\(\int \cos ^5(c+d x) (a+b \sin (c+d x)) \, dx\) [376]

Optimal result

Integrand size = 19, antiderivative size = 60 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {b \cos ^6(c+d x)}{6 d}+\frac {a \sin (c+d x)}{d}-\frac {2 a \sin ^3(c+d x)}{3 d}+\frac {a \sin ^5(c+d x)}{5 d} \]

[Out]

-1/6*b*cos(d*x+c)^6/d+a*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 = 0.03 (sec) , antiderivative size = 60, 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 = {2747, 655, 200} \[ \int \cos ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {a \sin ^5(c+d x)}{5 d}-\frac {2 a \sin ^3(c+d x)}{3 d}+\frac {a \sin (c+d x)}{d}-\frac {b \cos ^6(c+d x)}{6 d} \]

[In]

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

[Out]

-1/6*(b*Cos[c + d*x]^6)/d + (a*Sin[c + d*x])/d - (2*a*Sin[c + d*x]^3)/(3*d) + (a*Sin[c + d*x]^5)/(5*d)

Rule 200

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

Rule 655

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[e*((a + c*x^2)^(p + 1)/(2*c*(p + 1))),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

Rule 2747

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int (a+x) \left (b^2-x^2\right )^2 \, dx,x,b \sin (c+d x)\right )}{b^5 d} \\ & = -\frac {b \cos ^6(c+d x)}{6 d}+\frac {a \text {Subst}\left (\int \left (b^2-x^2\right )^2 \, dx,x,b \sin (c+d x)\right )}{b^5 d} \\ & = -\frac {b \cos ^6(c+d x)}{6 d}+\frac {a \text {Subst}\left (\int \left (b^4-2 b^2 x^2+x^4\right ) \, dx,x,b \sin (c+d x)\right )}{b^5 d} \\ & = -\frac {b \cos ^6(c+d x)}{6 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} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {b \cos ^6(c+d x)}{6 d}+\frac {a \sin (c+d x)}{d}-\frac {2 a \sin ^3(c+d x)}{3 d}+\frac {a \sin ^5(c+d x)}{5 d} \]

[In]

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

[Out]

-1/6*(b*Cos[c + d*x]^6)/d + (a*Sin[c + d*x])/d - (2*a*Sin[c + d*x]^3)/(3*d) + (a*Sin[c + d*x]^5)/(5*d)

Maple [A] (verified)

Time = 0.55 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.15

method	result	size
derivativedivides	\(\frac {\frac {b \left (\sin ^{6}\left (d x +c \right )\right )}{6}+\frac {a \left (\sin ^{5}\left (d x +c \right )\right )}{5}-\frac {b \left (\sin ^{4}\left (d x +c \right )\right )}{2}-\frac {2 a \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (\sin ^{2}\left (d x +c \right )\right ) b}{2}+a \sin \left (d x +c \right )}{d}\)	\(69\)
default	\(\frac {\frac {b \left (\sin ^{6}\left (d x +c \right )\right )}{6}+\frac {a \left (\sin ^{5}\left (d x +c \right )\right )}{5}-\frac {b \left (\sin ^{4}\left (d x +c \right )\right )}{2}-\frac {2 a \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (\sin ^{2}\left (d x +c \right )\right ) b}{2}+a \sin \left (d x +c \right )}{d}\)	\(69\)
risch	\(\frac {5 a \sin \left (d x +c \right )}{8 d}-\frac {b \cos \left (6 d x +6 c \right )}{192 d}+\frac {a \sin \left (5 d x +5 c \right )}{80 d}-\frac {b \cos \left (4 d x +4 c \right )}{32 d}+\frac {5 a \sin \left (3 d x +3 c \right )}{48 d}-\frac {5 b \cos \left (2 d x +2 c \right )}{64 d}\)	\(89\)
parallelrisch	\(\frac {2 \left (a \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +\frac {7 a \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\frac {26 a \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {10 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b}{3}+\frac {26 a \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {7 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}\)	\(139\)
norman	\(\frac {\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {14 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {52 a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}+\frac {52 a \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}+\frac {14 a \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 a \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {20 b \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 b \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}\)	\(169\)

