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

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

[Out]

(a*Log[a + c*Cot[(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 = 0.0531722, 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, 3121, 31} \[ \frac{a \log \left (a+c \cot \left (\frac{1}{2} (d+e x)\right )\right )}{4 c^3 e}-\frac{a \sin (d+e x)+c \cos (d+e x)}{4 c^2 e (a (-\cos (d+e x))+a+c \sin (d+e x))} \]

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*Cot[(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 3121

Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^(-1), x_Symbol] :> Module[{f = Free
Factors[Cot[(d + e*x)/2], x]}, -Dist[f/e, Subst[Int[1/(a + c*f*x), x], x, Cot[(d + e*x)/2]/f], x]] /; FreeQ[{a
, b, c, d, e}, x] && EqQ[a + b, 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}{2 a+2 c x} \, dx,x,\cot \left (\frac{1}{2} (d+e x)\right )\right )}{2 c^2 e}\\ &=\frac{a \log \left (a+c \cot \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 [B]  time = 0.413908, size = 229, normalized size = 3.05 \[ -\frac{\sin \left (\frac{1}{2} (d+e x)\right ) \left (a \sin \left (\frac{1}{2} (d+e x)\right )+c \cos \left (\frac{1}{2} (d+e x)\right )\right ) \left (\cos (d+e x) \left (2 a^2 \log \left (a \sin \left (\frac{1}{2} (d+e x)\right )+c \cos \left (\frac{1}{2} (d+e x)\right )\right )-2 a^2 \log \left (\sin \left (\frac{1}{2} (d+e x)\right )\right )+a^2+2 c^2\right )+a \left (a \left (-2 \log \left (a \sin \left (\frac{1}{2} (d+e x)\right )+c \cos \left (\frac{1}{2} (d+e x)\right )\right )+2 \log \left (\sin \left (\frac{1}{2} (d+e x)\right )\right )-1\right )+c \sin (d+e x) \left (-2 \log \left (a \sin \left (\frac{1}{2} (d+e x)\right )+c \cos \left (\frac{1}{2} (d+e x)\right )\right )+2 \log \left (\sin \left (\frac{1}{2} (d+e x)\right )\right )+1\right )\right )\right )}{4 c^3 e (a (-\cos (d+e x))+a+c \sin (d+e x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sin[(d + e*x)/2]*(c*Cos[(d + e*x)/2] + a*Sin[(d + e*x)/2])*(Cos[d + e*x]*(a^2 + 2*c^2 - 2*a^2*Log[Sin[(d + e
*x)/2]] + 2*a^2*Log[c*Cos[(d + e*x)/2] + a*Sin[(d + e*x)/2]]) + a*(a*(-1 + 2*Log[Sin[(d + e*x)/2]] - 2*Log[c*C
os[(d + e*x)/2] + a*Sin[(d + e*x)/2]]) + c*(1 + 2*Log[Sin[(d + e*x)/2]] - 2*Log[c*Cos[(d + e*x)/2] + a*Sin[(d
+ e*x)/2]])*Sin[d + e*x])))/(4*c^3*e*(a - a*Cos[d + e*x] + c*Sin[d + e*x])^2)

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Maple [A]  time = 0.151, size = 110, normalized size = 1.5 \begin{align*} -{\frac{a}{8\,{c}^{2}e} \left ( c+a\tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) \right ) ^{-1}}-{\frac{1}{8\,ae} \left ( c+a\tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) \right ) ^{-1}}+{\frac{a}{4\,{c}^{3}e}\ln \left ( c+a\tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) \right ) }-{\frac{1}{8\,{c}^{2}e} \left ( \tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) \right ) ^{-1}}-{\frac{a}{4\,{c}^{3}e}\ln \left ( \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*c*sin(e*x+d))^2,x)

[Out]

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

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Maxima [A]  time = 1.01679, size = 185, normalized size = 2.47 \begin{align*} -\frac{\frac{a c + \frac{{\left (2 \, a^{2} + c^{2}\right )} \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1}}{\frac{a c^{3} \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} + \frac{a^{2} c^{2} \sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}}} - \frac{2 \, a \log \left (c + \frac{a \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1}\right )}{c^{3}} + \frac{2 \, a \log \left (\frac{\sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1}\right )}{c^{3}}}{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*c + (2*a^2 + c^2)*sin(e*x + d)/(cos(e*x + d) + 1))/(a*c^3*sin(e*x + d)/(cos(e*x + d) + 1) + a^2*c^2*s
in(e*x + d)^2/(cos(e*x + d) + 1)^2) - 2*a*log(c + a*sin(e*x + d)/(cos(e*x + d) + 1))/c^3 + 2*a*log(sin(e*x + d
)/(cos(e*x + d) + 1))/c^3)/e

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Fricas [B]  time = 2.34212, 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 = 1.15568, size = 155, normalized size = 2.07 \begin{align*} \frac{1}{8} \,{\left (\frac{2 \, a \log \left ({\left | a \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) + c \right |}\right )}{c^{3}} - \frac{2 \, a \log \left ({\left | \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) \right |}\right )}{c^{3}} - \frac{2 \, a^{2} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) + c^{2} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) + a c}{{\left (a \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{2} + c \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )\right )} a c^{2}}\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(a*tan(1/2*x*e + 1/2*d) + c))/c^3 - 2*a*log(abs(tan(1/2*x*e + 1/2*d)))/c^3 - (2*a^2*tan(1/2*x*
e + 1/2*d) + c^2*tan(1/2*x*e + 1/2*d) + a*c)/((a*tan(1/2*x*e + 1/2*d)^2 + c*tan(1/2*x*e + 1/2*d))*a*c^2))*e^(-
1)