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

Optimal. Leaf size=34 \[ -\frac{2 \cos (c+d x)}{d \left (a^2 \sin (c+d x)+a^2\right )}-\frac{x}{a^2} \]

[Out]

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

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Rubi [A]  time = 0.0427926, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2680, 8} \[ -\frac{2 \cos (c+d x)}{d \left (a^2 \sin (c+d x)+a^2\right )}-\frac{x}{a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2680

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \frac{\cos ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx &=-\frac{2 \cos (c+d x)}{d \left (a^2+a^2 \sin (c+d x)\right )}-\frac{\int 1 \, dx}{a^2}\\ &=-\frac{x}{a^2}-\frac{2 \cos (c+d x)}{d \left (a^2+a^2 \sin (c+d x)\right )}\\ \end{align*}

Mathematica [B]  time = 0.180156, size = 104, normalized size = 3.06 \[ \frac{2 \left (\sin ^{-1}\left (\frac{\sqrt{1-\sin (c+d x)}}{\sqrt{2}}\right ) \sqrt{1-\sin (c+d x)} (\sin (c+d x)+1)+\sqrt{\sin (c+d x)+1} (\sin (c+d x)-1)\right ) \cos ^3(c+d x)}{a^2 d (\sin (c+d x)-1)^2 (\sin (c+d x)+1)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Cos[c + d*x]^3*((-1 + Sin[c + d*x])*Sqrt[1 + Sin[c + d*x]] + ArcSin[Sqrt[1 - Sin[c + d*x]]/Sqrt[2]]*Sqrt[1
- Sin[c + d*x]]*(1 + Sin[c + d*x])))/(a^2*d*(-1 + Sin[c + d*x])^2*(1 + Sin[c + d*x])^(5/2))

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Maple [A]  time = 0.076, size = 41, normalized size = 1.2 \begin{align*} -2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{2}}}-4\,{\frac{1}{d{a}^{2} \left ( \tan \left ( 1/2\,dx+c/2 \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))^2,x)

[Out]

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

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Maxima [A]  time = 1.43919, size = 76, normalized size = 2.24 \begin{align*} -\frac{2 \,{\left (\frac{2}{a^{2} + \frac{a^{2} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}} + \frac{\arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\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))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.80153, size = 151, normalized size = 4.44 \begin{align*} -\frac{d x +{\left (d x + 2\right )} \cos \left (d x + c\right ) +{\left (d x - 2\right )} \sin \left (d x + c\right ) + 2}{a^{2} d \cos \left (d x + c\right ) + a^{2} d \sin \left (d x + c\right ) + a^{2} 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))^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 8.42848, size = 143, normalized size = 4.21 \begin{align*} \begin{cases} - \frac{5 d x \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{5 a^{2} d \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 5 a^{2} d} - \frac{5 d x}{5 a^{2} d \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 5 a^{2} d} + \frac{12 \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{5 a^{2} d \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 5 a^{2} d} - \frac{8}{5 a^{2} d \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 5 a^{2} d} & \text{for}\: d \neq 0 \\\frac{x \cos ^{2}{\left (c \right )}}{\left (a \sin{\left (c \right )} + a\right )^{2}} & \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))**2,x)

[Out]

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

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Giac [A]  time = 1.12626, size = 45, normalized size = 1.32 \begin{align*} -\frac{\frac{d x + c}{a^{2}} + \frac{4}{a^{2}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\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))^2,x, algorithm="giac")

[Out]

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