\(\int \cos ^3(c+d x) (a+b \sec (c+d x)) (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [771]

Optimal result

Integrand size = 38, antiderivative size = 52 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{2} (a B+2 b C) x+\frac {(b B+a C) \sin (c+d x)}{d}+\frac {a B \cos (c+d x) \sin (c+d x)}{2 d} \]

[Out]

1/2*(B*a+2*C*b)*x+(B*b+C*a)*sin(d*x+c)/d+1/2*a*B*cos(d*x+c)*sin(d*x+c)/d

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {4157, 4081, 3872, 2717, 8} \[ \int \cos ^3(c+d x) (a+b \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {(a C+b B) \sin (c+d x)}{d}+\frac {1}{2} x (a B+2 b C)+\frac {a B \sin (c+d x) \cos (c+d x)}{2 d} \]

[In]

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

[Out]

((a*B + 2*b*C)*x)/2 + ((b*B + a*C)*Sin[c + d*x])/d + (a*B*Cos[c + d*x]*Sin[c + d*x])/(2*d)

Rule 8

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

Rule 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3872

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 4081

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

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]

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.98 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 a B c+2 a B d x+4 b C d x+4 (b B+a C) \sin (c+d x)+a B \sin (2 (c+d x))}{4 d} \]

[In]

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

[Out]

(2*a*B*c + 2*a*B*d*x + 4*b*C*d*x + 4*(b*B + a*C)*Sin[c + d*x] + a*B*Sin[2*(c + d*x)])/(4*d)

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.90

method	result	size
parallelrisch	\(\frac {a B \sin \left (2 d x +2 c \right )+\left (4 B b +4 C a \right ) \sin \left (d x +c \right )+2 \left (a B +2 C b \right ) x d}{4 d}\)	\(47\)
risch	\(\frac {a B x}{2}+x C b +\frac {\sin \left (d x +c \right ) B b}{d}+\frac {\sin \left (d x +c \right ) C a}{d}+\frac {a B \sin \left (2 d x +2 c \right )}{4 d}\)	\(51\)
derivativedivides	\(\frac {a B \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B b \sin \left (d x +c \right )+C a \sin \left (d x +c \right )+C b \left (d x +c \right )}{d}\)	\(57\)
default	\(\frac {a B \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B b \sin \left (d x +c \right )+C a \sin \left (d x +c \right )+C b \left (d x +c \right )}{d}\)	\(57\)
norman	\(\frac {\left (\frac {a B}{2}+C b \right ) x +\left (\frac {a B}{2}+C b \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (\frac {a B}{2}+C b \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (\frac {a B}{2}+C b \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (-a B -2 C b \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-a B -2 C b \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\frac {\left (a B +2 B b +2 C a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {4 \left (B b +C a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-\frac {\left (a B -2 B b -2 C a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}-\frac {2 a B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}+\frac {2 a B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2}}\)	\(262\)

[In]

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

[Out]

1/4*(a*B*sin(2*d*x+2*c)+(4*B*b+4*C*a)*sin(d*x+c)+2*(B*a+2*C*b)*x*d)/d

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.81 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {{\left (B a + 2 \, C b\right )} d x + {\left (B a \cos \left (d x + c\right ) + 2 \, C a + 2 \, B b\right )} \sin \left (d x + c\right )}{2 \, d} \]

[In]

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

[Out]

1/2*((B*a + 2*C*b)*d*x + (B*a*cos(d*x + c) + 2*C*a + 2*B*b)*sin(d*x + c))/d

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.06 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a + 4 \, {\left (d x + c\right )} C b + 4 \, C a \sin \left (d x + c\right ) + 4 \, B b \sin \left (d x + c\right )}{4 \, d} \]

[In]

integrate(cos(d*x+c)^3*(a+b*sec(d*x+c))*(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 + 4*(d*x + c)*C*b + 4*C*a*sin(d*x + c) + 4*B*b*sin(d*x + c))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 121 vs. \(2 (48) = 96\).

Time = 0.29 (sec) , antiderivative size = 121, normalized size of antiderivative = 2.33 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {{\left (B a + 2 \, C b\right )} {\left (d x + c\right )} - \frac {2 \, {\left (B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \]

[In]

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

[Out]

1/2*((B*a + 2*C*b)*(d*x + c) - 2*(B*a*tan(1/2*d*x + 1/2*c)^3 - 2*C*a*tan(1/2*d*x + 1/2*c)^3 - 2*B*b*tan(1/2*d*
x + 1/2*c)^3 - B*a*tan(1/2*d*x + 1/2*c) - 2*C*a*tan(1/2*d*x + 1/2*c) - 2*B*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 = 17.26 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.96 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {B\,a\,x}{2}+C\,b\,x+\frac {B\,b\,\sin \left (c+d\,x\right )}{d}+\frac {C\,a\,\sin \left (c+d\,x\right )}{d}+\frac {B\,a\,\sin \left (2\,c+2\,d\,x\right )}{4\,d} \]

[In]

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

[Out]

(B*a*x)/2 + C*b*x + (B*b*sin(c + d*x))/d + (C*a*sin(c + d*x))/d + (B*a*sin(2*c + 2*d*x))/(4*d)