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

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

[Out]

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

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Rubi [A]  time = 0.0487294, 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} \[ -\frac{a \log \left (a+c \tan \left (\frac{1}{2} (d+e x)\right )\right )}{4 c^3 e}-\frac{c \cos (d+e x)-a \sin (d+e x)}{4 c^2 e (a \cos (d+e x)+a+c \sin (d+e x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*Log[a + c*Tan[(d + e*x)/2]])/(4*c^3*e) - (c*Cos[d + e*x] - a*Sin[d + e*x])/(4*c^2*e*(a + a*Cos[d + e*x] +
c*Sin[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 c \sin (d+e x))^2} \, dx &=-\frac{c \cos (d+e x)-a \sin (d+e x)}{4 c^2 e (a+a \cos (d+e x)+c \sin (d+e x))}+\frac{\int -\frac{2 a}{2 a+2 a \cos (d+e x)+2 c \sin (d+e x)} \, dx}{4 c^2}\\ &=-\frac{c \cos (d+e x)-a \sin (d+e x)}{4 c^2 e (a+a \cos (d+e x)+c \sin (d+e x))}-\frac{a \int \frac{1}{2 a+2 a \cos (d+e x)+2 c \sin (d+e x)} \, dx}{2 c^2}\\ &=-\frac{c \cos (d+e x)-a \sin (d+e x)}{4 c^2 e (a+a \cos (d+e x)+c \sin (d+e x))}-\frac{a \operatorname{Subst}\left (\int \frac{1}{4 a+4 c x} \, dx,x,\tan \left (\frac{1}{2} (d+e x)\right )\right )}{c^2 e}\\ &=-\frac{a \log \left (a+c \tan \left (\frac{1}{2} (d+e x)\right )\right )}{4 c^3 e}-\frac{c \cos (d+e x)-a \sin (d+e x)}{4 c^2 e (a+a \cos (d+e x)+c \sin (d+e x))}\\ \end{align*}

Mathematica [A]  time = 0.538032, size = 115, normalized size = 1.53 \[ \frac{\frac{c \left (a^2+c^2\right ) \sin \left (\frac{1}{2} (d+e x)\right )}{a \left (a \cos \left (\frac{1}{2} (d+e x)\right )+c \sin \left (\frac{1}{2} (d+e x)\right )\right )}+2 a \left (\log \left (\cos \left (\frac{1}{2} (d+e x)\right )\right )-\log \left (a \cos \left (\frac{1}{2} (d+e x)\right )+c \sin \left (\frac{1}{2} (d+e x)\right )\right )\right )+c \tan \left (\frac{1}{2} (d+e x)\right )}{8 c^3 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(Log[Cos[(d + e*x)/2]] - Log[a*Cos[(d + e*x)/2] + c*Sin[(d + e*x)/2]]) + (c*(a^2 + c^2)*Sin[(d + e*x)/2])
/(a*(a*Cos[(d + e*x)/2] + c*Sin[(d + e*x)/2])) + c*Tan[(d + e*x)/2])/(8*c^3*e)

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Maple [A]  time = 0.137, size = 91, normalized size = 1.2 \begin{align*}{\frac{1}{8\,{c}^{2}e}\tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) }-{\frac{a}{4\,{c}^{3}e}\ln \left ( a+c\tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) \right ) }-{\frac{{a}^{2}}{8\,{c}^{3}e} \left ( a+c\tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) \right ) ^{-1}}-{\frac{1}{8\,ce} \left ( a+c\tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/e/c^2*tan(1/2*d+1/2*e*x)-1/4*a*ln(a+c*tan(1/2*d+1/2*e*x))/c^3/e-1/8/e/c^3/(a+c*tan(1/2*d+1/2*e*x))*a^2-1/8
/e/c/(a+c*tan(1/2*d+1/2*e*x))

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Maxima [A]  time = 1.03015, size = 122, normalized size = 1.63 \begin{align*} -\frac{\frac{a^{2} + c^{2}}{a c^{3} + \frac{c^{4} \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1}} + \frac{2 \, a \log \left (a + \frac{c \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1}\right )}{c^{3}} - \frac{\sin \left (e x + d\right )}{c^{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*c*sin(e*x+d))^2,x, algorithm="maxima")

[Out]

-1/8*((a^2 + c^2)/(a*c^3 + c^4*sin(e*x + d)/(cos(e*x + d) + 1)) + 2*a*log(a + c*sin(e*x + d)/(cos(e*x + d) + 1
))/c^3 - sin(e*x + d)/(c^2*(cos(e*x + d) + 1)))/e

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Fricas [B]  time = 2.2152, size = 398, normalized size = 5.31 \begin{align*} -\frac{2 \, c^{2} \cos \left (e x + d\right ) - 2 \, a c \sin \left (e x + d\right ) +{\left (a^{2} \cos \left (e x + d\right ) + a c \sin \left (e x + d\right ) + a^{2}\right )} \log \left (a c \sin \left (e x + d\right ) + \frac{1}{2} \, a^{2} + \frac{1}{2} \, c^{2} + \frac{1}{2} \,{\left (a^{2} - c^{2}\right )} \cos \left (e x + d\right )\right ) -{\left (a^{2} \cos \left (e x + d\right ) + a c \sin \left (e x + d\right ) + a^{2}\right )} \log \left (\frac{1}{2} \, \cos \left (e x + d\right ) + \frac{1}{2}\right )}{8 \,{\left (a c^{3} e \cos \left (e x + d\right ) + c^{4} e \sin \left (e x + d\right ) + a c^{3} 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*c*sin(e*x+d))^2,x, algorithm="fricas")

[Out]

-1/8*(2*c^2*cos(e*x + d) - 2*a*c*sin(e*x + d) + (a^2*cos(e*x + d) + a*c*sin(e*x + d) + a^2)*log(a*c*sin(e*x +
d) + 1/2*a^2 + 1/2*c^2 + 1/2*(a^2 - c^2)*cos(e*x + d)) - (a^2*cos(e*x + d) + a*c*sin(e*x + d) + a^2)*log(1/2*c
os(e*x + d) + 1/2))/(a*c^3*e*cos(e*x + d) + c^4*e*sin(e*x + d) + a*c^3*e)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.14521, size = 116, normalized size = 1.55 \begin{align*} -\frac{1}{8} \,{\left (\frac{2 \, a \log \left ({\left | c \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) + a \right |}\right )}{c^{3}} - \frac{\tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )}{c^{2}} - \frac{2 \, a c \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) + a^{2} - c^{2}}{{\left (c \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) + a\right )} c^{3}}\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*c*sin(e*x+d))^2,x, algorithm="giac")

[Out]

-1/8*(2*a*log(abs(c*tan(1/2*x*e + 1/2*d) + a))/c^3 - tan(1/2*x*e + 1/2*d)/c^2 - (2*a*c*tan(1/2*x*e + 1/2*d) +
a^2 - c^2)/((c*tan(1/2*x*e + 1/2*d) + a)*c^3))*e^(-1)