\(\int (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^3(c+d x) \, dx\) [297]

Optimal result

Integrand size = 29, antiderivative size = 51 \[ \int \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {(A+2 C) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {B \tan (c+d x)}{d}+\frac {A \sec (c+d x) \tan (c+d x)}{2 d} \] Output:

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.16 \[ \int \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {C \coth ^{-1}(\sin (c+d x))}{d}+\frac {A \text {arctanh}(\sin (c+d x))}{2 d}+\frac {B \tan (c+d x)}{d}+\frac {A \sec (c+d x) \tan (c+d x)}{2 d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.06, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3500

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4254

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

\(\Big \downarrow \) 24

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

\(\Big \downarrow \) 4257

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

Input:

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

Output:

(A*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (((A + 2*C)*ArcTanh[Sin[c + d*x]])/d 
 + (2*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 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 = 0.56 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.24

method	result	size
derivativedivides	\(\frac {A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B \tan \left (d x +c \right )+C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(63\)
default	\(\frac {A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B \tan \left (d x +c \right )+C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(63\)
parts	\(\frac {A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {B \tan \left (d x +c \right )}{d}+\frac {C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(68\)
parallelrisch	\(\frac {-\left (\cos \left (2 d x +2 c \right )+1\right ) \left (A +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (\cos \left (2 d x +2 c \right )+1\right ) \left (A +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+2 A \sin \left (d x +c \right )+2 B \sin \left (2 d x +2 c \right )}{2 d \left (\cos \left (2 d x +2 c \right )+1\right )}\)	\(100\)
risch	\(-\frac {i \left (A \,{\mathrm e}^{3 i \left (d x +c \right )}-2 B \,{\mathrm e}^{2 i \left (d x +c \right )}-A \,{\mathrm e}^{i \left (d x +c \right )}-2 B \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}-\frac {C \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}+\frac {C \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}\)	\(135\)
norman	\(\frac {\frac {\left (A -2 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}+\frac {\left (A +2 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {\left (3 A -2 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}+\frac {\left (3 A +2 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2}}-\frac {\left (A +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {\left (A +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(160\)

Input:

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.61 \[ \int \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {{\left (A + 2 \, C\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (A + 2 \, C\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, B \cos \left (d x + c\right ) + A\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \] Input:

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3,x, algorithm="frica 
s")
 

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.61 \[ \int \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=-\frac {A {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 2 \, C {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 4 \, B \tan \left (d x + c\right )}{4 \, d} \] Input:

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3,x, algorithm="maxim 
a")
 

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 113 vs. \(2 (47) = 94\).

Time = 0.16 (sec) , antiderivative size = 113, normalized size of antiderivative = 2.22 \[ \int \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {{\left (A + 2 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (A + 2 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \] Input:

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3,x, algorithm="giac" 
)
 

Output:

1/2*((A + 2*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (A + 2*C)*log(abs(tan( 
1/2*d*x + 1/2*c) - 1)) + 2*(A*tan(1/2*d*x + 1/2*c)^3 - 2*B*tan(1/2*d*x + 1 
/2*c)^3 + A*tan(1/2*d*x + 1/2*c) + 2*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 = 0.60 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.67 \[ \int \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {\left (A-2\,B\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (A+2\,B\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (A+2\,C\right )}{d} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 259, normalized size of antiderivative = 5.08 \[ \int \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {-\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a -2 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} c +\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a +2 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) c +\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a +2 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} c -\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a -2 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) c -\cos \left (d x +c \right ) \sin \left (d x +c \right ) a +2 \sin \left (d x +c \right )^{3} b -2 \sin \left (d x +c \right ) b}{2 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*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*cos(c + d 
*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*c + cos(c + d*x)*log(tan((c 
+ d*x)/2) - 1)*a + 2*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*c + cos(c + d* 
x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a + 2*cos(c + d*x)*log(tan((c 
 + d*x)/2) + 1)*sin(c + d*x)**2*c - cos(c + d*x)*log(tan((c + d*x)/2) + 1) 
*a - 2*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*c - cos(c + d*x)*sin(c + d*x 
)*a + 2*sin(c + d*x)**3*b - 2*sin(c + d*x)*b)/(2*cos(c + d*x)*d*(sin(c + d 
*x)**2 - 1))