\(\int \cos ^2(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [874]

Optimal result

Integrand size = 41, antiderivative size = 118 \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{2} \left (2 A b^2+4 a b B+a^2 (A+2 C)\right ) x+\frac {b (b B+2 a C) \text {arctanh}(\sin (c+d x))}{d}+\frac {a (A b+a B) \sin (c+d x)}{d}+\frac {A \cos (c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}-\frac {b^2 (A-2 C) \tan (c+d x)}{2 d} \]

[Out]

1/2*(2*A*b^2+4*B*a*b+a^2*(A+2*C))*x+b*(B*b+2*C*a)*arctanh(sin(d*x+c))/d+a*(A*b+B*a)*sin(d*x+c)/d+1/2*A*cos(d*x
+c)*(a+b*sec(d*x+c))^2*sin(d*x+c)/d-1/2*b^2*(A-2*C)*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.146, Rules used = {4179, 4161, 4132, 8, 4130, 3855} \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{2} x \left (a^2 (A+2 C)+4 a b B+2 A b^2\right )+\frac {a (a B+A b) \sin (c+d x)}{d}+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^2}{2 d}+\frac {b (2 a C+b B) \text {arctanh}(\sin (c+d x))}{d}-\frac {b^2 (A-2 C) \tan (c+d x)}{2 d} \]

[In]

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

[Out]

((2*A*b^2 + 4*a*b*B + a^2*(A + 2*C))*x)/2 + (b*(b*B + 2*a*C)*ArcTanh[Sin[c + d*x]])/d + (a*(A*b + a*B)*Sin[c +
d*x])/d + (A*Cos[c + d*x]*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(2*d) - (b^2*(A - 2*C)*Tan[c + d*x])/(2*d)

Rule 8

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

Rule 3855

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

Rule 4130

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] + Dist[(C*m + A*(m + 1))/(b^2*m), Int[(b*Csc[e + f*x])^(m + 2), x], x] /
; FreeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]

Rule 4132

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[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 4161

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[(-b)*C*Csc[e + f*x]*Cot[e + f*x]*((d*Csc[e + f
*x])^n/(f*(n + 2))), x] + Dist[1/(n + 2), Int[(d*Csc[e + f*x])^n*Simp[A*a*(n + 2) + (B*a*(n + 2) + b*(C*(n + 1
) + A*(n + 2)))*Csc[e + f*x] + (a*C + B*b)*(n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C
, n}, x] &&  !LtQ[n, -1]

Rule 4179

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] - Dist[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)))*Csc[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]

Rubi steps \begin{align*} \text {integral}& = \frac {A \cos (c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac {1}{2} \int \cos (c+d x) (a+b \sec (c+d x)) \left (2 (A b+a B)+(2 b B+a (A+2 C)) \sec (c+d x)-b (A-2 C) \sec ^2(c+d x)\right ) \, dx \\ & = \frac {A \cos (c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}-\frac {b^2 (A-2 C) \tan (c+d x)}{2 d}+\frac {1}{2} \int \cos (c+d x) \left (2 a (A b+a B)+\left (2 A b^2+4 a b B+a^2 (A+2 C)\right ) \sec (c+d x)+2 b (b B+2 a C) \sec ^2(c+d x)\right ) \, dx \\ & = \frac {A \cos (c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}-\frac {b^2 (A-2 C) \tan (c+d x)}{2 d}+\frac {1}{2} \int \cos (c+d x) \left (2 a (A b+a B)+2 b (b B+2 a C) \sec ^2(c+d x)\right ) \, dx+\frac {1}{2} \left (2 A b^2+4 a b B+a^2 (A+2 C)\right ) \int 1 \, dx \\ & = \frac {1}{2} \left (2 A b^2+4 a b B+a^2 (A+2 C)\right ) x+\frac {a (A b+a B) \sin (c+d x)}{d}+\frac {A \cos (c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}-\frac {b^2 (A-2 C) \tan (c+d x)}{2 d}+(b (b B+2 a C)) \int \sec (c+d x) \, dx \\ & = \frac {1}{2} \left (2 A b^2+4 a b B+a^2 (A+2 C)\right ) x+\frac {b (b B+2 a C) \text {arctanh}(\sin (c+d x))}{d}+\frac {a (A b+a B) \sin (c+d x)}{d}+\frac {A \cos (c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}-\frac {b^2 (A-2 C) \tan (c+d x)}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.07 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.30 \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \left (2 A b^2+4 a b B+a^2 (A+2 C)\right ) (c+d x)-4 b (b B+2 a C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 b (b B+2 a C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\left (a^2 A+4 b^2 C+4 a (2 A b+a B) \cos (c+d x)+a^2 A \cos (2 (c+d x))\right ) \tan (c+d x)}{4 d} \]

