\(\int \cos (c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [412]

Optimal result

Integrand size = 37, antiderivative size = 46 \[ \int \cos (c+d x) (a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=a (A+B) x+\frac {a (B+C) \text {arctanh}(\sin (c+d x))}{d}+\frac {a A \sin (c+d x)}{d}+\frac {a C \tan (c+d x)}{d} \] Output:

a*(A+B)*x+a*(B+C)*arctanh(sin(d*x+c))/d+a*A*sin(d*x+c)/d+a*C*tan(d*x+c)/d
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.54 \[ \int \cos (c+d x) (a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=a A x+a B x+\frac {a B \coth ^{-1}(\sin (c+d x))}{d}+\frac {a C \coth ^{-1}(\sin (c+d x))}{d}+\frac {a A \cos (d x) \sin (c)}{d}+\frac {a A \cos (c) \sin (d x)}{d}+\frac {a C \tan (c+d x)}{d} \] Input:

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

Output:

a*A*x + a*B*x + (a*B*ArcCoth[Sin[c + d*x]])/d + (a*C*ArcCoth[Sin[c + d*x]] 
)/d + (a*A*Cos[d*x]*Sin[c])/d + (a*A*Cos[c]*Sin[d*x])/d + (a*C*Tan[c + d*x 
])/d
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.243, Rules used = {3042, 4564, 3042, 4535, 24, 3042, 4533, 3042, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4564

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4535

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

\(\Big \downarrow \) 24

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4533

\(\displaystyle a (B+C) \int \sec (c+d x)dx+a x (A+B)+\frac {a A \sin (c+d x)}{d}+\frac {a C \tan (c+d x)}{d}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4257

\(\displaystyle a x (A+B)+\frac {a A \sin (c+d x)}{d}+\frac {a (B+C) \text {arctanh}(\sin (c+d x))}{d}+\frac {a C \tan (c+d x)}{d}\)

Input:

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

Output:

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

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 4257

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

rule 4533

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) 
+ (A_)), x_Symbol] :> Simp[A*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*m)), x] + 
Simp[(C*m + A*(m + 1))/(b^2*m)   Int[(b*Csc[e + f*x])^(m + 2), x], x] /; Fr 
eeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]
 

rule 4535

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]* 
(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.)), x_Symbol] :> Simp[B/b   Int[(b*Cs 
c[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x]^2) 
, x] /; FreeQ[{b, e, f, A, B, C, m}, x]
 

rule 4564

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. 
))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_)), x_Symbol] :> Simp[(-b)*C*Csc[e + f*x]*Cot[e + f*x]*((d*Csc[e + f*x])^ 
n/(f*(n + 2))), x] + Simp[1/(n + 2)   Int[(d*Csc[e + f*x])^n*Simp[A*a*(n + 
2) + (B*a*(n + 2) + b*(C*(n + 1) + A*(n + 2)))*Csc[e + f*x] + (a*C + B*b)*( 
n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] 
&&  !LtQ[n, -1]
 

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.61

method	result	size
derivativedivides	\(\frac {a A \left (d x +c \right )+a B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C a \tan \left (d x +c \right )+a A \sin \left (d x +c \right )+a B \left (d x +c \right )+C a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(74\)
default	\(\frac {a A \left (d x +c \right )+a B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C a \tan \left (d x +c \right )+a A \sin \left (d x +c \right )+a B \left (d x +c \right )+C a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(74\)
parallelrisch	\(\frac {a \left (-2 \cos \left (d x +c \right ) \left (C +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+2 \cos \left (d x +c \right ) \left (C +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+A \sin \left (2 d x +2 c \right )+2 d x \left (A +B \right ) \cos \left (d x +c \right )+2 C \sin \left (d x +c \right )\right )}{2 \cos \left (d x +c \right ) d}\)	\(95\)
risch	\(a A x +a B x -\frac {i a A \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {i a A \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {2 i C a}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{d}\)	\(143\)
norman	\(\frac {\left (a A +a B \right ) x +\left (-a A -a B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (-a A -a B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (a A +a B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-\frac {4 a A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}+\frac {2 a \left (A -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}+\frac {2 a \left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2}}+\frac {a \left (C +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a \left (C +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(206\)

