\(\int \cos (c+d x) \sin (c+d x) (a+a \sin (c+d x))^4 \, dx\) [220]

Optimal result

Integrand size = 25, antiderivative size = 45 \[ \int \cos (c+d x) \sin (c+d x) (a+a \sin (c+d x))^4 \, dx=-\frac {(a+a \sin (c+d x))^5}{5 a d}+\frac {(a+a \sin (c+d x))^6}{6 a^2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2912, 12, 45} \[ \int \cos (c+d x) \sin (c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {(a \sin (c+d x)+a)^6}{6 a^2 d}-\frac {(a \sin (c+d x)+a)^5}{5 a d} \]

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 2912

Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)
])^(n_.), x_Symbol] :> Dist[1/(b*f), Subst[Int[(a + x)^m*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[
{a, b, c, d, e, f, m, n}, x]

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.67 \[ \int \cos (c+d x) \sin (c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {a^4 (1+\sin (c+d x))^5 (-1+5 \sin (c+d x))}{30 d} \]

[In]

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

[Out]

(a^4*(1 + Sin[c + d*x])^5*(-1 + 5*Sin[c + d*x]))/(30*d)

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.58

method	result	size
derivativedivides	\(\frac {\frac {a^{4} \left (\sin ^{6}\left (d x +c \right )\right )}{6}+\frac {4 a^{4} \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {3 a^{4} \left (\sin ^{4}\left (d x +c \right )\right )}{2}+\frac {4 a^{4} \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {a^{4} \left (\sin ^{2}\left (d x +c \right )\right )}{2}}{d}\)	\(71\)
default	\(\frac {\frac {a^{4} \left (\sin ^{6}\left (d x +c \right )\right )}{6}+\frac {4 a^{4} \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {3 a^{4} \left (\sin ^{4}\left (d x +c \right )\right )}{2}+\frac {4 a^{4} \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {a^{4} \left (\sin ^{2}\left (d x +c \right )\right )}{2}}{d}\)	\(71\)
parallelrisch	\(\frac {a^{4} \left (-560 \sin \left (3 d x +3 c \right )+1440 \sin \left (d x +c \right )-1035 \cos \left (2 d x +2 c \right )-5 \cos \left (6 d x +6 c \right )+210 \cos \left (4 d x +4 c \right )+48 \sin \left (5 d x +5 c \right )+830\right )}{960 d}\)	\(74\)
risch	\(\frac {3 a^{4} \sin \left (d x +c \right )}{2 d}-\frac {a^{4} \cos \left (6 d x +6 c \right )}{192 d}+\frac {a^{4} \sin \left (5 d x +5 c \right )}{20 d}+\frac {7 a^{4} \cos \left (4 d x +4 c \right )}{32 d}-\frac {7 a^{4} \sin \left (3 d x +3 c \right )}{12 d}-\frac {69 a^{4} \cos \left (2 d x +2 c \right )}{64 d}\)	\(101\)
norman	\(\frac {\frac {32 a^{4} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {32 a^{4} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {32 a^{4} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {288 a^{4} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}+\frac {288 a^{4} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}+\frac {32 a^{4} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 a^{4} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 a^{4} \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {212 a^{4} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}\)	\(189\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 85 vs. \(2 (41) = 82\).

Time = 0.26 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.89 \[ \int \cos (c+d x) \sin (c+d x) (a+a \sin (c+d x))^4 \, dx=-\frac {5 \, a^{4} \cos \left (d x + c\right )^{6} - 60 \, a^{4} \cos \left (d x + c\right )^{4} + 120 \, a^{4} \cos \left (d x + c\right )^{2} - 8 \, {\left (3 \, a^{4} \cos \left (d x + c\right )^{4} - 11 \, a^{4} \cos \left (d x + c\right )^{2} + 8 \, a^{4}\right )} \sin \left (d x + c\right )}{30 \, d} \]

[In]

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

[Out]

-1/30*(5*a^4*cos(d*x + c)^6 - 60*a^4*cos(d*x + c)^4 + 120*a^4*cos(d*x + c)^2 - 8*(3*a^4*cos(d*x + c)^4 - 11*a^
4*cos(d*x + c)^2 + 8*a^4)*sin(d*x + c))/d

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 97 vs. \(2 (34) = 68\).

Time = 0.32 (sec) , antiderivative size = 97, normalized size of antiderivative = 2.16 \[ \int \cos (c+d x) \sin (c+d x) (a+a \sin (c+d x))^4 \, dx=\begin {cases} \frac {a^{4} \sin ^{6}{\left (c + d x \right )}}{6 d} + \frac {4 a^{4} \sin ^{5}{\left (c + d x \right )}}{5 d} + \frac {3 a^{4} \sin ^{4}{\left (c + d x \right )}}{2 d} + \frac {4 a^{4} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {a^{4} \sin ^{2}{\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right )^{4} \sin {\left (c \right )} \cos {\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((a**4*sin(c + d*x)**6/(6*d) + 4*a**4*sin(c + d*x)**5/(5*d) + 3*a**4*sin(c + d*x)**4/(2*d) + 4*a**4*s
in(c + d*x)**3/(3*d) + a**4*sin(c + d*x)**2/(2*d), Ne(d, 0)), (x*(a*sin(c) + a)**4*sin(c)*cos(c), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.58 \[ \int \cos (c+d x) \sin (c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {5 \, a^{4} \sin \left (d x + c\right )^{6} + 24 \, a^{4} \sin \left (d x + c\right )^{5} + 45 \, a^{4} \sin \left (d x + c\right )^{4} + 40 \, a^{4} \sin \left (d x + c\right )^{3} + 15 \, a^{4} \sin \left (d x + c\right )^{2}}{30 \, d} \]

[In]

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

[Out]

1/30*(5*a^4*sin(d*x + c)^6 + 24*a^4*sin(d*x + c)^5 + 45*a^4*sin(d*x + c)^4 + 40*a^4*sin(d*x + c)^3 + 15*a^4*si
n(d*x + c)^2)/d

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.58 \[ \int \cos (c+d x) \sin (c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {5 \, a^{4} \sin \left (d x + c\right )^{6} + 24 \, a^{4} \sin \left (d x + c\right )^{5} + 45 \, a^{4} \sin \left (d x + c\right )^{4} + 40 \, a^{4} \sin \left (d x + c\right )^{3} + 15 \, a^{4} \sin \left (d x + c\right )^{2}}{30 \, d} \]

[In]

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

[Out]

1/30*(5*a^4*sin(d*x + c)^6 + 24*a^4*sin(d*x + c)^5 + 45*a^4*sin(d*x + c)^4 + 40*a^4*sin(d*x + c)^3 + 15*a^4*si
n(d*x + c)^2)/d

Mupad [B] (verification not implemented)

Time = 9.07 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.56 \[ \int \cos (c+d x) \sin (c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {\frac {a^4\,{\sin \left (c+d\,x\right )}^6}{6}+\frac {4\,a^4\,{\sin \left (c+d\,x\right )}^5}{5}+\frac {3\,a^4\,{\sin \left (c+d\,x\right )}^4}{2}+\frac {4\,a^4\,{\sin \left (c+d\,x\right )}^3}{3}+\frac {a^4\,{\sin \left (c+d\,x\right )}^2}{2}}{d} \]

[In]

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

[Out]

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