\(\int \cos (c+d x) (a+b \sec (c+d x))^3 (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [881]

Optimal result

Integrand size = 39, antiderivative size = 192 \[ \int \cos (c+d x) (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=a^2 (3 A b+a B) x+\frac {\left (6 a^2 b B+b^3 B+2 a^3 C+3 a b^2 (2 A+C)\right ) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {A (a+b \sec (c+d x))^3 \sin (c+d x)}{d}+\frac {b \left (9 a b B-a^2 (6 A-8 C)+b^2 (3 A+2 C)\right ) \tan (c+d x)}{3 d}-\frac {b^2 (6 a A-3 b B-5 a C) \sec (c+d x) \tan (c+d x)}{6 d}-\frac {b (3 A-C) (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d} \] Output:

a^2*(3*A*b+B*a)*x+1/2*(6*B*a^2*b+B*b^3+2*a^3*C+3*a*b^2*(2*A+C))*arctanh(si 
n(d*x+c))/d+A*(a+b*sec(d*x+c))^3*sin(d*x+c)/d+1/3*b*(9*B*a*b-a^2*(6*A-8*C) 
+b^2*(3*A+2*C))*tan(d*x+c)/d-1/6*b^2*(6*A*a-3*B*b-5*C*a)*sec(d*x+c)*tan(d* 
x+c)/d-1/3*b*(3*A-C)*(a+b*sec(d*x+c))^2*tan(d*x+c)/d
                                                                                    
                                                                                    
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(509\) vs. \(2(192)=384\).

Time = 8.57 (sec) , antiderivative size = 509, normalized size of antiderivative = 2.65 \[ \int \cos (c+d x) (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\cos ^5(c+d x) (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (12 a^2 (3 A b+a B) (c+d x)-6 \left (6 a^2 b B+b^3 B+2 a^3 C+3 a b^2 (2 A+C)\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+6 \left (6 a^2 b B+b^3 B+2 a^3 C+3 a b^2 (2 A+C)\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {b^2 (9 a C+b (3 B+C))}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {2 b^3 C \sin \left (\frac {1}{2} (c+d x)\right )}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {4 b \left (3 A b^2+9 a b B+9 a^2 C+2 b^2 C\right ) \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {2 b^3 C \sin \left (\frac {1}{2} (c+d x)\right )}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}-\frac {b^2 (9 a C+b (3 B+C))}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 b \left (3 A b^2+9 a b B+9 a^2 C+2 b^2 C\right ) \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )}+12 a^3 A \sin (c+d x)\right )}{6 d (b+a \cos (c+d x))^3 (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x)))} \] Input:

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

Output:

(Cos[c + d*x]^5*(a + b*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x 
]^2)*(12*a^2*(3*A*b + a*B)*(c + d*x) - 6*(6*a^2*b*B + b^3*B + 2*a^3*C + 3* 
a*b^2*(2*A + C))*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 6*(6*a^2*b*B + 
 b^3*B + 2*a^3*C + 3*a*b^2*(2*A + C))*Log[Cos[(c + d*x)/2] + Sin[(c + d*x) 
/2]] + (b^2*(9*a*C + b*(3*B + C)))/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 
 + (2*b^3*C*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3 + (4 
*b*(3*A*b^2 + 9*a*b*B + 9*a^2*C + 2*b^2*C)*Sin[(c + d*x)/2])/(Cos[(c + d*x 
)/2] - Sin[(c + d*x)/2]) + (2*b^3*C*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + 
Sin[(c + d*x)/2])^3 - (b^2*(9*a*C + b*(3*B + C)))/(Cos[(c + d*x)/2] + Sin[ 
(c + d*x)/2])^2 + (4*b*(3*A*b^2 + 9*a*b*B + 9*a^2*C + 2*b^2*C)*Sin[(c + d* 
x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + 12*a^3*A*Sin[c + d*x]))/(6* 
d*(b + a*Cos[c + d*x])^3*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)]) 
)
 

