\(\int \frac {(B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx\) [800]

Optimal result

Integrand size = 40, antiderivative size = 143 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx=-\frac {2 b^2 (b B-a C) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 \sqrt {a-b} \sqrt {a+b} d}+\frac {\left (a^2 B+2 b^2 B-2 a b C\right ) \text {arctanh}(\sin (c+d x))}{2 a^3 d}-\frac {(b B-a C) \tan (c+d x)}{a^2 d}+\frac {B \sec (c+d x) \tan (c+d x)}{2 a d} \] Output:

-2*b^2*(B*b-C*a)*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/a^3/(a 
-b)^(1/2)/(a+b)^(1/2)/d+1/2*(B*a^2+2*B*b^2-2*C*a*b)*arctanh(sin(d*x+c))/a^ 
3/d-(B*b-C*a)*tan(d*x+c)/a^2/d+1/2*B*sec(d*x+c)*tan(d*x+c)/a/d
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(300\) vs. \(2(143)=286\).

Time = 2.69 (sec) , antiderivative size = 300, normalized size of antiderivative = 2.10 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {\frac {8 b^2 (b B-a C) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}-2 \left (a^2 B+2 b^2 B-2 a b C\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 \left (a^2 B+2 b^2 B-2 a b C\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {a^2 B}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 a (-b B+a C) \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}-\frac {a^2 B}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 a (-b B+a C) \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )}}{4 a^3 d} \] Input:

Integrate[((B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^4)/(a + b*Cos[ 
c + d*x]),x]
 

Output:

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

Rubi [A] (verified)

Time = 1.12 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.08, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {3042, 3508, 3042, 3479, 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.

\(\displaystyle \int \frac {\sec ^4(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^4 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx\)

\(\Big \downarrow \) 3508

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3534

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3480

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

\(\Big \downarrow \) 3042

\(\displaystyle \frac {B \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 (b B-a C) \tan (c+d x)}{a d}-\frac {\frac {\left (a^2 B-2 a b C+2 b^2 B\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {2 b^2 (b B-a C) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a}}{a}}{2 a}\)

\(\Big \downarrow \) 3138

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 4257

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

Input:

Int[((B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^4)/(a + b*Cos[c + d* 
x]),x]
 

Output:

(B*Sec[c + d*x]*Tan[c + d*x])/(2*a*d) - (-(((-4*b^2*(b*B - a*C)*ArcTan[(Sq 
rt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*Sqrt[a + b]*d) + 
((a^2*B + 2*b^2*B - 2*a*b*C)*ArcTanh[Sin[c + d*x]])/(a*d))/a) + (2*(b*B - 
a*C)*Tan[c + d*x])/(a*d))/(2*a)
 

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 3508

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[1/b^2   Int[(a + b*Sin[e + f*x])^(m 
+ 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ 
[{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - 
a*b*B + a^2*C, 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 = 0.62 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.60

method	result	size
derivativedivides	\(\frac {\frac {B}{2 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-B a -2 B b +2 C a}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-B \,a^{2}-2 B \,b^{2}+2 a b C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{3}}-\frac {2 b^{2} \left (B b -C a \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {B}{2 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {-B a -2 B b +2 C a}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (B \,a^{2}+2 B \,b^{2}-2 a b C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{3}}}{d}\)	\(229\)
default	\(\frac {\frac {B}{2 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-B a -2 B b +2 C a}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-B \,a^{2}-2 B \,b^{2}+2 a b C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{3}}-\frac {2 b^{2} \left (B b -C a \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {B}{2 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {-B a -2 B b +2 C a}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (B \,a^{2}+2 B \,b^{2}-2 a b C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{3}}}{d}\)	\(229\)
risch	\(-\frac {i \left (B a \,{\mathrm e}^{3 i \left (d x +c \right )}+2 B b \,{\mathrm e}^{2 i \left (d x +c \right )}-2 C a \,{\mathrm e}^{2 i \left (d x +c \right )}-B a \,{\mathrm e}^{i \left (d x +c \right )}+2 B b -2 C a \right )}{d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {B \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 a d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B \,b^{2}}{a^{3} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b C}{a^{2} d}-\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) B}{\sqrt {-a^{2}+b^{2}}\, d \,a^{3}}+\frac {b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, d \,a^{2}}+\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) B}{\sqrt {-a^{2}+b^{2}}\, d \,a^{3}}-\frac {b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, d \,a^{2}}+\frac {B \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 a d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B \,b^{2}}{a^{3} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b C}{a^{2} d}\)	\(524\)

Input:

int((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+b*cos(d*x+c)),x,method=_ 
RETURNVERBOSE)
 

Output:

1/d*(1/2*B/a/(tan(1/2*d*x+1/2*c)-1)^2-1/2*(-B*a-2*B*b+2*C*a)/a^2/(tan(1/2* 
d*x+1/2*c)-1)+1/2/a^3*(-B*a^2-2*B*b^2+2*C*a*b)*ln(tan(1/2*d*x+1/2*c)-1)-2* 
b^2*(B*b-C*a)/a^3/((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*B/a/(tan(1/2*d*x+1/2*c)+1)^2-1/2*(-B*a-2*B*b+2*C*a)/a 
^2/(tan(1/2*d*x+1/2*c)+1)+1/2*(B*a^2+2*B*b^2-2*C*a*b)/a^3*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. 260 vs. \(2 (129) = 258\).

