\(\int \sec (c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [57]

Optimal result

Integrand size = 27, antiderivative size = 51 \[ \int \sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {(2 A+C) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {B \tan (c+d x)}{d}+\frac {C \sec (c+d x) \tan (c+d x)}{2 d} \] Output:

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.16 \[ \int \sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {A \coth ^{-1}(\sin (c+d x))}{d}+\frac {C \text {arctanh}(\sin (c+d x))}{2 d}+\frac {B \tan (c+d x)}{d}+\frac {C \sec (c+d x) \tan (c+d x)}{2 d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {3042, 4535, 3042, 4254, 24, 4534, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4535

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4254

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

\(\Big \downarrow \) 24

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

\(\Big \downarrow \) 4534

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4257

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

Input:

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

Output:

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

rule 4534

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

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.24

method	result	size
derivativedivides	\(\frac {A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \tan \left (d x +c \right )+C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(63\)
default	\(\frac {A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \tan \left (d x +c \right )+C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(63\)
parts	\(\frac {A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {B \tan \left (d x +c \right )}{d}+\frac {C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(68\)
parallelrisch	\(\frac {-\left (1+\cos \left (2 d x +2 c \right )\right ) \left (A +\frac {C}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (1+\cos \left (2 d x +2 c \right )\right ) \left (A +\frac {C}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+B \sin \left (2 d x +2 c \right )+C \sin \left (d x +c \right )}{d \left (1+\cos \left (2 d x +2 c \right )\right )}\)	\(97\)
norman	\(\frac {\frac {\left (2 B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {\left (2 B -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2}}-\frac {\left (2 A +C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {\left (2 A +C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(104\)
risch	\(-\frac {i \left (C \,{\mathrm e}^{3 i \left (d x +c \right )}-2 B \,{\mathrm e}^{2 i \left (d x +c \right )}-C \,{\mathrm e}^{i \left (d x +c \right )}-2 B \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 d}\)	\(135\)

Input:

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

Output:

1/d*(A*ln(sec(d*x+c)+tan(d*x+c))+B*tan(d*x+c)+C*(1/2*sec(d*x+c)*tan(d*x+c) 
+1/2*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 \[ \int \sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {{\left (2 \, A + C\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (2 \, A + C\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, B \cos \left (d x + c\right ) + C\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.47 \[ \int \sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {C {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 4 \, A \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) - 4 \, B \tan \left (d x + c\right )}{4 \, d} \] Input:

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

Output:

-1/4*(C*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log 
(sin(d*x + c) - 1)) - 4*A*log(sec(d*x + c) + tan(d*x + c)) - 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. 115 vs. \(2 (47) = 94\).

Time = 0.31 (sec) , antiderivative size = 115, normalized size of antiderivative = 2.25 \[ \int \sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {{\left (2 \, A + C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (2 \, A + C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \] Input:

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

Output:

1/2*((2*A + C)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (2*A + C)*log(abs(tan( 
1/2*d*x + 1/2*c) - 1)) - 2*(2*B*tan(1/2*d*x + 1/2*c)^3 - C*tan(1/2*d*x + 1 
/2*c)^3 - 2*B*tan(1/2*d*x + 1/2*c) - C*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 = 12.30 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.75 \[ \int \sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (2\,A+C\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,B-C\right )-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,B+C\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 192, normalized size of antiderivative = 3.76 \[ \int \sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {-2 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b -2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} c +2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) c +2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} c -2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) c -\sin \left (d x +c \right ) c}{2 d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:

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

Output:

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