Rubi [A] (verified)

Time = 0.79 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3042, 4582, 3042, 4544, 3042, 4536, 2009}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4582

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4544

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4536

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(A*(a + b*Sec[c + d*x])^3*Sin[c + d*x])/d - (b*(3*A - C)*(a + b*Sec[c + d* 
x])^2*Tan[c + d*x])/(3*d) + (-1/2*(b^2*(6*a*A - 3*b*B - 5*a*C)*Sec[c + d*x 
]*Tan[c + d*x])/d + (6*a^2*(3*A*b + a*B)*x + (3*(6*a^2*b*B + b^3*B + 2*a^3 
*C + 3*a*b^2*(2*A + C))*ArcTanh[Sin[c + d*x]])/d + (2*b*(9*a*b*B - a^2*(6* 
A - 8*C) + b^2*(3*A + 2*C))*Tan[c + d*x])/d)/2)/3
 

Defintions of rubi rules used

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 4536

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

rule 4544

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

rule 4582

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 
_))^(m_), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e 
 + f*x])^n/(f*n)), x] - Simp[1/(d*n)   Int[(a + b*Csc[e + f*x])^(m - 1)*(d* 
Csc[e + f*x])^(n + 1)*Simp[A*b*m - a*B*n - (b*B*n + a*(C*n + A*(n + 1)))*Cs 
c[e + f*x] - b*(C*n + A*(m + n + 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a 
, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && LeQ[n, -1]
 

Maple [A] (verified)

