\(\int \frac {\cos ^5(c+d x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx\) [535]

Optimal result

Integrand size = 27, antiderivative size = 55 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\cos ^4(c+d x)}{4 a d}-\frac {\sin ^3(c+d x)}{3 a d}+\frac {\sin ^5(c+d x)}{5 a d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2914, 2645, 30, 2644, 14} \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\sin ^5(c+d x)}{5 a d}-\frac {\sin ^3(c+d x)}{3 a d}-\frac {\cos ^4(c+d x)}{4 a d} \]

[In]

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

[Out]

-1/4*Cos[c + d*x]^4/(a*d) - Sin[c + d*x]^3/(3*a*d) + Sin[c + d*x]^5/(5*a*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 2914

Int[(cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]
), x_Symbol] :> Dist[1/a, Int[Cos[e + f*x]^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[1/(b*d), Int[Cos[e + f*x]
^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n, p}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2
- b^2, 0] && IntegerQ[n] && (LtQ[0, n, (p + 1)/2] || (LeQ[p, -n] && LtQ[-n, 2*p - 3]) || (GtQ[n, 0] && LeQ[n,
-p]))

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.87 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\sin ^2(c+d x) \left (30-20 \sin (c+d x)-15 \sin ^2(c+d x)+12 \sin ^3(c+d x)\right )}{60 a d} \]

[In]

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

[Out]

(Sin[c + d*x]^2*(30 - 20*Sin[c + d*x] - 15*Sin[c + d*x]^2 + 12*Sin[c + d*x]^3))/(60*a*d)

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.89

method	result	size
derivativedivides	\(\frac {\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}}{d a}\)	\(49\)
default	\(\frac {\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}}{d a}\)	\(49\)
parallelrisch	\(\frac {-15 \cos \left (4 d x +4 c \right )+75-60 \cos \left (2 d x +2 c \right )+10 \sin \left (3 d x +3 c \right )-60 \sin \left (d x +c \right )+6 \sin \left (5 d x +5 c \right )}{480 d a}\)	\(63\)
risch	\(-\frac {\sin \left (d x +c \right )}{8 a d}+\frac {\sin \left (5 d x +5 c \right )}{80 d a}-\frac {\cos \left (4 d x +4 c \right )}{32 a d}+\frac {\sin \left (3 d x +3 c \right )}{48 d a}-\frac {\cos \left (2 d x +2 c \right )}{8 a d}\)	\(84\)
norman	\(\frac {\frac {2 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {2 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}-\frac {2 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {12 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d a}+\frac {12 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d a}+\frac {12 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d a}+\frac {12 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d a}+\frac {4 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {4 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}\)	\(221\)

[In]

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

[Out]

1/d/a*(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 = 0.27 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.89 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {15 \, \cos \left (d x + c\right )^{4} - 4 \, {\left (3 \, \cos \left (d x + c\right )^{4} - \cos \left (d x + c\right )^{2} - 2\right )} \sin \left (d x + c\right )}{60 \, a d} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 741 vs. \(2 (39) = 78\).

Time = 11.43 (sec) , antiderivative size = 741, normalized size of antiderivative = 13.47 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx=\begin {cases} \frac {30 \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{15 a d \tan ^{10}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 75 a d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 150 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 150 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 75 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 15 a d} - \frac {40 \tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{15 a d \tan ^{10}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 75 a d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 150 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 150 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 75 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 15 a d} + \frac {30 \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{15 a d \tan ^{10}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 75 a d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 150 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 150 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 75 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 15 a d} + \frac {16 \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{15 a d \tan ^{10}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 75 a d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 150 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 150 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 75 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 15 a d} + \frac {30 \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{15 a d \tan ^{10}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 75 a d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 150 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 150 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 75 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 15 a d} - \frac {40 \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{15 a d \tan ^{10}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 75 a d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 150 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 150 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 75 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 15 a d} + \frac {30 \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{15 a d \tan ^{10}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 75 a d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 150 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 150 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 75 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 15 a d} & \text {for}\: d \neq 0 \\\frac {x \sin {\left (c \right )} \cos ^{5}{\left (c \right )}}{a \sin {\left (c \right )} + a} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((30*tan(c/2 + d*x/2)**8/(15*a*d*tan(c/2 + d*x/2)**10 + 75*a*d*tan(c/2 + d*x/2)**8 + 150*a*d*tan(c/2
+ d*x/2)**6 + 150*a*d*tan(c/2 + d*x/2)**4 + 75*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) - 40*tan(c/2 + d*x/2)**7/(15*
a*d*tan(c/2 + d*x/2)**10 + 75*a*d*tan(c/2 + d*x/2)**8 + 150*a*d*tan(c/2 + d*x/2)**6 + 150*a*d*tan(c/2 + d*x/2)
**4 + 75*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) + 30*tan(c/2 + d*x/2)**6/(15*a*d*tan(c/2 + d*x/2)**10 + 75*a*d*tan(
c/2 + d*x/2)**8 + 150*a*d*tan(c/2 + d*x/2)**6 + 150*a*d*tan(c/2 + d*x/2)**4 + 75*a*d*tan(c/2 + d*x/2)**2 + 15*
a*d) + 16*tan(c/2 + d*x/2)**5/(15*a*d*tan(c/2 + d*x/2)**10 + 75*a*d*tan(c/2 + d*x/2)**8 + 150*a*d*tan(c/2 + d*
x/2)**6 + 150*a*d*tan(c/2 + d*x/2)**4 + 75*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) + 30*tan(c/2 + d*x/2)**4/(15*a*d*
tan(c/2 + d*x/2)**10 + 75*a*d*tan(c/2 + d*x/2)**8 + 150*a*d*tan(c/2 + d*x/2)**6 + 150*a*d*tan(c/2 + d*x/2)**4
+ 75*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) - 40*tan(c/2 + d*x/2)**3/(15*a*d*tan(c/2 + d*x/2)**10 + 75*a*d*tan(c/2
+ d*x/2)**8 + 150*a*d*tan(c/2 + d*x/2)**6 + 150*a*d*tan(c/2 + d*x/2)**4 + 75*a*d*tan(c/2 + d*x/2)**2 + 15*a*d)
+ 30*tan(c/2 + d*x/2)**2/(15*a*d*tan(c/2 + d*x/2)**10 + 75*a*d*tan(c/2 + d*x/2)**8 + 150*a*d*tan(c/2 + d*x/2)
**6 + 150*a*d*tan(c/2 + d*x/2)**4 + 75*a*d*tan(c/2 + d*x/2)**2 + 15*a*d), Ne(d, 0)), (x*sin(c)*cos(c)**5/(a*si
n(c) + a), True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.89 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {12 \, \sin \left (d x + c\right )^{5} - 15 \, \sin \left (d x + c\right )^{4} - 20 \, \sin \left (d x + c\right )^{3} + 30 \, \sin \left (d x + c\right )^{2}}{60 \, a d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.42 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.89 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {12 \, \sin \left (d x + c\right )^{5} - 15 \, \sin \left (d x + c\right )^{4} - 20 \, \sin \left (d x + c\right )^{3} + 30 \, \sin \left (d x + c\right )^{2}}{60 \, a d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.04 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {{\sin \left (c+d\,x\right )}^2}{2\,a}-\frac {{\sin \left (c+d\,x\right )}^3}{3\,a}-\frac {{\sin \left (c+d\,x\right )}^4}{4\,a}+\frac {{\sin \left (c+d\,x\right )}^5}{5\,a}}{d} \]

[In]

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

[Out]

(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*a))/d