\(\int \sec ^2(c+d x) (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [38]

Optimal result

Integrand size = 28, antiderivative size = 63 \[ \int \sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {B \text {arctanh}(\sin (c+d x))}{2 d}+\frac {C \tan (c+d x)}{d}+\frac {B \sec (c+d x) \tan (c+d x)}{2 d}+\frac {C \tan ^3(c+d x)}{3 d} \] Output:

1/2*B*arctanh(sin(d*x+c))/d+C*tan(d*x+c)/d+1/2*B*sec(d*x+c)*tan(d*x+c)/d+1 
/3*C*tan(d*x+c)^3/d
 

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.95 \[ \int \sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {B \text {arctanh}(\sin (c+d x))}{2 d}+\frac {B \sec (c+d x) \tan (c+d x)}{2 d}+\frac {C \left (\tan (c+d x)+\frac {1}{3} \tan ^3(c+d x)\right )}{d} \] Input:

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

Output:

(B*ArcTanh[Sin[c + d*x]])/(2*d) + (B*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (C 
*(Tan[c + d*x] + Tan[c + d*x]^3/3))/d
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.02, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {3042, 4535, 27, 3042, 4254, 2009, 4255, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4535

\(\displaystyle B \int \sec ^3(c+d x)dx+\int C \sec ^4(c+d x)dx\)

\(\Big \downarrow \) 27

\(\displaystyle B \int \sec ^3(c+d x)dx+C \int \sec ^4(c+d x)dx\)

\(\Big \downarrow \) 3042

\(\displaystyle B \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+C \int \csc \left (c+d x+\frac {\pi }{2}\right )^4dx\)

\(\Big \downarrow \) 4254

\(\displaystyle B \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {C \int \left (\tan ^2(c+d x)+1\right )d(-\tan (c+d x))}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle B \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {C \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\)

\(\Big \downarrow \) 4255

\(\displaystyle B \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {C \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle B \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {C \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\)

\(\Big \downarrow \) 4257

\(\displaystyle B \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {C \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\)

Input:

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

Output:

B*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d)) - (C*( 
-Tan[c + d*x] - Tan[c + d*x]^3/3))/d
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

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 4255

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) 
  Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] 
&& IntegerQ[2*n]
 

rule 4257

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

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 = 0.25 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.95

method	result	size
derivativedivides	\(\frac {B \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-C \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(60\)
default	\(\frac {B \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-C \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(60\)
parts	\(\frac {B \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}-\frac {C \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(62\)
risch	\(-\frac {i \left (3 B \,{\mathrm e}^{5 i \left (d x +c \right )}-12 C \,{\mathrm e}^{2 i \left (d x +c \right )}-3 B \,{\mathrm e}^{i \left (d x +c \right )}-4 C \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {B \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}-\frac {B \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}\)	\(99\)
norman	\(\frac {\frac {\left (B -2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}+\frac {4 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}-\frac {\left (B +2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3}}-\frac {B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(111\)
parallelrisch	\(\frac {-9 \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+9 \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+6 B \sin \left (2 d x +2 c \right )+4 C \left (\sin \left (3 d x +3 c \right )+3 \sin \left (d x +c \right )\right )}{6 d \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\)	\(126\)

Input:

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

Output:

1/d*(B*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))-C*(-2/3-1 
/3*sec(d*x+c)^2)*tan(d*x+c))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.40 \[ \int \sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, B \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, B \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (4 \, C \cos \left (d x + c\right )^{2} + 3 \, B \cos \left (d x + c\right ) + 2 \, C\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \] Input:

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

Output:

1/12*(3*B*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*B*cos(d*x + c)^3*log(-s 
in(d*x + c) + 1) + 2*(4*C*cos(d*x + c)^2 + 3*B*cos(d*x + c) + 2*C)*sin(d*x 
 + c))/(d*cos(d*x + c)^3)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.11 \[ \int \sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C - 3 \, B {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{12 \, d} \] Input:

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

Output:

1/12*(4*(tan(d*x + c)^3 + 3*tan(d*x + c))*C - 3*B*(2*sin(d*x + c)/(sin(d*x 
 + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)))/d
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 122 vs. \(2 (57) = 114\).

Time = 0.35 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.94 \[ \int \sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, B \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, B \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (3 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \] Input:

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

Output:

1/6*(3*B*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*B*log(abs(tan(1/2*d*x + 1/ 
2*c) - 1)) + 2*(3*B*tan(1/2*d*x + 1/2*c)^5 - 6*C*tan(1/2*d*x + 1/2*c)^5 + 
4*C*tan(1/2*d*x + 1/2*c)^3 - 3*B*tan(1/2*d*x + 1/2*c) - 6*C*tan(1/2*d*x + 
1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d
 

Mupad [B] (verification not implemented)

Time = 13.85 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.73 \[ \int \sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {B\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {\left (B-2\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\frac {4\,C\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+\left (-B-2\,C\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \] Input:

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

Output:

(B*atanh(tan(c/2 + (d*x)/2)))/d + ((4*C*tan(c/2 + (d*x)/2)^3)/3 - tan(c/2 
+ (d*x)/2)*(B + 2*C) + tan(c/2 + (d*x)/2)^5*(B - 2*C))/(d*(3*tan(c/2 + (d* 
x)/2)^2 - 3*tan(c/2 + (d*x)/2)^4 + tan(c/2 + (d*x)/2)^6 - 1))
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 161, normalized size of antiderivative = 2.56 \[ \int \sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {-3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} b +3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b +3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} b -3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b -3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b +4 \sin \left (d x +c \right )^{3} c -6 \sin \left (d x +c \right ) c}{6 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:

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

Output:

( - 3*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b + 3*cos(c + 
 d*x)*log(tan((c + d*x)/2) - 1)*b + 3*cos(c + d*x)*log(tan((c + d*x)/2) + 
1)*sin(c + d*x)**2*b - 3*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*b - 3*cos( 
c + d*x)*sin(c + d*x)*b + 4*sin(c + d*x)**3*c - 6*sin(c + d*x)*c)/(6*cos(c 
 + d*x)*d*(sin(c + d*x)**2 - 1))