\(\int \cos ^3(c+d x) (a+b \sec (c+d x))^3 (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [789]

Optimal result

Integrand size = 40, antiderivative size = 124 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{2} a \left (a^2 B+6 b^2 B+6 a b C\right ) x+\frac {b^2 (b B+3 a C) \text {arctanh}(\sin (c+d x))}{d}+\frac {a^2 (2 b B+a C) \sin (c+d x)}{d}+\frac {a B \cos (c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}-\frac {b^2 (a B-2 b C) \tan (c+d x)}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.175, Rules used = {4157, 4110, 4161, 4132, 8, 4130, 3855} \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{2} a x \left (a^2 B+6 a b C+6 b^2 B\right )+\frac {a^2 (a C+2 b B) \sin (c+d x)}{d}+\frac {b^2 (3 a C+b B) \text {arctanh}(\sin (c+d x))}{d}-\frac {b^2 (a B-2 b C) \tan (c+d x)}{2 d}+\frac {a B \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^2}{2 d} \]

[In]

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

[Out]

(a*(a^2*B + 6*b^2*B + 6*a*b*C)*x)/2 + (b^2*(b*B + 3*a*C)*ArcTanh[Sin[c + d*x]])/d + (a^2*(2*b*B + a*C)*Sin[c +
d*x])/d + (a*B*Cos[c + d*x]*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(2*d) - (b^2*(a*B - 2*b*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 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 4157

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] :> Dist[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 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}& = \int \cos ^2(c+d x) (a+b \sec (c+d x))^3 (B+C \sec (c+d x)) \, dx \\ & = \frac {a B \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 b B+a C)-\left (a^2 B+2 b^2 B+4 a b C\right ) \sec (c+d x)+b (a B-2 b C) \sec ^2(c+d x)\right ) \, dx \\ & = \frac {a B \cos (c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}-\frac {b^2 (a B-2 b C) \tan (c+d x)}{2 d}-\frac {1}{2} \int \cos (c+d x) \left (-2 a^2 (2 b B+a C)-a \left (a^2 B+6 b^2 B+6 a b C\right ) \sec (c+d x)-2 b^2 (b B+3 a C) \sec ^2(c+d x)\right ) \, dx \\ & = \frac {a B \cos (c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}-\frac {b^2 (a B-2 b C) \tan (c+d x)}{2 d}-\frac {1}{2} \int \cos (c+d x) \left (-2 a^2 (2 b B+a C)-2 b^2 (b B+3 a C) \sec ^2(c+d x)\right ) \, dx+\frac {1}{2} \left (a \left (a^2 B+6 b^2 B+6 a b C\right )\right ) \int 1 \, dx \\ & = \frac {1}{2} a \left (a^2 B+6 b^2 B+6 a b C\right ) x+\frac {a^2 (2 b B+a C) \sin (c+d x)}{d}+\frac {a B \cos (c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}-\frac {b^2 (a B-2 b C) \tan (c+d x)}{2 d}+\left (b^2 (b B+3 a C)\right ) \int \sec (c+d x) \, dx \\ & = \frac {1}{2} a \left (a^2 B+6 b^2 B+6 a b C\right ) x+\frac {b^2 (b B+3 a C) \text {arctanh}(\sin (c+d x))}{d}+\frac {a^2 (2 b B+a C) \sin (c+d x)}{d}+\frac {a B \cos (c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}-\frac {b^2 (a B-2 b C) \tan (c+d x)}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.05 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.75 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 a \left (a^2 B+6 b^2 B+6 a b C\right ) (c+d x)-4 b^2 (b B+3 a C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 b^2 (b B+3 a C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {4 b^3 C \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {4 b^3 C \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )}+4 a^2 (3 b B+a C) \sin (c+d x)+a^3 B \sin (2 (c+d x))}{4 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.57 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.06

