\(\int \cos ^4(c+d x) (a+b \sec (c+d x))^3 (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [790]

Optimal result

Integrand size = 40, antiderivative size = 145 \[ \int \cos ^4(c+d x) (a+b \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{2} \left (3 a^2 b B+2 b^3 B+a^3 C+6 a b^2 C\right ) x+\frac {b^3 C \text {arctanh}(\sin (c+d x))}{d}+\frac {a \left (2 a^2 B+8 b^2 B+9 a b C\right ) \sin (c+d x)}{3 d}+\frac {a^2 (5 b B+3 a C) \cos (c+d x) \sin (c+d x)}{6 d}+\frac {a B \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{3 d} \] Output:

1/2*(3*B*a^2*b+2*B*b^3+C*a^3+6*C*a*b^2)*x+b^3*C*arctanh(sin(d*x+c))/d+1/3* 
a*(2*B*a^2+8*B*b^2+9*C*a*b)*sin(d*x+c)/d+1/6*a^2*(5*B*b+3*C*a)*cos(d*x+c)* 
sin(d*x+c)/d+1/3*a*B*cos(d*x+c)^2*(a+b*sec(d*x+c))^2*sin(d*x+c)/d
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 1.25 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.10 \[ \int \cos ^4(c+d x) (a+b \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {6 \left (3 a^2 b B+2 b^3 B+a^3 C+6 a b^2 C\right ) (c+d x)-12 b^3 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+12 b^3 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+9 a \left (a^2 B+4 b^2 B+4 a b C\right ) \sin (c+d x)+3 a^2 (3 b B+a C) \sin (2 (c+d x))+a^3 B \sin (3 (c+d x))}{12 d} \] Input:

Integrate[Cos[c + d*x]^4*(a + b*Sec[c + d*x])^3*(B*Sec[c + d*x] + C*Sec[c 
+ d*x]^2),x]
 

Output:

(6*(3*a^2*b*B + 2*b^3*B + a^3*C + 6*a*b^2*C)*(c + d*x) - 12*b^3*C*Log[Cos[ 
(c + d*x)/2] - Sin[(c + d*x)/2]] + 12*b^3*C*Log[Cos[(c + d*x)/2] + Sin[(c 
+ d*x)/2]] + 9*a*(a^2*B + 4*b^2*B + 4*a*b*C)*Sin[c + d*x] + 3*a^2*(3*b*B + 
 a*C)*Sin[2*(c + d*x)] + a^3*B*Sin[3*(c + d*x)])/(12*d)
 

Rubi [A] (verified)

Time = 1.13 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.05, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {3042, 4560, 3042, 4513, 25, 3042, 4562, 25, 3042, 4535, 24, 3042, 4533, 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.

\(\displaystyle \int \cos ^4(c+d x) (a+b \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4560

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4513

\(\displaystyle \frac {a B \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}-\frac {1}{3} \int -\cos ^2(c+d x) (a+b \sec (c+d x)) \left (3 b^2 C \sec ^2(c+d x)+\left (2 B a^2+6 b C a+3 b^2 B\right ) \sec (c+d x)+a (5 b B+3 a C)\right )dx\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{3} \int \cos ^2(c+d x) (a+b \sec (c+d x)) \left (3 b^2 C \sec ^2(c+d x)+\left (2 B a^2+6 b C a+3 b^2 B\right ) \sec (c+d x)+a (5 b B+3 a C)\right )dx+\frac {a B \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \left (3 b^2 C \csc \left (c+d x+\frac {\pi }{2}\right )^2+\left (2 B a^2+6 b C a+3 b^2 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+a (5 b B+3 a C)\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a B \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 4562

\(\displaystyle \frac {1}{3} \left (\frac {a^2 (3 a C+5 b B) \sin (c+d x) \cos (c+d x)}{2 d}-\frac {1}{2} \int -\cos (c+d x) \left (6 C \sec ^2(c+d x) b^3+2 a \left (2 B a^2+9 b C a+8 b^2 B\right )+3 \left (C a^3+3 b B a^2+6 b^2 C a+2 b^3 B\right ) \sec (c+d x)\right )dx\right )+\frac {a B \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \int \cos (c+d x) \left (6 C \sec ^2(c+d x) b^3+2 a \left (2 B a^2+9 b C a+8 b^2 B\right )+3 \left (C a^3+3 b B a^2+6 b^2 C a+2 b^3 B\right ) \sec (c+d x)\right )dx+\frac {a^2 (3 a C+5 b B) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {a B \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \int \frac {6 C \csc \left (c+d x+\frac {\pi }{2}\right )^2 b^3+2 a \left (2 B a^2+9 b C a+8 b^2 B\right )+3 \left (C a^3+3 b B a^2+6 b^2 C a+2 b^3 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a^2 (3 a C+5 b B) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {a B \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 4535

