\(\int \frac {(B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx\) [799]

Optimal result

Integrand size = 40, antiderivative size = 99 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {2 b (b B-a C) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 \sqrt {a-b} \sqrt {a+b} d}-\frac {(b B-a C) \text {arctanh}(\sin (c+d x))}{a^2 d}+\frac {B \tan (c+d x)}{a d} \] Output:

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

Mathematica [A] (verified)

Time = 1.12 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.30 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {-\frac {2 b (b B-a C) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}+(b B-a C) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+a B \tan (c+d x)}{a^2 d} \] Input:

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

Output:

((-2*b*(b*B - a*C)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/S 
qrt[-a^2 + b^2] + (b*B - a*C)*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 
Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + a*B*Tan[c + d*x])/(a^2*d)
 

Rubi [A] (verified)

Time = 0.67 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.97, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3042, 3508, 3042, 3479, 25, 27, 3042, 3226, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3508

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3226

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3138

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 4257

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

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

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 3226

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)])), x_Symbol] :> Simp[b/(b*c - a*d)   Int[1/(a + b*Sin[e + f*x]), 
 x], x] - Simp[d/(b*c - a*d)   Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[ 
{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 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 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 4257

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

Maple [A] (verified)

Time = 0.47 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.45

method	result	size
derivativedivides	\(\frac {-\frac {B}{a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (B b -C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{2}}+\frac {2 b \left (B b -C a \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {B}{a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (-B b +C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{2}}}{d}\)	\(144\)
default	\(\frac {-\frac {B}{a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (B b -C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{2}}+\frac {2 b \left (B b -C a \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {B}{a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (-B b +C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{2}}}{d}\)	\(144\)
risch	\(\frac {2 i B}{d a \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B b}{d \,a^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d a}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B b}{d \,a^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d a}-\frac {b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) B}{\sqrt {-a^{2}+b^{2}}\, d \,a^{2}}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, d a}+\frac {b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) B}{\sqrt {-a^{2}+b^{2}}\, d \,a^{2}}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, d a}\)	\(411\)

Input:

int((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+b*cos(d*x+c)),x,method=_ 
RETURNVERBOSE)
 

Output:

1/d*(-B/a/(tan(1/2*d*x+1/2*c)-1)+(B*b-C*a)/a^2*ln(tan(1/2*d*x+1/2*c)-1)+2* 
b*(B*b-C*a)/a^2/((a-b)*(a+b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a-b) 
*(a+b))^(1/2))-B/a/(tan(1/2*d*x+1/2*c)+1)+1/a^2*(-B*b+C*a)*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. 196 vs. \(2 (89) = 178\).

Time = 0.19 (sec) , antiderivative size = 460, normalized size of antiderivative = 4.65 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx=\left [\frac {{\left (C a b - B b^{2}\right )} \sqrt {-a^{2} + b^{2}} \cos \left (d x + c\right ) \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) + {\left (C a^{3} - B a^{2} b - C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (C a^{3} - B a^{2} b - C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (B a^{3} - B a b^{2}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{4} - a^{2} b^{2}\right )} d \cos \left (d x + c\right )}, -\frac {2 \, {\left (C a b - B b^{2}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right ) - {\left (C a^{3} - B a^{2} b - C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (C a^{3} - B a^{2} b - C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (B a^{3} - B a b^{2}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{4} - a^{2} b^{2}\right )} d \cos \left (d x + c\right )}\right ] \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(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 [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.77 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {\frac {{\left (C a - B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac {{\left (C a - B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} - \frac {2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a} + \frac {2 \, {\left (C a b - B b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} a^{2}}}{d} \] Input:

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

Output:

((C*a - B*b)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 - (C*a - B*b)*log(abs( 
tan(1/2*d*x + 1/2*c) - 1))/a^2 - 2*B*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 
1/2*c)^2 - 1)*a) + 2*(C*a*b - B*b^2)*(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))/s 
qrt(a^2 - b^2)))/(sqrt(a^2 - b^2)*a^2))/d
 

Mupad [B] (verification not implemented)

Time = 1.62 (sec) , antiderivative size = 675, normalized size of antiderivative = 6.82 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx =\text {Too large to display} \] Input:

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

Output:

(B*b*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*2i)/(d*(a^2 - b^2)) 
- (C*a*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*2i)/(d*(a^2 - b^2) 
) + (B*a*tan(c + d*x))/(d*(a^2 - b^2)) - (B*b^3*atan((sin(c/2 + (d*x)/2)*1 
i)/cos(c/2 + (d*x)/2))*2i)/(a^2*d*(a^2 - b^2)) + (C*b^2*atan((sin(c/2 + (d 
*x)/2)*1i)/cos(c/2 + (d*x)/2))*2i)/(a*d*(a^2 - b^2)) - (B*b^2*tan(c + d*x) 
)/(a*d*(a^2 - b^2)) - (C*b*atan(((a^5*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) 
 + 2*b^3*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(3/2) - 2*b^5*sin(c/2 + (d*x)/2)*( 
b^2 - a^2)^(1/2) + 3*a^2*b^3*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) - a^3*b^ 
2*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) - a^4*b*sin(c/2 + (d*x)/2)*(b^2 - a 
^2)^(1/2))*1i)/(cos(c/2 + (d*x)/2)*(a*b^2 - a^3)^2))*(-(a + b)*(a - b))^(1 
/2)*2i)/(a*d*(a^2 - b^2)) + (B*b^2*atan(((a^5*sin(c/2 + (d*x)/2)*(b^2 - a^ 
2)^(1/2) + 2*b^3*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(3/2) - 2*b^5*sin(c/2 + (d 
*x)/2)*(b^2 - a^2)^(1/2) + 3*a^2*b^3*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) 
- a^3*b^2*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) - a^4*b*sin(c/2 + (d*x)/2)* 
(b^2 - a^2)^(1/2))*1i)/(cos(c/2 + (d*x)/2)*(a*b^2 - a^3)^2))*(-(a + b)*(a 
- b))^(1/2)*2i)/(a^2*d*(a^2 - b^2))
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 360, normalized size of antiderivative = 3.64 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {-2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) \cos \left (d x +c \right ) a b c +2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) \cos \left (d x +c \right ) b^{3}-\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{3} c +\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2} b^{2}+\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a \,b^{2} c -\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b^{4}+\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{3} c -\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{2} b^{2}-\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a \,b^{2} c +\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b^{4}+\sin \left (d x +c \right ) a^{3} b -\sin \left (d x +c \right ) a \,b^{3}}{\cos \left (d x +c \right ) a^{2} d \left (a^{2}-b^{2}\right )} \] Input:

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

Output:

( - 2*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)*a*b*c + 2*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**3 - cos(c 
 + d*x)*log(tan((c + d*x)/2) - 1)*a**3*c + cos(c + d*x)*log(tan((c + d*x)/ 
2) - 1)*a**2*b**2 + cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a*b**2*c - cos( 
c + d*x)*log(tan((c + d*x)/2) - 1)*b**4 + cos(c + d*x)*log(tan((c + d*x)/2 
) + 1)*a**3*c - cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a**2*b**2 - cos(c + 
 d*x)*log(tan((c + d*x)/2) + 1)*a*b**2*c + cos(c + d*x)*log(tan((c + d*x)/ 
2) + 1)*b**4 + sin(c + d*x)*a**3*b - sin(c + d*x)*a*b**3)/(cos(c + d*x)*a* 
*2*d*(a**2 - b**2))