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

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Rubi [A]  time = 0.04, antiderivative size = 34, normalized size of antiderivative = 1.00, 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 8

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

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*}

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Mathematica [B]  time = 0.18, 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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fricas [A]  time = 0.86, size = 61, normalized size = 1.79 \[ -\frac {d x + {\left (d x + 2\right )} \cos \left (d x + c\right ) + {\left (d x - 2\right )} \sin \left (d x + c\right ) + 2}{a^{2} d \cos \left (d x + c\right ) + a^{2} d \sin \left (d x + c\right ) + a^{2} d} \]

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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giac [A]  time = 0.49, size = 33, normalized size = 0.97 \[ -\frac {\frac {d x + c}{a^{2}} + \frac {4}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}}}{d} \]

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

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maple [A]  time = 0.20, size = 41, normalized size = 1.21 \[ -\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2} d}-\frac {4}{a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]

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

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maxima [A]  time = 0.76, size = 56, normalized size = 1.65 \[ -\frac {2 \, {\left (\frac {2}{a^{2} + \frac {a^{2} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}} + \frac {\arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )}}{d} \]

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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mupad [B]  time = 4.64, size = 28, normalized size = 0.82 \[ -\frac {x}{a^2}-\frac {4}{a^2\,d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 6.93, size = 95, normalized size = 2.79 \[ \begin {cases} - \frac {d x \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a^{2} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{2} d} - \frac {d x}{a^{2} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{2} d} - \frac {4}{a^{2} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{2} d} & \text {for}\: d \neq 0 \\\frac {x \cos ^{2}{\relax (c )}}{\left (a \sin {\relax (c )} + a\right )^{2}} & \text {otherwise} \end {cases} \]

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

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