Time = 2.39 (sec) , antiderivative size = 589, normalized size of antiderivative = 4.12 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx=\left [\frac {2 \, {\left (C a b^{2} - B b^{3}\right )} \sqrt {-a^{2} + b^{2}} \cos \left (d x + c\right )^{2} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) + {\left (B a^{4} - 2 \, C a^{3} b + B a^{2} b^{2} + 2 \, C a b^{3} - 2 \, B b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (B a^{4} - 2 \, C a^{3} b + B a^{2} b^{2} + 2 \, C a b^{3} - 2 \, B b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (B a^{4} - B a^{2} b^{2} + 2 \, {\left (C a^{4} - B a^{3} b - C a^{2} b^{2} + B a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{5} - a^{3} b^{2}\right )} d \cos \left (d x + c\right )^{2}}, \frac {4 \, {\left (C a b^{2} - B b^{3}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right )^{2} + {\left (B a^{4} - 2 \, C a^{3} b + B a^{2} b^{2} + 2 \, C a b^{3} - 2 \, B b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (B a^{4} - 2 \, C a^{3} b + B a^{2} b^{2} + 2 \, C a b^{3} - 2 \, B b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (B a^{4} - B a^{2} b^{2} + 2 \, {\left (C a^{4} - B a^{3} b - C a^{2} b^{2} + B a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{5} - a^{3} b^{2}\right )} d \cos \left (d x + c\right )^{2}}\right ] \] Input:

integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+b*cos(d*x+c)),x, a 
lgorithm="fricas")
 

Output:

[1/4*(2*(C*a*b^2 - B*b^3)*sqrt(-a^2 + b^2)*cos(d*x + c)^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)) + (B*a^4 - 2*C*a^3*b + B*a^2*b^2 + 2*C*a*b^3 - 2*B*b^4)*cos(d*x 
+ c)^2*log(sin(d*x + c) + 1) - (B*a^4 - 2*C*a^3*b + B*a^2*b^2 + 2*C*a*b^3 
- 2*B*b^4)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(B*a^4 - B*a^2*b^2 + 
2*(C*a^4 - B*a^3*b - C*a^2*b^2 + B*a*b^3)*cos(d*x + c))*sin(d*x + c))/((a^ 
5 - a^3*b^2)*d*cos(d*x + c)^2), 1/4*(4*(C*a*b^2 - B*b^3)*sqrt(a^2 - b^2)*a 
rctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c)))*cos(d*x + c)^2 
 + (B*a^4 - 2*C*a^3*b + B*a^2*b^2 + 2*C*a*b^3 - 2*B*b^4)*cos(d*x + c)^2*lo 
g(sin(d*x + c) + 1) - (B*a^4 - 2*C*a^3*b + B*a^2*b^2 + 2*C*a*b^3 - 2*B*b^4 
)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(B*a^4 - B*a^2*b^2 + 2*(C*a^4 
- B*a^3*b - C*a^2*b^2 + B*a*b^3)*cos(d*x + c))*sin(d*x + c))/((a^5 - a^3*b 
^2)*d*cos(d*x + c)^2)]
 

Sympy [F]

\[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx=\int \frac {\left (B + C \cos {\left (c + d x \right )}\right ) \cos {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}}{a + b \cos {\left (c + d x \right )}}\, dx \] Input:

integrate((B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**4/(a+b*cos(d*x+c)),x)
 

Output:

Integral((B + C*cos(c + d*x))*cos(c + d*x)*sec(c + d*x)**4/(a + b*cos(c + 
d*x)), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+b*cos(d*x+c)),x, a 
lgorithm="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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 269 vs. \(2 (129) = 258\).

