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

Optimal. Leaf size=30 \[ \frac{\sin ^2(c+d x)}{2 a d (a \sin (c+d x)+a)^2} \]

[Out]

Sin[c + d*x]^2/(2*a*d*(a + a*Sin[c + d*x])^2)

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Rubi [A]  time = 0.0458973, antiderivative size = 30, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {2833, 12, 37} \[ \frac{\sin ^2(c+d x)}{2 a d (a \sin (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sin[c + d*x]^2/(2*a*d*(a + a*Sin[c + d*x])^2)

Rule 2833

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

Rule 12

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

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0289561, size = 30, normalized size = 1. \[ \frac{\sin ^2(c+d x)}{2 a d (a \sin (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sin[c + d*x]^2/(2*a*d*(a + a*Sin[c + d*x])^2)

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Maple [A]  time = 0.032, size = 33, normalized size = 1.1 \begin{align*}{\frac{1}{{a}^{3}d} \left ({\frac{1}{2\, \left ( 1+\sin \left ( dx+c \right ) \right ) ^{2}}}- \left ( 1+\sin \left ( dx+c \right ) \right ) ^{-1} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)*sin(d*x+c)/(a+a*sin(d*x+c))^3,x)

[Out]

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

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Maxima [A]  time = 1.06835, size = 59, normalized size = 1.97 \begin{align*} -\frac{2 \, \sin \left (d x + c\right ) + 1}{2 \,{\left (a^{3} \sin \left (d x + c\right )^{2} + 2 \, a^{3} \sin \left (d x + c\right ) + a^{3}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.33102, size = 111, normalized size = 3.7 \begin{align*} \frac{2 \, \sin \left (d x + c\right ) + 1}{2 \,{\left (a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \sin \left (d x + c\right ) - 2 \, a^{3} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 2.0969, size = 99, normalized size = 3.3 \begin{align*} \begin{cases} - \frac{2 \sin{\left (c + d x \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin{\left (c + d x \right )} + 2 a^{3} d} - \frac{1}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin{\left (c + d x \right )} + 2 a^{3} d} & \text{for}\: d \neq 0 \\\frac{x \sin{\left (c \right )} \cos{\left (c \right )}}{\left (a \sin{\left (c \right )} + a\right )^{3}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-2*sin(c + d*x)/(2*a**3*d*sin(c + d*x)**2 + 4*a**3*d*sin(c + d*x) + 2*a**3*d) - 1/(2*a**3*d*sin(c +
d*x)**2 + 4*a**3*d*sin(c + d*x) + 2*a**3*d), Ne(d, 0)), (x*sin(c)*cos(c)/(a*sin(c) + a)**3, True))

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Giac [A]  time = 1.25707, size = 38, normalized size = 1.27 \begin{align*} -\frac{2 \, \sin \left (d x + c\right ) + 1}{2 \, a^{3} d{\left (\sin \left (d x + c\right ) + 1\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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