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (\int \cos (c+d x) \left (6 C \sec ^2(c+d x) b^3+2 a \left (2 B a^2+9 b C a+8 b^2 B\right )\right )dx+3 \left (a^3 C+3 a^2 b B+6 a b^2 C+2 b^3 B\right ) \int 1dx\right )+\frac {a^2 (3 a C+5 b B) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {a B \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 24

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (\int \cos (c+d x) \left (6 C \sec ^2(c+d x) b^3+2 a \left (2 B a^2+9 b C a+8 b^2 B\right )\right )dx+3 x \left (a^3 C+3 a^2 b B+6 a b^2 C+2 b^3 B\right )\right )+\frac {a^2 (3 a C+5 b B) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {a B \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (\int \frac {6 C \csc \left (c+d x+\frac {\pi }{2}\right )^2 b^3+2 a \left (2 B a^2+9 b C a+8 b^2 B\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx+3 x \left (a^3 C+3 a^2 b B+6 a b^2 C+2 b^3 B\right )\right )+\frac {a^2 (3 a C+5 b B) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {a B \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 4533

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (6 b^3 C \int \sec (c+d x)dx+\frac {2 a \left (2 a^2 B+9 a b C+8 b^2 B\right ) \sin (c+d x)}{d}+3 x \left (a^3 C+3 a^2 b B+6 a b^2 C+2 b^3 B\right )\right )+\frac {a^2 (3 a C+5 b B) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {a B \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (6 b^3 C \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {2 a \left (2 a^2 B+9 a b C+8 b^2 B\right ) \sin (c+d x)}{d}+3 x \left (a^3 C+3 a^2 b B+6 a b^2 C+2 b^3 B\right )\right )+\frac {a^2 (3 a C+5 b B) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {a B \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {1}{3} \left (\frac {a^2 (3 a C+5 b B) \sin (c+d x) \cos (c+d x)}{2 d}+\frac {1}{2} \left (\frac {2 a \left (2 a^2 B+9 a b C+8 b^2 B\right ) \sin (c+d x)}{d}+3 x \left (a^3 C+3 a^2 b B+6 a b^2 C+2 b^3 B\right )+\frac {6 b^3 C \text {arctanh}(\sin (c+d x))}{d}\right )\right )+\frac {a B \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}\)

Input:

Int[Cos[c + d*x]^4*(a + b*Sec[c + d*x])^3*(B*Sec[c + d*x] + C*Sec[c + d*x] 
^2),x]
 

Output:

(a*B*Cos[c + d*x]^2*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(3*d) + ((a^2*(5* 
b*B + 3*a*C)*Cos[c + d*x]*Sin[c + d*x])/(2*d) + (3*(3*a^2*b*B + 2*b^3*B + 
a^3*C + 6*a*b^2*C)*x + (6*b^3*C*ArcTanh[Sin[c + d*x]])/d + (2*a*(2*a^2*B + 
 8*b^2*B + 9*a*b*C)*Sin[c + d*x])/d)/2)/3
 

Defintions of rubi rules used

rule 24

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4257

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

rule 4513

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[a*A*Cot 
[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*n)), x] + Sim 
p[1/(d*n)   Int[(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^(n + 1)*Simp[ 
a*(a*B*n - A*b*(m - n - 1)) + (2*a*b*B*n + A*(b^2*n + a^2*(1 + n)))*Csc[e + 
 f*x] + b*(b*B*n + a*A*(m + n))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, 
d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] & 
& LeQ[n, -1]
 

rule 4533

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) 
+ (A_)), x_Symbol] :> Simp[A*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*m)), x] + 
Simp[(C*m + A*(m + 1))/(b^2*m)   Int[(b*Csc[e + f*x])^(m + 2), x], x] /; Fr 
eeQ[{b, e, f, A, C}, 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 4560

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_. 
)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.) 
*(x_)]*(d_.))^(n_.), x_Symbol] :> Simp[1/b^2   Int[(a + b*Csc[e + f*x])^(m 
+ 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ 
[{a, b, c, d, e, f, A, B, C, m, n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
 

rule 4562

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

Maple [A] (verified)

