\(\int \cos ^4(c+d x) (a+a \sec (c+d x)) (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [312]

Optimal result

Integrand size = 38, antiderivative size = 77 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{2} a (B+C) x+\frac {a (2 B+3 C) \sin (c+d x)}{3 d}+\frac {a (B+C) \cos (c+d x) \sin (c+d x)}{2 d}+\frac {a B \cos ^2(c+d x) \sin (c+d x)}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {4157, 4081, 3872, 2715, 8, 2717} \[ \int \cos ^4(c+d x) (a+a \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a (2 B+3 C) \sin (c+d x)}{3 d}+\frac {a (B+C) \sin (c+d x) \cos (c+d x)}{2 d}+\frac {a B \sin (c+d x) \cos ^2(c+d x)}{3 d}+\frac {1}{2} a x (B+C) \]

[In]

Int[Cos[c + d*x]^4*(a + a*Sec[c + d*x])*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(a*(B + C)*x)/2 + (a*(2*B + 3*C)*Sin[c + d*x])/(3*d) + (a*(B + C)*Cos[c + d*x]*Sin[c + d*x])/(2*d) + (a*B*Cos[
c + d*x]^2*Sin[c + d*x])/(3*d)

Rule 8

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

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

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 ^3(c+d x) (a+a \sec (c+d x)) (B+C \sec (c+d x)) \, dx \\ & = \frac {a B \cos ^2(c+d x) \sin (c+d x)}{3 d}-\frac {1}{3} \int \cos ^2(c+d x) (-3 a (B+C)-a (2 B+3 C) \sec (c+d x)) \, dx \\ & = \frac {a B \cos ^2(c+d x) \sin (c+d x)}{3 d}+(a (B+C)) \int \cos ^2(c+d x) \, dx+\frac {1}{3} (a (2 B+3 C)) \int \cos (c+d x) \, dx \\ & = \frac {a (2 B+3 C) \sin (c+d x)}{3 d}+\frac {a (B+C) \cos (c+d x) \sin (c+d x)}{2 d}+\frac {a B \cos ^2(c+d x) \sin (c+d x)}{3 d}+\frac {1}{2} (a (B+C)) \int 1 \, dx \\ & = \frac {1}{2} a (B+C) x+\frac {a (2 B+3 C) \sin (c+d x)}{3 d}+\frac {a (B+C) \cos (c+d x) \sin (c+d x)}{2 d}+\frac {a B \cos ^2(c+d x) \sin (c+d x)}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.84 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a (6 B c+6 c C+6 B d x+6 C d x+3 (3 B+4 C) \sin (c+d x)+3 (B+C) \sin (2 (c+d x))+B \sin (3 (c+d x)))}{12 d} \]

[In]

Integrate[Cos[c + d*x]^4*(a + a*Sec[c + d*x])*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(a*(6*B*c + 6*c*C + 6*B*d*x + 6*C*d*x + 3*(3*B + 4*C)*Sin[c + d*x] + 3*(B + C)*Sin[2*(c + d*x)] + B*Sin[3*(c +
d*x)]))/(12*d)

