\(\int \frac {\sec ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx\) [467]

Optimal result

Integrand size = 21, antiderivative size = 217 \[ \int \frac {\sec ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx=-\frac {2 b^3 \left (4 a^2-3 b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 (a-b)^{3/2} (a+b)^{3/2} d}+\frac {\left (a^2+6 b^2\right ) \text {arctanh}(\sin (c+d x))}{2 a^4 d}-\frac {b \left (2 a^2-3 b^2\right ) \tan (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac {\left (a^2-3 b^2\right ) \sec (c+d x) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac {b^2 \sec (c+d x) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))} \] Output:

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

Mathematica [A] (verified)

Time = 6.24 (sec) , antiderivative size = 285, normalized size of antiderivative = 1.31 \[ \int \frac {\sec ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {\frac {8 b^3 \left (-4 a^2+3 b^2\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{3/2}}-2 a^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-12 b^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 a^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+12 b^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {a^2}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {a^2}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 a b^4 \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))}-8 a b \tan (c+d x)}{4 a^4 d} \] Input:

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

Output:

((8*b^3*(-4*a^2 + 3*b^2)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^ 
2]])/(-a^2 + b^2)^(3/2) - 2*a^2*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 
 12*b^2*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 2*a^2*Log[Cos[(c + d*x) 
/2] + Sin[(c + d*x)/2]] + 12*b^2*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] 
+ a^2/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 - a^2/(Cos[(c + d*x)/2] + Si 
n[(c + d*x)/2])^2 + (4*a*b^4*Sin[c + d*x])/((a - b)*(a + b)*(a + b*Cos[c + 
 d*x])) - 8*a*b*Tan[c + d*x])/(4*a^4*d)
 

Rubi [A] (verified)

Time = 1.44 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.05, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3042, 3281, 3042, 3534, 25, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3281

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3534

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3534

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3480

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3138

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 4257

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

Input:

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

Output:

(b^2*Sec[c + d*x]*Tan[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Cos[c + d*x])) + ( 
((a^2 - 3*b^2)*Sec[c + d*x]*Tan[c + d*x])/(2*a*d) - (-(((-4*b^3*(4*a - (3* 
b^2)/a)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*S 
qrt[a + b]*d) + ((a^4 + 5*a^2*b^2 - 6*b^4)*ArcTanh[Sin[c + d*x]])/(a*d))/a 
) + (2*b*(2*a^2 - 3*b^2)*Tan[c + d*x])/(a*d))/(2*a))/(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 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 = 1.84 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.26

method	result	size
derivativedivides	\(\frac {-\frac {2 b^{3} \left (-\frac {a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )}+\frac {\left (4 a^{2}-3 b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{4}}-\frac {1}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {-4 b -a}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (a^{2}+6 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{4}}+\frac {1}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-4 b -a}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-a^{2}-6 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{4}}}{d}\)	\(273\)
default	\(\frac {-\frac {2 b^{3} \left (-\frac {a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )}+\frac {\left (4 a^{2}-3 b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{4}}-\frac {1}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {-4 b -a}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (a^{2}+6 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{4}}+\frac {1}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-4 b -a}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-a^{2}-6 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{4}}}{d}\)	\(273\)
risch	\(-\frac {i \left (a^{3} b \,{\mathrm e}^{5 i \left (d x +c \right )}-3 a \,b^{3} {\mathrm e}^{5 i \left (d x +c \right )}+2 a^{4} {\mathrm e}^{4 i \left (d x +c \right )}+2 a^{2} b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-6 b^{4} {\mathrm e}^{4 i \left (d x +c \right )}+8 a^{3} b \,{\mathrm e}^{3 i \left (d x +c \right )}-12 a \,b^{3} {\mathrm e}^{3 i \left (d x +c \right )}-2 a^{4} {\mathrm e}^{2 i \left (d x +c \right )}+10 a^{2} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-12 b^{4} {\mathrm e}^{2 i \left (d x +c \right )}+7 a^{3} b \,{\mathrm e}^{i \left (d x +c \right )}-9 b^{3} a \,{\mathrm e}^{i \left (d x +c \right )}+4 a^{2} b^{2}-6 b^{4}\right )}{d \,a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2} \left (a^{2}-b^{2}\right ) \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 a^{2} d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2}}{a^{4} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 a^{2} d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2}}{a^{4} d}-\frac {4 b^{3} \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 ) \left (a -b \right ) d \,a^{2}}+\frac {3 b^{5} \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 ) \left (a -b \right ) d \,a^{4}}+\frac {4 b^{3} \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 ) \left (a -b \right ) d \,a^{2}}-\frac {3 b^{5} \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 ) \left (a -b \right ) d \,a^{4}}\)	\(693\)

Input:

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

Output:

1/d*(-2*b^3/a^4*(-a*b/(a^2-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)+(4*a^2-3*b^2)/(a-b)/(a+b)/((a-b)*(a+b))^(1/2) 
*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a-b)*(a+b))^(1/2)))-1/2/a^2/(tan(1/2*d* 
x+1/2*c)+1)^2-1/2*(-4*b-a)/a^3/(tan(1/2*d*x+1/2*c)+1)+1/2*(a^2+6*b^2)/a^4* 
ln(tan(1/2*d*x+1/2*c)+1)+1/2/a^2/(tan(1/2*d*x+1/2*c)-1)^2-1/2*(-4*b-a)/a^3 
/(tan(1/2*d*x+1/2*c)-1)+1/2/a^4*(-a^2-6*b^2)*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. 415 vs. \(2 (204) = 408\).

