm e}^{i \left (d x +c \right )}+21 a \,b^{5} {\mathrm e}^{i \left (d x +c \right )}+2 a^{4} b^{2}-11 b^{4} a^{2}+6 b^{6}\right )}{\left (a^{2}-b^{2}\right )^{2} d \,a^{3} \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 ({\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 {6 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) b^{2}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}+\frac {15 b^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d \,a^{2}}-\frac {3 b^{6} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d \,a^{4}}+\frac {6 b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}-\frac {15 b^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d \,a^{2}}+\frac {3 b^{6} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d \,a^{4}}-\frac {3 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a^{4} d}\)	\(880\)

Input:

int(sec(d*x+c)^2/(a+cos(d*x+c)*b)^3,x,method=_RETURNVERBOSE)
 

Output:

1/d*(2*b^2/a^4*((-1/2*(8*a^2+a*b-4*b^2)*a*b/(a-b)/(a^2+2*a*b+b^2)*tan(1/2* 
d*x+1/2*c)^3-1/2*(8*a^2-a*b-4*b^2)*a*b/(a+b)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1 
/2*c))/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^2+3/2*(4*a^4-5* 
a^2*b^2+2*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)))-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)-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. 638 vs. \(2 (217) = 434\).

Time = 0.65 (sec) , antiderivative size = 1346, normalized size of antiderivative = 5.80 \[ \int \frac {\sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Too large to display} \] Input:

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

Output:

[-1/4*(3*((4*a^4*b^4 - 5*a^2*b^6 + 2*b^8)*cos(d*x + c)^3 + 2*(4*a^5*b^3 - 
5*a^3*b^5 + 2*a*b^7)*cos(d*x + c)^2 + (4*a^6*b^2 - 5*a^4*b^4 + 2*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)) + 6*((a^6*b^3 - 3 
*a^4*b^5 + 3*a^2*b^7 - b^9)*cos(d*x + c)^3 + 2*(a^7*b^2 - 3*a^5*b^4 + 3*a^ 
3*b^6 - a*b^8)*cos(d*x + c)^2 + (a^8*b - 3*a^6*b^3 + 3*a^4*b^5 - a^2*b^7)* 
cos(d*x + c))*log(sin(d*x + c) + 1) - 6*((a^6*b^3 - 3*a^4*b^5 + 3*a^2*b^7 
- b^9)*cos(d*x + c)^3 + 2*(a^7*b^2 - 3*a^5*b^4 + 3*a^3*b^6 - a*b^8)*cos(d* 
x + c)^2 + (a^8*b - 3*a^6*b^3 + 3*a^4*b^5 - a^2*b^7)*cos(d*x + c))*log(-si 
n(d*x + c) + 1) - 2*(2*a^9 - 6*a^7*b^2 + 6*a^5*b^4 - 2*a^3*b^6 + (2*a^7*b^ 
2 - 13*a^5*b^4 + 17*a^3*b^6 - 6*a*b^8)*cos(d*x + c)^2 + (4*a^8*b - 20*a^6* 
b^3 + 25*a^4*b^5 - 9*a^2*b^7)*cos(d*x + c))*sin(d*x + c))/((a^10*b^2 - 3*a 
^8*b^4 + 3*a^6*b^6 - a^4*b^8)*d*cos(d*x + c)^3 + 2*(a^11*b - 3*a^9*b^3 + 3 
*a^7*b^5 - a^5*b^7)*d*cos(d*x + c)^2 + (a^12 - 3*a^10*b^2 + 3*a^8*b^4 - a^ 
6*b^6)*d*cos(d*x + c)), 1/2*(3*((4*a^4*b^4 - 5*a^2*b^6 + 2*b^8)*cos(d*x + 
c)^3 + 2*(4*a^5*b^3 - 5*a^3*b^5 + 2*a*b^7)*cos(d*x + c)^2 + (4*a^6*b^2 - 5 
*a^4*b^4 + 2*a^2*b^6)*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^6*b^3 - 3*a^4*b^5 + 3*a^2*b 
^7 - b^9)*cos(d*x + c)^3 + 2*(a^7*b^2 - 3*a^5*b^4 + 3*a^3*b^6 - a*b^8)*...
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f 
or more de
 

Giac [A] (verification not implemented)

Time = 0.58 (sec) , antiderivative size = 380, normalized size of antiderivative = 1.64 \[ \int \frac {\sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=-\frac {\frac {3 \, {\left (4 \, a^{4} b^{2} - 5 \, a^{2} b^{4} + 2 \, b^{6}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{8} - 2 \, a^{6} b^{2} + a^{4} b^{4}\right )} \sqrt {a^{2} - b^{2}}} + \frac {8 \, a^{3} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 7 \, a^{2} b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, a b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, b^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 8 \, a^{3} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 7 \, a^{2} b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 5 \, a b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, b^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )}^{2}} + \frac {3 \, b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac {3 \, b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} + \frac {2 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{3}}}{d} \] Input:

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

Output:

