\(\int \cos ^5(c+d x) (a+b \sec (c+d x)) \, dx\) [455]

Optimal result

Integrand size = 19, antiderivative size = 92 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x)) \, dx=\frac {3 b x}{8}+\frac {a \sin (c+d x)}{d}+\frac {3 b \cos (c+d x) \sin (c+d x)}{8 d}+\frac {b \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {2 a \sin ^3(c+d x)}{3 d}+\frac {a \sin ^5(c+d x)}{5 d} \]

[Out]

3/8*b*x+a*sin(d*x+c)/d+3/8*b*cos(d*x+c)*sin(d*x+c)/d+1/4*b*cos(d*x+c)^3*sin(d*x+c)/d-2/3*a*sin(d*x+c)^3/d+1/5*
a*sin(d*x+c)^5/d

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {3872, 2713, 2715, 8} \[ \int \cos ^5(c+d x) (a+b \sec (c+d x)) \, dx=\frac {a \sin ^5(c+d x)}{5 d}-\frac {2 a \sin ^3(c+d x)}{3 d}+\frac {a \sin (c+d x)}{d}+\frac {b \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3 b \sin (c+d x) \cos (c+d x)}{8 d}+\frac {3 b x}{8} \]

[In]

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

[Out]

(3*b*x)/8 + (a*Sin[c + d*x])/d + (3*b*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (b*Cos[c + d*x]^3*Sin[c + d*x])/(4*d)
- (2*a*Sin[c + d*x]^3)/(3*d) + (a*Sin[c + d*x]^5)/(5*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]

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.97 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x)) \, dx=\frac {3 b (c+d x)}{8 d}+\frac {a \sin (c+d x)}{d}-\frac {2 a \sin ^3(c+d x)}{3 d}+\frac {a \sin ^5(c+d x)}{5 d}+\frac {b \sin (2 (c+d x))}{4 d}+\frac {b \sin (4 (c+d x))}{32 d} \]

[In]

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

[Out]

(3*b*(c + d*x))/(8*d) + (a*Sin[c + d*x])/d - (2*a*Sin[c + d*x]^3)/(3*d) + (a*Sin[c + d*x]^5)/(5*d) + (b*Sin[2*
(c + d*x)])/(4*d) + (b*Sin[4*(c + d*x)])/(32*d)

Maple [A] (verified)

Time = 0.83 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.75

method	result	size
parallelrisch	\(\frac {180 b x d +6 a \sin \left (5 d x +5 c \right )+15 b \sin \left (4 d x +4 c \right )+50 a \sin \left (3 d x +3 c \right )+120 b \sin \left (2 d x +2 c \right )+300 a \sin \left (d x +c \right )}{480 d}\)	\(69\)
derivativedivides	\(\frac {\frac {a \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+b \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\)	\(70\)
default	\(\frac {\frac {a \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+b \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\)	\(70\)
risch	\(\frac {3 b x}{8}+\frac {5 a \sin \left (d x +c \right )}{8 d}+\frac {a \sin \left (5 d x +5 c \right )}{80 d}+\frac {b \sin \left (4 d x +4 c \right )}{32 d}+\frac {5 a \sin \left (3 d x +3 c \right )}{48 d}+\frac {b \sin \left (2 d x +2 c \right )}{4 d}\)	\(78\)
norman	\(\frac {\frac {3 b x}{8}+\frac {116 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d}+\frac {15 b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{8}+\frac {15 b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{4}+\frac {15 b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{4}+\frac {15 b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{8}+\frac {3 b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{8}+\frac {\left (8 a -5 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{4 d}+\frac {\left (8 a +5 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (16 a -3 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{6 d}+\frac {\left (16 a +3 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5}}\)	\(204\)

[In]

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

[Out]

1/480*(180*b*x*d+6*a*sin(5*d*x+5*c)+15*b*sin(4*d*x+4*c)+50*a*sin(3*d*x+3*c)+120*b*sin(2*d*x+2*c)+300*a*sin(d*x
+c))/d

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.70 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x)) \, dx=\frac {45 \, b d x + {\left (24 \, a \cos \left (d x + c\right )^{4} + 30 \, b \cos \left (d x + c\right )^{3} + 32 \, a \cos \left (d x + c\right )^{2} + 45 \, b \cos \left (d x + c\right ) + 64 \, a\right )} \sin \left (d x + c\right )}{120 \, d} \]

[In]

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

[Out]

1/120*(45*b*d*x + (24*a*cos(d*x + c)^4 + 30*b*cos(d*x + c)^3 + 32*a*cos(d*x + c)^2 + 45*b*cos(d*x + c) + 64*a)
*sin(d*x + c))/d

Sympy [F]

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

[In]

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

[Out]

Integral((a + b*sec(c + d*x))*cos(c + d*x)**5, x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.75 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x)) \, dx=\frac {32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} a + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} b}{480 \, d} \]

[In]

integrate(cos(d*x+c)^5*(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 + 15*(12*d*x + 12*c + sin(4*d*x + 4*c) +
8*sin(2*d*x + 2*c))*b)/d

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.67 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x)) \, dx=\frac {45 \, {\left (d x + c\right )} b + \frac {2 \, {\left (120 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 75 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 160 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 30 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 464 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 160 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 30 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 75 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5}}}{120 \, d} \]

[In]

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

[Out]

1/120*(45*(d*x + c)*b + 2*(120*a*tan(1/2*d*x + 1/2*c)^9 - 75*b*tan(1/2*d*x + 1/2*c)^9 + 160*a*tan(1/2*d*x + 1/
2*c)^7 - 30*b*tan(1/2*d*x + 1/2*c)^7 + 464*a*tan(1/2*d*x + 1/2*c)^5 + 160*a*tan(1/2*d*x + 1/2*c)^3 + 30*b*tan(
1/2*d*x + 1/2*c)^3 + 120*a*tan(1/2*d*x + 1/2*c) + 75*b*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 = 17.84 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.23 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x)) \, dx=\frac {3\,b\,x}{8}+\frac {\left (2\,a-\frac {5\,b}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {8\,a}{3}-\frac {b}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\frac {116\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{15}+\left (\frac {8\,a}{3}+\frac {b}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,a+\frac {5\,b}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^5} \]

[In]

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

[Out]

(3*b*x)/8 + (tan(c/2 + (d*x)/2)*(2*a + (5*b)/4) + tan(c/2 + (d*x)/2)^3*((8*a)/3 + b/2) + tan(c/2 + (d*x)/2)^9*
(2*a - (5*b)/4) + tan(c/2 + (d*x)/2)^7*((8*a)/3 - b/2) + (116*a*tan(c/2 + (d*x)/2)^5)/15)/(d*(tan(c/2 + (d*x)/
2)^2 + 1)^5)