Time = 1.82 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.22

method	result	size
derivativedivides	\(\frac {a^{3} A \sin \left (d x +c \right )+B \,a^{3} \left (d x +c \right )+a^{3} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 A \,a^{2} b \left (d x +c \right )+3 B \,a^{2} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 C \,a^{2} b \tan \left (d x +c \right )+3 a A \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 B \tan \left (d x +c \right ) a \,b^{2}+3 C a \,b^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+A \,b^{3} \tan \left (d x +c \right )+B \,b^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-C \,b^{3} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(235\)
default	\(\frac {a^{3} A \sin \left (d x +c \right )+B \,a^{3} \left (d x +c \right )+a^{3} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 A \,a^{2} b \left (d x +c \right )+3 B \,a^{2} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 C \,a^{2} b \tan \left (d x +c \right )+3 a A \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 B \tan \left (d x +c \right ) a \,b^{2}+3 C a \,b^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+A \,b^{3} \tan \left (d x +c \right )+B \,b^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-C \,b^{3} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(235\)
parallelrisch	\(\frac {-6 \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right ) \left (\frac {B \,b^{3}}{6}+a \left (A +\frac {C}{2}\right ) b^{2}+B \,a^{2} b +\frac {a^{3} C}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+6 \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right ) \left (\frac {B \,b^{3}}{6}+a \left (A +\frac {C}{2}\right ) b^{2}+B \,a^{2} b +\frac {a^{3} C}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+6 x \left (A b +\frac {B a}{3}\right ) a^{2} d \cos \left (3 d x +3 c \right )+2 b \left (\left (A +\frac {2 C}{3}\right ) b^{2}+3 B a b +3 C \,a^{2}\right ) \sin \left (3 d x +3 c \right )+2 \left (a^{3} A +B \,b^{3}+3 C a \,b^{2}\right ) \sin \left (2 d x +2 c \right )+a^{3} A \sin \left (4 d x +4 c \right )+18 x \left (A b +\frac {B a}{3}\right ) a^{2} d \cos \left (d x +c \right )+2 b \sin \left (d x +c \right ) \left (b^{2} \left (A +2 C \right )+3 B a b +3 C \,a^{2}\right )}{2 d \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\)	\(300\)
risch	\(3 a^{2} A b x +a^{3} B x -\frac {i a^{3} A \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {i a^{3} A \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}-\frac {i b \left (3 B \,b^{2} {\mathrm e}^{5 i \left (d x +c \right )}+9 C a b \,{\mathrm e}^{5 i \left (d x +c \right )}-6 A \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-18 B a b \,{\mathrm e}^{4 i \left (d x +c \right )}-18 C \,a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-12 A \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-36 B a b \,{\mathrm e}^{2 i \left (d x +c \right )}-36 C \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-12 C \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-3 B \,b^{2} {\mathrm e}^{i \left (d x +c \right )}-9 C a b \,{\mathrm e}^{i \left (d x +c \right )}-6 A \,b^{2}-18 B a b -18 C \,a^{2}-4 C \,b^{2}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a A \,b^{2}}{d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B \,a^{2} b}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B \,b^{3}}{2 d}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}-\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C \,b^{2}}{2 d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a A \,b^{2}}{d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B \,a^{2} b}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B \,b^{3}}{2 d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}+\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C \,b^{2}}{2 d}\)	\(484\)
norman	\(\frac {\left (3 A \,a^{2} b +B \,a^{3}\right ) x +\left (-9 A \,a^{2} b -3 B \,a^{3}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (-9 A \,a^{2} b -3 B \,a^{3}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (3 A \,a^{2} b +B \,a^{3}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (6 A \,a^{2} b +2 B \,a^{3}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (6 A \,a^{2} b +2 B \,a^{3}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\frac {\left (2 a^{3} A -2 A \,b^{3}-6 B a \,b^{2}+B \,b^{3}-6 a^{2} b C +3 C a \,b^{2}-2 C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}+\frac {\left (2 a^{3} A +2 A \,b^{3}+6 B a \,b^{2}+B \,b^{3}+6 a^{2} b C +3 C a \,b^{2}+2 C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 \left (6 a^{3} A -B \,b^{3}-3 C a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-\frac {4 \left (6 a^{3} A -3 A \,b^{3}-9 B a \,b^{2}-9 a^{2} b C -C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d}-\frac {4 \left (6 a^{3} A +3 A \,b^{3}+9 B a \,b^{2}+9 a^{2} b C +C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4}}-\frac {\left (6 a A \,b^{2}+6 B \,a^{2} b +B \,b^{3}+2 a^{3} C +3 C a \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {\left (6 a A \,b^{2}+6 B \,a^{2} b +B \,b^{3}+2 a^{3} C +3 C a \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(534\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.17 \[ \int \cos (c+d x) (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {12 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} d x \cos \left (d x + c\right )^{3} + 3 \, {\left (2 \, C a^{3} + 6 \, B a^{2} b + 3 \, {\left (2 \, A + C\right )} a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (2 \, C a^{3} + 6 \, B a^{2} b + 3 \, {\left (2 \, A + C\right )} a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (6 \, A a^{3} \cos \left (d x + c\right )^{3} + 2 \, C b^{3} + 2 \, {\left (9 \, C a^{2} b + 9 \, B a b^{2} + {\left (3 \, A + 2 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.46 \[ \int \cos (c+d x) (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {12 \, {\left (d x + c\right )} B a^{3} + 36 \, {\left (d x + c\right )} A a^{2} b + 4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C b^{3} - 9 \, C a b^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 3 \, B b^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, C a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 18 \, B a^{2} b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 18 \, A a b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, A a^{3} \sin \left (d x + c\right ) + 36 \, C a^{2} b \tan \left (d x + c\right ) + 36 \, B a b^{2} \tan \left (d x + c\right ) + 12 \, A b^{3} \tan \left (d x + c\right )}{12 \, d} \] Input:

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

Output:

1/12*(12*(d*x + c)*B*a^3 + 36*(d*x + c)*A*a^2*b + 4*(tan(d*x + c)^3 + 3*ta 
n(d*x + c))*C*b^3 - 9*C*a*b^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(s 
in(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 3*B*b^3*(2*sin(d*x + c)/(sin(d 
*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 6*C*a^3* 
(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 18*B*a^2*b*(log(sin(d*x 
+ c) + 1) - log(sin(d*x + c) - 1)) + 18*A*a*b^2*(log(sin(d*x + c) + 1) - l 
og(sin(d*x + c) - 1)) + 12*A*a^3*sin(d*x + c) + 36*C*a^2*b*tan(d*x + c) + 
36*B*a*b^2*tan(d*x + c) + 12*A*b^3*tan(d*x + c))/d
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 438 vs. \(2 (185) = 370\).

