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

Optimal. Leaf size=55 \[ \frac{\sin ^5(c+d x)}{5 a^2 d}-\frac{\sin ^4(c+d x)}{2 a^2 d}+\frac{\sin ^3(c+d x)}{3 a^2 d} \]

[Out]

Sin[c + d*x]^3/(3*a^2*d) - Sin[c + d*x]^4/(2*a^2*d) + Sin[c + d*x]^5/(5*a^2*d)

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Rubi [A]  time = 0.104194, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {2836, 12, 43} \[ \frac{\sin ^5(c+d x)}{5 a^2 d}-\frac{\sin ^4(c+d x)}{2 a^2 d}+\frac{\sin ^3(c+d x)}{3 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sin[c + d*x]^3/(3*a^2*d) - Sin[c + d*x]^4/(2*a^2*d) + Sin[c + d*x]^5/(5*a^2*d)

Rule 2836

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

Rule 12

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

Rule 43

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])

Rubi steps

\begin{align*} \int \frac{\cos ^5(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{(a-x)^2 x^2}{a^2} \, dx,x,a \sin (c+d x)\right )}{a^5 d}\\ &=\frac{\operatorname{Subst}\left (\int (a-x)^2 x^2 \, dx,x,a \sin (c+d x)\right )}{a^7 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (a^2 x^2-2 a x^3+x^4\right ) \, dx,x,a \sin (c+d x)\right )}{a^7 d}\\ &=\frac{\sin ^3(c+d x)}{3 a^2 d}-\frac{\sin ^4(c+d x)}{2 a^2 d}+\frac{\sin ^5(c+d x)}{5 a^2 d}\\ \end{align*}

Mathematica [A]  time = 0.672682, size = 38, normalized size = 0.69 \[ -\frac{\sin ^3(c+d x) (15 \sin (c+d x)+3 \cos (2 (c+d x))-13)}{30 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sin[c + d*x]^3*(-13 + 3*Cos[2*(c + d*x)] + 15*Sin[c + d*x]))/(30*a^2*d)

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Maple [A]  time = 0.075, size = 39, normalized size = 0.7 \begin{align*}{\frac{1}{d{a}^{2}} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{5}}-{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{2}}+{\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))^2,x)

[Out]

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

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Maxima [A]  time = 1.12476, size = 53, normalized size = 0.96 \begin{align*} \frac{6 \, \sin \left (d x + c\right )^{5} - 15 \, \sin \left (d x + c\right )^{4} + 10 \, \sin \left (d x + c\right )^{3}}{30 \, a^{2} 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))^2,x, algorithm="maxima")

[Out]

1/30*(6*sin(d*x + c)^5 - 15*sin(d*x + c)^4 + 10*sin(d*x + c)^3)/(a^2*d)

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Fricas [A]  time = 1.09539, size = 155, normalized size = 2.82 \begin{align*} -\frac{15 \, \cos \left (d x + c\right )^{4} - 30 \, \cos \left (d x + c\right )^{2} - 2 \,{\left (3 \, \cos \left (d x + c\right )^{4} - 11 \, \cos \left (d x + c\right )^{2} + 8\right )} \sin \left (d x + c\right )}{30 \, a^{2} 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))^2,x, algorithm="fricas")

[Out]

-1/30*(15*cos(d*x + c)^4 - 30*cos(d*x + c)^2 - 2*(3*cos(d*x + c)^4 - 11*cos(d*x + c)^2 + 8)*sin(d*x + c))/(a^2
*d)

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Sympy [A]  time = 132.556, size = 588, normalized size = 10.69 \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))**2,x)

[Out]

Piecewise((40*tan(c/2 + d*x/2)**7/(15*a**2*d*tan(c/2 + d*x/2)**10 + 75*a**2*d*tan(c/2 + d*x/2)**8 + 150*a**2*d
*tan(c/2 + d*x/2)**6 + 150*a**2*d*tan(c/2 + d*x/2)**4 + 75*a**2*d*tan(c/2 + d*x/2)**2 + 15*a**2*d) - 120*tan(c
/2 + d*x/2)**6/(15*a**2*d*tan(c/2 + d*x/2)**10 + 75*a**2*d*tan(c/2 + d*x/2)**8 + 150*a**2*d*tan(c/2 + d*x/2)**
6 + 150*a**2*d*tan(c/2 + d*x/2)**4 + 75*a**2*d*tan(c/2 + d*x/2)**2 + 15*a**2*d) + 176*tan(c/2 + d*x/2)**5/(15*
a**2*d*tan(c/2 + d*x/2)**10 + 75*a**2*d*tan(c/2 + d*x/2)**8 + 150*a**2*d*tan(c/2 + d*x/2)**6 + 150*a**2*d*tan(
c/2 + d*x/2)**4 + 75*a**2*d*tan(c/2 + d*x/2)**2 + 15*a**2*d) - 120*tan(c/2 + d*x/2)**4/(15*a**2*d*tan(c/2 + d*
x/2)**10 + 75*a**2*d*tan(c/2 + d*x/2)**8 + 150*a**2*d*tan(c/2 + d*x/2)**6 + 150*a**2*d*tan(c/2 + d*x/2)**4 + 7
5*a**2*d*tan(c/2 + d*x/2)**2 + 15*a**2*d) + 40*tan(c/2 + d*x/2)**3/(15*a**2*d*tan(c/2 + d*x/2)**10 + 75*a**2*d
*tan(c/2 + d*x/2)**8 + 150*a**2*d*tan(c/2 + d*x/2)**6 + 150*a**2*d*tan(c/2 + d*x/2)**4 + 75*a**2*d*tan(c/2 + d
*x/2)**2 + 15*a**2*d), Ne(d, 0)), (x*sin(c)**2*cos(c)**5/(a*sin(c) + a)**2, True))

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Giac [A]  time = 1.22469, size = 53, normalized size = 0.96 \begin{align*} \frac{6 \, \sin \left (d x + c\right )^{5} - 15 \, \sin \left (d x + c\right )^{4} + 10 \, \sin \left (d x + c\right )^{3}}{30 \, a^{2} 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))^2,x, algorithm="giac")

[Out]

1/30*(6*sin(d*x + c)^5 - 15*sin(d*x + c)^4 + 10*sin(d*x + c)^3)/(a^2*d)