3.371 \(\int \frac{1}{(2 a+2 a \cos (d+e x)+2 a \sin (d+e x))^2} \, dx\)

Optimal. Leaf size=75 \[ -\frac{\log \left (\tan \left (\frac{1}{2} (d+e x)\right )+1\right )}{4 a^2 e}-\frac{a \cos (d+e x)-a \sin (d+e x)}{4 e \left (a^3 \sin (d+e x)+a^3 \cos (d+e x)+a^3\right )} \]

[Out]

-Log[1 + Tan[(d + e*x)/2]]/(4*a^2*e) - (a*Cos[d + e*x] - a*Sin[d + e*x])/(4*e*(a^3 + a^3*Cos[d + e*x] + a^3*Si
n[d + e*x]))

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Rubi [A]  time = 0.0479758, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3129, 12, 3124, 31} \[ -\frac{\log \left (\tan \left (\frac{1}{2} (d+e x)\right )+1\right )}{4 a^2 e}-\frac{a \cos (d+e x)-a \sin (d+e x)}{4 e \left (a^3 \sin (d+e x)+a^3 \cos (d+e x)+a^3\right )} \]

Antiderivative was successfully verified.

[In]

Int[(2*a + 2*a*Cos[d + e*x] + 2*a*Sin[d + e*x])^(-2),x]

[Out]

-Log[1 + Tan[(d + e*x)/2]]/(4*a^2*e) - (a*Cos[d + e*x] - a*Sin[d + e*x])/(4*e*(a^3 + a^3*Cos[d + e*x] + a^3*Si
n[d + e*x]))

Rule 3129

Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^(n_), x_Symbol] :> Simp[((-(c*Cos[d
+ e*x]) + b*Sin[d + e*x])*(a + b*Cos[d + e*x] + c*Sin[d + e*x])^(n + 1))/(e*(n + 1)*(a^2 - b^2 - c^2)), x] +
Dist[1/((n + 1)*(a^2 - b^2 - c^2)), Int[(a*(n + 1) - b*(n + 2)*Cos[d + e*x] - c*(n + 2)*Sin[d + e*x])*(a + b*C
os[d + e*x] + c*Sin[d + e*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0] && LtQ[n
, -1] && NeQ[n, -3/2]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3124

Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^(-1), x_Symbol] :> Module[{f = Free
Factors[Tan[(d + e*x)/2], x]}, Dist[(2*f)/e, Subst[Int[1/(a + b + 2*c*f*x + (a - b)*f^2*x^2), x], x, Tan[(d +
e*x)/2]/f], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int \frac{1}{(2 a+2 a \cos (d+e x)+2 a \sin (d+e x))^2} \, dx &=-\frac{a \cos (d+e x)-a \sin (d+e x)}{4 e \left (a^3+a^3 \cos (d+e x)+a^3 \sin (d+e x)\right )}+\frac{\int -\frac{2 a}{2 a+2 a \cos (d+e x)+2 a \sin (d+e x)} \, dx}{4 a^2}\\ &=-\frac{a \cos (d+e x)-a \sin (d+e x)}{4 e \left (a^3+a^3 \cos (d+e x)+a^3 \sin (d+e x)\right )}-\frac{\int \frac{1}{2 a+2 a \cos (d+e x)+2 a \sin (d+e x)} \, dx}{2 a}\\ &=-\frac{a \cos (d+e x)-a \sin (d+e x)}{4 e \left (a^3+a^3 \cos (d+e x)+a^3 \sin (d+e x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{1}{4 a+4 a x} \, dx,x,\tan \left (\frac{1}{2} (d+e x)\right )\right )}{a e}\\ &=-\frac{\log \left (1+\tan \left (\frac{1}{2} (d+e x)\right )\right )}{4 a^2 e}-\frac{a \cos (d+e x)-a \sin (d+e x)}{4 e \left (a^3+a^3 \cos (d+e x)+a^3 \sin (d+e x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.183596, size = 93, normalized size = 1.24 \[ \frac{\tan \left (\frac{1}{2} (d+e x)\right )+2 \log \left (\cos \left (\frac{1}{2} (d+e x)\right )\right )+\frac{2 \sin \left (\frac{1}{2} (d+e x)\right )}{\sin \left (\frac{1}{2} (d+e x)\right )+\cos \left (\frac{1}{2} (d+e x)\right )}-2 \log \left (\sin \left (\frac{1}{2} (d+e x)\right )+\cos \left (\frac{1}{2} (d+e x)\right )\right )}{8 a^2 e} \]

Antiderivative was successfully verified.

[In]

Integrate[(2*a + 2*a*Cos[d + e*x] + 2*a*Sin[d + e*x])^(-2),x]

[Out]

(2*Log[Cos[(d + e*x)/2]] - 2*Log[Cos[(d + e*x)/2] + Sin[(d + e*x)/2]] + (2*Sin[(d + e*x)/2])/(Cos[(d + e*x)/2]
+ Sin[(d + e*x)/2]) + Tan[(d + e*x)/2])/(8*a^2*e)

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Maple [A]  time = 0.086, size = 60, normalized size = 0.8 \begin{align*}{\frac{1}{8\,{a}^{2}e}\tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) }-{\frac{1}{4\,{a}^{2}e} \left ( 1+\tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) \right ) ^{-1}}-{\frac{1}{4\,{a}^{2}e}\ln \left ( 1+\tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2*a+2*a*cos(e*x+d)+2*a*sin(e*x+d))^2,x)