Time = 0.92 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.94

method	result	size
parallelrisch	\(\frac {-12 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b^{3}+12 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b^{3}+\left (9 B \,a^{2} b +3 a^{3} C \right ) \sin \left (2 d x +2 c \right )+B \,a^{3} \sin \left (3 d x +3 c \right )+9 a \left (B \,a^{2}+4 B \,b^{2}+4 C a b \right ) \sin \left (d x +c \right )+18 \left (B \,a^{2} b +\frac {2}{3} B \,b^{3}+\frac {1}{3} a^{3} C +2 C a \,b^{2}\right ) x d}{12 d}\)	\(137\)
derivativedivides	\(\frac {\frac {B \,a^{3} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+a^{3} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 B \,a^{2} b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 C \,a^{2} b \sin \left (d x +c \right )+3 B a \,b^{2} \sin \left (d x +c \right )+3 C a \,b^{2} \left (d x +c \right )+B \,b^{3} \left (d x +c \right )+C \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(151\)
default	\(\frac {\frac {B \,a^{3} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+a^{3} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 B \,a^{2} b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 C \,a^{2} b \sin \left (d x +c \right )+3 B a \,b^{2} \sin \left (d x +c \right )+3 C a \,b^{2} \left (d x +c \right )+B \,b^{3} \left (d x +c \right )+C \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(151\)
risch	\(\frac {3 B \,a^{2} b x}{2}+x B \,b^{3}+\frac {a^{3} x C}{2}+3 x C a \,b^{2}-\frac {3 i B \,a^{3} {\mathrm e}^{i \left (d x +c \right )}}{8 d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} B a \,b^{2}}{2 d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} a^{2} b C}{2 d}+\frac {3 i B \,a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{8 d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} B a \,b^{2}}{2 d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} a^{2} b C}{2 d}+\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}-\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}+\frac {\sin \left (3 d x +3 c \right ) B \,a^{3}}{12 d}+\frac {3 \sin \left (2 d x +2 c \right ) B \,a^{2} b}{4 d}+\frac {\sin \left (2 d x +2 c \right ) a^{3} C}{4 d}\)	\(247\)
norman	\(\frac {\left (\frac {3}{2} B \,a^{2} b +B \,b^{3}+\frac {1}{2} a^{3} C +3 C a \,b^{2}\right ) x +\left (\frac {3}{2} B \,a^{2} b +B \,b^{3}+\frac {1}{2} a^{3} C +3 C a \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{16}+\left (-6 B \,a^{2} b -4 B \,b^{3}-2 a^{3} C -12 C a \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-6 B \,a^{2} b -4 B \,b^{3}-2 a^{3} C -12 C a \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+\left (9 B \,a^{2} b +6 B \,b^{3}+3 a^{3} C +18 C a \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\frac {a \left (2 B \,a^{2}-3 B a b -18 B \,b^{2}-C \,a^{2}-18 C a b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{d}+\frac {a \left (2 B \,a^{2}-3 B a b +6 B \,b^{2}-C \,a^{2}+6 C a b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{d}+\frac {a \left (2 B \,a^{2}+3 B a b -18 B \,b^{2}+C \,a^{2}-18 C a b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}+\frac {a \left (2 B \,a^{2}+3 B a b +6 B \,b^{2}+C \,a^{2}+6 C a b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {a \left (2 B \,a^{2}-45 B a b +54 B \,b^{2}-15 C \,a^{2}+54 C a b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{3 d}+\frac {a \left (2 B \,a^{2}+45 B a b +54 B \,b^{2}+15 C \,a^{2}+54 C a b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d}-\frac {a \left (14 B \,a^{2}-27 B a b +18 B \,b^{2}-9 C \,a^{2}+18 C a b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{3 d}-\frac {a \left (14 B \,a^{2}+27 B a b +18 B \,b^{2}+9 C \,a^{2}+18 C a b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4}}+\frac {C \,b^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {C \,b^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(622\)

Input:

int(cos(d*x+c)^4*(a+b*sec(d*x+c))^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x,method 
=_RETURNVERBOSE)
 

Output:

1/12*(-12*C*ln(tan(1/2*d*x+1/2*c)-1)*b^3+12*C*ln(tan(1/2*d*x+1/2*c)+1)*b^3 
+(9*B*a^2*b+3*C*a^3)*sin(2*d*x+2*c)+B*a^3*sin(3*d*x+3*c)+9*a*(B*a^2+4*B*b^ 
2+4*C*a*b)*sin(d*x+c)+18*(B*a^2*b+2/3*B*b^3+1/3*a^3*C+2*C*a*b^2)*x*d)/d
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.90 \[ \int \cos ^4(c+d x) (a+b \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, C b^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, C b^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (C a^{3} + 3 \, B a^{2} b + 6 \, C a b^{2} + 2 \, B b^{3}\right )} d x + {\left (2 \, B a^{3} \cos \left (d x + c\right )^{2} + 4 \, B a^{3} + 18 \, C a^{2} b + 18 \, B a b^{2} + 3 \, {\left (C a^{3} + 3 \, B a^{2} b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, d} \] Input:

integrate(cos(d*x+c)^4*(a+b*sec(d*x+c))^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, 
 algorithm="fricas")
 