Time = 0.16 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.88 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {\frac {{\left (B a^{2} - 2 \, C a b + 2 \, B b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {{\left (B a^{2} - 2 \, C a b + 2 \, B b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} - \frac {4 \, {\left (C a b^{2} - B b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} a^{3}} + \frac {2 \, {\left (B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{2}}}{2 \, d} \] Input:

integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+b*cos(d*x+c)),x, a 
lgorithm="giac")
 

Output:

1/2*((B*a^2 - 2*C*a*b + 2*B*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 
(B*a^2 - 2*C*a*b + 2*B*b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^3 - 4*(C* 
a*b^2 - B*b^3)*(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)))/(sqrt 
(a^2 - b^2)*a^3) + 2*(B*a*tan(1/2*d*x + 1/2*c)^3 - 2*C*a*tan(1/2*d*x + 1/2 
*c)^3 + 2*B*b*tan(1/2*d*x + 1/2*c)^3 + B*a*tan(1/2*d*x + 1/2*c) + 2*C*a*ta 
n(1/2*d*x + 1/2*c) - 2*B*b*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 
- 1)^2*a^2))/d
 

Mupad [B] (verification not implemented)

Time = 2.86 (sec) , antiderivative size = 4051, normalized size of antiderivative = 28.33 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx=\text {Too large to display} \] Input:

int((B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^4*(a + b*cos(c + d*x 
))),x)
 

Output:

(C*a*sin(2*c + 2*d*x))/(2*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) - (B*b 
*sin(2*c + 2*d*x))/(2*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) + (B*a*sin 
(c + d*x))/(2*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) - (B*a*atan((sin(c 
/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*1i)/(2*d*(a^2 - b^2)*(cos(2*c + 2*d* 
x)/2 + 1/2)) + (C*b*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*1i)/( 
d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) - (B*b^2*atan((sin(c/2 + (d*x)/2 
)*1i)/cos(c/2 + (d*x)/2))*1i)/(2*a*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2 
)) + (B*b^4*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*1i)/(a^3*d*(a 
^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) - (C*b^3*atan((sin(c/2 + (d*x)/2)*1i 
)/cos(c/2 + (d*x)/2))*1i)/(a^2*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) + 
 (B*b^3*sin(2*c + 2*d*x))/(2*a^2*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) 
 - (C*b^2*sin(2*c + 2*d*x))/(2*a*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) 
 - (B*a*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*cos(2*c + 2*d*x)* 
1i)/(2*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) + (C*b*atan((sin(c/2 + (d 
*x)/2)*1i)/cos(c/2 + (d*x)/2))*cos(2*c + 2*d*x)*1i)/(d*(a^2 - b^2)*(cos(2* 
c + 2*d*x)/2 + 1/2)) - (B*b^2*sin(c + d*x))/(2*a*d*(a^2 - b^2)*(cos(2*c + 
2*d*x)/2 + 1/2)) - (B*b^2*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2)) 
*cos(2*c + 2*d*x)*1i)/(2*a*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) + (B* 
b^4*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*cos(2*c + 2*d*x)*1i)/ 
(a^3*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) - (C*b^3*atan((sin(c/2 +...
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 827, normalized size of antiderivative = 5.78 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx =\text {Too large to display} \] Input:

int((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+b*cos(d*x+c)),x)
 

Output:

(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*a*b**2*c - 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**4 - 
 4*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**2*c + 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**4 - 2*cos(c + d*x)*sin(c + d*x)*a** 
4*c + 2*cos(c + d*x)*sin(c + d*x)*a**3*b**2 + 2*cos(c + d*x)*sin(c + d*x)* 
a**2*b**2*c - 2*cos(c + d*x)*sin(c + d*x)*a*b**4 - log(tan((c + d*x)/2) - 
1)*sin(c + d*x)**2*a**4*b + 2*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a* 
*3*b*c - log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**2*b**3 - 2*log(tan(( 
c + d*x)/2) - 1)*sin(c + d*x)**2*a*b**3*c + 2*log(tan((c + d*x)/2) - 1)*si 
n(c + d*x)**2*b**5 + log(tan((c + d*x)/2) - 1)*a**4*b - 2*log(tan((c + d*x 
)/2) - 1)*a**3*b*c + log(tan((c + d*x)/2) - 1)*a**2*b**3 + 2*log(tan((c + 
d*x)/2) - 1)*a*b**3*c - 2*log(tan((c + d*x)/2) - 1)*b**5 + log(tan((c + d* 
x)/2) + 1)*sin(c + d*x)**2*a**4*b - 2*log(tan((c + d*x)/2) + 1)*sin(c + d* 
x)**2*a**3*b*c + log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**2*b**3 + 2*l 
og(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a*b**3*c - 2*log(tan((c + d*x)/2) 
 + 1)*sin(c + d*x)**2*b**5 - log(tan((c + d*x)/2) + 1)*a**4*b + 2*log(tan( 
(c + d*x)/2) + 1)*a**3*b*c - log(tan((c + d*x)/2) + 1)*a**2*b**3 - 2*log(t 
an((c + d*x)/2) + 1)*a*b**3*c + 2*log(tan((c + d*x)/2) + 1)*b**5 - sin(...