\(\int \cos ^2(c+d x) (a+b \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx\) [298]

Optimal result

Integrand size = 31, antiderivative size = 124 \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {1}{2} a \left (a^2 A+6 A b^2+6 a b B\right ) x+\frac {b^2 (A b+3 a B) \text {arctanh}(\sin (c+d x))}{d}+\frac {a^2 (2 A b+a B) \sin (c+d x)}{d}+\frac {a A \cos (c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}-\frac {b^2 (a A-2 b B) \tan (c+d x)}{2 d} \]

[Out]

1/2*a*(A*a^2+6*A*b^2+6*B*a*b)*x+b^2*(A*b+3*B*a)*arctanh(sin(d*x+c))/d+a^2*(2*A*b+B*a)*sin(d*x+c)/d+1/2*a*A*cos
(d*x+c)*(a+b*sec(d*x+c))^2*sin(d*x+c)/d-1/2*b^2*(A*a-2*B*b)*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {4110, 4161, 4132, 8, 4130, 3855} \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {1}{2} a x \left (a^2 A+6 a b B+6 A b^2\right )+\frac {a^2 (a B+2 A b) \sin (c+d x)}{d}+\frac {b^2 (3 a B+A b) \text {arctanh}(\sin (c+d x))}{d}-\frac {b^2 (a A-2 b B) \tan (c+d x)}{2 d}+\frac {a A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^2}{2 d} \]

[In]

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

[Out]

(a*(a^2*A + 6*A*b^2 + 6*a*b*B)*x)/2 + (b^2*(A*b + 3*a*B)*ArcTanh[Sin[c + d*x]])/d + (a^2*(2*A*b + a*B)*Sin[c +
d*x])/d + (a*A*Cos[c + d*x]*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(2*d) - (b^2*(a*A - 2*b*B)*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 4110

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]
+ Dist[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] /; Free
Q[{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 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]

Rubi steps \begin{align*} \text {integral}& = \frac {a 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 (2 A b+a B)-\left (a^2 A+2 A b^2+4 a b B\right ) \sec (c+d x)+b (a A-2 b B) \sec ^2(c+d x)\right ) \, dx \\ & = \frac {a A \cos (c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}-\frac {b^2 (a A-2 b B) \tan (c+d x)}{2 d}-\frac {1}{2} \int \cos (c+d x) \left (-2 a^2 (2 A b+a B)-a \left (a^2 A+6 A b^2+6 a b B\right ) \sec (c+d x)-2 b^2 (A b+3 a B) \sec ^2(c+d x)\right ) \, dx \\ & = \frac {a A \cos (c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}-\frac {b^2 (a A-2 b B) \tan (c+d x)}{2 d}-\frac {1}{2} \int \cos (c+d x) \left (-2 a^2 (2 A b+a B)-2 b^2 (A b+3 a B) \sec ^2(c+d x)\right ) \, dx+\frac {1}{2} \left (a \left (a^2 A+6 A b^2+6 a b B\right )\right ) \int 1 \, dx \\ & = \frac {1}{2} a \left (a^2 A+6 A b^2+6 a b B\right ) x+\frac {a^2 (2 A b+a B) \sin (c+d x)}{d}+\frac {a A \cos (c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}-\frac {b^2 (a A-2 b B) \tan (c+d x)}{2 d}+\left (b^2 (A b+3 a B)\right ) \int \sec (c+d x) \, dx \\ & = \frac {1}{2} a \left (a^2 A+6 A b^2+6 a b B\right ) x+\frac {b^2 (A b+3 a B) \text {arctanh}(\sin (c+d x))}{d}+\frac {a^2 (2 A b+a B) \sin (c+d x)}{d}+\frac {a A \cos (c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}-\frac {b^2 (a A-2 b B) \tan (c+d x)}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.12 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.75 \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {2 a \left (a^2 A+6 A b^2+6 a b B\right ) (c+d x)-4 b^2 (A b+3 a B) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 b^2 (A b+3 a B) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {4 b^3 B \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {4 b^3 B \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )}+4 a^2 (3 A b+a B) \sin (c+d x)+a^3 A \sin (2 (c+d x))}{4 d} \]

[In]

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

[Out]

(2*a*(a^2*A + 6*A*b^2 + 6*a*b*B)*(c + d*x) - 4*b^2*(A*b + 3*a*B)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 4*
b^2*(A*b + 3*a*B)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (4*b^3*B*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Si
n[(c + d*x)/2]) + (4*b^3*B*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + 4*a^2*(3*A*b + a*B)*Sin[c
+ d*x] + a^3*A*Sin[2*(c + d*x)])/(4*d)

Maple [A] (verified)