Time = 0.32 (sec) , antiderivative size = 438, normalized size of antiderivative = 2.28 \[ \int \cos (c+d x) (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\frac {12 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} + 6 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} {\left (d x + c\right )} + 3 \, {\left (2 \, C a^{3} + 6 \, B a^{2} b + 6 \, A a b^{2} + 3 \, C a b^{2} + B b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (2 \, C a^{3} + 6 \, B a^{2} b + 6 \, A a b^{2} + 3 \, C a b^{2} + B b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (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} - 9 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 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} - 12 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 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 ) + 9 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C b^{3} \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)*(a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, 
 algorithm="giac")
 

Output:

1/6*(12*A*a^3*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 + 1) + 6*(B*a^3 
 + 3*A*a^2*b)*(d*x + c) + 3*(2*C*a^3 + 6*B*a^2*b + 6*A*a*b^2 + 3*C*a*b^2 + 
 B*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(2*C*a^3 + 6*B*a^2*b + 6*A* 
a*b^2 + 3*C*a*b^2 + B*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(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 - 9*C*a*b^2 
*tan(1/2*d*x + 1/2*c)^5 + 6*A*b^3*tan(1/2*d*x + 1/2*c)^5 - 3*B*b^3*tan(1/2 
*d*x + 1/2*c)^5 + 6*C*b^3*tan(1/2*d*x + 1/2*c)^5 - 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 - 12*A*b^3*tan(1/2*d*x + 1/ 
2*c)^3 - 4*C*b^3*tan(1/2*d*x + 1/2*c)^3 + 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) + 9*C*a*b^2*tan(1/2*d*x + 1/2*c) + 6*A*b 
^3*tan(1/2*d*x + 1/2*c) + 3*B*b^3*tan(1/2*d*x + 1/2*c) + 6*C*b^3*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 = 16.50 (sec) , antiderivative size = 2437, normalized size of antiderivative = 12.69 \[ \int \cos (c+d x) (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \] Input:

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

Output:

