\(\int \cos ^3(c+d x) \sin (c+d x) (a+a \sin (c+d x)) \, dx\) [352]

Optimal result

Integrand size = 25, antiderivative size = 49 \[ \int \cos ^3(c+d x) \sin (c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \cos ^4(c+d x)}{4 d}+\frac {a \sin ^3(c+d x)}{3 d}-\frac {a \sin ^5(c+d x)}{5 d} \]

[Out]

-1/4*a*cos(d*x+c)^4/d+1/3*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 = 49, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2913, 2645, 30, 2644, 14} \[ \int \cos ^3(c+d x) \sin (c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \sin ^5(c+d x)}{5 d}+\frac {a \sin ^3(c+d x)}{3 d}-\frac {a \cos ^4(c+d x)}{4 d} \]

[In]

Int[Cos[c + d*x]^3*Sin[c + d*x]*(a + a*Sin[c + d*x]),x]

[Out]

-1/4*(a*Cos[c + d*x]^4)/d + (a*Sin[c + d*x]^3)/(3*d) - (a*Sin[c + d*x]^5)/(5*d)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2644

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 2645

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[-(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 2913

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]),
x_Symbol] :> Dist[a, Int[Cos[e + f*x]^p*(d*Sin[e + f*x])^n, x], x] + Dist[b/d, Int[Cos[e + f*x]^p*(d*Sin[e +
f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n, p}, x] && IntegerQ[(p - 1)/2] && IntegerQ[n] && ((LtQ[p, 0]
&& NeQ[a^2 - b^2, 0]) || LtQ[0, n, p - 1] || LtQ[p + 1, -n, 2*p + 1])

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.18 \[ \int \cos ^3(c+d x) \sin (c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a (45+60 \cos (2 (c+d x))+15 \cos (4 (c+d x))-60 \sin (c+d x)+10 \sin (3 (c+d x))+6 \sin (5 (c+d x)))}{480 d} \]

[In]

Integrate[Cos[c + d*x]^3*Sin[c + d*x]*(a + a*Sin[c + d*x]),x]

[Out]

-1/480*(a*(45 + 60*Cos[2*(c + d*x)] + 15*Cos[4*(c + d*x)] - 60*Sin[c + d*x] + 10*Sin[3*(c + d*x)] + 6*Sin[5*(c
+ d*x)]))/d

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.98

method	result	size
derivativedivides	\(-\frac {a \left (\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}\right )}{d}\)	\(48\)
default	\(-\frac {a \left (\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}\right )}{d}\)	\(48\)
parallelrisch	\(-\frac {a \left (10 \sin \left (3 d x +3 c \right )-60 \sin \left (d x +c \right )+6 \sin \left (5 d x +5 c \right )+15 \cos \left (4 d x +4 c \right )-75+60 \cos \left (2 d x +2 c \right )\right )}{480 d}\)	\(61\)
risch	\(\frac {a \sin \left (d x +c \right )}{8 d}-\frac {a \sin \left (5 d x +5 c \right )}{80 d}-\frac {a \cos \left (4 d x +4 c \right )}{32 d}-\frac {a \sin \left (3 d x +3 c \right )}{48 d}-\frac {a \cos \left (2 d x +2 c \right )}{8 d}\)	\(74\)
norman	\(\frac {\frac {2 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 a \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {8 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {16 a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}+\frac {8 a \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 a \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 a \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}\)	\(137\)

[In]

int(cos(d*x+c)^3*sin(d*x+c)*(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.04 \[ \int \cos ^3(c+d x) \sin (c+d x) (a+a \sin (c+d x)) \, dx=-\frac {15 \, a \cos \left (d x + c\right )^{4} + 4 \, {\left (3 \, a \cos \left (d x + c\right )^{4} - a \cos \left (d x + c\right )^{2} - 2 \, a\right )} \sin \left (d x + c\right )}{60 \, d} \]

[In]

integrate(cos(d*x+c)^3*sin(d*x+c)*(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/60*(15*a*cos(d*x + c)^4 + 4*(3*a*cos(d*x + c)^4 - a*cos(d*x + c)^2 - 2*a)*sin(d*x + c))/d

Sympy [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.35 \[ \int \cos ^3(c+d x) \sin (c+d x) (a+a \sin (c+d x)) \, dx=\begin {cases} \frac {2 a \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {a \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} - \frac {a \cos ^{4}{\left (c + d x \right )}}{4 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right ) \sin {\left (c \right )} \cos ^{3}{\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)**3*sin(d*x+c)*(a+a*sin(d*x+c)),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.02 \[ \int \cos ^3(c+d x) \sin (c+d x) (a+a \sin (c+d x)) \, dx=-\frac {12 \, a \sin \left (d x + c\right )^{5} + 15 \, a \sin \left (d x + c\right )^{4} - 20 \, a \sin \left (d x + c\right )^{3} - 30 \, a \sin \left (d x + c\right )^{2}}{60 \, d} \]

[In]

integrate(cos(d*x+c)^3*sin(d*x+c)*(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/60*(12*a*sin(d*x + c)^5 + 15*a*sin(d*x + c)^4 - 20*a*sin(d*x + c)^3 - 30*a*sin(d*x + c)^2)/d

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.02 \[ \int \cos ^3(c+d x) \sin (c+d x) (a+a \sin (c+d x)) \, dx=-\frac {12 \, a \sin \left (d x + c\right )^{5} + 15 \, a \sin \left (d x + c\right )^{4} - 20 \, a \sin \left (d x + c\right )^{3} - 30 \, a \sin \left (d x + c\right )^{2}}{60 \, d} \]

[In]

integrate(cos(d*x+c)^3*sin(d*x+c)*(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/60*(12*a*sin(d*x + c)^5 + 15*a*sin(d*x + c)^4 - 20*a*sin(d*x + c)^3 - 30*a*sin(d*x + c)^2)/d

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00 \[ \int \cos ^3(c+d x) \sin (c+d x) (a+a \sin (c+d x)) \, dx=\frac {-\frac {a\,{\sin \left (c+d\,x\right )}^5}{5}-\frac {a\,{\sin \left (c+d\,x\right )}^4}{4}+\frac {a\,{\sin \left (c+d\,x\right )}^3}{3}+\frac {a\,{\sin \left (c+d\,x\right )}^2}{2}}{d} \]

[In]

int(cos(c + d*x)^3*sin(c + d*x)*(a + a*sin(c + d*x)),x)

[Out]

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