3.1.4 \(\int (a+a \cos (c+d x)) (A+C \cos ^2(c+d x)) \sec (c+d x) \, dx\) [4]

3.1.4.1 Optimal result

Integrand size = 29, antiderivative size = 58 \[ \int (a+a \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {1}{2} a (2 A+C) x+\frac {a A \text {arctanh}(\sin (c+d x))}{d}+\frac {a C \sin (c+d x)}{d}+\frac {a C \cos (c+d x) \sin (c+d x)}{2 d} \]

output

1/2*a*(2*A+C)*x+a*A*arctanh(sin(d*x+c))/d+a*C*sin(d*x+c)/d+1/2*a*C*cos(d*x 
+c)*sin(d*x+c)/d
 

3.1.4.2 Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.90 \[ \int (a+a \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {a (2 c C+4 A d x+2 C d x+4 A \text {arctanh}(\sin (c+d x))+4 C \sin (c+d x)+C \sin (2 (c+d x)))}{4 d} \]

input

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

output

(a*(2*c*C + 4*A*d*x + 2*C*d*x + 4*A*ArcTanh[Sin[c + d*x]] + 4*C*Sin[c + d* 
x] + C*Sin[2*(c + d*x)]))/(4*d)
 

3.1.4.3 Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.07, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {3042, 3513, 3042, 3502, 3042, 3214, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3513

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

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

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \left (\int \frac {2 a A+a (2 A+C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 a C \sin (c+d x)}{d}\right )+\frac {a C \sin (c+d x) \cos (c+d x)}{2 d}\)

\(\Big \downarrow \) 3214

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

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \left (2 a A \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+a x (2 A+C)+\frac {2 a C \sin (c+d x)}{d}\right )+\frac {a C \sin (c+d x) \cos (c+d x)}{2 d}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {1}{2} \left (\frac {2 a A \text {arctanh}(\sin (c+d x))}{d}+a x (2 A+C)+\frac {2 a C \sin (c+d x)}{d}\right )+\frac {a C \sin (c+d x) \cos (c+d x)}{2 d}\)

input

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

output

(a*C*Cos[c + d*x]*Sin[c + d*x])/(2*d) + (a*(2*A + C)*x + (2*a*A*ArcTanh[Si 
n[c + d*x]])/d + (2*a*C*Sin[c + d*x])/d)/2
 

3.1.4.3.1 Defintions of rubi rules used

rule 3042

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

rule 3214

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. 
)*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/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 3502

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co 
s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m 
+ 2))   Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m 
 + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] 
 &&  !LtQ[m, -1]
 

rule 3513

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[ 
(-C)*d*Cos[e + f*x]*Sin[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 3) 
)), x] + Simp[1/(b*(m + 3))   Int[(a + b*Sin[e + f*x])^m*Simp[a*C*d + A*b*c 
*(m + 3) + b*d*(C*(m + 2) + A*(m + 3))*Sin[e + f*x] - (2*a*C*d - b*c*C*(m + 
 3))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, m}, x] && 
 NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]
 

rule 4257

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

3.1.4.4 Maple [A] (verified)

Time = 2.97 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.12

method	result	size
derivativedivides	\(\frac {a A \left (d x +c \right )+a C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a C \sin \left (d x +c \right )}{d}\)	\(65\)
default	\(\frac {a A \left (d x +c \right )+a C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a C \sin \left (d x +c \right )}{d}\)	\(65\)
parallelrisch	\(-\frac {\left (A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {\sin \left (2 d x +2 c \right ) C}{4}-\sin \left (d x +c \right ) C -d x \left (A +\frac {C}{2}\right )\right ) a}{d}\)	\(67\)
parts	\(\frac {a A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {a A \left (d x +c \right )}{d}+\frac {a C \sin \left (d x +c \right )}{d}+\frac {a C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(73\)
risch	\(a x A +\frac {a C x}{2}-\frac {i a C \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {i a C \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {a A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {a A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {\sin \left (2 d x +2 c \right ) a C}{4 d}\)	\(100\)
norman	\(\frac {\left (a A +\frac {1}{2} a C \right ) x +\left (a A +\frac {1}{2} a C \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (3 a A +\frac {3}{2} a C \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (3 a A +\frac {3}{2} a C \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {a C \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {3 a C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {4 a C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {a A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(182\)

input

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

output

1/d*(a*A*(d*x+c)+a*C*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+a*A*ln(sec( 
d*x+c)+tan(d*x+c))+a*C*sin(d*x+c))
 

3.1.4.5 Fricas [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.09 \[ \int (a+a \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {{\left (2 \, A + C\right )} a d x + A a \log \left (\sin \left (d x + c\right ) + 1\right ) - A a \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (C a \cos \left (d x + c\right ) + 2 \, C a\right )} \sin \left (d x + c\right )}{2 \, d} \]

input

integrate((a+a*cos(d*x+c))*(A+C*cos(d*x+c)^2)*sec(d*x+c),x, algorithm="fri 
cas")
 

output

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

3.1.4.6 Sympy [F]

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

input

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

output

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

3.1.4.7 Maxima [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.09 \[ \int (a+a \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {4 \, {\left (d x + c\right )} A a + {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a + 4 \, A a \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 4 \, C a \sin \left (d x + c\right )}{4 \, d} \]

input

integrate((a+a*cos(d*x+c))*(A+C*cos(d*x+c)^2)*sec(d*x+c),x, algorithm="max 
ima")
 

output

1/4*(4*(d*x + c)*A*a + (2*d*x + 2*c + sin(2*d*x + 2*c))*C*a + 4*A*a*log(se 
c(d*x + c) + tan(d*x + c)) + 4*C*a*sin(d*x + c))/d
 

3.1.4.8 Giac [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.71 \[ \int (a+a \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {2 \, A a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 2 \, A a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + {\left (2 \, A a + C a\right )} {\left (d x + c\right )} + \frac {2 \, {\left (C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \]

input

integrate((a+a*cos(d*x+c))*(A+C*cos(d*x+c)^2)*sec(d*x+c),x, algorithm="gia 
c")
 

output

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

3.1.4.9 Mupad [B] (verification not implemented)

Time = 1.31 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.98 \[ \int (a+a \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {C\,a\,\sin \left (c+d\,x\right )}{d}+\frac {2\,A\,a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,A\,a\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {C\,a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {C\,a\,\sin \left (2\,c+2\,d\,x\right )}{4\,d} \]

input

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

output

(C*a*sin(c + d*x))/d + (2*A*a*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))) 
/d + (2*A*a*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (C*a*atan(si 
n(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (C*a*sin(2*c + 2*d*x))/(4*d)