\(\int (a+b \cos (c+d x)) (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx\) [220]

Optimal result

Integrand size = 29, antiderivative size = 61 \[ \int (a+b \cos (c+d x)) (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {(a A+2 b B) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {(A b+a B) \tan (c+d x)}{d}+\frac {a A \sec (c+d x) \tan (c+d x)}{2 d} \] Output:

1/2*(A*a+2*B*b)*arctanh(sin(d*x+c))/d+(A*b+B*a)*tan(d*x+c)/d+1/2*a*A*sec(d 
*x+c)*tan(d*x+c)/d
 

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.23 \[ \int (a+b \cos (c+d x)) (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {b B \coth ^{-1}(\sin (c+d x))}{d}+\frac {a A \text {arctanh}(\sin (c+d x))}{2 d}+\frac {A b \tan (c+d x)}{d}+\frac {a B \tan (c+d x)}{d}+\frac {a A \sec (c+d x) \tan (c+d x)}{2 d} \] Input:

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

Output:

(b*B*ArcCoth[Sin[c + d*x]])/d + (a*A*ArcTanh[Sin[c + d*x]])/(2*d) + (A*b*T 
an[c + d*x])/d + (a*B*Tan[c + d*x])/d + (a*A*Sec[c + d*x]*Tan[c + d*x])/(2 
*d)
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.05, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.345, Rules used = {3042, 3447, 3042, 3500, 3042, 3227, 3042, 4254, 24, 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.

\(\displaystyle \int \sec ^3(c+d x) (a+b \cos (c+d x)) (A+B \cos (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3447

\(\displaystyle \int \sec ^3(c+d x) \left ((a B+A b) \cos (c+d x)+a A+b B \cos ^2(c+d x)\right )dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3500

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4254

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

\(\Big \downarrow \) 24

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

\(\Big \downarrow \) 4257

\(\displaystyle \frac {1}{2} \left (\frac {(a A+2 b B) \text {arctanh}(\sin (c+d x))}{d}+\frac {2 (a B+A b) \tan (c+d x)}{d}\right )+\frac {a A \tan (c+d x) \sec (c+d x)}{2 d}\)

Input:

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

Output:

(a*A*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (((a*A + 2*b*B)*ArcTanh[Sin[c + d* 
x]])/d + (2*(A*b + a*B)*Tan[c + d*x])/d)/2
 

Defintions of rubi rules used

rule 24

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3227

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[c   Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b   Int 
[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
 

rule 3447

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a 
 + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), 
 x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
 

rule 3500

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((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)/(b*f*(m + 1)* 
(a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x 
])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A 
*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, 
B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
 

rule 4254

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1)   Subst[Int[Exp 
andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, 
 d}, x] && IGtQ[n/2, 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 = 5.04 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.23

method	result	size
derivativedivides	\(\frac {A a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B a \tan \left (d x +c \right )+A b \tan \left (d x +c \right )+B b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(75\)
default	\(\frac {A a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B a \tan \left (d x +c \right )+A b \tan \left (d x +c \right )+B b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(75\)
parts	\(\frac {A a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {\left (A b +B a \right ) \tan \left (d x +c \right )}{d}+\frac {B b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(76\)
parallelrisch	\(\frac {-\left (\cos \left (2 d x +2 c \right )+1\right ) \left (A a +2 B b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (\cos \left (2 d x +2 c \right )+1\right ) \left (A a +2 B b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (2 A b +2 B a \right ) \sin \left (2 d x +2 c \right )+2 A a \sin \left (d x +c \right )}{2 d \left (\cos \left (2 d x +2 c \right )+1\right )}\)	\(114\)
risch	\(-\frac {i \left (A a \,{\mathrm e}^{3 i \left (d x +c \right )}-2 A b \,{\mathrm e}^{2 i \left (d x +c \right )}-2 B a \,{\mathrm e}^{2 i \left (d x +c \right )}-A a \,{\mathrm e}^{i \left (d x +c \right )}-2 A b -2 B a \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {A a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B b}{d}-\frac {A a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B b}{d}\)	\(160\)
norman	\(\frac {\frac {\left (A a -2 A b -2 B a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}+\frac {\left (A a +2 A b +2 B a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {\left (3 A a -2 A b -2 B a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}+\frac {\left (3 A a +2 A b +2 B a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2}}-\frac {\left (A a +2 B b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {\left (A a +2 B b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(192\)

