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

Optimal. Leaf size=27 \[ -\frac{\cos ^3(c+d x)}{3 d (a \sin (c+d x)+a)^3} \]

[Out]

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

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Rubi [A]  time = 0.038762, antiderivative size = 27, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.048, Rules used = {2671} \[ -\frac{\cos ^3(c+d x)}{3 d (a \sin (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2671

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b*(g*
Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*m), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^
2, 0] && EqQ[Simplify[m + p + 1], 0] &&  !ILtQ[p, 0]

Rubi steps

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

Mathematica [A]  time = 0.0170647, size = 28, normalized size = 1.04 \[ -\frac{\cos ^3(c+d x)}{3 a^3 d (\sin (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Cos[c + d*x]^3/(3*a^3*d*(1 + Sin[c + d*x])^3)

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Maple [B]  time = 0.089, size = 55, normalized size = 2. \begin{align*} 2\,{\frac{1}{d{a}^{3}} \left ( 2\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-2}-4/3\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-3}- \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-1} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 0.974754, size = 134, normalized size = 4.96 \begin{align*} -\frac{2 \,{\left (\frac{3 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}}{3 \,{\left (a^{3} + \frac{3 \, a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{3 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.76474, size = 238, normalized size = 8.81 \begin{align*} -\frac{\cos \left (d x + c\right )^{2} +{\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}{3 \,{\left (a^{3} d \cos \left (d x + c\right )^{2} - a^{3} d \cos \left (d x + c\right ) - 2 \, a^{3} d -{\left (a^{3} d \cos \left (d x + c\right ) + 2 \, a^{3} d\right )} \sin \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 29.2513, size = 298, normalized size = 11.04 \begin{align*} \begin{cases} \frac{\tan ^{3}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{3 a^{3} d \tan ^{3}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 9 a^{3} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 9 a^{3} d \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 3 a^{3} d} - \frac{3 \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{3 a^{3} d \tan ^{3}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 9 a^{3} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 9 a^{3} d \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 3 a^{3} d} + \frac{3 \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{3 a^{3} d \tan ^{3}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 9 a^{3} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 9 a^{3} d \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 3 a^{3} d} - \frac{1}{3 a^{3} d \tan ^{3}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 9 a^{3} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 9 a^{3} d \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 3 a^{3} d} & \text{for}\: d \neq 0 \\\frac{x \cos ^{2}{\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)**2/(a+a*sin(d*x+c))**3,x)

[Out]

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

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Giac [A]  time = 1.14986, size = 49, normalized size = 1.81 \begin{align*} -\frac{2 \,{\left (3 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}}{3 \, a^{3} d{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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