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]

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

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Rubi [A]  time = 0.05, antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, 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 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]

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]

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*}

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Mathematica [A]  time = 0.03, size = 30, normalized size = 1.00 \[ \frac {\sin ^2(c+d x)}{2 a d (a \sin (c+d x)+a)^2} \]

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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fricas [A]  time = 0.51, size = 46, normalized size = 1.53 \[ \frac {2 \, \sin \left (d x + c\right ) + 1}{2 \, {\left (a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \sin \left (d x + c\right ) - 2 \, a^{3} d\right )}} \]

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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giac [A]  time = 0.16, size = 28, normalized size = 0.93 \[ -\frac {2 \, \sin \left (d x + c\right ) + 1}{2 \, a^{3} d {\left (\sin \left (d x + c\right ) + 1\right )}^{2}} \]

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)

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maple [A]  time = 0.20, size = 33, normalized size = 1.10 \[ \frac {\frac {1}{2 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {1}{1+\sin \left (d x +c \right )}}{d \,a^{3}} \]

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 = 0.39, size = 44, normalized size = 1.47 \[ -\frac {2 \, \sin \left (d x + c\right ) + 1}{2 \, {\left (a^{3} \sin \left (d x + c\right )^{2} + 2 \, a^{3} \sin \left (d x + c\right ) + a^{3}\right )} d} \]

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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mupad [B]  time = 0.05, size = 37, normalized size = 1.23 \[ \frac {1}{2\,a^3\,d\,{\left (\sin \left (c+d\,x\right )+1\right )}^2}-\frac {1}{a^3\,d\,\left (\sin \left (c+d\,x\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 1.70, size = 99, normalized size = 3.30 \[ \begin {cases} - \frac {2 \sin {\left (c + d x \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} - \frac {1}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} & \text {for}\: d \neq 0 \\\frac {x \sin {\relax (c )} \cos {\relax (c )}}{\left (a \sin {\relax (c )} + a\right )^{3}} & \text {otherwise} \end {cases} \]

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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