[In]

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

[Out]

1/d*(1/6*b*sin(d*x+c)^6+1/5*a*sin(d*x+c)^5-1/2*b*sin(d*x+c)^4-2/3*a*sin(d*x+c)^3+1/2*sin(d*x+c)^2*b+a*sin(d*x+
c))

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.85 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {5 \, b \cos \left (d x + c\right )^{6} - 2 \, {\left (3 \, a \cos \left (d x + c\right )^{4} + 4 \, a \cos \left (d x + c\right )^{2} + 8 \, a\right )} \sin \left (d x + c\right )}{30 \, d} \]

[In]

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

[Out]

-1/30*(5*b*cos(d*x + c)^6 - 2*(3*a*cos(d*x + c)^4 + 4*a*cos(d*x + c)^2 + 8*a)*sin(d*x + c))/d

Sympy [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.38 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x)) \, dx=\begin {cases} \frac {8 a \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 a \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {a \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} - \frac {b \cos ^{6}{\left (c + d x \right )}}{6 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin {\left (c \right )}\right ) \cos ^{5}{\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((8*a*sin(c + d*x)**5/(15*d) + 4*a*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) + a*sin(c + d*x)*cos(c + d*x
)**4/d - b*cos(c + d*x)**6/(6*d), Ne(d, 0)), (x*(a + b*sin(c))*cos(c)**5, True))

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.17 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {5 \, b \sin \left (d x + c\right )^{6} + 6 \, a \sin \left (d x + c\right )^{5} - 15 \, b \sin \left (d x + c\right )^{4} - 20 \, a \sin \left (d x + c\right )^{3} + 15 \, b \sin \left (d x + c\right )^{2} + 30 \, a \sin \left (d x + c\right )}{30 \, d} \]

[In]

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

[Out]

1/30*(5*b*sin(d*x + c)^6 + 6*a*sin(d*x + c)^5 - 15*b*sin(d*x + c)^4 - 20*a*sin(d*x + c)^3 + 15*b*sin(d*x + c)^
2 + 30*a*sin(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.47 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {b \cos \left (6 \, d x + 6 \, c\right )}{192 \, d} - \frac {b \cos \left (4 \, d x + 4 \, c\right )}{32 \, d} - \frac {5 \, b \cos \left (2 \, d x + 2 \, c\right )}{64 \, d} + \frac {a \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {5 \, a \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {5 \, a \sin \left (d x + c\right )}{8 \, d} \]

[In]

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

[Out]

-1/192*b*cos(6*d*x + 6*c)/d - 1/32*b*cos(4*d*x + 4*c)/d - 5/64*b*cos(2*d*x + 2*c)/d + 1/80*a*sin(5*d*x + 5*c)/
d + 5/48*a*sin(3*d*x + 3*c)/d + 5/8*a*sin(d*x + c)/d

Mupad [B] (verification not implemented)

Time = 4.47 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.13 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {\frac {b\,{\sin \left (c+d\,x\right )}^6}{6}+\frac {a\,{\sin \left (c+d\,x\right )}^5}{5}-\frac {b\,{\sin \left (c+d\,x\right )}^4}{2}-\frac {2\,a\,{\sin \left (c+d\,x\right )}^3}{3}+\frac {b\,{\sin \left (c+d\,x\right )}^2}{2}+a\,\sin \left (c+d\,x\right )}{d} \]

[In]

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

[Out]

(a*sin(c + d*x) - (2*a*sin(c + d*x)^3)/3 + (a*sin(c + d*x)^5)/5 + (b*sin(c + d*x)^2)/2 - (b*sin(c + d*x)^4)/2
+ (b*sin(c + d*x)^6)/6)/d