[In]

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

[Out]

(2*(2*A*b^2 + 4*a*b*B + a^2*(A + 2*C))*(c + d*x) - 4*b*(b*B + 2*a*C)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]
+ 4*b*(b*B + 2*a*C)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (a^2*A + 4*b^2*C + 4*a*(2*A*b + a*B)*Cos[c + d*
x] + a^2*A*Cos[2*(c + d*x)])*Tan[c + d*x])/(4*d)

Maple [A] (verified)

Time = 0.65 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.14

method	result	size
derivativedivides	\(\frac {a^{2} A \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \,a^{2} \sin \left (d x +c \right )+C \,a^{2} \left (d x +c \right )+2 a A b \sin \left (d x +c \right )+2 B a b \left (d x +c \right )+2 C a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+A \,b^{2} \left (d x +c \right )+B \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C \,b^{2} \tan \left (d x +c \right )}{d}\)	\(134\)
default	\(\frac {a^{2} A \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \,a^{2} \sin \left (d x +c \right )+C \,a^{2} \left (d x +c \right )+2 a A b \sin \left (d x +c \right )+2 B a b \left (d x +c \right )+2 C a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+A \,b^{2} \left (d x +c \right )+B \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C \,b^{2} \tan \left (d x +c \right )}{d}\)	\(134\)
parallelrisch	\(\frac {-8 b \cos \left (d x +c \right ) \left (B b +2 C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+8 b \cos \left (d x +c \right ) \left (B b +2 C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (8 a A b +4 B \,a^{2}\right ) \sin \left (2 d x +2 c \right )+a^{2} A \sin \left (3 d x +3 c \right )+4 d \left (2 A \,b^{2}+4 B a b +a^{2} \left (A +2 C \right )\right ) x \cos \left (d x +c \right )+\sin \left (d x +c \right ) \left (a^{2} A +8 C \,b^{2}\right )}{8 d \cos \left (d x +c \right )}\)	\(159\)
risch	\(\frac {a^{2} A x}{2}+x A \,b^{2}+2 x B a b +a^{2} x C -\frac {i a^{2} A \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {i a A b \,{\mathrm e}^{i \left (d x +c \right )}}{d}-\frac {i B \,a^{2} {\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} a A b}{d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} B \,a^{2}}{2 d}+\frac {i a^{2} A \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {2 i C \,b^{2}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B \,b^{2}}{d}-\frac {2 a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B \,b^{2}}{d}+\frac {2 a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}\)	\(247\)
norman	\(\frac {\left (-\frac {1}{2} a^{2} A -A \,b^{2}-2 B a b -C \,a^{2}\right ) x +\left (-\frac {1}{2} a^{2} A -A \,b^{2}-2 B a b -C \,a^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (\frac {1}{2} a^{2} A +A \,b^{2}+2 B a b +C \,a^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (\frac {1}{2} a^{2} A +A \,b^{2}+2 B a b +C \,a^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (-a^{2} A -2 A \,b^{2}-4 B a b -2 C \,a^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (a^{2} A +2 A \,b^{2}+4 B a b +2 C \,a^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\frac {2 \left (3 a^{2} A -2 C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-\frac {\left (a^{2} A -4 a A b -2 B \,a^{2}+2 C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}-\frac {\left (a^{2} A +4 a A b +2 B \,a^{2}+2 C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {4 a \left (a A -2 A b -a B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}+\frac {4 a \left (a A +2 A b +a B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3}}+\frac {b \left (B b +2 C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {b \left (B b +2 C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(453\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.25 \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {{\left ({\left (A + 2 \, C\right )} a^{2} + 4 \, B a b + 2 \, A b^{2}\right )} d x \cos \left (d x + c\right ) + {\left (2 \, C a b + B b^{2}\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (2 \, C a b + B b^{2}\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (A a^{2} \cos \left (d x + c\right )^{2} + 2 \, C b^{2} + 2 \, {\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.25 \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} + 4 \, {\left (d x + c\right )} C a^{2} + 8 \, {\left (d x + c\right )} B a b + 4 \, {\left (d x + c\right )} A b^{2} + 4 \, C a b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, B b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, B a^{2} \sin \left (d x + c\right ) + 8 \, A a b \sin \left (d x + c\right ) + 4 \, C b^{2} \tan \left (d x + c\right )}{4 \, d} \]

