\(\int \cos ^5(c+d x) (a+a \sin (c+d x)) \, dx\) [3]

Optimal result

Integrand size = 19, antiderivative size = 64 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {(a+a \sin (c+d x))^4}{a^3 d}-\frac {4 (a+a \sin (c+d x))^5}{5 a^4 d}+\frac {(a+a \sin (c+d x))^6}{6 a^5 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2746, 45} \[ \int \cos ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {(a \sin (c+d x)+a)^6}{6 a^5 d}-\frac {4 (a \sin (c+d x)+a)^5}{5 a^4 d}+\frac {(a \sin (c+d x)+a)^4}{a^3 d} \]

[In]

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

[Out]

(a + a*Sin[c + d*x])^4/(a^3*d) - (4*(a + a*Sin[c + d*x])^5)/(5*a^4*d) + (a + a*Sin[c + d*x])^6/(6*a^5*d)

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2746

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 + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int (a-x)^2 (a+x)^3 \, dx,x,a \sin (c+d x)\right )}{a^5 d} \\ & = \frac {\text {Subst}\left (\int \left (4 a^2 (a+x)^3-4 a (a+x)^4+(a+x)^5\right ) \, dx,x,a \sin (c+d x)\right )}{a^5 d} \\ & = \frac {(a+a \sin (c+d x))^4}{a^3 d}-\frac {4 (a+a \sin (c+d x))^5}{5 a^4 d}+\frac {(a+a \sin (c+d x))^6}{6 a^5 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.94 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \cos ^6(c+d x)}{6 d}+\frac {a \sin (c+d x)}{d}-\frac {2 a \sin ^3(c+d x)}{3 d}+\frac {a \sin ^5(c+d x)}{5 d} \]

[In]

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

[Out]

-1/6*(a*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 = 0.29 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.98

method	result	size
derivativedivides	\(\frac {a \left (\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{6}+\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{2}-\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}+\sin \left (d x +c \right )\right )}{d}\)	\(63\)
default	\(\frac {a \left (\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{6}+\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{2}-\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}+\sin \left (d x +c \right )\right )}{d}\)	\(63\)
risch	\(\frac {5 a \sin \left (d x +c \right )}{8 d}-\frac {a \cos \left (6 d x +6 c \right )}{192 d}+\frac {a \sin \left (5 d x +5 c \right )}{80 d}-\frac {a \cos \left (4 d x +4 c \right )}{32 d}+\frac {5 a \sin \left (3 d x +3 c \right )}{48 d}-\frac {5 a \cos \left (2 d x +2 c \right )}{64 d}\)	\(89\)
parallelrisch	\(\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (15 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+15 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+35 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+78 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+50 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+78 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+35 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+15\right )}{15 d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}\)	\(135\)
norman	\(\frac {\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {14 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {52 a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}+\frac {52 a \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}+\frac {14 a \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 a \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {20 a \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 a \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}\)	\(169\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.80 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {5 \, a \cos \left (d x + c\right )^{6} - 2 \, {\left (3 \, a \cos \left (d x + c\right )^{4} + 4 \, a \cos \left (d x + c\right )^{2} + 8 \, a\right )} \sin \left (d x + c\right )}{30 \, d} \]

[In]

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

[Out]

-1/30*(5*a*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 = 0.36 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.30 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x)) \, dx=\begin {cases} \frac {8 a \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 a \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {a \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} - \frac {a \cos ^{6}{\left (c + d x \right )}}{6 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right ) \cos ^{5}{\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)**5*(a+a*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 - a*cos(c + d*x)**6/(6*d), Ne(d, 0)), (x*(a*sin(c) + a)*cos(c)**5, True))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.09 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {5 \, a \sin \left (d x + c\right )^{6} + 6 \, a \sin \left (d x + c\right )^{5} - 15 \, a \sin \left (d x + c\right )^{4} - 20 \, a \sin \left (d x + c\right )^{3} + 15 \, a \sin \left (d x + c\right )^{2} + 30 \, a \sin \left (d x + c\right )}{30 \, d} \]

[In]

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

[Out]

1/30*(5*a*sin(d*x + c)^6 + 6*a*sin(d*x + c)^5 - 15*a*sin(d*x + c)^4 - 20*a*sin(d*x + c)^3 + 15*a*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 = 1.38 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \cos \left (6 \, d x + 6 \, c\right )}{192 \, d} - \frac {a \cos \left (4 \, d x + 4 \, c\right )}{32 \, d} - \frac {5 \, a \cos \left (2 \, d x + 2 \, c\right )}{64 \, d} + \frac {a \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {5 \, a \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {5 \, a \sin \left (d x + c\right )}{8 \, d} \]

[In]

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

[Out]

-1/192*a*cos(6*d*x + 6*c)/d - 1/32*a*cos(4*d*x + 4*c)/d - 5/64*a*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 = 0.06 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.06 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {\frac {a\,{\sin \left (c+d\,x\right )}^6}{6}+\frac {a\,{\sin \left (c+d\,x\right )}^5}{5}-\frac {a\,{\sin \left (c+d\,x\right )}^4}{2}-\frac {2\,a\,{\sin \left (c+d\,x\right )}^3}{3}+\frac {a\,{\sin \left (c+d\,x\right )}^2}{2}+a\,\sin \left (c+d\,x\right )}{d} \]

[In]

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

[Out]

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