3.534 \(\int \frac{\cos ^5(c+d x) \sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=73 \[ \frac{\sin ^6(c+d x)}{6 a d}-\frac{\sin ^5(c+d x)}{5 a d}-\frac{\sin ^4(c+d x)}{4 a d}+\frac{\sin ^3(c+d x)}{3 a d} \]

[Out]

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

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Rubi [A]  time = 0.155688, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {2835, 2564, 14} \[ \frac{\sin ^6(c+d x)}{6 a d}-\frac{\sin ^5(c+d x)}{5 a d}-\frac{\sin ^4(c+d x)}{4 a d}+\frac{\sin ^3(c+d x)}{3 a d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^5*Sin[c + d*x]^2)/(a + a*Sin[c + d*x]),x]

[Out]

Sin[c + d*x]^3/(3*a*d) - Sin[c + d*x]^4/(4*a*d) - Sin[c + d*x]^5/(5*a*d) + Sin[c + d*x]^6/(6*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 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 ^2(c+d x)}{a+a \sin (c+d x)} \, dx &=\frac{\int \cos ^3(c+d x) \sin ^2(c+d x) \, dx}{a}-\frac{\int \cos ^3(c+d x) \sin ^3(c+d x) \, dx}{a}\\ &=\frac{\operatorname{Subst}\left (\int x^2 \left (1-x^2\right ) \, dx,x,\sin (c+d x)\right )}{a d}-\frac{\operatorname{Subst}\left (\int x^3 \left (1-x^2\right ) \, dx,x,\sin (c+d x)\right )}{a d}\\ &=\frac{\operatorname{Subst}\left (\int \left (x^2-x^4\right ) \, dx,x,\sin (c+d x)\right )}{a d}-\frac{\operatorname{Subst}\left (\int \left (x^3-x^5\right ) \, dx,x,\sin (c+d x)\right )}{a d}\\ &=\frac{\sin ^3(c+d x)}{3 a d}-\frac{\sin ^4(c+d x)}{4 a d}-\frac{\sin ^5(c+d x)}{5 a d}+\frac{\sin ^6(c+d x)}{6 a d}\\ \end{align*}

Mathematica [A]  time = 0.207874, size = 48, normalized size = 0.66 \[ \frac{\sin ^3(c+d x) \left (10 \sin ^3(c+d x)-12 \sin ^2(c+d x)-15 \sin (c+d x)+20\right )}{60 a d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^5*Sin[c + d*x]^2)/(a + a*Sin[c + d*x]),x]

[Out]

(Sin[c + d*x]^3*(20 - 15*Sin[c + d*x] - 12*Sin[c + d*x]^2 + 10*Sin[c + d*x]^3))/(60*a*d)

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Maple [A]  time = 0.069, size = 49, normalized size = 0.7 \begin{align*}{\frac{1}{da} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{6}}-{\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}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*sin(d*x+c)^2/(a+a*sin(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 1.13324, size = 66, normalized size = 0.9 \begin{align*} \frac{10 \, \sin \left (d x + c\right )^{6} - 12 \, \sin \left (d x + c\right )^{5} - 15 \, \sin \left (d x + c\right )^{4} + 20 \, \sin \left (d x + c\right )^{3}}{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)^2/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/60*(10*sin(d*x + c)^6 - 12*sin(d*x + c)^5 - 15*sin(d*x + c)^4 + 20*sin(d*x + c)^3)/(a*d)

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Fricas [A]  time = 1.66327, size = 149, normalized size = 2.04 \begin{align*} -\frac{10 \, \cos \left (d x + c\right )^{6} - 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)^2/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/60*(10*cos(d*x + c)^6 - 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 = 79.2048, size = 862, normalized size = 11.81 \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)**2/(a+a*sin(d*x+c)),x)

[Out]

Piecewise((40*tan(c/2 + d*x/2)**9/(15*a*d*tan(c/2 + d*x/2)**12 + 90*a*d*tan(c/2 + d*x/2)**10 + 225*a*d*tan(c/2
+ d*x/2)**8 + 300*a*d*tan(c/2 + d*x/2)**6 + 225*a*d*tan(c/2 + d*x/2)**4 + 90*a*d*tan(c/2 + d*x/2)**2 + 15*a*d
) - 60*tan(c/2 + d*x/2)**8/(15*a*d*tan(c/2 + d*x/2)**12 + 90*a*d*tan(c/2 + d*x/2)**10 + 225*a*d*tan(c/2 + d*x/
2)**8 + 300*a*d*tan(c/2 + d*x/2)**6 + 225*a*d*tan(c/2 + d*x/2)**4 + 90*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) + 24*
tan(c/2 + d*x/2)**7/(15*a*d*tan(c/2 + d*x/2)**12 + 90*a*d*tan(c/2 + d*x/2)**10 + 225*a*d*tan(c/2 + d*x/2)**8 +
300*a*d*tan(c/2 + d*x/2)**6 + 225*a*d*tan(c/2 + d*x/2)**4 + 90*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) + 40*tan(c/2
+ d*x/2)**6/(15*a*d*tan(c/2 + d*x/2)**12 + 90*a*d*tan(c/2 + d*x/2)**10 + 225*a*d*tan(c/2 + d*x/2)**8 + 300*a*
d*tan(c/2 + d*x/2)**6 + 225*a*d*tan(c/2 + d*x/2)**4 + 90*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) + 24*tan(c/2 + d*x/
2)**5/(15*a*d*tan(c/2 + d*x/2)**12 + 90*a*d*tan(c/2 + d*x/2)**10 + 225*a*d*tan(c/2 + d*x/2)**8 + 300*a*d*tan(c
/2 + d*x/2)**6 + 225*a*d*tan(c/2 + d*x/2)**4 + 90*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) - 60*tan(c/2 + d*x/2)**4/(
15*a*d*tan(c/2 + d*x/2)**12 + 90*a*d*tan(c/2 + d*x/2)**10 + 225*a*d*tan(c/2 + d*x/2)**8 + 300*a*d*tan(c/2 + d*
x/2)**6 + 225*a*d*tan(c/2 + d*x/2)**4 + 90*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)**12 + 90*a*d*tan(c/2 + d*x/2)**10 + 225*a*d*tan(c/2 + d*x/2)**8 + 300*a*d*tan(c/2 + d*x/2)**6
+ 225*a*d*tan(c/2 + d*x/2)**4 + 90*a*d*tan(c/2 + d*x/2)**2 + 15*a*d), Ne(d, 0)), (x*sin(c)**2*cos(c)**5/(a*si
n(c) + a), True))

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Giac [A]  time = 1.14886, size = 66, normalized size = 0.9 \begin{align*} \frac{10 \, \sin \left (d x + c\right )^{6} - 12 \, \sin \left (d x + c\right )^{5} - 15 \, \sin \left (d x + c\right )^{4} + 20 \, \sin \left (d x + c\right )^{3}}{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)^2/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

1/60*(10*sin(d*x + c)^6 - 12*sin(d*x + c)^5 - 15*sin(d*x + c)^4 + 20*sin(d*x + c)^3)/(a*d)