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.70

method	result	size
parallelrisch	\(\frac {a \left (\frac {\left (B +C \right ) \sin \left (2 d x +2 c \right )}{2}+\frac {B \sin \left (3 d x +3 c \right )}{6}+\left (\frac {3 B}{2}+2 C \right ) \sin \left (d x +c \right )+\left (B +C \right ) x d \right )}{2 d}\)	\(54\)
derivativedivides	\(\frac {\frac {a B \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+a B \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C a \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C a \sin \left (d x +c \right )}{d}\)	\(85\)
default	\(\frac {\frac {a B \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+a B \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C a \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C a \sin \left (d x +c \right )}{d}\)	\(85\)
risch	\(\frac {a B x}{2}+\frac {a x C}{2}+\frac {3 a B \sin \left (d x +c \right )}{4 d}+\frac {\sin \left (d x +c \right ) C a}{d}+\frac {a B \sin \left (3 d x +3 c \right )}{12 d}+\frac {a B \sin \left (2 d x +2 c \right )}{4 d}+\frac {\sin \left (2 d x +2 c \right ) C a}{4 d}\)	\(85\)
norman	\(\frac {\left (\frac {1}{2} a B +\frac {1}{2} C a \right ) x +\left (-\frac {1}{2} a B -\frac {1}{2} C a \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-\frac {1}{2} a B -\frac {1}{2} C a \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (\frac {1}{2} a B +\frac {1}{2} C a \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+\left (-2 a B -2 C a \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (a B +C a \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (a B +C a \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\frac {a \left (B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{d}+\frac {2 a \left (B -3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d}+\frac {3 a \left (B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {a \left (B +9 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{3 d}-\frac {a \left (5 B -3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}-\frac {2 a \left (5 B +9 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2}}\)	\(301\)

[In]

int(cos(d*x+c)^4*(a+a*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

1/2*a*(1/2*(B+C)*sin(2*d*x+2*c)+1/6*B*sin(3*d*x+3*c)+(3/2*B+2*C)*sin(d*x+c)+(B+C)*x*d)/d

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.73 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (B + C\right )} a d x + {\left (2 \, B a \cos \left (d x + c\right )^{2} + 3 \, {\left (B + C\right )} a \cos \left (d x + c\right ) + 2 \, {\left (2 \, B + 3 \, C\right )} a\right )} \sin \left (d x + c\right )}{6 \, d} \]

[In]

integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \cos ^4(c+d x) (a+a \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=a \left (\int B \cos ^{4}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int B \cos ^{4}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int C \cos ^{4}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int C \cos ^{4}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx\right ) \]

[In]

integrate(cos(d*x+c)**4*(a+a*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

a*(Integral(B*cos(c + d*x)**4*sec(c + d*x), x) + Integral(B*cos(c + d*x)**4*sec(c + d*x)**2, x) + Integral(C*c
os(c + d*x)**4*sec(c + d*x)**2, x) + Integral(C*cos(c + d*x)**4*sec(c + d*x)**3, x))

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.03 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a - 12 \, C a \sin \left (d x + c\right )}{12 \, d} \]

[In]

integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

-1/12*(4*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a - 3*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a - 3*(2*d*x + 2*c + sin
(2*d*x + 2*c))*C*a - 12*C*a*sin(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.61 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (B a + C a\right )} {\left (d x + c\right )} + \frac {2 \, {\left (3 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \]

[In]

integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/6*(3*(B*a + C*a)*(d*x + c) + 2*(3*B*a*tan(1/2*d*x + 1/2*c)^5 + 3*C*a*tan(1/2*d*x + 1/2*c)^5 + 4*B*a*tan(1/2*
d*x + 1/2*c)^3 + 12*C*a*tan(1/2*d*x + 1/2*c)^3 + 9*B*a*tan(1/2*d*x + 1/2*c) + 9*C*a*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 = 15.44 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.09 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {B\,a\,x}{2}+\frac {C\,a\,x}{2}+\frac {3\,B\,a\,\sin \left (c+d\,x\right )}{4\,d}+\frac {C\,a\,\sin \left (c+d\,x\right )}{d}+\frac {B\,a\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,a\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {C\,a\,\sin \left (2\,c+2\,d\,x\right )}{4\,d} \]

[In]

int(cos(c + d*x)^4*(B/cos(c + d*x) + C/cos(c + d*x)^2)*(a + a/cos(c + d*x)),x)

[Out]

(B*a*x)/2 + (C*a*x)/2 + (3*B*a*sin(c + d*x))/(4*d) + (C*a*sin(c + d*x))/d + (B*a*sin(2*c + 2*d*x))/(4*d) + (B*
a*sin(3*c + 3*d*x))/(12*d) + (C*a*sin(2*c + 2*d*x))/(4*d)