3.252 \(\int \frac {\cos (c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^4} \, dx\)

Optimal. Leaf size=30 \[ \frac {\sin ^3(c+d x)}{3 a d (a \sin (c+d x)+a)^3} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2833, 12, 37} \[ \frac {\sin ^3(c+d x)}{3 a d (a \sin (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]*Sin[c + d*x]^2)/(a + a*Sin[c + d*x])^4,x]

[Out]

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

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

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Mathematica [A]  time = 0.18, size = 53, normalized size = 1.77 \[ \frac {-6 \sin (c+d x)+3 \cos (2 (c+d x))-5}{6 a^4 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^6} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]*Sin[c + d*x]^2)/(a + a*Sin[c + d*x])^4,x]

[Out]

(-5 + 3*Cos[2*(c + d*x)] - 6*Sin[c + d*x])/(6*a^4*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^6)

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fricas [B]  time = 0.48, size = 72, normalized size = 2.40 \[ -\frac {3 \, \cos \left (d x + c\right )^{2} - 3 \, \sin \left (d x + c\right ) - 4}{3 \, {\left (3 \, a^{4} d \cos \left (d x + c\right )^{2} - 4 \, a^{4} d + {\left (a^{4} d \cos \left (d x + c\right )^{2} - 4 \, a^{4} d\right )} \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*sin(d*x+c)^2/(a+a*sin(d*x+c))^4,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.22, size = 38, normalized size = 1.27 \[ -\frac {3 \, \sin \left (d x + c\right )^{2} + 3 \, \sin \left (d x + c\right ) + 1}{3 \, a^{4} d {\left (\sin \left (d x + c\right ) + 1\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*sin(d*x+c)^2/(a+a*sin(d*x+c))^4,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.22, size = 43, normalized size = 1.43 \[ \frac {-\frac {1}{3 \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {1}{\left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {1}{1+\sin \left (d x +c \right )}}{d \,a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)*sin(d*x+c)^2/(a+a*sin(d*x+c))^4,x)

[Out]

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

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maxima [B]  time = 0.40, size = 67, normalized size = 2.23 \[ -\frac {3 \, \sin \left (d x + c\right )^{2} + 3 \, \sin \left (d x + c\right ) + 1}{3 \, {\left (a^{4} \sin \left (d x + c\right )^{3} + 3 \, a^{4} \sin \left (d x + c\right )^{2} + 3 \, a^{4} \sin \left (d x + c\right ) + a^{4}\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*sin(d*x+c)^2/(a+a*sin(d*x+c))^4,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 8.48, size = 54, normalized size = 1.80 \[ \frac {1}{a^4\,d\,{\left (\sin \left (c+d\,x\right )+1\right )}^2}-\frac {1}{a^4\,d\,\left (\sin \left (c+d\,x\right )+1\right )}-\frac {1}{3\,a^4\,d\,{\left (\sin \left (c+d\,x\right )+1\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)*sin(c + d*x)^2)/(a + a*sin(c + d*x))^4,x)

[Out]

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

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sympy [A]  time = 3.26, size = 192, normalized size = 6.40 \[ \begin {cases} - \frac {3 \sin ^{2}{\left (c + d x \right )}}{3 a^{4} d \sin ^{3}{\left (c + d x \right )} + 9 a^{4} d \sin ^{2}{\left (c + d x \right )} + 9 a^{4} d \sin {\left (c + d x \right )} + 3 a^{4} d} - \frac {3 \sin {\left (c + d x \right )}}{3 a^{4} d \sin ^{3}{\left (c + d x \right )} + 9 a^{4} d \sin ^{2}{\left (c + d x \right )} + 9 a^{4} d \sin {\left (c + d x \right )} + 3 a^{4} d} - \frac {1}{3 a^{4} d \sin ^{3}{\left (c + d x \right )} + 9 a^{4} d \sin ^{2}{\left (c + d x \right )} + 9 a^{4} d \sin {\left (c + d x \right )} + 3 a^{4} d} & \text {for}\: d \neq 0 \\\frac {x \sin ^{2}{\relax (c )} \cos {\relax (c )}}{\left (a \sin {\relax (c )} + a\right )^{4}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*sin(d*x+c)**2/(a+a*sin(d*x+c))**4,x)

[Out]

Piecewise((-3*sin(c + d*x)**2/(3*a**4*d*sin(c + d*x)**3 + 9*a**4*d*sin(c + d*x)**2 + 9*a**4*d*sin(c + d*x) + 3
*a**4*d) - 3*sin(c + d*x)/(3*a**4*d*sin(c + d*x)**3 + 9*a**4*d*sin(c + d*x)**2 + 9*a**4*d*sin(c + d*x) + 3*a**
4*d) - 1/(3*a**4*d*sin(c + d*x)**3 + 9*a**4*d*sin(c + d*x)**2 + 9*a**4*d*sin(c + d*x) + 3*a**4*d), Ne(d, 0)),
(x*sin(c)**2*cos(c)/(a*sin(c) + a)**4, True))

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