\(\int \cos ^5(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx\) [52]

Optimal result

Integrand size = 29, antiderivative size = 125 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {3}{8} a (A+B) x+\frac {a (4 A+5 B) \sin (c+d x)}{5 d}+\frac {3 a (A+B) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a (A+B) \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {a A \cos ^4(c+d x) \sin (c+d x)}{5 d}-\frac {a (4 A+5 B) \sin ^3(c+d x)}{15 d} \]

[Out]

3/8*a*(A+B)*x+1/5*a*(4*A+5*B)*sin(d*x+c)/d+3/8*a*(A+B)*cos(d*x+c)*sin(d*x+c)/d+1/4*a*(A+B)*cos(d*x+c)^3*sin(d*
x+c)/d+1/5*a*A*cos(d*x+c)^4*sin(d*x+c)/d-1/15*a*(4*A+5*B)*sin(d*x+c)^3/d

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {4081, 3872, 2715, 8, 2713} \[ \int \cos ^5(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=-\frac {a (4 A+5 B) \sin ^3(c+d x)}{15 d}+\frac {a (4 A+5 B) \sin (c+d x)}{5 d}+\frac {a (A+B) \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3 a (A+B) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {3}{8} a x (A+B)+\frac {a A \sin (c+d x) \cos ^4(c+d x)}{5 d} \]

[In]

Int[Cos[c + d*x]^5*(a + a*Sec[c + d*x])*(A + B*Sec[c + d*x]),x]

[Out]

(3*a*(A + B)*x)/8 + (a*(4*A + 5*B)*Sin[c + d*x])/(5*d) + (3*a*(A + B)*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a*(A
+ B)*Cos[c + d*x]^3*Sin[c + d*x])/(4*d) + (a*A*Cos[c + d*x]^4*Sin[c + d*x])/(5*d) - (a*(4*A + 5*B)*Sin[c + d*
x]^3)/(15*d)

Rule 8

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

Rule 2713

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

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 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]

Rubi steps \begin{align*} \text {integral}& = \frac {a A \cos ^4(c+d x) \sin (c+d x)}{5 d}-\frac {1}{5} \int \cos ^4(c+d x) (-5 a (A+B)-a (4 A+5 B) \sec (c+d x)) \, dx \\ & = \frac {a A \cos ^4(c+d x) \sin (c+d x)}{5 d}+(a (A+B)) \int \cos ^4(c+d x) \, dx+\frac {1}{5} (a (4 A+5 B)) \int \cos ^3(c+d x) \, dx \\ & = \frac {a (A+B) \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {a A \cos ^4(c+d x) \sin (c+d x)}{5 d}+\frac {1}{4} (3 a (A+B)) \int \cos ^2(c+d x) \, dx-\frac {(a (4 A+5 B)) \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{5 d} \\ & = \frac {a (4 A+5 B) \sin (c+d x)}{5 d}+\frac {3 a (A+B) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a (A+B) \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {a A \cos ^4(c+d x) \sin (c+d x)}{5 d}-\frac {a (4 A+5 B) \sin ^3(c+d x)}{15 d}+\frac {1}{8} (3 a (A+B)) \int 1 \, dx \\ & = \frac {3}{8} a (A+B) x+\frac {a (4 A+5 B) \sin (c+d x)}{5 d}+\frac {3 a (A+B) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a (A+B) \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {a A \cos ^4(c+d x) \sin (c+d x)}{5 d}-\frac {a (4 A+5 B) \sin ^3(c+d x)}{15 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.62 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {a \left (480 (A+B) \sin (c+d x)-160 (2 A+B) \sin ^3(c+d x)+96 A \sin ^5(c+d x)+15 (A+B) (12 (c+d x)+8 \sin (2 (c+d x))+\sin (4 (c+d x)))\right )}{480 d} \]

[In]

