3.3.32 \(\int (a+a \cos (c+d x)) (B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^4(c+d x) \, dx\) [232]

3.3.32.1 Optimal result

Integrand size = 38, antiderivative size = 56 \[ \int (a+a \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {a (B+2 C) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a (B+C) \tan (c+d x)}{d}+\frac {a B \sec (c+d x) \tan (c+d x)}{2 d} \]

output

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

3.3.32.2 Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.34 \[ \int (a+a \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {a B \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a C \text {arctanh}(\sin (c+d x))}{d}+\frac {a B \tan (c+d x)}{d}+\frac {a C \tan (c+d x)}{d}+\frac {a B \sec (c+d x) \tan (c+d x)}{2 d} \]

input

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

output

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

3.3.32.3 Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.05, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {3042, 3508, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3508

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3447

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3500

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4254

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

\(\Big \downarrow \) 24

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

\(\Big \downarrow \) 4257

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

input

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

output

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

3.3.32.3.1 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 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 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]
 

3.3.32.4 Maple [A] (verified)

Time = 6.58 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.34

method	result	size
derivativedivides	\(\frac {a C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B a \tan \left (d x +c \right )+a C \tan \left (d x +c \right )+B a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(75\)
default	\(\frac {a C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B a \tan \left (d x +c \right )+a C \tan \left (d x +c \right )+B a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(75\)
parts	\(\frac {\left (B a +a C \right ) \tan \left (d x +c \right )}{d}+\frac {B a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {a C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(76\)
parallelrisch	\(-\frac {\left (\left (1+\cos \left (2 d x +2 c \right )\right ) \left (B +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\left (1+\cos \left (2 d x +2 c \right )\right ) \left (B +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (-2 B -2 C \right ) \sin \left (2 d x +2 c \right )-2 B \sin \left (d x +c \right )\right ) a}{2 d \left (1+\cos \left (2 d x +2 c \right )\right )}\)	\(106\)
risch	\(-\frac {i a \left (B \,{\mathrm e}^{3 i \left (d x +c \right )}-2 B \,{\mathrm e}^{2 i \left (d x +c \right )}-2 C \,{\mathrm e}^{2 i \left (d x +c \right )}-B \,{\mathrm e}^{i \left (d x +c \right )}-2 B -2 C \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {B a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}-\frac {B a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}\)	\(155\)
norman	\(\frac {\frac {a \left (B -2 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 a \left (B +2 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a \left (B +2 C \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a \left (3 B +2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 a \left (3 B +2 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a \left (5 B +2 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {a \left (B +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a \left (B +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(215\)

input

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

output

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

3.3.32.5 Fricas [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.59 \[ \int (a+a \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {{\left (B + 2 \, C\right )} a \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (B + 2 \, C\right )} a \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (B + C\right )} a \cos \left (d x + c\right ) + B a\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]

input

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

output

1/4*((B + 2*C)*a*cos(d*x + c)^2*log(sin(d*x + c) + 1) - (B + 2*C)*a*cos(d* 
x + c)^2*log(-sin(d*x + c) + 1) + 2*(2*(B + C)*a*cos(d*x + c) + B*a)*sin(d 
*x + c))/(d*cos(d*x + c)^2)
 

3.3.32.6 Sympy [F]

\[ \int (a+a \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=a \left (\int B \cos {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int B \cos ^{2}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int C \cos ^{2}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int C \cos ^{3}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx\right ) \]

input

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

output

a*(Integral(B*cos(c + d*x)*sec(c + d*x)**4, x) + Integral(B*cos(c + d*x)** 
2*sec(c + d*x)**4, x) + Integral(C*cos(c + d*x)**2*sec(c + d*x)**4, x) + I 
ntegral(C*cos(c + d*x)**3*sec(c + d*x)**4, x))
 

3.3.32.7 Maxima [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.70 \[ \int (a+a \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=-\frac {B a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 2 \, C a {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 4 \, B a \tan \left (d x + c\right ) - 4 \, C a \tan \left (d x + c\right )}{4 \, d} \]

input

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

output

-1/4*(B*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*C*a*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 
1)) - 4*B*a*tan(d*x + c) - 4*C*a*tan(d*x + c))/d
 

3.3.32.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 124 vs. \(2 (52) = 104\).

Time = 0.31 (sec) , antiderivative size = 124, normalized size of antiderivative = 2.21 \[ \int (a+a \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {{\left (B a + 2 \, C a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (B a + 2 \, C a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]

input

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

output

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

3.3.32.9 Mupad [B] (verification not implemented)

Time = 1.99 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.68 \[ \int (a+a \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,B\,a+2\,C\,a\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (B\,a+2\,C\,a\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (B+2\,C\right )}{d} \]

input

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

output

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