3.357 \(\int \frac{\cos ^3(c+d x) \sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=37 \[ \frac{\sin ^3(c+d x)}{3 a d}-\frac{\sin ^4(c+d x)}{4 a d} \]

[Out]

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

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Rubi [A]  time = 0.0970588, antiderivative size = 37, 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} \[ \frac{\sin ^3(c+d x)}{3 a d}-\frac{\sin ^4(c+d x)}{4 a d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^3*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)

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

Mathematica [A]  time = 0.123074, size = 28, normalized size = 0.76 \[ \frac{(4-3 \sin (c+d x)) \sin ^3(c+d x)}{12 a d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^3*Sin[c + d*x]^2)/(a + a*Sin[c + d*x]),x]

[Out]

((4 - 3*Sin[c + d*x])*Sin[c + d*x]^3)/(12*a*d)

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Maple [A]  time = 0.023, size = 30, normalized size = 0.8 \begin{align*} -{\frac{1}{da} \left ({\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)^3*sin(d*x+c)^2/(a+a*sin(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 0.98732, size = 39, normalized size = 1.05 \begin{align*} -\frac{3 \, \sin \left (d x + c\right )^{4} - 4 \, \sin \left (d x + c\right )^{3}}{12 \, a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*sin(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/12*(3*sin(d*x + c)^4 - 4*sin(d*x + c)^3)/(a*d)

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Fricas [A]  time = 1.41939, size = 120, normalized size = 3.24 \begin{align*} -\frac{3 \, \cos \left (d x + c\right )^{4} - 6 \, \cos \left (d x + c\right )^{2} + 4 \,{\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right )}{12 \, a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*sin(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/12*(3*cos(d*x + c)^4 - 6*cos(d*x + c)^2 + 4*(cos(d*x + c)^2 - 1)*sin(d*x + c))/(a*d)

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Sympy [A]  time = 27.4755, size = 277, normalized size = 7.49 \begin{align*} \begin{cases} \frac{8 \tan ^{5}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{3 a d \tan ^{8}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 12 a d \tan ^{6}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 18 a d \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 12 a d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 3 a d} - \frac{12 \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{3 a d \tan ^{8}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 12 a d \tan ^{6}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 18 a d \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 12 a d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 3 a d} + \frac{8 \tan ^{3}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{3 a d \tan ^{8}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 12 a d \tan ^{6}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 18 a d \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 12 a d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 3 a d} & \text{for}\: d \neq 0 \\\frac{x \sin ^{2}{\left (c \right )} \cos ^{3}{\left (c \right )}}{a \sin{\left (c \right )} + a} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3*sin(d*x+c)**2/(a+a*sin(d*x+c)),x)

[Out]

Piecewise((8*tan(c/2 + d*x/2)**5/(3*a*d*tan(c/2 + d*x/2)**8 + 12*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*
x/2)**4 + 12*a*d*tan(c/2 + d*x/2)**2 + 3*a*d) - 12*tan(c/2 + d*x/2)**4/(3*a*d*tan(c/2 + d*x/2)**8 + 12*a*d*tan
(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 12*a*d*tan(c/2 + d*x/2)**2 + 3*a*d) + 8*tan(c/2 + d*x/2)**3/(3
*a*d*tan(c/2 + d*x/2)**8 + 12*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 12*a*d*tan(c/2 + d*x/2)**
2 + 3*a*d), Ne(d, 0)), (x*sin(c)**2*cos(c)**3/(a*sin(c) + a), True))

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Giac [A]  time = 1.38205, size = 39, normalized size = 1.05 \begin{align*} -\frac{3 \, \sin \left (d x + c\right )^{4} - 4 \, \sin \left (d x + c\right )^{3}}{12 \, a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*sin(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/12*(3*sin(d*x + c)^4 - 4*sin(d*x + c)^3)/(a*d)