\(\int \cos ^4(c+d x) (a+b \sin (c+d x)) \, dx\) [382]

Optimal result

Integrand size = 19, antiderivative size = 65 \[ \int \cos ^4(c+d x) (a+b \sin (c+d x)) \, dx=\frac {3 a x}{8}-\frac {b \cos ^5(c+d x)}{5 d}+\frac {3 a \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a \cos ^3(c+d x) \sin (c+d x)}{4 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2748, 2715, 8} \[ \int \cos ^4(c+d x) (a+b \sin (c+d x)) \, dx=\frac {a \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3 a \sin (c+d x) \cos (c+d x)}{8 d}+\frac {3 a x}{8}-\frac {b \cos ^5(c+d x)}{5 d} \]

[In]

Int[Cos[c + d*x]^4*(a + b*Sin[c + d*x]),x]

[Out]

(3*a*x)/8 - (b*Cos[c + d*x]^5)/(5*d) + (3*a*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a*Cos[c + d*x]^3*Sin[c + d*x])
/(4*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 2748

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-b)*((g*Co
s[e + f*x])^(p + 1)/(f*g*(p + 1))), x] + Dist[a, Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x
] && (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.95 \[ \int \cos ^4(c+d x) (a+b \sin (c+d x)) \, dx=\frac {3 a (c+d x)}{8 d}-\frac {b \cos ^5(c+d x)}{5 d}+\frac {a \sin (2 (c+d x))}{4 d}+\frac {a \sin (4 (c+d x))}{32 d} \]

[In]

Integrate[Cos[c + d*x]^4*(a + b*Sin[c + d*x]),x]

[Out]

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

Maple [A] (verified)

Time = 1.22 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.80

method	result	size
derivativedivides	\(\frac {-\frac {b \left (\cos ^{5}\left (d x +c \right )\right )}{5}+a \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\)	\(52\)
default	\(\frac {-\frac {b \left (\cos ^{5}\left (d x +c \right )\right )}{5}+a \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\)	\(52\)
parallelrisch	\(\frac {60 a x d -10 b \cos \left (3 d x +3 c \right )-20 \cos \left (d x +c \right ) b -2 b \cos \left (5 d x +5 c \right )+5 a \sin \left (4 d x +4 c \right )+40 a \sin \left (2 d x +2 c \right )-32 b}{160 d}\)	\(72\)
risch	\(\frac {3 a x}{8}-\frac {b \cos \left (d x +c \right )}{8 d}-\frac {b \cos \left (5 d x +5 c \right )}{80 d}+\frac {a \sin \left (4 d x +4 c \right )}{32 d}-\frac {b \cos \left (3 d x +3 c \right )}{16 d}+\frac {a \sin \left (2 d x +2 c \right )}{4 d}\)	\(78\)
norman	\(\frac {\frac {3 a x}{8}-\frac {2 b}{5 d}+\frac {5 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {a \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {5 a \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {15 a x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {15 a x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {15 a x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {15 a x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {3 a x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}-\frac {4 b \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 b \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}\)	\(203\)

[In]

int(cos(d*x+c)^4*(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.78 \[ \int \cos ^4(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {8 \, b \cos \left (d x + c\right )^{5} - 15 \, a d x - 5 \, {\left (2 \, a \cos \left (d x + c\right )^{3} + 3 \, a \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{40 \, d} \]

[In]

integrate(cos(d*x+c)^4*(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/40*(8*b*cos(d*x + c)^5 - 15*a*d*x - 5*(2*a*cos(d*x + c)^3 + 3*a*cos(d*x + c))*sin(d*x + c))/d

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 124 vs. \(2 (60) = 120\).

Time = 0.22 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.91 \[ \int \cos ^4(c+d x) (a+b \sin (c+d x)) \, dx=\begin {cases} \frac {3 a x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 a x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 a x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 a \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {5 a \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} - \frac {b \cos ^{5}{\left (c + d x \right )}}{5 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin {\left (c \right )}\right ) \cos ^{4}{\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)**4*(a+b*sin(d*x+c)),x)

[Out]

Piecewise((3*a*x*sin(c + d*x)**4/8 + 3*a*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + 3*a*x*cos(c + d*x)**4/8 + 3*a*s
in(c + d*x)**3*cos(c + d*x)/(8*d) + 5*a*sin(c + d*x)*cos(c + d*x)**3/(8*d) - b*cos(c + d*x)**5/(5*d), Ne(d, 0)
), (x*(a + b*sin(c))*cos(c)**4, True))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.74 \[ \int \cos ^4(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {32 \, b \cos \left (d x + c\right )^{5} - 5 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a}{160 \, d} \]

[In]

integrate(cos(d*x+c)^4*(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/160*(32*b*cos(d*x + c)^5 - 5*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*a)/d

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.18 \[ \int \cos ^4(c+d x) (a+b \sin (c+d x)) \, dx=\frac {3}{8} \, a x - \frac {b \cos \left (5 \, d x + 5 \, c\right )}{80 \, d} - \frac {b \cos \left (3 \, d x + 3 \, c\right )}{16 \, d} - \frac {b \cos \left (d x + c\right )}{8 \, d} + \frac {a \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {a \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} \]

[In]

integrate(cos(d*x+c)^4*(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

3/8*a*x - 1/80*b*cos(5*d*x + 5*c)/d - 1/16*b*cos(3*d*x + 3*c)/d - 1/8*b*cos(d*x + c)/d + 1/32*a*sin(4*d*x + 4*
c)/d + 1/4*a*sin(2*d*x + 2*c)/d

Mupad [B] (verification not implemented)

Time = 8.38 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.71 \[ \int \cos ^4(c+d x) (a+b \sin (c+d x)) \, dx=\frac {3\,a\,x}{8}-\frac {\frac {5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{4}+2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{2}+4\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2}-\frac {5\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}+\frac {2\,b}{5}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^5} \]

[In]

int(cos(c + d*x)^4*(a + b*sin(c + d*x)),x)

[Out]

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