Time = 0.52 (sec) , antiderivative size = 899, normalized size of antiderivative = 4.14 \[ \int \frac {\sec ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate(sec(d*x+c)^3/(a+b*cos(d*x+c))^2,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 = 0.56 (sec) , antiderivative size = 293, normalized size of antiderivative = 1.35 \[ \int \frac {\sec ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {\frac {4 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{5} - a^{3} b^{2}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )}} + \frac {4 \, {\left (4 \, a^{2} b^{3} - 3 \, b^{5}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{6} - a^{4} b^{2}\right )} \sqrt {a^{2} - b^{2}}} + \frac {{\left (a^{2} + 6 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac {{\left (a^{2} + 6 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} + \frac {2 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{3}}}{2 \, d} \] Input:

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

Output:

1/2*(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)) + 4*(4*a^2*b^3 - 3*b^5)*(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) 
) + (a^2 + 6*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - (a^2 + 6*b^2)*l 
og(abs(tan(1/2*d*x + 1/2*c) - 1))/a^4 + 2*(a*tan(1/2*d*x + 1/2*c)^3 + 4*b* 
tan(1/2*d*x + 1/2*c)^3 + a*tan(1/2*d*x + 1/2*c) - 4*b*tan(1/2*d*x + 1/2*c) 
)/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*a^3))/d
 

Mupad [B] (verification not implemented)

Time = 48.66 (sec) , antiderivative size = 3699, normalized size of antiderivative = 17.05 \[ \int \frac {\sec ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

- ((tan(c/2 + (d*x)/2)*(3*a*b^3 - 3*a^3*b + a^4 + 6*b^4 - 5*a^2*b^2))/((a^ 
3*b - a^4)*(a + b)) + (tan(c/2 + (d*x)/2)^5*(3*a^3*b - 3*a*b^3 + a^4 + 6*b 
^4 - 5*a^2*b^2))/((a^3*b - a^4)*(a + b)) + (2*tan(c/2 + (d*x)/2)^3*(a^4 - 
6*b^4 + 3*a^2*b^2))/(a*(a^2*b - a^3)*(a + b)))/(d*(a + b - tan(c/2 + (d*x) 
/2)^2*(a + 3*b) - tan(c/2 + (d*x)/2)^4*(a - 3*b) + tan(c/2 + (d*x)/2)^6*(a 
 - b))) - (atan((((a^2 + 6*b^2)*((8*tan(c/2 + (d*x)/2)*(a^10 - 2*a^9*b - 7 
2*a*b^9 + 72*b^10 - 120*a^2*b^8 + 120*a^3*b^7 + 17*a^4*b^6 - 26*a^5*b^5 + 
23*a^6*b^4 - 20*a^7*b^3 + 11*a^8*b^2))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) - 
 ((a^2 + 6*b^2)*((8*(2*a^15 - 12*a^8*b^7 + 6*a^9*b^6 + 28*a^10*b^5 - 14*a^ 
11*b^4 - 16*a^12*b^3 + 6*a^13*b^2))/(a^11*b + a^12 - a^9*b^3 - a^10*b^2) - 
 (4*tan(c/2 + (d*x)/2)*(a^2 + 6*b^2)*(8*a^13*b - 8*a^8*b^6 + 8*a^9*b^5 + 1 
6*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))))/(2*a^4))*1i)/(2*a^4) + ((a^2 + 6*b^2)*((8*tan(c/2 + (d*x)/2)*(a^10 
 - 2*a^9*b - 72*a*b^9 + 72*b^10 - 120*a^2*b^8 + 120*a^3*b^7 + 17*a^4*b^6 - 
 26*a^5*b^5 + 23*a^6*b^4 - 20*a^7*b^3 + 11*a^8*b^2))/(a^8*b + a^9 - a^6*b^ 
3 - a^7*b^2) + ((a^2 + 6*b^2)*((8*(2*a^15 - 12*a^8*b^7 + 6*a^9*b^6 + 28*a^ 
10*b^5 - 14*a^11*b^4 - 16*a^12*b^3 + 6*a^13*b^2))/(a^11*b + a^12 - a^9*b^3 
 - a^10*b^2) + (4*tan(c/2 + (d*x)/2)*(a^2 + 6*b^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))))/(2*a^4))*1i)/(2*a^4))/((16*(108*b^11 - 54*a*b^10 ...
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 1637, normalized size of antiderivative = 7.54 \[ \int \frac {\sec ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 16*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sq 
rt(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 + 16*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((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 
 - b**2))*cos(c + d*x)*b**6 - 16*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 + 12* 
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 + 16*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 - 12*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 - cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**6*b - 4* 
cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**4*b**3 + 11*cos( 
c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**2*b**5 - 6*cos(c + d 
*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b**7 + cos(c + d*x)*log(tan( 
(c + d*x)/2) - 1)*a**6*b + 4*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a**4*b 
**3 - 11*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a**2*b**5 + 6*cos(c + d*x) 
*log(tan((c + d*x)/2) - 1)*b**7 + cos(c + d*x)*log(tan((c + d*x)/2) + 1)*s 
in(c + d*x)**2*a**6*b + 4*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c ...