Output:

1/6*(3*C*b^3*log(sin(d*x + c) + 1) - 3*C*b^3*log(-sin(d*x + c) + 1) + 3*(C 
*a^3 + 3*B*a^2*b + 6*C*a*b^2 + 2*B*b^3)*d*x + (2*B*a^3*cos(d*x + c)^2 + 4* 
B*a^3 + 18*C*a^2*b + 18*B*a*b^2 + 3*(C*a^3 + 3*B*a^2*b)*cos(d*x + c))*sin( 
d*x + c))/d
 

Sympy [F(-1)]

Timed out. \[ \int \cos ^4(c+d x) (a+b \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**4*(a+b*sec(d*x+c))**3*(B*sec(d*x+c)+C*sec(d*x+c)**2) 
,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.05 \[ \int \cos ^4(c+d x) (a+b \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{3} - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} - 9 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} b - 36 \, {\left (d x + c\right )} C a b^{2} - 12 \, {\left (d x + c\right )} B b^{3} - 6 \, C b^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 36 \, C a^{2} b \sin \left (d x + c\right ) - 36 \, B a b^{2} \sin \left (d x + c\right )}{12 \, d} \] Input:

integrate(cos(d*x+c)^4*(a+b*sec(d*x+c))^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, 
 algorithm="maxima")
 

Output:

-1/12*(4*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a^3 - 3*(2*d*x + 2*c + sin(2* 
d*x + 2*c))*C*a^3 - 9*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^2*b - 36*(d*x + 
 c)*C*a*b^2 - 12*(d*x + c)*B*b^3 - 6*C*b^3*(log(sin(d*x + c) + 1) - log(si 
n(d*x + c) - 1)) - 36*C*a^2*b*sin(d*x + c) - 36*B*a*b^2*sin(d*x + c))/d
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 314 vs. \(2 (137) = 274\).

Time = 0.32 (sec) , antiderivative size = 314, normalized size of antiderivative = 2.17 \[ \int \cos ^4(c+d x) (a+b \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {6 \, C b^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 6 \, C b^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + 3 \, {\left (C a^{3} + 3 \, B a^{2} b + 6 \, C a b^{2} + 2 \, B b^{3}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (6 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 18 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 18 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 36 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 36 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 18 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 18 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \] Input:

integrate(cos(d*x+c)^4*(a+b*sec(d*x+c))^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, 
 algorithm="giac")
 

Output:

1/6*(6*C*b^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 6*C*b^3*log(abs(tan(1/2* 
d*x + 1/2*c) - 1)) + 3*(C*a^3 + 3*B*a^2*b + 6*C*a*b^2 + 2*B*b^3)*(d*x + c) 
 + 2*(6*B*a^3*tan(1/2*d*x + 1/2*c)^5 - 3*C*a^3*tan(1/2*d*x + 1/2*c)^5 - 9* 
B*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 18*C*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 18*B* 
a*b^2*tan(1/2*d*x + 1/2*c)^5 + 4*B*a^3*tan(1/2*d*x + 1/2*c)^3 + 36*C*a^2*b 
*tan(1/2*d*x + 1/2*c)^3 + 36*B*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 6*B*a^3*tan( 
1/2*d*x + 1/2*c) + 3*C*a^3*tan(1/2*d*x + 1/2*c) + 9*B*a^2*b*tan(1/2*d*x + 
1/2*c) + 18*C*a^2*b*tan(1/2*d*x + 1/2*c) + 18*B*a*b^2*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 = 14.25 (sec) , antiderivative size = 1924, normalized size of antiderivative = 13.27 \[ \int \cos ^4(c+d x) (a+b \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \] Input:

int(cos(c + d*x)^4*(B/cos(c + d*x) + C/cos(c + d*x)^2)*(a + b/cos(c + d*x) 
)^3,x)
 

Output:

(tan(c/2 + (d*x)/2)^5*(2*B*a^3 - C*a^3 + 6*B*a*b^2 - 3*B*a^2*b + 6*C*a^2*b 
) + tan(c/2 + (d*x)/2)*(2*B*a^3 + C*a^3 + 6*B*a*b^2 + 3*B*a^2*b + 6*C*a^2* 
b) + tan(c/2 + (d*x)/2)^3*((4*B*a^3)/3 + 12*B*a*b^2 + 12*C*a^2*b))/(d*(3*t 
an(c/2 + (d*x)/2)^2 + 3*tan(c/2 + (d*x)/2)^4 + tan(c/2 + (d*x)/2)^6 + 1)) 
+ (atan((((B*b^3*1i + (C*a^3*1i)/2 + (B*a^2*b*3i)/2 + C*a*b^2*3i)*(32*B*b^ 
3 + 16*C*a^3 + 32*C*b^3 + 48*B*a^2*b + 96*C*a*b^2) + tan(c/2 + (d*x)/2)*(3 
2*B^2*b^6 + 8*C^2*a^6 + 32*C^2*b^6 + 96*B^2*a^2*b^4 + 72*B^2*a^4*b^2 + 288 
*C^2*a^2*b^4 + 96*C^2*a^4*b^2 + 192*B*C*a*b^5 + 48*B*C*a^5*b + 320*B*C*a^3 
*b^3))*(B*b^3*1i + (C*a^3*1i)/2 + (B*a^2*b*3i)/2 + C*a*b^2*3i)*1i - ((B*b^ 
3*1i + (C*a^3*1i)/2 + (B*a^2*b*3i)/2 + C*a*b^2*3i)*(32*B*b^3 + 16*C*a^3 + 
32*C*b^3 + 48*B*a^2*b + 96*C*a*b^2) - tan(c/2 + (d*x)/2)*(32*B^2*b^6 + 8*C 
^2*a^6 + 32*C^2*b^6 + 96*B^2*a^2*b^4 + 72*B^2*a^4*b^2 + 288*C^2*a^2*b^4 + 
96*C^2*a^4*b^2 + 192*B*C*a*b^5 + 48*B*C*a^5*b + 320*B*C*a^3*b^3))*(B*b^3*1 
i + (C*a^3*1i)/2 + (B*a^2*b*3i)/2 + C*a*b^2*3i)*1i)/(((B*b^3*1i + (C*a^3*1 
i)/2 + (B*a^2*b*3i)/2 + C*a*b^2*3i)*(32*B*b^3 + 16*C*a^3 + 32*C*b^3 + 48*B 
*a^2*b + 96*C*a*b^2) + tan(c/2 + (d*x)/2)*(32*B^2*b^6 + 8*C^2*a^6 + 32*C^2 
*b^6 + 96*B^2*a^2*b^4 + 72*B^2*a^4*b^2 + 288*C^2*a^2*b^4 + 96*C^2*a^4*b^2 
+ 192*B*C*a*b^5 + 48*B*C*a^5*b + 320*B*C*a^3*b^3))*(B*b^3*1i + (C*a^3*1i)/ 
2 + (B*a^2*b*3i)/2 + C*a*b^2*3i) + ((B*b^3*1i + (C*a^3*1i)/2 + (B*a^2*b*3i 
)/2 + C*a*b^2*3i)*(32*B*b^3 + 16*C*a^3 + 32*C*b^3 + 48*B*a^2*b + 96*C*a...
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.36 \[ \int \cos ^4(c+d x) (a+b \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{3} c +9 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2} b^{2}-6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b^{3} c +6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b^{3} c -2 \sin \left (d x +c \right )^{3} a^{3} b +6 \sin \left (d x +c \right ) a^{3} b +18 \sin \left (d x +c \right ) a^{2} b c +18 \sin \left (d x +c \right ) a \,b^{3}+3 a^{3} c^{2}+3 a^{3} c d x +9 a^{2} b^{2} c +9 a^{2} b^{2} d x +18 a \,b^{2} c^{2}+18 a \,b^{2} c d x +6 b^{4} c +6 b^{4} d x}{6 d} \] Input:

int(cos(d*x+c)^4*(a+b*sec(d*x+c))^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x)
 

Output:

(3*cos(c + d*x)*sin(c + d*x)*a**3*c + 9*cos(c + d*x)*sin(c + d*x)*a**2*b** 
2 - 6*log(tan((c + d*x)/2) - 1)*b**3*c + 6*log(tan((c + d*x)/2) + 1)*b**3* 
c - 2*sin(c + d*x)**3*a**3*b + 6*sin(c + d*x)*a**3*b + 18*sin(c + d*x)*a** 
2*b*c + 18*sin(c + d*x)*a*b**3 + 3*a**3*c**2 + 3*a**3*c*d*x + 9*a**2*b**2* 
c + 9*a**2*b**2*d*x + 18*a*b**2*c**2 + 18*a*b**2*c*d*x + 6*b**4*c + 6*b**4 
*d*x)/(6*d)