[In]

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

[Out]

1/4*((2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^2 + 4*(d*x + c)*C*a^2 + 8*(d*x + c)*B*a*b + 4*(d*x + c)*A*b^2 + 4*C*
a*b*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 2*B*b^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1))
+ 4*B*a^2*sin(d*x + c) + 8*A*a*b*sin(d*x + c) + 4*C*b^2*tan(d*x + c))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 229 vs. \(2 (112) = 224\).

Time = 0.33 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.94 \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {\frac {4 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} - {\left (A a^{2} + 2 \, C a^{2} + 4 \, B a b + 2 \, A b^{2}\right )} {\left (d x + c\right )} - 2 \, {\left (2 \, C a b + B b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) + 2 \, {\left (2 \, C a b + B b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \]

[In]

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

[Out]

-1/2*(4*C*b^2*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) - (A*a^2 + 2*C*a^2 + 4*B*a*b + 2*A*b^2)*(d*x +
c) - 2*(2*C*a*b + B*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) + 2*(2*C*a*b + B*b^2)*log(abs(tan(1/2*d*x + 1/2*c
) - 1)) + 2*(A*a^2*tan(1/2*d*x + 1/2*c)^3 - 2*B*a^2*tan(1/2*d*x + 1/2*c)^3 - 4*A*a*b*tan(1/2*d*x + 1/2*c)^3 -
A*a^2*tan(1/2*d*x + 1/2*c) - 2*B*a^2*tan(1/2*d*x + 1/2*c) - 4*A*a*b*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 = 17.34 (sec) , antiderivative size = 274, normalized size of antiderivative = 2.32 \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {B\,a^2\,\sin \left (c+d\,x\right )}{d}+\frac {C\,b^2\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {2\,A\,a\,b\,\sin \left (c+d\,x\right )}{d}+\frac {A\,a^2\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d}-\frac {A\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,1{}\mathrm {i}}{d}-\frac {A\,b^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}}{d}-\frac {B\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}}{d}-\frac {C\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}}{d}-\frac {B\,a\,b\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,4{}\mathrm {i}}{d}-\frac {C\,a\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,4{}\mathrm {i}}{d} \]

[In]

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

[Out]

(B*a^2*sin(c + d*x))/d - (A*a^2*atanh((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*1i)/d - (A*b^2*atanh((sin(c/
2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*2i)/d - (B*b^2*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*2i)/d - (
C*a^2*atanh((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*2i)/d + (C*b^2*sin(c + d*x))/(d*cos(c + d*x)) + (2*A*a
*b*sin(c + d*x))/d - (B*a*b*atanh((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*4i)/d - (C*a*b*atan((sin(c/2 + (
d*x)/2)*1i)/cos(c/2 + (d*x)/2))*4i)/d + (A*a^2*cos(c + d*x)*sin(c + d*x))/(2*d)