Input:

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

Output:

1/d*(A*a*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+B*a*tan 
(d*x+c)+A*b*tan(d*x+c)+B*b*ln(sec(d*x+c)+tan(d*x+c)))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.57 \[ \int (a+b \cos (c+d x)) (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {{\left (A a + 2 \, B b\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (A a + 2 \, B b\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (A a + 2 \, {\left (B a + A b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \] Input:

integrate((a+b*cos(d*x+c))*(A+B*cos(d*x+c))*sec(d*x+c)^3,x, algorithm="fri 
cas")
 

Output:

1/4*((A*a + 2*B*b)*cos(d*x + c)^2*log(sin(d*x + c) + 1) - (A*a + 2*B*b)*co 
s(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(A*a + 2*(B*a + A*b)*cos(d*x + c)) 
*sin(d*x + c))/(d*cos(d*x + c)^2)
 

Sympy [F]

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

integrate((a+b*cos(d*x+c))*(A+B*cos(d*x+c))*sec(d*x+c)**3,x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.56 \[ \int (a+b \cos (c+d x)) (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=-\frac {A a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 2 \, B b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 4 \, B a \tan \left (d x + c\right ) - 4 \, A b \tan \left (d x + c\right )}{4 \, d} \] Input:

integrate((a+b*cos(d*x+c))*(A+B*cos(d*x+c))*sec(d*x+c)^3,x, algorithm="max 
ima")
 

Output:

-1/4*(A*a*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + l 
og(sin(d*x + c) - 1)) - 2*B*b*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 
1)) - 4*B*a*tan(d*x + c) - 4*A*b*tan(d*x + c))/d
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 151 vs. \(2 (57) = 114\).

Time = 0.19 (sec) , antiderivative size = 151, normalized size of antiderivative = 2.48 \[ \int (a+b \cos (c+d x)) (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {{\left (A a + 2 \, B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (A a + 2 \, B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \] Input:

integrate((a+b*cos(d*x+c))*(A+B*cos(d*x+c))*sec(d*x+c)^3,x, algorithm="gia 
c")
 

Output:

1/2*((A*a + 2*B*b)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (A*a + 2*B*b)*log( 
abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(A*a*tan(1/2*d*x + 1/2*c)^3 - 2*B*a*tan 
(1/2*d*x + 1/2*c)^3 - 2*A*b*tan(1/2*d*x + 1/2*c)^3 + A*a*tan(1/2*d*x + 1/2 
*c) + 2*B*a*tan(1/2*d*x + 1/2*c) + 2*A*b*tan(1/2*d*x + 1/2*c))/(tan(1/2*d* 
x + 1/2*c)^2 - 1)^2)/d
 

Mupad [B] (verification not implemented)

Time = 41.66 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.70 \[ \int (a+b \cos (c+d x)) (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A\,a+2\,A\,b+2\,B\,a\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,A\,b-A\,a+2\,B\,a\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (A\,a+2\,B\,b\right )}{d} \] Input:

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

Output:

(tan(c/2 + (d*x)/2)*(A*a + 2*A*b + 2*B*a) - tan(c/2 + (d*x)/2)^3*(2*A*b - 
A*a + 2*B*a))/(d*(tan(c/2 + (d*x)/2)^4 - 2*tan(c/2 + (d*x)/2)^2 + 1)) + (a 
tanh(tan(c/2 + (d*x)/2))*(A*a + 2*B*b))/d
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 279, normalized size of antiderivative = 4.57 \[ \int (a+b \cos (c+d x)) (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {-\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a^{2}-2 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} b^{2}+\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2}+2 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b^{2}+\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a^{2}+2 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} b^{2}-\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{2}-2 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b^{2}-\cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2}+4 \sin \left (d x +c \right )^{3} a b -4 \sin \left (d x +c \right ) a b}{2 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:

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

Output:

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