(2*a^2*atan((a^2*(tan(c/2 + (d*x)/2)*(32*B^2*a^6 + 8*B^2*b^6 + 32*C^2*a^6 
+ 288*A^2*a^2*b^4 + 288*A^2*a^4*b^2 + 96*B^2*a^2*b^4 + 288*B^2*a^4*b^2 + 7 
2*C^2*a^2*b^4 + 96*C^2*a^4*b^2 + 96*A*B*a*b^5 + 192*A*B*a^5*b + 48*B*C*a*b 
^5 + 192*B*C*a^5*b + 576*A*B*a^3*b^3 + 288*A*C*a^2*b^4 + 192*A*C*a^4*b^2 + 
 320*B*C*a^3*b^3) - a^2*(3*A*b + B*a)*(32*B*a^3 + 16*B*b^3 + 32*C*a^3 + 96 
*A*a*b^2 + 96*A*a^2*b + 96*B*a^2*b + 48*C*a*b^2)*1i)*(3*A*b + B*a) + a^2*( 
tan(c/2 + (d*x)/2)*(32*B^2*a^6 + 8*B^2*b^6 + 32*C^2*a^6 + 288*A^2*a^2*b^4 
+ 288*A^2*a^4*b^2 + 96*B^2*a^2*b^4 + 288*B^2*a^4*b^2 + 72*C^2*a^2*b^4 + 96 
*C^2*a^4*b^2 + 96*A*B*a*b^5 + 192*A*B*a^5*b + 48*B*C*a*b^5 + 192*B*C*a^5*b 
 + 576*A*B*a^3*b^3 + 288*A*C*a^2*b^4 + 192*A*C*a^4*b^2 + 320*B*C*a^3*b^3) 
+ a^2*(3*A*b + B*a)*(32*B*a^3 + 16*B*b^3 + 32*C*a^3 + 96*A*a*b^2 + 96*A*a^ 
2*b + 96*B*a^2*b + 48*C*a*b^2)*1i)*(3*A*b + B*a))/(64*B*C^2*a^9 - 64*B^2*C 
*a^9 - 192*B^3*a^8*b + 1728*A^3*a^4*b^5 - 1728*A^3*a^5*b^4 + 16*B^3*a^3*b^ 
6 + 192*B^3*a^5*b^4 - 32*B^3*a^6*b^3 + 576*B^3*a^7*b^2 + a^2*(tan(c/2 + (d 
*x)/2)*(32*B^2*a^6 + 8*B^2*b^6 + 32*C^2*a^6 + 288*A^2*a^2*b^4 + 288*A^2*a^ 
4*b^2 + 96*B^2*a^2*b^4 + 288*B^2*a^4*b^2 + 72*C^2*a^2*b^4 + 96*C^2*a^4*b^2 
 + 96*A*B*a*b^5 + 192*A*B*a^5*b + 48*B*C*a*b^5 + 192*B*C*a^5*b + 576*A*B*a 
^3*b^3 + 288*A*C*a^2*b^4 + 192*A*C*a^4*b^2 + 320*B*C*a^3*b^3) - a^2*(3*A*b 
 + B*a)*(32*B*a^3 + 16*B*b^3 + 32*C*a^3 + 96*A*a*b^2 + 96*A*a^2*b + 96*B*a 
^2*b + 48*C*a*b^2)*1i)*(3*A*b + B*a)*1i - a^2*(tan(c/2 + (d*x)/2)*(32*B...
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 704, normalized size of antiderivative = 3.67 \[ \int \cos (c+d x) (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx =\text {Too large to display} \] Input:

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

Output:

( - 6*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**3*c - 36*c 
os(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**2*b**2 - 9*cos(c 
+ d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a*b**2*c - 3*cos(c + d*x) 
*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b**4 + 6*cos(c + d*x)*log(tan(( 
c + d*x)/2) - 1)*a**3*c + 36*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a**2*b 
**2 + 9*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a*b**2*c + 3*cos(c + d*x)*l 
og(tan((c + d*x)/2) - 1)*b**4 + 6*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*s 
in(c + d*x)**2*a**3*c + 36*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + 
d*x)**2*a**2*b**2 + 9*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)* 
*2*a*b**2*c + 3*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b** 
4 - 6*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a**3*c - 36*cos(c + d*x)*log( 
tan((c + d*x)/2) + 1)*a**2*b**2 - 9*cos(c + d*x)*log(tan((c + d*x)/2) + 1) 
*a*b**2*c - 3*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*b**4 + 6*cos(c + d*x) 
*sin(c + d*x)**3*a**4 + 24*cos(c + d*x)*sin(c + d*x)**2*a**3*b*c + 24*cos( 
c + d*x)*sin(c + d*x)**2*a**3*b*d*x - 6*cos(c + d*x)*sin(c + d*x)*a**4 - 9 
*cos(c + d*x)*sin(c + d*x)*a*b**2*c - 3*cos(c + d*x)*sin(c + d*x)*b**4 - 2 
4*cos(c + d*x)*a**3*b*c - 24*cos(c + d*x)*a**3*b*d*x + 18*sin(c + d*x)**3* 
a**2*b*c + 24*sin(c + d*x)**3*a*b**3 + 4*sin(c + d*x)**3*b**3*c - 18*sin(c 
 + d*x)*a**2*b*c - 24*sin(c + d*x)*a*b**3 - 6*sin(c + d*x)*b**3*c)/(6*cos( 
c + d*x)*d*(sin(c + d*x)**2 - 1))