-(3*(4*a^4*b^2 - 5*a^2*b^4 + 2*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))/sq 
rt(a^2 - b^2)))/((a^8 - 2*a^6*b^2 + a^4*b^4)*sqrt(a^2 - b^2)) + (8*a^3*b^3 
*tan(1/2*d*x + 1/2*c)^3 - 7*a^2*b^4*tan(1/2*d*x + 1/2*c)^3 - 5*a*b^5*tan(1 
/2*d*x + 1/2*c)^3 + 4*b^6*tan(1/2*d*x + 1/2*c)^3 + 8*a^3*b^3*tan(1/2*d*x + 
 1/2*c) + 7*a^2*b^4*tan(1/2*d*x + 1/2*c) - 5*a*b^5*tan(1/2*d*x + 1/2*c) - 
4*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) + 3*b*log(abs(tan(1/2*d*x 
+ 1/2*c) + 1))/a^4 - 3*b*log(abs(tan(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 = 48.40 (sec) , antiderivative size = 5347, normalized size of antiderivative = 23.05 \[ \int \frac {\sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Too large to display} \] Input:

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

Output:

(b*atan(((b*((8*tan(c/2 + (d*x)/2)*(72*b^12 - 72*a*b^11 - 288*a^2*b^10 + 2 
88*a^3*b^9 + 441*a^4*b^8 - 432*a^5*b^7 - 288*a^6*b^6 + 288*a^7*b^5 + 36*a^ 
8*b^4 - 72*a^9*b^3 + 36*a^10*b^2))/(a^12*b + a^13 - a^6*b^7 - a^7*b^6 + 3* 
a^8*b^5 + 3*a^9*b^4 - 3*a^10*b^3 - 3*a^11*b^2) - (3*b*((24*(4*a^17*b - 4*a 
^8*b^10 + 2*a^9*b^9 + 18*a^10*b^8 - 8*a^11*b^7 - 32*a^12*b^6 + 14*a^13*b^5 
 + 26*a^14*b^4 - 12*a^15*b^3 - 8*a^16*b^2))/(a^15*b + a^16 - a^9*b^7 - a^1 
0*b^6 + 3*a^11*b^5 + 3*a^12*b^4 - 3*a^13*b^3 - 3*a^14*b^2) - (24*b*tan(c/2 
 + (d*x)/2)*(8*a^17*b - 8*a^8*b^10 + 8*a^9*b^9 + 32*a^10*b^8 - 32*a^11*b^7 
 - 48*a^12*b^6 + 48*a^13*b^5 + 32*a^14*b^4 - 32*a^15*b^3 - 8*a^16*b^2))/(a 
^4*(a^12*b + a^13 - a^6*b^7 - a^7*b^6 + 3*a^8*b^5 + 3*a^9*b^4 - 3*a^10*b^3 
 - 3*a^11*b^2))))/a^4)*3i)/a^4 + (b*((8*tan(c/2 + (d*x)/2)*(72*b^12 - 72*a 
*b^11 - 288*a^2*b^10 + 288*a^3*b^9 + 441*a^4*b^8 - 432*a^5*b^7 - 288*a^6*b 
^6 + 288*a^7*b^5 + 36*a^8*b^4 - 72*a^9*b^3 + 36*a^10*b^2))/(a^12*b + a^13 
- a^6*b^7 - a^7*b^6 + 3*a^8*b^5 + 3*a^9*b^4 - 3*a^10*b^3 - 3*a^11*b^2) + ( 
3*b*((24*(4*a^17*b - 4*a^8*b^10 + 2*a^9*b^9 + 18*a^10*b^8 - 8*a^11*b^7 - 3 
2*a^12*b^6 + 14*a^13*b^5 + 26*a^14*b^4 - 12*a^15*b^3 - 8*a^16*b^2))/(a^15* 
b + a^16 - a^9*b^7 - a^10*b^6 + 3*a^11*b^5 + 3*a^12*b^4 - 3*a^13*b^3 - 3*a 
^14*b^2) + (24*b*tan(c/2 + (d*x)/2)*(8*a^17*b - 8*a^8*b^10 + 8*a^9*b^9 + 3 
2*a^10*b^8 - 32*a^11*b^7 - 48*a^12*b^6 + 48*a^13*b^5 + 32*a^14*b^4 - 32*a^ 
15*b^3 - 8*a^16*b^2))/(a^4*(a^12*b + a^13 - a^6*b^7 - a^7*b^6 + 3*a^8*b...
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 2086, normalized size of antiderivative = 8.99 \[ \int \frac {\sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx =\text {Too large to display} \] Input:

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

Output:

(24*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( 
a**2 - b**2))*cos(c + d*x)*sin(c + d*x)**2*a**4*b**4 - 30*sqrt(a**2 - b**2 
)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*cos(c 
+ d*x)*sin(c + d*x)**2*a**2*b**6 + 12*sqrt(a**2 - b**2)*atan((tan((c + d*x 
)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*cos(c + d*x)*sin(c + d*x)* 
*2*b**8 - 24*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2) 
*b)/sqrt(a**2 - b**2))*cos(c + d*x)*a**6*b**2 + 6*sqrt(a**2 - b**2)*atan(( 
tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*cos(c + d*x)*a 
**4*b**4 + 18*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2 
)*b)/sqrt(a**2 - b**2))*cos(c + d*x)*a**2*b**6 - 12*sqrt(a**2 - b**2)*atan 
((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*cos(c + d*x) 
*b**8 + 48*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**5*b**3 - 60*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**5 + 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**7 - 48*sqrt(a**2 - b**2) 
*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*a**5*b* 
*3 + 60*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/s 
qrt(a**2 - b**2))*a**3*b**5 - 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**7 + 6*cos(c + d*x)*log(...