[Out]

1/8/e/a^2*tan(1/2*d+1/2*e*x)-1/4/e/a^2/(1+tan(1/2*d+1/2*e*x))-1/4*ln(1+tan(1/2*d+1/2*e*x))/a^2/e

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Maxima [A]  time = 1.01651, size = 108, normalized size = 1.44 \begin{align*} -\frac{\frac{2}{a^{2} + \frac{a^{2} \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1}} + \frac{2 \, \log \left (\frac{\sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} + 1\right )}{a^{2}} - \frac{\sin \left (e x + d\right )}{a^{2}{\left (\cos \left (e x + d\right ) + 1\right )}}}{8 \, e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*a+2*a*cos(e*x+d)+2*a*sin(e*x+d))^2,x, algorithm="maxima")

[Out]

-1/8*(2/(a^2 + a^2*sin(e*x + d)/(cos(e*x + d) + 1)) + 2*log(sin(e*x + d)/(cos(e*x + d) + 1) + 1)/a^2 - sin(e*x
+ d)/(a^2*(cos(e*x + d) + 1)))/e

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Fricas [A]  time = 2.12157, size = 285, normalized size = 3.8 \begin{align*} \frac{{\left (\cos \left (e x + d\right ) + \sin \left (e x + d\right ) + 1\right )} \log \left (\frac{1}{2} \, \cos \left (e x + d\right ) + \frac{1}{2}\right ) -{\left (\cos \left (e x + d\right ) + \sin \left (e x + d\right ) + 1\right )} \log \left (\sin \left (e x + d\right ) + 1\right ) - 2 \, \cos \left (e x + d\right ) + 2 \, \sin \left (e x + d\right )}{8 \,{\left (a^{2} e \cos \left (e x + d\right ) + a^{2} e \sin \left (e x + d\right ) + a^{2} e\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*a+2*a*cos(e*x+d)+2*a*sin(e*x+d))^2,x, algorithm="fricas")

[Out]

1/8*((cos(e*x + d) + sin(e*x + d) + 1)*log(1/2*cos(e*x + d) + 1/2) - (cos(e*x + d) + sin(e*x + d) + 1)*log(sin
(e*x + d) + 1) - 2*cos(e*x + d) + 2*sin(e*x + d))/(a^2*e*cos(e*x + d) + a^2*e*sin(e*x + d) + a^2*e)

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Sympy [A]  time = 2.48721, size = 168, normalized size = 2.24 \begin{align*} \begin{cases} - \frac{2 \log{\left (\tan{\left (\frac{d}{2} + \frac{e x}{2} \right )} + 1 \right )} \tan{\left (\frac{d}{2} + \frac{e x}{2} \right )}}{8 a^{2} e \tan{\left (\frac{d}{2} + \frac{e x}{2} \right )} + 8 a^{2} e} - \frac{2 \log{\left (\tan{\left (\frac{d}{2} + \frac{e x}{2} \right )} + 1 \right )}}{8 a^{2} e \tan{\left (\frac{d}{2} + \frac{e x}{2} \right )} + 8 a^{2} e} + \frac{\tan ^{2}{\left (\frac{d}{2} + \frac{e x}{2} \right )}}{8 a^{2} e \tan{\left (\frac{d}{2} + \frac{e x}{2} \right )} + 8 a^{2} e} - \frac{3}{8 a^{2} e \tan{\left (\frac{d}{2} + \frac{e x}{2} \right )} + 8 a^{2} e} & \text{for}\: e \neq 0 \\\frac{x}{\left (2 a \sin{\left (d \right )} + 2 a \cos{\left (d \right )} + 2 a\right )^{2}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*a+2*a*cos(e*x+d)+2*a*sin(e*x+d))**2,x)

[Out]

Piecewise((-2*log(tan(d/2 + e*x/2) + 1)*tan(d/2 + e*x/2)/(8*a**2*e*tan(d/2 + e*x/2) + 8*a**2*e) - 2*log(tan(d/
2 + e*x/2) + 1)/(8*a**2*e*tan(d/2 + e*x/2) + 8*a**2*e) + tan(d/2 + e*x/2)**2/(8*a**2*e*tan(d/2 + e*x/2) + 8*a*
*2*e) - 3/(8*a**2*e*tan(d/2 + e*x/2) + 8*a**2*e), Ne(e, 0)), (x/(2*a*sin(d) + 2*a*cos(d) + 2*a)**2, True))

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Giac [A]  time = 1.14342, size = 92, normalized size = 1.23 \begin{align*} -\frac{1}{8} \,{\left (\frac{2 \, \log \left ({\left | \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) + 1 \right |}\right )}{a^{2}} - \frac{\tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )}{a^{2}} - \frac{2 \, \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )}{a^{2}{\left (\tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) + 1\right )}}\right )} e^{\left (-1\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*a+2*a*cos(e*x+d)+2*a*sin(e*x+d))^2,x, algorithm="giac")

[Out]

-1/8*(2*log(abs(tan(1/2*x*e + 1/2*d) + 1))/a^2 - tan(1/2*x*e + 1/2*d)/a^2 - 2*tan(1/2*x*e + 1/2*d)/(a^2*(tan(1
/2*x*e + 1/2*d) + 1)))*e^(-1)