Time = 1.81 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.06

method	result	size
derivativedivides	\(\frac {a^{3} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \,a^{3} \sin \left (d x +c \right )+3 A \,a^{2} b \sin \left (d x +c \right )+3 B \,a^{2} b \left (d x +c \right )+3 A a \,b^{2} \left (d x +c \right )+3 B a \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+A \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \tan \left (d x +c \right ) b^{3}}{d}\)	\(132\)
default	\(\frac {a^{3} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \,a^{3} \sin \left (d x +c \right )+3 A \,a^{2} b \sin \left (d x +c \right )+3 B \,a^{2} b \left (d x +c \right )+3 A a \,b^{2} \left (d x +c \right )+3 B a \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+A \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \tan \left (d x +c \right ) b^{3}}{d}\)	\(132\)
parallelrisch	\(\frac {-8 b^{2} \cos \left (d x +c \right ) \left (A b +3 B a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+8 b^{2} \cos \left (d x +c \right ) \left (A b +3 B a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+4 \left (3 A \,a^{2} b +B \,a^{3}\right ) \sin \left (2 d x +2 c \right )+a^{3} A \sin \left (3 d x +3 c \right )+4 a d x \left (A \,a^{2}+6 A \,b^{2}+6 B a b \right ) \cos \left (d x +c \right )+\sin \left (d x +c \right ) \left (a^{3} A +8 B \,b^{3}\right )}{8 d \cos \left (d x +c \right )}\)	\(162\)
risch	\(\frac {a^{3} A x}{2}+3 A a \,b^{2} x +3 B \,a^{2} b x -\frac {i a^{3} A \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} A \,a^{2} b}{2 d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} B \,a^{3}}{2 d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} A \,a^{2} b}{2 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} B \,a^{3}}{2 d}+\frac {i a^{3} A \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {2 i B \,b^{3}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A \,b^{3}}{d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B a \,b^{2}}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A \,b^{3}}{d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B a \,b^{2}}{d}\)	\(253\)
norman	\(\frac {\left (-\frac {1}{2} a^{3} A -3 A a \,b^{2}-3 B \,a^{2} b \right ) x +\left (-\frac {1}{2} a^{3} A -3 A a \,b^{2}-3 B \,a^{2} b \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (\frac {1}{2} a^{3} A +3 A a \,b^{2}+3 B \,a^{2} b \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (\frac {1}{2} a^{3} A +3 A a \,b^{2}+3 B \,a^{2} b \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (-a^{3} A -6 A a \,b^{2}-6 B \,a^{2} b \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (a^{3} A +6 A a \,b^{2}+6 B \,a^{2} b \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\frac {2 \left (3 a^{3} A -2 B \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-\frac {\left (a^{3} A -6 A \,a^{2} b -2 B \,a^{3}+2 B \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}-\frac {\left (a^{3} A +6 A \,a^{2} b +2 B \,a^{3}+2 B \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {4 a^{2} \left (a A -3 A b -B a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}+\frac {4 a^{2} \left (a A +3 A b +B a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} \left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}+\frac {b^{2} \left (A b +3 B a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {b^{2} \left (A b +3 B a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(451\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.23 \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {{\left (A a^{3} + 6 \, B a^{2} b + 6 \, A a b^{2}\right )} d x \cos \left (d x + c\right ) + {\left (3 \, B a b^{2} + A b^{3}\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (3 \, B a b^{2} + A b^{3}\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (A a^{3} \cos \left (d x + c\right )^{2} + 2 \, B b^{3} + 2 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.16 \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} + 12 \, {\left (d x + c\right )} B a^{2} b + 12 \, {\left (d x + c\right )} A a b^{2} + 6 \, B a b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, A b^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, B a^{3} \sin \left (d x + c\right ) + 12 \, A a^{2} b \sin \left (d x + c\right ) + 4 \, B b^{3} \tan \left (d x + c\right )}{4 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.89 \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=-\frac {\frac {4 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} - {\left (A a^{3} + 6 \, B a^{2} b + 6 \, A a b^{2}\right )} {\left (d x + c\right )} - 2 \, {\left (3 \, B a b^{2} + A b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) + 2 \, {\left (3 \, B a b^{2} + A b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \]

[In]

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

[Out]

-1/2*(4*B*b^3*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) - (A*a^3 + 6*B*a^2*b + 6*A*a*b^2)*(d*x + c) -
2*(3*B*a*b^2 + A*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) + 2*(3*B*a*b^2 + A*b^3)*log(abs(tan(1/2*d*x + 1/2*c)
- 1)) + 2*(A*a^3*tan(1/2*d*x + 1/2*c)^3 - 2*B*a^3*tan(1/2*d*x + 1/2*c)^3 - 6*A*a^2*b*tan(1/2*d*x + 1/2*c)^3 -
A*a^3*tan(1/2*d*x + 1/2*c) - 2*B*a^3*tan(1/2*d*x + 1/2*c) - 6*A*a^2*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 = 15.55 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.90 \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {A\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )-A\,b^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}+6\,A\,a\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+6\,B\,a^2\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )-B\,a\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,6{}\mathrm {i}}{d}+\frac {\frac {A\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{8}+\frac {B\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {A\,a^3\,\sin \left (c+d\,x\right )}{8}+B\,b^3\,\sin \left (c+d\,x\right )+\frac {3\,A\,a^2\,b\,\sin \left (2\,c+2\,d\,x\right )}{2}}{d\,\cos \left (c+d\,x\right )} \]

[In]

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

[Out]

(A*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) - A*b^3*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*2i
+ 6*A*a*b^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + 6*B*a^2*b*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2
)) - B*a*b^2*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*6i)/d + ((A*a^3*sin(3*c + 3*d*x))/8 + (B*a^3*sin
(2*c + 2*d*x))/2 + (A*a^3*sin(c + d*x))/8 + B*b^3*sin(c + d*x) + (3*A*a^2*b*sin(2*c + 2*d*x))/2)/(d*cos(c + d*
x))