3.7.39 \(\int \sec (c+d x) (a+b \sec (c+d x)) (A+C \sec ^2(c+d x)) \, dx\) [639]

3.7.39.1 Optimal result

Integrand size = 29, antiderivative size = 86 \[ \int \sec (c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a (2 A+C) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {b (3 A+2 C) \tan (c+d x)}{3 d}+\frac {a C \sec (c+d x) \tan (c+d x)}{2 d}+\frac {b C \sec ^2(c+d x) \tan (c+d x)}{3 d} \]

output

1/2*a*(2*A+C)*arctanh(sin(d*x+c))/d+1/3*b*(3*A+2*C)*tan(d*x+c)/d+1/2*a*C*s 
ec(d*x+c)*tan(d*x+c)/d+1/3*b*C*sec(d*x+c)^2*tan(d*x+c)/d
 

3.7.39.2 Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.69 \[ \int \sec (c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 a (2 A+C) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (6 b (A+C)+3 a C \sec (c+d x)+2 b C \tan ^2(c+d x)\right )}{6 d} \]

input

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

output

(3*a*(2*A + C)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(6*b*(A + C) + 3*a*C*S 
ec[c + d*x] + 2*b*C*Tan[c + d*x]^2))/(6*d)
 

3.7.39.3 Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.02, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.345, Rules used = {3042, 4565, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4565

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4535

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4254

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

\(\Big \downarrow \) 24

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

\(\Big \downarrow \) 4534

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4257

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

input

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

output

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

3.7.39.3.1 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 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]
 

rule 4565

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_. 
))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(-b)*C*C 
sc[e + f*x]*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*(n + 2))), x] + Simp[1/(n + 
 2)   Int[(d*Csc[e + f*x])^n*Simp[A*a*(n + 2) + b*(C*(n + 1) + A*(n + 2))*C 
sc[e + f*x] + a*C*(n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, 
f, A, C, n}, x] &&  !LtQ[n, -1]
 

3.7.39.4 Maple [A] (verified)

Time = 0.73 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.02

method	result	size
derivativedivides	\(\frac {a A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+A \tan \left (d x +c \right ) b -C b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(88\)
default	\(\frac {a A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+A \tan \left (d x +c \right ) b -C b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(88\)
parts	\(\frac {A b \tan \left (d x +c \right )}{d}+\frac {C a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}-\frac {C b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {a A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(96\)
norman	\(\frac {-\frac {\left (2 A b -C a +2 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-\frac {\left (2 A b +C a +2 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {4 b \left (3 A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3}}-\frac {a \left (2 A +C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a \left (2 A +C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(142\)
parallelrisch	\(\frac {-3 a \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \left (A +\frac {C}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+3 a \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \left (A +\frac {C}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+b \left (A +\frac {2 C}{3}\right ) \sin \left (3 d x +3 c \right )+C a \sin \left (2 d x +2 c \right )+\sin \left (d x +c \right ) b \left (A +2 C \right )}{d \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\)	\(143\)
risch	\(-\frac {i \left (3 C a \,{\mathrm e}^{5 i \left (d x +c \right )}-6 A b \,{\mathrm e}^{4 i \left (d x +c \right )}-12 A b \,{\mathrm e}^{2 i \left (d x +c \right )}-12 C b \,{\mathrm e}^{2 i \left (d x +c \right )}-3 C a \,{\mathrm e}^{i \left (d x +c \right )}-6 A b -4 C b \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 d}\)	\(174\)

input

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

output

1/d*(a*A*ln(sec(d*x+c)+tan(d*x+c))+C*a*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(s 
ec(d*x+c)+tan(d*x+c)))+A*tan(d*x+c)*b-C*b*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+ 
c))
 

3.7.39.5 Fricas [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.24 \[ \int \sec (c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (2 \, A + C\right )} a \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (2 \, A + C\right )} a \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (3 \, A + 2 \, C\right )} b \cos \left (d x + c\right )^{2} + 3 \, C a \cos \left (d x + c\right ) + 2 \, C b\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \]

input

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

output

1/12*(3*(2*A + C)*a*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(2*A + C)*a*c 
os(d*x + c)^3*log(-sin(d*x + c) + 1) + 2*(2*(3*A + 2*C)*b*cos(d*x + c)^2 + 
 3*C*a*cos(d*x + c) + 2*C*b)*sin(d*x + c))/(d*cos(d*x + c)^3)
 

3.7.39.6 Sympy [F]

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

input

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

output

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

3.7.39.7 Maxima [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.16 \[ \int \sec (c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C b - 3 \, C a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, A a \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 12 \, A b \tan \left (d x + c\right )}{12 \, d} \]

input

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

output

1/12*(4*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*b - 3*C*a*(2*sin(d*x + c)/(sin 
(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 12*A*a 
*log(sec(d*x + c) + tan(d*x + c)) + 12*A*b*tan(d*x + c))/d
 

3.7.39.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 184 vs. \(2 (78) = 156\).

Time = 0.33 (sec) , antiderivative size = 184, normalized size of antiderivative = 2.14 \[ \int \sec (c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (2 \, A a + C a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (2 \, A a + C a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (3 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \]

input

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

output

1/6*(3*(2*A*a + C*a)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(2*A*a + C*a)* 
log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(3*C*a*tan(1/2*d*x + 1/2*c)^5 - 6*A 
*b*tan(1/2*d*x + 1/2*c)^5 - 6*C*b*tan(1/2*d*x + 1/2*c)^5 + 12*A*b*tan(1/2* 
d*x + 1/2*c)^3 + 4*C*b*tan(1/2*d*x + 1/2*c)^3 - 3*C*a*tan(1/2*d*x + 1/2*c) 
 - 6*A*b*tan(1/2*d*x + 1/2*c) - 6*C*b*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 
 1/2*c)^2 - 1)^3)/d
 

3.7.39.9 Mupad [B] (verification not implemented)

Time = 18.12 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.59 \[ \int \sec (c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (2\,A+C\right )}{d}-\frac {\left (2\,A\,b-C\,a+2\,C\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-4\,A\,b-\frac {4\,C\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A\,b+C\,a+2\,C\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

input

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

output

(a*atanh(tan(c/2 + (d*x)/2))*(2*A + C))/d - (tan(c/2 + (d*x)/2)*(2*A*b + C 
*a + 2*C*b) - tan(c/2 + (d*x)/2)^3*(4*A*b + (4*C*b)/3) + tan(c/2 + (d*x)/2 
)^5*(2*A*b - C*a + 2*C*b))/(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))