∫ cos5 (c+dx) sin(c+dx)______________

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

Optimal. Leaf size=55 \[ \frac{\sin ^5(c+d x)}{5 a d}-\frac{\sin ^3(c+d x)}{3 a d}-\frac{\cos ^4(c+d x)}{4 a d} \]

[Out]

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

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Rubi [A] time = 0.107202, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {2835, 2565, 30, 2564, 14} \[ \frac{\sin ^5(c+d x)}{5 a d}-\frac{\sin ^3(c+d x)}{3 a d}-\frac{\cos ^4(c+d x)}{4 a d} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 2835

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]))

Rule 2565

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 30

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

Rule 2564

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 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]]

Rubi steps

\begin{align*} \int \frac{\cos ^5(c+d x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx &=\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{\operatorname{Subst}\left (\int x^3 \, dx,x,\cos (c+d x)\right )}{a d}-\frac{\operatorname{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{\operatorname{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] time = 0.154122, size = 48, normalized size = 0.87 \[ \frac{\sin ^2(c+d x) \left (12 \sin ^3(c+d x)-15 \sin ^2(c+d x)-20 \sin (c+d x)+30\right )}{60 a d} \]

Antiderivative was successfully veriﬁed.

[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)

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Maple [A] time = 0.058, size = 49, normalized size = 0.9 \begin{align*}{\frac{1}{da} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{5}}-{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{4}}-{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{3}}+{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{2}} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[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)

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Maxima [A] time = 1.0715, size = 66, normalized size = 1.2 \begin{align*} \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} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A] time = 1.49926, size = 122, normalized size = 2.22 \begin{align*} -\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} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [A] time = 45.0321, size = 741, normalized size = 13.47 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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))

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Giac [A] time = 1.17198, size = 66, normalized size = 1.2 \begin{align*} \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} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)

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