Integrate[Cos[c + d*x]^5*(a + a*Sec[c + d*x])*(A + B*Sec[c + d*x]),x]

[Out]

(a*(480*(A + B)*Sin[c + d*x] - 160*(2*A + B)*Sin[c + d*x]^3 + 96*A*Sin[c + d*x]^5 + 15*(A + B)*(12*(c + d*x) +
8*Sin[2*(c + d*x)] + Sin[4*(c + d*x)])))/(480*d)

Maple [A] (verified)

Time = 3.30 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.70

method	result	size
parallelrisch	\(\frac {\left (8 \left (A +B \right ) \sin \left (2 d x +2 c \right )+\frac {2 \left (5 A +4 B \right ) \sin \left (3 d x +3 c \right )}{3}+\left (A +B \right ) \sin \left (4 d x +4 c \right )+\frac {2 A \sin \left (5 d x +5 c \right )}{5}+4 \left (5 A +6 B \right ) \sin \left (d x +c \right )+12 \left (A +B \right ) x d \right ) a}{32 d}\)	\(87\)
derivativedivides	\(\frac {\frac {a A \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+a A \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+B a \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {B a \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}}{d}\)	\(128\)
default	\(\frac {\frac {a A \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+a A \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+B a \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {B a \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}}{d}\)	\(128\)
risch	\(\frac {3 a A x}{8}+\frac {3 a x B}{8}+\frac {5 a A \sin \left (d x +c \right )}{8 d}+\frac {3 \sin \left (d x +c \right ) B a}{4 d}+\frac {a A \sin \left (5 d x +5 c \right )}{80 d}+\frac {a A \sin \left (4 d x +4 c \right )}{32 d}+\frac {\sin \left (4 d x +4 c \right ) B a}{32 d}+\frac {5 a A \sin \left (3 d x +3 c \right )}{48 d}+\frac {\sin \left (3 d x +3 c \right ) B a}{12 d}+\frac {a A \sin \left (2 d x +2 c \right )}{4 d}+\frac {\sin \left (2 d x +2 c \right ) B a}{4 d}\)	\(150\)
norman	\(\frac {-\frac {3 a \left (A +B \right ) x}{8}+\frac {a \left (A -31 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 d}-\frac {13 a \left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {3 a \left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{4 d}-\frac {3 a \left (A +B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2}-\frac {15 a \left (A +B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{8}+\frac {15 a \left (A +B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{8}+\frac {3 a \left (A +B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{2}+\frac {3 a \left (A +B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{8}+\frac {a \left (17 A +49 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{12 d}-\frac {a \left (137 A +25 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{30 d}+\frac {a \left (167 A +55 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{30 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5} \left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\)	\(262\)

[In]

int(cos(d*x+c)^5*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/32*(8*(A+B)*sin(2*d*x+2*c)+2/3*(5*A+4*B)*sin(3*d*x+3*c)+(A+B)*sin(4*d*x+4*c)+2/5*A*sin(5*d*x+5*c)+4*(5*A+6*B
)*sin(d*x+c)+12*(A+B)*x*d)*a/d

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.70 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {45 \, {\left (A + B\right )} a d x + {\left (24 \, A a \cos \left (d x + c\right )^{4} + 30 \, {\left (A + B\right )} a \cos \left (d x + c\right )^{3} + 8 \, {\left (4 \, A + 5 \, B\right )} a \cos \left (d x + c\right )^{2} + 45 \, {\left (A + B\right )} a \cos \left (d x + c\right ) + 16 \, {\left (4 \, A + 5 \, B\right )} a\right )} \sin \left (d x + c\right )}{120 \, d} \]

[In]

integrate(cos(d*x+c)^5*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

1/120*(45*(A + B)*a*d*x + (24*A*a*cos(d*x + c)^4 + 30*(A + B)*a*cos(d*x + c)^3 + 8*(4*A + 5*B)*a*cos(d*x + c)^
2 + 45*(A + B)*a*cos(d*x + c) + 16*(4*A + 5*B)*a)*sin(d*x + c))/d