method	result	size
derivativedivides	\(\frac {B \,a^{3} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{3} C \sin \left (d x +c \right )+3 B \,a^{2} b \sin \left (d x +c \right )+3 a^{2} b C \left (d x +c \right )+3 B a \,b^{2} \left (d x +c \right )+3 C a \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C \,b^{3} \tan \left (d x +c \right )}{d}\)	\(132\)
default	\(\frac {B \,a^{3} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{3} C \sin \left (d x +c \right )+3 B \,a^{2} b \sin \left (d x +c \right )+3 a^{2} b C \left (d x +c \right )+3 B a \,b^{2} \left (d x +c \right )+3 C a \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C \,b^{3} \tan \left (d x +c \right )}{d}\)	\(132\)
parallelrisch	\(\frac {-8 b^{2} \cos \left (d x +c \right ) \left (B b +3 C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+8 b^{2} \cos \left (d x +c \right ) \left (B b +3 C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+4 \left (3 B \,a^{2} b +a^{3} C \right ) \sin \left (2 d x +2 c \right )+B \,a^{3} \sin \left (3 d x +3 c \right )+4 a d x \left (B \,a^{2}+6 B \,b^{2}+6 C a b \right ) \cos \left (d x +c \right )+\sin \left (d x +c \right ) \left (B \,a^{3}+8 C \,b^{3}\right )}{8 d \cos \left (d x +c \right )}\)	\(162\)
risch	\(\frac {a^{3} B x}{2}+3 B a \,b^{2} x +3 C \,a^{2} b x -\frac {i B \,a^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} B \,a^{2} b}{2 d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} a^{3} C}{2 d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} B \,a^{2} b}{2 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} a^{3} C}{2 d}+\frac {i B \,a^{3} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {2 i C \,b^{3}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B \,b^{3}}{d}-\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C \,b^{2}}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B \,b^{3}}{d}+\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C \,b^{2}}{d}\)	\(253\)
norman	\(\frac {\left (\frac {1}{2} B \,a^{3}+3 B a \,b^{2}+3 a^{2} b C \right ) x +\left (-\frac {3}{2} B \,a^{3}-9 B a \,b^{2}-9 a^{2} b C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-\frac {3}{2} B \,a^{3}-9 B a \,b^{2}-9 a^{2} b C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (-\frac {1}{2} B \,a^{3}-3 B a \,b^{2}-3 a^{2} b C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (-\frac {1}{2} B \,a^{3}-3 B a \,b^{2}-3 a^{2} b C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+\left (\frac {1}{2} B \,a^{3}+3 B a \,b^{2}+3 a^{2} b C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}+\left (\frac {3}{2} B \,a^{3}+9 B a \,b^{2}+9 a^{2} b C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (\frac {3}{2} B \,a^{3}+9 B a \,b^{2}+9 a^{2} b C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\frac {\left (B \,a^{3}+6 B \,a^{2} b +2 a^{3} C +2 C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {\left (5 B \,a^{3}-6 B \,a^{2} b -2 a^{3} C -6 C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-\frac {\left (B \,a^{3}-6 B \,a^{2} b -2 a^{3} C +2 C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{d}-\frac {\left (5 B \,a^{3}+6 B \,a^{2} b +2 a^{3} C -6 C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}+\frac {8 a^{2} \left (3 B b +C a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}+\frac {4 a^{2} \left (a B -3 B b -C a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{d}-\frac {4 a^{2} \left (a B +3 B b +C 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 )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4}}+\frac {b^{2} \left (B b +3 C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {b^{2} \left (B b +3 C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(600\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.23 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {{\left (B a^{3} + 6 \, C a^{2} b + 6 \, B a b^{2}\right )} d x \cos \left (d x + c\right ) + {\left (3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (B a^{3} \cos \left (d x + c\right )^{2} + 2 \, C b^{3} + 2 \, {\left (C a^{3} + 3 \, B a^{2} b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \]

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.16 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} + 12 \, {\left (d x + c\right )} C a^{2} b + 12 \, {\left (d x + c\right )} B a b^{2} + 6 \, C a b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, B b^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, C a^{3} \sin \left (d x + c\right ) + 12 \, B a^{2} b \sin \left (d x + c\right ) + 4 \, C b^{3} \tan \left (d x + c\right )}{4 \, d} \]

[In]

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

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.89 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {\frac {4 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} - {\left (B a^{3} + 6 \, C a^{2} b + 6 \, B a b^{2}\right )} {\left (d x + c\right )} - 2 \, {\left (3 \, C a b^{2} + B b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) + 2 \, {\left (3 \, C a b^{2} + B b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \]

[In]

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

[Out]

-1/2*(4*C*b^3*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) - (B*a^3 + 6*C*a^2*b + 6*B*a*b^2)*(d*x + c) -
2*(3*C*a*b^2 + B*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) + 2*(3*C*a*b^2 + B*b^3)*log(abs(tan(1/2*d*x + 1/2*c)
- 1)) + 2*(B*a^3*tan(1/2*d*x + 1/2*c)^3 - 2*C*a^3*tan(1/2*d*x + 1/2*c)^3 - 6*B*a^2*b*tan(1/2*d*x + 1/2*c)^3 -
B*a^3*tan(1/2*d*x + 1/2*c) - 2*C*a^3*tan(1/2*d*x + 1/2*c) - 6*B*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 = 17.85 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.90 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {B\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )-B\,b^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}+6\,B\,a\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+6\,C\,a^2\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )-C\,a\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,6{}\mathrm {i}}{d}+\frac {\frac {B\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{8}+\frac {C\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {B\,a^3\,\sin \left (c+d\,x\right )}{8}+C\,b^3\,\sin \left (c+d\,x\right )+\frac {3\,B\,a^2\,b\,\sin \left (2\,c+2\,d\,x\right )}{2}}{d\,\cos \left (c+d\,x\right )} \]

[In]

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

[Out]

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