Input:

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

Output:

1/d*(a*A*(d*x+c)+a*B*ln(sec(d*x+c)+tan(d*x+c))+C*a*tan(d*x+c)+a*A*sin(d*x+ 
c)+a*B*(d*x+c)+C*a*ln(sec(d*x+c)+tan(d*x+c)))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 92, normalized size of antiderivative = 2.00 \[ \int \cos (c+d x) (a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (A + B\right )} a d x \cos \left (d x + c\right ) + {\left (B + C\right )} a \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (B + C\right )} a \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (A a \cos \left (d x + c\right ) + C a\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \] Input:

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

Output:

1/2*(2*(A + B)*a*d*x*cos(d*x + c) + (B + C)*a*cos(d*x + c)*log(sin(d*x + c 
) + 1) - (B + C)*a*cos(d*x + c)*log(-sin(d*x + c) + 1) + 2*(A*a*cos(d*x + 
c) + C*a)*sin(d*x + c))/(d*cos(d*x + c))
 

Sympy [F]

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

integrate(cos(d*x+c)*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)
 

Output:

a*(Integral(A*cos(c + d*x), x) + Integral(A*cos(c + d*x)*sec(c + d*x), x) 
+ Integral(B*cos(c + d*x)*sec(c + d*x), x) + Integral(B*cos(c + d*x)*sec(c 
 + d*x)**2, x) + Integral(C*cos(c + d*x)*sec(c + d*x)**2, x) + Integral(C* 
cos(c + d*x)*sec(c + d*x)**3, x))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 92, normalized size of antiderivative = 2.00 \[ \int \cos (c+d x) (a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (d x + c\right )} A a + 2 \, {\left (d x + c\right )} B a + B a {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + C a {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, A a \sin \left (d x + c\right ) + 2 \, C a \tan \left (d x + c\right )}{2 \, d} \] Input:

integrate(cos(d*x+c)*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, a 
lgorithm="maxima")
 

Output:

1/2*(2*(d*x + c)*A*a + 2*(d*x + c)*B*a + B*a*(log(sin(d*x + c) + 1) - log( 
sin(d*x + c) - 1)) + C*a*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 
 2*A*a*sin(d*x + c) + 2*C*a*tan(d*x + c))/d
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 134 vs. \(2 (46) = 92\).

Time = 0.28 (sec) , antiderivative size = 134, normalized size of antiderivative = 2.91 \[ \int \cos (c+d x) (a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {{\left (A a + B a\right )} {\left (d x + c\right )} + {\left (B a + C a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (B a + C a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 1}}{d} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 12.84 (sec) , antiderivative size = 159, normalized size of antiderivative = 3.46 \[ \int \cos (c+d x) (a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {C\,a\,\mathrm {tan}\left (c+d\,x\right )}{d}+\frac {2\,A\,a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,B\,a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {A\,a\,\sin \left (2\,c+2\,d\,x\right )}{2\,d\,\cos \left (c+d\,x\right )}-\frac {B\,a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}}{d}-\frac {C\,a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}}{d} \] Input:

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

Output:

(C*a*tan(c + d*x))/d + (2*A*a*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))) 
/d + (2*B*a*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d - (B*a*atan((si 
n(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*2i)/d - (C*a*atan((sin(c/2 + (d*x 
)/2)*1i)/cos(c/2 + (d*x)/2))*2i)/d + (A*a*sin(2*c + 2*d*x))/(2*d*cos(c + d 
*x))
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 156, normalized size of antiderivative = 3.39 \[ \int \cos (c+d x) (a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a \left (-\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b -\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) c +\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b +\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) c +\cos \left (d x +c \right ) \sin \left (d x +c \right ) a +\cos \left (d x +c \right ) a c +\cos \left (d x +c \right ) a d x +\cos \left (d x +c \right ) b c +\cos \left (d x +c \right ) b d x +\sin \left (d x +c \right ) c \right )}{\cos \left (d x +c \right ) d} \] Input:

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

Output:

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