Sympy [F]

\[ \int \cos ^5(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=a \left (\int A \cos ^{5}{\left (c + d x \right )}\, dx + \int A \cos ^{5}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int B \cos ^{5}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int B \cos ^{5}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx\right ) \]

[In]

integrate(cos(d*x+c)**5*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x)

[Out]

a*(Integral(A*cos(c + d*x)**5, x) + Integral(A*cos(c + d*x)**5*sec(c + d*x), x) + Integral(B*cos(c + d*x)**5*s
ec(c + d*x), x) + Integral(B*cos(c + d*x)**5*sec(c + d*x)**2, x))

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.99 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} A a + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a - 160 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a}{480 \, d} \]

[In]

integrate(cos(d*x+c)^5*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

1/480*(32*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*A*a + 15*(12*d*x + 12*c + sin(4*d*x + 4*c)
+ 8*sin(2*d*x + 2*c))*A*a - 160*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a + 15*(12*d*x + 12*c + sin(4*d*x + 4*c) +
8*sin(2*d*x + 2*c))*B*a)/d

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.47 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {45 \, {\left (A a + B a\right )} {\left (d x + c\right )} + \frac {2 \, {\left (45 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 45 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 130 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 290 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 464 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 400 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 190 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 350 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 195 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 195 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5}}}{120 \, d} \]

[In]

integrate(cos(d*x+c)^5*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

1/120*(45*(A*a + B*a)*(d*x + c) + 2*(45*A*a*tan(1/2*d*x + 1/2*c)^9 + 45*B*a*tan(1/2*d*x + 1/2*c)^9 + 130*A*a*t
an(1/2*d*x + 1/2*c)^7 + 290*B*a*tan(1/2*d*x + 1/2*c)^7 + 464*A*a*tan(1/2*d*x + 1/2*c)^5 + 400*B*a*tan(1/2*d*x
+ 1/2*c)^5 + 190*A*a*tan(1/2*d*x + 1/2*c)^3 + 350*B*a*tan(1/2*d*x + 1/2*c)^3 + 195*A*a*tan(1/2*d*x + 1/2*c) +
195*B*a*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d

Mupad [B] (verification not implemented)

Time = 16.31 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.70 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {\left (\frac {3\,A\,a}{4}+\frac {3\,B\,a}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {13\,A\,a}{6}+\frac {29\,B\,a}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {116\,A\,a}{15}+\frac {20\,B\,a}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {19\,A\,a}{6}+\frac {35\,B\,a}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {13\,A\,a}{4}+\frac {13\,B\,a}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {3\,a\,\mathrm {atan}\left (\frac {3\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A+B\right )}{4\,\left (\frac {3\,A\,a}{4}+\frac {3\,B\,a}{4}\right )}\right )\,\left (A+B\right )}{4\,d} \]

[In]

int(cos(c + d*x)^5*(A + B/cos(c + d*x))*(a + a/cos(c + d*x)),x)

[Out]

(tan(c/2 + (d*x)/2)*((13*A*a)/4 + (13*B*a)/4) + tan(c/2 + (d*x)/2)^9*((3*A*a)/4 + (3*B*a)/4) + tan(c/2 + (d*x)
/2)^7*((13*A*a)/6 + (29*B*a)/6) + tan(c/2 + (d*x)/2)^3*((19*A*a)/6 + (35*B*a)/6) + tan(c/2 + (d*x)/2)^5*((116*
A*a)/15 + (20*B*a)/3))/(d*(5*tan(c/2 + (d*x)/2)^2 + 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)/2)^6 + 5*tan(
c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 + 1)) + (3*a*atan((3*a*tan(c/2 + (d*x)/2)*(A + B))/(4*((3*A*a)/4 + (3
*B*a)/4)))*(A + B))/(4*d)